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https://quarknet.org/data-portfolio/activity/dice-histograms-probability	1,721,451,256,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763514981.25/warc/CC-MAIN-20240720021925-20240720051925-00629.warc.gz	407,194,152	8,539	Dice, Histograms & Probability Students roll dice, record the resulting individual values as well as the sum of the values, create histograms of the data and develop insight into the concept of “degrees of freedom." Students roll provided dice (> 1 die per roll) and record the resulting individual values as well as the sum of the values. They create histograms of the data. Students develop insight into the concept of “degrees of freedom” by rolling differing numbers of dice and noting how the histograms change. Students can extend the activity by using a spreadsheet to simulate pseudo-random numbers, comparable to the random numbers they would obtain with dice rolls. Teachers can further extend the activity by providing loaded dice; sufficiently large numbers of rolls with these can create a “bump” in the histogram.	157	832	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.53125	3	CC-MAIN-2024-30	latest	en	0.89007
http://physics.stackexchange.com/questions/91583/what-is-the-direction-of-static-friction	1,466,876,938,000,000,000	text/html	crawl-data/CC-MAIN-2016-26/segments/1466783393463.1/warc/CC-MAIN-20160624154953-00081-ip-10-164-35-72.ec2.internal.warc.gz	242,058,237	19,138	# What is the direction of static friction? Note: My question is duplicate of the following I've gone through many related questions especially the first. As I understand the static friction is always opposite to the force applied on the object as shown: But in the case when front wheels of a vehichal are turned the force of static friction is not opposite to the applied force. For example consider a car accelerating forward. The net force on the car is in forward direction which is provided from the rear tyres, if eventually break is pressed static friction(assuming tyres aren't skidding) comes into picture. This friction should be and is opposite in direction to the direction of force applied by the rear tyres. When the front tyres are turned the direction of static friction is changed(radially inward) means the direction of static friction is not opposite to the direction of applied force as shown: Question: Is the force of static friction is always opposite to the applied force ? If not then what determines its direction? - Definitely not opposite to direction of force, it depends on the tendency of direction of motion of point of contact, if you push a tyre from top then the bottom point tends to move backwards and then the friction is also forward and in the same direction as that of applied force. – Rijul Gupta Dec 28 '13 at 14:41 What about net force ? In my example only 2 forces are applied, one weight of body, which conveniently does not account for direction in this case and the other force applied at the top of a wheel, the tire tends to turn and the bottom point tends to move in opposite direction of applied force, so friction is applied in the direction of force to resist motion of bottom point. – Rijul Gupta Dec 28 '13 at 14:47 Read your question again, you have asked is friction always opposite to applied force ? I just explained it is not, why all this skidding about ? – Rijul Gupta Dec 28 '13 at 14:54 Isn't this broadly the same question as physics.stackexchange.com/q/87976 – John Rennie Dec 28 '13 at 15:04 @rijulgupta can you cite any reference from where you got this example. If friction is in same direction then work done by friction will be positive and this would be like creating energy from friction which is not possible thermodynamically. >On my comment about skidding: If there's net force at the bottom of the tyre then the bottom will accelerate hence rolling will be converted into skidding(assuming tyre was at rest initially). – user Dec 29 '13 at 2:37 Static friction always opposes relative motion at the point of contact. There are two cases possible: 1)It orients itself in direction and magnitude in such a way that the relative acceleration of the contact point is zero. 2)If this is not possible(such as in friction is too small to prevent motion),it tries to minimize the relative acceleration. - Red: Direction of motion of the top of the wheel relative to road Orange: Direction of motion of the bottom of the wheel relative to road Green: Direction of force of friction acting on the wheel If the wheels are still rolling: friction at the ground/wheel interface simply opposes the rolling of the wheel, slowing it down. In the reference frame of the chassis, this friction has both x & y components for any nonzero angle of steer, and also acts on the chassis because the wheel is connected to it by axle (thereby allowing the car to simultaneously advance and turn). If the wheels are locked due to braking, then it doesn't matter what their orientation is, as a per the extreme example in the second diagram, and the reason why frozen roads are so dangerous. - The "friction has both x & y components"... is the y component providing the centripetal force? Also, as there is an x component, will magnitude of velocity increase(this goes against what I learned in uniform circular motion)? – Eliza Dec 28 '13 at 17:55	838	3,916	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.828125	4	CC-MAIN-2016-26	latest	en	0.940852
https://fiveminutelessons.com/learn-microsoft-excel/using-logical-operators-and-functions-excel?page=0	1,713,102,233,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296816879.72/warc/CC-MAIN-20240414130604-20240414160604-00292.warc.gz	223,207,626	27,067	# Using logical operators and functions in Excel A lot of work in Excel involves comparing data in different cells. When you make a comparison between two values, you want to know one of these things: • Is value A equal to value B (A=B) • Is A greater than B (A>B) • Is A less than B (A<B) • Is A greater than or equal to B (A>=B) • Is A less than or equal to B (A<=B) • Is A not equal to B (A<>B) These are called logical or boolean operators because there can only be two possible answers in any given case - TRUE or FALSE. ## Using Logical Operators in your formulas Excel is very flexible in the way that these logical operators can be used. For example, you can use them to compare two cells, or compare the results of one or more formulas. For example: • =A1=A2 • =A1=(A25) • =(A110)<=(A2/5) As these examples suggest, you can type these directly into a cell in Excel and have Excel calculate the results of the formula just as it would do with any formula. With these formulas, Excel will always return either TRUE or FALSE as the result in the cell. A common use of logical operators is found in Excel's IF function (you can read more about the IF function here). The IF function works like this: • =IF(logical_test,value_if_TRUE,value_if_FALSE) In essence, the IF function carries out a logical test (the three examples above are all logical tests) and then return the appropriate result depending on whether the result of the test is true or false. For example: • =IF(A1>A2,"Greater than","Less than") • =IF(A1>A2,A110%,A15%) However, you don't always need to use an IF formula. Here's a version of this formula that uses a logical operator, and also demonstrates another useful feature of logical operators in general: • =(A1>A2)(A110)+(A1<=A2)(A15%) It looks confusing, but in fact it is very logical (excuse the pun). However, it helps to know that in Excel, TRUE is the same as 1, and FALSE is the same as 0. So, in this example: • If A1>A2 is TRUE, then the formula will multiple (A110%) by 1. • Because A1>A2 is TRUE then A1<=A2 is false, so it will then multiply (A15%) by 0. • It will then add the results together: (A110%)1 + (A15%)0. • The final result is whatever (A110%) equals in the specific example. Obviously, if A1 is less than A2, then the reverse of this would occur. ## Using Multiple Logical Operators In some cases, you may want to perform more than one comparison as part of your formula. For example: • (Today is Wednesday) and (Sky is Blue) • (Today is Wednesday) or (Sky is Blue) • (Today is Wednesday and (Sky is NOT Blue) • (Today is Wednesday) or (Sky is NOT Blue) In Excel, you can use one of three logical functions to construct these formulas: • AND • OR • NOT The AND function works by performing multiple comparison tests and then returning TRUE if all of the tests were true, and FALSE if one or more of the tests were false. Here are a couple of examples: • =AND(A1>A2,A1<A3) (if A1 is greater than A2 AND less than A3, then return TRUE otherwise return FALSE) • IF(AND(A2>A2,A1<A3),"Both are true","At least one is false") (this IF function will return one of the two values depending on whether the AND function returns TRUE or FALSE). The OR function works in a very similar way to the AND function. However, whereas AND requires that all tests return true, the OR function will return TRUE if only one of the tests return true. For example: • =OR(A1>A2,A1<A3) (if either A1>A2 OR A1>A3 is true, then return TRUE. If neither are true, return FALSE). • =IF(OR(A1>A2,A1<A3),"One or both are true","Neither are true") It is important to note that the AND and the IF functions can both incorporate up to 255 logical tests (my examples here have only used 2). Regardless of the number of tests you include, the same rules apply as they did in my simple examples. It is also worth noting that you can combine the AND and OR functions in a single formula. For example: • =AND(OR(A1>A2,A1<A3),A1>A4) In this example, the AND function will only return TRUE if either (A1>A2 OR A1<A3) AND A1>A4 The final logical function you can use is the NOT function. The NOT function is somewhat self-explanatory - it takes any logical test result and does the opposite. For example: • =(Sky is Blue) - will return TRUE if the sky is blue, and FALSE if the sky is not blue. • =NOT(Sky is Blue) will return FALSE if the sky is blue, and TRUE if the sky is not blue. Note that this example doesn't care what other colors the sky might be! Of course, you can use the NOT function with the AND, OR and IF functions: • =NOT(AND(A1>A2,A1<A3)) - if A1>A2 AND A1<A3, then return FALSE • =AND(NOT(A1>A2),A1<A3) - if A1 is NOT >A2 AND A1<A3 then return TRUE. Note that writing NOT(A1>A2) is another way of writing (A1<=A2). In this simple example, using a NOT function didn't add much value, but in some cases the NOT function can be very handy. In summary, a lot of what you do in Excel, particularly once you start using IF functions, involves using logical operators. The logical functions, AND, OR and NOT are a great way to extend your use of logical operators to perform more complex calculations. ## Our Comment Policy. Add a comment to this lesson ### typo in Using logical operators •=(A1>A2)(A110)+(A1<=A2)(A15%) It looks confusing, but in fact it is very logical (excuse the pun). However, it helps to know that in Excel, TRUE is the same as 1, and FALSE is the same as 0. So, in this example: •If A1>A2 is TRUE, then the formula will multiple (A110%) by 1. •Because A1>A2 is TRUE then A1<=A2 is false, so it will then multiply (A15%) by 0. •It will then add the results together: (A110%)1 + (A15%)0. •The final result is whatever (A110%) equals in the specific example. Obviously, if A1 is less than A2, then the reverse of this would occur.	1,581	5,841	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18	latest	en	0.866323
http://www.slideshare.net/csbaker/linear-approx-differentials-newton-s-method	1,469,280,019,000,000,000	text/html	crawl-data/CC-MAIN-2016-30/segments/1469257822598.11/warc/CC-MAIN-20160723071022-00252-ip-10-185-27-174.ec2.internal.warc.gz	717,143,563	27,968	Upcoming SlideShare × # Linear Approx, Differentials, Newton S Method 1,090 views 936 views Published on Published in: Technology, Education 0 Likes Statistics Notes • Full Name Comment goes here. Are you sure you want to Yes No • Be the first to comment • Be the first to like this Views Total views 1,090 On SlideShare 0 From Embeds 0 Number of Embeds 10 Actions Shares 0 14 0 Likes 0 Embeds 0 No embeds No notes for slide ### Linear Approx, Differentials, Newton S Method 1. 1. 4.5: Linear Approximations, Differentials and Newton’s Method Greg Kelly, Hanford High School, Richland, Washington 2. 2. For any function f ( x ), the tangent is a close approximation of the function for some small distance from the tangent point. We call the equation of the tangent the linearization of the function. 3. 3. The linearization is the equation of the tangent line, and you can use the old formulas if you like. Start with the point/slope equation: linearization of f at a is the standard linear approximation of f at a. 4. 4. Important linearizations for x near zero: This formula also leads to non-linear approximations: 5. 5. Differentials: When we first started to talk about derivatives, we said that becomes when the change in x and change in y become very small. dy can be considered a very small change in y . dx can be considered a very small change in x . 6. 6. Let be a differentiable function. The differential is an independent variable. The differential is: 7. 7. Example: Consider a circle of radius 10. If the radius increases by 0.1, approximately how much will the area change? very small change in A very small change in r (approximate change in area) 8. 8. (approximate change in area) Compare to actual change: New area: Old area: 9. 9. Newton’s Method Finding a root for: We will use Newton’s Method to find the root between 2 and 3. 10. 10. Guess: (not drawn to scale) (new guess) 11. 11. Guess: (new guess) 12. 12. Guess: (new guess) 13. 13. Guess: Amazingly close to zero! This is Newton’s Method of finding roots. It is an example of an algorithm (a specific set of computational steps.) It is sometimes called the Newton-Raphson method This is a recursive algorithm because a set of steps are repeated with the previous answer put in the next repetition. Each repetition is called an iteration . 14. 14. This is Newton’s Method of finding roots. It is an example of an algorithm (a specific set of computational steps.) It is sometimes called the Newton-Raphson method Guess: Amazingly close to zero! This is a recursive algorithm because a set of steps are repeated with the previous answer put in the next repetition. Each repetition is called an iteration . Newton’s Method: 15. 15. Find where crosses . 16. 16. There are some limitations to Newton’s method: Wrong root found Looking for this root. Bad guess. Failure to converge 17. 17. Newton’s method is built in to the Calculus Tools application on the TI-89. Of course if you have a TI-89, you could just use the root finder to answer the problem. The only reason to use the calculator for Newton’s Method is to help your understanding or to check your work. It would not be allowed in a college course, on the AP exam or on one of my tests. 18. 18. Now let’s do one on the TI-89: APPS Select and press . Calculus Tools ENTER If you see this screen, press , change the mode settings as necessary, and press again. ENTER APPS Approximate the positive root of: 19. 19. Now let’s do one on the TI-89: Select and press . Calculus Tools Press (Deriv) Enter the equation. (You will have to unlock the alpha mode.) Set the initial guess to 1. Approximate the positive root of: Set the iterations to 3. APPS ENTER F2 Press (Newton’s Method) 3 Press . ENTER 20. 20. Press to see the summary screen. ESC Press to see each iteration. ENTER 21. 21.  Press and then to return your calculator to normal. ESC HOME	977	3,890	{"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}	4.40625	4	CC-MAIN-2016-30	latest	en	0.885998
https://byjus.com/question-answer/235-is/	1,674,868,569,000,000,000	text/html	crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00071.warc.gz	170,518,005	18,299	Question # $2.35$ is: (a) an integer (b) a rational number (c) an irrational number (d) a natural number A an integer No worries! We‘ve got your back. Try BYJU‘S free classes today! B a rational number Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses C an irrational number No worries! We‘ve got your back. Try BYJU‘S free classes today! D a natural number No worries! We‘ve got your back. Try BYJU‘S free classes today! Open in App Solution ## The correct option is B a rational numberGiven is $2.35$,$2.35$is a rational number because it is a terminating decimal. Note: Terminating decimal numbers has a finite number of digits after the decimal point. A number with a terminating decimal is always a rational number. If the denominator of a rational number can be expressed in form ${2}^{p}{5}^{q}$or ${2}^{p}$ or ${5}^{q}$, where $p,q\in N$, then the decimal expansion of the rational number terminates.Hence, the correct option is (b). Suggest Corrections 0	275	998	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.25	3	CC-MAIN-2023-06	latest	en	0.798445
https://mathematiques-web.fr/en/fractions-corrected-5th-grade-math-exercises-in-pdf-15983	1,685,782,300,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224649177.24/warc/CC-MAIN-20230603064842-20230603094842-00616.warc.gz	430,535,889	39,618	# Fractions: corrected 5th grade math exercises in PDF. Fractions with corrected 5th grade math exercises will be of great help. The student must know the definition and be able to graph the fraction of a figure or quantity and also know how to represent it on a graduated line. The goal is also to develop computational skills by knowing how to reduce two fractions to the same denominator in order to compare them or to arrange them in ascending or descending order. The correction of these exercises allows the student to identify his or her various errors and then to correct his or her difficulties in order to progress in mathematics throughout the school year. Many exercises similar to those in your textbook with increasing difficulty. We will end this lesson on fractions with real life problems to solve in the fifth grade. ## Exercise 1: How much of the area of this square is colored… 1.in green ? 2.in red ? 3.in green and red? ## Exercise 2: The cocktail “fruits of the islands” is composed : • 6 cL of lychee juice; • 8 cL of kiwi juice; • of 12 cL of passion fruit; • 10 cL of guava juice. 1.What is the proportion of each fruit juice in this cocktail? 2.simplify fractions. ## Exercise 3: For each piece of this tangram, indicate the proportion that represents its surface in relation to the total surface of the square. ## Exercise 4: Complete the following dotted lines: ## Exercise 5: Connect by a line the figures whose colored surface proportions are equal. Write then, the corresponding fractions equalities. ## Exercise 6: 1.color the equal squares with the same color. What is the number of the uncolored square? ## Exercise 7: We propose to compare the two fractions and . 1. Compare the fractions A and B to 1. 2. Deduce a comparison between A and B. ## Exercise 8: In each case below, compare the two fractions, comparing each fraction to 1. Cette publication est également disponible en : Français (French) العربية (Arabic) Télécharger puis imprimer cette fiche en PDF Télécharger ou imprimer cette fiche «fractions: corrected 5th grade math exercises in PDF.» au format PDF afin de pouvoir travailler en totale autonomie. ## D'autres fiches dans la section 5th grade math exercises Télécharger nos applications gratuites Mathématiques Web avec tous les cours,exercices corrigés. ## D'autres articles analogues à fractions: corrected 5th grade math exercises in PDF. • 98 The corrected math exercises on geometry in space in 5th grade will help you understand the chapter. The student should be able to perform volume conversions but also know by heart the formula for the volume of a cube, a rectangle, a right block and a cylinder. In addition, many… • 97 Figure areas with corrected exercises in fifth grade are always essential to the student's understanding. Thus, the latter will have to be able to convert quantities but also to calculate the area usual figures such as the square, the rectangle, the rhombus, the parallelogram, the trapezoid or the area of… • 96 Proportionality with corrected 5th grade math exercises will be of great help to you. The student will need to know how to demonstrate whether a table is proportional but also how to calculate the value of the proportionality coefficient. The student must also determine the value of the fourth… Les dernières fiches mises à jour Voici la liste des derniers cours et exercices ajoutés au site ou mis à jour et similaire à fractions: corrected 5th grade math exercises in PDF. . Mathématiques Web c'est 2 146 116 fiches de cours et d'exercices téléchargées. Copyright © 2008 - 2023 Mathématiques Web Tous droits réservés \| Mentions légales \| Signaler une Erreur \| Contact . Scroll to Top Mathématiques Web FREE VIEW	856	3,755	{"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": 2, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.5625	5	CC-MAIN-2023-23	latest	en	0.910313
https://www.bookbenefits.com/2022/10/electricity-electronic-workbooks.html	1,716,208,876,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058278.91/warc/CC-MAIN-20240520111411-20240520141411-00226.warc.gz	598,029,438	61,819	U3F1ZWV6ZTQ5ODMwNTcxNDIxX0FjdGl2YXRpb241NjQ1MTE2MjE1OTI= recent Hot news # Electricity & Electronic Workbooks Download PDF Electricity & Electronic Workbooks Picture Of The Book : About Of The Book : This Student Workbook provides a unit-by-unit outline of the Fault Assisted Circuits for Electronics Training (F.A.C.E.T.) curriculum. The following information is included together with space to take notes as you move through the curriculum. ♦ The unit objective ♦ Unit fundamentals ♦ A list of new terms and words for the unit ♦ Equipment required for the unit ♦ The exercise objectives ♦ Exercise discussion ♦ Exercise notes The Appendix includes safety information. Contents Of The Book : Unit 1 – Trainer Familiarization Exercise 1 – Instrument Familiarization Exercise 2 – F.A.C.E.T. Base Unit Familiarization Exercise 3 – DC Board Familiarization Unit 2 – Safety Exercise 1 – Basic Safety Rules Exercise 2 – Electrical Safety Rules Unit 3 – Electronic Quantities Exercise 1 – Circuit Voltages Exercise 2 – Circuit Current Exercise 3 – Circuit Resistance Unit 4 – DC Power Sources Exercise 1 – Series and Parallel Battery Circuits Exercise 2 – Series-Opposing DC Sources Unit 5 – Switches and Switching Concepts Exercise 1 – Identify Types of Switches Exercise 2 – Switching Concepts Unit 6 – Ohm's Law Exercise 1 – Ohm's Law - Circuit Resistance Exercise 2 – Ohm's Law - Circuit Current Exercise 3 – Ohm's Law - Circuit Voltage Unit 7 – Series Resistive Circuits Exercise 1 – Resistance in a Series Resistive Circuit Exercise 2 – Current in a Series Resistive Circuit Exercise 3 – Voltage in a Series Resistive Circuit Unit 8 – Parallel Resistive Circuits. Exercise 1 – Resistance in a Parallel Circuit. Exercise 2 – Voltage/Current in a Parallel Circuit. Unit 9 – Series/Parallel Resistive Circuits Exercise 1 – Resistance in a Series/Parallel Circuit Exercise 2 – Voltage in a Series/Parallel Circuit Exercise 3 – Current in a Series/Parallel Circuit Unit 10 – Power in DC Circuits Exercise 1 – Power in a Series Resistive Circuit Exercise 2 – Power in a Parallel Resistive Circuit Exercise 3 – Power in a Series/Parallel Circuit Unit 11 – Potentiometers and Rheostats Exercise 1 – The Rheostat Exercise 2 – The Potentiometer Unit 12 – Voltage and Current Divider Circuits Exercise 1 – Voltage Dividers Exercise 2 – Current Dividers Unit 13 – Direct Current Meters Exercise 1 – The DC Ammeter Exercise 2 – The DC Ohmmeter Exercise 3 – The DC Voltmeter Information Of The Book : Title: Electricity & Electronic Workbooks PDF Language: English. Size: 5,6 Mb. Pages: 1028. Format: PDF. Author:	687	2,608	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22	latest	en	0.780787
https://citizenmaths.com/frequency/cycles-per-picosecond-to-frames-per-second	1,675,338,847,000,000,000	text/html	crawl-data/CC-MAIN-2023-06/segments/1674764500017.27/warc/CC-MAIN-20230202101933-20230202131933-00377.warc.gz	187,748,530	13,181	# Cycle per Picosecond to Frame per Second Conversions From Cycle per Picosecond • Action per Minute • Attohertz • Centihertz • Cycle per Day • Cycle per Hour • Cycle per Microsecond • Cycle per Millisecond • Cycle per Minute • Cycle per Month • Cycle per Nanosecond • Cycle per Picosecond • Cycle per Second • Cycle per Year • Decahertz • Decihertz • Degree per Hour • Degree per Millisecond • Degree per Minute • Degree per Second • Exahertz • Femtohertz • Frame per Second • Fresnel • Gigahertz • Hectohertz • Hertz • Kilohertz • Megahertz • Microhertz • Millihertz • Nanohertz • Petahertz • Picohertz • Revolution per Minute • Terahertz • Yoctohertz • Yottahertz • Zeptohertz • Zettahertz To Frame per Second • Action per Minute • Attohertz • Centihertz • Cycle per Day • Cycle per Hour • Cycle per Microsecond • Cycle per Millisecond • Cycle per Minute • Cycle per Month • Cycle per Nanosecond • Cycle per Picosecond • Cycle per Second • Cycle per Year • Decahertz • Decihertz • Degree per Hour • Degree per Millisecond • Degree per Minute • Degree per Second • Exahertz • Femtohertz • Frame per Second • Fresnel • Gigahertz • Hectohertz • Hertz • Kilohertz • Megahertz • Microhertz • Millihertz • Nanohertz • Petahertz • Picohertz • Revolution per Minute • Terahertz • Yoctohertz • Yottahertz • Zeptohertz • Zettahertz Formula 9,599 cpps = 9599 x 1e+12 FPS = 9.6e+15 FPS ## How To Convert From Cycle per Picosecond to Frame per Second 1 Cycle per Picosecond is equivalent to 1.0e+12 Frame per Second: 1 cpps = 1.0e+12 FPS For example, if the Cycle per Picosecond number is (8.2), then its equivalent Frame per Second number would be (8.2e+12). Formula: 8.2 cpps = 8.2 x 1e+12 FPS = 8.2e+12 FPS ## Cycle per Picosecond to Frame per Second conversion table Cycle per Picosecond (cpps) Frame per Second (FPS) 0.1 cpps 1.0e+11 FPS 0.2 cpps 2.0e+11 FPS 0.3 cpps 3.0e+11 FPS 0.4 cpps 4.0e+11 FPS 0.5 cpps 5.0e+11 FPS 0.6 cpps 6.0e+11 FPS 0.7 cpps 7.0e+11 FPS 0.8 cpps 8.0e+11 FPS 0.9 cpps 9.0e+11 FPS 1 cpps 1.0e+12 FPS 1.1 cpps 1.1e+12 FPS 1.2 cpps 1.2e+12 FPS 1.3 cpps 1.3e+12 FPS 1.4 cpps 1.4e+12 FPS 1.5 cpps 1.5e+12 FPS 1.6 cpps 1.6e+12 FPS 1.7 cpps 1.7e+12 FPS 1.8 cpps 1.8e+12 FPS 1.9 cpps 1.9e+12 FPS 2 cpps 2.0e+12 FPS 2.1 cpps 2.1e+12 FPS 2.2 cpps 2.2e+12 FPS 2.3 cpps 2.3e+12 FPS 2.4 cpps 2.4e+12 FPS 2.5 cpps 2.5e+12 FPS 2.6 cpps 2.6e+12 FPS 2.7 cpps 2.7e+12 FPS 2.8 cpps 2.8e+12 FPS 2.9 cpps 2.9e+12 FPS 3 cpps 3.0e+12 FPS 3.1 cpps 3.1e+12 FPS 3.2 cpps 3.2e+12 FPS 3.3 cpps 3.3e+12 FPS 3.4 cpps 3.4e+12 FPS 3.5 cpps 3.5e+12 FPS 3.6 cpps 3.6e+12 FPS 3.7 cpps 3.7e+12 FPS 3.8 cpps 3.8e+12 FPS 3.9 cpps 3.9e+12 FPS 4 cpps 4.0e+12 FPS 4.1 cpps 4.1e+12 FPS 4.2 cpps 4.2e+12 FPS 4.3 cpps 4.3e+12 FPS 4.4 cpps 4.4e+12 FPS 4.5 cpps 4.5e+12 FPS 4.6 cpps 4.6e+12 FPS 4.7 cpps 4.7e+12 FPS 4.8 cpps 4.8e+12 FPS 4.9 cpps 4.9e+12 FPS 5 cpps 5.0e+12 FPS 5.1 cpps 5.1e+12 FPS 5.2 cpps 5.2e+12 FPS 5.3 cpps 5.3e+12 FPS 5.4 cpps 5.4e+12 FPS 5.5 cpps 5.5e+12 FPS 5.6 cpps 5.6e+12 FPS 5.7 cpps 5.7e+12 FPS 5.8 cpps 5.8e+12 FPS 5.9 cpps 5.9e+12 FPS 6 cpps 6.0e+12 FPS 6.1 cpps 6.1e+12 FPS 6.2 cpps 6.2e+12 FPS 6.3 cpps 6.3e+12 FPS 6.4 cpps 6.4e+12 FPS 6.5 cpps 6.5e+12 FPS 6.6 cpps 6.6e+12 FPS 6.7 cpps 6.7e+12 FPS 6.8 cpps 6.8e+12 FPS 6.9 cpps 6.9e+12 FPS 7 cpps 7.0e+12 FPS 7.1 cpps 7.1e+12 FPS 7.2 cpps 7.2e+12 FPS 7.3 cpps 7.3e+12 FPS 7.4 cpps 7.4e+12 FPS 7.5 cpps 7.5e+12 FPS 7.6 cpps 7.6e+12 FPS 7.7 cpps 7.7e+12 FPS 7.8 cpps 7.8e+12 FPS 7.9 cpps 7.9e+12 FPS 8 cpps 8.0e+12 FPS 8.1 cpps 8.1e+12 FPS 8.2 cpps 8.2e+12 FPS 8.3 cpps 8.3e+12 FPS 8.4 cpps 8.4e+12 FPS 8.5 cpps 8.5e+12 FPS 8.6 cpps 8.6e+12 FPS 8.7 cpps 8.7e+12 FPS 8.8 cpps 8.8e+12 FPS 8.9 cpps 8.9e+12 FPS 9 cpps 9.0e+12 FPS 9.1 cpps 9.1e+12 FPS 9.2 cpps 9.2e+12 FPS 9.3 cpps 9.3e+12 FPS 9.4 cpps 9.4e+12 FPS 9.5 cpps 9.5e+12 FPS 9.6 cpps 9.6e+12 FPS 9.7 cpps 9.7e+12 FPS 9.8 cpps 9.8e+12 FPS 9.9 cpps 9.9e+12 FPS 10 cpps 1.0e+13 FPS 20 cpps 2.0e+13 FPS 30 cpps 3.0e+13 FPS 40 cpps 4.0e+13 FPS 50 cpps 5.0e+13 FPS 60 cpps 6.0e+13 FPS 70 cpps 7.0e+13 FPS 80 cpps 8.0e+13 FPS 90 cpps 9.0e+13 FPS 100 cpps 1.0e+14 FPS 110 cpps 1.1e+14 FPS	2,108	4,103	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.15625	3	CC-MAIN-2023-06	latest	en	0.393259
https://simplywall.st/news/calculating-the-fair-value-of-come-sure-group-holdings-limited-hkg794/	1,571,481,595,000,000,000	text/html	crawl-data/CC-MAIN-2019-43/segments/1570986692723.54/warc/CC-MAIN-20191019090937-20191019114437-00289.warc.gz	692,129,650	16,236	latest # Calculating The Fair Value Of Come Sure Group (Holdings) Limited (HKG:794) In this article we are going to estimate the intrinsic value of Come Sure Group (Holdings) Limited (HKG:794) by taking the foreast future cash flows of the company and discounting them back to today’s value. This is done using the Discounted Cash Flow (DCF) model. It may sound complicated, but actually it is quite simple! Remember though, that there are many ways to estimate a company’s value, and a DCF is just one method. If you want to learn more about discounted cash flow, the rationale behind this calculation can be read in detail in the Simply Wall St analysis model. ### Step by step through the calculation As Come Sure Group (Holdings) operates in the packaging sector, we need to calculate the intrinsic value slightly differently. Instead of using free cash flows, which are hard to estimate and often not reported by analysts in this industry, dividends per share (DPS) payments are used. Unless a company pays out the majority of its FCF as a dividend, this method will typically underestimate the value of the stock. The ‘Gordon Growth Model’ is used, which simply assumes that dividend payments will continue to increase at a sustainable growth rate forever. The dividend is expected to growth at an annual growth rate equal to the 10-year government bond rate of 2%. We then discount this figure to today’s value at a cost of equity of 9.7%. This results in an intrinsic value estimate of HK\$0.52. Relative to the current share price of HK\$0.57, the company appears around fair value at the time of writing. The assumptions in any calculation have a big impact on the valuation, so it is better to view this as a rough estimate, not precise down to the last cent. Value Per Share = Expected Dividend Per Share / (Discount Rate – Perpetual Growth Rate) = HK\$0.040 / (9.7% – 2%) = HK\$0.52 ### Important assumptions Now the most important inputs to a discounted cash flow are the discount rate, and of course, the actual cash flows. Part of investing is coming up with your own evaluation of a company’s future performance, so try the calculation yourself and check your own assumptions. The DCF also does not consider the possible cyclicality of an industry, or a company’s future capital requirements, so it does not give a full picture of a company’s potential performance. Given that we are looking at Come Sure Group (Holdings) as potential shareholders, the cost of equity is used as the discount rate, rather than the cost of capital (or weighted average cost of capital, WACC) which accounts for debt. In this calculation we’ve used 9.7%, which is based on a levered beta of 1.15. Beta is a measure of a stock’s volatility, compared to the market as a whole. We get our beta from the industry average beta of globally comparable companies, with an imposed limit between 0.8 and 2.0, which is a reasonable range for a stable business. ### Next Steps: Valuation is only one side of the coin in terms of building your investment thesis, and it shouldn’t be the only metric you look at when researching a company. The DCF model is not a perfect stock valuation tool. Rather it should be seen as a guide to “what assumptions need to be true for this stock to be under/overvalued?” If a company grows at a different rate, or if its cost of equity or risk free rate changes sharply, the output can look very different. For Come Sure Group (Holdings), I’ve compiled three additional factors you should further examine: 1. Financial Health: Does 794 have a healthy balance sheet? Take a look at our free balance sheet analysis with six simple checks on key factors like leverage and risk. 2. Other High Quality Alternatives: Are there other high quality stocks you could be holding instead of 794? Explore our interactive list of high quality stocks to get an idea of what else is out there you may be missing! PS. The Simply Wall St app conducts a discounted cash flow valuation for every stock on the HKG every day. If you want to find the calculation for other stocks just search here. We aim to bring you long-term focused research analysis driven by fundamental data. Note that our analysis may not factor in the latest price-sensitive company announcements or qualitative material. If you spot an error that warrants correction, please contact the editor at editorial-team@simplywallst.com. This article by Simply Wall St is general in nature. It does not constitute a recommendation to buy or sell any stock, and does not take account of your objectives, or your financial situation. Simply Wall St has no position in the stocks mentioned. Thank you for reading.	1,005	4,690	{"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}	3.171875	3	CC-MAIN-2019-43	longest	en	0.92162
https://www.physicsforums.com/threads/why-does-an-airplane-fly.237423/	1,675,468,153,000,000,000	text/html	crawl-data/CC-MAIN-2023-06/segments/1674764500076.87/warc/CC-MAIN-20230203221113-20230204011113-00379.warc.gz	952,184,547	29,997	# Why does an airplane fly? Gold Member For as long as I can remember the anwer I've heard to this question was the "Equal Transit" theory - that since the top of the wing is longer that the bottom the air has to go more quickly over the top inorder to "keep up" with the bottom causing a lower pressure on the top than the bottom. But this doesn't really make sense - why should the air on top "care" about the air on the bottom? When I looked on Nasa's website I found a different explanation: http://www.grc.nasa.gov/WWW/K-12/airplane/right2.html and a debunking of the "Equal Transit" theory: http://www.grc.nasa.gov/WWW/K-12/airplane/wrong1.html So why have I never come accross the explanation Nasa gives and everyone gives the "Equal Transit" theory? Which explanation is true? Or is it both? Thanks. Mentor The "equal transit" explanation, while popular, is complete nonsense. Gold Member Even if the "equal transit" explanation of the velocity difference is silly, the speed of the air over the top of the wing in greater than on the bottom due to the shape. But what lifts the plane - the pressure difference, Nasa's "turning of air" or a combination of the two? Thanks. Mentor I didn't look at the NASA links (yet), but here's a decent description of how airplanes fly: http://www.aviation-history.com/theory/lift.htm" [Broken] Last edited by a moderator: Staff Emeritus Ultimately it is the turning of the airflow downward: Newton's third law. However, you can push air down using any old slab of plywood that is angled with respect to the airflow. This will of course generate a lot of drag; it is not a good airfoil. In a good airfoil it is the upper surface of the wing that does the lions share of the turning the airflow rather than the lower surface. To get a good picture of this it is good to look at things from the perspective of Bernoulli's principle. Mentor That said, the low pressure (plus the coanda effect) above the wing causes the air above the wing to be pushed down. So it's just different ways of saying the same thing. Perhaps it is easier to quantify or explain using Newton's Third, but I don't know. I prefer the pressure/coanda effect explanation because the Newton's Third is a "what", but not a "why". Mentor The "equal transit" explanation, while popular, is complete nonsense. Though I realize it is wrong, too-emphatic rejections of it can lead to misnomers as well - they imply that the Bernoulli equation is inapplicable and there is no area of low pressure above the wing. I don't like that NASA link for that reason. The NASA link combines the "longer path" and "equal transit time" explanations into one, and I don't think that that is correct. IE: Let's use the information we've just learned to evaluate the various parts of the "Equal Transit" Theory. {Lifting airfoils are designed to have the upper surface longer than the bottom.} This is not always correct. The symmetric airfoil in our experiment generates plenty of lift and its upper surface is the same length as the lower surface. That's wrong. As can clearly be seen from giving the airfoil some positive angle of attack, the stagnation point moves down on the leading edge of the airfoil, so the air going over the top surface does take a longer path than the air on the bottom surface. The other two bullet points (including the one addresses Bernoulli's equation) are correct, though one is just pointing out what those two misnomers get right (the way Bernoulli's eq can tell us the pressure gradient). Last edited: Homework Helper After visiting a large number of web sites, my conclusion is that lift is a combination of Coanda effect, "void effect" and simple deflection, all of which result in the "downwards" acceleration of air. Coanda effect explains how laminar flow follows a convex suface. "Void effect" explains how turbulent flow follows a convex surface. Concave surfaces simply deflect airflow. The curvature of air flow accelerates the air and generates lift. "Void effect" explains how drag is developed "behind" a wing, while direct forward deflection of air accounts for the drag in front of a wing, along with friction along the surface of a wing. Except for a special class of airfoils, most of the air flow over and under a wing is turbulent, with only a portion of the air flow being laminar near the leading edge. For most wings, the flow transitions from laminar to turbulent flow above and below a wing, detaching during the transition, but reattaching after the transition. This happens in the first 30% of the chord length or sooner on a "normal" airfoil, and between the first 30% to 70% of chord lengh for a "laminar" airfoil (by definition). In some cases, rough surfaces and/or turbalators are used to cause the transition to occur at a specific position on an air foil. In the case of gliders, an "oil flow test" is done to visualise this transition. A bead of oil is placed on the leading edge of the wings, the glider is flown for a while at a fixed speed, then landed and the oil pattern observed. It's common practice to do this in glider magazine reviews. http://www.standardcirrus.org/Turbulators.html [Broken] Oil flow testing is also done in wind tunnels: http://www.hisacproject.com/news.html [Broken] At this web site, pages 4 and 5 discuss how little air flow is laminar over many wings, and how "laminar" air foils increase laminar flow to 30% or more over the chord length of a wing. In the case of gliders, laminar "bubbles" result in either more drag or less lift so the laminar air flow is deliberately broken up sooner than it normally would via rougher surfaces or turbulators (this is mentioned in the article). The laminar section starts mid way down page 4: http://www.dreesecode.com/primer/airfoil4.html "All airfoils must have adverse pressure gradients on their aft end. The usual definition of a laminar flow airfoil is that the favorable pressure gradient ends somewhere between 30% and 75% of chord." http://www.aviation-history.com/theory/lam-flow.htm The next website does a descent job of explaining lift, but with a bit too much emphasis on Coanda effect, ignoring void effect and turbulent flow, but towards the end of this web page, there's a diagram of a wind blowing over a roof, and although the air downwind of the roof is turbulent, it's also at lower pressure, due to void effect. Since both laminar and tubulent air flows contribute to lift, both cases should have been covered better than it was at this web page: "The physical cause of low or high pressure is the forced normal (perpendicular) acceleration of streaming air caused by obstacles or curved planes in combination with the Coanda-effect.": http://user.uni-frankfurt.de/~weltner/Mis6/mis6.html [Broken] videos Assuming this next video isn't GGI, it appears to be a series of pictures of a flame aimed at various angles over an glowing (from the heat) airfoil at a fixed angle of about 45 degrees. As the flame angle is made more horizontal, the effective angle of attack becomes higher. What I call "void" effect is more evident here, as the flame flow is detaches from the aft end of the airfoil at low effective angle of attack. At higher effective angle of attack, the flame flow detaches from the "upper" surface of the airfoil, but it's still accelerated (curved) "downwards", while below the airfoil there is significant direct deflection. About 28 seconds into this video (you can hold it at this position), the downwards curvature of the flame over the "top" of the wing is still evident, in spite of the large amount of apparent detachment. Next is a link to a small wind tunnel video, considered a "2d" airflow (equivalent to a 3d wing with infinite wingspan). Air speed is slow, chord length is small, so the Reynolds number is quite low, and the air flow is much more laminar and the angle of attack before stall is much higher than it would be if everything were scaled up to a faster speed and a larger size. The smallish wind tunnel also prevents any significant upwards or downwards flow of air, so the air flow is not the same as it would be in an open environment. Wind tunnels that are much larger than the wing or model being tested, such as the one linked to above showing oil flow testing are much closer to "real world" environment. The transition into the stalled condition is very abrupt. In the segment annotated as "stall", there's virtually no lift, but near the end of the video, that starts off "flow attached", then "stall", there's still significant lift although there is a stall. For this model, the stalling angle of attack is fairly small: Another wind tunnel, slow air speed, short chord, but not as much as the first video. Again the nature of the wing tunnel (proably drawing air inwards from the right), prevents the air flow from remaining deflected, and skews what would happened in an open environment: Regarding equal transit times, here are a couple of links to pictures of a flat top, curved bottom, pre-shuttle prototype: M2-F2 glider with F104 chase plane: m2-f2.jpg M2-F3 rocket powered model (reached a speed of Mach 1.6) with B52: m2-f3.jpg[/QUOTE] Last edited by a moderator: JonnyV70 Would the same physics apply to a paper airplane or does the weight difference in this case change the physics? I always thought paper airplanes flew or glided due to the large surface area to weight ratio. Last edited by a moderator: Renge Ishyo So why have I never come accross the explanation Nasa gives and everyone gives the "Equal Transit" theory? Which explanation is true? Or is it both? Thanks. The equal transit theory is good enough for teaching people beginning physics. Its value lies in its simplicity. Yes, it is a crude approximation to the "truth," but so is most of beginning physics (if not all of it). The equal transit theory doesn't gives you a completely inaccurate view of how things work. Calling it "incorrect" is extremely unfair, it implies that the theory is completely false. This is not so. As NASA mentions, it is still a fact that "flow over the top of a lifting airfoil does travel faster than the flow beneath the airfoil". It is still a fact that "The upper flow is faster and from Bernoulli's equation the pressure is lower. The difference in pressure across the airfoil produces the lift". These are the two main points of the equal transit theory, and they hold up just fine. It is just that the equal transit theory makes many simplifying approximations about the flow of the air itself that cause it to deviate slightly from the real situation. For one thing the theory assumes laminar flow and so assumes that the fluid will recombine at both sides of the wing without "breaking" structure. This approximation turns out to be too simple a view (for one thing, we probably can guess that real air does not flow over a wing in straight uniform lines that never cross each other and never break). In fact, the proof that the equal transit theory is just an approximation is that under careful observation, the air over the top of the wing surpasses the flow over the bottom as they emerge from the other side of the wing. This can be seen clearly in this video yanked from the post above (watch from 25s to 50s): The question is, does this small difference justify coming up with a more complicated model of airflow, or does the simplified model come close enough to reality to be of use? It probably depends on what your needs are. If you are a student who is just trying to get the general idea of how wings generate lift the crude approximation is probably just fine (and this is probably the reason this theory predominates in education), but if you are an engineer for NASA building wings for a new airplane you might have to use more complicated airflow models to get more accurate calculations. Homework Helper The equal transit theory is good enough for teaching people beginning physics. Equal transit theory doesn't exist in the real world. There is no tendency for separated air flows to remerge with the same relative postion they had before being separated. These curved bottom flat top lifting bodies have the "hump" on the bottom, it's a much longer path to go around the bottom than the top and yet they glide (M2-F2) or fly (M2-F3) very well: M2-F2 glider with F104 chase plane: m2-f2.jpg M2-F3 rocket powered model (reached a speed of Mach 1.6) with B52: m2-f3.jpg As NASA mentions, it is still a fact that "flow over the top of a lifting airfoil does travel faster than the flow beneath the airfoil". But as shown in the case of the M2-F2 and M2-F3, this has nothing to do with the distance the air has to travel across a wing. The air over the top of the wing surpasses the flow over the bottom as they emerge from the other side of the wing. This can be seen clearly in this video yanked from the post above (watch from 25s to 50s): Note my comment: Again the nature of the wind tunnel prevents the air flow from remaining deflected, and skews what would happen in an open environment. Does this small difference justify coming up with a more complicated model of airflow, or does the simplified model come close enough to reality to be of use? It's not a small difference, it is an incorrect explanation versus a correct explanation. If you want simple, then use the Newton approach: wings produce lift by applying a downwards force to the air, which reacts with an equal and opposite upwards force to the wing, in accordance with Newton's third law. Newton's third law holds true regardless of wing effeciency. Bernoulli effects are related to the efficiency of an air foil. The acceleration of air results in an increase in kinetic energy of the air. Effecient airfoils obtain most of this increase in kinetic energy of air through a Bernoulli like conversion of pressure energy into kinetic energy. Inefficient airfoils have less of this Bernoulli like conversion of energy and consume more energy in order to produce lift. Since wings aren't 100% efficient, there is always some non-Bernoulli related increase in the total energy of the air while producing lift. Bernoulli doesn't cover this aspect of producing lift, where overall work is done on the air increasing it's total energy. The downwards force occurs due to the combination of an effective angle of attack and a forwards speed. The downwards force results in a downwards acceleration of air. The lower surface of a wing simply deflects air downwards. The upper surface draws air downwards, to fill in the void left behind by the upper surface of a wing, and because of friction and viscosity, all of which create a Coanada like effect, even in the case of turbulent flow. Aerodynamic forces are the result of accelerations of the air, not relative velocities. Regardless of the frame of reference, the wing itself or the air itself, most of the acceleration of the air is downwards corresponding to lift, and some of it is forwards corresponding to drag. The lower pressure area above a wing accelerates air toward that low pressure area in all directions, except the air can't flow through (it can flow around) the wing itself, resulting in a net downwards (and forwards) acceleration of air. The higher pressure below a wing causes air to accelerate away from that high pressure area in all directions, except that air can't flow through the wing, also resulting in a net downwards and forwards acceleration of air. Things get more complicated because air can flow around a wing. Vortices are created at the wing tips. Air flow near the leading edge of a wing is diverted somewhat over the wing, reducing the overall pressure differential. questions that I don't have good answers for: It seems that it should be more efficient to draw air downwards from above a wing than to deflect it from below. The total acceleration of the air results in an average terminal velocity of the air mostly forwards and somewhat downwards, and a large increase in kinetic energy of the air. For efficient airfoils, most of this increase in kinetic energy occurs through a Bernoulli like conversion of pressure energy into kinetic energy, reducing the amount of energy required to produce lift. Above a wing, the pressure is reduced below ambient, and air will accelerate towards this low pressure zone, exchanging pressure energy for kinetic energy as it accelerates towards this moving low pressure zone, with the wing just moving past before the downwards component of air flow "catches up". I don't fully understand the process of deflection below a wing; mechanical defelection of air would seem to increase both pressure and kinetic energy, consuming a significant amount of energy, but the M2-F2 lifting body glider, a deflection based airfoil, glides reasonably well. Using a wing based reference, the air stream is slowed down and the pressure is increased, a Bernoulli like conversion. However, an air based frame of reference has to work just as well, and in this case the wing accelerates the air downwards (and somewhat forwards), and this acceleration is due to a moving high pressure area under the wing. The work done could be based on the integral of the downwards force across the vertical component of distance that the lower surface of the wing is in "contact" with the air, which is the wing chord distance times the sin of the angle of attack. However this same concept of force times vertical component of distance of the wing in "contact" with the air could also be applied to the upper surface of a wing. The camber in air foils helps in that the net vertical component of distance is reduced (the distance is "up" for the leading portion, then "down" for the trailing portion), and fully cambered airfoils are generally more efficient than partially cambered airfoils. Last edited: Renge Ishyo It's not a small difference, it is an incorrect explanation versus a correct explanation. The books I have read that have used this theory have been very careful to stipulate that the model that gives rise to the equal transit theory is based on the assumption that there is laminar flow and that the flow lines never cross or break. If these assumptions are accepted (nevermind whether or not they carry over into "actual" reality) then the conclusions of the theory follow. Again I don't like the word "incorrect" versus "correct" explanation because by assuming laminar flow the model never pretended to be anything more than a crude approximation anyway. The kinetic theory of gases is also a crude approximation that is used in education as a teaching tool. So is it wise to substitute that out for the much more difficult models because the assumptions that bring about the kinetic theory are "incorrect"? Crude simplified approximations still have a place in Physics even if you personally get to leave them behind as you become more of an expert in the field. If you want simple, then use the Newton approach: wings produce lift by applying a downwards force to the air, which reacts with an equal and opposite upwards force to the wing, in accordance with Newton's third law. Newton's third law holds true regardless of wing effeciency. If you tried that explanation in a beginning physics class the first thing a student would ask in response to "wings produce lift by applying a downwards force to the air" would be "how does it do this"? Then you are still left with the problem of trying to explain how the airflow across the wing generates this force. As a physicist you may want to give the more accurate explanation as above, but as a teacher of beginning physics you probably wouldn't succeed if you did so... Holocene Why does an airplane fly? Magic. Seriously, as I pilot, I was actually taught the so-called "equal transit" theory. Later I learned it was false, and I was pissed that not only was I taught nonsene, but that publishers wrere willing to include this garbage in textbooks. Seriously, why has the bogus explanation persisted? Mentor Why does an airplane fly? Because driving is too slow. (Sorry, couldn't resist the temptation! ) Homework Helper If you want simple, then use the Newton approach: wings produce lift by applying a downwards force to the air, which reacts with an equal and opposite upwards force to the wing, in accordance with Newton's third law. The downwards force occurs due to the combination of an effective angle of attack and a forwards speed. If you tried that explanation in a beginning physics class the first thing a student would ask in response to "wings produce lift by applying a downwards force to the air" would be "how does it do this"? The downwards force occurs due to the combination of an effective angle of attack and a forwards speed.This can be demonstrated with a fan blowing air and a flat piece of cardboard, changing the angle on the cardboard changes the amount of lift and drag. Or the students, as passengers in car or a school bus, can stick their arms out the window of the moving car and by angling their hand, make their arms "fly". The books I have read that have used this theory have been very careful to stipulate that the model that gives rise to the equal transit theory is based on the assumption that there is laminar flow and that the flow lines never cross or break. If these assumptions are accepted (nevermind whether or not they carry over into "actual" reality) then the conclusions of the theory follow. Again I don't like the word "incorrect" versus "correct" explanation because by assuming laminar flow the model never pretended to be anything more than a crude approximation anyway. The issue isn't about an inverse relationship between airspeed^2 (kinetic energy) and pressure (energy), since this relationship does exist in a theortical no work situation and approximated in a real world low work situation. The "incorrect" part is implying that there is some tendency for air molecules to regroup back to their previous relative positions after separation by an air foil, as if there was some type of memory based attraction between air molecules separated into independent streams by a solid object. An example I saw on one website (will try to find it later) included a picture of a hoop (circular cylinder) cut open with one end facing into the wind and the other end facing away from the wind. The air molecules that flow inside and around this hoop will be displaced far away from the molecules that they were previously adjancent to. As already mentioned, the longer surface on the M2-F2 and M2-F3 is on the bottom, the top is a flat surface, so the distance the air has to travel is much longer under (big curve) these lifting bodies than over (a straight path); if equal transit theory were true, then the M2-F2 and M2-F3 couldn't glide or fly. The Bernoulli relationship also needs to work in the case air is used as a frame of reference. It's a dynamic situation, the wing passes horizontally through the air (horizontal by definition in this case, with lift perpendicular to the direction of travel, and drag in the direction of travel), with a stream of air flowing downwards and somwhat forwards towards the wing and passing behind the wing as it passes forwards through a volume of air. Last edited: rwjefferson As fluid air mass stream passes height of contour, fluid mass inertia creates lower pressure over wing. Low pressure over wing accelerates air stream downward as per Bernoulli. In verse (equal and opposite), wing is forced upward. Peace rwj Type_R For as long as I can remember the anwer I've heard to this question was the "Equal Transit" theory - that since the top of the wing is longer that the bottom the air has to go more quickly over the top inorder to "keep up" with the bottom causing a lower pressure on the top than the bottom. But this doesn't really make sense - why should the air on top "care" about the air on the bottom? When I looked on Nasa's website I found a different explanation: http://www.grc.nasa.gov/WWW/K-12/airplane/right2.html and a debunking of the "Equal Transit" theory: http://www.grc.nasa.gov/WWW/K-12/airplane/wrong1.html So why have I never come accross the explanation Nasa gives and everyone gives the "Equal Transit" theory? Which explanation is true? Or is it both? Thanks. Easy...stick your hand out the window while riding in a car and make it flat with palm facing down. Now direct it upwards you'll feel a lift. Now direct it downwards... you'll feel a drag. This is caused by the different pressure of wind underneath your palm versus the pressure of air above your hand. Homework Helper Since someone else blew the dust off this thread ... I thought I'd do a bit more cleaning. As fluid air mass stream passes height of contour, fluid mass inertia creates lower pressure over wing. Inertia allows the lower pressure zone to exist, but it doesn't create the low pressure areas. Most of the lowering of pressure is due to what I call "void effect", from Wiki on wings: In that case a low pressure region is generated on the upper surface of the wing which draws the air above the wing downwards towards what would otherwise be a void after the wing had passed. http://en.wikipedia.org/wiki/Wing I thought "void theory" was more common, but I've only seen it used at a few sites in refernce to lift. I seem to be the main user of the term, except for the wiki article and this one The plate scoops out a void ... http://www.terrycolon.com/1features/fly.html The issue is that fluids and gasses prevent the creation of voids, so what would be a void ends up as a low pressure area. A better term would be "void abhorence theory". The upper surface of a wing "attempts" to create a void, but this can't be done in a fluid or gas, and the result is a low pressure zone. Usually "void theory" is used to explain why most of the drag on a land vehicle, like a bus on a highway, occurs at the back of the bus. Here is a link about propellers, where the amount of work done on air is significant: http://www.grc.nasa.gov/WWW/K-12/airplane/propanl.html Note that both a wing and a propeller operate in their own induced wash, and perform work (a non-Bernoulli interaction) on the air, but the amounts are much less in the case of a wing. Since wings are so close to being 100% efficient (some are over 99% efficient), the work related aspects are often ignored (except for drag related effects). Another tid-bit for thought. The downforce applied by the wing to the air propagates through the air and eventually ends up as downforce applied by the air to the surface of the Earth (coexistant with an upwards force from the surface of the Earth onto the air). This effect is more commonly presented as birds or a model flying inside a closed container, but the Earth and it's atomosphere are part of a closed system. I don't see how "void abhorance" could contribute to lift. It will imply a horizontal drag on the wing, or a vertical drag contributing to a lower terminal velocity if the wing is falling. If the wing is flying and not stalled, I don't see that lowered pressure from relatively static air helps add a positive vertical force. Homework Helper I don't see how "void abhorance" could contribute to lift. It will imply a horizontal drag on the wing. Note that the uppper surface of a wing is cambered (or angled downwards for a flat wing) at a slight angle of attack. Using the air as a frame of reference, the surface is moving "away" from the air downwards much more slowly than that surface is moving forwards, so the air will take the path of least acceleration, which will be mostly downwards (lift) and only a tiny amount forwards (drag). This is why streamlining the back end of land speed vehicles with long tapered tails works. It minimizes the amount of forwards acceleration of air (drag) by letting the surrounding air accelerate inwards at a relatively slow rate towards the receding tapered tail as opposed to accelerating forwards at a very fast rate to chase the tail. For example, a tear drop shape is very efficient at reducing drag, and many air foil cross sections are basically thinned and cambered tear drop like shapes. Getting back to a wing, the top half is similar to the tapered tail, but here the goal is to generate significant downwards acceleration of air, with minimal forward acceleration. This works even with a flat board wing if the angle of attack is reasonable. In my opinion, what occurs at the bottom of an effcient wing is also tricky, as the air diverted downwards from under an efficient wing also experiences a reduction in pressure. Part of this is because the lower still pressure air that is flowing downwards from above the wing past the aft end of the wing is helping to draw the air from under the wing into that flow. Last edited: Steve Stone Hi everyone, I have always been amazed how a jumbo jet can get tanks landrovers and men into the air because the wings look tiny compared to the hold. There have been some interesting points here on how lift is acieved; but all that said how is it possible for a stunt plane to fly upside down? I thought at first it was an optical illusion in that the plane had forward movement but was in reality falling because wings simply would not work inverted. But then a few years ago I saw a stunt plane fly over us (the crowd) very low with smoke coming from the tail. It was easy to see the plane was flying in a straight line by lining up the smoke trail with the horizon by eye. Thanks, Steve rwjefferson Inertia allows the lower pressure zone to exist, but it doesn't create the low pressure areas. The wing and fluid mass inertia creates the lower pressure zone. Although it is the wing sweeps air out from underneath the air over the wing, if there were no inertia, there would be no lower pressure behind the height of contour. Peace rwj rwjefferson …but all that said how is it possible for a stunt plane to fly upside down? Thanks, Steve You seem to be confused by the dogma that says a wing creates lift because the upper surface is longer than the lower. By reality, any wing that deflects air can create lift. A wing lifts by forcing air mass downward. The shape of the wing is based on how to most efficiently accomplish this with minimum drag. Most stunt planes have symmetrical wings. Peace rwj Homework Helper How is it possible for a stunt plane to fly upside down? As a passenger in moving car, stick your hand out the window, straighten it out and angle it up to get lift, and down to get downforce. Although somewhat rare these days, hobby stores have these very cheap small balsa gliders that use flat board wings. As stated above, the curved surfaces on a wing aren't required in order to produce lift, they just consume less energy in the process. Symmetrical airfoils are used on stunt planes, the curve on top and bottom are identical. Symmetrical airfoils are also commonly used on helicopters, to prevent the downwards pitching torque on the blades that results from using cambered air foils (the rotor blades are designed to flex a bit to optimize peformance over a range of pitch angles, and a pitching torque would interfere with this design strategy). Regarding inverted flight, perhaps you missed the links to these lifting bodel NASA models, flat tops, very curved bottoms, great real world examples that disprove equal transit theory: M2-F2 glider (next to F104): http://www.dfrc.nasa.gov/Gallery/Photo/M2-F2/Medium/EC66-1567.jpg M2-F3 rocket powered model (top speed was Mach 1.6): http://www.dfrc.nasa.gov/Gallery/Photo/M2-F3/Medium/EC71-2774.jpg Last edited: Steve Stone Thanks for that Steve	6,936	31,735	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06	latest	en	0.936321
https://ruwix.com/competition/entry/yohaan-master/	1,544,522,002,000,000,000	text/html	crawl-data/CC-MAIN-2018-51/segments/1544376823614.22/warc/CC-MAIN-20181211083052-20181211104552-00562.warc.gz	720,205,966	4,454	# Yohaan Master ## A funny story about the cube ### Yohaan Master (10, Singapore) This incident happened on the sixth of March, earlier this year. 4th grade was reaching its end, and we were more or less passing time till the end of school. Our teachers handed assignments that we finished in the wink of an eye, and finally resigned to giving us free time. Chaos reigned. Bottles were flipped, airplanes were thrown and for a few of us, Rubik’s cubes were solved. At that time my best friend was reigning champion at solving the cube, and I was out to get a better time than him. I was still using the beginner’s method, i.e. I didn’t know OLL and PLL yet. I was on a lucky streak, and I could skip corner placing. I had only a few moves left and I would have been the new champion. Then my luck ran out. I cut the corner too far and it popped. I reached out to catch it…. And I missed. Instead, I punched a boy playing ‘hand cricket’ next to me in the face. The boy glared at me. With a sinking feeling I realized he was on the school martial arts team. He grabbed my cube and just when I thought he would break it, he started to solve the (now jumbled up) cube. But here’s the thing- he was faster than my friend! In 10 seconds flat he solved the cube (and fixed the corner and all) and tossed it at my stomach. Later (after he had forgiven me) I asked him how he did it. He told me all about OLL and PLL and wrote down the algorithms in my notebook. He taught me three look last layer, with two OLL’s. That got me thinking- could i really memorise 57 algorithms and do it in one step? Then I started on memorising all 21 PLL cases. Currently I have memorised 46 of the 57 cases and working on more. So, I guess the reason all this happened is thanks to the bad quality of my first Rubik’s cube!	439	1,808	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.65625	3	CC-MAIN-2018-51	latest	en	0.989118
https://gis.stackexchange.com/questions/245908/creating-prediction-map-by-combining-several-rasters-and-vectors-in-arcgis-deskt	1,720,910,184,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763514517.94/warc/CC-MAIN-20240713212202-20240714002202-00067.warc.gz	233,030,060	39,770	# Creating prediction map by combining several rasters and vectors in ArcGIS Desktop? I have three 'variables': 1. 6000 GPS points that represent where my focal species (a carnivore) is present. Attached to these points is one value representing the activity level of the animal at the time (it ranges between 0 and 200, thus 0 being inactive and 200 being very active). 2. Slope (standard slope raster) 3. NDVI (also raster) I ran a GLMM in R Studio, and the fixed effects show that as NDVI and Slope decrease, activity decreases. This is important because it may also help with an additional analysis to assess areas of risk for my other focal species (an herbivore that also happens to be a prey species of my focal carnivore). I am curious that given such a relationship (between activity points, NDVI, and Slope) whether it is possible to map out 'hotspots of activity' throughout the landscape while accounting for NDVI and Slope as a means to create a 'risk map' or 'probability of activity map' (same thing in this context) I have been reading about several methods including kriging, hotspot analysis, Residual trend mapping, etc. but I am afraid that I am still unsure whether either of these methods are the right way to go (probably because I am still relatively new when it comes to GIS). Would anyone be able to provide some advice on how to carry on with this? • I think you should start by turning the 6000 points to a raster surface with the activity level as the raster value. After all, your carnivores are a continuos surface and not discrete data, they don't appear and disappear like magic. Commented Jun 29, 2017 at 23:56 • Then, with the 3 raster you would be able to start doing map algebra and spatial statistics. Commented Jun 29, 2017 at 23:58 • thank you for your advice, when you mean map algebra, do you mean taking the fixed effects from the GLMM that I ran and plugging it into the raster calculator? (i.e. ndvi.304+slope123 etc.) or do you mean a different method. If so, I am not sure whether that would account for the actual activity points. Commented Jun 30, 2017 at 9:52 • Thank you! Yes, that's what I meant. You would like to test the similarities between the three surfaces as you were saying. Commented Jun 30, 2017 at 15:24 • Different methods to compare raster are listed here: researchgate.net/post/How_to_statistically_compare_two_maps Commented Jun 30, 2017 at 15:25	578	2,423	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.75	3	CC-MAIN-2024-30	latest	en	0.942583
http://www.chegg.com/homework-help/questions-and-answers/2-kg-bag-flour-dropped-roof-25-m-building-net-force-exerted-person-18m-tall-assuming-bag-s-q209771	1,472,506,565,000,000,000	text/html	crawl-data/CC-MAIN-2016-36/segments/1471982967784.68/warc/CC-MAIN-20160823200927-00209-ip-10-153-172-175.ec2.internal.warc.gz	372,508,262	13,348	A 2 kg bag of flour is dropped from the roof of a 25 m building. What is the net force exerted on a person 1.8m tall? Assuming the bag is spherical, falls freely and stops in a distance equal to its diameter, which is 15 cm.	62	225	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.75	3	CC-MAIN-2016-36	latest	en	0.946632
https://www.mapleprimes.com/questions/99434-Error-That-I-Dont-Know-How-To-Fix	1,604,161,573,000,000,000	text/html	crawl-data/CC-MAIN-2020-45/segments/1603107919459.92/warc/CC-MAIN-20201031151830-20201031181830-00311.warc.gz	779,932,496	24,283	# Question:error that I don't know how to fix ## Question:error that I don't know how to fix Maple Hi I attempted to plot this but failed... my instructor gave us a list of functions according to our name and when I put them together into an equation to plot I kept getting an error. I included the function list at the bottom. I am using DEtools: with(DEtools): #tangent field for dy/dx = AB + CD where A -first letter of my name, B - second letter of my first name, C - first letter of my surname, D- second letter of my surname, which in my case (Tran Lam) gives using the Name-function list; t=y^(2) +5 y, r= 3(x^(2) +y^(2)),l=x^(1/(3)), a= sin(xy), dy/dx = (sin (xy) +(3 x^(2)+x^(1/(3))+4 y^(2)+5 y)) dfieldplot(diff(y(x), x) = sin(xy)+3x^2+x^(1/3)+4y^2+5y, y(x), x = -5 .. 5, y = -5 .. 5, title = `tangent field`, color = black); %;Warning, y is present as both a dependent variable and a name. Inconsistent specification of the dependent variable is deprecated, and it is assumed that the name is being used in place of the dependent variable.Error, (in DEtools/dfieldplot) extra unknowns found: xy dfieldplot(diff(y(x), x) = sin(xy)+3x^2+x^(1/3)+4y^2+5y, y(x), x = -5 .. 5, y = -5 .. 5, title = `tangent field`, color = black); Warning, y is present as both a dependent variable and a name. Inconsistent specification of the dependent variable is deprecated, and it is assumed that the name is being used in place of the dependent variable.Error, (in DEtools/dfieldplot) extra unknowns found: xy ------------------------------------------------------------------------------------------------------------------------------ In case the function list is needed. I don't know why there are two different functions for m and n but in my case I don't use them so it doesn't matter. Letter Function Letter Function a sin(xy) m xy b x3 +y n sin(3x-y) c 2x-3y o cos(x+y) d x2 +cos(y) p y3 e x4- 2sin(x+y) q sin(x2-2y) f 5 r 3(x2 +y2) g -7y2 s cos(x) – sin(4y) h sin(x2 –x) t y2 +5y i - 15cos(2xy2) u -4x3y j sin(y2 –x2) v -3xy3 k 2y – sin(y) w x + y1/3 l x1/3 x 2y + x m x+ y sin(x) y x cos(y)- y2 n xy – cos(x) z y2 – cos(y)	826	3,654	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.796875	3	CC-MAIN-2020-45	latest	en	0.800216
http://www.instructables.com/id/Arduino-basics-No2-photoresistor/	1,498,331,241,000,000,000	text/html	crawl-data/CC-MAIN-2017-26/segments/1498128320323.17/warc/CC-MAIN-20170624184733-20170624204733-00128.warc.gz	558,051,747	16,168	In this guide i will show you how yo use the analog input of the Arduino and to control an led according to the values the analog inputs receives. ## Step 1: Gathering Materials What you'll need: 1) Arduino board (here is the one i baught http://www.ebay.com/itm/291113266836?ssPageName=ST... i connected it as an Uno board). 2)LED. 3) Photoresistor ( a photoresistor is a resistor the changes current according to the light, the higher the light level is the lower the resistance). 4) 330 Ω resistor. 5) Arduino cable. 6) Alligator cable. 7) Wires. ## Step 2: Building the Chassis First, connect the photo resistor to the resistor, the resistor goes to Gnd (Ground), from the connection between them connect a wire that goes to Analog input 0. Now connect the Arduino to the computer. We now want to see the values that the Arduino reads on A0 (refer to pic No.3). The purpose of the code is to print (Serial.println) the values the we get from A0. Notice that as i cover the photoresistor with my finger the values are droping (refer to pic No.4). ## Step 3: Values to Analog Now in order to control the led with our values we want to make sure that we output values in range of from 0 to 255. In the code (pic No.1) we see the A0 (senespin) reading values, than we make sure that our values are between 32 and 118 anything higher than 118 will become 118 or anything lower than 32 will become 32 (these are the values my arduino reads, if your range is higher or lower change the code according to it). After the "constrain" command I used the "map" one, this command changes 32 to 255 and 118 to 0, everything between 118 and 32 will scale proportionality between 0 and 255 (notice that as the light level is lower the photoresistor outputs more current the val is lower and the ledval gets higher, and the opposite). ## Step 4: Led Now connect the led to pin 12 and to Gnd and change the code according to picture No.1 the "analogWrite" command outputs a higher current as the ledval gets higher and so the led gets brighter. As you can see in video No.1 the led gets brighter i shut down my desk lamp. **Notice that the code are in the next step. Hope you enjoyed this guide, any comments/ grammar correction will be accepted gladly. Don't forget to view my other Arduino guides. ## Step 5: Codes `void setup()//code 1` ```{ Serial.begin(9600); }``` `void loop() ` ```{ delay(500); } // // // <p>int sens = 0; //code 2<br>void setup() { Serial.begin(9600);</p><p>} void loop() { val = constrain( val,32,118); int ledval = map(val,32,118,255,0); Serial.println(ledval); delay(500); }</p>// // // <p>int led = 12; // code 3<br>int sens = 0; void setup() { pinMode(led,OUTPUT); } void loop() { val = constrain( val,20,190); int ledval = map(val,20,190,255,0); analogWrite(led,ledval); }</p> ``` ## Step 6: That's It I hope you enjoyed this guide, if you have any questions I will answer them happily. Like this project? Check out my profile, my Facebook page and my YouTube Channel. <p>Looks pretty simple. </p> <p>It is..:P</p> 8,198views 11favorites Bio: I always liked making stuff and when my grandfather showed me Instructables I became an addict. I don't have any degrees yet but I ... More » More by crashzoom:	867	3,256	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.59375	3	CC-MAIN-2017-26	longest	en	0.889596
https://math.stackexchange.com/questions/670649/efficient-diagonal-update-of-matrix-inverse/727002	1,708,736,292,000,000,000	text/html	crawl-data/CC-MAIN-2024-10/segments/1707947474470.37/warc/CC-MAIN-20240223221041-20240224011041-00429.warc.gz	401,453,169	35,632	Efficient diagonal update of matrix inverse I am computing $(kI + A)^{-1}$ in an iterative algorithm where $k$ changes in each iteration. $I$ is an $n$-by-$n$ identity matrix, $A$ is an $n$-by-$n$ precomputed symmetric positive-definite matrix. Since $A$ is precomputed I may invert, factor, decompose, or do anything to $A$ before the algorithm starts. $k$ will converge (not monotonically) to the sought output. Now, my question is if there is an efficient way to compute the inverse that does not involve computing the inverse of a full $n$-by-$n$ matrix? • Turns out this was very simple. I'll just write it here if someone else has need for it: $(kI+A)^{-1}=(kI+PDP^{-1})^{-1}=(P(D + kI)P^{-1})^{-1}=P(D + kI)^{-1}P^{-1}$, where $A=PDP^{-1}$ is the eigenvalue decomposition. And the inverse of a diagonal matrix is quickly computed as the matrix with diagonal elements the reciprocal of the diagonal elements of the original matrix. This is roughly 50 times faster than the original solution for my data on my computer. Feb 17, 2014 at 14:22 • I have a similar question, math.stackexchange.com/q/2667566/10063. But my update of the diagonal has distinct entries along the diagonal, so your method doesn't work. – a06e Feb 26, 2018 at 15:16 Turns out this was very simple. I'll just write it here if someone else has need for it: $$(kI+A)^{-1}=(kI+PDP^{-1})^{-1}=(P(D + kI)P^{-1})^{-1}=P(D + kI)^{-1}P^{-1}$$ where $A=PDP^{-1}$ is the eigenvalue decomposition. And the inverse of a diagonal matrix is quickly computed as the matrix with diagonal elements the reciprocal of the diagonal elements of the original matrix. This is roughly $50$ times faster than the original solution for my data on my computer. -- Tommy L EDIT. 1. The Tommy L 's method is not better than the naive method. Indeed, the complexity of the calculation of $$(kI+A)^{-1}$$ is $$\approx n^3$$ blocks (addition-multiplication). About the complexity of $$P(D+kI)^{-1}P^{-1}=QP^{-1}$$ (when we know $$P,D,P^{-1}$$); the complexity of the calculation of $$Q$$ is $$O(n^2)$$ and the one of $$QP^{-1}$$ is $$\approx n^3$$ blocks as above. Note that one works with a fixed number of digits. 1. When the real $$k$$ can take many (more than $$n$$) values, one idea is to do the following The problem is equivalent to calculate the resolvent of $$A$$: $$R(x)=(xI-A)^{-1}=\dfrac{Adjoint(xI-A)}{\det(xI-A)}=$$ $$\dfrac{P_0+\cdots+P_{n-1}x^{n-1}}{a_0+\cdots+a_{n-1}x^{n-1}+x^n}$$. STEP 1. Using the Leverrier iteration, we can calculate the matrices $$P_i$$ and the scalars $$a_j$$ $$P_{n-1}:=I:a_{n-1}:=-Trace(A):$$ for $$k$$ from $$n-2$$ by $$-1$$ to $$0$$ do $$P_k:=P_{k+1}A+a_{k+1}I:a_k:=-\dfrac{1}{n-k}Trace(P_kA):$$ od: During this step, we must calculate the exact values of the $$P_i,a_j$$ (assuming that $$A$$ is exactly known), which requires a very large number of digits. Then the complexity of the calculation is $$O(n^4)$$ (at least) but it's done only one time. STEP 2. We put $$x:=-k_1,-k_2,\cdots$$. Unfortunately, the time of calculation -with a fixed number of significative digits- of $$R(-k)$$, is larger than the time of calculation of $$(-kI-A)^{-1}$$ !!!	983	3,161	{"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": 27, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.796875	4	CC-MAIN-2024-10	latest	en	0.859282
https://math.stackexchange.com/questions/3049164/can-we-obtain-an-explicit-and-efficient-analytic-interpolation-of-tetration-by-t/3049421	1,560,638,183,000,000,000	text/html	crawl-data/CC-MAIN-2019-26/segments/1560627997501.61/warc/CC-MAIN-20190615222657-20190616004600-00080.warc.gz	512,977,558	41,638	Can we obtain an explicit and efficient analytic interpolation of tetration by this method? I am curious about this. It has been a very long time since I have ever toyed with this topic but it was an old interest of mine quite some time ago - maybe 8 years (250 megaseconds) ago at least, and I never really got to a conclusion, nor do I think anyone did entirely satisfactorily, and some comments on a Youtube video I was watching inspired me to dust it off and try once more to give it another thwack. And the question is, basically, how can one construct a "reasonable" interpolation of the "tetration" operation, also called a "power tower", which for those who have not heard of it is defined for natural $$n > 1$$ and real $$a > 1$$ by $$^{n} a = a \uparrow \uparrow n := \underbrace{a^{a^{a^{...^a}}}}_{\mbox{n copies of a}}$$ where the nested exponentiations on the left are evaluated in rightassociative fashion, so the deepest ("highest") layer is done first, e.g. $$^{3} 3 = 3^{3^3} = 3^{27} = 7\ 625\ 597\ 484\ 987$$ and not left-associative, i.e. $$^{3} 3 \ne (3^3)^3 = 3^{3 \cdot 3} = 3^9 = 19\ 683$$ What the "interpolation" part means is basically, given that this definition clearly only works for values of the second argument, $$n$$ (called as such by analogy with exponentiation even though it is written first in the "left-superscript" notation just introduced), usually called the "height" of the tetration or power tower for obvious reasons, that are natural numbers at least 1, since we have to have a whole number of "copies of $$a$$" - half a copy, say, wouldn't make much sense, as while you can literally write a half-written "$$a$$", that is formally nonsense and has no mathematical meaning, although it may have other forms of meaning from other angles of human understanding and analysis, e.g. perhaps as a form of artsey commentary. Mathematical meaning is, though, of course, what we're interested in. We can, of course, naturally extend this in a way similar to extending exponentiation to the integers by noting that $$^{n+1} a = a^{^n a}$$ and thus $$^{n-1} a = \log_a(^n a)$$ and if we do this we can at least extend that $$^0 a = 1$$, similar to exponentiation, and $$^{(-1)} a = 0$$, a rather interesting result when viewed in contrast to exponentiation given that the first negative exponentiation of a number is not a constant but instead its reciprocal. Of course, we cannot extend now to $$^{(-2)} a$$, as then we get $$\log_a(0)$$ which is undefined (though of course if you want to stretch the maths a bit and expand the codomain to the extended reals, you can say $$^{(-2)}a = -\infty$$. In any case though, $$^{(-3)} a$$ and further are definitely, really undefined, since no real exponential can be negative, much less negative infinity!). So this peters out. Of course, the most interesting bit - as hinted at with the "half a copy" business above - is trying to extend the height $$n$$ to real values, presumably in $$(-2, \infty)$$ at least. And there have been a number of methods fielded in those past epochs which attempt to do this as well as some interesting ones regarding the conditions which are required to produce a suitably "natural" extension, given that it is trivially obvious that one can, of course, "interpolate" a given sparse sequence of points in any way that one desires and, moreover, even with the identities $$^{n+1} a = a^{^n a}$$, they only suffice to make it unique insofar as whole-number increments of the tower are concerned - fill any unit interval with anything you like, and the identity will extend to provide an interpolant that will satisfy it. For exponentiation, this non-uniqueness is much less of a problem because we also have the additional identity $$a^{n+m} = a^n a^m$$, which lets us extend to rational values, however no such identity exists for tetration. In this regard, extension of tetration is similar to the question of extension of the factorial, which is similarly impoverished of identities, with new ones interestingly only coming about after the extension was done by Euler in the form of the gamma function, to meet a challenge originally proposed by Bernoulli to do exactly this. The gamma function, however, is still ostensibly more "natural" simply because a) it often crops up and b) it has some very cute integral representations, esp. the darlin' $$\Gamma(x) = \int_{0}^{1} [-\log(u)]^{n-1}\ du$$ (Though with regard to objection a), one could say this may be simply because we have not yet found such an expression, and thus places where it might be useful, could be currently written off as "unsolvable".) Yet clearly, that doesn't seem to have been the case for tetration, either. Moreover, in all these past discussions, many of the extension methods proposed are in fact extremely cumbersome and computationally intensive to approximate, involving elaborate constructs like Riemann mappings, infinite limits of integral equations, and so forth - all things that are, while mathematically valid, both inelegant and also not something you're going be able to program into a software pack like Mathematica and have it spit out 2500 digits of $$^{1/2} 2$$ in the blink of an eye. But nonetheless, one particular method out of these proposed methods seems be both fairly simple and like that it might possible be amenable to more detailed analysis, and that is the "Carleman matrix" operator method. This method is most succinctly expressed for the specific case $$a = e$$, to construct the "natural tetrational" $$\mathrm{tet}(x) :=\ ^x e$$ with real height $$x$$, so we'll just focus on that for now. But basically it is based on the following two observations. The first is that one can consider the result of the height-$$n$$ power tower of $$e$$ as the iterated exponential evaluated at $$1$$, namely $$^n e = \exp^n(1)$$ or perhaps more nicely for what we're about to do, $$^{n-1} e = \exp^n(0)$$ which has some interesting gamma-function like quality about it with the offset. And the second one is the following. If we let $$\exp$$ be given by its power series, $$\exp(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$ then we can actually represent such iterated exponentials using what is called its Carleman matrix, basically the infinite-order "matrix" with entries $$C[\exp]_{ij} = \frac{i^j}{j!}$$ such that if we have the infinite vector of exponential coefficients $$\mathbf{a} = \left[ \frac{1}{1!}\ \frac{1}{2!}\ \frac{1}{3!}\ \cdots \right]^T$$ then the vector $$\mathbf{b}_n = (\mathbf{C}[\exp])^n \mathbf{a}$$ is the coefficients of $$\exp^n$$. In particular, if we sum the top row, we get exactly what we want: $$^{n-1} e$$. Now the question is, though, how can we compute this matrix power for fractional $$n$$ in some explicit form? It seems one possible way to do this, and the way that I saw when this method was suggested (by Gottfried Helms, who was here a long time ago, not sure if they're still so) was to try to diagonalize the matrix, so that you can use the fact that if a matrix $$\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}$$ for some diagonal matrix $$\mathbf{D}$$ (which happens to be the matrix of eigenvalues) then $$\mathbf{A}^t = \mathbf{P}\mathbf{D}^t\mathbf{P}^{-1}$$ where that the inner power is easy to compute as you just take exponents of the diagonal terms. And numerically this seems to work for at least finite truncations of the matrix $$\mathbf{C}[\exp]$$, but analytically it may be on shaky ground, as I see with this thread here that I just saw when trying to dig into this once more: Diagonalization of an infinite matrix and moreover it's still not super efficient as we'd like - we're not computing 2500 digits of $$^{1/2} e$$ and I want them computed, dammit! However, it seems that in this case some of the objections raised in that thread do not perhaps apply here. In particular, it is mentioned how that an infinite matrix diagonalization is ambiguous (seems to mirror the situation with the tetration interpolation generally where there is great freedom) due to choice of suitable normed vector space over which to make sense of it, and moreover in that question it was pointed that the usual most "natural" space, $$\ell^2$$, did not work for that questioner's particular matrix, because in particular it would "map" most "vectors" of $$\ell^2$$ to effectively outside of the space. However, this Carleman matrix seems better behaved - in particular, due to the fact that any vector $$[\ a_0\ a_1\ a_2\ \cdots\ ]^T \in \ell^2$$ by definition must have terms that converge absolutely as an infinite sum, then that owing to the factorials in the matrix when we multiply by it we should also get an (even more) convergent series, as is illustrated by considering the bounding "vector" $$[\ 1\ 1\ 1\ \cdots\ ]^T$$ as "formally" acted upon by $$\mathbf{C}[\exp]$$. So it seems that in that regard, we are in better shape against at least the objections raised in that thread in this case than for that poster's scenario. Thus the question I have is, if we take the relevant target space as $$\ell^2$$, can we find a "natural" infinite diagonalization and thus matrix-power for this case, and moreover express it somehow in terms of at least some sort of infinite combinatorial sums or otherwise easily-manipulated expressions for its coefficients? Moreover, another interesting and seemingly natural question provoked by this is, exactly how sensitive is the method to the choice of at least norm used to interpret the matrix power, i.e. can we have absolute freedom to interpolate $$^n e$$ in any way we please, and if not, then just how much do we get? I suspect "a lot", but there are some "lots" that are more than others in mathematics, even when infinities are concerned, thanks to Cantor. And can we do the inverse - i.e. if I fill in $$^{n} e$$ with some freely-chosen interpolant in the interval $$[0, 1]$$ (for the purpose of making things easy I'll assume it's continuous and moreover equals $$1$$ at $$x = 0$$ and $$e$$ at $$x = 1$$) - can we find a norm such that the associated Carleman matrix power will produce that interpolant? Does it have to be analytic (seems right, but keep in mind that we are summing the top row, not necessarily creating a power series valid for all or even any inputs, though again it also "seems" right that if not analytic, it'll diverge)? If so, what's the proof? ADD (epoch time 1545.46 Ms): In the quest for an at least summatory-formula shot at the diagonalization, I note this matrix has the interesting relations among rows and columns given by $$C[\exp]_{(i+1)j} = \sum_{k=0}^{j} \frac{1}{(j - k)!} C_{ik}$$ and $$C[\exp]_{i(j+1)} = \frac{i}{j+1} C_{ij}$$ Not sure if this helps anything, though. But at least it shows there is structure and thus we're not just dealing with effectively purely random matrices and thus in theory it might somehow be exploited in some fashion to simplify things. ADD 2 (same time): You should actually sum the second row of the matrix power to get the tetration $$^n e$$, not the first to get $$^{n-1} e$$. The first row always sums to 1. ADD 3 (ue+1545.47 Ms): The first row formula above allows us to derive the interesting property that if $$\mathbf{C}[\exp]$$ is right-multiplied by the factorial matrix $$\mathbf{D} := \begin{bmatrix} 1 & \frac{1}{1!} & \frac{1}{2!} & \frac{1}{3!} & \cdots \\ 0 & 1 & \frac{1}{1!} & \frac{1}{2!} & \cdots \\ 0 & 0 & 1 & \frac{1}{1!} & \cdots \\ 0 & 0 & 0 & 1 & \cdots \\ & & \cdots & & \end{bmatrix}$$ where $$D_{kj} = \frac{1}{(j-k)!}$$ (with negative-argument factorials "naturally" extended as $$\infty$$ so these $$D$$ terms are $$0$$), it shifts up by one row. ADD 4 (sim. time): It looks like we can move both up and down, in particular the matrix $$\mathbf{D}$$ with entries $$D_{kj} = (-1)^{k-j} \frac{1}{(k-j)!}$$ will shift down, while the other matrix above, perhaps better called $$\mathbf{U}$$ instead will shift up. Shifting left and right is possible as well but via the Hadamard product and not the ordinary product, as the second set of relations between columns indicates. In particular, the Hadamard product with the matrix $$\mathbf{L}$$ with entries $$L_{ij} = \frac{i}{j+1}$$ will shift left, and the matrix $$\mathbf{R}$$ with entries $$R_{ij} = \frac{j}{i}$$ will shift right. Thus we have some interesting system for "moving" the matrix about like some kind of tableau or grid - not sure what these symmetry properties do though for easing the analytic solution/explicit formula of the matrixpower as a summation. • By my numerical tests, the method using truncated Carleman-matrices and try to approach to a sensical result by increasing the size of the matrix, it seems as if we implement an approximation to the Kneser method. The unfortunate thing is that the diagonalizations lead to sets of eigenvalues which vary much with the matrix-size, so for instance use of a 64x64-matrix approximated the Kneser-solution to some error of 1e-6 or so. The advantage(?) of the method is to have a real-to-real approximation for real fractional "heights". One other method (...) – Gottfried Helms Dec 22 '18 at 10:52 • (...) One other method tries to do the analysis with the ansatz of the Schröder-mechanism which introduces power series for the exponentiation shifted around the attracting or repelling fixpoint. But this works nicely only with bases $b$ from the range $1 \lt b \lt e^{1/e}$, for general bases we get power series/Carlemanmatrices with complex coefficients and difficult to evaluate for fractional heights - and in the general case, complex values even for real fractional heights. I've compared 5 basic methods in a small essay go.helms-net.de/math/tetdocs/ComparisionOfInterpolations.pdf – Gottfried Helms Dec 22 '18 at 10:58 • @Gottfried Helms : Yupp! What I'm trying to do is see if we can't break it down analytically to find a combinatorial formula (infinite summation) for the Kneser tetration, which may enable both its fast and efficient evaluation and its mathematical analysis for interesting properties and relations to other areas of maths. However that's a bit discouraging, since it suggests that a direct attack by diagonalization may not work in the infinite case even if we can fangle some kind of formula. – The_Sympathizer Dec 22 '18 at 10:59 • The coefficient seem so nice and simple that it feels as if there's almost gotta be some kind of explicit form, but I don't know, it's no proof... (or disproof, either :((( ) – The_Sympathizer Dec 22 '18 at 10:59 • Last comment at the moment. At the problem of analytical diagonalization of the infinite Carlemanmatrix: I've really invested much time to do this - found an analytical expression via the $\exp(x)-1$ case and then seeing that this was identical to the Schröder-solution (without knowing it at all and being pointed to it only later). There might be one more possible alternative path here. I've read an article of Eri Jabotinsky who extended the Carlemanmatrix to the twoside infinite index and thus including the $f(x)^{-k}$-case which reminded me of Ramanujan-integrals. But I couldn't get in... – Gottfried Helms Dec 22 '18 at 11:12 Just a short answer what I've found concerning to express analytically the matrix-diagonalization for $$C[exp]$$ which I called $$B$$ for $$\exp(x)$$ with base $$b=e$$ and $$B_b$$ for the general $$b^x$$ . Using the notation $$V(x)=[1,x,x^2,x^3,...]$$ for the infinite "Vandermonde" row-vector of consecutive powers of $$x$$, $$P$$ for the upper-triangular Pascalmatrix, $$S2$$ for the lower-triangular matrix of Stirling-numbers 2'nd kind, $$F$$ and $$f=F^{-1}$$ for the diagonal matrix of factorials and its inverse we can write $$V(x) \cdot B =V(e^x)$$ By simple $$LDU$$-decomposition of $$B$$ we find $$\begin{array}{} &B = (f \cdot S2 \cdot F) \cdot P \\ V(x) \cdot ( (f \cdot S2 \cdot F) \cdot P) &= V(e^x) \\ \end{array}$$ What I now found (by numerical tests inspired by some hypothese) was that a pair of suitable powers of $$P$$ would triangularize $$B$$: $$\begin{array}{} B &= P^{-t} \cdot U_t \cdot P^{t} \\ \end{array}$$ where $$t$$ is the (complex) fixpoint of $$\exp()$$ and $$U_t$$ was lower triangular. Let's write $$u$$ for $$u=\log(t)$$ and $$\;^dV(\cdot)$$ for the diagonally written $$V(\cdot)$$. Then this is simply $$U_t = \;^dV(u) \cdot fS2F$$ This can be diagonalized (but with complex coefficients if $$t$$ or more precisely $$u (=\log(t))$$ is complex as in our current example) $$\begin{array}{} U_t &= W_t \cdot \;^dV(u) \cdot W_t^{-1} \\ \end{array}$$ So all in all we could write $$\begin{array}{} V(x) \cdot ( P^{-t} \cdot (W_t \cdot \;^dV(u) \cdot W_t^{-1} ) \cdot P^{t}) &= V(e^x) \\ V(x) \cdot ( P^{-t} \cdot (W_t \cdot \;^dV(u^h) \cdot W_t^{-1} ) \cdot P^{t}) &= V(\exp°^h(x)) \\ \text{ and conveniantly } \\ (V(x) \cdot P^{-t}) \cdot (W_t \cdot \;^dV(u^h) \cdot W_t^{-1} ) &= V(\exp°^h(x))\cdot P^{-t} \\ \end{array}$$ Ironically, by the functionality of $$P$$ and its powers $$P^{-t}$$ we have $$V(x) \cdot P^{-t} = V(x-t)$$ and our formula reduces by further cosmetic to $$\begin{array}{} V(x-t) \cdot (W_t \cdot \;^dV(u) \cdot W_t^{-1} ) &= V(e^x-t) \\ \end{array}$$ and one recognizes the mechanism of shift of the powerseries towards its fixpoint $$t$$! It needed not much time to recognize that $$W_t$$ is just the Carlemanmatrix for the Schröder-function, having complex coefficients (in our current example), thus this all gives complex resulting values for fractional real heights on real startingvalues $$x$$. This all is simpler (and has only real numbers) if the fixpoint $$t$$ is real, and so the base $$b$$ is $$1 \lt b \lt e^{1/e}$$ thus $$1 \lt t \lt e$$ and $$0 \lt u (=\log(t)) \lt 1$$. For instance, if $$b=\sqrt{2}$$ then $$t=2$$ and $$u=\log(2) \approx 0.693$$ then also $$W_t$$ has only real coefficients. Moreover, all coefficients in $$U_t = W_t \cdot \;^dV(u^h)\cdot W_t^{-1}$$ (where $$h$$ is the iteration-height, possibly fractional) are real and are descibable in terms of polynomials in $$u$$ and $$h$$, with the order depending of the index in the powerseries (see text, see index-of-pages) Additional remark: W. Reshetnikow in mathoverflow presented this year(don't have the link at hand) an ansatz which avoided the Carleman-matrix-method introducing essentially the $$q$$-factorials and -binomials. This made it really good looking! However, by my own earlier analyses I had found out that this leads to the same polynomials in $$u$$ and $$h$$ as I had found them via the just described analysis of the infinite Carlemanmatrices. So the analytical access via infinite Carlemanmatrices and their diagonalization leads naturally to the Schröder-function and the Schröder-mechanism (including shift towards fixpoints) - which is, what you did not want... • Hmm. This is interesting, as it suggests that the matrixpower is not unique, as thought. What you really want is then a matrixpower which ends up as somehow the limit of the finite matrixpowers for truncated matrix. – The_Sympathizer Dec 22 '18 at 12:48 • @The_Sympathizer : hmm, only to avoid a possible misunderstanding: what/where do you mean that refers to "matrix-power is not unique"? – Gottfried Helms Dec 22 '18 at 12:51 • It seems (and I might have read it wrong) that you constructed a sort of matrixpower which causes the Carleman matrix to power like the Schroder equation solution. – The_Sympathizer Dec 22 '18 at 12:52 • Though now I'm not sure... – The_Sympathizer Dec 22 '18 at 12:53 • Yes, it "implements" exactly the Schröder-mechanism, and so the fractional powers of matrix and by this the fractional iterates of function. Note that the eigenmatrix $W$ can in general be understood as (formal) infinite power of the basic matrix-operator for the function and this reflects then as well the term of the infinite iterated basic-function in the Schröder-mechanism. So $W_t$ is the Carlemanmatrix of the Schröder-function $\sigma()$ – Gottfried Helms Dec 22 '18 at 12:57	5,418	20,013	{"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": 127, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.9375	4	CC-MAIN-2019-26	latest	en	0.961971
https://jp.mathworks.com/matlabcentral/answers/594505-about-epwm-block-ti-c2000-dc-dc-buck-converter-example	1,719,133,439,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198862464.38/warc/CC-MAIN-20240623064523-20240623094523-00263.warc.gz	288,505,827	29,979	# About EPWM BLOCK - TI C2000 DC/DC Buck Converter Example 5 ビュー (過去 30 日間) Burak Caykenari 2020 年 9 月 16 日コメント済み: Venkatesh Chilapur 2020 年 9 月 18 日 Hello Everyone, I have basic question about f28379d EPWM block in this example -> (https://www.mathworks.com/help/supportpkg/texasinstrumentsc2000/ug/dc-dc-buck-converter.html) According to working principle of ADC , We don't make the ADC start of conversion,while pwm transitioning . For this reason ADC start of conversion launchs the conversion before the X(any time) of period. In buck example with f28379d , and its EPWM block as following: This is provided via CMPB value. Period is 1000. When counter is equai to CMPB value , ADC will start conversion as you can see at CMPB value. But I didn't understand how to choose CMPB value as (15+54)/2 ? What are the mean 15 and 54 values ? Where do they come from ? Can you explain briefly ? Manythanks. Burak . サインインしてコメントする。 ### 回答 (1 件) Venkatesh Chilapur 2020 年 9 月 16 日 Hi, The first factor i.e 15 is related to the ADC SOCx acquisition window. The second factor i.e 54 is time from the end of the S+H window until the ADC results latch in the ADCRESULTx register. Refer the ADC block in the model and the controller TRM for more details. Since we have 2 SOC's configured here, hence the mutiplication factor of 2. HTH, Regards, Venkatesh C ##### 6 件のコメント4 件の古いコメントを表示4 件の古いコメントを非表示 burak caykenari 2020 年 9 月 18 日 Hi Venkatesh , My final question, I'm asking for confirmation :) 'But lets say, you have one SOC in ADC-A module and other in ADC-C, since they are different ADC module, they can run parallel.' In this time equation is (1000-(15+54)). ? These informations which you gave was very useful , I really appreciate you . Manythanks. Venkatesh Chilapur 2020 年 9 月 18 日 Correct! サインインしてコメントする。 ### カテゴリ Help Center および File ExchangeTexas Instruments C2000 Processors についてさらに検索 R2019b ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting! Translated by	585	2,047	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.59375	3	CC-MAIN-2024-26	latest	en	0.661191
http://oeis.org/A268192	1,556,160,884,000,000,000	text/html	crawl-data/CC-MAIN-2019-18/segments/1555578678807.71/warc/CC-MAIN-20190425014322-20190425040322-00463.warc.gz	129,052,098	4,613	This site is supported by donations to The OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A268192 Triangle read by rows: T(n,k) is the number of partitions of weight k among the complements of the partitions of n. 2 1, 2, 2, 1, 3, 0, 2, 2, 2, 0, 2, 1, 4, 0, 2, 1, 2, 0, 2, 2, 2, 2, 2, 0, 4, 0, 0, 2, 1, 4, 1, 2, 0, 6, 0, 2, 2, 1, 0, 2, 0, 2, 3, 2, 0, 6, 0, 2, 4, 4, 0, 2, 0, 2, 2, 0, 0, 2, 1, 4, 0, 6, 0, 2, 4, 5, 0, 6, 0, 4, 2, 0, 0, 4, 1, 0, 0, 2, 0, 2, 2, 4, 0, 2, 6, 5, 0, 6, 0, 8 (list; graph; refs; listen; history; text; internal format) OFFSET 1,2 COMMENTS The complement of a partition p[1] >= p[2] >=...>= p[k] is p[1]-p[2], p[1]-p[3], ..., p[1]-p[k]. Its Ferrers board emerges naturally from the Ferrers board of the given partition. The weight of a partition of n is n. Sum of entries in row n is A000041(n) (the partition numbers). Apparently, number of entries in row n is A033638(n-1) = 1 + floor((n-1)^2/4). T(n,0) = A000005(n) = number of divisors of n. T(n,1) = A070824(n+1). Sum(kT(n,k),k>0) = A188814(n). LINKS Alois P. Heinz, Rows n = 1..70, flattened FORMULA The weight of the complement of a partition p is (number of parts of p)(largest part of p) - weight of p. For a given q, the Maple program yields the generating polynomial of row q. EXAMPLE Row 4 is 3,0,2; indeed, the complements of [4], [3,1], [2,2], [2,1,1], [1,1,1,1] are: empty, [2], empty, [1,1], empty; their weights are 0, 2, 0, 2, 0, respectively. Triangle starts: 1; 2; 2,1; 3,0,2; 2,2,0,2,1; 4,0,2,1,2,0,2; MAPLE q := 10: with(combinat): a := proc (i, j) options operator, arrow: partition(i)[j] end proc: P[q] := 0: for j to numbpart(q) do P[q] := sort(P[q]+t^(nops(a(q, j))max(a(q, j))-q)) end do: P[q] := P[q]; # second Maple program: b:= proc(n, i, l) option remember; expand(`if`(n=0 or i=1, x^(`if`(l=0, 0, n(l-i))), b(n, i-1, l)+`if`(i>n, 0, x^(`if`(l=0, 0, l-i))b(n-i, i, `if`(l=0, i, l))))) end: T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n\$2, 0)): seq(T(n), n=1..15); # Alois P. Heinz, Feb 12 2016 MATHEMATICA b[n_, i_, l_] := b[n, i, l] = Expand[If[n == 0 \|\| i == 1, x^(If[l == 0, 0, n(l - i)]), b[n, i - 1, l] + If[i > n, 0, x^(If[l == 0, 0, l - i])b[n - i, i, If[l == 0, i, l]]]]]; T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n, n, 0]]; Table[T[n], {n, 1, 15}] // Flatten ( Jean-François Alcover, Dec 22 2016, after Alois P. Heinz ) CROSSREFS Cf. A000005, A000041, A033638, A070824, A188814. Sequence in context: A124839 A294076 A117046 A077653 A077889 A305805 Adjacent sequences: A268189 A268190 A268191 * A268193 A268194 A268195 KEYWORD nonn,tabf AUTHOR Emeric Deutsch, Feb 12 2016 STATUS approved Lookup \| Welcome \| Wiki \| Register \| Music \| Plot 2 \| Demos \| Index \| Browse \| More \| WebCam Contribute new seq. or comment \| Format \| Style Sheet \| Transforms \| Superseeker \| Recent The OEIS Community \| Maintained by The OEIS Foundation Inc. Last modified April 24 22:26 EDT 2019. Contains 322446 sequences. (Running on oeis4.)	1,320	3,050	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.640625	4	CC-MAIN-2019-18	latest	en	0.619308
https://za.pinterest.com/pin/361976888773841900/	1,516,146,323,000,000,000	text/html	crawl-data/CC-MAIN-2018-05/segments/1516084886758.34/warc/CC-MAIN-20180116224019-20180117004019-00339.warc.gz	1,050,218,245	88,369	# Explore Classroom Freebies, Math Classroom, and more! ### Doubles plus and minus one using number bonds Help your students learn the strategy of doubles plus and minus one by using two and three part number bonds. ### Subtraction Common Core Number Bond Practice Pages Subtraction Common Core Number Bond Practice Pages: Moving towards the Common Core? Number bonds create the "deeper" thinking with numbers! Inside the Packet: 4 to a page number bonds: Numbers versions} Numbers number bonds practice sheets versions} ### Common Core Number Bond Practice Pages Common Core Number Bond Practice Pages: Moving towards the Common Core? Number bonds create the "deeper" thinking with numbers! The story of numbers Number bonds practice pages per number) Number bonds with beans- recording sheets Click {here} ### Number Bonds (Math Facts Families) Chart and Worksheet Number Bonds (Math Facts Families) Chart and Worksheet ~ Free Printable ### Space Bonds {Number Bonds Board Game} Your Math-stronauts are on a mission to fix the satellite. They must find the missing number in each bond to move along the intergalactic highway. They will meet some new space friends along the way, but watch out for space junk, that will send you to the garage! This game is designed to help students develop awareness of how number values are related to each other, forming number bonds within 10. \$ ### Picture Addition Number Bonds Task Cards Owl Style (Common Core Aligned) I like this visual for fact families ### Number Bonds Lesson A Kindergarten Smorgasboard Number Bond Lesson Pinterest	320	1,594	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05	latest	en	0.786411
https://access.openupresources.org/curricula/our-k5-math/en/grade-1/unit-4/section-b/lesson-8/student.html	1,716,391,521,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058557.6/warc/CC-MAIN-20240522132433-20240522162433-00762.warc.gz	66,025,175	8,319	# Lesson 8 Different Representations of Tens and Ones • Let’s think about how two-digit numbers can be shown. ## Warm-up Estimation Exploration: How Many? 1. How many do you see? Record an estimate that is: too low too high 2. How many do you see? Record an estimate that is: too low too high ## Activity 1 Compare Representations of a Collection Each student counted and showed a collection. • Clare drew • Han drew • Kiran wrote 3 ones and 7 tens. • Priya wrote . Did the students count the same number of objects? How do you know? Show your thinking using drawings, numbers, or words. ## Activity 2 Card Sort: Base-ten Representations Your teacher will give you a set of cards that show different representations of a two-digit number. Find the cards that match. Be ready to explain your reasoning. ## Activity 3 Introduce Grab and Count: Ones Cubes We are going to learn a new center called Grab and Count. ## Problem 1 Circle 3 representations of 63. 1. 6 tens and 3 tens 2. 6 tens and 3 ones	251	1,022	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22	latest	en	0.896478
https://www.jiskha.com/questions/1789291/The-control-group-score-of-47-26-on-the-pretest-put-it-at-the-26th-percentile-does	1,558,793,304,000,000,000	text/html	crawl-data/CC-MAIN-2019-22/segments/1558232258058.61/warc/CC-MAIN-20190525124751-20190525150751-00011.warc.gz	797,005,054	4,345	# Statistics The control group score of 47.26 on the pretest put it at the 26th percentile does this percentile score represent nominal oridinL or interval scale data 1. 👍 0 2. 👎 0 3. 👁 152 1. Interval, it does more than rank. 1. 👍 0 2. 👎 0 posted by PsyDAG 2. Answer this question 1. 👍 0 2. 👎 0 posted by Suleman ## Similar Questions 1. ### Statistics The control group's score of 47.26 on the pretest put it at the 26th percentile. Does this percentile score represent nominal, ordinal, or interval scale data? asked by Sajjad on April 27, 2019 2. ### Statistics The control group's score of 47.26 on the pretest put it at the 26th percentile. Does this percentile score represent nominal, ordinal, or interval scale data? asked by Annaa on February 16, 2019 3. ### statistics The control group's score of 47.26 on the pretest put it at the 26th percentile. Does this percentile score represent nominal, ordinal, or interval scale data? asked by ali on April 4, 2019 4. ### Math the control group score of 47.26 on the pretest put it at the 26th percentile. Does this percentile score represent nominal, ordinal, or interval scale data? asked by Gul on March 12, 2019 5. ### math The control group 47.26 on the pretest put it at the 26th percentile. does this percentile score represent nominal,ordinal or interval scale data? asked by sadaqat khan on March 29, 2019 6. ### Algebra The control score of the 47.26 on the pretest put it at the 26th percentile.Does this percentile score represent nomina,ordinal,or interval scale data? asked by Bilal on April 17, 2019 7. ### algebra the control groups score of 47.26 on the pretest put it at the 26th percentile. does this percentile score represent nominal ordinal or interval scale data? asked by Alein on March 2, 2019 8. ### Mathematiics The control group score 47.26 onthe pretest it at the percentile. Does this percentile score represent nominal ordinal or interval scale data. asked by Annaa on March 20, 2019 9. ### Mathematics The control groups ,score of 46: Algebra27•26 on the prestet put it at the 26th percentile .Does this percentile score nominal ,ordinal,or interval scale data? asked by M Asif baloch on May 4, 2019 10. ### Math Score Percentile Score Percentile Score Percentile Score Percentile -3.5 0.02 -1 15.87 0 50 1.1 86.43 -3 0.13 -0.95 17.11 0.05 51.99 1.2 88.49 -2.9 0.19 -0.9 18.41 0.1 53.98 1.3 90.32 -2.8 0.26 -0.85 19.77 0.15 55.96 1.4 91.92 asked by Need Help!!! on May 12, 2012 More Similar Questions	775	2,513	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22	latest	en	0.853457
https://programming.vip/docs/python-data-structure-bubble-selection-insertion-sorting-algorithm.html	1,716,828,075,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971059044.17/warc/CC-MAIN-20240527144335-20240527174335-00410.warc.gz	410,194,455	4,710	# python data structure bubble selection insertion sorting algorithm ## Sorting algorithm Sorting algorithm (English: Sorting algorithm) is an algorithm that can arrange a string of data in a specific order. ### Stability of sorting algorithm Stability: the stable sorting algorithm will maintain the relative order of records with equal key values. That is, if -- sorting algorithm is stable, when there are two records R and S with equal key values, and R appears before S in the original list, R will also be before S in the sorted list. When equal elements are indistinguishable, such as integers, stability is not a problem. However, suppose that the following pairs of numbers will be sorted by their first number. ### Bubble sorting Bubble sort is a simple sort algorithm. It repeatedly traverses the sequence to be sorted, - compares two elements, and swaps them if they are in the wrong order. The work of traversing the sequence is repeated until there is no need to exchange, that is, the sequence has been sorted. The name of this algorithm comes from the fact that smaller elements will slowly "float" to the top of the sequence through exchange. Illustration realization ```# coding: utf-8 def bubble_sort(alist): """Bubble sorting""" n = len(alist) for j in range(n-1): # for j in rang(len(alist)-1,0,-1): count = 0 for i in range(0, n-1-j): # for i in range(j): # From beginning to end if alist[i] > alist[i+1]: alist[i], alist[i+1] = alist[i+1], alist[i] count += 1 if 0 == count: return if __name__ == "__main__": li = [45, 65, 54, 87, 9, 4, 5, 55] print(li) bubble_sort(li) print(li) ``` Operation results Time complexity Optimal time complexity: O(n) (indicates traversal -- no elements can be exchanged for each discovery) (end of sorting.) Worst time complexity: o(n^2) Stability: stable ### Select sort Selection sort is a simple and intuitive sorting algorithm. Its working principle is as follows. First, find the smallest (large) element in the unordered sequence and store it at the beginning of the sorted sequence. Then, continue to find the smallest (large) element from the remaining unordered elements, and then put it at the end of the sorted sequence. And so on until all elements are sorted. The main advantage of selective sorting is related to data movement. If a dollar; If the element is in the correct final position, it will not be moved. Select sort to exchange a pair of elements each time, and at least - of them will be moved to their final position. Therefore, sort the table of n elements and exchange at most n-1 times in total. Among all the sorting methods that completely rely on exchange to move elements, selective sorting is a very good one. Illustration realization ``` def select_sort(alist): """Select sort""" n = len(alist) for j in range(n-1): # j: 0~ n-2 min_index = j for i in range(j+1, n): if alist[min_index] > alist[i]: min_index = i alist[j], alist[min_index] = alist[min_index], alist[j] if __name__ == "__main__": li = [45, 65, 54, 87, 9, 4, 5, 55] print(li) select_sort(li) print(li) ``` Operation results Time complexity Optimal time complexity: O(n^2) Worst time complexity: O(n^2) Stability: unstable (considering the maximum selection of ascending order each time) ### Insert sort Insertion sort (English: Insertion Sort) is a simple and intuitive sorting algorithm. Its working principle is to build an ordered sequence, scan the unordered data from back to front in the sorted sequence, find the corresponding position and insert it. In the implementation of insertion sorting, in the process of scanning from back to front, it is necessary to repeatedly move the sorted elements backward step by step to provide insertion space for the latest elements. Illustration realization ```# coding : utf-8 def insert_sort(alist): """Insert sort""" n = len(alist) for j in range(1, n): # How many elements are taken from the unordered sequence on the right to perform such a process i = j # i represents the starting value of inner loop while i > 0: # Execute to take the first element from the unordered sequence on the right, that is, the element at i position, and then insert it into the previous position if alist[i] < alist[i-1]: alist[i], alist[i-1] = alist[i-1], alist[i] i -= 1 else: break if __name__ == "__main__": li = [25, 65, 54, 87, 19, 4, 35, 55] print(li) insert_sort(li) print(li) ``` Operation results Time complexity Optimal time complexity: O(n) (in ascending order, the sequence is already in ascending order Status) Worst time complexity: 0(n^2) Stability: stable Keywords: Python Algorithm data structure Added by codygoodman on Fri, 04 Feb 2022 07:03:53 +0200	1,169	4,723	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.59375	4	CC-MAIN-2024-22	latest	en	0.858321
https://mail.python.org/pipermail/python-list/2002-October/126910.html	1,529,668,822,000,000,000	text/html	crawl-data/CC-MAIN-2018-26/segments/1529267864391.61/warc/CC-MAIN-20180622104200-20180622124200-00196.warc.gz	672,286,142	2,415	# [Python-Dev] PEP239 (Rational Numbers) Reference Implementation and new issues Eric S. Raymond esr at thyrsus.com Thu Oct 3 00:14:21 CEST 2002 ```python-pep at ccraig.org <python-pep at ccraig.org>: > I just uploaded a reference implementation of how rationals might look > in Python as patch 617779 [1]. I do have some new issues for > discussion that I'd like to get some comments on before I change the > PEP. > > 1) Should future division return rationals rather than floats. I had > sort of assumed this would happen, but I just had a discussion with > Kirby Urner and couldn't convince him it was a good idea, so I guess > it isn't so clear. > > Arguments for: > - you don't lose precision on divides > - It provides a really nice way to specify rationals (i.e. 1/3) > - It allows you to eventually unify int/long/rationals so that > rationals with a denominator of 1 are automagically upcast. > > Arguments against: > - people who have already changed their code to expect floats will > have to change it again > - rationals are slow +1 for returning rationals. It's the right thing -- and if it fails, it will fail noisily, right? > 2) Should floats compare equal with rationals only when they are > equal, or whenever the are the closest float? (i.e. will .2 > compare equal to rational(1, 5)) > > 3) Should rationals try to hash the same as floats? My leaning on > this is that it will be decided by (2). If they compare equal when > 'close enough' then they should hash the same, if not then they should > only hash the same when both are integral. I would rather not see .5 > hash with rational(1, 2) but not .2 with rational(1, 5). APL faced this problem twenty-five years ago. I like its solution; a `fuzz' variable defining the close-enough-for-equality range. -- <a href="http://www.tuxedo.org/~esr/">Eric S. Raymond</a> ```	522	1,901	{"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}	3.015625	3	CC-MAIN-2018-26	latest	en	0.939572
https://algorithms.rocks/how-to-use-algorithm-in-a-sentence/	1,685,709,081,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224648635.78/warc/CC-MAIN-20230602104352-20230602134352-00470.warc.gz	122,897,133	46,158	Mastering the Art of Algorithmic Sentences: A Comprehensive Guide to Using Algorithms in Language and Text Processing Fernando Velarde June 1, 2023 2:17 pm Title: Unveiling the Mystery: How to Use Algorithm in a Sentence Are you ready to embark on a journey to discover the secret behind using algorithms in sentences? In this informative article, we will unravel the mystery step-by-step, and by the end, you will be able to use the term “algorithm” with ease while mastering the art of crafting thought-provoking sentences. ## What exactly is an algorithm? Before we dive into the world of algorithms, let’s first understand what they are. An algorithm is a step-by-step procedure or set of rules that help solve a problem, complete a task, or make a decision. These algorithms play a vital role in our daily lives, from Google search results to the recipes we follow while cooking. Now that we have a grasp on the concept, let’s begin our adventure. ## Breaking down the keyword: How to use algorithm in a sentence You may have noticed that our main focus is to answer the question, “how to use algorithm in a sentence?”. To do so, we will cleverly integrate secondary keywords derived from the main one as we go along. This will help us provide a comprehensive understanding of the topic. ### Algorithm usage in English sentences Using “algorithm” in a sentence might seem tricky at first, but with a little practice, you’ll find that it’s easy to incorporate into your everyday conversations. Here are a few examples for your reference: 1. The computer scientist developed a new algorithm to optimize search engine results. 2. Social media platforms rely on complex algorithms to curate content for users based on their interests and engagement. 3. The success of his invention was a result of careful research and a well-crafted algorithm. As you can see, in each example, “algorithm” refers to a specific method, technique, or system that helps solve a problem or achieve a desired outcome. ### Using algorithm in different sentence structures Now that you have a better understanding of how to use “algorithm” in a sentence, let’s explore how it fits into different sentence structures: Simple sentences: A new algorithm was created for facial recognition software. Compound sentences: The team studied various algorithms, and they eventually developed a more efficient solution. Complex sentences: Despite several attempts to improve the algorithm, the results remained inaccurate in certain situations. Narrative sentences: After months of hard work, the programmer finally discovered the perfect algorithm that would revolutionize machine learning applications. Remember, using “algorithm” effectively in your sentences will require practice, but by following these examples and keeping the context in mind, you’ll be an expert in no time. ## Exploring algorithm-related vocabulary To further enhance your understanding of algorithms and their usage in sentences, it’s important to explore related vocabulary. Familiarizing yourself with these terms will not only help you use “algorithm” more confidently but also broaden your overall knowledge. Some common algorithm-related terms include: 1. Heuristic: A problem-solving technique that uses shortcuts to speed up the process. 2. Data structure: A way of organizing and storing data for efficient access and modification. 3. Optimization: The process of finding the best solution to a problem by analyzing various possibilities. ## Conclusion: Becoming an Algorithm Sentence Master You’ve made it to the end of our journey! By now, you should have a thorough understanding of how to use “algorithm” in a sentence and the confidence to incorporate it into your daily conversations. Remember to practice using the suggested sentence structures and explore related vocabulary to truly master the art of using algorithm-related language. Now, go forth and impress your friends and colleagues with your newfound expertise! ## How are algorithms utilized in our daily lives? Algorithms play a crucial role in our daily lives, often without us even realizing it. These sets of instructions or rules are designed to perform specific tasks or solve particular problems, making our lives more convenient and efficient. Here are some common ways algorithms are utilized in our daily lives: 1. Search engines: When we search for something online, algorithms help us find the most relevant information by analyzing and ranking web pages based on their content, structure, and other factors. 2. Social media: Algorithms determine what content you see on your social media feeds by analyzing your interests, interactions, and the popularity of posts shared by others. 3. Online shopping: E-commerce platforms use algorithms to recommend products you might be interested in based on your browsing history, purchases, and customer reviews. 4. Navigation: GPS systems rely on algorithms to calculate the shortest, quickest, or most efficient routes between two locations while taking into account real-time traffic data. 5. Banking and finance: Algorithmic trading uses complex algorithms to analyze market data, make investment decisions, and execute trades faster than any human could. 6. Healthcare: In the medical field, algorithms are used to analyze patient data, help with diagnosis, and recommend treatment plans based on a vast database of medical knowledge. 7. Entertainment: Algorithms are essential for personalizing content recommendations in streaming services like Netflix or Spotify, based on your viewing or listening history and preferences. 8. Artificial intelligence (AI): From virtual assistants like Siri or Google Assistant to more advanced AI systems, algorithms play a key role in enabling machines to learn, analyze data, and perform tasks. 9. Data encryption: Encryption algorithms are used to secure sensitive information, like passwords and personal data, ensuring that our online communications remain private and protected. 10. Weather forecasting: Algorithms process vast amounts of data from satellites, weather stations, and other sources to predict weather patterns and generate accurate forecasts. In conclusion, algorithms significantly impact almost every aspect of our lives, streamlining processes, improving decision-making, and providing more personalized experiences. The continued development of advanced algorithms will undoubtedly lead to even greater efficiency and convenience in the future. ## Can you provide an example to illustrate what an algorithm is? An algorithm is a well-defined, step-by-step procedure to solve a specific problem or perform a particular task. It is a set of rules or instructions that a computer, or a person, can follow to achieve the desired result. Example: Let’s consider an algorithm for finding the greatest common divisor (GCD) of two numbers using the Euclidean algorithm, which is one of the oldest and most well-known algorithms in existence. 1. Input: Given two positive integers A and B. 2. Step 1: Check if B is equal to 0. If so, the GCD is A, and the algorithm terminates. 3. Step 2: Divide A by B and find the remainder, R. 4. Step 3: Set A equal to B and set B equal to R. 5. Step 4: Go back to Step 1. The algorithm iteratively reduces the problem size until it reaches the base case, where B is equal to 0. This process continues until the GCD is found. Here, the key components of the algorithm include step-by-step instructions, a clear input, and a termination condition. ## What does the term “algorithm” signify when used within a sentence? The term “algorithm” signifies a step-by-step procedure or set of rules to be followed in order to solve a problem or perform a specific task, especially within the context of computer programming and mathematical calculations. Algorithms are essential for efficient and effective problem-solving and form the basis for most computer programs and applications. ## What is a synonym for the term “algorithm”? A synonym for the term “algorithm” in the context of algorithms is procedure or set of rules. These terms refer to a step-by-step process or a series of instructions that are designed to solve a specific problem, perform a specific task, or achieve a certain goal. ### What are the key components to consider when constructing a sentence that effectively demonstrates the usage of an algorithm? When constructing a sentence that effectively demonstrates the usage of an algorithm, consider the following key components: 1. Purpose: Clearly state the goal or problem the algorithm is designed to solve. 2. Algorithm’s Name: If applicable, mention the specific name or type of the algorithm being used. 3. Input: Describe the data or information that will be processed by the algorithm. 4. Process: Briefly explain how the algorithm works or manipulates the input data to achieve its purpose. 5. Output: Mention the expected result or output produced by the algorithm. Here’s an example sentence: The Dijkstra’s algorithm efficiently finds the shortest path between two nodes in a weighted graph by iteratively selecting the node with the smallest known distance from the source node and updating its neighbors’ distances. ### Can you provide examples of sentences that highlight the practical applications of various algorithms in everyday situations or problem-solving tasks? 1. Pathfinding algorithms, such as Dijkstra’s and A* algorithms, play a crucial role in applications like GPS navigation systems, enabling users to find the shortest or quickest route from point A to point B. 2. Online shopping platforms utilize recommendation algorithms, including collaborative filtering and content-based filtering, to suggest products that match a customer’s interests and preferences, improving the overall shopping experience. 3. Sorting algorithms, such as quicksort, mergesort, and insertion sort, are fundamental to database management, allowing for rapid organization and searchability of large volumes of data in various industries. 4. In finance, optimization algorithms like linear programming and integer programming are employed to solve complex problems, including portfolio optimization, risk management, and resource allocation. 5. Machine learning algorithms, encompassing supervised learning, unsupervised learning, and reinforcement learning, have become integral to the development of artificial intelligence, powering applications like image recognition, natural language processing, and autonomous vehicles. 6. The modern cryptography heavily relies on cryptographic algorithms such as RSA, AES, and SHA, which enable secure communication, digital signatures, and data protection in various online services, including e-commerce, online banking, and cloud computing. 7. Search algorithms, including depth-first search and breadth-first search, play a significant role in network analysis, allowing the study of connections and relationships between various entities in graphs, such as social networks, computer networks, and transportation systems. 8. Video streaming platforms, like YouTube and Netflix, employ video compression algorithms such as H.264 and VP9 to reduce the size of video files while maintaining visual quality, allowing for smoother streaming experiences for users with varying internet connection speeds. 9. In healthcare, clustering algorithms like K-means and hierarchical clustering have proven valuable for interpreting complex medical data, assisting in tasks such as diagnostics, treatment personalization, and disease outbreak prediction. 10. Text analysis algorithms, including sentiment analysis and topic modeling, are used in natural language processing for applications like chatbots, spam filtering, and targeted advertising, improving user interactions and experiences on various platforms. ### How can we best incorporate algorithm-related terminology and concepts into sentences to accurately convey its significance and functionality? To best incorporate algorithm-related terminology and concepts into sentences, it is important to focus on accurately conveying the significance and functionality of the specific algorithm being discussed. Here are some tips for doing so: 1. Begin by clearly defining the algorithm’s purpose and the problem it aims to solve. This helps set the context and allows readers to better understand the relevance of the algorithm. 2. Use accurate and consistent terminology throughout your content. For instance, always use the term “algorithm” instead of interchanging it randomly with “formula” or “method.” 3. Explain key concepts and terms associated with the algorithm. For example, if discussing a search algorithm, define terms such as “time complexity,” “space complexity,” and “efficiency” to ensure readers are completely familiar with how the algorithm operates. 4. Utilize concrete examples or analogies when explaining complex concepts, as this can make them easier to grasp for both novice and advanced readers. 5. Focus on the practical applications of the algorithm, highlighting its importance and usefulness in real-world scenarios. 6. Provide comparisons between different algorithms that solve the same problem, discussing their respective pros and cons. This will help readers develop an informed understanding of the algorithm’s performance and practicality. 7. Lastly, consider using visual aids, such as diagrams or flowcharts, to depict crucial aspects of the algorithm’s functionality. This can aid in reinforcing the concepts presented through written explanations. In summary, effectively incorporating algorithm-related terminology and concepts into your content requires a clear focus on the significance and functionality of the algorithm, consistent use of accurate terminology, and employing examples, comparisons, and visual aids to enhance understanding. #### Author Profile Fernando Velarde I am a passionate tech enthusiast with a deep-seated love for all things digital. As a seasoned blogger, SEO expert, programmer, and graphic designer, I thrive in the intersection of creativity and technology. My journey began with a fascination for coding and graphic design, sparking a drive to create, innovate, and share my insights with a wider audience. June 1, 2023 2:17 pm	2,646	14,357	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.640625	3	CC-MAIN-2023-23	latest	en	0.914656
https://www.worldwidefaqs.com/what-causes-the-shielding-effect-to-remain-constant/	1,680,167,246,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296949107.48/warc/CC-MAIN-20230330070451-20230330100451-00011.warc.gz	1,154,248,863	14,246	# What Causes The Shielding Effect To Remain Constant? Answer and Explanation: The number of inner shell electronsshell electronsEach shell can contain only a fixed number of electrons: the first shell can hold up to two electrons, the second shell can hold up to eight (2 + 6) electrons, the third shell can hold up to 18 (2 + 6 + 10) and so on. The general formula is that the nth shell can in principle hold up to 2(n2) electrons.https://en.wikipedia.org › wiki › Electron_shellElectron shell – Wikipedia being constant causes the shielding effect to remain constant across a period. ## Why does the shielding effect remain constant as you move from left to right across the periodic table? I believe that electron shielding remains constant because when you move across a period, you are essentially adding more valence electrons, not shielding electrons, in your valence shell. Therefore, your valence-electron-count increases from left to right in a period, but your shielding-electron-count stays the same. ## Why is shielding effect constant across a period? It is also referred to as the screening effect or atomic shielding. Shielding effect is the same across periods but increases across groups. This is because across the periods all the outermost electrons are in the same shell meaning they're all the same distance from the nucleus. ## What happens to the shielding effect when you move from left to right on the periodic table? So the amount of shielding is increasing as we move left to right. The apparent contradiction with the ionization energy comes about because you have not considered the increase in the actual nuclear charge. Each time we add a 2p electron, we also add a proton to the nucleus. ## How is the shielding constant determined? Hint: Slater's rule is used to calculate shielding constant. Formula used- $= (0.35 \times n) + (0.85 \times m) + (1.00 \times p)$ where n is number of electrons in n shell, m is number of electrons in n-1 shell, p is number of electrons in the remaining inner shells. Find the number of electrons in n shell. ## What happens to the shielding effect across a period? Shielding effect is the same across periods but increases across groups. This is because across the periods all the outermost electrons are in the same shell meaning they're all the same distance from the nucleus. ## What cause the shielding effect to? Shielding is caused by the combination of partial neutralization of nuclear charge by core electrons, and by electron-electron repulsion. The amount of charge felt by an electron depends on its distance from the nucleus. ## Why does shielding increase across a period? Across a period, effective nuclear charge increases as electron shielding remains constant. This pulls the electron cloud closer to the nucleus, strengthening the nuclear attraction to the outer-most electron, and is more difficult to remove (requires more energy).	599	2,936	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.421875	3	CC-MAIN-2023-14	latest	en	0.899942
https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_20&oldid=190680	1,708,829,033,000,000,000	text/html	crawl-data/CC-MAIN-2024-10/segments/1707947474573.20/warc/CC-MAIN-20240225003942-20240225033942-00004.warc.gz	110,783,751	13,211	# 2001 AMC 10 Problems/Problem 20 ## Problem A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length $2000$. What is the length of each side of the octagon? $\textbf{(A) } \frac{1}{3}(2000) \qquad \textbf{(B) } {2000(\sqrt{2}-1)} \qquad \textbf{(C) } {2000(2-\sqrt{2})} \qquad \textbf{(D) } {1000} \qquad \textbf{(E) } {1000\sqrt{2}}$ ## Solution 2 Let $x$ represent the length of each side of the octagon, which is also the length of the hypotenuse of each of the right triangles. Each leg of the right triangles has length $x\sqrt{2}/2$, so $$2 \cdot \frac{x\sqrt{2}}{2} +x=2000, \text{ and } x = \frac{2000}{\sqrt{2}+1}=\boxed{\textbf{(B) }2000(\sqrt{2}-1)}.$$ ## Solution 3 (Longer solution-credit: Ileytyn) First, realize that each triangle is congruent, a right triangle and that the two legs are equal. Also, each side of the octagon is equal, because of the definition of regular shapes. Let $s$ be the length of a leg of the isosceles right triangle. In terms of $s$, the hypotenuse of the isosceles right triangle, which is also the length of a side of the regular octagon, is $s \sqrt{2}$. Since the length of each side of the square is 2000, the length of each side of the regular octagon is equal to the length of a side of the square ($2000$) subtracted by $2$ times the length of a leg of the isosceles right triangle ( the total length of the side is $2s+ o$, $o$ being the length of a side of the regular octagon), which is the same as $2s$. As an expression, this is $2000-2s$, which we can equate to $s \sqrt{2}$, ( since the octagon is regular, meaning all of the side's lengths are congruent) giving us the following equation:$2000-2s = s \sqrt{2}$. By isolating the variable and simplifying the right side, we get the following: $2000 = s(2 + \sqrt{2})$. Dividing both sides by $(2 + \sqrt{2})$, we arrive with $\frac{2000}{2 + \sqrt{2}} = s$, now, to find the length of the side of the octagon, we can plug in $s$ and use the equation $2000-2s = o$, $o$ being the length of a side of the octagon, to derive the value of a side of the octagon. After plugging in the values, we derive $2000-2(\frac{2000}{2 + \sqrt{2}})$, which is the same as $2000-(\frac{4000}{2 + \sqrt{2}})$, factoring out a $2000$, we derive the following: $2000(1-(\frac{2}{2 + \sqrt{2}}))$, by rationalizing the denominator of $\frac{2}{2 + \sqrt{2}}$, we get $2000(1-(2 - \sqrt{2}))$, after expanding, finally, we get $\boxed{\textbf{(B) }2000(\sqrt{2} -1)}$ !(not a factorial symbol, just an exclamation point)	833	2,582	{"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": 29, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.71875	5	CC-MAIN-2024-10	latest	en	0.866748
https://nz.education.com/science-fair/article/model-earth-water-cycle/?source=related_materials&order=12	1,591,268,940,000,000,000	text/html	crawl-data/CC-MAIN-2020-24/segments/1590347439928.61/warc/CC-MAIN-20200604094848-20200604124848-00548.warc.gz	469,021,556	31,402	Premium # The Hydrologic Cycle ### Problem Observe evaporation and condensation in different environments. ### Materials • Glass jar with metal lid (a mason jar would work well) • Ice cubes • Water • Stopwatch • Thermometer ### Procedure 1. Fill the glass jar with 1 cm of water covering the bottom. Screw on the metal lid. 2. Place an ice cube on the top of the metal lid. 3. Take the temperature of the room. 4. Star the stop watch and time how long it takes for the ice cube to melt completely. 5. Carefully lift the jar so it remains level (parallel to the floor) and look under the bottom of the lid. Record your observations. What do you observe? This would be a good opportunity to take a picture or make a quick sketch. 6. Tilt the jar slightly so some of the droplets on the lid of the jar group together. What happens? 7. Repeat the experiment outside in the sun, or in a cool garage or basement. ### Results Water will evaporate fastest in warm environments, and slower in cool environments. Water will condense and form droplets on the underside of the metal lid. Tilting the jar will cause condensed droplets to join together, and once they are large enough, they will drop back down into the jar. ### Why? In a warmer environment, the ice cube will melt faster because there will be a greater temperature difference between the surroundings and the ice cube. In warmer environments, the water will also evaporate faster. Evaporation occurs when a molecule has enough energy to turn in to a vapor, or a gas. In warmer surroundings, the air has more heat energy that can be transferred to the water molecules, thus making more of the water molecules have enough energy to vaporize. When the water in the jar evaporates, it will be condensed on the cooled metal lid. Condensation, which is when a gas returns back to a liquid through cooling, is the opposite of vaporization (which evaporation can be considered). The ice cools the metal lid, which causes the gaseous water to condense upon. Room temperature and a cooler room will have similar results, but the ice cube will melt more slowly and there will be less water evaporation and condensation. Tilting the jar until the condensed droplets join together is representative of precipitation. In precipitation, condensed water droplets, which we see in the form of clouds, eventually become heavy enough that they fall towards earth in the form of rain, or if it’s cold enough, hail or snow. Disclaimer and Safety Precautions Education.com provides the Science Fair Project Ideas for informational purposes only. Education.com does not make any guarantee or representation regarding the Science Fair Project Ideas and is not responsible or liable for any loss or damage, directly or indirectly, caused by your use of such information. By accessing the Science Fair Project Ideas, you waive and renounce any claims against Education.com that arise thereof. In addition, your access to Education.com's website and Science Fair Project Ideas is covered by Education.com's Privacy Policy and site Terms of Use, which include limitations on Education.com's liability. Warning is hereby given that not all Project Ideas are appropriate for all individuals or in all circumstances. Implementation of any Science Project Idea should be undertaken only in appropriate settings and with appropriate parental or other supervision. Reading and following the safety precautions of all materials used in a project is the sole responsibility of each individual. For further information, consult your state's handbook of Science Safety. ### Add to collection Create new collection 0 ### New Collection> 0Items What could we do to improve Education.com? Please note: Use the Contact Us link at the bottom of our website for account-specific questions or issues. What would make you love Education.com?	784	3,874	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.765625	3	CC-MAIN-2020-24	latest	en	0.884761
https://www.braincells.com/perl/2020/02/perl_weekly_challenge_week_47.html	1,596,954,092,000,000,000	text/html	crawl-data/CC-MAIN-2020-34/segments/1596439738425.43/warc/CC-MAIN-20200809043422-20200809073422-00261.warc.gz	608,654,717	4,520	### Perl Weekly Challenge: Week 47 #### Challenge 1: Roman Calculator Write a script that accepts two roman numbers and operation. It should then perform the operation on the givenn Roman numbers and print the result. For example, `````` perl ch-1.pl V + VI `````` It should print `````` XI `````` At first I was going to chicken out and just use my existing code from Challenge 10 to convert the roman numerals to Hindu-Arabic, do the calculation and convert them back but I did some fascinating research into how Romans actually did arithmetic. In particular this blog gave an algorithm that was easy to understand and implement. I did run into problems along the way and as a result I missed the deadline for the challenge but I had a lot of fun nevertheless. Addition is easy. I implemented it in Perl like this: ``````say normalize(reorder(unprefix(\$num1) . unprefix(\$num2))); `````` A peculiar feature of Roman numerals is that instead of being laid out strictly by place, some of them are prefixed. For example 8 is VIII but 9 isn't VIIII it's IX. So the `unprefix()` function converts these prefixes to suffixes. ``````sub unprefix { my (\$num) = @_; my @from = qw/ CM CD XC XL IX IV /; my @to = qw/ DCCCC CCCC LXXXX XXXX VIIII IIII /; for my \$i (0 .. scalar @from - 1) { \$num =~ s/\$from[\$i]/\$to[\$i]/g; } return \$num; } `````` Then both numbers are concatenated to each other. The digits in resulting string are then sorted by size. ``````sub reorder { my (\$num) = @_; my %order = ( 'M' => 0, 'D' => 1, 'C' => 2, 'L' => 3, 'X' => 4, 'V' => 5, 'I' => 6 ); return join q{}, sort { \$order{\$a} <=> \$order{\$b} } split //, \$num; } `````` The last step is to turn the string back into a proper Roman number with the right characters and prefixes in the right place etc. ``````sub normalize { my (\$num) = @_; my @from = qw/ IIIII IIII VV VIV XXXXX XXXX LL LXL CCCCC CCCC DD DCD /; my @to = qw/ V IV X IX L XL C XC D CD M CM /; for my \$i (0 .. scalar @from - 1) { \$num =~ s/\$from[\$i]/\$to[\$i]/g; } return \$num; } `````` ...and that's it for addition. Subtraction is more complicated. The first step as with addition, is to `unprefix()` the operands. ``````my \$un1 = unprefix(\$num1); my \$un2 = unprefix(\$num2); `````` Then as long as the second number has digits, we have to: 1. remove common substrings from the two numbers. 2. For the largest digit in the second number (i.e. the first one), take the first digit in the first number that is larger, and expand it. I expressed this with the following loop: ``````while (length \$un2) { (\$un1, \$un2) = expandLargest(removeCommon(\$un1, \$un2)); } `````` To remove substrings first we need to generate them all. The appropriately named `substrings()` function does that. ``````sub substrings { my (\$num) = @_; my %substrings; for my \$i (0 .. (length \$num) - 1) { for my \$j (1 .. (length \$num) - \$i) { my \$ss = substr(\$num, \$i, \$j); \$substrings{\$ss}++; } } return sort { length \$b <=> length \$a } keys %substrings; } `````` One error I initially made was to not sort the list of substrings by length. This led to subtle errors where a shorter sequence than possible got matched. In the `removeCommon()` function we take the list of substrings of the first number and if a substring is present in both numbers it is removed from both. ``````sub removeCommon { my (\$num1, \$num2) = @_; for my \$ss (substrings(\$num1)) { if (\$num1 =~ /\$ss/ && \$num2 =~ /\$ss/) { \$num1 =~ s/\$ss//; \$num2 =~ s/\$ss//; } } return (\$num1, \$num2); } `````` I made another mistake here. In my first attempt, I only checked if the substring is present in the second number as it is already part of the first number. But I failed to account for the fact that each time a substring is matched, the size of the strings changes and the substring may not be valid anymore. Step 2, expanding the largest digit was the most difficult bit for me to wrap my head around. ``````sub expandLargest { my (\$num1, \$num2) = @_; my %order = ( 'M' => 0, 'D' => 1, 'C' => 2, 'L' => 3, 'X' => 4, 'V' => 5, 'I' => 6 ); `````` It would seem to be easier to define `@reverseOrder` as `keys @order` but keys does not guarantee any particular order. `````` my @reverseOrder = qw/ M D C L X V I /; my %expansion = ( 'M' => 'DCCCCC', 'D' => 'CCCCC', 'C' => 'LXXXXX', 'L' => 'XXXXX', 'X' => 'VIIIII', 'V' => 'IIIII', 'I' => q{} ); `````` `\$first` might be empty if the digits of the second number have been exhausted in which case we can skip the rest of the function. `````` my \$first = substr(\$num2, 0, 1); if (\$first) { my \$i = (\$first eq 'M') ? 0 : \$order{\$first} - 1; `````` This is yet another place where I hit a problem. Take the sum M - I. The next largest unit after I is V but V does not exist in the first number (i.e M). What should be done in this circumstance is to keep trying the next larger unit until one that does match is found and that's what this while loop does. `````` while (\$i >= 0 && \$num1 !~ /\$reverseOrder[\$i]/) { \$i--; } \$num1 =~ s/\$reverseOrder[\$i]/\$expansion{\$reverseOrder[\$i]}/; } `````` Because the expansion may have caused `\$num1`s digits to get out of order, it is run through `reorder()` before being returned. `````` return (reorder(\$num1), \$num2); } `````` Finally, after there are no more digits left in the second number, the result is `normalize()`ed as with addition and printed out. ``````say normalize(\$un1); `````` After all this I didn't dare attempt multiplication or division! (Full code on Github.) The Raku version is similar minus the usual small syntactic differences so I'm not going to reproduce the whole thing here. But there are a few noteworthy details. Take some code from the `unprefix()` function for example. A function or method parameter is immutable by default. So if you wanted to change it, you would have to make a copy and change that. sub unprefix(Str \$num) { my \$unprefixed = \$num; In perl I did `s/\$from[\$i]/\$to[\$i]/g;`. Regex susbstitution works the same in Raku (except `g` and other flags go after the `s`) but you run into problems with array elements because the subscript is interpreted as part of the regex. The simply way to get around this is to use a strings `.subst()` method instead like this: ``````\$unprefixed = \$unprefixed.subst(@from[\$i], @to[\$i], :g); `````` But you have to remember to assign it back to the string variable if you actually want the substitution to occur. Embarrassingly, I forgot this at first. I like to nest functions so that the output of one becomes the input of another. `removeCommon()` returns an array with two elements. `largestOrder()` wants two separate arguments so it raises an error when given only one array. Perl would automagically "flatten" the array in this context but Raku doesn't unless you put th `\|` operator before `removeCommon()`. ``````(\$un1, \$un2) = largestOrder(\|removeCommon(\$un1, \$un2)); `````` Now it does the right thing. (Full Raku code on Github.) #### Challenge 2: Gapful Number Write a script to print first 20 Gapful Numbers greater than or equal to 100. Please check out the page for more information about Gapful Numbers. After all the excitement of the previous task, this one was easy. Here is the Perl solution. ``````my @gapfuls; my \$number = 100; while (scalar @gapfuls != 20) { my @digits = split //, \$number; my \$divisor = join q{}, (\$digits[0], \$digits[-1]); if (\$number % \$divisor == 0) { push @gapfuls, \$number; } \$number++; } say join ', ', @gapfuls; `````` (Full code on Github.) And this is Raku. ``````(gather { for (100 .. ∞) -> \$number { my @digits = \$number.comb; if \$number %% (@digits[0], @digits[*-1]).join(q{}) { take \$number; } } })[0 .. 19].join(', ').say; `````` (Full code on Github.)	2,279	7,893	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.203125	3	CC-MAIN-2020-34	latest	en	0.872508
http://www.mathisfunforum.com/viewtopic.php?pid=306909	1,485,064,776,000,000,000	text/html	crawl-data/CC-MAIN-2017-04/segments/1484560281353.56/warc/CC-MAIN-20170116095121-00012-ip-10-171-10-70.ec2.internal.warc.gz	579,816,866	5,859	Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ ° You are not logged in. ## #401 2014-01-20 04:28:56 gAr Member Registered: 2011-01-09 Posts: 3,479 ### Re: Probability problem. Hi bobbym, "Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha? "Data! Data! Data!" he cried impatiently. "I can't make bricks without clay." Offline ## #402 2014-01-20 04:30:40 bobbym From: Bumpkinland Registered: 2009-04-12 Posts: 107,182 ### Re: Probability problem. Hi; In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #403 2014-01-20 04:35:59 gAr Member Registered: 2011-01-09 Posts: 3,479 ### Re: Probability problem. Hi, "Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha? "Data! Data! Data!" he cried impatiently. "I can't make bricks without clay." Offline ## #404 2014-01-20 04:46:35 bobbym From: Bumpkinland Registered: 2009-04-12 Posts: 107,182 ### Re: Probability problem. Hi; In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #405 2014-01-20 04:55:39 gAr Member Registered: 2011-01-09 Posts: 3,479 ### Re: Probability problem. Hi, "Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha? "Data! Data! Data!" he cried impatiently. "I can't make bricks without clay." Offline ## #406 2014-01-20 05:02:48 bobbym From: Bumpkinland Registered: 2009-04-12 Posts: 107,182 ### Re: Probability problem. Hi; In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #407 2014-01-20 05:17:52 gAr Member Registered: 2011-01-09 Posts: 3,479 ### Re: Probability problem. Hi, "Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha? "Data! Data! Data!" he cried impatiently. "I can't make bricks without clay." Offline ## #408 2014-01-20 05:18:40 bobbym From: Bumpkinland Registered: 2009-04-12 Posts: 107,182 ### Re: Probability problem. Hi; In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #409 2014-03-30 21:54:53 gAr Member Registered: 2011-01-09 Posts: 3,479 ### Re: Probability problem. New problem: There are 2 blue balls, 4 green balls, 8 white balls, 16 red balls and 32 black balls. What is the probability of picking 5 balls, such that at most two balls of each color is to be included (e.g. [red, red, green, green, black] is fine; but [red, red, red, green, black] is forbidden) i) Without replacement? ii) With replacement? "Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha? "Data! Data! Data!" he cried impatiently. "I can't make bricks without clay." Offline ## #410 2014-03-30 23:33:28 bobbym From: Bumpkinland Registered: 2009-04-12 Posts: 107,182 ### Re: Probability problem. Hi gAr; In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #411 2014-03-31 01:21:59 gAr Member Registered: 2011-01-09 Posts: 3,479 ### Re: Probability problem. Hi bobbym, "Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha? "Data! Data! Data!" he cried impatiently. "I can't make bricks without clay." Offline ## #412 2014-03-31 01:27:28 bobbym From: Bumpkinland Registered: 2009-04-12 Posts: 107,182 ### Re: Probability problem. Hi gAr; In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #413 2014-03-31 01:45:44 gAr Member Registered: 2011-01-09 Posts: 3,479 ### Re: Probability problem. Yes. "Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha? "Data! Data! Data!" he cried impatiently. "I can't make bricks without clay." Offline ## #414 2014-03-31 01:52:06 bobbym From: Bumpkinland Registered: 2009-04-12 Posts: 107,182 ### Re: Probability problem. Hi; Okay, I will check the calculations. Could have made the mistake right there. Yep, you are right. Calculation error. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #415 2014-03-31 02:22:28 gAr Member Registered: 2011-01-09 Posts: 3,479 ### Re: Probability problem. Hi, "Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha? "Data! Data! Data!" he cried impatiently. "I can't make bricks without clay." Offline ## #416 2014-03-31 02:26:26 bobbym From: Bumpkinland Registered: 2009-04-12 Posts: 107,182 ### Re: Probability problem. Hi gAr; Sorry, for the first answer. Seems I can not copy and paste correctly into calculation boxes, been a bad week. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #417 2014-03-31 02:30:25 gAr Member Registered: 2011-01-09 Posts: 3,479 ### Re: Probability problem. No problem, things will get better. "Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha? "Data! Data! Data!" he cried impatiently. "I can't make bricks without clay." Offline ## #418 2014-03-31 02:33:02 bobbym From: Bumpkinland Registered: 2009-04-12 Posts: 107,182 ### Re: Probability problem. They sure better not get worse. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #419 2014-05-08 00:54:35 ElainaVW Member Registered: 2013-04-29 Posts: 571 ### Re: Probability problem. Three balls are thrown into eight boxes that are lined up in a row. What is the probability of two balls landing in the same box and the next one landing in another box? Balls always land in a box. Offline ## #420 2014-05-08 00:59:07 bobbym From: Bumpkinland Registered: 2009-04-12 Posts: 107,182 ### Re: Probability problem. Hi; In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline	2,290	7,717	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.921875	3	CC-MAIN-2017-04	latest	en	0.860618
http://tucows.com/macintosh/home-education/mathematics/fractions/?f=hig	1,527,279,436,000,000,000	text/html	crawl-data/CC-MAIN-2018-22/segments/1526794867217.1/warc/CC-MAIN-20180525200131-20180525220131-00335.warc.gz	307,606,938	5,283	# Fractions downloads Fractions Currently Browsing: Highest Rated Filters: Shareware \| Freeware \| Demo \| OpenSource \| Other \| All « Prev [Page 1 of 1] Next » Kids join their Blaster Pals at the Intergalactic Fairgrounds, where creatures from all... Demo Classic Added Aug 9, 1999 Popularity: 27% This cool cow makes learning fractions fun! He will thoroughly explain how fractions are... See Home Page Classic Added Jun 6, 1999 Popularity: 100% Fraction Tiles provides eight fractions on tiles, and you must select three that add up to... Freeware Classic Added Aug 12, 2003 Popularity: 18% This program performs standard mathematical operations on fractions and mixed numbers. Freeware Classic / OS X Added Jul 20, 2009 Popularity: 55% Having trouble with fractions? Fractions don't have to be hard and this program makes... See Home Page Classic Added Jul 1, 1998 Popularity: 36% Fraction Puzzles shows a series of 150 preset problems. Freeware Classic Added Aug 22, 2001 Popularity: 9% Pizza Party Puzzles uses a pizza to create fraction problems. Freeware Classic Added Mar 13, 2001 Popularity: 45% Mathville Waterway is a perfect program for helping you better understand fractions. Demo Classic Added Sep 11, 1998 Popularity: 73% So you're ready to learn how to multiply and divide fractions? See Home Page Classic Added Jul 31, 1998 Popularity: 82% In this program, students must identify the fractions of various types of models,... Demo Classic / OS X Added Jan 4, 2003 Popularity: 91% Students in grades five through seven can use these squares to practice basic fractions. Freeware Classic Added Jun 11, 2002 Popularity: 64% « Prev [Page 1 of 1] Next »	457	1,698	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.65625	3	CC-MAIN-2018-22	latest	en	0.865331
https://www.physicsforums.com/threads/using-two-positions-on-a-rigid-body-to-calculate-rotation.669839/	1,726,302,514,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00567.warc.gz	877,077,231	20,825	# Using two positions on a rigid body to calculate rotation • gge In summary: But consider that the boat can only translate and rotate in a plane (no change in roll or pitch). The coordinates of two points on opposite ends of the boat are being measured every minute or so....would have a relationship to the period of oscillation, wouldn't it?Yes, I believe so. gge Hello all, I'm currently a undergrad university student doing research and I'm anaylising some position data. The data is a time-series' of Eastings (x) and Northings (y) for two points (P1, P2) on a rigid body in motion (T, E1, N1, E2, N2), with a position reported every 1 minute. I know that P1 and P2 are separated by a pre-determined distance of 10 m. My goal is to use the two positions to calculate the rotation of the rigid body over time (which I need every 1 second). The two sets of positions contain noise (not correlated with each other) which I first need to "remove" to best model the motion. My first thought was to bestfit (in a least squares sense) a different polynomial to all four time series (E1-T, N1-T, E2-T, N2-T) and use those polynomials to then interpolate and calculate the rotation. However, this has lead to me wondering the following: 1. Would it be better to propagate the noise in the positions through the rotation calculation and then bestfit the rotation data (instead of removing the noise at the position stage and then calculating a rotation)? 2. Is it valid for me to separate the E and N from each other and model them separately? 3. I've noticed since I best fit them separately, and calculated the distance between P1 and P2 using the polynomials, the distance between the two points (which was very close around 10 m before) now has a larger standard deviation. Any help would be appreciated! Cheers! Because there are several different defintions of "best estimate" in statistics and because your problem appears complicated, I think you should write a computer simulation of the situation (if you haven't done so already). With a simulation, you can specify the true values, simulate the noise, generate simulated data, apply various analysis methods to the data and compare the statistical behavior of the estimated values to the true values. One question is whether the "true" rotation involes a stationary axis of rotation? Are yo trying to estimate only the rate of rotation and is the true rate of rotation constant? Or are you trying to estimate a possibly varying rate of rotation? Questions like whether the E and N noises can be treated independently are questions about the physics of the problem and the measurement devices. They don't have a definite mathematical answer until the physics is modeled in a definite way. Last edited: Hi Stephen, thanks for your reponse. Thanks for the suggestions about simulations. I have been considering starting down this path. The true rotation does not involve a stationary axis, the axis is experiencing translational motion as well. The rate and even direction (CW or CCW) of rotation is varying over time as the body translates and rotates in space. The physics of the problem is basically this: picture a boat in the water with no one at the helm, floating freely. But consider that the boat can only translate and rotate in a plane (no change in roll or pitch). The coordinates of two points on opposite ends of the boat are being measured every minute or so. Cheers! How can you measure every minute but want to draw a conclusion about successive seconds? That sounds unrealistic to me. I am not sure if I have understood your problem correctly, but wouldn't the perpendicular bisectors of two successive positions of the chord p1 p2 intersect at an instantaneous centre of rotation? Hi Pongo I'm having to model the motion based on the 1-minute interval data (I'm using a polynomial) and then interpolate using the functional model to get down to the second-level. It's not ideal, but it's the best data I have. Yes, the perpendicular bisectors would intersect at a pole of planer displacement. My apologies, the Physics of the problem is where I'm likely falling down. I'm wondering if you have any thoughts on how I could use this fact? So far, to calculate the rotation between two instants, I've been calculating the azimuth of the p1 p2 chord at t1 and the same at t2, then subtracting them. Is this valid? Sorry. I now realize I was mistaken about the perp bisectors. I mis-applied a principle I use in the collapse of building frames, and my case doesn't include translation. I 'll think more about it. Diagrams help to clarify. Whether your 1 minute readings to 1 sec predictions are sufficiently accurate would have a relationship to the period of oscillation, wouldn't it? gge said: The physics of the problem is basically this: picture a boat in the water with no one at the helm, floating freely. But consider that the boat can only translate and rotate in a plane (no change in roll or pitch). The coordinates of two points on opposite ends of the boat are being measured every minute or so. The physics of the problem should include the physics of the measurement. if the boat is of constant length do the endpoints in the data always produce points that are that length apart? ## 1. How do you determine the two positions on a rigid body for calculating rotation? The two positions on a rigid body are typically chosen as two points that are fixed and do not move during the rotation. These points are often referred to as pivot points or reference points. They can be located anywhere on the body, but it is important that they are accurately measured and clearly defined to ensure accurate calculations. ## 2. What is the importance of using two positions on a rigid body for calculating rotation? Using two positions on a rigid body allows for a more accurate and precise calculation of rotation. It takes into account the distance between the two points and their relative positions, allowing for a more comprehensive understanding of the body's movement. It also helps to eliminate any errors or discrepancies that may occur when using only one point for calculations. ## 3. How do you calculate rotation using two positions on a rigid body? The most common method for calculating rotation using two positions on a rigid body is by using the distance between the two points and the angle of rotation. This can be done using trigonometric functions such as sine, cosine, and tangent. It is important to also take into account the direction of rotation, which can be determined by the order in which the points are measured. ## 4. Can you use more than two positions on a rigid body for calculating rotation? Yes, it is possible to use more than two positions on a rigid body for calculating rotation. However, this may result in more complex calculations and may not necessarily provide a more accurate result. It is important to carefully consider the placement and number of points used in order to minimize errors and ensure accurate calculations. ## 5. Are there any limitations to using two positions on a rigid body for calculating rotation? While using two positions on a rigid body is a common and effective method for calculating rotation, it does have some limitations. This method assumes that the body is rigid and does not take into account any deformations that may occur during rotation. It also does not account for any external forces acting on the body, which may affect its rotation. Therefore, it is important to use this method in conjunction with other techniques and to consider these limitations when interpreting the results. Replies 3 Views 538 Replies 5 Views 1K Replies 3 Views 1K Replies 8 Views 1K Replies 13 Views 2K Replies 8 Views 1K Replies 30 Views 2K Replies 21 Views 493 Replies 2 Views 10K Replies 3 Views 1K	1,678	7,872	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38	latest	en	0.949574
http://www.slideserve.com/fallon-waller/denormalization	1,488,066,907,000,000,000	text/html	crawl-data/CC-MAIN-2017-09/segments/1487501171900.13/warc/CC-MAIN-20170219104611-00085-ip-10-171-10-108.ec2.internal.warc.gz	602,750,226	15,148	Sponsored Links This presentation is the property of its rightful owner. 1 / 10 # Denormalization PowerPoint PPT Presentation Denormalization. - Causes redundancy, but fast performance & no referential integrity - Denormalize when • specific queries occur frequently, • a strict performance is required and • it is not heavily updated ### Download Presentation Denormalization An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - ### Denormalization - Causes redundancy, but fast performance & no referential integrity - Denormalize when • specific queries occur frequently, • a strict performance is required and • it is not heavily updated -So, denormalize only when there is a very clear advantage to doing so and document carefully the reason for doing so ### typical denormalization techniques • Flatten a repeating group in one table Instead of EMP (E#, Ename) SKILL (E#, Skill) UseEMP (E#, Skill, Ename) when Emp has a smaller # of attributes. - This means use Method 2 of 1NF algorithm. But know the danger of this method as we discussed in MVD. ### Cont’ (2) Embed stable Code-Interpretation (Reference) Table. Instead of FLIGHT (F#, Departs, From_Code, To_Code) CODE (Code, Airport_Name) Use FLIGHT (F#, Departs, From_AP, From_Code, To_AP, To_Code) ### Cont’ Combine1:1 or 1:N (a) when N is small and (b) the record on the "one" side is small (thus the amount of redundancy will be small) Instead of SALE (S#, SPName, SaleDate), SALE_ITEMS (S#, Line#, Code, Qty) Use SALE(S#, Line#, SPName, SaleDate, Code, Qty) -- "How many T179's did we sell yeaterday?" can be answered without join. • Another example: Order_Item(O#, I#, C#, Cname, I_Desc, Qty, I_Price) ### Cont’ (4) When the other entity in is not interesting by itself Order(O#, ODate, OShipTerms, PmtTerms, Cname, CAddr) (5) Replicate non-frequently updated attributes to avoid JOIN WORK_ON (ESSN, P_NUM, PName, Hours) ### Problems of denormalization • Makes row longer • Makes data transfer longer • Needs more memory for memory processing • Cause redundancy and expensive update ### Adding redundant data - Add summary attributes or derived attributes - Redundant relationships can improve performance with the cost of update overhead ### Schema translation • Reduce #of relations for JOIN by using mapped translation • Handling null values • Combine 1:1 relationships • Relax participation constraints • Divide the big table into two, if A & B are distinct in R(A, B) • Ignore FDs based on co-occurring attributes, which are not updated ZIP --> CITY ### Primary key - Most frequently used attributes - Prefer small sized attributes (used in indexes, Ref. integrity) ### Index - Create a set of appropriate indexes optimzing queries (This will be discussed more in physical DB chapters.)	802	3,283	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3	3	CC-MAIN-2017-09	longest	en	0.843283
https://essaycreek.com/investigation-into-stress-capacitythis-physics-experiment-was-conducted-to-investigate-stress-capacity-or-known-as-tensile-strength-we-had-to-determine-the-relationship-between-the-force-applied-2/	1,660,307,041,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882571692.3/warc/CC-MAIN-20220812105810-20220812135810-00685.warc.gz	248,496,115	19,020	# Investigation into Stress Capacity This physics experiment was conducted to investigate stress capacity or known as tensile strength. We had to determine the relationship between the force applied on an object and the extension of the elastic material. The link between the force applied and the extension of the spring is governed by Hooke’s Law. To establish this connection, we need to measure the relationship between the force applied and the extension. We did this by adding a series of masses, which builds the force on the spring (The Professor, 2016). Hooke’s Law: When an elastic object is stretched, the expanded length is called extension. The extension of the elastic object is directly proportional to the force applied to it also known as F=kx. F is Force and expressed in Newtons (N). k is the spring constant expressed in newtons per metre (N/m) x is the extension expressed in metres (m). (BBC Bitesize, 2016) A force causes an object to undergo changes, forces that are unbalanced change speed, shape or direction. Gravitational forces are only in effect if there are two or more masses while friction opposes motion (BBC Bitesize,2017). Forces that affected this experiment are: Gravitational Force, Applied Force, Tensional Force. Gravitational Force: A force of attraction between all objects, it depends on the masses of the object and the distance between them. The larger object, the stronger the gravitational force. Applied Force: A force when an object is being pushed or pulled by another object. Applied forces causes an object to undergo changes in acceleration, velocity and direction. The force equation is F=ma. Tensional Force: It is a force that is transmitted through a string, cable, elastic band, rope or a wire. Tensional force is directed along the length of the object when it is pulled by forces from opposite ends. The relationship between stress and strain is known as the stress-strain curve. It is different for each material and is found by recording the amount of deformation at distinct intervals of tensile or compressive loading. Strain is relative change in shape or size of an object due to externally applied forces. This means it becomes dimensionless and has no units. Stress is the internal force (per unit area) associated with strain. It is predicted that the two materials will show different conclusions. I believe that the silicon tube has higher capability of holding larger masses. It depends on the length and thickness of the material. The prediction is that the silicon tube will have the highest stress capacity and won’t have a long extension while the elastic band will stretch the furthest but won’s have much stress capacity. Equipment Retort Stand Clamp 1 Metre Ruler Mass Hanger Silicon Tube Elastic Band 50g and 100g Masses Notebook and Pen Procedure Setup Retort stand and clamp. To start with, place the material/object on without any stress. Measure the weight of the mass hanger. Measure the length of the material/object. Place mass on it and measure the length to figure out extension Record results on table Repeat step 5, three times for each weight to allow results to be reliable then find average. Keep repeating step 5,6,7 until enough data has been recorded for each material Milan Abraham Elastic Band Mass (kg) Force Original Length 1st Extension 2nd Extension 3rd Extension Average 0 0 14.9 0.061 0.60N 14.9 +1.1 (16) +0.9 (15.8) +1 (15.9) +1 (15.9) 0.161 1.58N 14.9 +1.5 (16.4) +1.4 (16.3) +1.5 (16.4) +1.5 (16.4) 0.261 2.56N 14.9 +1.8 (16.7) +1.7 (16.6) +1.9 (16.8) +1.8 (16.7) 0.361 3.54N 14.9 +2.1 (17) +2 (16.9) +2.2 (17.1) +2.1 (17) 0.461 4.52N 14.9 +2.4 (17.3) +2.3 (17.2) +2.6 (17.5) +2.4 (17.3) 0.561 5.50N 14.9 +2.9 (17.8) +2.9 (17.8) +2.7 (17.6) +2.8 (17.7) Milan Abraham Silicon Tube Mass (kg) Force Original Length 1st Extension 2nd Extension 3rd Extension Average 0 0 72.2 0.061 0.60N 72.2 +2.3 (74.5) +2.3 (74.5) +2.8 (75) +2.5 (74.7) 0.161 1.58N 72.2 +4.3 (76.5) +4.3 (76.5) +4.3 (76.5) +4.3 (76.5) 0.261 2.56N 72.2 +5.3 (77.5) +6.3 (78.5) +5.8 (78) +5.8 (78) 0.361 3.54N 72.2 +7.8 (80) +7.8 (80) +8.3 (80.5) +8 (80.2) 0.461 4.52N 72.2 +8.8 (81) +8.3 (80.5) +8.8 (81) +8.6 (80.8) 0.561 5.50N 72.2 +10.3 (82.5) +9.8 (82) +9.8 (82) +10 (82.2) 0.98 1.4 0.7 The Elastic band and Silicon tube was hung vertically with a mass hanger to the end of the material. Masses from 61g to 561g were added. The length of the elastic band was measured once it was at rest. In this structure, certain forces were in effect. Gravitational force directed the hanging masses downwards. The Applied Force in this case is the masses pushes the elastic material downwards with gravity supporting it. Restoring Force directs the elastic material upwards, in the opposite direction of displacement. Tension is directed through each material pulling each end. Using Newton’s Law the spring constant was calculated for each material. F=ma was transferred into W=mg to calculate gravity. To find the spring constant for the elastic band, mass was turned into dynes. Gravity times mass= dynes. 9836.07 times 0.561 = 5518.04 dynes. Then it is dynes divided by extension(cm) which is 5518.07 divided by 2.8 = 1970.74 dynes/cm which is 1.97N/m. The spring constant for the silicon tube was which is 0.55N/m. The intercept for the best fit straight line is close to the origin and is also consistent with Hooke’s Law. To minimise errors, we should have viewed the ruler from specific angles to make sure it was vertical. The scale should be viewed at eye level to avoid parallax error. There were some issues with the retort stand so some equipment was not fit for the experiment and were faulty. In some instances, we needed to make sure that when the person was adding mass, he didn’t stretch the material while he was doing it so the results could be accurate. We reduced the decimal places to two so that results could be easy to substitute while being accurate. The hypothesis was predicted to be that the elastic band while have a longer extension than the silicon tube and the silicon tube would be stronger than the elastic band. The prediction wasn’t correct as from results it says that silicon tube had a bigger extension which was a surprising result as the material was much thicker. The elastic band had the best result because it showed results that were somewhat predicted and accurate. Through this experiment investigated stress capacity, Hooke’s Law and certain forces which have influenced the experiment. The hypothesis wasn’t as predicted. For this experiment, we determine the spring constant and Young’s Module, we had to interpret the behaviour of two different materials. The results will be analysed to determine what is happening to the materials physically and which one is more capable of the masses. The two types of materials used for this experiment were an elastic band and a tube of silicon rubber. ## Order a Unique Copy of this Paper Essay Creek is an academic writing service provided to you by, a London-based company. • Experience Helping students successfully for 11 years. • Confidentiality & Security Be sure your information will be kept confidential due to our secure service. • Quality & Reliability 8.5 out of 10 average quality score according to our customers' feedback. 97.45% of orders delivered on time. • Versatility 478 active writers in 68 disciplines. • 100% money back guarantee You can always request a refund if you are not satisfied with the result. 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View samples • Our Free Essay Tools Even more opportunities to improve your academic papers.	2,151	8,412	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.640625	4	CC-MAIN-2022-33	latest	en	0.959062
http://gmatclub.com/forum/in-a-certain-infinite-sequence-17725.html	1,435,955,192,000,000,000	text/html	crawl-data/CC-MAIN-2015-27/segments/1435375096209.79/warc/CC-MAIN-20150627031816-00115-ip-10-179-60-89.ec2.internal.warc.gz	117,464,034	46,110	Find all School-related info fast with the new School-Specific MBA Forum It is currently 03 Jul 2015, 12:26 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # In a certain infinite sequence Author Message TAGS: Senior Manager Joined: 30 May 2005 Posts: 374 Followers: 1 Kudos [?]: 7 [0], given: 0 In a certain infinite sequence [#permalink] 02 Jul 2005, 21:56 00:00 Difficulty: (N/A) Question Stats: 0% (00:00) correct 0% (00:00) wrong based on 1 sessions In a certain infinite sequence n(1)=7,n(2)=70,n(3)=700,n(x)=10n(x-1) Is the integer j a factor of every member of n? S1. j is a prime number. S2. j is a factor of more than one member of the sequence Director Joined: 18 Apr 2005 Posts: 548 Location: Canuckland Followers: 1 Kudos [?]: 8 [0], given: 0 Re: DS - Infinite sequence divisior [#permalink] 02 Jul 2005, 23:31 AJB77 wrote: In a certain infinite sequence n(1)=7,n(2)=70,n(3)=700,n(x)=10n(x-1) Is the integer j a factor of every member of n? S1. j is a prime number. S2. j is a factor of more than one member of the sequence 1) insuff - 7 yes, 5 and 2 no 2) 1 and 7 factors of every member, 2,5, 10, 70 etc are factors of more than one member but not all 1)+2) insuff, 7 yes and 5 and 2 no again E. Senior Manager Joined: 17 Apr 2005 Posts: 375 Location: India Followers: 1 Kudos [?]: 17 [0], given: 0 Re: DS - Infinite sequence divisior [#permalink] 03 Jul 2005, 00:13 AJB77 wrote: In a certain infinite sequence n(1)=7,n(2)=70,n(3)=700,n(x)=10n(x-1) Is the integer j a factor of every member of n? S1. j is a prime number. S2. j is a factor of more than one member of the sequence Shouldn't the general Xth term be n(x) = [10^(x-1)]n. Agree with sparky, E. HMTG. Current Student Joined: 28 Dec 2004 Posts: 3387 Location: New York City Schools: Wharton'11 HBS'12 Followers: 14 Kudos [?]: 186 [0], given: 2 i think the n(x)=n*(10^(x-1)) E it should be... we have 3 primes, 7, 5 and 2 it could be anyone... Similar topics Replies Last post Similar Topics: An infinite sequence of positive integers is called a 3 06 Jun 2012, 18:33 15 An infinite sequence of positive integers is called a 10 16 Feb 2012, 15:11 13 S is an infinite sequence, in which the value of a 9 23 Sep 2011, 22:23 If S is an infinite sequence of positive integers, is S an 1 04 Sep 2006, 07:15 Display posts from previous: Sort by	918	2,871	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.859375	4	CC-MAIN-2015-27	longest	en	0.869312
https://cs.stackexchange.com/questions/103112/how-does-one-know-what-statements-in-coq-require-induction	1,701,948,492,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100651.34/warc/CC-MAIN-20231207090036-20231207120036-00813.warc.gz	220,439,408	49,302	# How does one know what statements in Coq require Induction? I was trying to learn Coq using the famous book Software Foundations. In it I found the following: Theorem mult_0_r : forall n:nat, n * 0 = 0. Proof. induction n as [\| n IHn]. - simpl. reflexivity. - simpl. rewrite -> IHn. reflexivity. Qed. which I understand perfectly but find rather unintuitive. I understand perfectly how each step works but it would have never occurred to me to use induction to prove such a trivial fact. In fact in the mathematical proof I had in mind that would have been a fact/property (or I guess an axiom) of 0. i.e. $$\forall n \in N, n \cdot 0 = 0$$ is true by definition. I guess in Coq (or the way we set up numbers? thats not true). My biggest complaint or worry is that if such a trivial thing requires induction I feel now I am unable to recognize what needs induction (at least in Coq). I know it needs it here because I am in the induction chapter. But in normal maths its usually quite obvious because the problem is obviously recursive. But I wouldn't have really thought of that proposition as recursive. For example it goes on to prove more things as exercises: Theorem plus_n_Sm : ∀n m : nat, S (n + m) = n + (S m). Proof. (* FILL IN HERE ) Admitted. Theorem plus_comm : ∀n m : nat, n + m = m + n. Proof. ( FILL IN HERE ) Admitted. Theorem plus_assoc : ∀n m p : nat, n + (m + p) = (n + m) + p. Proof. ( FILL IN HERE ) Admitted. which I am sure are not too difficult, but my worry is that in isolation I would have never thought such trivial statements required something as sophisticated as induction. I didn't even learn induction until last 2 years of highschool and didn't really do it seriously until college. So now I see trivial statements requiring what seems to me, sophisticated mathematics. I just feel my intuition got really lost. When I do Coq proofs (in isolation), how do I know when to use induction and on what? I doubt there is a general procedure (of course) but proofs do exist. So there must be something guiding us to use induction in Coq. • Every mathematical book that actually bothers to prove $n \cdot 0 = 0$ from Peano axioms does so by induction on $n$. This all looks unintuitive to you because you have never seen any proofs of such facts, and now you're looking at their formal incarnations. (Note: the relevant Peano axiom $0 \cdot n = 0$ does not imply $n \cdot 0 = 0$, you first need to prove $n \cdot m = m \cdot n$. However, most proofs of $n \cdot m = m \cdot n$ rely on $n \cdot 0 = 0$.) Jan 21, 2019 at 7:42 • @AndrejBauer seems theorem provers lack a severe case of common sense mathematics! As do computers in all AI fields. Jan 22, 2019 at 23:31 Coq allows one to prove mathematical theorems in a completely formal way. At first, this copes with our experience of doing maths, which is far more informal. Most of the time, people doing maths are not exposed to mathematical logic. They do not know, for instance, how to define the set of real numbers. Or even the set of natural numbers. Numeric sets are simply taken for granted, together with their operations and related properties. On top of these "basic facts" people build more complex theorems. Essentially, the "basic facts" are used as if they were axioms. Note that one could work at a similar level in Coq, if one wishes. One can import the Arith Coq library, which provides the basic facts of arithmetics, and exploit those. However, the underlying logic of Coq does not take those properties for granted. The type of natural numbers, for instance, is not a primitive concept in the logic, but rather something which is defined using induction. Since that makes nat an inductive type, virtually every proof about naturals will require induction. (This also holds in set theory, e.g. Zermelo-Fraenkel, where naturals are also defined in a sort-of inductive way.) Arithmetical operators are also defined using induction. Using a simplified syntax, one could define multiplication as x 0 = 0 x * (S y) = x + (x * y) or, alternatively, 0 * x = 0 (S y) * x = x + (y * x) Both definitions are equivalent. However, using the first one will result in us being able to prove x * 0 = 0 "without induction" -- it follows by definition. In Coq, we can simply write reflexivity and let the system expand the definition for us. By comparison, 0 * x = 0 is not a trivial equation which follows from the definition: to expand the definition we need to know what is the second argument of the multiplication, but here it's merely an unknown x. Hence, for this we do need induction over x. There's no way around that. Note that, if we used the second definition instead, we would prove 0 * x = 0 immediately, and instead require induction for x * 0 = 0. So, no definition is strictly better than the other one. • just to study this specific example more in depth and hopefully gain an intuition when to use induction (in contrast to destruct for example). I understand that in either case x is unknown. But what seems to be a coincidence or even lucky that works is induction on x. Why exactly did we use induction on the unknown? How did it help? this remains a bit mysterious to me. It's still unclear to me when induction would be used or what the intuition for it is (especially contrasted with destruct for example). Jan 20, 2019 at 20:02 • @Pinocchio Well, if * was defined by induction on the first argument, and our goal to be proved is property (x * something), inducing on x looks promising, since that will make x * something simplify, unrolling the definition. Actually, destruct also does that, but it does not provide any induction hypothesis. Very roughly, you might look at destruct as a weak form of induction which does not provide induction hypotheses. – chi Jan 20, 2019 at 22:06 • is that why we use induction on n when dealing with something like (n + m) * p = (n * p) + (m * p). I tried p but the proof didn't move forward when I got to the inductive case. Jan 20, 2019 at 22:24 • @Pinocchio Yes, + and * are defined by induction on their first argument in the standard library of Coq. – chi Jan 20, 2019 at 22:36 As the others have mentioned, in Coq's standard library (or typical presentations of naturals in Coq), naturals are defined inductively, usually a la Peano. We could make other choices, e.g. one could imagine defining naturals as the free semiring on one generator which would give you many of the "axiomatic" facts for "free". (Well, you'd have to prove them about general rings...) For any inductively defined type, T, whenever you want to prove a statement of the form forall x:T, P(x) you will need to use induction on T unless the proof doesn't actually depend on T, i.e. it is parametric, e.g. forall x:T, x = x. In fact, all proofs of this form can be written as an induction, they may just be a trivial induction. Often the induction will be hidden away in some lemma. However, the key concept it seems you are unfamiliar/uncomfortable with is the notion of definitional equality in intensional dependent type theories like Coq. Let's say you state that $$2$$ means $$1+1$$. If you then ask whether $$2=1+1$$ there are no steps of proof required. You simply substituted the meaning of $$2$$ and see that $$1+1=1+1$$ which is true by reflexivity. More generally, (closed) terms in Coq have normal forms, and Coq treats any two terms with the same normal form as identical. This is definitional equality. Reducing to that normal form may involve large amounts of computation, but there is no proof rule that to tell Coq to reduce a term to normal form. It just does it in the background as necessary to perform type checking. Definitions, such as the one for multiplication reproduced below: Fixpoint mul n m := match n with \| 0 => 0 \| S p => m + p * m end where "n * m" := (mul n m) : nat_scope. introduce new (non-normal) terms and new computational rules for normalizing them. In this case, it adds a rule that says 0 * x = 0. These follow from the generic rules for Fixpoint and match for the Coq type theory. If you were to ask whether forall x:nat, 0 * x = 0, Coq sees forall x:nat, 0 = 0 since 0 is the normal form of 0 * x. There is no computational rule saying how x * 0 should be reduced when x is a variable, so the "normal" form of x * 0 is x * 0. With this understanding, it is much easier to see why and how induction helps in this particular case. When we do induction, we get two new problems: 0 * 0 = 0 and (S p) * 0 = 0 which allows steps of computation to be performed. In the latter case, we get 0 + p * 0 which then leans on the computation rules induced by the definition of natural number addition, but also leaves us with n * 0 which requires the induction hypothesis to resolve. At more advanced levels of proof engineering, it is common to try to set up definitions so that a decent amount of work can be done by this implicit computation mechanism. Extreme forms of this are reflection principles where we prove an implementation of a decision procedure correct after which we can simply run it and appeal to this result to prove other results. Techniques like this are behind things like the ring solver and ssreflect. To see how you can prove something about natural numbers, you must know what you know about natural numbers. Usually, we learn all sorts of 'basic facts' about numbers in high school that are pretty much given without further proof, which we can use later to prove more interesting statements about numbers. In most mathematical disciplines, this 'naive' approach to numbers is fine. However, when we care about the formalization of numbers, we want a 'minimal definition' of what a number (and related concepts such as successors, predecessors and arithmetic) is and prove all 'basic facts' from them. So, in the book you're reading, all we know about numbers is the following definition: Inductive nat : Type := \| O \| S (n : nat). That is, natural numbers are an inductive type, defined as either the natural number O or the successor of a natural number. This is the reason why we do proofs by induction: it is all we know about the natural numbers in this setting. So, where did your intuition let you get lost? When learning mathematics, you get taught that 'basic facts' about numbers early on, which therefore seem easy and get taught induction fairly late (and when you actually have to prove things!) which makes it seem hard. What is actually the case is that, from a viewpoint of formalisation, induction is the one fundamental property of natural numbers! The best way to see this is that, in fact, all other properties of natural numbers follow from it. My final advice is not to worry. The fact that all 'basic facts' are suddenly not as basic as they may have seemed so far can be intimidating, but this is simply what formalisation is about: having rigid constructions for 'basic facts'. Once you have these in place, you can prove more complicated statements as usual, by relying on the 'basic facts' just proven. • the thing that still concerns me is if you see the remaining 3 facts/theorems I wrote, I know they should use induction cuz they are in the induction chapter. But how would you have suspected to use induction in the first place? This is what still remains a but mysterious to me. Jan 20, 2019 at 20:04 • As I said, the fact that nat is defined as an inductive type (or recursively, if you like) and the fact that we know nearly nothing else about nat is something that indicates induction is a useful approach here. Be sure that you understand how inductive types related to proofs by induction! If you do, and still feel lost, then I'm afraid that your question goes into the "how should I prove things" territory, which is far to general to answer. Consider delaying this question until you actually get stuck proving something in a later chapter. Jan 20, 2019 at 20:54 • Ok if I get stuck on later chapters I shall come back. Though, my current strategy seems to try destruct and induction sort of randomly. Though, my biggest confusion is that in highschool things that needed induction were so obvious. But say, the commutativity property doesn't seem like an recursive statement (except for the fact that nat is inductive in Coq). It's just I never had though of commutativity as recursive before which is bothering me, but I will follow your advice and see what happens alter... Jan 20, 2019 at 21:08 I want to share my own experience of learning Coq and theorem proving in general. Most of the time, the proof of a statement largely depends on the recursive structure of the function or operation at hand. IMHO it is even more important than the recursive structure of data types. Again, let's look at multiplication of natural numbers. Fixpoint mul n m := match n with \| 0 => 0 \| S p => m + p * m end where "n * m" := (mul n m) : nat_scope. Theorem mult_0_r : forall n, mul n 0 = 0. (Before you write a proof, just focus on the definition of the function, and ignore what it means as a number operation.) The definition of mul matches on the structure of n, then, if n = S p, recursively calls itself with p. The structure is the strong sign that you'll need induction on the first argument of mul. On the other hand, you won't need to do induction on the second argument, or even destruct, simply because it's not pattern-matched inside mul. This applies to more complex statements as well. Say we were to prove distributivity of multiplication over addition: Theorem mult_distr_plus_r : forall n m p, (n + m) * p = n * p + m * p. And suppose that we know both plus and mult are recursive on the first argument. In this case, the natural choice to do induction on is n because, by its positions, destructing n gives the largest amount of expansion (three times in total): (S n + m) * p = S (n + m) * p = p + (n + m) * p S n * p + m * p = (p + n * p) + m * p It's easy to see that you can complete this proof by IH and plus_assoc. In contrast, destructing m gives just one, and p gives none, so you'll need more lemmas to complete the proof (e.g. plus_n_Sm, plus_comm if you destruct m, mult_n_Sm in addition if you destruct p). (n + S m) * p = ? n * p + S m * p = n * p + (p + m * p) (n + m) * S p = ? n * S p + m * S p = ? Here's another example to show that proof by induction depends on the function structure. Suppose we have the Fibonacci function: Fixpoint fib n := match n with \| 0 => 0 \| 1 => 1 \| S (S p) => fib p + fib (S p) end. Then we want to prove some property on it, say it gives the same result as another definition of Fibonacci. Theorem fib_is_fib' : forall n, fib n = fib' n. Now, the recursive structure of fib is more complicated than mul above. It has two base cases, and the result of fib (S (S p)) depends on both fib p and fib (S p). If you try good ol' induction (with tactic induction n.) on it, you immediately fail. In order to prove a property of fib, you actually need a stronger induction principle: • Given a property P on natural numbers, if all of the following holds: • P holds for 0. • P holds for 1. • If P holds on n and S n, it holds on S (S n). • Then P holds on all natural numbers. In Coq: Theorem nat_ind_fib : forall P : nat -> Prop, (* For a given property P, ) P 0 -> ( if P holds for 0 ) P 1 -> ( and 1, ) (forall n : nat, P n -> P (S n) -> P (S (S n))) -> ( and P n and P (S n) implies P (S (S n)) ) forall n : nat, P n. ( then P holds for all n. *) Now we can prove a property on fib by induction using tactic induction n using nat_ind_fib.	3,847	15,650	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.78125	4	CC-MAIN-2023-50	longest	en	0.963799
https://pdfcoffee.com/load-calculation-for-trussesxlsx-pdf-free.html	1,716,533,195,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058684.66/warc/CC-MAIN-20240524060414-20240524090414-00502.warc.gz	384,025,168	9,604	# LOAD CALCULATION FOR TRUSSES.xlsx ##### Citation preview LOAD CALCULATION FOR RECTANGULAR CLADED BUILDING WITH PITCHED ROOFS(As per IS875-1987) Project : 22-Jul-2013 Date : INPUT BUILDING HEIGHT BUILDING WIDTH BUILDING LENGTH TRUSS ROOF ANGLE TRUSS SPACING RUNNER SPACING PURLIN SPACING WIND SPEED : : : : : : : : 10.00 m 10.00 m 15.00 m 20 deg 5.00 m 1.30 m 0.750 m 39 m/s 0.084 0.125 0.084 0.063 kN/m^2 kN/m Weight of sheet = Weight of runner = Dead load on purlins = Dead load on purlin on edge= x 0.750 /2 = = = = = = Dist between outerface of column flange to center of runner Moment on corner column due to bottom runner = due to top runner = due to runner below top runner = all other runners = Moment on other columns due to bottom runner = due to top runner = due to runner below top runner = all other runners = Dist between outerface of column web to center of runner Moment on corner column due to bottom runner = due to top runner = due to runner below top runner = all other runners = 0.063 kN/m 0.0315 kN/m (0.650 x (0.450 x (1.100 x (1.300 x 0.084 ) 0.084 ) 0.084 ) 0.084 ) + + + + = 0.125 0.125 0.125 0.125 = = = = 0.1796 0.1628 0.2174 0.2342 kN/m kN/m kN/m kN/m 0.120 m 2.500 x 0.120 x 2.500 x 0.120 x 2.500 x 0.120 x 2.500 x 0.120 x 0.180 0.163 0.217 0.234 = = = = 0.05388 0.04884 0.06522 0.07026 kNm kNm kNm kNm 5.000 x 0.120 x 5.000 x 0.120 x 5.000 x 0.120 x 5.000 x 0.120 x 0.180 0.163 0.217 0.234 = = = = 0.10776 0.09768 0.13044 0.14052 kNm kNm kNm kNm 0.08531 0.07733 0.103265 0.111245 kNm kNm kNm kNm = 2.500 x 0.190 x 2.500 x 0.190 x 2.500 x 0.190 x 2.500 x 0.190 x 0.190 m 0.180 0.163 0.217 0.234 = = = = Live load per m^2 = Live load on purlins Live load on purlin on edge = = 0.75 - [(20 - 10 ) x 0.55 x 0.750 0.4125 /2 = = 0.02] = 0.55 0.4125 kN/m 0.20625 kN/m kN/m^2 k1 Uptp 10m height 1 = k2 = Vz = 39 x 1 x Pz = (0.6x 39 x On rectangular cladded portion 1 k3 = 1 x 1 = 39 ) / 1000 39 m/s = h/w Elevation B C D 0.9 -0.8 -0.45 -0.8 -0.8 0.9 -0.8 -0.45 Wind angle in 0 degree (in X direction) On side A Force on corner column due to bottom runner due to top runner due to runner below top runner all other runners Force on other columns due to bottom runner due to top runner due to runner below top runner all other runners On side B Force on corner column due to bottom runner due to top runner due to runner below top runner all other runners Force on other columns due to bottom runner due to top runner due to runner below top runner all other runners On side C Force on corner column due to bottom runner due to top runner due to runner below top runner = 1 Cpe for surface l/w = 1.5 A B C D 0 90 0.7 -0.6 -0.25 -0.6 -0.6 0.7 -0.6 -0.25 Cpi A 0.9126 Kn/m^2 wind angle in degree Plan Cpe +Cpi 1 = Local Cpe -1.1 -0.2 or 0.2 Local Cpe+Cpi -1.3 = = = = 0.90x 0.90x 0.90x 0.90x 0.650 x 2.500 x 0.450 x 2.500 x 1.100 x 2.500 x 1.300 x 2.500 x 0.913 0.913 0.913 0.913 = = = = 1.334678 0.924008 2.258685 2.669355 kN kN kN kN = = = = 0.90x 0.90x 0.90x 0.90x 0.650 x 5.000 x 0.450 x 5.000 x 1.100 x 5.000 x 1.300 x 5.000 x 0.913 0.913 0.913 0.913 = = = = 2.669355 1.848015 4.51737 5.33871 kN kN kN kN = = = = -0.45x -0.45x -0.45x -0.45x 0.650 x 2.500 x 0.450 x 2.500 x 1.100 x 2.500 x 1.300 x 2.500 x 0.913 0.913 0.913 0.913 = = = = -0.66734 -0.462 -1.12934 -1.33468 kN kN kN kN = = = = -0.45x -0.45x -0.45x -0.45x 0.650 x 5.000 x 0.450 x 5.000 x 1.100 x 5.000 x 1.300 x 5.000 x 0.913 0.913 0.913 0.913 = = = = -1.33468 -0.92401 -2.25869 -2.66936 kN kN kN kN = = = -0.80x -0.80x -0.80x 0.650 x 2.500 x 0.913 = 0.450 x 2.500 x 0.913 = 1.100 x 2.500 x 0.913 = -1.18638 kN -0.82134 kN -2.00772 kN all other runners Force on other columns due to bottom runner due to top runner due to runner below top runner all other runners On side D Force on corner column due to bottom runner due to top runner due to runner below top runner all other runners Force on other columns due to bottom runner due to top runner due to runner below top runner all other runners Wind angle in 90 degree (in Z direction) On side A Force on corner column due to bottom runner due to top runner due to runner below top runner all other runners Force on other columns due to bottom runner due to top runner due to runner below top runner all other runners On side B Force on corner column due to bottom runner due to top runner due to runner below top runner all other runners Force on other columns due to bottom runner due to top runner due to runner below top runner all other runners On side C Force on corner column due to bottom runner due to top runner due to runner below top runner all other runners Force on other columns due to bottom runner due to top runner due to runner below top runner = -0.80x 1.300 x = = = = -0.80x -0.80x -0.80x -0.80x = = = = 2.500 x 0.913 = -2.37276 kN 0.650 x 5.000 x 0.450 x 5.000 x 1.100 x 5.000 x 1.300 x 5.000 x 0.913 0.913 0.913 0.913 = = = = -2.37276 -1.64268 -4.01544 -4.74552 kN kN kN kN -0.80x -0.80x -0.80x -0.80x 0.650 x 2.500 x 0.450 x 2.500 x 1.100 x 2.500 x 1.300 x 2.500 x 0.913 0.913 0.913 0.913 = = = = -1.18638 -0.82134 -2.00772 -2.37276 kN kN kN kN = = = = -0.80x -0.80x -0.80x -0.80x 0.650 x 5.000 x 0.450 x 5.000 x 1.100 x 5.000 x 1.300 x 5.000 x 0.913 0.913 0.913 0.913 = = = = -2.37276 -1.64268 -4.01544 -4.74552 kN kN kN kN = = = = -0.80x -0.80x -0.80x -0.80x 0.650 x 2.500 x 0.450 x 2.500 x 1.100 x 2.500 x 1.300 x 2.500 x 0.913 0.913 0.913 0.913 = = = = -1.18638 -0.82134 -2.00772 -2.37276 kN kN kN kN = = = = -0.80x -0.80x -0.80x -0.80x 0.650 x 5.000 x 0.450 x 5.000 x 1.100 x 5.000 x 1.300 x 5.000 x 0.913 0.913 0.913 0.913 = = = = -2.37276 -1.64268 -4.01544 -4.74552 kN kN kN kN = = = = -0.80x -0.80x -0.80x -0.80x 0.650 x 2.500 x 0.450 x 2.500 x 1.100 x 2.500 x 1.300 x 2.500 x 0.913 0.913 0.913 0.913 = = = = -1.18638 -0.82134 -2.00772 -2.37276 kN kN kN kN = = = = -0.80x -0.80x -0.80x -0.80x 0.650 x 5.000 x 0.450 x 5.000 x 1.100 x 5.000 x 1.300 x 5.000 x 0.913 0.913 0.913 0.913 = = = = -2.37276 -1.64268 -4.01544 -4.74552 kN kN kN kN = = = = 0.90x 0.90x 0.90x 0.90x 0.650 x 2.500 x 0.450 x 2.500 x 1.100 x 2.500 x 1.300 x 2.500 x 0.913 0.913 0.913 0.913 = = = = 1.334678 0.924008 2.258685 2.669355 kN kN kN kN = = = 0.90x 0.90x 0.90x 0.650 x 5.000 x 0.913 = 0.450 x 5.000 x 0.913 = 1.100 x 5.000 x 0.913 = 2.669355 kN 1.848015 kN 4.51737 kN all other runners On side D Force on corner column due to bottom runner due to top runner due to runner below top runner all other runners Force on other columns due to bottom runner due to top runner due to runner below top runner all other runners On roof Building height Roof angle ratio 1 20 deg h/w = 0.90x 1.300 x = = = = -0.45x -0.45x -0.45x -0.45x 0.650 x 2.500 x 0.450 x 2.500 x 1.100 x 2.500 x 1.300 x 2.500 x 0.913 0.913 0.913 0.913 = = = = -0.66734 -0.462 -1.12934 -1.33468 kN kN kN kN = = = = -0.45x -0.45x -0.45x -0.45x 0.650 x 5.000 x 0.450 x 5.000 x 1.100 x 5.000 x 1.300 x 5.000 x 0.913 0.913 0.913 0.913 = = = = -1.33468 -0.92401 -2.25869 -2.66936 kN kN kN kN 1 Cpe for roof Wind angle in degree 0 90 EF GH EG FH -0.7 -0.5 -0.8 -0.6 0.913 = 5.33871 kN = Local coefficients -1.5 Cpi Building height Roof angle ratio 1 kN/m^2 5.000 x (Cpe+Cpi) for roof Wind angle in degree 0 90 EF GH EG FH -0.9 -0.7 -1 -0.8 = -1.5 -1.5 -1.5 -0.2 or 0.2 Local coefficients -1.7 -1.7 -1.7 -1.7 = = -0.90x 0.750 x -0.616 /2 0.913 x = = -0.61601 kN/m -0.308 kN/m = = = = -0.70x 0.750 x -0.479 /2 ( -0.9 + -1.7 ) x ( 0.9 + -1.7 ) x 0.913 x = = 0.913 0.913 -0.47912 kN/m -0.23956 kN/m sin70= -0.6886 cos70= -0.17339 = = = -1.00x 0.750 x -0.684 /2 ( -1.0 + -1.0 ) x 0.913 x = = 0.913 -0.68445 kN/m -0.34223 kN/m sin70= -0.52969 = = = -0.80x 0.750 x -0.548 /2 ( -0.8 + -0.8 ) x 0.913 x = = 0.913 -0.54756 kN/m -0.27378 kN/m sin70= -0.42375 0.375 x 0.375 x 0.375 x 0.375 x In STAAD MODEL, In addition to these loads self weight of the structure is considered in dead load The following load combinations are taken in STAAD 1) 1.5(DL+LL) 2) 1.2(DL+LL+WLX) 3) 1.2(DL+LL+WLZ) 4) 0.9DL+1.5WLZ 5) 0.9DL+1.5WLX per IS875-1987) 22-Jul-2013 kN/m kN/m kN/m kN/m	3,793	8,311	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.765625	3	CC-MAIN-2024-22	latest	en	0.526253
https://www.tensorflow.org/versions/r2.1/api_docs/python/tf/math/reduce_logsumexp?hl=TR	1,632,687,446,000,000,000	text/html	crawl-data/CC-MAIN-2021-39/segments/1631780057913.34/warc/CC-MAIN-20210926175051-20210926205051-00086.warc.gz	1,036,042,624	44,757	# tf.math.reduce_logsumexp Computes log(sum(exp(elements across dimensions of a tensor))). Reduces `input_tensor` along the dimensions given in `axis`. Unless `keepdims` is true, the rank of the tensor is reduced by 1 for each entry in `axis`. If `keepdims` is true, the reduced dimensions are retained with length 1. If `axis` has no entries, all dimensions are reduced, and a tensor with a single element is returned. This function is more numerically stable than log(sum(exp(input))). It avoids overflows caused by taking the exp of large inputs and underflows caused by taking the log of small inputs. #### For example: ``````x = tf.constant([[0., 0., 0.], [0., 0., 0.]]) tf.reduce_logsumexp(x) # log(6) tf.reduce_logsumexp(x, 0) # [log(2), log(2), log(2)] tf.reduce_logsumexp(x, 1) # [log(3), log(3)] tf.reduce_logsumexp(x, 1, keepdims=True) # [[log(3)], [log(3)]] tf.reduce_logsumexp(x, [0, 1]) # log(6) `````` `input_tensor` The tensor to reduce. Should have numeric type. `axis` The dimensions to reduce. If `None` (the default), reduces all dimensions. Must be in the range ```[-rank(input_tensor), rank(input_tensor))```. `keepdims` If true, retains reduced dimensions with length 1. `name` A name for the operation (optional). The reduced tensor. [{ "type": "thumb-down", "id": "missingTheInformationINeed", "label":"Missing the information I need" },{ "type": "thumb-down", "id": "tooComplicatedTooManySteps", "label":"Too complicated / too many steps" },{ "type": "thumb-down", "id": "outOfDate", "label":"Out of date" },{ "type": "thumb-down", "id": "samplesCodeIssue", "label":"Samples / code issue" },{ "type": "thumb-down", "id": "otherDown", "label":"Other" }] [{ "type": "thumb-up", "id": "easyToUnderstand", "label":"Easy to understand" },{ "type": "thumb-up", "id": "solvedMyProblem", "label":"Solved my problem" },{ "type": "thumb-up", "id": "otherUp", "label":"Other" }]	554	1,907	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.734375	3	CC-MAIN-2021-39	latest	en	0.717025
https://edurev.in/t/79876/NCERT-Solution-Chapter-3--Recording-of-Transaction	1,721,899,403,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00032.warc.gz	183,379,492	70,092	NCERT Solution (Part - 3) Recording of Transactions-I # NCERT Solution (Part - 3) Recording of Transactions-I \| Accountancy Class 11 - Commerce PDF Download Page No 91: Question 9: Transactions of M/s. Vipin Traders are given below. Show the effects on Assets, Liabilities and Capital with the help of accounting Equation. Question 10: Bobby opened a consulting firm and completed these transactions during November, 2005: (a) Invested Rs 4,00,000 cash and office equipment with Rs 1,50,000 in a business called Bobbie Consulting. (b) Purchased land and a small office building. The land was worth Rs 1,50,000 and the building worth Rs 3,50,000. The purchase price was paid with Rs 2,00,000 cash and a long term note payable for Rs 8,00,000. (c) Purchased office supplies on credit for Rs 12,000. (d) Bobbie transferred title of motor car to the business. The motor car was worth Rs 90,000. (e) Purchased for Rs 30,000 additional office equipment on credit. (f) Paid Rs 75,00 salary to the office manager. (g) Provided services to a client and collected Rs 30,000 (h) Paid Rs 4,000 for the month’s utilities. (i) Paid supplier created in transaction (c). (j) Purchase new office equipment by paying Rs 93,000 cash and trading in old equipment with a recorded cost of Rs 7,000. (k) Completed services of a client for Rs 26,000. This amount is to be paid within 30 days (l) Received Rs 19,000 payment from the client created in transaction (k). (m) Bobby withdrew Rs 20,000 from the business. Analyse the above stated transactions and open the following T-accounts: Cash, client, office supplies, motor car, building, land, long term payables, capital, withdrawals, salary, expense and utilities expense. a) The transaction (a) increases assets by Rs 5,50,000 (cash Rs 4,00,000 and office equipment Rs 1,5,000) it will be debited and on the other hand it will increase the capital by Rs 5,50,000, so it will be credited in capital account. b) Purchase of land and small office building are assets. On one hand, the purchase of these items will increase their individual accounts and this will increase the total amount of the assets in the business; so, both the accounts will be debited. On the other hand, payment in cash on the purchase of these assets will decrease the cash balance, so cash account will be credited to the extent of amount paid. After payment for building in cash, the balance of building account will be transferred to creditors for building account. This will increase the amount of the creditors, which in turn will increase the total liabilities of the business. Long term payables are regarded as loan to the business that will increase both cash balance (due to intake of loan) as well as liabilities of the business. c) Here ‘office supplies’ is an expense. So, according to the golden rule, ‘All expenses are debited’, it will be debited on one hand while on the other hand, office supplies has been purchased on credit, so it will increase the liability, on account of which, supplier’s account will be credited. d) Amount invested (motor car) by the proprietor in the business would increase both the capital and assets. e) Purchase of additional equipment increases the assets; hence, offices equipment account will be debited. Further as the office equipment was purchased on credit, it increases the amount of the creditors for office equipment and the creditors account will be credited. f) Salary is an expense and as all the expenses are debited, so the payment of salary to the manager will be debited to the salary account. And on the other hand the payment of the salary in cash decreases the cash balance (Assets) so the cash account would be credited (as decrease in assets is credited). g) Amount received or receivable from services rendered to the client is revenue for the business. All revenues are to be credited so client service account will be credited. On the other hand, cash received in exchange of services would increase the cash balance. It would be debited to the cash account. h) The ‘utilities’ has been treated as a revenue expense. All expenses are to be debited. Amount paid for utilities would be debited to Utilities account. Utilities have been paid in cash so the cash account will be credited (as this decreases assets). i) Payment to the supplier (creditors) will be debited. It results in the decrease in liabilities. Further as the payment has been made in cash, so it results in decrease in the cash balance (assets) and hence the cash account will be credited. j) Purchase of the equipments will be debited in the Equipment Account (as there is increase in the assets). Also as the equipments of worth Rs 1,00,000 and Rs 93,000 have been purchased for cash and old equipments of worth Rs 7,000 have been exchanged so the purchase of the equipments will be debited in the Office Equipment account and equipment of Rs 7,000 will be credited in the same account. k) Receipt from ‘Client services’ is revenue. All revenues are credited. The client services account will be credited and client is considered as debtors, so the client account will be debited. l) The client has been considered as Debtors. The amount received from the client will lead to the decrease in the debtors balance and the client account will be credited. Receipts from the client will increase the cash balance (asset), and hence the cash account will be debited. m) The amount withdrawn by the proprietor is considered as ‘drawings’. According to the Business Entity Concept, drawings decrease the owner’s capital,) Thus the drawings account will be debited (as decrease in capital is debited). On the other hand as drawings have been made in cash, decrease in cash means cash account will be credited with the amount of drawings. T – Accounts Page No 92: Question 11: Journalise the following transactions in the books of Himanshu: Question 12: Enter the following Transactions in the Journal of Mudit : Page No 93: Question 13: Journalise the following transactions: Question 14: Jouranlise the following transactions in the books of Harpreet Bros.: (a) Rs 1,000 due from Rohit are now bad debts. (b) Goods worth Rs 2,000 were used by the proprietor. (c) Charge depreciation @ 10% p.a for two month on machine costing Rs 30,000. (d) Provide interest on capital of Rs 1,50,000 at 6% p.a. for 9 months. (e) Rahul become insolvent, who owed is Rs 2,000 a final dividend of 60 paise in a rupee is received from his estate. Question 15: Prepare Journal from the transactions given below: The document NCERT Solution (Part - 3) Recording of Transactions-I \| Accountancy Class 11 - Commerce is a part of the Commerce Course Accountancy Class 11. All you need of Commerce at this link: Commerce ## Accountancy Class 11 82 videos\|105 docs\|42 tests ## FAQs on NCERT Solution (Part - 3) Recording of Transactions-I - Accountancy Class 11 - Commerce 1. What is the purpose of recording transactions in accounting? Ans. The purpose of recording transactions in accounting is to ensure that all financial activities of a business are accurately and consistently documented. This helps in maintaining a reliable record of the company's financial activities, facilitating the preparation of financial statements, and ensuring compliance with legal and regulatory requirements. 2. What are the different methods of recording transactions? Ans. There are two main methods of recording transactions in accounting - the single-entry system and the double-entry system. In the single-entry system, only one account is affected by each transaction, while in the double-entry system, at least two accounts are affected, with debits and credits being recorded to maintain the balance. 3. How does the recording of transactions help in financial analysis? Ans. The recording of transactions helps in financial analysis by providing a detailed and organized record of all financial activities. This data can be used to analyze the company's financial performance, identify trends, and make informed decisions. Financial ratios, such as liquidity ratios and profitability ratios, can also be calculated using the recorded transaction data to assess the company's financial health. 4. What are the essential elements of a transaction that need to be recorded? Ans. The essential elements of a transaction that need to be recorded include the date of the transaction, the parties involved, a description of the transaction, the amount involved, and the method of payment. These details ensure that the transaction is properly identified and can be easily traced and analyzed in the future. 5. What are the consequences of not recording transactions accurately? Ans. Not recording transactions accurately can have several consequences for a business. It can lead to incorrect financial statements, which can misrepresent the company's financial position and performance. It can also result in non-compliance with legal and regulatory requirements, leading to penalties and fines. Inaccurate recording of transactions can also hinder financial analysis and decision-making, as the data used for analysis will be unreliable. ## Accountancy Class 11 82 videos\|105 docs\|42 tests ### Up next Explore Courses for Commerce exam ### Top Courses for Commerce Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests. 10M+ students study on EduRev Related Searches , , , , , , , , , , , , , , , , , , , , , ;	2,077	9,526	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.65625	3	CC-MAIN-2024-30	latest	en	0.938228
https://setscholars.net/mastering-complex-calculations-with-the-wolfram-chatgpt-plugin-a-comprehensive-guide/	1,716,993,622,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971059246.67/warc/CC-MAIN-20240529134803-20240529164803-00008.warc.gz	438,315,070	24,363	Mastering Complex Calculations with the Wolfram ChatGPT Plugin: A Comprehensive Guide In the realm of computational mathematics, the Wolfram ChatGPT plugin stands as a beacon of innovation and utility. This tool, which marries the computational prowess of Wolfram Alpha with the conversational capabilities of ChatGPT, has revolutionized the way we approach complex calculations. Whether you’re a student grappling with intricate mathematical concepts, an engineer working on a challenging problem, or a math enthusiast looking to explore the depths of numerical complexity, the Wolfram ChatGPT plugin is a tool you can’t afford to overlook. This comprehensive guide aims to provide an in-depth exploration of this plugin, offering step-by-step instructions, visualizations, and explanations to help you harness its full potential. A Deep Dive into the Wolfram ChatGPT Plugin The Wolfram ChatGPT plugin is a remarkable tool that brings together the computational knowledge engine of Wolfram Alpha and the conversational abilities of ChatGPT. This fusion results in a tool that can solve complex mathematical problems, evaluate intricate expressions, perform advanced calculus operations, convert complex numbers between different forms, and calculate properties of complex numbers, all within a conversational interface. The plugin’s ability to provide step-by-step solutions and visualizations makes it an invaluable resource for anyone seeking to understand and solve complex mathematical problems. Solving Complex Equations with the Wolfram ChatGPT Plugin One of the primary uses of the Wolfram ChatGPT plugin is to solve equations involving complex numbers. Complex numbers, which are numbers that consist of a real part and an imaginary part, often pose a significant challenge to many individuals. However, with the Wolfram ChatGPT plugin, these equations can be solved with relative ease. For instance, consider the equation “3x^2 + 2ix – 1 = 0”. This equation involves both real and imaginary components, making it a complex equation. To solve this equation using the Wolfram ChatGPT plugin, you would simply need to ask the plugin to solve the equation. The plugin will then provide a step-by-step solution, breaking down the process into understandable steps and making it easier to grasp the underlying mathematical principles. Similarly, you can use the plugin to solve equations like “2z + 3i = 5 + 4i”. Again, the plugin will provide a step-by-step solution, allowing you to understand the process and learn from it. Evaluating Mathematical Expressions with the Wolfram ChatGPT Plugin Beyond solving complex equations, the Wolfram ChatGPT plugin can also evaluate complex mathematical expressions. This feature is particularly useful when dealing with expressions that involve both real and imaginary components. Consider the expression “(3 + 2i) * (1 – i)”. This expression involves the multiplication of two complex numbers. To evaluate this expression using the Wolfram ChatGPT plugin, you would simply need to ask the plugin to evaluate the expression. The plugin will then simplify the expression and provide the result, saving you the time and effort of manually calculating the result. Similarly, you can use the plugin to evaluate expressions like “(2 + 3i)^2”. Again, the plugin will simplify the expression and provide the result, allowing you to focus on understanding the principles behind the calculation rather than getting bogged down in the calculation itself. Performing Calculus Operations with the Wolfram ChatGPT Plugin Calculus, with its derivatives, integrals, and limits, often poses a significant challenge to many individuals. However, the Wolfram ChatGPT plugin can simplify these operations, making calculus more accessible and understandable. For instance, consider the function “f(z) = z^2 + 2iz”. To compute the derivative of this function using the Wolfram ChatGPT plugin, you would simply need to ask the plugin to compute the derivative. The plugin will then provide a step-by-step solution, breaking down the process into manageable steps and making it easier to understand. Similarly, you can use the plugin to find the integral of a function like “f(z) = e^(iz) over the range 0 to 2π”. Again, the plugin will provide a step-by-step solution, allowing you to understand the process and learn from it. Conversion Between Forms with the Wolfram ChatGPT Plugin Complex numbers can be represented in different forms, including rectangular form (a + bi) and polar form (r(cos θ + i sin θ)). Converting between these forms can often be a tedious process. However, the Wolfram ChatGPT plugin can simplify this process, allowing you to convert complex numbers between different forms with ease. For instance, consider the complex number “3 + 4i”. This number is in rectangular form. To convert this number to polar form using the Wolfram ChatGPT plugin, you would simply need to ask the plugin to perform the conversion. The plugin will then provide the result, saving you the time and effort of manually performing the conversion. Similarly, you can use the plugin to find the rectangular form of a complex number in polar form, such as “5e^(iπ/4)”. Again, the plugin will provide the result, allowing you to focus on understanding the principles behind the conversion rather than getting bogged down in the conversion itself. Calculating Complex Number Properties with the Wolfram ChatGPT Plugin Complex numbers have various properties, including magnitude (or modulus), phase (or argument), real part, and imaginary part. Calculating these properties can often be a complex process. However, the Wolfram ChatGPT plugin can simplify this process, allowing you to calculate these properties with ease. For instance, consider the complex number “3 + 4i”. To find the magnitude and phase of this number using the Wolfram ChatGPT plugin, you would simply need to ask the plugin to calculate these properties. The plugin will then provide the results, saving you the time and effort of manually calculating these properties. Similarly, you can use the plugin to find the real and imaginary parts of a complex number, such as “2e^(iπ/2)”. Again, the plugin will provide the results, allowing you to focus on understanding the principles behind these properties rather than getting bogged down in the calculations themselves. Relevant Prompts for the Wolfram ChatGPT Plugin To help you get started with the Wolfram ChatGPT plugin, here are some prompts that you can use: 1. “Can you use Wolfram to solve this equation for me? 3x^2 + 2ix – 1 = 0.” 2. “What’s the solution to 2z + 3i = 5 + 4i using Wolfram?” 3. “Can you help me evaluate this expression using Wolfram: (3 + 2i) * (1 – i)?” 4. “What is the value of the expression (2 + 3i)^2 using Wolfram?” 5. “Can you help me compute the derivative of f(z) = z^2 + 2iz using Wolfram?” 6. “What’s the integral of f(z) = e^(iz) over the range 0 to 2π using Wolfram?” 7. “Can you convert 3 + 4i to polar form using Wolfram?” 8. “What’s the rectangular form of the complex number 5e^(iπ/4) using Wolfram?” 9. “What’s the magnitude and phase of the complex number 3 + 4i using Wolfram?” 10. “Can you find the real and imaginary parts of the number 2e^(iπ/2) using Wolfram?” The Impact of the Wolfram ChatGPT Plugin The Wolfram ChatGPT plugin has had a profound impact on the way we approach complex calculations. By providing step-by-step solutions, visualizations, and explanations, it has made complex mathematical concepts more accessible and understandable. This has had a significant impact on education, particularly in the fields of mathematics and engineering, where complex calculations are commonplace. For students, the plugin has been a game-changer. It has provided them with a tool that can assist them in their studies, helping them to understand complex mathematical concepts and solve intricate problems. This has not only improved their understanding of these concepts but has also boosted their confidence, encouraging them to tackle more challenging problems. For engineers, the plugin has been an invaluable resource. It has provided them with a tool that can assist them in their work, helping them to solve complex equations, evaluate intricate expressions, and perform advanced calculus operations. This has not only improved their efficiency and productivity but has also enhanced the quality of their work. For math enthusiasts, the plugin has been a source of endless fascination. It has provided them with a tool that can assist them in their exploration of complex mathematical concepts, allowing them to delve deeper into the world of numbers and equations. This has not only enriched their understanding of these concepts but has also provided them with countless hours of intellectual stimulation. Conclusion The Wolfram ChatGPT plugin is a powerful tool that has revolutionized the way we approach complex calculations. By providing step-by-step solutions, visualizations, and explanations, it has made complex mathematical concepts more accessible and understandable. Whether you’re a student, an engineer, or a math enthusiast, the Wolfram ChatGPT plugin is a tool you can’t afford to overlook. So, start using the Wolfram ChatGPT plugin today. Explore its features, experiment with its capabilities, and experience the power of computational mathematics at your fingertips. With the Wolfram ChatGPT plugin, complex calculations are no longer a source of frustration, but a gateway to understanding and discovery. Personal Career & Learning Guide for Data Analyst, Data Engineer and Data Scientist Applied Machine Learning & Data Science Projects and Coding Recipes for Beginners A list of FREE programming examples together with eTutorials & eBooks @ SETScholars Projects and Coding Recipes, eTutorials and eBooks: The best All-in-One resources for Data Analyst, Data Scientist, Machine Learning Engineer and Software Developer Topics included:Classification, Clustering, Regression, Forecasting, Algorithms, Data Structures, Data Analytics & Data Science, Deep Learning, Machine Learning, Programming Languages and Software Tools & Packages. (Discount is valid for limited time only) Find more … … Kotlin example for Beginners – Kotlin Program to Add Two Complex Numbers C Example for Beginners: C Program to Add Two Complex Numbers by Passing Structure to a Function Tableau for Data Analyst – Tableau Table Calculations	2,161	10,482	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3	3	CC-MAIN-2024-22	latest	en	0.880456
https://pastecode.io/s/mvrtvkfb	1,725,811,328,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00395.warc.gz	441,223,085	57,435	# Untitled unknown plain_text 2 years ago 2.7 kB 3 Indexable Never import java.util.Scanner; public class Main { final static int maxNumber=50; static Scanner scanner = new Scanner(System.in); static int contGuess = 0; static int fewestGuess; static String option = "Y"; static int numberPlayGames = 0; static int totalGuesses = 0; public static void initGame(){ System.out.println("This program allows you to play a guessing game.\n"+ "It will think of a number between 1 and 100\n "+ "and will allow you to guess until you get the number.\n" + "For each guess, it will tell you whether you have\n"+ "guessed too high or too low\n"); } public static void statistics(){ System.out.println("Game Statistics"); System.out.println("Games: "+numberPlayGames); System.out.println("Guesses: "+totalGuesses); System.out.println("Guesses/Games: "+(float) totalGuesses/numberPlayGames); System.out.println("Fewest guesses: "+fewestGuess); } public static void playGame(){ contGuess = 0; numberPlayGames += 1; int numberToGuess = (int) (Math.random()*maxNumber+1); System.out.println("I'm thinking of a number between 1 and 100..."); do { int myNumber = scanner.nextInt(); contGuess += 1; totalGuesses += 1; if ( numberToGuess == myNumber){ if (contGuess == 1) System.out.println("You got it right on the first guess!"); else System.out.printf("You got it right in %d guesses\n",contGuess); if ( numberPlayGames == 1){ fewestGuess = contGuess; }else { if (contGuess < fewestGuess) fewestGuess = contGuess; } System.out.print("Do you want to play again?"); option = scanner.next(); if (option.toLowerCase().charAt(0) == 'y'){ playGame(); }else{ statistics(); } }else { if (myNumber>numberToGuess){ System.out.println("Too high"); }else{ System.out.println("Too low"); } } }while(option.toLowerCase().charAt(0) == 'y'); } public static void main(String args[]) { initGame(); playGame(); } }	494	1,899	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.796875	3	CC-MAIN-2024-38	latest	en	0.600104
https://www.physicsforums.com/threads/proving-in-linear-algebra.323507/	1,508,228,987,000,000,000	text/html	crawl-data/CC-MAIN-2017-43/segments/1508187820930.11/warc/CC-MAIN-20171017072323-20171017092323-00725.warc.gz	985,196,191	16,348	# Proving in Linear Algebra 1. Jul 5, 2009 ### putongren 1. The problem statement, all variables and given/known data I am studying linear algebra independently using a pdf book online. The question I have difficulty is this: Prove that, where a, b... e are real numbers and a does not equal 0, if (1) ax + by = c has the same solution set as (2) ax + dy = e then they are the same equation. 2. Relevant equations 3. The attempt at a solution Since they are a couple of linear equations, I performed a Gaussian operation on it (1) ax + by = c (2) ax + dy = e (1) - (2) y = c-e/b-d , x = c/a - (b/a)(c-e/b-d) So basically I subtracted (1) from the (2) to get the intercept point between the equations. That should be the solution set. Since I'm trying to prove that (1) and (2) are the same, then I should let b=d and c=e. However, if b = d and c = e, then y would be 0/0. So I'm not sure what I did wrong. Last edited: Jul 5, 2009 2. Jul 5, 2009 ### jgens Alright, if the two equations have the same solution set, that means that an ordered pair (x1,y1) must be a solution to each equation. Clearly (x2,y2) is a solution to each equation as well. Using this principle, subtract the two equations. [a(x1) + b(y1)] - [a(x1) + d(y1)] = c - e c - e = (b - d)y1 Use a similar analysis for the pair (x2,y2) to finish the proof. I'd imagine there is a simple way to do this but I can't think of anything. Edit: Just thought of a much simpler way to do it but this should work. Last edited: Jul 5, 2009 3. Jul 5, 2009 ### putongren I've rethought the question (graphically this time on a x-y axis) and realize that the question is asking that if (1) and (2) intersect, they must be the same line. 4. Jul 5, 2009 ### jgens Not quite. If they only intersect at one point, then they are not the same line. It's saying that if the lines intersect at infinitely many points (they have the same solution set) then they are the same line. Here, a simpler way to do it is to consider the solution (x1,0). Then we have, a(x1) = c and a(x1) = e. Can you take it from there? 5. Jul 7, 2009 ### putongren Actually, I have the answer key to the online textbook. But I just don't like the answer from the book. The answer to the book is the following: Let the first solution set to be x=(c-by)/a, let y=0 gives the solution set (c/a,0) Substitute that solution set to the second equations and you will get a(c/a) + d*0 = e so c = e. Let y =1 gives the solution ((c-b)/a,1) and substitute the solution to the second equation and get d = b. The problem I have with this method is that you have to pick 2 arbitrary points and then solve for the answer. How do you know that the 2 equations intersect at least two points? You have to make the assumption that the 2 equations does intersect at y= 1 and y = 2. I guess this is a reasonable assumption since we know that they are linear equations and that a does not equal zero, so the both equations span the entire number line. I feel that there should be a more general approach to this problem. The most important thing for me is that to identify the my flaw in understanding the problem. If ax + by = c and ax + dy = e both share the same solution set, then it means that the solution set is located at where the equations intersect (whether they intersect once or more times). The solution set they both share is ((c-e)/(b-d)), (b/a)((c-e)/(b-d)).This is the general expression for where the 2 equations intersect. The question tells us to prove that c = e, d = b. After that I have no idea what to do. 6. Jul 8, 2009 ### jgens The book's solution is almost identical to the second solution I suggested. Since the equations have the same solution set - they intersect at infinitely many points - if (x1,0) is a solution to the first equation, it is also a solution to the second equation!	1,052	3,845	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.21875	4	CC-MAIN-2017-43	longest	en	0.944091
https://www.convertunits.com/from/kilotonne/to/firkin	1,670,166,434,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446710974.36/warc/CC-MAIN-20221204140455-20221204170455-00627.warc.gz	759,235,930	24,079	## ››Convert kilotonne to firkin [butter, soap] kilotonne firkin How many kilotonne in 1 firkin? The answer is 2.540117272E-5. We assume you are converting between kilotonne and firkin [butter, soap]. You can view more details on each measurement unit: kilotonne or firkin The SI base unit for mass is the kilogram. 1 kilogram is equal to 1.0E-6 kilotonne, or 0.039368261104442 firkin. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between kilotonnes and firkin [butter, soap]. Type in your own numbers in the form to convert the units! ## ››Quick conversion chart of kilotonne to firkin 1 kilotonne to firkin = 39368.2611 firkin 2 kilotonne to firkin = 78736.52221 firkin 3 kilotonne to firkin = 118104.78331 firkin 4 kilotonne to firkin = 157473.04442 firkin 5 kilotonne to firkin = 196841.30552 firkin 6 kilotonne to firkin = 236209.56663 firkin 7 kilotonne to firkin = 275577.82773 firkin 8 kilotonne to firkin = 314946.08884 firkin 9 kilotonne to firkin = 354314.34994 firkin 10 kilotonne to firkin = 393682.61104 firkin ## ››Want other units? You can do the reverse unit conversion from firkin to kilotonne, or enter any two units below: ## Enter two units to convert From: To: ## ››Definition: Kilotonne The SI prefix "kilo" represents a factor of 103, or in exponential notation, 1E3. So 1 kilotonne = 103 tonnes. The definition of a tonne is as follows: A tonne (also called metric ton) is a non-SI unit of mass, accepted for use with SI, defined as: 1 tonne = 1000 kg (= 106 g). ## ››Metric conversions and more ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!	603	2,055	{"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}	2.9375	3	CC-MAIN-2022-49	latest	en	0.76674
https://support.khanacademy.org/hc/bg/profiles/4079749517-Forgotten	1,719,209,497,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198865074.62/warc/CC-MAIN-20240624052615-20240624082615-00879.warc.gz	490,728,391	10,821	Как да помогнем? # Forgotten • Forgotten коментира, • Forgotten коментира, List of KA built-in fonts: Lato Source Serif Pro FontAwesome FreeMono PjsEmptyFont Source: I used my browser font tool to see what font came from where. What fonts my browser used: Name (alternat... • Forgotten създаде постинг, ### P3D performace improvement. On my computer, the normal P3D runs with regular frame rate drops down to 8.I worked on a solution to fix this and my best one works much better. The fix was binding an array to a filp function, in... • Forgotten коментира, Algebraically you can do a simple proof as follows. Let h be the height of a triangle ABC perpendicular to the side AB. Let a, b, c be BC, AC, AB respectively. It follows that sin A = h/b and sin B... • Forgotten коментира, SinA / a * b = b * SinA / a = b / a * SinA = b/a SinA They omitted the multiplication operator here. If you want to understand the formula. Do a geometric proof of it. • Forgotten коментира, I guess these are the built in ones, if you can call them that: sans-serif, serif, monospace, fantasy, cursive. source: exec-pjs.js println(PFont);/*function PFont(e,t){ if(e===qr){e=""} this... • Общо активност 13 • Последна дейност • Член от • Следи 0 потребители • Следен от 0 потребители • Гласове 7 • Абонаменти 0 • Общо активност 13 • Последна дейност • Член от • Следи 0 потребители • Следен от 0 потребители • Гласове 7 • Абонаменти 0	418	1,436	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.640625	3	CC-MAIN-2024-26	latest	en	0.43931
http://www.jiskha.com/display.cgi?id=1294871854	1,498,265,218,000,000,000	text/html	crawl-data/CC-MAIN-2017-26/segments/1498128320206.44/warc/CC-MAIN-20170623235306-20170624015306-00681.warc.gz	569,775,758	3,875	# Math posted by . The domain of t(x) = -3.8x - 4.2 is {-3, -1.4, 0, 8}. What is the range? • Math - I'll do one -3.8x - 4.2 for x = -3 y = -3.8(-3) - 4.2 y = 11.4 - 4.2 y = 7.2 so y = 7.2 is incl. in the range repeat for each x value • Math - idk	127	254	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.53125	4	CC-MAIN-2017-26	latest	en	0.803025
http://lingsprimaryblogs.net/2020/05/20/year-4-maths-task-wednesday-20th-may/	1,721,934,698,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763861452.88/warc/CC-MAIN-20240725175545-20240725205545-00412.warc.gz	16,371,766	9,195	# Year 4 Maths task- Wednesday 20th May Good morning! Today you will need a pen and paper. Email in the photo of your work with the correct fraction written. ### 9 thoughts on “Year 4 Maths task- Wednesday 20th May” • 20th May 2020 at 9:20 am Permalink i still cant do this i dont have any paper and my brother is using my pen for the whole day he is greedy with pens Reply • 20th May 2020 at 4:58 pm Permalink Kristers I have said I would pop round and deliver paper and pens! Mrs Denny Reply • 20th May 2020 at 10:01 am Permalink I’m going to do this maths on some paper and when my mum gets back she is going to take a photo of my work and send it in for you to see it. Reply • 20th May 2020 at 12:39 pm Permalink If 10 is the whole, then 2 is part of the whole. The whole has been divided into 2 equal parts. The whole has been divided into 3 equal parts. One of the part has been shaded. The whole has been divided into 5 equal parts. One of the part has been shaded. The whole has been divided into 6 equal parts. One of the part has been shaded. The whole has been divided into 2 equal parts. One of the part has been shaded. The whole has been divided into 4 equal parts. One of the part has been shaded. The whole has been divided into 3 equal parts. 1/3. The whole has been divided into 5 equal parts 1/5. 1 of the parts has been shaded. The whole has been divided into 4 equal parts 1/4. 1 of the part has been shaded. The whole has been divided into 2 equal parts 1/2. 1 of the part has been shaded. The whole has been divided into 3 equal parts 1/3. 1 of the part has been shaded. 1/2,1/1,1/5 and 1/11 Reply • 20th May 2020 at 4:56 pm Permalink Great understanding Regina Reply • 20th May 2020 at 6:22 pm Permalink If 10 is the whole then 2 is part of the whole. The whole has been divided into 2 equal parts. The whole has been divided into 3 equal parts. One of the parts have been shaded. The whole has been divided into 5 equal parts . Reply • 21st May 2020 at 1:26 pm Permalink Task: 1/3, 1/2, 1/6, 1/12 Reply • 21st May 2020 at 2:53 pm Permalink Thank you Damian, Hope you are enjoying the sunshine. Reply	642	2,144	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5625	4	CC-MAIN-2024-30	latest	en	0.979753
http://www.xwordinfo.com/Crossword?date=8/2/2009&g=17&d=D	1,472,569,280,000,000,000	text/html	crawl-data/CC-MAIN-2016-36/segments/1471982984973.91/warc/CC-MAIN-20160823200944-00018-ip-10-153-172-175.ec2.internal.warc.gz	786,669,805	11,704	# GROUP FORMATION ### New York Times, Sunday, August 2, 2009 Author: Patrick Berry Editor: Will Shortz TotalDebutLatestCollabs 2147/11/19998/13/20162 SunMonTueWedThuFriSatVariety 711241669492 ScrabbleRebusCirclePangram 1.54980 ## This puzzle: Rows: 21, Columns: 21, Words: 140, Blocks: 74 Missing: {QXZ} This is puzzle # 122 for Mr. Berry. NYT links: Across Lite JimH notes: I have shown the Greek letters in the grid but to read the crossing answers, you must spell those letters out. So, for example, 39 Down is β θ π but the crossing answers are TI[BETA]NS, ON[THETA]KE, and AMERICAN[PI]E. 1M 2O 3U 4T 5H 6S 7T 8R 9I 10P 11E 12A 13L 14G 15A 16A 17M 18Y 19S U S H I 20H A I K U S 21S O R T 22V I A 23G R E E K 24L E T T E R S 25K N E E 26S O C K 27F R E E B E E 28P E 29R S I A 30O N E S 31E 32L O I S E 33I O N I A 34TAU 35DELTA PHI 36M I R V 37T 38I 39BETA N S 40P R 41I N C E 42D 43O 44M 45M A C E 46D 47O N THETA K E 48P O R T O 49R U E 50A M E R I 51C A N PI E 52B L U E 53P I T T 54V A S E 55D 56R A I N S 57C E S A R 58R 59E 60G I S T R 61Y 62I C E D 63C O S T C O 64M O M O N E Y 65O 66I N K S 67T O M C A T S 68A M I N E S 69B U N S 70T 71R A D E I N S 72L U N G S 73W A R D E 74D 75O L E O 76A L E S 77H I H O 78R 79ALPHA B E R N 80A 81T 82H 83Y 84W A N 85P 86A O L O 87D U CHI E S 88S P R E E 89I N T 90H E M A I N 91A I RHO S E 92P E E N 93OMEGA PSI PHI 94W O 95R L D 96J 97U L E P S 98F 99O 100A M 101B 102E I R U T 103E 104M 105P A N E L 106A M N E 107S I A C 108F R 109A T E R N I T I 110E 111S 112C A D 113M A S K 114F E D O R A 115T O N T O 116T R Y 117U N E S 118S Y D N E Y 119S N E A D © 2009, The New York TimesNo. 21,817	822	1,651	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-36	longest	en	0.620232
http://www.ehow.co.uk/how_5057033_calculate-volume-water-tank.html	1,527,012,838,000,000,000	text/html	crawl-data/CC-MAIN-2018-22/segments/1526794864837.40/warc/CC-MAIN-20180522170703-20180522190703-00602.warc.gz	371,893,557	10,532	DISCOVER # How to calculate the volume of a water tank Updated April 17, 2017 Determining the volume of a cylindrical water tank can be a challenge. This requires you to use the mathematical figure known as pi in computing the volume. Completing the calculation in several steps will help limit your risk of making a mistake in computing the volume of a water tank. The math to compute the volume of a square or rectangular water tank is simpler. Measure the height of your water tank. This should be from the bottom of the tank to the top of the tank, regardless of the amount of water in the tank. For the purpose of this example the height is 20 feet. Measure the diameter of your water tank. The diameter of the tank is a straight line that goes from one side of the circle forming the base of the tank to the other side, passing through the exact centre of the circular base. Once you have calculated the diameter of your water tank, divide it by 2 to get the radius. The radius is the measurement from the centre of the circle that forms the base of the tank in a straight line to any edge of the tank. In this example the diameter is 10 feet and the radius is 5 feet. Calculate the volume of the water tank using the following formula for the volume of a cylinder: Volume = pi times the radius squared (the radius times itself) times the height of the tank, or volume = pi x r squared x height. Calculate the total volume in our example using the formula in Step 3. For this example multiply pi (which is 3.14159) x 5 x 5 x 20. This equals about 1,570 cubic feet, which is the volume of the water tank. Use a calculator that has the pi symbol on it to get a more accurate solution. The pi symbol is the Greek letter "p." Measure the length, width and height of the square or rectangular water tank. In this example the length will be 20m, the width 10m and the height 40m. Calculate the volume of the rectangular tank using this formula: Volume = length x width x height. Compute the total volume in our example using the formula in Step 2. For this example, multiply 20 x 10 x 40. This would provide a measurement of 8,000 sq. m as the volume of the water tank. #### Things You'll Need • Measuring tape • Calculator with pi symbol on it	513	2,256	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.59375	5	CC-MAIN-2018-22	latest	en	0.889997
https://books.google.no/books?id=5CkEAAAAQAAJ&q=square&dq=editions:UOM39015067252117&lr=&hl=no&output=html&source=gbs_word_cloud_r&cad=5	1,632,501,267,000,000,000	text/html	crawl-data/CC-MAIN-2021-39/segments/1631780057558.23/warc/CC-MAIN-20210924140738-20210924170738-00245.warc.gz	194,627,479	10,065	# The Elements of Euclid, the parts read in the University of Cambridge [book 1-6 and parts of book 11,12] with geometrical problems, by J.W. Colenso ### Hva folk mener -Skriv en omtale Vi har ikke funnet noen omtaler på noen av de vanlige stedene. ### Populære avsnitt Side 42 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Side 4 - Let it be granted that a straight line may be drawn from any one point to any other point. Side 33 - F, which is the common vertex of the triangles: that is », together with four right angles. Therefore all the angles of the figure, together with four right angles are equal to twice as many right angles as the figure has sides. Side 62 - If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line. Side 22 - If from the ends of a side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle. Side 58 - If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line. Side 146 - ... may be demonstrated from what has been said of the pentagon : and likewise a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by a method like to that used for the pentagon. Side 194 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar. Side 2 - A circle is a plane figure contained by one line, which is called the circumference, and is such, that all straight lines drawn from a certain point within the figure to the circumference are equal to one another : 16. Side 69 - To divide a given straight line into two parts, so that the rectangle contained by the whole, and one of the parts, may be equal to the square of the other part.	533	2,294	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.21875	4	CC-MAIN-2021-39	latest	en	0.911593
https://examobjective.in/strength-of-material-mcq-70/	1,670,009,178,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00174.warc.gz	271,524,746	31,661	SECTION-5 #### Strength of Material MCQ Q51. Two materials are having modulus of elasticities, modulus of rigidities and bulk modulus as (E1, E2), (C1, CV), and (K1, K2). The modular ratio is given by $\text { (a) } \frac{E_{1}}{C_{2}}$ $\text { (b) } \frac{E_{1}}{K_{2}}$ $\text { (c) } \frac{K_{1}}{E_{2}}$ $\text { (d) } \frac{C_{1}}{K_{2}}$ Ans:$\text { (c) } \frac{K_{1}}{E_{2}}$ ##### Q52. For uniform strength, a bar which is fixed at the upper end and is subjected to an external load P at the lower end, the area (A) at any section at a distance x from the lower end is given by $(a) A_{2} e^{\frac{w}{p x}}$ $\text { (b) } A_{2} e^{\frac{p}{w x}}$ $\text { (c) } A_{2} e^{\frac{w x}{p}}$ $\text { (d) } A_{2} \times \frac{p}{w x}$ where A2 = Area at the lower end, w = Weight per unit volume, p = Uniform stress intensity in the bar. Ans:$\text { (c) } A_{2} e^{\frac{w x}{p}}$ ##### Q53. A steel tyre of internal diameter d is heated so that it can be shrunk on a wheel of diameter D. If now the steel tyre be cooled, it will grip the wheel. The tensile stress induced circumferentially is called (a) shear stress (b) hoop stress (c) longitudinal stress (d) ultimate stress. Ans:(a) shear stress ##### Q54. The diameter of a cast iron round bar, on which tensile test is performed, at fracture will (a) increase (b) decrease (c) be approximately same (d) none of the above. Ans:(c) be approximately same ##### Q55. The diameter of a mild steel round bar, on which tensile test is performed, at fracture will (a) increase (b) decrease (c) be same (d) none of the above. Ans:(b) decrease (a) M/R= E/ I (b)M/I= E/ R (c) M/I= R/ E (d) M/E= RI. Ans:(b)M/I= E/ R ##### Q57. The expression EI $\frac{d^{2} y}{d x^{2}}$at a section of a member represents (a) shearing force (c) bending moment (d) slope. Ans:(c) bending moment ##### Q58. The expression EI$\frac{d^{3} y}{d x^{3}}$ at a section of a member represents (a) shearing force (c) bending moment (d) slope. Ans:(a) shearing force ##### Q59. The expression EI $\frac{d^{4} y}{d x^{4}}$at a section of a member represents (a) shearing force (c) bending moment (d) slope. Ans:(a) shearing force ##### Q60. A fixed beam of span (l) carries a uniformly distributed load of w per unit length over the whole span. The deflection at the centre is (a) equal of the central deflection of a simply supported beam (b) half of the central deflection for a simply supported beam (c) one-fourth of the central deflection for a simply supported beam (d) one-fifth of the central deflection of the simply supported beam. Ans:(d) one-fifth of the central deflection of the simply supported beam. 1 2 3 4 6 8 DEAR READERS IF YOU FIND ANY ERROR/TYPING MISTAKE PLEASE LET ME KNOW IN THE COMMENT SECTIONS OR E-MAIL: [email protected] ##### Read More Sections of Strength of Material Each section contains maximum 80 Questions. To practice more questions visit other sections. Strength of Material MCQ – Section-1 Strength of Material MCQ – Section-2 Strength of Material MCQ – Section-3 Strength of Material MCQ – Section-4 Strength of Material MCQ – Section-5 Strength of Material MCQ – Section-6	986	3,219	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.0625	3	CC-MAIN-2022-49	latest	en	0.725795
https://quickhomeworksolutions.com/product/fin-534-fin534-fin-534-week-11-final-exam-part-2/	1,566,347,255,000,000,000	text/html	crawl-data/CC-MAIN-2019-35/segments/1566027315695.36/warc/CC-MAIN-20190821001802-20190821023802-00487.warc.gz	609,534,895	12,785	+1835 731 5494 Email: instantessays65@gmail.com # FIN/534 FIN534 FIN 534 Week 11 Final Exam Part 2 \$12.99 Week 11 Final Exam Part 2 â€¢ Question 1 A U.S.-based importer, Zarb Inc., makes a purchase of crystal glassware from a firm in Switzerland for 39,960 Swiss francs, or \$24,000, at the spot rate of 1.665 francs per dollar. The terms of the purchase are net 90 days, and the U.S. firm wants to cover this trade payable with a forward market hedge to eliminate its exchange rate risk. Suppose the firm completes a forward hedge at the 90-day forward rate of 1.682 francs. If the spot rate in 90 days is actually 1.638 francs, how much will the U.S. firm have saved or lost in U.S. dollars by hedging its exchange rate exposure? â€¢ Question 2 Suppose a foreign investor who holds tax-exempt Eurobonds paying 9% is considering investing in an equivalent-risk domestic bond in a country with a 28% withholding tax on interest paid to foreigners. If 9% after-tax is the investor’s required return, what before-tax rate would the domestic bond need to pay to provide the required after-tax return? â€¢ Question 3 Suppose the exchange rate between U.S. dollars and Swiss francs is SF 1.41 = \$1.00, and the exchange rate between the U.S. dollar and the euro is \$1.00 = 1.64 euros. What is the cross-rate of Swiss francs to euros? â€¢ Question 4 Suppose Yates Inc., a U.S. exporter, sold a consignment of antique American muscle-cars to a Japanese customer at a price of 143.5 million yen, when the exchange rate was 140 yen per dollar. In order to close the sale, Yates agreed to make the bill payable in yen, thus agreeing to take some exchange rate risk for the transaction. The terms were net 6 months. If the yen fell against the dollar such that one dollar would buy 154.4 yen when the invoice was paid, what dollar amount would Yates actually receive after it exchanged yen for U.S. dollars? â€¢ Question 5 If the spot rate of the Israeli shekel is 5.51 shekels per dollar and the 180-day forward rate is 5.97 shekels per dollar, then the forward rate for the Israeli shekel is selling at a ____ to the spot rate. â€¢ Question 6 If 1.64 Canadian dollars can purchase one U.S. dollar, how many U.S. dollars can you purchase for one Canadian dollar? â€¢ Question 7 Suppose a carton of hockey pucks sell in Canada for 105 Canadian dollars, and 1 Canadian dollar equals 0.71 U.S. dollars. If purchasing power parity (PPP) holds, what is the price of hockey pucks in the United States? â€¢ Question 8 If the inflation rate in the United States is greater than the inflation rate in Britain, other things held constant, the British pound will â€¢ Question 9 Which of the following statements is correct? â€¢ Question 10 Consider two very different firms, M and N. Firm M is a mature firm in a mature industry. Its annual net income and net cash flows are both consistently high and stable. However, M’s growth prospects are quite limited, so its capital budget is small relative to its net income. Firm N is a relatively new firm in a new and growing industry. Its markets and products have not stabilized, so its annual operating income fluctuates considerably. However, N has substantial growth opportunities, and its capital budget is expected to be large relative to its net income for the foreseeable future. Which of the following statements is correct? â€¢ Question 11 Rohter Galeano Inc. is considering how to set its dividend policy. It has a capital budget of \$3,000,000. The company wants to maintain a target capital structure that is 15% debt and 85% equity. The company forecasts that its net income this year will be \$3,500,000. If the company follows a residual dividend policy, what will be its total dividend payment? â€¢ Question 12 Which of the following statements is correct? â€¢ Question 13 The capital budget of Creative Ventures Inc. is \$1,000,000. The company wa ## Reviews There are no reviews yet.	956	3,934	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35	latest	en	0.937134
http://www.vb-helper.com/howto_intersect_lines.html	1,685,279,382,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224643784.62/warc/CC-MAIN-20230528114832-20230528144832-00707.warc.gz	95,805,579	4,199	Home Search What's New Index Books Links Q & A Newsletter Banners Feedback Tip Jar MSDN Visual Basic Community Title See where two lines intersect line, intersect Graphics, Algorithms Treat the lines as parametric where line 1 is: ``` X = x11 + dx1 * t1 Y = y11 + dy1 * t1``` and line 2 is: ``` X = x21 + dx2 * t2 Y = y21 + dy2 * t2``` Setting these equal gives: ``` x11 + dx1 * t1 = x21 + dx2 * t2 y11 + dy1 * t1 = y21 + dy2 * t2``` Rearranging: ``` x11 - x21 + dx1 * t1 = dx2 * t2 y11 - y21 + dy1 * t1 = dy2 * t2``` ``` (x11 - x21 + dx1 * t1) * dy2 = dx2 * t2 * dy2 (y11 - y21 + dy1 * t1) * (-dx2) = dy2 * t2 * (-dx2)``` ``` (x11 - x21) * dy2 + ( dx1 * dy2) * t1 + (y21 - y11) * dx2 + (-dy1 * dx2) * t1 = 0``` Solving for t1 gives: ``` t1 * (dy1 * dx2 - dx1 * dy2) = (x11 - x21) * dy2 + (y21 - y11) * dx2``` ``` t1 = ((x11 - x21) * dy2 + (y21 - y11) * dx2) / (dy1 * dx2 - dx1 * dy2)``` Similarly solve for t2. Notes: • If 0 ≤ t1 ≤ 1, then the point lies on segment 1. • If 0 ≤ t2 ≤ 1, then the point lies on segment 1. • If dy1 * dx2 - dx1 * dy2 = 0 then the lines are parallel. • If the point of intersection is not on both segments, then this is almost certainly not the point where the two segments are closest. ```' Find the point where two segments intersect. Public Sub FindLineIntersection( _ ByVal x11 As Single, ByVal y11 As Single, _ ByVal x12 As Single, ByVal y12 As Single, _ ByVal x21 As Single, ByVal y21 As Single, _ ByVal x22 As Single, ByVal y22 As Single, _ ByRef inter_x As Single, ByRef inter_y As Single, _ ByRef inter_x1 As Single, ByRef inter_y1 As Single, _ ByRef inter_x2 As Single, ByRef inter_y2 As Single) Dim dx1 As Single Dim dy1 As Single Dim dx2 As Single Dim dy2 As Single Dim t1 As Single Dim t2 As Single Dim denominator As Single ' Get the segments' parameters. dx1 = x12 - x11 dy1 = y12 - y11 dx2 = x22 - x21 dy2 = y22 - y21 ' Solve for t1 and t2. On Error Resume Next denominator = (dy1 * dx2 - dx1 * dy2) t1 = ((x11 - x21) * dy2 + (y21 - y11) * dx2) / _ denominator If Err.Number <> 0 Then ' The lines are parallel. inter_x = 1E+38: inter_y = 1E+38 inter_x1 = 1E+38: inter_y1 = 1E+38 inter_x2 = 1E+38: inter_y2 = 1E+38 Exit Sub End If On Error GoTo 0 t2 = ((x21 - x11) * dy1 + (y11 - y21) * dx1) / _ -denominator ' Find the point of intersection. inter_x = x11 + dx1 * t1 inter_y = y11 + dy1 * t1 ' Find the closest points on the segments. If t1 < 0 Then t1 = 0 ElseIf t1 > 1 Then t1 = 1 End If If t2 < 0 Then t2 = 0 ElseIf t2 > 1 Then t2 = 1 End If inter_x1 = x11 + dx1 * t1 inter_y1 = y11 + dy1 * t1 inter_x2 = x21 + dx2 * t2 inter_y2 = y21 + dy2 * t2 End Sub```	1,035	2,685	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.953125	4	CC-MAIN-2023-23	latest	en	0.634038
http://www.dreamincode.net/forums/topic/319749-collision-detection-between-two-objects/	1,524,498,260,000,000,000	text/html	crawl-data/CC-MAIN-2018-17/segments/1524125946077.4/warc/CC-MAIN-20180423144933-20180423164933-00134.warc.gz	404,655,575	18,227	# Collision Detection between two objects Page 1 of 1 ## 1 Replies - 1122 Views - Last Post: 28 April 2013 - 04:40 AMRate Topic: //<![CDATA[ rating = new ipb.rating( 'topic_rate_', { url: 'http://www.dreamincode.net/forums/index.php?app=forums&module=ajax&section=topics&do=rateTopic&t=319749&s=4c227890ddc8d1f4677707c472d993f9&md5check=' + ipb.vars['secure_hash'], cur_rating: 0, rated: 0, allow_rate: 0, multi_rate: 1, show_rate_text: true } ); //]]> ### #1 qobacha Reputation: 0 • Posts: 9 • Joined: 27-April 12 # Collision Detection between two objects Posted 28 April 2013 - 04:17 AM Can someone help me on how i can detect collision between two objects. At first i thought IntersectsWith would help but i get the error saying my solution doesnot contain the definition for IntersectsWith. Here is my code that gives me errors: ```class Car { private int _x,_y, _width, _height,_xvel; public Car(Random r, int y, int width, int height, int xvel) { this._x = r.Next(500)+20; //this._x = x; this._y = y; this._width = width; this._height = height; this._xvel = xvel; } public int X { get { return _x; } } public int Y { get { return _y; } } public int Width { get { return _width; } } public int Height{ get { return _height; } } public int Xvel { get { return _xvel; } } public void DrawCar(Graphics g) { g.DrawRectangle(Pens.Blue, new Rectangle(X, Y, Width, Height)); } public void MoveCar(int gamewidth) { if( _x + _width >= gamewidth) { _x = 0; } _x = _x + _xvel; } } class Player { private int _x, _y, _width, _height, _xvel,_yvel; public Player(int x, int y, int width,int height, int xvel, int yvel) { this._x = x; this._y = y; this._width = width; this._height = height; this._xvel = xvel; this._yvel = yvel; } public int X { get { return _x; } } public int Y { get { return _y; } } public int Width { get { return _width; } } public int Height{ get { return _height; } } public int Xvel { get { return _xvel; } } public int Yvel { get { return _yvel; } } public void DrawPlayer(Graphics g) { g.DrawRectangle(Pens.Red, new Rectangle(X, Y, Width, Height)); } public void CollisionDetection(Rectangle player, Rectangle car) { if (player.IntersectsWith(car)) { MessageBox.Show("You Lose!"); } } public void MovePlayerLeft(int gamewidth) { if (_x > 0) { _x -= _xvel; } } } public partial class Form1 : Form { Car cars; Player player; public Form1() { InitializeComponent(); player = new Player(((Width / 2) - 15), (Height - 75),30, 30, 10, 10); Random r = new Random(); cars = new Car(r, 200, 30, 30, 10); } private void Form1_Paint(object sender, PaintEventArgs e) { Graphics g = e.Graphics; g.Clear(Color.White); cars.DrawCar(g); player.DrawPlayer(g); player.CollisionDetection(player, cars); //This is the part that implements collision detection } private void timer1_Tick(object sender, EventArgs e) { cars.MoveCar(this.Width); //foreach (Car c in cars) //{ // c.MoveCar(this.Width); //} this.Invalidate(); } } ``` Is This A Good Question/Topic? 0 ## Replies To: Collision Detection between two objects ### #2 Skydiver • Code herder Reputation: 6113 • Posts: 21,044 • Joined: 05-May 12 ## Re: Collision Detection between two objects Posted 28 April 2013 - 04:40 AM You need to let your Car (and possibly Player) class return a Rectangle property the represents their bounds. When you have the Rectangle, you can use IntersectsWith(): http://msdn.microsof...rsectswith.aspx The C# compiler would just automatically deduce that you've got the bit of data that represents a Rectangle and automatically make a Rectangle for you. You'll need to write code to create that Rectangle.	1,064	3,635	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.765625	3	CC-MAIN-2018-17	latest	en	0.546587
http://www.cse.unsw.edu.au/~cs1011/05s2/ass1/	1,542,590,642,000,000,000	text/html	crawl-data/CC-MAIN-2018-47/segments/1542039744803.68/warc/CC-MAIN-20181119002432-20181119024432-00153.warc.gz	395,013,648	7,995	COMP1011 COMP1011 Assignment 1 - 05s2 Computing 1A 05s2 Last updated Mon 22 Aug 2005 10:14 Mail cs1011@cse.unsw.edu.au # John Conway's Game of Life Version 1.2 This document has been changed since it was first released. To see the list of changes see section entitled Changes at the end of this specification. ## Introduction John Conway's Game of Life is one of the best known examples of emergent behaviour; complex behaviour that arises from deceptively simple rules. The game itself is based on a handful of very simple rules. The results, however, are reminiscent of microscopic life breeding and dying within a petri dish. Your task in this assignment will be to implement the functions that evolve this "life" from generation to generation. ## The Game Life is played on a rectangular grid of arbitrary size which we shall henceforth refer to as the board. Each square of the board is known as a cell and these can be either live or dead. Each cell has eight adjacent cells, that is orthogonally as well as diagonally opposed cells included. Figure 1 illustrates this; the eight white squares in the diagram below are adjacent to the black one. Figure 1 ### The Rules There are only two rules: 1. A live cell with exactly two or three adjacent cells stays alive. Otherwise the cell dies. 2. A dead cell with exactly three adjacent cells comes to life (i.e. becomes a live cell). Otherwise it remains dead. If you like, you can justify the birth and death of cells in the following manner. A live cell with one or no adjacent cell dies as if by loneliness or starvation, while one with four or more also dies but this time from overpopulation. The birth of cells with exactly three adjacent live cells can be thought of as reproduction under precise circumstances. #### Boundary cases A cell at a corner of the board only has three adjacent cells, while one on the edge but not in a corner has only five. The cells which lie "outside the board" so to speak can be though of as being dead. That is, for the purposes of applying the rules you can think of the board has being surrounded by a border of dead cells that is one cell in thickness. ### Playing the Game Life is a zero player game. It's behaviour is completely determined by the initial state of the board. Each turn in the game is known as a generation and applying the rules is known as evolving the board. The generations are numbered from zero so that the initial state of the board is generation zero. The rules are applied in no particular order to the individual cells. They may be thought of as being updated simultaneously. #### Example This figure shows the initial board and the subsequent two generations. Interestingly, Generation 2 is what is known as a stable configuration. Applying the rules to this arrangement of cells will lead to the same configuration. Try it and see. Figure 2 #### Example It is also possible for a collection of cells to oscillate between two configurations endlessly (called a cycle of period 2.) All even generations will look like Generation 0 while odd ones will look like Generation 1. Figure 3 Cells are represented by booleans. `True` and ```False ``` represent live and dead cells respectively. `type Cell = Bool` The board is represented as a list of lists of cells. Each list represents a column of the board and all such lists are equal in length. `type Board = [[Cell]]` The `Location` type represents the coordinates of a cell in the world. The location (0,0) represents the lower left corner of the board, the origin. `type Location = (Int,Int)` The world is represented by a pair. The first component is the location of the top right hand corner of the board. The locations of cells range from (0,0) to this value. The second component of the pair is the board itself. As mentioned above, the board is as a list of lists of cells, each such list representing a column of the board. The first element of the first list corresponds to the location (0,0). The last element of the last list corresponds to the top right hand corner of the board. `type World = (Location, Board)` The figure below clarifies locations and representation of the world somewhat. Figure 4 It would be represented as: `((1,2), [[True, True, False], [False, True, True]]` ## Implementation In this assignment you will implement the logic that drives the Game of Life. We present the functions in a rough order of difficulty. We provide example invocations of each function. We shall use the following worlds in all examples, save `initWorld`. They represent the world in Figure 4 and Generation 0 of Figure 3 respectively. ```world = ((1,2), [[True, True, False], [False, True, True]]) world2 = ((4,4), [[False, False, False, False, False], [False, False, True, False, False], [False, False, True, False, False], [False, False, True, False, False], [False, False, False, False, False]])``` ### Functions #### `getCell` `getCell :: World -> Location -> Cell` Given a game world and a location, yield the liveness at that location. ``` GameOfLife> getCell world (0,0) True GameOfLife> getCell world (1,0) False GameOfLife> getCell world (1,2) True ``` If the location argument to `getCell` is out of bounds (i.e. refer to a cell which is not present in the world) the function should signal an error using Haskell's `error` function. An example invocation is: ``` GameOfLife> getCell world (50,21) * Exception: Cell (50,21) is out of bounds ``` The error message must be exactly of this form. i.e. `Cell (`x,y```) is out of bounds``` where (x,y) is the location argument given to `getCell`. Note that `error` will automatically print `"* Exception: "` in front of the string you pass to it. Make sure that you do not add this yourself. #### `setCell` `setCell :: World -> Location -> Cell -> World` Update a location in the world with the given liveness. ``` GameOfLife> setCell world (1,1) False ((1,2), [[True, True, False], [False, False, True]]) GameOfLife> setCell world (1,0) False ((1,2), [[True, True, False], [False, True, True]]) ``` Like `getCell`, `setCell` should also signal an error when it is given a location argument that is out of bounds. The format is the same as for `getCell`. #### `evolveCell` `evolveCell :: World -> Location -> Cell` Compute the next-generation state for the cell at the given location. This is done according the rules presented earlier. ``` GameOfLife>evolveCell world (0,0) True GameOfLife>evolveCell world (0,2) True ``` #### `evolve` `evolve :: World -> World` Given a world in the game, produce the next state of the world. ``` GameOfLife>evolve world ((1,2), [[True, True, True], [True, True, True]]) GameOfLife>evolve world2 ((4,4), [[False, False, False, False, False], [False, False, False, False, False], [False, True, True, True, False], [False, False, False, False, False], [False, False, False, False, False]]) ``` #### `initWorld` `initWorld :: String -> World` Initialises the world from a string representation. • The '' character represents live cells • The '.' character represents a dead cell • Each row of the board is terminated by a newline An example of a valid world is: ```..... .. .. ** .. .... ``` which as a string looks like: `".....\n.*.\n..\n**\n..\n....\n"` Hint: This is a good world to test your function on. If you look carefully it looks like a large arrow pointing downwards. Make sure your function parses the world so that it has correct orientation. ``` GameOfLife> initWorld ".....\n.*.\n..\n**\n..\n....\n" ((4,5),[[False, False, True, False, False, False],[False, True, True, True, True, False],[True, True, True, True, True, False], [False, True, True, True, True, False],[False, False, True, False, False, False]]) ``` While testing your other functions you may want to type in the value of a world directly into a `ghci` window. Be careful that you get the order of values correct. Remember, the first element of the first list represents the bottom left hand corner of the board. ### Bonus If you find this assignment easy and would like some additional challenge, this optional bonus part is for you. Otherwise, you can safely ignore this section of the specification. If you do solve the bonus part, you can get up to two extra marks. However, before starting on the bonus, you must first complete and submit the basic part of the assignment. An `evolve` function that repeatedly uses `evolveCell` to yield the next generation is likely to be quite inefficient. It should be remembered that indexing into a list using the `!!` operator has a cost proportional to the length of the list. If this operation is used for each cell, the resulting solution will run in time proportional to the square of the number of cells. This is not as efficient as it could be. For bonus marks, implement a two pass approach to evolution. The first pass should yield a representation of the board where each cell is annotated with the number of adjacent cells. The following pass should evolve each annotated cell in turn. Such a solution will run in a time proportional to the number of cells. For large boards this can make quite a difference. It should be noted that bonus marks will only be awarded to well designed and well written solutions. It will not be enough for them to simply work. ## Stub files We are providing you with the skeleton of the solution (which also compiles). It contains: • Type definitions of `Cell`,`Board` etc • Type signatures for each function that you are required to implement • Dummy definitions for those functions IMPORTANT: You must not alter any of the function signatures in the stub file. This is because your assignments will be automatically tested against a test harness. This harness expects that the functions will have exactly the types specified above and will fail if any of them do not. Naturally the function bodies may be changed at will. ## Simulator The Game of Life is made a lot more interesting when you can see your creations creep and crawl across the screen. To aid in this endeavour we have provided a simulator in which to test your final solution. It is bundled as a gzipped tarball and contains the files necessary to compile the simulator against your solution for testing. For further instructions see the README file contained in the tarball. Note that you can develop the functions in `GameOfLife.hs` independent of this simulator, just with GHCi. However, once you have implemented all the functions, you will find the simulator useful to see how your worlds evolve. LifeClient.tar.gz - to unpack, use the command `tar xzf LifeClient.tar.gz` on Linux If you plan to work on your assignment on your own machine (instead of CSE lab computers), you must install the library `wxHaskell` in addition to GHC; see our software downloads. Important: It seems as if the `LifeClient` does not work properly on Windows. Hence, please use it on the CSE lab machines if you don't have Linux at home. The `LifeClient` is not essential to solve this assignment. So, you can still solve the assignment at home; however, if you want to see the animation, you need to do that at uni if you haven't got Linux at home. In the absence of the `LifeClient`, you can use the following function to convert a `World` to a character representation. ```import Data.List showWorld :: World -> String showWorld (_, board) = unlines (map (map toChar) board') where board' = reverse (transpose board) toChar True = '*' toChar False ='.'``` ## Diary Every time you work at your assignment solution document what you did and what the outcomes where in your assignment diary. The diary must be a plain text file called `diary.txt`. Create and edit it in Xemacs, or some other text editor, not with some word processing software. The file must be saved in ASCII format. Although this part of the assignment is assessable it is actually designed to help you. Writing down how you plan to design something and problems you encountered along the way can be of enormous benefit in arriving at a working solution. As such, you should not write this document at the very end. Not only will it show, you will also have missed out on any benefit it might have provided. You must submit your assignment by 23:58 on Sunday 28 August, 2005 10:00AM on Monday 29 August, 2005. If you wish to submit an assignment late, you may do so, but the maximum available mark for late assignments is reduced by 10% per day for up to five days. Assignments that are more than 5 days late will be awarded zero marks. So if your assignment is worth 85% and you submit it one day late you still get 85%, but if you submit it two days late you get 80%, and so on. Assignment extensions are only awarded for serious and unforeseeable events. Having a cold for a few days, deleting your assignment by mistake, going on holiday, going abroad, work commitments, etc do not qualify. Therefore aim to complete your assignments before the due date in case of last minute illness, and make regular backups of your work. We cannot stress this last point enough. "Regular backups" does not mean every few days - it should be done after every significant change to your code. No extension will be granted without a fully documented submission for special consideration at the Student Centre. Remember that we check for plagiarism (i.e., copied code and team work) and there are serious penalties. See the plagiarism information in the Course Outline. ## Submission Your solution should be submitted using the `give` command. Type the following at the unix command line: `give cs1011 ass1 GameOfLife.hs diary.txt` For those of you who wish to submit the solution outlined in the Bonus section also type the following: `give cs1011 ass1_bonus GameOfLife.hs diary.txt` Remember, this is a separate solution. The original must still be submitted ### Early submission There is a single mark for early submission of a draft. To receive this mark your solution must: • compile, • have `getCell` and `setCell` fully implemented, and • be submitted by 23:58 28 August, 2005 11:00AM 22 August, 2005 using the command `give cs1011 ass1_draft GameOfLife.hs diary.txt` Note how the assignment specifier for the early submissions and final submission are different (i.e., they are `ass1_draft` and `ass1`, respectively). Important hint: The `give` program allows you to submit your assignment multiple times and only your last submission will be marked. So, there is no harm in submitting partial solutions before the deadline. Do not leave your first submission until the last minute. Always submit some intermediate versions early. ## Marking Category Marks Early submission 1 Diary 2 Automarking (functionality) 8 Subjective (style & structure) 4 Bonus (optional) 2 It is a generally accepted software engineering principle that it is not simply enough for a program to be understood by the machine; it must also be human readable. For the subjective part of the marking we will be awarding marks for programs that have good comments, useful names, and good decomposition and overall design. Good decomposition practices are things such as placing repeatedly used computations in `where` clauses and breaking complex functions up into calls to small, less complex functions. Source files should also be no wider than 80 characters. ## Questions or Problems? Date Change 17 August, 2005 Added `showWorld` function 21 August, 2005 New deadline	3,536	15,551	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5625	4	CC-MAIN-2018-47	latest	en	0.951527
https://www.gamedev.net/forums/topic/330860-two-things/	1,537,844,361,000,000,000	text/html	crawl-data/CC-MAIN-2018-39/segments/1537267160923.61/warc/CC-MAIN-20180925024239-20180925044639-00199.warc.gz	748,110,230	33,754	Public Group # two things This topic is 4826 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic. ## Recommended Posts Hi! I'm new here and kinda new to pragramming. I've two questions. First of all, I want to show my first program I've just finished and kinda hope for some feedback on my code. #include <cstdlib> #include <iostream> using namespace std; bool stillRunning = true; // is program still running? int randomNumber = 0; // this will contain the "random" number int guess = 0; // this will contain the user's guess bool guessed = false; // is the rand guessed? char yesNo; // for when the user has to make a y/n decision // this will generate the "random" number... I want a real random number! :( int GenNumber() { randomNumber = (rand()%25); cout << "Ok, I have made up a number between 0 and 25" << endl; return 0; } // getting the guess from the user... int GuessNumber() { cout << "What's your guess?" << endl; cin >> guess; return 0; } // check if the user has guessed it or if he's close (or not) int Check() { if ( guess == randomNumber ) { cout << "Yay, you guessed the number! Congrats!" << endl; guessed = true; return 0; } if ( guess - randomNumber <= 5 && guess - randomNumber >= -5 ) { cout << "OWW, you're close!" << endl; } else cout << "Not even close, man." << endl; return 0; } // ask if the user wants to play again (looky looky, it's fool proof ^^) int PlayAgain() { bool correctInput = false; //there are always people who don't wanna listen while ( correctInput != true ) { cout << "Play again? y/n" << endl; cin >> yesNo; if ( yesNo == 'n' ) { stillRunning = false; // the user wants to quit correctInput = true; // but atleast he got the char right cout << "Ok, bye bye!" << endl; } if ( yesNo == 'y' ) { guessed = false; // we reset it for another round correctInput = true; cout << "Great!" << endl; } } correctInput = false; // reset for next time return 0; } //nice and simple loop int main() { while ( stillRunning != false ) { GenNumber(); while ( guessed != true ) { GuessNumber(); Check(); } PlayAgain(); } return 0; // ok, we're done } I'm so proud ^^ Anyway, any sort of feedback on the current code is appreciated. The second is.... pointers. I'm currently reading C++ Primer, recommended in the "for beginners guide" and am just overwhelmed by it's explanation of pointers. What I do know is the basic reference/dereference stuff... I think :/ int thing = 0; int pthing; pthing = &thing; pthing = pthing + 1; // which makes thing = 1 Ok, that seems pretty straightforward. Is this all I need to know for the moment? Another thing that I just don't get is the actual use of it. For example, where in my number guess program could this be of use? Ok, that was it for now :D edit: hmm.. how do I make code-like text? [.code][./code] is not the answer, so it seems. edit2: ah got it. [Edited by - Deere on July 8, 2005 9:19:45 AM] ##### Share on other sites Quote: Is this all I need to know for the moment? That is pretty much the basics, yes. Quote: Another thing that I just don't get is the actual use of it. For example, where in my number guess program could this be of use? In a number guessing program it would not be of much use. One use is that you can use it to 'return' multiple values from a function. For example: void f( int pIntOne, int* pIntTwo ){ pIntOne = 1; pIntTwo = 2;}int i, j;f( &i, &j ); The more useful use comes when dealing with things bigger than single integeres. Consider a structure of which you know little, only that it takes up a lot of memory: struct SBig{ ...}; Suppose you want to do something with it in a separate function. Without pointers you would need to pass it to the function (i.e. copy it) and return it (i.e. copy it again). More efficient is pass-by-reference: void f( SBig* pBig ){ ...}; There is much more to it, of course. Also, in C++ there are references which many coders find better in such cases. Greetz, Illco ##### Share on other sites As for the code, I think it's very nicely written, and adequately commented. Very well done! ##### Share on other sites Quote: Original post by IllcoThere is much more to it, of course. Also, in C++ there are references which many coders find better in such cases. Yes. In C++ you would do: void f(SBig& big) { ... }, and modifying big within the function would directly modify the variable you passed to it. Quote: First of all, I want to show my first program I've just finished and kinda hope for some feedback on my code. Not bad at all for a first program. Now try to get rid of all those global variables. I notice you don't really make use of the return values of your functions (why return 0 ?) PlayAgain() could return true or false based on whether the player wants to play again. GuessNumber() could return the number picked by the player. And so on and so forth. ##### Share on other sites wow, quick replies. Ok, The need for pointer is clear now. Thanks for the explanation. And I will try to modify the code a bit, according to your suggestions. Thanks again :) ##### Share on other sites Yes it's a very good program for a first one.As everybody said rid of the globals and use function's return value.Like int GenNumber() //some func that returns int without parameters { int randomNumber = (rand()%25); cout << "Ok, I have made up a number between 0 and 25" << endl; return randomNumber; } int main() { int number //local variable that will hold the value of GenNumber while ( stillRunning != false ) { number = GenNumber(); ..... This will help you if you want to just copy-paste your GenNumber function in another program which right now you cant because in that another prog you must have global variable named randomNumber.This makes your prog more modular. And for the pointers , yeah you got it right.I got a lot of problem understanding them until i realised they are like shortcut to some file. Anyway good job!!! ##### Share on other sites EDIT: Sorry page loading has stoped when i was replaying to topic so i send it again ##### Share on other sites Ok, prog has changed a bit... #include <cstdlib>#include <iostream>using namespace std;// this will generate the "random" number... I want a real random number! :(int GenNumber(){ int randomNumber; randomNumber = (rand()%25); cout << "Ok, I have made up a number between 0 and 25" << endl; return randomNumber;}// getting the guess from the user...int GuessNumber(){ int userGuess; cout << "What's your guess?" << endl; cin >> userGuess; return userGuess;}// check if the user has guessed it or if he's close (or not)bool Guessed(int* pnumber, int* pguess){ if ( pguess == pnumber ) { cout << "Yay, you guessed the number! Congrats!" << endl; return true; } if ( pguess - pnumber <= 5 && pguess - pnumber >= -5 ) { cout << "OWW, you're close!" << endl; } else cout << "Not even close, man." << endl; return false;}// ask if the user wants to play again (looky looky, it's fool proof ^^)bool PlayAgain(){ bool correctInput = false; //there are always people who don't wanna listen char yesNo; // for when the user has to make a y/n decision while ( correctInput != true ) { cout << "Play again? y/n" << endl; cin >> yesNo; if ( yesNo == 'n' ) { return false; // the user wants to quit correctInput = true; // but atleast he got the char right cout << "Ok, bye bye!" << endl; } if ( yesNo == 'y' ) { correctInput = true; cout << "Great!" << endl; } } correctInput = false; // reset for next time return true;}//nice and simple loopint main(){ bool stillRunning = true; // is program still running? int number; // container for the rand int guess; // this will contain the user's guess bool guessed = false; // is it guessed or not? bool playAgain; // if player wants to play again or not while ( stillRunning != false ) { guessed = false; // this is just a reset number = GenNumber(); // get our rand while ( guessed != true ) { guess = GuessNumber(); // get our guess guessed = Guessed(&number, &guess); // compare and react accordingly } playAgain = PlayAgain(); if ( playAgain == false ) // when player wants to quit stillRunning = false; // we tell the prog to quit } return 0; // ok, we're done} What I noticed is that this forces you to take a different point of view. You actually have to keep track of the "dataflow" or whatever you call it. It was a little harder than the first one, but it works. Notice the use of a pointer ^^ This was for testing purposes, more or less, but still satisfying. One problem I do have with this type of pragramming is that it can get a little messy (similar var names). I guess when I have some more experience it will solve itself. ##### Share on other sites Very well done, I like it. The new design you implemented is very effective and keeps things nice looking. An even better enhancement to the design would be to implement the use of classes. Although it would seem kind of meaningless in this simple project (much like the pointer example), it would greatly improve the design of the program. If you make it through that step, you'll be well on your way [smile]. ##### Share on other sites Yes, that seems to be the next logical step, although I'm still not sure how to implement classes into a prog. My first thoughts are that I make a class out of the vars in main and pass them to the functions with pointers. It seems like that would be kind of useless at the moment, but I realise that this is a very small program. I'll just look into classes and see if I can enhance my prog even more. Thanks for the kind words. 1. 1 2. 2 3. 3 Rutin 18 4. 4 5. 5 JoeJ 13 • 14 • 10 • 25 • 9 • 57 • ### Forum Statistics • Total Topics 632644 • Total Posts 3007625 • ### Who's Online (See full list) There are no registered users currently online ×	2,555	9,961	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.53125	3	CC-MAIN-2018-39	latest	en	0.790142
https://medical-dictionary.thefreedictionary.com/Microamp	1,556,034,643,000,000,000	text/html	crawl-data/CC-MAIN-2019-18/segments/1555578605555.73/warc/CC-MAIN-20190423154842-20190423180842-00453.warc.gz	470,817,406	13,459	# ampere (redirected from Microamp) Also found in: Dictionary, Thesaurus, Encyclopedia. ## ampere (A) [am´pēr] the base SI unit of electric current strength, defined in terms of the force of attraction between two parallel conductors carrying current. ## Am·père (ahm'pēr), André-Marie, French physicist, 1775-1836. See: ampere, statampere, Ampère postulate. ## am·pere (A), (am'pēr), 1. The practical unit of electrical current; the absolute, practical ampere originally was defined as having the value of one tenth of the electromagnetic unit. 2. Legal definition: the current that, flowing for 1 second, will deposit 1.118 mg of silver from silver nitrate solution. 3. Scientific (SI) definition: the current that, if maintained in two straight parallel conductors of infinite length and of negligible circular cross-sections and placed 1 m apart in a vacuum, produces between them a force of 2 × 10-7 N/m of length. [André-Marie Ampère] ## ampere /am·pere/ (A) (am´pēr) the base SI unit of electric current strength, defined in terms of the force of attraction between two parallel conductors carrying current. ## ampere (A) [am′pēr] Etymology: André-Marie Ampère, French physicist, 1775-1836 a unit of measurement of the amount of electric current. An ampere, according to the meter-kilogram-second (MKS) system, is the amount of current passed through a resistance of 1 ohm by an electric potential of 1 volt; in the International System (SI) of Units, an ampere is a unit of electric current that carries a charge of 1 coulomb through a conductor in 1 second. The standard international ampere is the amount of current that deposits 0.001118 g of silver per second when passed, according to certain specifications, through a silver nitrate solution. See also ohm, volt, watt. ## am·pere (A) (am'pēr) 1. The practical unit of electrical current; the absolute, practical ampere originally was defined as having the value of1/10 of the electromagnetic unit (see abampere and coulomb). 2. Legal definition: the current that, flowing for 1 second, will deposit 1.118 mg of silver from silver nitrate solution. 3. Scientific (SI) definition: the current that, if maintained in two straight parallel conductors of infinite length and of negligible circular cross-sections and placed 1 m apart in a vacuum, produces between them a force of 2 × 10-7 N/m of length. [André-Marie Ampère] ## Ampère, André-Marie, French physicist, 1775-1836. ampere - the practical unit of electrical current. Ampère postulate - Synonym(s): Avogadro law statampere - the electrostatic unit of current, equal to 3.335641 X 10-10 ampere. ## am·pere (A) (am'pēr) 1. The practical unit of electrical current. 2. Scientific (SI) definition: the current that, if maintained in two straight parallel conductors of infinite length and of negligible circular cross-sections and placed 1 m apart in a vacuum, produces between them a force of 2 × 10-7 N/m of length. [André-Marie Ampère] ## ampere (am´pir), n (Amp), a unit of measurement of the quantity of electric current, equal to a flow of 1 coulomb per second or 6.25 time 1018 electrons per second. The current produced by 1 volt acting through a resistance of 1 ohm. ## ampere a unit of electric current strength, the current yielded by one volt of electromotive force against one ohm of resistance. References in periodicals archive ? In addition to consuming the lowest current per MHz, a common industry specification, the new MCUs contain an integrated low drop-out (LDO) regulator that keeps the current constant at 150 microamps per MHz over the entire operating range of 1. Thermal cycling and monitoring of fluorescence during PCR and data analysis were carried out using the ABI PRISM[R] 7700 Sequence Detection System (SDS; PE Biosystems), MicroAmp[R] optical 96-well reaction plates, and MicroAmp optical caps (PE Biosystems). The TagMan PCR Core Reagent Kit, MicroAmp Optical Tubes [TM], and MicroAmp Optical Caps [TM] were from PE Biosystems. During qualification of the 30 kA fuse, all models passed HIPOT testing at 2000 VAC and 2000 VDC between the microamp, milliamp, and ampere terminals and common ground. MicroAmp optical tubes and caps (PE) were used to prevent light scattering or reflection. Battery- and solar-powered applications benefit from the device's low current consumption at 1 microamp in sleep mode and 5 microamps in active mode. PCR was carried out in singlicate in Microamp thin-walled tubes on a Geneamp 9600 thermal cycler (Perkin-Elmer, Norwalk, CT). It uses less than 1 microamp of write current per cell, has a 10x write performance and better endurance compared to NAND, and at a much lower cost," stated Alan Niebel, CEO of Web-Feet Research, Monterey, CA. after two years of intensive research and development, Podio consists of a proprietary MicroAmp and a full range driver, housed in what is best described as a cylinder-shaped metal micro speaker only 88 millimeters long. In this state, the DSCs are engineered to consume one microamp in lowest power mode while preserving key state information in RAM. A sleep mode and a very low current stand-by mode (20 microamp (oA)) with remote wake up via the bus provide engineers with significant flexibility in reducing power consumption. AEC-Q100-qualified LDOs with Power-On Reset Options, Built-in Protection and Quiescent Currents Down to 22 Microamp Target Automotive Modules Requiring Very Low Ignition-off Currents Site: Follow: Share: Open / Close	1,351	5,495	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18	longest	en	0.853209
https://www.mrexcel.com/board/threads/overcomplicating-a-simple-problem.277413/	1,680,344,628,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296949958.54/warc/CC-MAIN-20230401094611-20230401124611-00622.warc.gz	976,291,771	18,480	# Overcomplicating a simple problem #### storm79 ##### New Member I'm certain there's a simple solution to this problem if only I knew the correct function(s) to use! I've got a spreadsheet with a list of events and the date they occured. There's simply two colums, Date and Description. The description can only be one of four possibilities, so say "Alpha", "Bravo", "Tango" or "Foxtrot", and the sheet looks a little like this: 01/01/2007 Alpha 02/01/2007 Alpha 02/01/2007 Bravo 02/01/2007 Bravo 02/01/2007 Foxtrot 03/01/2007 Tango 03/01/2007 Alpha 04/01/2007 Foxtrot 07/01/2007 Tango 07/01/2007 Alpha What I need to do is count the number of events for each day. So for example I'd like the spreadsheet would give me a second column of data listing each unique day and the number of events, by type that occured then. E.G: 01/01/2007 Alpha x 1 02/01/2007 Alpha x 1, Bravo x 2, Foxtrot x 1 03/01/2007 Alpha x 1, Tango x 1 04/01/2007 Foxtrot x 1 05/01/2007 No data 06/01/2007 No data 07/01/2007 Alpha x 1, Tango x 1 I wouldn't need the "Alpha x 1", that's just for illustration, but if it could return a count in an appropriate column for the number of events that day then that would be ideal. So start at Row 5, in Cell A put the date, Cell B put the Alpha count, cell C put the Bravo count, etc. etc. then move to a new row for a new day. I've tried various different ways to work this out by trying to combine the "Countif" function with other functions, but have failed. I'm sure I'm overcomplicating the method needed, but I'm really stuck. Has anyone got any ideas? ### Excel Facts Show numbers in thousands? Use a custom number format of #,##0,K. Each comma after the final 0 will divide the displayed number by another thousand #### SteveO59L ##### Well-known Member Try a Pivot Table #### goblin ##### Active Member Pivot table is what you are looking for. Put the date as row data and the description as column data. Then drop the description field into the Data section as well. That will give you exactly the table you are looking for. #### storm79 ##### New Member Thanks for the replies. That's exactly what I needed. I couldn't get it working at first, but the problem is with my data rather than the pivot table. You see, I'm pulling the "date" information from a different workbook (basically by saying Cell A5 = Cell A5 in a different workbook). However, the data in the other workbook also contains time information in the same cell (01/01/07 15:34). Even though I've got the date column formated to just display the date the pivot table is still using the time information and so the the result isn't quite right. Is there an easy way to either get the pivot table to ignore the time information, or not to copy it across in the first place? #### Norie ##### Well-known Member Try INT. =INT([Book2]Sheet1!A5) Replies 8 Views 82 Replies 10 Views 82 Replies 13 Views 303 Replies 9 Views 212 Replies 11 Views 354 1,191,687 Messages 5,988,019 Members 440,125 Latest member vincentchu2369 ### We've detected that you are using an adblocker. We have a great community of people providing Excel help here, but the hosting costs are enormous. You can help keep this site running by allowing ads on MrExcel.com. ### Which adblocker are you using? 1)Click on the icon in the browser’s toolbar. 2)Click on the icon in the browser’s toolbar. 2)Click on the "Pause on this site" option. Go back 1)Click on the icon in the browser’s toolbar. 2)Click on the toggle to disable it for "mrexcel.com". Go back ### Disable uBlock Origin Follow these easy steps to disable uBlock Origin 1)Click on the icon in the browser’s toolbar. 2)Click on the "Power" button. 3)Click on the "Refresh" button. Go back ### Disable uBlock Follow these easy steps to disable uBlock 1)Click on the icon in the browser’s toolbar. 2)Click on the "Power" button. 3)Click on the "Refresh" button. Go back	1,069	3,922	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14	latest	en	0.900026
https://electronics.stackexchange.com/questions/100713/where-did-i-go-wrong-transistor-burning-up	1,718,793,734,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861817.10/warc/CC-MAIN-20240619091803-20240619121803-00109.warc.gz	195,544,633	39,828	# Where did I go wrong? Transistor burning up I'm trying to control 3 LED downlights using an arduino. I'm very inexperienced so I'll go through my thought process and you might be able to catch what I'm doing wrong... I've chosen a 2N5551 NPN transistor based on the following information: NPN transistor because I will be switching higher voltage than what is supplied to the Arduino. Downlights: http://www.malmbergs.com/frmProductDisplay_new.aspx?item=9974006 They are rated at 350 mA, 3x1.2W / downlight equals a total of 10.8W 10.8W / 0.35 A = 30.86 V (to make sure my math is correct I measured the circuit, 28V and 350 mA) They are driven by a 350 mA CC driver. So I'm searching for an NPN transistor capable of handling 31V and 350 mA. According to the datasheet: https://www.fairchildsemi.com/ds/MM/MMBT5551.pdf the VCEO is 160V and Ic 600 mA, so far so good? Here comes the difficult part, calculating the minimum current needed to keep the transistor in saturation (for a newbie like me, transistor datasheets seem to be incredibly inconsistent in their layout and overall hard to read). The guides I've used are telling me to look for Hfe in saturation, Ic/Ib or if it can't be found use a worst case value of 10. Since I can't seem to find the information in the datasheet, I'll use 10. Base current: 350 mA / 10 = 35 mA (safe value from Arduino seems to be around 40 mA) Next step, find Vbe: Figure 3 shows a peak of 0.9V at 200? mA (why is the graph not expanded to show up to 600 mA, does the Vbe not rise above 0.9V?) Output voltage of the arduino is 5V, giving us a voltage drop over the resistor of: 5V-0.9V = 4.1V Base resistor: 4.1V/0.035 = 117 ohm Feeding the output of the arduino through two 220 ohm resistors in parallel to the transistor base, and the emitter and collector switching the ground for the LEDs, everything is working as expected for 20~ seconds, and then there was smoke... Any help to solving my problem is greatly appreciated. I have an off-topic question as well, is it possible to PWM a CC driver on the secondary side? Would be real nice to be able to dim these LEDs...	574	2,125	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26	latest	en	0.933658
https://www.cravencountryjamboree.com/helpful-tips/how-fast-is-a-100w-motor/	1,680,133,305,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296949035.66/warc/CC-MAIN-20230329213541-20230330003541-00587.warc.gz	804,260,030	14,836	# How fast is a 100W motor? ## How fast is a 100W motor? In average, a human can deliver around 100W of power. Let’s consider you can deliver the whole power to the pedals. This is enough to any biker to reach 10~15 miles per hour. ### What is the maximum voltage you can use on a 12v DC motor? 12 V motor will work well with 12 V battery, and will be fine up to the maximum voltage the battery goes when fully charged. This means it will work well up to 14 V. #### Can a 12v motor run on 12v? In a nutshell, there is nothing wrong with running a 12V motor on anything below 12V, but it will be weaker, and therefore slower, or even may fail to start depending on the load. How much power does a 12v motor use? 0.5 HP = 42 amperes. Does higher wattage motor mean more power? The wattage rating is usually (read the manufacturer’s specs very carefully) the measure of how much power the motor can draw. It doesn’t tell you anything more than that. Wattage does not tell you whether the motor is efficient at converting the input electrical energy into output mechanical energy. ## Is a higher voltage motor better? A higher voltage system is more efficient than a lower voltage since it experiences less energy loss from resistance given the same amount of power draw. You get the same exact voltage—but with 80 amps of current. That’s 80% more energy! ### What is the output of a DC motor? Typical DC motors may operate on as few as 1.5 Volts or up to 100 Volts or more. Roboticists often use motors that operate on 6, 12, or 24 volts because most robots are battery powered, and batteries are typically available with these values. #### Can a 9V battery run on a 12V motor? The 9V batteries are weak because the other half of the story is how much current they can supply. It’s possible to run a 12V motor on 9V, but it will run slower and with less Torque. Will a 6 volt battery run a 12 volt motor? If you give a 6 V to 12 V motor, the motor will run at half of its rated speed and the Torque will also reduce. Is this true? The motor is on half of its speed. What is the power of DC motor? DC motors use magnetic fields that occur from the electrical currents generated, which powers the movement of a rotor fixed within the output shaft. The output torque and speed depends upon both the electrical input and the design of the motor.	567	2,358	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.46875	3	CC-MAIN-2023-14	latest	en	0.93391
https://www.jplcomputer.co.uk/triathlon	1,624,030,707,000,000,000	text/html	crawl-data/CC-MAIN-2021-25/segments/1623487637721.34/warc/CC-MAIN-20210618134943-20210618164943-00630.warc.gz	768,556,247	13,011	## Triathlon ### How Many Features in Excel do you recognise in my Triathlon story? – When I first decided in 2011 to enter my first sprint triathlon I knew I had to get myself sorted and into condition. There is so much data out there about how to train and how much to eat so surely there was sum kind of formula I could match to consolidate my own thoughts into one. My background strength was always the run so I knew that the swim and the cycle would raise some comments as I started to train. However, with all my connections I had forecast a reasonable first effort. What surprised me was the advanced array of equipment on the day. My pivotal moment was in transition trying to find my bike amongst so many others looking the same! As I crossed the finish line all the effort was worthwhile and I managed to lookup to see my time. Now for rest till next time! This year I am attempting my first half Ironman distance. Fancy joining me? Sorted Sorting is a standard feature on the Data Tab in Excel. You can also access Simple sorts from the Sort Ascending and sort Descending buttons on that Same ribbon Condition A condition is the first part of an IF function which results in either it being True or False Data This Tab contains all kinds of features such as Filtering, Sub Totalling, connecting to other packages SUM Probably the best known of all formulas. Adds up a range of cells Match This feature returns the position in a range of cells matching to the value Consolidate On the Data Tab and used to add up typically different sheet ranges of data when laid out in the same way Background Purely formatting of the back of the worksheet grid to personalise it Run One the 5 ways to initiate a Macro program is to Run it from the View Macros tab Comments Simply right click on a cell to add a description to that cell Connections Found on the Data Tab to connect to say a database like Access to read data from it into Excel Forecast Again one of the many functions found in the fx area of Excel Advanced Found in terms of filtering and allows the AND OR combination of questions to be resolved Array A term used I the Match and Index functions to suggest a range of data Pivotal Bit cheeky this one as its Pivot really. Relating to a Pivot Table – a brilliant way to analyse data on any which way you like. Find To search for a work or formula use the Home Tab and Find button Lookup Used typically as a Vlookup which looks up a value to match to in a table and returns one of its column values We hope you found this article informative, you might also like to read our previous blog on how our new customers enjoy training. You might also like to read more about our Microsoft excel training. We have placed cookies on your device to help make this website better. I'm fine with this We use cookies to give you the best online experience. Please let us know if you agree to all of these cookies. Some of the cookies we use are essential for the site to work. We also use some non-essential cookies to collect information for making reports and to help us improve the site. The cookies collect information in an anonymous form.	684	3,344	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.546875	3	CC-MAIN-2021-25	longest	en	0.93433
stairsideas.com	1,701,357,183,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100227.61/warc/CC-MAIN-20231130130218-20231130160218-00450.warc.gz	619,104,162	16,647	When designing Straight staircase, Quarter Landing Stairs, or Half Landing Stairs we cope with directly flights and generally speaking it’s not tough to create a calculation of setup, as well as of the arrangement of measures in the flight. ### Example Calculation for Stairs with a Straight Flight Let’s describe the calculation procedure for the staircase with a straight flight for instance. Suppose we construct a staircase from the floor to the second floor. Floor to Floor Height is equivalent to 2600 millimeter . Consequently we would like to obtain the staircase with significance of the riser height and moving 17/29 (you may read how pick a correlation of the riser height and moving from the area How to build stairs). To estimate the amount of risers, we should split the Floor to Floor Height from the riser height, that we take from the significance of the riser height and moving 2600 millimeter ÷ 170 mm = 15.29. Since the amount of risers will be a lot, we round down the value obtained and get 15 risers. Now to find out the precise value of the riser height, we will split the Floor to Floor Height from the amount of risers, that’s 2,600 millimeter ÷ 15 = 173.33 millimeter . We must round down the value received to 173 millimeter , as it’s not possible to measure out the value of 173.33 millimeter by hand marking. But today if we multiple 173 millimeter by 15 risers we get the value of 2,595 mm. As you’ve already understood, the precision in dimension is of crucial significance . Ignoring even tenths following the decimal point at the same step leads from the aggregate to the gap between the Floor to Floor Height and the staircase elevation. In the case set above the value of 5 millimeter can be paid if you marginally level up the stairs, or when you increase the height of the last riser. However, the best thing is to avoid these situations. When laying out the steps and picking the significance of the riser height and heading to your staircase, you will always take into account the dimensions of the present aperture and recall about the headroom of stairs and its minimum dimensions. With this purpose, the measurements of the present aperture are plotted on the design of the staircase. Then we decide on which measure the edge of the aperture and the staircase overlap. Then subtract the complete height of steps up to the overlapping out of the floor-to-ceiling height. As you see, it rarely occurs to obtain the specific match of values of significance of the riser height and moving given in the section How to build stairs. You can only get them if the floor-to-floor elevation is even divisible from the riser height. Therefore we must round the obtained values in favor of those correlations. You mark the obtained values on the closed string or open series upward and sidewards, and remember to stick to parallel and vertical alignment of the measures. When designing Single Winder Stairs or Dual Winder Stairs, to compute and determine the form of treads from the flight is quite tough. The form of winder treads cannot be chosen randomly. To provide for security, convenience of usage and to be given a harmonic bend of the flight, particular care must be used to evenly increase the thickness of the treads in the tread with the narrowest up and going to another square tread in the flight. At the going of the most significant winder tread in the distance of 305 mm from the border will not exceed the moving of the tiniest winder tread for over 10 mm. The centre winder tread, that’s the tread with the narrowest going, shall be positioned symmetrically to the axis extending from the vertex of the flip angle of the flight to the stage constituting the middle of rounding of the interior area of the flight. The minimum going for any tread in the narrowest point will make 152 mm based on American standards, and 120 mm based on European standards. To be given a harmonic bend and to adhere to the minimum sizes mentioned previously, the amount of winder treads from the flight shall be from 7 to 9. Everything about carpentry! I'm a blogger who love woodwork, home improvement tips and anything to make life cozier.	889	4,180	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.34375	4	CC-MAIN-2023-50	latest	en	0.900801
https://www.justanswer.com/multiple-problems/8kp13-help-these-total-25-quest.html	1,624,402,468,000,000,000	text/html	crawl-data/CC-MAIN-2021-25/segments/1623488525399.79/warc/CC-MAIN-20210622220817-20210623010817-00025.warc.gz	770,753,980	53,886	Multiple Problems For Help with Multiple Homework Problems, Ask an Expert Related Multiple Problems Questions Chapter 16 Problems 1) Shapland Inc. has fixed operating quot;For Business Tutor" Chapter 16 Problems 1) Shapland Inc. has fixed operating costs of \$500,000 and variable costs of \$50 per unit. If it sells the product for \$75 per unit, what is the break-even… read more GLENN GILONGOS Master of Mathematics 637 satisfied customers Stock Valuation: A stock has an initial price of \$100 per Stock Valuation: A stock has an initial price of \$100 per share, paid a dividend of \$2.00 per share during the year, and had an ending share price of \$125. Compute the percentage total return, capital… read more Manal Elkhoshkhany Bachelor's Degree 1,420 satisfied customers 1. (TCO A) Below you will find selected information (in 1. (TCO A) Below you will find selected information (in millions) from Coca-Cola Co.'s 2012 Annual Report: Income Taxes Payable \$471 Short-term Investments and Marketable Securities 8,109 Cash 8,442 O… read more CEO Bachelor's Degree 158 satisfied customers 1. (TCO A) Below you will find selected information (in millions) 1. (TCO A) Below you will find selected information (in millions) from Coca-Cola Co.'s 2012 Annual Report: Income Taxes Payable \$471 Short-term Investments and Marketable Securities 8,109 Cash 8,442 O… read more Mr. Gregory White Master's Degree 886 satisfied customers (TCO A) Below you will find selected information (in millions) (TCO A) Below you will find selected information (in millions) from Coca-Cola Co.'s 2012 Annual Report: Income Taxes Payable \$471 Short-term Investments and Marketable Securities 8,109 Cash 8,442 Othe… read more F. Naz B.Com CA Finalist & Completed B.com 1,145 satisfied customers "for fivestarlaws only Question 51 1 points Save The "for fivestarlaw's only Question 51 1 points Save The first phase of the National Labor Relations Act is known as the: a. Clayton Act b. TaftAct c. NorrisGuardia Act d. Wagner Act e. 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D. the dealer market. 2. The sale of pre… read more Mr. Gregory White Master's Degree 886 satisfied customers Perfectly Competitive Firms Must Make All Of The Following 1. Perfectly competitive firms must make all of the following decisions EXCEPT (Points : 1) how much output to supply. which production technology to use. how much of each input to demand. what price … read more Mr. Gregory White Master's Degree 886 satisfied customers Produce as much as you can because there is a limitless market. 1.The business philosophy of "produce as much as you can because there is a limitless market" is consistent with which of the following marketing eras? A) production era B) selling era C) marketing er… read more hrguru206 HR Manager Bachelor's Degree 151 satisfied customers Assignment 8.3: Case Study, O'Grady Apparel Company O'Grady Apparel Company O'Grady Apparel Company was founded nearly 160 years ago when an Irish merchant named Garrett O'Grady landed in Los Angeles … read more linda_us Master's Degree Master's Degree 1,402 satisfied customers 51. (Points: 1.0) Benny, an employee of Parrott Company, 51. (Points: 1.0) Benny, an employee of Parrott Company, used company assets for his own personal gain. This is an example of a. embezzlement. b. fraud. c. internal control. d. ethics. e. employment p… read more Neo 13,706 satisfied customers I forgot we got one more 13 mins. 1. (Points: 1.0) I forgot we got one more 13 mins. 1. (Points: 1.0) The costs of bringing a corporation into existence, including legal fees, promoter fees, and amounts paid to obtain a charter are called: a. Minimum … read more Neo 13,706 satisfied customers Question 13 1 points Save A __________ is a potential Question 13 1 points Save A __________ is a potential problem in service flow analysis. fissure fail point critical incident hub critical path Question 14 1 points Save Which of the following is the p… read more Mr. Gregory White Master's Degree 886 satisfied customers Which country has the highest annual rate of new business start-ups?An unvrs Lecturer of Mathematics Master's Degree 36 satisfied customers 74) At the break-even point of 2,000 units, variable costs 74) At the break-even point of 2,000 units, variable costs are \$55,000, and fixed costs are \$32,000. How much is the selling price per unit? A. Not enough information B. \$16.00. C. \$43.50. 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Education / Educational Administration Bizhelp CPA 539 satisfied customers BA degree and Certified Public Accountant MyVirtualCPA Master's Degree 155 satisfied customers Brittany Professional 33 satisfied customers	2,116	8,803	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.53125	3	CC-MAIN-2021-25	latest	en	0.85709
https://algebra-help.com/math-tutorials/clep-college-algebra-free-test.html	1,548,287,262,000,000,000	text/html	crawl-data/CC-MAIN-2019-04/segments/1547584431529.98/warc/CC-MAIN-20190123234228-20190124020228-00033.warc.gz	419,623,117	7,633	# Algebrator can start solving your homework in the next 5 minutes Get Algebrator Now! only \$49.00 \$35.00 2checkout.com is an authorized reseller of goods provided by Softmath January 23rd ## It is easy as 1-2-3! Step 1 : Enter your algebra problem straight from the book: Step 2 : Let Algebrator solve it: Step 3 : Ask for an explanation for the steps you don't understand: • solving a system of two and three linear equations (including Cramer's rule) • simplification of algebraic expressions (polynomials (simplifying, degree, division...), exponential expressions, fractions, radicals (roots), absolute values...) • factoring and expanding expressions • finding LCM and GCF • basic step-by-step arithmetics operations (adding, subtracting, multiplying and dividing) • operations with complex numbers (simplifying, rationalizing complex denominators...) • solving linear, quadratic and many other equations and inequalities (including log. and exponential) • graphing general functions • graphing curves (lines, parabolas, hyperbolas, circles, ellipses, equation and inequality solutions) • operations with functions (composition, inverse, range, domain...) • simplifying logarithms • sequences (classifying progressions, find the nth term of an arithmetic progression...) • basic geometry and trigonometry (similarity, calculating trig functions, right triangle...) • arithmetic and other pre-algebra topics (ratios, proportions, measurements...) • linear algebra (operations with matrices, inverse matrix, determinants...) • statistics (mean, median, mode, range...) ## System requirements: Any PC running any version of Windows. Any Intel Mac running OS X. Get Algebrator Now! only \$49.00 \$35.00 2checkout.com is an authorized reseller of goods provided by Softmath # Our users: I recommend the Algebrator to students who need help with fractions, equations and algebra. The program is a great tool! Not only does it give you the answers but it also shows you how and why you come up with those answers. Ive shown my students how to use the program during some of our lessons. A couple of them even bought the program to help them out with their algebra homework. Lisa Schuster, NY Vow! What a teacher! Thanks for making Algebra easy! The software provides amazing ways to deal with complex problems. Anyone caught up there and finds it hard to solve out must buy a copy. Youll get a great tool at a reasonable price. Tina Washington, TX I recommend the Algebrator to students who need help with fractions, equations and algebra. The program is a great tool! 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The GCF of two non-zero integers a and b is the largest positive integer that divides both a and b without remainder. • numeric expressions grade 4 worksheet • power of a fraction • use the ladder method to find lcm • grade 9 end of year math review sheets • nth term fractions calculator • rationAL exponent solver • simplifying radical expressions calculator free • algebra with pizzazz • the distance and midpoint formula • how to teach my son houghton mifflin science grade 5 • polynomial word problems • online absolute value equation calculator • DEMO ALGEBRATOR • how to find scale factor measurements • binomal theorem in c++ • balancing equation sheets • fractions excel formula • new converts class printables • Matlab solution nonlinear differential equations • A Hungerford s Algebra Solutions Manual • practice test in advance algebra • Practice Sats Papers Printable for year 4	1,380	6,267	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.640625	3	CC-MAIN-2019-04	longest	en	0.849014
https://www.jiskha.com/display.cgi?id=1249089262	1,516,131,700,000,000,000	text/html	crawl-data/CC-MAIN-2018-05/segments/1516084886639.11/warc/CC-MAIN-20180116184540-20180116204540-00666.warc.gz	919,781,625	3,921	# geometry posted by . M is the midpoint of segment AB, AM=2x+4 and AB=12x+4. What is the value of MB? Show your work for credit. • geometry - I don't need the credit, you do. If M is the midpoint, AM must equal MB, and both must be half of AB. Tey tell you that AB = 12x = 4. Therefore 2x + 4 = 6x + 2 Solve for x and complete the problem. • geometry (typo correction) - They tell you that AB = 12x + 4. Therefore 2x + 4 = 6x + 2 x = 0.5 AM = MB = 5 ## Similar Questions 1. ### geometry User: Courtney Thompson In Course: Geometry V8 ( 2167) Instructor: Tracie Morgan -------------------------------------------------------------------------------- WARNING: You must not leave this exam form! If you try to click back … 2. ### geometry B is the midpoint of Symbol for segment AC. AB=3x-6 BC=2x+2 What is the value of AB? 3. ### Geometry B is the midpoint of Symbol for segment A C. . AB=5x+2 AC=12x-2 What is the value of BC? 4. ### geometry M is the midpoint of Symbol for Segment A B.. Find the coordinates of B given A(7,-2) and M(10,10). Show your work for credit 5. ### Geometry Please write a paragraph proof for this statement. 30: Point Y is the midpoint of segment XZ. Z is the midpoint of segment YW. PRove that segment or line XY is congruent to segment or line ZW. 6. ### Geometry Please write a paragraph proof for this statement. Point Y is the midpoint of segment XZ. Z is the midpoint of segment YW. PRove that segment or line XY is congruent to segment or line ZW. 7. ### geometry if a pointis onthe perpendicular bisector of a segment,then it is:A. the midpoint of the segment,B.equidistant from the endpoints of the segment. C. on the segment. D.equidistant from the midpoint and one endpoint of the segment. 8. ### geometry Draw and label the following: segment AB intersects segment CD at point M. M is the midpoint of both segments. If AM = x2 + 8x + 17 and MB = 12x + 14, find AB. 9. ### geometry One endpoint and the midpoint of a segment are given. Find the coordinates of the other endpoint. Show all your work. A. Endpoint: (-1,9) Midpoint: (-9,-10) Endpoint: _______ B. Endpoint: (9,-10) Midpoint: (4,8) Endpoint: ________ 10. ### Geometry 1) Explain how you can use a straightedge and a compass to construct an angle that is both congruent and adjacent to a given angle. 2) When constructing a perpendicular bisector, why must the compass opening be greater than the 1/2 … More Similar Questions	667	2,447	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05	latest	en	0.826509
http://www.reference.com/browse/acceleration	1,432,505,144,000,000,000	text/html	crawl-data/CC-MAIN-2015-22/segments/1432207928078.25/warc/CC-MAIN-20150521113208-00282-ip-10-180-206-219.ec2.internal.warc.gz	699,569,909	19,854	Definitions Nearby Words # acceleration [ak-sel-uh-rey-shuhn] acceleration, change in the velocity of a body with respect to time. Since velocity is a vector quantity, involving both magnitude and direction, acceleration is also a vector. In order to produce an acceleration, a force must be applied to the body. The magnitude of the force F must be directly proportional to both the mass of the body m and the desired acceleration a, according to Newton's second law of motion, F=ma. The exact nature of the acceleration produced depends on the relative directions of the original velocity and the force. A force acting in the same direction as the velocity changes only the speed of the body. An appropriate force acting always at right angles to the velocity changes the direction of the velocity but not the speed. An example of such an accelerating force is the gravitational force exerted by a planet on a satellite moving in a circular orbit. A force may also act in the opposite direction from the original velocity. In this case the speed of the body is decreased. Such an acceleration is often referred to as a deceleration. If the acceleration is constant, as for a body falling near the earth, the following formulas may be used to compute the acceleration a of a body from knowledge of the elapsed time t, the distance s through which the body moves in that time, the initial velocity vi, and the final velocity vf: a=(vf2-vi2)/2s a=2(s-vit)/t2 a=(vf-vi)/t For the waltz composed by Johann Strauss, see Accelerationen. In kinematics, acceleration is defined as the first derivative of velocity with respect to time (that is, the rate of change of velocity), or equivalently as the second derivative of position. It is a vector quantity with dimension L T−2. In SI units, acceleration is measured in metres per second squared (m/s2). In common speech, the term acceleration is only used for an increase in speed (the magnitude of velocity); a decrease in speed is called deceleration. In physics, any increase or decrease in speed is referred to as acceleration, and also a change in the direction of velocity is an acceleration (the centripetal acceleration; whereas the rate of change of speed is the tangential acceleration). In classical mechanics, the acceleration of a body is proportional to the resultant (total) force acting on it (Newton's second law): $mathbf\left\{F\right\} = mmathbf\left\{a\right\} quad to quad mathbf\left\{a\right\} = mathbf\left\{F\right\}/m$ where F is the resultant force acting on the body, m is the mass of the body, and a is its acceleration. ## Tangential and centripetal acceleration The acceleration of a particle can be written as: $mathbf\left\{a\right\} = frac\left\{mathrm\left\{d\right\}v\right\}\left\{mathrm\left\{d\right\}t\right\} mathbf\left\{u\right\}_mathrm\left\{t\right\} + frac\left\{v^2\right\}\left\{R\right\}mathbf\left\{u\right\}_mathrm\left\{n\right\}$ where ut and un are (respectively) the unit tangent vector and the unit normal vector to the particle's trajectory, and R is its radius of curvature. These components are called the tangential acceleration and the centripetal acceleration, respectively. ## Relation to relativity After completing his theory of special relativity, Albert Einstein realized that forces felt by objects undergoing constant proper acceleration are indistinguishable from those in a gravitational field. This was the basis for his development of general relativity, a relativistic theory of gravity. This is also the basis for the popular twin paradox, which asks why one twin ages less when moving away from his sibling at near light-speed and then returning, since the non-aging twin can say that it is the other twin that was moving. General relativity solved the "why does only one object feel accelerated?" problem which had plagued philosophers and scientists since Newton's time (and caused Newton to endorse absolute space). In special relativity, only inertial frames of reference (non-accelerated frames) can be used and are equivalent; general relativity considers all frames, even accelerated ones, to be equivalent. (The path from these considerations to the full theory of general relativity is traced in the introduction to general relativity.)	974	4,275	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-22	latest	en	0.926675
http://applevisaservices.com/cuba/question-how-does-k-relate-to-delta-g.html	1,721,848,541,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763518427.68/warc/CC-MAIN-20240724171328-20240724201328-00377.warc.gz	2,038,104	21,728	# Question: How Does K Relate To Delta G? ΔG° is related to K by the equation ΔG°=−RTlnK. If ΔG° 0, then K 1, and products are favored over reactants at equilibrium. If ΔG° = 0, then K = 1, and the amount of products will be roughly equal to the amount of reactants at equilibrium. ## What is K when Delta G is negative? If ΔG is negative, then K>1, which means that the reaction will be spontaneous in the forward direction when all species are present in standard concentrations (1 bar for gases, 1 M for solutes). ## Is K positive delta G? When delta Go is positive, the reaction is not spontaneous because it requires the input of energy at standard conditions. K is therefore less than one because the reaction favors the reactants. If delta Go is 0, than the reaction is at equilibrium, and k must equal 1. ## How does equilibrium constant relate to Gibbs free energy? You will recall that the relative concentrations of reactants and products in the equilibrium state is expressed by the equilibrium constant. If the sum of the standard Gibbs energies of the products is less than that of the reactants, ΔG° for the reaction will be negative and the reaction will proceed to the right. You might be interested: FAQ: What Is Ecs Cluster? ## Why is an equilibrium constant in direct relationship with Delta G? A spontaneous reaction has a negative delta G and a large K value. A non-spontaneous reaction has a positive delta G and a small K value. When delta G is equal to zero and K is around one, the reaction is at equilibrium. This relationship allows us to relate the standard free energy change to the equilibrium constant. ## What is K in a spontaneous reaction? The change in free energy of a reaction can be expressed in terms of the standard free-energy change and the equilibrium constant K or K p and indicates whether a reaction will occur spontaneously under a given set of conditions. ## How does G affect K? The smaller the magnitude of ΔG‡, the higher the rate constant k, the faster the reaction. ## What is K at equilibrium? In a reaction at equilibrium, the equilibrium concentrations of all reactants and products can be measured. The equilibrium constant (K) is a mathematical relationship that shows how the concentrations of the products vary with the concentration of the reactants. ## Can K be negative? Another way to think of it is that reactions rates are always positive and since k is a proportionality constant that relates some given concentration(s) (which are always positive) with the rate (also always positive), it’s not possible for k to be negative. ## Why is k1 spontaneous? At standard conditions, Q will always be 1. If K > 1, this means K > Q. This means you want to go forward in the reaction to achieve equilibrium. If K < 1, this means K < Q, and the system will spontaneously convert products to reactants. ## Does K 1 at equilibrium? The general rules are: If K>>1, the mixture will be mostly product. If K<<1, the mixture will be mostly reactant. If K is about 1, the reaction will reach equilibrium at some intermediate mixture. You might be interested: Question: What Does Ce Mean On A Hayward Heater? ## Why is Gibbs free energy at equilibrium? The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is not spontaneous as written but is spontaneous in the reverse direction, ΔG > 0. ## How does free energy relate to equilibrium? The balance between reactants and products in a reaction will be determined by the free energy difference between the two sides of the reaction. The greater the free energy difference, the more the reaction will favor one side or the other. ## Is K positive or negative in a spontaneous reaction? The freezing of water is an example of this type of process. It is spontaneous only at a relatively low temperature. Above 273. K, the larger TΔS value causes the sign of ΔG to be positive, and freezing does not occur. ## What is ∆ G when ∆ G is − 257.2 kJ and the pressure of each gas is 0.0358 atm at 25 C CO G ½ o2 G ⇌ co2 G? When ∆G° is −257.2 kJ and the pressure of each gas is 0.0358 atm, ∆G equals −253.1 kJ.	976	4,301	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30	latest	en	0.912236
https://byjusexamprep.com/gate-me/for-actual-lifting-machine	1,720,957,880,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00158.warc.gz	133,053,376	29,195	# For Actual Lifting Machine By BYJU'S Exam Prep Updated on: September 25th, 2023 ## For actual lifting machine, the mechanical advantage is 1. greater than the velocity ratio. 2. less than the velocity ratio 3. equal to the velocity ratio. 4. None of these Ans: option B ## For actual lifting machine, the mechanical advantage is less than the velocity ratio. Solution: Mechanical advantage: In a lifting machine, the ratio of load lifted to the effort applied is known as mechanical advantage. M.A. = Load lifted (W) / Effort applied (P) The efficiency of a lifting machine: it is the ratio of mechanical advantage to the velocity ratio of the lifting machine. Efficiency(η) = Mechanical advantage (M.A.) / Velocity ratio (V.R.) For an ideal lifting machine, the efficiency is 100% i.e. equal to one. Hence, for an ideal lifting machine; Mechanical advantage (M.A.) = Velocity ratio (M.A.) For an actual lifting machine, the efficiency is less than one. Hence; 1>Efficiency(η) 1 > Mechanical advantage (M.A.) / Velocity ratio (V.R.) Mechanical advantage (M.A.) < Velocity ratio (M.A.) From the above expression it is clear that, for the actual lifting machine, the Mechanical advantage is less than the velocity ratio. ### ☛ Related Questions: POPULAR EXAMS SSC and Bank Other Exams GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 help@byjusexamprep.com	366	1,439	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.203125	3	CC-MAIN-2024-30	latest	en	0.819993
https://converter.ninja/time/years-to-weeks/731-yr-to-wk/	1,610,744,919,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703496947.2/warc/CC-MAIN-20210115194851-20210115224851-00729.warc.gz	285,083,912	5,743	# 731 years in weeks ## Conversion 731 years is equivalent to 38141.7210441301 weeks.[1] ## Conversion formula How to convert 731 years to weeks? We know (by definition) that: $1\mathrm{yr}\approx 52.177457\mathrm{wk}$ We can set up a proportion to solve for the number of weeks. $1 ⁢ yr 731 ⁢ yr ≈ 52.177457 ⁢ wk x ⁢ wk$ Now, we cross multiply to solve for our unknown $x$: $x\mathrm{wk}\approx \frac{731\mathrm{yr}}{1\mathrm{yr}}*52.177457\mathrm{wk}\to x\mathrm{wk}\approx 38141.721067\mathrm{wk}$ Conclusion: $731 ⁢ yr ≈ 38141.721067 ⁢ wk$ ## Conversion in the opposite direction The inverse of the conversion factor is that 1 week is equal to 2.62180093772642e-05 times 731 years. It can also be expressed as: 731 years is equal to $\frac{1}{\mathrm{2.62180093772642e-05}}$ weeks. ## Approximation An approximate numerical result would be: seven hundred and thirty-one years is about thirty-eight thousand, one hundred and forty-one point seven two weeks, or alternatively, a week is about zero times seven hundred and thirty-one years. ## Footnotes [1] The precision is 15 significant digits (fourteen digits to the right of the decimal point). Results may contain small errors due to the use of floating point arithmetic.	351	1,245	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.890625	4	CC-MAIN-2021-04	latest	en	0.794699
https://www.cheenta.com/math-kangaroo-benjamin-2016-problem-24/	1,713,148,430,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296816939.51/warc/CC-MAIN-20240415014252-20240415044252-00221.warc.gz	659,473,928	28,349	How Cheenta works to ensure student success? Explore the Back-Story # Math Kangaroo (Benjamin) 2016 Problem 24 \| Play With Numbers Try this Problem based on Playing With Numbers from Math Kangaroo (Benjamin) 2016 Problem 24 ## Playing With Numbers \| Math Kangaroo (Benjamin) 2016 \| Problem 24 Two three-digit numbers are made up of six different digits. The first digit of the second number is twice as big as the last digit of the first number. (Note: 0 is also a digit but cannot be the first digit of a number!) How big is the smallest possible sum of the two numbers? Numbers Arithmetic Counting ## Suggested Book \| Source \| Answer Mathematical Circle Math Kangaroo (Benjamin) 2016 \| Problem 24 537 ## Try with Hints Let us assume these three digit numbers are $ABC$, $DEF$. According to the question $D=2C$. Let's follow the given condition and try to construct the smallest numbers. So here $ABC=102$. And if I follow the given condition then $DEF= 435$. We did this keeping in mind that repetitions are not allowed.	261	1,040	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.78125	4	CC-MAIN-2024-18	latest	en	0.873407
https://bobistheoilguy.com/forums/threads/true-or-false.109134/	1,716,118,876,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971057786.92/warc/CC-MAIN-20240519101339-20240519131339-00403.warc.gz	120,357,683	17,705	# True or false? Status Not open for further replies. #### wavinwayne Assuming that the clamping method, steel type, and all other conditions are equal, the amount of force required to bend a 1" diameter X 36" long steel shaft is the same, regardless of the case hardness of the shaft. (Shaft size given is just a number I picked at random) Originally Posted By: wavinwayne Assuming that the clamping method, steel type, and all other conditions are equal, the amount of force required to bend a 1" diameter X 36" long steel shaft is the same, regardless of the case hardness of the shaft. (Shaft size given is just a number I picked at random) Within the realm of elastic deformation for both, the same. The case hardened shaft will require more force to bend it into the plastic deformation region. I take it then that the offering is "false" as presented? ..or is it assumed that you're required to over think it? I don't know anything about the subject matter, but ever True:False test that I run into has two or three "trues" when you assume depth beyond the presented conditions. Originally Posted By: Gary Allan I take it then that the offering is "false" as presented? ..or is it assumed that you're required to over think it? I don't know anything about the subject matter, but ever True:False test that I run into has two or three "trues" when you assume depth beyond the presented conditions. The question wasn't asked in a way that allowed a meaningful true/false answer. I suspect the answer the questioner wanted was "false". XS650 is correct on both posts. There is no singular answer to the open-ended question. The real answer is "it depends". ..but it's not a question. It's a statement. The only question is whether it's true or false. You're saying that the statement is lacking in information. From XS650's knowledge on the topic. I'd say that it's true. The statement didn't offer the condition of being in the "once the threshold of deformation is breached" realm. Unless I misread what he responded with. Did the hardening and crystal structure change have zero impact on the density of the shaft? It's a horribly stated question. XS650 got it, but I'm trying to nit pick. Originally Posted By: oilyriser Did the hardening and crystal structure change have zero impact on the density of the shaft? Density has no direct effect on the topic being discussed. The property that was indirectly being questioned was Young's Modulus. It's quite constant for a wide variety of steels and even more independent of hardness in any real world sense. To answer your question, if you case harden a precision machined 1.000 shaft 0.100 inches deep, it will no longer be a precision machined shaft I don't have any numbers for you though. Edit: a brief Google encounter indicates that I may have been too pessimistic. I saw some allusions to fairly precise dimensional control in case hardening. I doubt that that is just packing a red hot part in ground up bones and dried urine though. Last edited: Would there not be a point at which the case hardening would cause the shaft to be brittle and then break upon bending? heat treat additional to case hardening would do that, but I think that case hardening is the only variable being discussed. I'd hypothesise that case hardening could add additional compressive stresses that would make the shaft a little bit stiffer than a non case hardened sample. Just a guess. And when did Kestas become a mod ? congrats mate. Does adding carbon to steel affect its modulus at all? I suspect it would make a very tiny difference. Carbon-iron bonds probably behave differently from iron-iron bonds. But if both rods are chemically identical, stiffness should be the same in the elastic region. Status Not open for further replies.	832	3,812	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.65625	3	CC-MAIN-2024-22	latest	en	0.954761
https://www.reference.com/science/convert-kilograms-centigrams-91e94e4322a504bc	1,547,937,671,000,000,000	text/html	crawl-data/CC-MAIN-2019-04/segments/1547583684033.26/warc/CC-MAIN-20190119221320-20190120003320-00504.warc.gz	923,089,628	17,270	How Do You Convert Kilograms to Centigrams? # How Do You Convert Kilograms to Centigrams? To convert kilograms to centigrams, simply multiply the number of kilograms by 100,000 to arrive at the equivalent value in centigrams. This may require either a calculator or a pen and paper. 1. Note the number of kilograms you wish to convert The easiest way is to start by writing (or inputting into a calculator) the value that represents the number of kilograms you wish to convert to centigrams. 2. Multiply the number of kilograms by the centigram conversion factor Multiply the number of kilograms by the centigram conversation factor of 100,000. Written as an equation, the process would look like this: (x kilograms)(100,000 centigrams per kilogram) = y centigrams.	167	771	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.296875	3	CC-MAIN-2019-04	latest	en	0.774954
https://www.doubtnut.com/question-answer/the-surface-are-of-a-balloon-being-inflated-changes-at-a-constant-rate-if-initially-its-radius-is3-u-642567382	1,656,747,116,000,000,000	text/html	crawl-data/CC-MAIN-2022-27/segments/1656103989282.58/warc/CC-MAIN-20220702071223-20220702101223-00426.warc.gz	799,490,867	92,947	Home > English > Class 12 > Maths > Chapter > Differential Equation > The surface are of a balloon b... Updated On: 27-06-2022 Get Answer to any question, just click a photo and upload the photo and get the answer completely free, Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams. Transcript the question is the surface area of a balloon being inflated changes at a constant rate if initially its registered 3 units and and after 2 seconds it is a 5 minutes find the radius of after 30 seconds to now latest new army the radius of the balloon and SP the surface area now the surface area of a square and we have given that surface area of balloon being infected changes at a constant rate so we DS by b equals to ke this implies DS = 2 ke chana integrate both sides and we get vs post to Ke Beti this implies equals to kehti + see now the surface areas for a square so we have four square equals to Kitty + c now we have given initially its radius is 3 units that is equals to zero Hour 3 these values in this equation so we get into 3 square is 9 = 2 ke in 20 + c is employees sequels 236 buy now we have given that after 2 seconds it it is a 5 units that is at equals to 2 hours equals to fine now we have if we put the value of c in this equations of b square equals to 80 + 36 5 and now we put these values we get over by 5 square equals to ke into 2 + 36 by Sobi have worked into 25 equals to 2 k + 36 by this implies took a = 200 - 36 by was 265 which implies cake was to active 2 price so we have four square equals to 32512 T + 36 by not divide the equation by 4.2 we get R squared equals to 1 + 9 now equals to square root 1 t + 9 India	435	1,715	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27	latest	en	0.953165
http://www.roseindia.net/discussion/49120-Conversion-of-Decimal-to-Binary--in-Java.html	1,406,116,504,000,000,000	text/html	crawl-data/CC-MAIN-2014-23/segments/1405997877881.80/warc/CC-MAIN-20140722025757-00217-ip-10-33-131-23.ec2.internal.warc.gz	1,083,902,753	9,847	Home Discussion Discuss: Conversion of Decimal to Binary in Java Subject (*): Related Articles Java Beginners BINARY TO DECIMAL - Java Beginners BINARY TO DECIMAL HELP ME PLSS. PROGRAM THAT CONVERTS BINARY TO DECIMAL Hi friend, Program to convert Binary To Decimal : import... " + decimal); } } For read information on Java : http Java Beginners Binary to decimal - Java Beginners Binary to decimal Need help pls.. i cannot run this program... pls... = Integer.parseInt(JOptionPane.showInputDialog("Input Binary:")); String c = args[0...; String value = JOptionPane.showInputDialog("Input Binary"); int len Java Tips and Examples Conversion of Decimal to Binary in Java Conversion of Decimal to Binary in Java In this section we will discuss about conversion of Decimal to Binary in Java. You know that number having...(Binary equivalent of 12). Example : Using this example we pass a decimal Java Tips and Examples Convert Decimal into Binary ; a decimal number into a binary number. Description of program: This program takes a decimal number from console and it converts it into binary format... the Decimal number into binary number use toBinaryString() method. General Java binary to decimal Java binary to decimal This Tutorial helps you to know the code for Java binary... you in understanding a how to get a 'Java binary to decimal'. For this we have Java Tips and Examples Convert Binary to Decimal Convert Binary to Decimal ... to convert the integer data into the binary to decimal. Code Description... the Binary value: 10010 Decimal:=18 C:\corejava> Programming Tutorials Java Convert decimal to binary using Stacks Java Convert decimal to binary using Stacks In this section, you will learn how to convert decimal into binary using stacks. You all are aware of Stacks... have converted a decimal number into binary. Here is the code: import Swing AWT swing program for decimal to binary swing program for decimal to binary swing program for decimal to binary IDE Questions binary binary Hi I want to write a program in pascal that ask a user to input a decimal number and then return its binary equivalent in the minimum number of bits required to repesent the number. Thks Java Beginners I converting binary to decimal and vice versa I converting binary to decimal and vice versa can i ask for the codes in converting binary to decimal and decimal to binary to be run in one program with repeat.. pls. help me! i badly need the codes by monday!! pls Java Code Examples Java Convert Octal to Binary Java Convert Octal to Binary In this tutorial, you will learn how to convert octal to binary. Java has provide different ways to change the number system of different numbers. You can convert and decimal to octal, decimal to binary Java Beginners Java Beginners convert to decimal convert to decimal f42a4 convert to decimal by using java Java Beginners Decimal Format Issue Java Decimal Format Issue Java Decimal Format Issue Java Java Tips and Examples Convert Number to Binary binary. Code Description: This program takes a decimal number at the console... Convert Number to Binary In this section, you will learn to convert a number to a binary (0,1 Java Tips and Examples :\corejava>java BinaryToHexa Enter the Binary number: 1111111 Hexa... Convert Binary to Hexadecimal In this section, you will learn to convert binary data General Java Conversion Java Conversion ... will learn to convert decimal number into binary. The java.lang package provides the functionality to convert a decimal number into a binary number. Java Beginners binary addition,subtraction and modulus operations - Java Beginners operations on these two binary numbers and finally dipaly the result in decimal... adder binary logic in this!!! please send me the java coding for these three...binary addition,subtraction and modulus operations i wanna perform IDE Questions program binary program binary Hi I want to write a program in pascal that ask a user to input a decimal number and then return its binary equivalent in the minimum number of bits required to repesent the number. In pascal language please Java Beginners binary tree ://www.roseindia.net/java/java-get-example/java-binary-tree-code.shtml http://www.roseindia.net/java/java-get-example/java-binary-tree-insert.shtml...binary tree can a binary tree be implemented with out comparing Java Beginners octal to binary octal to binary i want to know how to convert a octal to binary number Here is a java example that converts octal to binary. import... binary=Integer.toBinaryString(i); System.out.println(binary Java Beginners Binary Search - Java Beginners Binary Search how to use Binary think in java give me the Binary Search programm thx Java Beginners java program to convert decimal in words java program to convert decimal in words write a java program to convert a decimal no. in words.Ex: give input as 12 output must be twelve General Java Beginners Java read binary file I want Java read binary file example code... at Reading binary file into byte array in Java. Thanks Hi, There is many more examples at Java File - Learn how to handle files in Java with Examples Java Beginners How to round a number to n decimal places in Java How to round a number to n decimal places in Java How to round a number to n decimal places in Java Java Beginners java binary file io example java binary file io example java binary file io example Java Tips and Examples Convert Decimal to Hexadecimal In this section, you will learn to convert a decimal data into hexadecimal. The java.lang package provides the functionality to convert a decimal Java Tips and Examples	1,232	5,736	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-23	longest	en	0.671253
https://www.markedbyteachers.com/gcse/maths/edexcel-gcse-statistics-coursework.html	1,638,558,024,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964362918.89/warc/CC-MAIN-20211203182358-20211203212358-00105.warc.gz	977,798,074	26,251	• Join over 1.2 million students every month • Accelerate your learning by 29% • Unlimited access from just £6.99 per month Page 1. 1 1 2. 2 2 3. 3 3 4. 4 4 5. 5 5 6. 6 6 7. 7 7 8. 8 8 9. 9 9 10. 10 10 11. 11 11 12. 12 12 13. 13 13 14. 14 14 • Level: GCSE • Subject: Maths • Word count: 5834 Edexcel GCSE Statistics Coursework Extracts from this document... Introduction Edexcel GCSE Statistics 1389 PLANNING SHEET – MAYFIELD HIGH Student Name: Anya Sweilam Class: 11H3 This investigation is based upon the students of Mayfield High School, a fictitious school- there are 1182 students at Mayfield presented within 13 categories. I will be investigating the relationship between height and weight and how these statistics differ between females and males. I have chosen to look at height and weight mainly because in this line of enquiry my data will be numerical and continuous, meaning that I will be able to produce a more detailed analysis. For example, if I had chosen to look at eye colour and hair colour my analysis would be limited and therefore my investigation may be imprecise. My aim in this investigation is to query whether or not there is a correlation between height and weight and find out if this varies between genders. I believe that as a student becomes taller their weight will increase; due to this assumption I expect a graph of weight and height to show a rising trend. Listed below are my hypotheses. The height and weight of a person is affected by their age and gender. I assume that in years 7-9 girls will generally be taller than boys- this is because girls tend to grow faster than boys during the early stages of development. Boys will, however, eventually grow taller and so in years 10-11 it can be assumed the number boys taller than girls will be greater. This also applies to adults aged 20 and above. As for the weight, boys are generally heavier than girls; this is due to their body structure. I, therefore, predict that my results will produce a pattern which shows that boys weigh more than girls. ...read more. Middle 1 30 0.2 75 With the cumulative frequency graph displaying weight, the female’s data produces an almost perfect S-shape curve, whereas the male’s data has, what seems to be, an anomaly (the third point allocated at the weight of 45KG and cumulative frequency of 9) which affects its shape. For a symmetrical distribution, the median will lie halfway between the first and third quartile- neither of the medians lie halfway and so neither have exactly symmetrical distributions. The female’s median, however, is extremely close to being halfway between the two quartiles showing us a more symmetrical distribution than that of the males; this may explain the almost perfect curve on the frequency graph which the points plotted for females produce. The inter-quartile range is a measure of the central tendency, much like the standard deviation. The advantage of the inter-quartile range over the standard deviation, however, is that the inter-quartile range includes half of the points regardless of the shape of the distribution. The smaller the inter-quartile range, the more consistent the data is. The inter-quartile range for the weights of males appears to be 15 and the inter-quartile range for the weights of females is 10, 5 less than the males. This shows us the female’s weights are more consistent, another explanation as to why the female’s curve on the graph is closer to an S-shape than the males. Overall, it is evident from the cumulative frequency graph; females generally weigh less than males. Neither curves on the graph displaying height are perfect- nor near perfect, S-shape curves and neither median lies halfway between the first and third quartile, and so neither males nor females have symmetrical distributions. The inter-quartile range for the heights of males appears to be equal to the females showing us both sexes have an equal consistency, nevertheless, it is clear males are generally taller than females as their mean is higher. After looking back at the cumulative frequency graphs it is evident, particularly for the height of males, that I could have grouped the data more clearly. The third and fourth row in the group of male heights show a frequency of 0, which has an effect on the S-shape of the curve on my graph, and possibly having an effect on the lower quartile. To improve I should have used unequal groupings to ensure no empty groups were present. Box plots are an informative way to display a range of numerical data. It can show many things about a data set, like the lowest term in the set, the highest term in the set, the median, the upper quartile, and the lower quartile. Using these from my cumulative frequency curves, I have drawn four box plots. Outliers are not present in every box plot drawn, except one where there is an extreme value which deviates significantly from the rest of the sample. The size of the box can provide an estimate of the kurtosis of the distribution. A thin box relative to the whiskers indicates that a very high number of cases are contained within a very small segment of the sample indicating a distribution with a thinner peak whereas a wider box is indicative of a wider peak and so, the wider the box, the more U-shaped the distribution becomes. Looking at the box plots representing height, we can see the box plot for females is slightly more negatively skewed than that of the males, showing that most of the data are smaller values, proving females generally weigh less than males. The medians lie at the same point- 1.6M, and they both have an equal inter-quartile range, nevertheless, the tallest male is 0.5M taller than the tallest female. As both boxes are of equal size both distributions are equally U-shaped. The box represents the middle 50% of the data sample- half of all cases are contained within it. The 50% of data within the box for the males ranges between 1.55M and 1.7M whereas for the females it ranges between 1.5M and 1.65M, showing us females are generally shorter than males. Looking at the box plots for weight, we see that half the female's weights are between 45 and 55KG whereas half the men's weights lie between 45 and 60KG. The highest value for females is 70KG (ignoring the outlier) and for males: 75KG, the median for the males’ weight is 5KG higher than that of the females. The lowest value which appears on the box plot for males is 30KG and the highest is 75KG, giving us a range of 45KG. Looking at the same pieces of data for the females, we can work out that the range is in fact 5KG less than that of the males. It is evident that the distribution of the female’s box plot has a thinner peak than the males attributable to the simple fact that the box of the female’s weight is far thinner than the males’. The distribution for the weight of males is, therefore, more U-shaped. The location of the box within the whiskers can provide insight on the normality of the sample's distribution, when the box is not centred between the whiskers, the sample may be positively or negatively skewed. If the box is shifted significantly to the low end, it is positively skewed; if the box is shifted significantly to the high end, it is negatively skewed, however, none of the four box plots are shifted significantly to either the high end or the low end. Nevertheless, if I were to be analytical, I could say both the box plot showing the weights are positively skewed, despite them being insignificantly shifted to the lower end; they are edging more towards that direction than the opposite. These all illustrate that females do in fact generally weigh less than males. An outlier appears on the box plot showing the weights of females, this may be the result of an error in measurement, in which case it will distort the interpretation of the data, having undue influence on many summary statistics- for example: the mean, however, if the outlier is a genuine result, it is important because it may perhaps indicate an extreme of behaviour or may have been affected by external behaviour, for example, dietary habits. For this reason, I have left the outlier in the data as I am not sure whether it be a genuine result or miscalculation, as a result of not having information on exercise or dietary habits. To conclude, it is construable that my hypothesis was in fact correct. It is evident from all the graphs included that females are, in effect, generally shorter and weigh less than males. Whether this is attributable to, as studies show, the varied skeletons of the opposed sexes or the dissimilar hormones produced in both female and male bodies, it is known females are generally shorter and weigh less than males. When the average male and female both reach the age of 20 it is said ‘females are generally 10 percent shorter than males and 20 per cent lighter’ and between the ages of 11 and 16 ‘males appear to generally be 15 percent taller and heavier than the female sex’. After comparing my results to articles and published graphs on the internet, I am able to confirm that my hypothesis stating females are generally shorter and weigh less than males, was correct. Hypothesis 2: After calculating the frequency density for the male and female heights and weights, I created four histograms; the advantage of a histogram is that it shows the shape of the distribution for a large set of data and so was therefore able to show me the shape of the distributions for male and female heights and weights, however, when using histograms it is more difficult to compare two or more data sets as we are unable to read exact values as the data is grouped into categories. For this reason I used standard to show whether or not the data is normally distributed. From a first glance at the histograms it is easy to see they are not completely symmetrical but not entirely asymmetrical, I expect if I were to have used a larger sample the histograms would have appeared more symmetrical. Tables in which I used to create the histograms Females Height (M) Frequency FD Upper-Class 1 Weight (KG) Frequency FD Upper-Class 30 Males Height (M) Frequency FD Upper-Class 1 From looking at the histograms, it is clear only two of these encompass curves which are appropriate to super impose normal distribution curves, and so for this reason I will not calculate the normal distribution. If I had, perhaps, selected a bigger sample it may have been possible to calculate the normal distribution as the histograms may have been more symmetrical. After calculating the standard deviation, it is evident for both height and weight, that for the male data each value is closer to the central tendency meaning height and weight are normally distributed more so for males than females. Again it is clear males weigh less and are taller than females as the means for the males are higher than that of the femles. Gender Weight (x) x - 47.9 (x-47)SQD Gender Height (x) x - 1.56 (x-1.56)SQD Female 45 -2.9 8.41 Female 1.03 -0.53 1.0609 Female 51 3.1 9.61 Female 1.35 -0.21 1.8225 Female 30 -17.9 320.41 Female 1.42 -0.14 2.0164 Female 40 -7.9 62.41 Female 1.43 -0.13 2.0449 Female 45 -2.9 8.41 Female 1.43 -0.13 2.0449 Female 45 -2.9 8.41 Female 1.46 -0.1 2.1316 Female 44 -3.9 15.21 Female 1.47 -0.09 2.1609 Female 45 -2.9 8.41 Female 1.5 -0.06 2.25 Female 50 2.1 4.41 Female 1.55 -0.01 2.4025 Female 50 2.1 4.41 Female 1.55 -0.01 2.4025 Female 60 12.1 146.41 Female 1.55 -0.01 2.4025 Female 45 -2.9 8.41 Female 1.56 0 2.4336 Female 45 -2.9 8.41 Female 1.57 0.01 2.4649 Female 48 0.1 0.01 Female 1.59 0.03 2.5281 Female 56 8.1 65.61 Female 1.6 0.04 2.56 Female 52 4.1 16.81 Female 1.61 0.05 2.5921 Female 42 -5.9 34.81 Female 1.62 0.06 2.6244 Female 48 0.1 0.01 Female 1.62 0.06 2.6244 Female 49 1.1 1.21 Female 1.62 0.06 2.6244 Female 49 1.1 1.21 Female 1.62 0.06 2.6244 Female 54 6.1 37.21 Female 1.62 0.06 2.6244 Female 38 -9.9 98.01 Female 1.63 0.07 2.6569 Female 47 -0.9 0.81 Female 1.63 0.07 2.6569 Female 48 0.1 0.01 Female 1.63 0.07 2.6569 Female 49 1.1 1.21 Female 1.65 0.09 2.7225 Female 54 6.1 37.21 Female 1.65 1.65 2.7225 Female 50 2.1 4.41 Female 1.68 1.68 2.8224 Female 42 -5.9 34.81 Female 1.71 1.71 2.9241 Female 60 12.1 146.41 Female 1.72 1.72 2.9584 Female 56 8.1 65.61 Female 1.75 1.75 3.0625 Total 1437 0 1158.7 Total 46.82 7.82 73.6234 SD= 211.5 SD= 13.4 ...read more. Conclusion 54 3.3 10.89 Male 1.77 1.77 3.1329 Male 50 -0.7 0.49 Male 1.83 1.83 3.3489 Male 60 9.3 86.49 Male 1.85 1.85 3.4225 Male 75 24.3 590.49 Total 47.56 9.5 19.0888 Total 50.7 0 3422.3 SD= 3.49 SD= 624.9 Gender Height (M) Weight (Kg) Height Rank Weight Rank Diff Diff^2 female 1.42 30 3 1 2 4 female 1.63 38 23 2 21 441 female 1.43 40 4.5 3 1.5 2.25 female 1.62 42 19 4.5 14.5 210.25 female 1.71 42 28 4.5 23.5 552.25 female 1.47 44 7 6 1 1 female 1.03 45 1 9.5 -8.5 72.25 female 1.43 45 4.5 9.5 -5 25 female 1.46 45 6 9.5 -3.5 12.25 female 1.5 45 8 9.5 -1.5 2.25 female 1.56 45 12 9.5 2.5 6.25 female 1.57 45 13 9.5 3.5 12.25 female 1.63 47 23 13 10 100 female 1.59 48 14 15 -1 1 female 1.62 48 19 15 4 16 female 1.63 48 23 15 8 64 female 1.62 49 19 18 1 1 female 1.62 49 19 18 1 1 female 1.65 49 25.5 18 7.5 56.25 female 1.55 50 10 21 -11 121 female 1.55 50 10 21 -11 121 female 1.68 50 27 21 6 36 female 1.35 51 2 23 -21 441 female 1.61 52 16 24 -8 64 female 1.62 54 19 25 -6 36 female 1.65 54 25.5 25 0.5 0.25 female 1.6 56 15 27.5 -12.5 156.25 female 1.75 56 30 27.5 2.5 6.25 female 1.55 60 10 29.5 -19.5 380.25 female 1.72 60 29 29.5 -0.5 0.25 Summation Sum 2942.5 6 * sum 17655 Count n 30 n(n^2-1) 26970 Spearman's R 0.4 Gender Height (M) Weight (KG) Height Rank Weight Rank Diff Diff^2 male 1.42 26 5 1 4 16 male 1.41 31 4 2 2 4 male 1.32 38 2 3 -1 1 male 1.6 38 15 3 12 144 male 1.63 40 18 5 13 169 male 1.65 41 21 6 15 225 male 1.34 42 3 7.5 -4.5 20.25 male 1.44 42 6 7.5 -1.5 2.25 male 1.26 44 1 9 -8 64 male 1.46 45 7 10 -3 9 male 1.66 46 24 11 13 169 male 1.55 50 9 13 -4 16 male 1.58 50 13 13 0 0 male 1.68 50 25 13 12 144 male 1.59 52 14 17.5 -3.5 12.25 male 1.57 54 11.5 17.5 -6 36 male 1.57 54 11.5 17.5 -6 36 male 1.65 54 21 17.5 3.5 12.25 male 1.65 54 21 17.5 3.5 12.25 male 1.77 54 28 17.5 10.5 110.25 male 1.65 55 21 21.5 -0.5 0.25 male 1.85 55 30 21.5 8.5 72.25 male 1.62 56 17 23 -6 36 male 1.73 57 26.5 24 2.5 6.25 male 1.6 60 15 25.5 -10.5 110.25 male 1.73 60 26.5 25.5 1 1 male 1.55 64 9 27 -18 324 male 1.55 65 9 28 -19 361 male 1.65 69 21 29 -8 64 male 1.83 75 29 30 -1 1 Summation Sum 2178.5 6 * sum 13071 Count n 30 n(n^2-1) 26970 Spearman's R 0.5 After calculating the spearman’s rank it is evident there is a correlation between height and weight, and the taller the person is the heavier they are, vice versa. There is a weak positive correlation between height and weight for females and a moderate positive correlation for males as it is slightly stronger. The height and weight of a person is affected by their age and gender. I assumed that in years 7-9 girls will generally be taller than boys- due to the fact girls tend to grow faster than boys during the early stages of development. Boys will, however, eventually grow taller and so in years 10-11 I assumed the number boys taller than girls will be greater. I was correct. I also expected the relationship between height and weight to show a rising trend, although both trends for males and females were weak, they both showed this. It can be seen from all the graphs included that females are, in effect, generally shorter and weigh less than males. Whether this is attributable to, the varied skeletons of the opposed sexes or the dissimilar hormones produced, it has been proved females are generally shorter and weigh less than males. ...read more. This student written piece of work is one of many that can be found in our GCSE Height and Weight of Pupils and other Mayfield High School investigations section. Found what you're looking for? • Start learning 29% faster today • 150,000+ documents available • Just £6.99 a month Not the one? Search for your essay title... • Join over 1.2 million students every month • Accelerate your learning by 29% • Unlimited access from just £6.99 per month Related GCSE Height and Weight of Pupils and other Mayfield High School investigations essays 1. Maths Statistics Coursework - relationship between the weight and height The number that will be given, I multiply by the total number of either boys or girls for that particular year group, and for example, I would multiply the random number by 131 because this is the total number of girls in year 7. 2. Statistics GCSE Coursework. Height and weight of pupils. The sampling method I am ... so I do have some grounds for further investigation. The answer also proves I have a positive linear correlation between the two variables. Based on the value of r2 (0.42108310812 =0.177310984) therefore I can say that with a 17% chance that from any point on the line of best fit 1. HYPOTHESIS: Boys at Mayfield School are Taller and Weigh more average in comparison to ... whether my two sets of data are related from my sample: I will conclude after both graphs Boys Height vs Weight [Scatter diagram boy's height vs weight] Boys Height vs Weight [Scatter diagram girl's height vs weight] Conclusion of Scatter Graphs Boys & Girls Height vs. 2. mayfield course work -boys are generally heavier than girl. This has to do with ... In the same way, women respond to their problems by smoking. The survey showed that 33% of men and 35% of women currently smoke cigarettes. The percentage of women is more than that of men because more women said they did not have much social support. 1. Maths Data Handling Plan I am trying to show that there is a better correlation between height and weight when age is taken into consideration. I will need to create a new stratified sample from the whole population in Mayfield High School. I already know that this data is reliable and will enable me to get accurate results. 2. I will be testing the following hypothesis in my pilot study: ... Positive More than the mean Negative Less than the mean Zero Equal to the mean 5. Minimum, lower quartile, median, upper quartile, inter-quartile range and maximum for weights (or heights). * Inter-quartile Range - This is also a measure of spread but looks at the spread of the middle 50% of the data around the median. 1. GCSE maths statistics coursework was larger for boys than that of the girls as both went up to 1.9m although the frequency density between 1.7m to 1.9m was larger than the frequency density of the girls. I will now do frequency polygons of the girls and boys weights. 2. Statistics coursework part 1 0 0 1 1 1 1 1 1 2 2 2 2 3 3 3 3 3 44: 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 3 3 3 3 3 3 3 3 48: • Over 160,000 pieces of student written work • Annotated by experienced teachers • Ideas and feedback to improve your own work	5,554	19,043	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.453125	3	CC-MAIN-2021-49	latest	en	0.933164
http://www.longrangehunting.com/threads/clicks-or-moa.41362/	1,502,960,541,000,000,000	text/html	crawl-data/CC-MAIN-2017-34/segments/1502886102993.24/warc/CC-MAIN-20170817073135-20170817093135-00640.warc.gz	606,177,088	18,214	# Clicks or moa ?? Discussion in 'Long Range Scopes and Other Optics' started by JEREMY logan, Apr 16, 2009. 1. ### JEREMY loganGuest I use to count clicks and now I have an hand held that will give me clicks or moa adj. I was wandering what yall use and why , when my hand held says for example come up 30 3/4 moa or 30 1/4 moa and my leupold if I remember correctly 1 full rotation will be 15 moa adj. How do I figure mathematically the difference of 3/4 ,1/4 what's the fastest way in doing this? Thanks for any advice jeremy 2. ### ss7mmWriters Guild Messages: 3,707 Joined: Jun 11, 2005 Your scope is probably marked off in 1/4 moa increments so, 1/4 moa is 1 click past a full increment mark and 3/4 moa is 3 clicks past a full increment mark. If you have to go up 30 3/4 moa just go up 30 moa, which in the case of 15 moa per full turn, would be 2 full turns from zero and then go 3 clicks more and you should be there. For me it's way quicker to dial in moa increments, and full turns, than it is to count clicks. Don't even remember when the last time was that I counted clicks. I just like to do it quick and easy so I can get on with the shot and have fun. 3. ### JEREMY loganGuest Thanks for the reply, sounds easy now that you've explained it sometimes this stuff seems harder than it really is Messages: 2,595 Joined: Feb 4, 2005 5. ### JEREMY loganGuest my scope adjusts 1/4" @ 100yrds , i didn't think that their was a difference . Is their?? 6. ### ss7mmWriters Guild Messages: 3,707 Joined: Jun 11, 2005 1" of scope adjustment at 100 yards is 1". At 1000 yards it's 10" 1moa of scope adjustment is 1.047" at 100 yards. At 1000 yards, 1moa is 10.47". 7. ### JEREMY loganGuest okay ss7mm so when someone says their gun is a 10 MOA gun @ 1000 yrds then that gun shoots inside 10" if the shooter does his part correct. 8. ### ss7mmWriters Guild Messages: 3,707 Joined: Jun 11, 2005 Nope......1 moa at 1000 yards is 10.47" so 10 moa at 1000 yards would be 10x10.47" or 104.7". For a gun to shot 1 moa at 1000 yards it would have to be capable of groups of 10.47" or less. A lot of guns and a lot of people can do it occasionally but to do it every time is something else. Just check the long range target scores. 9. ### JEREMY loganGuest okay thanks ..... 10. ### BountyHunterWriters Guild Messages: 4,799 Joined: Jun 13, 2007 Use MOA. Clicks are iffy and not accurate unless you have checked to confirm what your scope actually is (not what it is marked). Too many people assume because it says 1/4 inch that means it actually is that. Same model scopes have been proven to vary quite a bit. check yours. Darrell Holland has a good article on the articles forum about problems discussing this very issue and exactly how to check and see what your scope is. Dan Lilja has a discussion on this also on his website in his LR article also. BH 11. ### MikecrWell-Known Member Messages: 4,262 Joined: Aug 10, 2003 Sounds like +/-10.47"/MOA from center of mark to me. (104.7inches for 10moa) Last edited: Apr 17, 2009 12. ### bigngreenWell-Known Member Messages: 5,836 Joined: Nov 24, 2008 I have been fighting with getting the trajectories more accurate in Laodbase 2.0 and this week the lightbulb came on that I don't have a MOA scope but a IPHY. Made the change and my predicted trajectories are much better. I need to check my exact amount of come up per rev. though. If one is trying to predict a shot for a hunting situation then using the right value for you scope will make a difference. 13. ### mattjWell-Known Member Messages: 213 Joined: Jun 27, 2007 Not to hijack the thread, but this is one of the reasons I'm working on switching over to all MIL/MIL based optics. While keeping track of 1/4, 1/2, 3/4 MOA isn't THAT hard, it's still one more little thing to keep your head wrapped around. Working in base-10 units is just easier to deal with... you don't have to think of MILs versus clicks, because converting between them in base 10 just doesn't require any mental energy. 3.1 MILs is 31 clicks, or 3 MILs plus one click -- but really you don't have to think about it, because it's just how we're used to seeing numbers -- dialing 3.1 MILs is just... dialing 3.1 MILs. I was resistant to MILs for a while, until I stopped trying to think of it in terms of MOA or tying it to Meters or something. All it is is a ratio, and it works just as well in yards or meters or inches. 1 yard is 1 MIL at 1000 yards. 1 meter is 1 MIL at 1000 meters. 1 inch is 1 inch at 1000 inches 1 yard is 2 MILs at 500 yards. 1 yard is 4 MILs at 250 yards. etc. Once I stopped trying to make it hard by thinking in terms of MOA, I've found it a much simpler system to work in. This article is what started winning me over: Using a Mil Based Scope - Easy Transition \| Sniper's Hide It's also nice not having to worry about "true MOA" vs. IPHY.	1,410	4,880	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.796875	3	CC-MAIN-2017-34	latest	en	0.948331
https://www.numbersaplenty.com/2461136	1,701,439,542,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100287.49/warc/CC-MAIN-20231201120231-20231201150231-00203.warc.gz	1,037,861,480	3,142	Search a number 2461136 = 24193797 BaseRepresentation bin1001011000110111010000 311122001001012 421120313100 51112224021 6124430052 726630216 oct11306720 94561035 102461136 1114310a7 129a8328 136822c2 14480cb6 1533935b hex258dd0 2461136 has 20 divisors (see below), whose sum is σ = 4799172. Its totient is φ = 1222656. The previous prime is 2461103. The next prime is 2461139. The reversal of 2461136 is 6311642. It is a happy number. It can be written as a sum of positive squares in 2 ways, for example, as 883600 + 1577536 = 940^2 + 1256^2 . It is a congruent number. It is not an unprimeable number, because it can be changed into a prime (2461139) by changing a digit. It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 2690 + ... + 3486. 22461136 is an apocalyptic number. It is an amenable number. 2461136 is a deficient number, since it is larger than the sum of its proper divisors (2338036). 2461136 is a wasteful number, since it uses less digits than its factorization. 2461136 is an evil number, because the sum of its binary digits is even. The sum of its prime factors is 998 (or 992 counting only the distinct ones). The product of its digits is 864, while the sum is 23. The square root of 2461136 is about 1568.8008159100. The cubic root of 2461136 is about 135.0139171752. Adding to 2461136 its reverse (6311642), we get a palindrome (8772778). The spelling of 2461136 in words is "two million, four hundred sixty-one thousand, one hundred thirty-six".	472	1,542	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.140625	3	CC-MAIN-2023-50	latest	en	0.862736
https://biosidmartin.com/what-is-classified-as-a-major-earthquake/	1,669,881,407,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446710801.42/warc/CC-MAIN-20221201053355-20221201083355-00636.warc.gz	164,574,414	11,250	# What is classified as a major earthquake? ## What is classified as a major earthquake? An earthquake of 7.0 magnitude and above is classified as a major earthquake. ### Is a magnitude 5 earthquake big? For example, a magnitude 5.3 is a moderate earthquake, and a 6.3 is a strong earthquake. Because of the logarithmic basis of the scale, each whole number increase in magnitude represents a tenfold increase in measured amplitude as measured on a seismogram. How strong is a category 5 earthquake? Richter magnitudes Magnitude Description Average frequency of occurrence globally (estimated) 5.0–5.9 Moderate 1,000 to 1,500 per year 6.0–6.9 Strong 100 to 150 per year 7.0–7.9 Major 10 to 20 per year 8.0–8.9 Great One per year Which is bigger an earthquake or an earthquake? Calculator How much bigger is a magnitude 8.7 earthquake than a magnitude 5.8 earthquake? An explanation of the magnitude of an earthquake versus the strength, or energy release, of an earthquake… with a little bit of math. How much bigger is a magnitude… than a magnitude… ????? The difference between these two magnitudes is… ## How often does a 5.5 earthquake cause damage? Often felt, but only causes minor damage. 30,000 5.5 to 6.0 Slight damage to buildings and other structures. 500 6.1 to 6.9 May cause a lot of damage in very populated areas. 100 7.0 to 7.9 Major earthquake. Serious damage. 20 8.0 or greater Great earthquake. Can totally destroy communities near the epicenter. One every 5 to 10 years ### Is the magnitude of an earthquake the same as the intensity? Whereas the magnitude of an earthquake is one value that describes the size, there are many intensity values for each earthquake that are distributed across the geographic area around the earthquake epicenter. What are the different classes of an earthquake? Earthquake Magnitude Classes. Earthquakes are also classified in categories ranging from minor to great, depending on their magnitude. Class. Magnitude. Great. 8 or more. Major. 7 – 7.9. Strong. Begin typing your search term above and press enter to search. Press ESC to cancel.	505	2,108	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49	latest	en	0.926158
https://www.askiitians.com/forums/AIPMT/a-mixture-of-cacl2-and-nacl-weighing-4-44g-is-trea_236798.htm	1,725,981,198,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00864.warc.gz	617,180,078	43,002	# A mixture of CaCl2 and NaCl weighing 4.44g is treated with sodium carbonate solution of precipitate all the Ca2+ ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to get 0.56g of CaO. The percentage of NaCl in the mixture is [ans:75] Arun 25750 Points 5 years ago Weight of CaO = 0.56 g 0.56 g of CaO contains 0.40 g of calcium. Mol. weight of CaCl2 = 40 + 71 = 111 40 g of calcium present in 111 g of CaCl2 ∴ 0.4 g of calcium present in 0.40 x 111/40 = 1.11 g of CaCl2 Amount of NaCl present in mixture = 4.44 – 1.11 = 3.33 g ∴ Percentage of NaCl = 3.33 /4.44x 100 = 75	214	603	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.78125	4	CC-MAIN-2024-38	latest	en	0.873692
https://discusstest.codechef.com/t/hackerearth-candy-problem-help/18923	1,685,583,011,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224647525.11/warc/CC-MAIN-20230601010402-20230601040402-00324.warc.gz	264,856,534	6,640	# Hackerearth \| Candy problem \| Help I got stuck in this questions, please suggest me any elegant approach to solve this problem. You are given a string S consisting of lowercase English letters denoting different types of candies. A substring of a string S is a string S’ that occurs in S. For example, “bam” is a substring of “babammm”. Each candy costs 1 unit. You can pick some continuous candies such that you can create a palindrome of length K by using some or all picked candies and rearranging them. Your task is to find the minimum cost to create a palindrome of length K. Input Format: First line contains string S. Next line contains an integer T denoting the number of test cases. Next T lines contain a single integer K. Output Format: For each test case, print minimum cost as mentioned above. If you cannot create a palindrome of length K then, simply print -1. Constraints 1 <= \|S\| <= 10^5 1 <= T <=10 1<=K<=10^5 ``````Sample Input babammm 2 2 5 Sample Output 2 5 `````` Explanation Test Case 1: You can pick candies denoted by “mm” and create a palindrome of size 2. So the cost will be 2 units. Test Case 2: You can pick candies denoted by “babam” and rearrange them, “bamab”, to create a palindrome of size 5. So the cost will be 5 units. Here you can use binary search+prefix sum. First store the prefix count of a-z in array dp[N][26]. Now just start binary search to find minimum length string from low=k to high=string.size().Now consider mid=(lo+hi)/2. Now for all substrings of length mi just check the count of characters from a-z occurring odd number of times. 1. if k is even just calculate cur_len=mi-count_of_odd_chars. If cur_len>=k then just store it as ans and set low=mi+1. Else check for remaining substrings of length mi. 2. if k is odd , 2.1 if count_of_odd_chars=0 skip this string 2.2 else find cur_len=mi-count_of_odd_chars+1 to include one extra odd character. Now just check for cur_len>=k if it holds change low=mid+1 else skip the string 1 Like Can you please elaborate more? Or you could write down pseudo code to clarify more. What is ‘mi’ at here? https://ideone.com/uhSNlG - My code, Out of 26 test cases, it gives AC in 20, and TLE in rest. Logic - Store for every ith index the number of a, b, c … z till that index. freq[i][26] Represented by freq[][] array in code. For every query do a binary search with low = k, and high = s.length in the beginning. Find mid. Check if we can generate palindrome from length = k, with “mid” characters. Done by checking all window of length mid from i = 0 to length - mid. In a particular window of length mid = [i, i + mid). We find frequency of characters (a…z) from i to i + mid, could be calculated using freq[][] array. Answer (for this window) = Count number of even frequency character using freq[][] array. for odd frequency sort them in increasing order. Add to Answer, odd frequency - 1 excluding the last odd frequency. Add to answer the last odd frequency. Basically in above step we are making a palindrome with maximum characters, we can take all even frequency character and split them in left and right half. For all odd frequency except the maximum one we take (odd - 1) for every such characters (make them even), in the end we take the maximum odd frequency character count (keeping it in centre). if Answer >= n, we found the palindrome with mid characters. so we do high = mid - 1 else low = mid + 1. Thanks 1 Like mi is length of current string considered. i.e. mi=(low+high)/2 see https://ideone.com/1ZS3no for further implementation	902	3,577	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.34375	3	CC-MAIN-2023-23	latest	en	0.81584
https://docs.juliahub.com/General/CounterfactualExplanations/stable/tutorials/generators/	1,709,277,480,000,000,000	text/html	crawl-data/CC-MAIN-2024-10/segments/1707947475203.41/warc/CC-MAIN-20240301062009-20240301092009-00570.warc.gz	212,558,963	5,408	# Handling Generators Generating Counterfactual Explanations can be seen as a generative modelling task because it involves generating samples in the input space: $x \sim \mathcal{X}$. In this tutorial, we will introduce how Counterfactual GradientBasedGenerators are used. They are discussed in more detail in the explanatory section of the documentation. ## Composable Generators Breaking Changes Expected Work on this feature is still in its very early stages and breaking changes should be expected. One of the key objectives for this package is Composability. It turns out that many of the various counterfactual generators that have been proposed in the literature, essentially do the same thing: they optimize an objective function. Formally we have, \begin{aligned} \mathbf{s}^\prime &= \arg \min_{\mathbf{s}^\prime \in \mathcal{S}} \left\{ {\text{yloss}(M(f(\mathbf{s}^\prime)),y^*)}+ \lambda {\text{cost}(f(\mathbf{s}^\prime)) } \right\} \end{aligned} \qquad(1) where $\text{yloss}$ denotes the main loss function and $\text{cost}$ is a penalty term (Altmeyer et al. 2023). Without going into further detail here, the important thing to mention is that Equation 1 very closely describes how counterfactual search is actually implemented in the package. In other words, all off-the-shelf generators currently implemented work with that same objective. They just vary in the way that penalties are defined, for example. This gives rise to an interesting idea: Why not compose generators that combine ideas from different off-the-shelf generators? The GradientBasedGenerator class provides a straightforward way to do this, without requiring users to build custom GradientBasedGenerators from scratch. It can be instantiated as follows: generator = GradientBasedGenerator() By default, this creates a generator that simply performs gradient descent without any penalties. To modify the behaviour of the generator, you can define the counterfactual search objective function using the @objective macro: @objective(generator, logitbinarycrossentropy + 0.1distance_l2 + 1.0ddp_diversity) Here we have essentially created a version of the DiCEGenerator: ce = generate_counterfactual(x, target, counterfactual_data, M, generator; num_counterfactuals=5) plot(ce) Multiple macros can be chained using Chains.jl making it easy to create entirely new flavours of counterfactual generators. The following generator, for example, combines ideas from DiCE (Mothilal, Sharma, and Tan 2020) and REVISE (Joshi et al. 2019): @chain generator begin @objective logitcrossentropy + 1.0ddp_diversity # DiCE (Mothilal et al. 2020) @search_latent_space # REVISE (Joshi et al. 2019) end Let’s take this generator to our MNIST dataset and generate a counterfactual explanation for turning a 0 into a 8. ## Off-the-Shelf Generators Off-the-shelf generators are just default recipes for counterfactual generators. Currently, the following off-the-shelf counterfactual generators are implemented in the package: generator_catalogue Dict{Symbol, Any} with 11 entries: :gravitational => GravitationalGenerator :growing_spheres => GrowingSpheresGenerator :revise => REVISEGenerator :clue => CLUEGenerator :probe => ProbeGenerator :dice => DiCEGenerator :feature_tweak => FeatureTweakGenerator :claproar => ClaPROARGenerator :wachter => WachterGenerator :generic => GenericGenerator :greedy => GreedyGenerator To specify the type of generator you want to use, you can simply instantiate it: # Search: generator = GenericGenerator() ce = generate_counterfactual(x, target, counterfactual_data, M, generator) plot(ce) We generally make an effort to follow the literature as closely as possible when implementing off-the-shelf generators. ## References Altmeyer, Patrick, Giovan Angela, Aleksander Buszydlik, Karol Dobiczek, Arie van Deursen, and Cynthia Liem. 2023. “Endogenous Macrodynamics in Algorithmic Recourse.” In First IEEE Conference on Secure and Trustworthy Machine Learning. https://doi.org/10.1109/satml54575.2023.00036. Joshi, Shalmali, Oluwasanmi Koyejo, Warut Vijitbenjaronk, Been Kim, and Joydeep Ghosh. 2019. “Towards Realistic Individual Recourse and Actionable Explanations in Black-Box Decision Making Systems.” https://arxiv.org/abs/1907.09615. Mothilal, Ramaravind K, Amit Sharma, and Chenhao Tan. 2020. “Explaining Machine Learning Classifiers Through Diverse Counterfactual Explanations.” In Proceedings of the 2020 Conference on Fairness, Accountability, and Transparency, 607–17. https://doi.org/10.1145/3351095.3372850.	1,129	4,661	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.515625	3	CC-MAIN-2024-10	latest	en	0.827618
http://us.metamath.org/ileuni/fnbrfvb.html	1,638,017,212,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964358180.42/warc/CC-MAIN-20211127103444-20211127133444-00563.warc.gz	82,553,364	5,302	Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > fnbrfvb GIF version Theorem fnbrfvb 5239 Description: Equivalence of function value and binary relation. (Contributed by NM, 19-Apr-2004.) (Revised by Mario Carneiro, 28-Apr-2015.) Assertion Ref Expression fnbrfvb ((𝐹 Fn 𝐴𝐵𝐴) → ((𝐹𝐵) = 𝐶𝐵𝐹𝐶)) Proof of Theorem fnbrfvb Dummy variable 𝑥 is distinct from all other variables. StepHypRef Expression 1 eqid 2054 . . . 4 (𝐹𝐵) = (𝐹𝐵) 2 funfvex 5217 . . . . . 6 ((Fun 𝐹𝐵 ∈ dom 𝐹) → (𝐹𝐵) ∈ V) 32funfni 5024 . . . . 5 ((𝐹 Fn 𝐴𝐵𝐴) → (𝐹𝐵) ∈ V) 4 eqeq2 2063 . . . . . . . 8 (𝑥 = (𝐹𝐵) → ((𝐹𝐵) = 𝑥 ↔ (𝐹𝐵) = (𝐹𝐵))) 5 breq2 3793 . . . . . . . 8 (𝑥 = (𝐹𝐵) → (𝐵𝐹𝑥𝐵𝐹(𝐹𝐵))) 64, 5bibi12d 228 . . . . . . 7 (𝑥 = (𝐹𝐵) → (((𝐹𝐵) = 𝑥𝐵𝐹𝑥) ↔ ((𝐹𝐵) = (𝐹𝐵) ↔ 𝐵𝐹(𝐹𝐵)))) 76imbi2d 223 . . . . . 6 (𝑥 = (𝐹𝐵) → (((𝐹 Fn 𝐴𝐵𝐴) → ((𝐹𝐵) = 𝑥𝐵𝐹𝑥)) ↔ ((𝐹 Fn 𝐴𝐵𝐴) → ((𝐹𝐵) = (𝐹𝐵) ↔ 𝐵𝐹(𝐹𝐵))))) 8 fneu 5028 . . . . . . 7 ((𝐹 Fn 𝐴𝐵𝐴) → ∃!𝑥 𝐵𝐹𝑥) 9 tz6.12c 5228 . . . . . . 7 (∃!𝑥 𝐵𝐹𝑥 → ((𝐹𝐵) = 𝑥𝐵𝐹𝑥)) 108, 9syl 14 . . . . . 6 ((𝐹 Fn 𝐴𝐵𝐴) → ((𝐹𝐵) = 𝑥𝐵𝐹𝑥)) 117, 10vtoclg 2628 . . . . 5 ((𝐹𝐵) ∈ V → ((𝐹 Fn 𝐴𝐵𝐴) → ((𝐹𝐵) = (𝐹𝐵) ↔ 𝐵𝐹(𝐹𝐵)))) 123, 11mpcom 36 . . . 4 ((𝐹 Fn 𝐴𝐵𝐴) → ((𝐹𝐵) = (𝐹𝐵) ↔ 𝐵𝐹(𝐹𝐵))) 131, 12mpbii 140 . . 3 ((𝐹 Fn 𝐴𝐵𝐴) → 𝐵𝐹(𝐹𝐵)) 14 breq2 3793 . . 3 ((𝐹𝐵) = 𝐶 → (𝐵𝐹(𝐹𝐵) ↔ 𝐵𝐹𝐶)) 1513, 14syl5ibcom 148 . 2 ((𝐹 Fn 𝐴𝐵𝐴) → ((𝐹𝐵) = 𝐶𝐵𝐹𝐶)) 16 fnfun 5021 . . . 4 (𝐹 Fn 𝐴 → Fun 𝐹) 17 funbrfv 5237 . . . 4 (Fun 𝐹 → (𝐵𝐹𝐶 → (𝐹𝐵) = 𝐶)) 1816, 17syl 14 . . 3 (𝐹 Fn 𝐴 → (𝐵𝐹𝐶 → (𝐹𝐵) = 𝐶)) 1918adantr 265 . 2 ((𝐹 Fn 𝐴𝐵𝐴) → (𝐵𝐹𝐶 → (𝐹𝐵) = 𝐶)) 2015, 19impbid 124 1 ((𝐹 Fn 𝐴𝐵𝐴) → ((𝐹𝐵) = 𝐶𝐵𝐹𝐶)) Colors of variables: wff set class Syntax hints: → wi 4 ∧ wa 101 ↔ wb 102 = wceq 1257 ∈ wcel 1407 ∃!weu 1914 Vcvv 2572 class class class wbr 3789 Fun wfun 4921 Fn wfn 4922 ‘cfv 4927 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 103 ax-ia2 104 ax-ia3 105 ax-io 638 ax-5 1350 ax-7 1351 ax-gen 1352 ax-ie1 1396 ax-ie2 1397 ax-8 1409 ax-10 1410 ax-11 1411 ax-i12 1412 ax-bndl 1413 ax-4 1414 ax-14 1419 ax-17 1433 ax-i9 1437 ax-ial 1441 ax-i5r 1442 ax-ext 2036 ax-sep 3900 ax-pow 3952 ax-pr 3969 This theorem depends on definitions: df-bi 114 df-3an 896 df-tru 1260 df-nf 1364 df-sb 1660 df-eu 1917 df-mo 1918 df-clab 2041 df-cleq 2047 df-clel 2050 df-nfc 2181 df-ral 2326 df-rex 2327 df-v 2574 df-sbc 2785 df-un 2947 df-in 2949 df-ss 2956 df-pw 3386 df-sn 3406 df-pr 3407 df-op 3409 df-uni 3606 df-br 3790 df-opab 3844 df-id 4055 df-xp 4376 df-rel 4377 df-cnv 4378 df-co 4379 df-dm 4380 df-iota 4892 df-fun 4929 df-fn 4930 df-fv 4935 This theorem is referenced by: fnopfvb 5240 funbrfvb 5241 dffn5im 5244 fnsnfv 5257 fndmdif 5297 dffo4 5340 dff13 5432 isoini 5482 1stconst 5867 2ndconst 5868 Copyright terms: Public domain W3C validator	1,943	2,855	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49	longest	en	0.162291
http://smlnj-gforge.cs.uchicago.edu/scm/viewvc.php/branches/lamont/test/implicit-surface/unit-circle.diderot?view=markup&revision=2237&root=diderot&sortby=rev&pathrev=2270	1,585,709,422,000,000,000	text/html	crawl-data/CC-MAIN-2020-16/segments/1585370505359.23/warc/CC-MAIN-20200401003422-20200401033422-00107.warc.gz	179,471,024	4,956	Home My Page Projects Code Snippets Project Openings diderot # SCM Repository [diderot] View of /branches/lamont/test/implicit-surface/unit-circle.diderot [diderot] / branches / lamont / test / implicit-surface / unit-circle.diderot # View of /branches/lamont/test/implicit-surface/unit-circle.diderot Sat Mar 2 02:00:56 2013 UTC (7 years, 1 month ago) by lamonts File size: 2145 byte(s) `Fixed error for ceil, had it taking in 2 arguments rather than 1` ```// unit-circle // // Demo of distributing particles on the unit circle // int numOfParticles = length(initPosns)/2; input real rr = 0.08; // actual particle radius input real RR = 0.15; // neighbor query radius (MUST be bigger than rr, but should get same results for any RR > rr) input real hh = 0.001; // integration step size to control how quickly iterations change position. input int MAX_ITER = 200; // maximum number of steps in the program vec2 xDom = [-1,1]; vec2 yDom = [-1,1]; real xSamples = floor((xDom[1] - xDom[0])/RR); // Lamont verify logic floor vs ceil here real ySamples = floor((yDom[1] - yDom[0])/RR); vec4 qWinDim = [xDom[0],xDom[1],yDom[0],yDom[1]]; // i.e. [XMIN, XMAX, YMIN, YMAX] (required for the query function) vec2 qGridDim = [xSamples,ySamples]; // how many grid cells you want in each direction for the uniform grid (required for the query function) vec2 qCellDim = [(xDom[1] - xDom[0])/xSamples,(yDom[1] - yDom[0])/ySamples]; // the length in each direction for a cell (required for the query function) int iter = 1; strand Particle (int ii, real posx, real posy) { vec2 pos = normalize([posx,posy]); output vec2 outPos = pos; int id = ii; update { /* print(atan2(pos[1],pos[0]), " "); if (i == numOfParticles-1) { print("\n"); } / real energy = 0; vec2 force = [0,0]; foreach (Particle p_j in sphere(RR)) { vec2 r_ij = (pos - p_j.pos)/rr; vec2 d_ij = normalize(r_ij); print("===== ", id, ": with ", p_j.id, "; \|r_ij\| = ", \|r_ij\|, "\n"); if (\|r_ij\| < 1) { energy += (1 - \|r_ij\|)^4; force += - (-4(1 - \|r_ij\|)^3) * d_ij; } } vec2 step = hhforce/rr; pos = normalize(pos + step); outPos = pos; if (iter >= MAX_ITER) { stabilize; } } } // can add statements in here to do some global computation global{ iter+=1; } initially {Particle(ii, initPosns{ii2}, initPosns{ii*2+1}) \| ii in 0 .. numOfParticles-1 }; ```	750	2,338	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.671875	3	CC-MAIN-2020-16	latest	en	0.478113
https://www.coursehero.com/tutors-problems/Physics/504545-A-75-kg-person-escapes-from-a-burning-building-by-jumping-from-a-windo/	1,545,038,544,000,000,000	text/html	crawl-data/CC-MAIN-2018-51/segments/1544376828501.85/warc/CC-MAIN-20181217091227-20181217113227-00384.warc.gz	867,604,188	20,829	View the step-by-step solution to: # A 75 kg person escapes from a burning building by jumping from a window 25 m above a catching net. A 75 kg person escapes from a burning building by jumping from a window 25 m above a catching net. Assuming that air resistance exerts a 95 N force on the person during the fall, determine the person's velocity just before hitting the net. My main problem is finding the force of the person... do you take 9.81*75 and minus 95 to get the downward force of the person?ok, thanks. I think I'm on the right track.Yes. The force down is mg minus air resistance. ### Why Join Course Hero? Course Hero has all the homework and study help you need to succeed! We’ve got course-specific notes, study guides, and practice tests along with expert tutors. ### - Educational Resources • ### - Study Documents Find the best study resources around, tagged to your specific courses. Share your own to gain free Course Hero access. Browse Documents	216	977	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51	latest	en	0.894963
https://oeis.org/A038728	1,722,874,699,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640451031.21/warc/CC-MAIN-20240805144950-20240805174950-00400.warc.gz	353,028,476	4,381	The OEIS is supported by the many generous donors to the OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A038728 Configurations of linear chains in a 5-dimensional hypercubic lattice. 1 0, 0, 0, 0, 2240, 35840, 433040, 4862560, 51759280, 527313040, 5218528800, 50434399280, 478624474160, 4473452644480 (list; graph; refs; listen; history; text; internal format) OFFSET 1,5 COMMENTS In the notation of Nemirovsky et al. (1992), a(n), the n-th term of the current sequence is C_{n,m} with m=2 (and d=5). Here, for a d-dimensional hypercubic lattice, C_{n,m} is "the number of configurations of an n-bond self-avoiding chain with m neighbor contacts." (Let n >= 1. For d=2, we have C(n,m=2) = A033323(n); for d=3, we have C(n,m=2) = A049230(n); and for d=4, we have C(n,m=2) = A046788(n).) - Petros Hadjicostas, Jan 05 2019 LINKS Table of n, a(n) for n=1..14. A. M. Nemirovsky, K. F. Freed, T. Ishinabe, and J. F. Douglas, Marriage of exact enumeration and 1/d expansion methods: lattice model of dilute polymers, J. Statist. Phys., 67 (1992), 1083-1108; see Table 1 on p. 1090. CROSSREFS Cf. A033323, A046788, A049230. Sequence in context: A372248 A079013 A186865 * A002520 A337416 A183771 Adjacent sequences: A038725 A038726 A038727 * A038729 A038730 A038731 KEYWORD nonn,more AUTHOR N. J. A. Sloane, May 02 2000 EXTENSIONS Terms a(10) and a(11) were copied from Table 1 (p. 1090) of Nemirovsky et al. (1992) by Petros Hadjicostas, Jan 05 2019 Name edited by Petros Hadjicostas, Jan 05 2019 a(12)-a(14) from Sean A. Irvine, Feb 01 2021 STATUS approved Lookup \| Welcome \| Wiki \| Register \| Music \| Plot 2 \| Demos \| Index \| Browse \| More \| WebCam Contribute new seq. or comment \| Format \| Style Sheet \| Transforms \| Superseeker \| Recents The OEIS Community \| Maintained by The OEIS Foundation Inc. Last modified August 5 12:17 EDT 2024. Contains 374950 sequences. (Running on oeis4.)	658	1,926	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.640625	3	CC-MAIN-2024-33	latest	en	0.67793
https://ccm.net/forum/affich-1091642-if-statement	1,669,949,728,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446710890.97/warc/CC-MAIN-20221202014312-20221202044312-00062.warc.gz	192,670,224	23,885	# IF statement Solved/Closed MCiort - Aug 30, 2019 at 07:22 AM Blocked Profile - Aug 30, 2019 at 09:03 AM Hello, I am really struggling to get the correct formula for below situation: example: Cell A2 is a "due date" Cell B2 is a "completion date" Cell C2 is a "current progress" In cell C2 I am trying without success to insert an IF formula so that: 1 - if A2 <= Today() then value in C2="Open - Late" 2 - if A2 > Today() then value in C2="Open - On time" the 2 above conditions, considering B2 is empty when B2 is filled in, what i am looking to include in the formula is that 3 - if A2 >= B2 then value in C2="Completed - On time" 4 - if A2 < B2 then value in C2="Completed - Late" the formula I tried so far looks like below: =IF(A2<=Today(),"Open-Late",IF(A2>Today(),"Open-On time",IF(A2>=B2,"Completed-On time",IF(A2<B2,"Completed-Late")))) However, using the above, only first 2 conditions are returned in C2, therefore not sure how best to include all conditions so that the value in C2 is returned as per above conditions Thanks a lot for your help! ## 1 reply Ok, just swap the order in which you check for status. So check for a completed late first, completed on time next, and so on. Also, on first check, use this: If (and (b2 <>"", a2 <b2),"COMP LATE", So on first logic test, use AND (). AND (FIRST TEST, SECOND TEST) IF (AND (FIRST, SECOND), TRUE, FALSE) Hi Mark, Thank you for your quick response. Just tried changing the order, however, still only first 2 conditions are returned. Maybe, I did not explained clearly what I needed the formula to do, but basically I need all 4 returned values available. i.e. if B2 is not filled, than the value in C2 is either Open Late / On Time and when B2 is filled, then the value returned to be based on condition between A2 and B2 --> Completed On time / Late. for some reason, the current formula only works for the first 2 conditions. Not sure what am I missing in the formula, or even if this is the right one to use Hi Mark, My apologies as only saw first 2 lines of your reply and completely omitted the second part of your reply stating to use AND(). In meantime sorted the issue with below: =IF(B2="",(IF(A2<Today(),"Open-Late",IF(A2>=Today(),"Open-On time"))),IF(A2>=B2,"Completed-On Time",IF(A2<B2,"Completed-Late",""))) Also just to confirm that using the formula suggested by you, also solved the issue. Thank you a lot for your quick response and help. Blocked Profile Aug 30, 2019 at 09:03 AM Thank you for the feedback. Nested if statements can be tricky!	692	2,549	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.921875	3	CC-MAIN-2022-49	latest	en	0.910645
http://slideplayer.com/slide/717196/	1,542,256,699,000,000,000	text/html	crawl-data/CC-MAIN-2018-47/segments/1542039742483.3/warc/CC-MAIN-20181115033911-20181115055911-00213.warc.gz	306,108,563	33,130	# Chapter 4 An Introduction to Finite Automata ## Presentation on theme: "Chapter 4 An Introduction to Finite Automata"— Presentation transcript: Chapter 4 An Introduction to Finite Automata Automata Theory CS 3313 Chapter 4 An Introduction to Finite Automata Finite State Systems: * The finite automaton is a mathematical model of a system, with discrete inputs and outputs. * The system can be in any one of a finite number of internal configurations or "states". * The control mechanism of an elevator is a good example of a finite state system. * That mechanism does not remember all previous requests for service but only the current floor, the direction of motion (up or down), and the collection of not yet satisfied requests for service. * In computer science we find many examples of finite state systems, and the theory of finite automata is a useful design tool for these systems. ·. A primary example is a switching circuit, such as the · A primary example is a switching circuit, such as the control unit of a computer. · A switching circuit is composed of a finite number of gates, each of which can be in one of two conditions, usually denoted 0 and 1. · Certain commonly used programs such as text editors and the lexical analyzers found in most compliers are often designed as finite state systems. · For example, a lexical analyzer scans the symbols of a computer program to locate the strings of characters corresponding to identifiers, numerical constants, reserved words, and so on. · In this process the lexical analyzer needs to remember only a finite amount of information, such as how long a prefix of a reserved word it has seen since startup. · An Example: · Before formally defining finite state systems let us consider an example. · A man with a wolf, goat, and cabbage is on the left bank of a river. · There is a boat large enough to carry the man and only one of the other three. · The man and his entourage wish to cross to the right bank, and the man can ferry each across, one at a time. · However, if the man leaves the wolf and goat unattended on either shore, the wolf will surely eat the goat. ·. Similarly, if the goat and cabbage are left unattended, · Similarly, if the goat and cabbage are left unattended, the goat will eat the cabbage. · Is it possible to cross the river without the goat or cabbage being eaten? · The problem is modeled by observing that the pertinent information is the occupants of each bank after a crossing. · There are 16 subsets of the man (M), wolf (W), goat (G), and cabbage (C). * Shown in Following figure: ·. There are two equally short solutions to the problem, · There are two equally short solutions to the problem, as can be seen by searching for paths form the initial state to the final state. · Before proceeding, we should note that there are at least two important ways in which the above example is a typical of finite state systems. · First, there is only one final state; in general there may be many. · Second, it happens that for each transition there is a reverse transition on the same symbol, which need not be the case in general. · Also, note that the term "final state", although traditional, does not mean that the computation needs halt when it is reached. · We may continue making transitions. Basic Definitions · A finite Automaton (FA) consists of a finite set of states and a set of transition from state to state that occur on input symbols chosen from analphabet Σ. · For each input symbol there is exactly one transition out of each state (possibly back to the state itself). · One state, usually denoted q0, is the initial state, in which the automaton starts. · Some states are designed as final or accepting states. · A directed graph, called a "transition diagram", is associated with an FA as follows: · The vertices of the graph correspond to the states of the FA. · If there is a transition from state q to state p on input a, then there is an arc labeled "a" from state q to state p in the transition diagram. · The FA accepts a string x if the sequence of transitions corresponding to the symbol of x leads from the start state to an accepting state. · [Example Figure-2.2 on page-16 of text book.] ·. We formally denote a finite automaton by a 5-tuples (Q, Σ, δ, · We formally denote a finite automaton by a 5-tuples (Q, Σ, δ, q0, F). · Where: · Q is a finite set of states, · Σ is a finite input alphabet, · q0 in Q is the initial state, · F Q is the set of final states and · δ is the transition function mapping Q X Σ to Q. * That is, δ(q0, a) is a state for each q and input symbol a. How DFA Works * A DFA can be seen as a state machine as follows: 1. The machine reads the leftmost symbol of the word ‘W’, in 1. The machine reads the leftmost symbol of the word ‘W’, in the initial state q0. 2. The automaton moves one symbol to the right and enters a new state as define by the transition function d . 3. If after reading the word ‘W’, the automaton stops in a state that belongs to ‘F’, then ‘W’ is accepted, otherwise it is rejected. d (q current , symbol ) = q next ·. To formally describe the behavior of an FA on a string, we · To formally describe the behavior of an FA on a string, we must extend the transition function δ to apply to a state and a string rather than a state and a symbol. · We define a function δ from Q X Σ* to Q. · The intention is that δ(q, w) is the state the FA will be in after reading w starting in state q. · Put another way, δ(q, w) is the unique state p such that there is a path in the transition diagram from q to p, labeled w. · Formally we define: 1) δ(q, ε) = q and 2) for all strings w and input symbols a, δ(q, wa) = δ(δ(q, w), a) ·. Thus (1) states that without reading an input symbol the FA cannot · * Thus (1) states that without reading an input symbol the FA cannot change state, and · Tells us how to find the state after reading a nonempty input string wa. · That is, find the state, p= δ(q, w), after reading w. · Then compute the state δ(p, a). · Since δ(q, a) = δ(δ(q, ε), a) = δ(q, a) [letter w = ε in rule (2) above, there can be no disagreement between δ and δ on arguments for which both are defined. · Thus we shall for convenience use δ instead of δ from here on. · Convention: We shall strive to use the same symbol to mean the same thing throughout the material on finite automata. · In particular, unless it is stated otherwise, the reader may assume: 1). Q is a set of states. Symbols q and p, with or without subscripts, 1) Q is a set of states. Symbols q and p, with or without subscripts, are states. q0 is the initial state. 2) Σ is an input alphabet. Symbols a and b, with or without subscripts and the digits are input symbols. 3) Δ is a transition function. 4) F is a set of final states. 5) w, x, y, and z with or without subscripts, are strings of input symbols. · A string x is said to be accepted by a finite automaton M = (Q, Σ, δ, q0, F) if δ(q0, x) = p for some p in F. · The language accepted by M, designated L(M), is the set {x \| (δ0, x) is in F}. · A language is a regular set (or just regular) if it is the set accepted by some finite automaton. [Example-2.2, page-18, of text book.] Non-deterministic Finite Automata · Any set accepted by a non-deterministic finite automaton can also be accepted by a deterministic finite automaton. · However, the non-deterministic finite automaton is a useful concept in proving theorems. · Also, the concept of non-determinism plays a central role in both the theory of languages and the theory of computation, and it is useful to understand this notion fully in a very simple context initially. · Later we shall meet automata whose deterministic and non- deterministic versions are known not to be equivalent. · Consider modifying the finite automaton model to allow zero, one or more transitions from a state on the same input symbol. · This new model is called non-deterministic finite automata (NFA). · [Figure-2.5, page 20, of text book.] · Formally we denote a non-deterministic finite automaton by a 5 tuple (Q, Σ, δ, q0, F), where Q, Σ, q0, and F (states, inputs, start state and final states) have the same meaning as for a DFA, but δ is a map from Q X Σ to 2Q. · Recall 2Q is the power set of Q, the set of all subsets of Q). · The intention is that δ(q, a) is the set of all states p such that there is a transition labeled a from q to p. · The function δ can be extended to a function δ mapping Q X Σ* to 2Q and reflecting sequences of inputs as follows: 1) δ(q, ε) = {q} 2) δ(q, wa) = {p \| for some state r in δ(q, w), p is in δ(r, a)} · Conditions (1) disallows a change in sate without an input. · Condition (2) indicates that starting in state q and reading the string w followed by input symbol "a" we can be in state p if and only if one possible state we can be in after reading w is r, and from r we may go to p upon reading a. · Note that δ(q, a) = δ(q, a) for a an input symbol. · Thus we may again use δ in place of δ. · It is also useful to extend δ to arguments in 2Q X Σ* by 3) δ(P, w) = U q in P δ(q, w) · For each set of states P Q. · L(M), where M is the NFA (Q, Σ, δ, q0, F), is {w \| δ(q0, w) contains a state in F} · [Example-2.4, page-21, of text book.] The equivalence of DFA's and NFA's · Since every DFA is an NFA, it is clear that the class of languages accepted by NFA's includes the regular sets (the language accepted by DFA's). · However, it turns out that these are the only sets accepted by NFA's. · For every NFA we can construct an equivalent DFA (one which accepts the same language) · [Assignment: Theorem-2.1, page 22 of the text book.] · [Assignment: Example-2.5, page 23 of the text book.] Finite Automata with ε-Moves · We may extend our model of the non-deterministic finite automaton to include transitions on the empty input ε. · The transition diagram of such an NFA accepting the language consisting of any number (including zero) of 0's followed by any number of 1's followed by any number of 2's is given in Figure-2.8 (page-24). · As always, we say an NFA accepts a string w if there is some path labeled w from the initial state to a final state. · Of course, edges labeled ε move be included in the path, although the ε's do not appear explicitly in w. · For example, the word 002 is accepted by the NFA of Figure- 2.8 by the path q0, q0, q0, q1, q2, q2 with arcs labeled 0, 0, ε, ε, 2. · [Assignment: example-2.7, example-2.8, example-2.9.] Regular Expressions · The language accepted by finite automata are easily described by simple expressions called regular expressions. · L0 = { ε } and L1 = L Li - 1 for i ≥ 1. · (Kleene Closure or just Closure) L* denotes words constructed by concatenating any number of words from L. · L+ is the same, but the case of zero words, whose concatenation is defined to be ε, is excluded. · Note that L+ contains ε if and only if L does. · [Example: 2.10, page 28 of the text book.] · [Example: 2.11, page 29 of the text book.] · [Theorem: 2.3, page 30 of the text book.] · [Example: 2.12, page 32 of the text book.] · [Assignment: Example-2.13, page 34 of the text book.] Two-Way Finite Automata · We have viewed the finite automaton as a control unit that reads a tape, moving one square right at each move. · We added non-determinism to the model, which allowed many "copies" of the control unit to exist and scan the tape simultaneously. · Next we added ε-transitions, which allowed change of state without reading the input symbol or moving the tape head. · Another interesting extension is to allow the tape head the ability to move left as well as right. · Such a finite automaton is called a two-way finite automaton [Assignment] · [Example: 2.14, page 37 of the text book.] * [Assignment: Crossing Sequences, and Figure:2.19, page 38 of the text book.] Finite Automata with Output ·One limitation of the finite automaton as we have defined it is that its output is limited to a binary signal: "accept" / " don't accept". ·There are two distinct approaches; the output may be associated with the state (called a Moore Machine) or with the transition (called a Mealy Machine). Moore Machines ·A Moore machine is a six-tuple (Q, Σ, Δ, δ, λ, q0), where Q, Σ, δ and q0 are as in the DFA. ·Δ is the output alphabet and λ is a mapping from Q to Δ giving the output associated with each state. ·Note that any Moore machine gives output λ(q0) in response to input ε. The DFA may be viewed as a special case of a Moore *The DFA may be viewed as a special case of a Moore machine where the output alphabet is {0, 1} and state q is "accepting" if and only if λ(q) = 1. ·[Example: 2.17, page 42 of the text book.] Mealy Machines ·A Mealy machine is also a six-tuple M = (Q, Σ, Δ, δ, λ, q0), where all is as in the Moore machine, except that λ maps Q x Σ to Δ. ·[Example: 2.18, page 43, of the text book.] Applications of Finite Automata · [Assignment: Note on Lexical Analyzer & Text Editors] Similar presentations	3,348	13,097	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.109375	3	CC-MAIN-2018-47	latest	en	0.895392
http://www1.rmit.edu.au/browse;ID=C6093CIVE56581345tafecombined	1,547,986,003,000,000,000	text/html	crawl-data/CC-MAIN-2019-04/segments/1547583705737.21/warc/CC-MAIN-20190120102853-20190120124853-00533.warc.gz	401,463,476	14,501	# Course Title: Use technical mathematics (basic) ## Part B: Course Detail Teaching Period: Term2 2013 Course Code: CIVE5658 Course Title: Use technical mathematics (basic) School: 130T Vocational Engineering Campus: City Campus Program: C6093 - Advanced Diploma of Engineering Design Course Contact: Program Manager Course Contact Phone: +61 3 9925 4468 Course Contact Email: engineering-tafe@rmit.edu.au Name and Contact Details of All Other Relevant Staff Program Manager Mr. Ahmet Ertuncay Tel. 9925 8375 Email: ahmet.ertuncay@rmit.edu.au Ms. Annabelle Lopez Tel. 9925 4823 Email: Annabelle.lopez@rmit.edu.au Nominal Hours: 60 Regardless of the mode of delivery, represent a guide to the relative teaching time and student effort required to successfully achieve a particular competency/module. This may include not only scheduled classes or workplace visits but also the amount of effort required to undertake, evaluate and complete all assessment requirements, including any non-classroom activities. Pre-requisites and Co-requisites None Course Description This unit of competency deals with the skills and knowledge required to apply the essential core skills in basic arithmetic, algebra and geometry to simple engineering problems, common to all engineering disciplines. National Codes, Titles, Elements and Performance Criteria National Element Code & Title: EDX130B Use technical mathematics (basic) Element: 1. Solve engineering mathematics problems using fractions and decimals Performance Criteria: 1.1 Calculations involving fractions and mixed numbers are performed. 1.2 Calculations involving decimals are performed. 1.3 Vocational mathematical problems involving fractions and decimals are solved Element: 2. Solve engineering problems using ratio proportion and percent. Performance Criteria: 2.1 A ratio can be formed from information in a practical problem and expressed in its lowest terms. 2.2 A quantity can be divided into its lowest terms. 2.3 Simple practical problems involving proportional quantities are solved. 2.4 Quantities are converted between fraction, percent and decimal forms. 2.5 Vocational problems are solved involving precent of a quantity, one quantity as a percent of another and a quantity when a percent is known. 2.6 Percentage increases and decreases of a quantity are calculated. Element: 3. Solve two and three dimensional engineering mathematics problems. Performance Criteria: 3.1 The lengths and perimeters of rectangles, circles and combined shapes are calculated. 3.2 The areas of rectangles, triangles, circles and combined shapes are calculated. 3.3 Elementary problems requiring the use of the concepts of measurement and mensuration are solved. Element: 4 Solve engineering mathematical problems using elementary geometric concepts. Performance Criteria: 4.1 Diagrams are drawn to illustrate the meaning of a line, line segment, ray, parallel and perpendicular lines and an angle. 4.2 Angles in a diagram are measured using a protractor and are correctly named and classified. 4.3 The size of an angle is determined in a diagram involving adjacent and vertically opposite angles and parallel lines. 4.4 A ruler and set square are used to construct a line parallel or perpendicular to another line through a given points not on the line. 4.5 A ruler and protractor are used to construct a diagram involving lines and angles, given a written description. 4.6 A ruler and a pair of compasses are used to construct the bisector of an angle, the perpendicular bisector of a line segment and an angle equal in size to another angle. 4.7 Non routine problems requiring the use of elementary geometric principles can be solved. Element: 5. Solve mathematical problems involving triangles. Performance Criteria: 5.1 Triangles are identified by side or angle. 5.2 Triangles are constructed from given data. 5.3 Medians and centroids, altitudes and orthocentres, circum-centre and circum-circle are identified and constructed. 5.4 The angle and side properties of a triangle are used to solve triangles. 5.5 Pythagorus theorem is used to find the length of an unknown side and to test whether a triangle is right angled. 5.6 The four criteria for congruent triangle are used. 5.7 The three criteria for similar triangles are used. 5.8 The areas of triangles are calculated using appropriate formulae. 5.9 Quadrilaterals are identified and classified. 5.10 Quadrilaterals are constructed form given data. 5.11 The properties of a quadrilateral are used to find unknown angles and sides in a quadrilateral. 5.12 The area of quadrilaterals is calculated. Element: 6. Solve engineering mathematical problems by determining the equations of straight lines and representing them graphically on the Cartesian Plane. Performance Criteria: 6.1 The equation of a straight line is determined by measuring the gradient and finding the y intercept. 6.2 The equation of a straight line is determined given the coordinates of two points on the line. 6.3 The graph of a straight line is sketched given in the form y = ax + b. 6.4 The simultaneous solution of a pair of linear equations is determined graphically. 6.5 Word expression are converted into mathematical statements that define relationships. 6.6 Interpolation and extrapolation are carried out for the line of best fit noting limitations. 6.7 The meaning of the gradient and the y-intercept of a straight line is interpreted. 6.8 Non-routine problems are solved using the concepts and techniques of coordinate geometry. 6.9 Empirical data is collected and a summary of results written when fitting a straight line to the data. Element: 7. Solve analytical and applied problems using the right-angled triangle definition of sine, cosine, tangent. Performance Criteria: 7.1 The unknown side or angle of a right-angled triangle is determined using sine, cosine or tangent of an angle. Element: 8. Solve engineering problems involving operations on real numbers and the manipulation eg algebraic terms leading to the solution of linear equations. Performance Criteria: 8.1 The number line is sketched and rational and irrational number location are indicated. 8.2 The number line is used to graphically establish the location of irrational numbers. 8.3 Arithmetic problems are solved involving the correct order of operations. 8.4 Problems involving algebraic functions are solved, grouping symbols and using the correct order of operations. 8.5 A graphics calculator is used to solve problems involving the use of grouping symbols. 8.6 Values are substituted into linear equations to solve simple practical engineering problems 8.7 Simple linear equations are derived and solved involving simple engineering problems. 8.8 Simple simultaneous equations are solved involving simple engineering problems. Element: 9. Transpose and evaluate engineering formulae. Performance Criteria: 9.1 Given values are substituted into simple non-linear formulae to find physical quantities. 9.2 Non-linear formulae are manipulated using the four mathematical operations and the root, in their correct order in simple cases where the subject occurs at most twice. Learning Outcomes Refer Elements above. Details of Learning Activities You will participate in individual and team problem solving activities related to typical engineering workplace problems. These activities involve class participation (discussions and oral presentations), completion of all assigned work, prescribed exercises, homework, tutorials, application of theory to engineering problems and completion of calculations to industry standard as well as completion of all other assessments to a satisfactory standard. Engineering employment requires the capacity to work effectively in teams, to communicate effectively in both oral and writing and to learn effectively. In order to prepare students for employment as graduates they will be provided a quality assured teaching and learning environment which is conductive to the development of adult learning. Adult learning is characterised by the students accepting responsibility for their own learning and actively participating in the learning process as individuals and as contributors to the teams. Adult learning is the hallmark of a professional. The specific responsibilities as adult learners in respect of this subject are: . to be aware of and to observe the regulations related to plagiarism . to submit (on time) all work for assessment as required . to complete all pre-reading and preparatory work prior to the class for which it will be used . to effectively use the academic staff resources provided (consultation time, tutors, e- mail etc) . to participate as an effective and honest member of a learning team . to contribute effectively to a group of peers in a climate of mutual respect and to question each other and the academic staff when uncertain. Hence, you will participate in individual and team problem solving activities related to typical engineering workplace problems. These activities involve class participation, discussions, prescribed exercises, assignments and other self-directed student activities. Lecture sessions are generally devoted to topic summaries and short quizzes whilst tutorial sessions are generally devoted to discussions (student queries) of assigned exercises and portfolio questions. PLEASE NOTE, IN THIS COURSE, LEARNING GUIDE IS USED AS REFERENCE ONLY. Learning and simulated work activities to demonstrate an understanding of typical problems encountered in meeting performance requirements and compliance standards are outlined below: * Classroom tutorial activities are achieved to consolidate theories Practical Exercises This course requires that students demonstrate highly practical skills. Underpinning knowledge is required before undertaking practical exercises. Research activities to undertake investigative activities are undertaken. It is expected that students would require approximately 50% of course hours to be allocated for independent study to do project research and problem solving activities. Assignment tasks involve applications of standards and codes whenever applicable and shall be as close as practicable to real work situations and include real work decisions by the learner. Teaching Schedule This is an indicative teaching schedule. Refer to Online Blackboard announcements for changes. For absences due to public holidays and other class cancellations, the topics & assessment tasks will be shifted accordingly. As teaching schedule is currently on contingency mode, whilst flexibility is offered, self directed learning is much called for on the part of the students. Week Number, Topic Delivered, Assessment Task 1 Introduction to course, course guide, assessment, topic breakdown, resources, OHS issues. 2 Course Summary OR otherwise, Introduction to course, course guide, assessment, topic breakdown, resources, OHS issues (for Late Starters) 3 Fractions and Decimals – Quiz (Part of Option 1) 4 Ratio, Proportion and Percent – Quiz (Part of Option 1) 5 Perimeters, Areas and Measurement – Quiz (Part of Option 1) 6 Introduction to Algebra – Quiz (Part of Option 1) 7 Algebra/Formulae Evaluation and Transposition 8 Formula Evaluation and Transposition – Quiz (Part of Option 1) 9 Introduction to Geometry– Quiz (Part of Option 1) 10 Geometry of Triangles and Quadrilaterals 11 Geometry of Triangles and Quadrilaterals– Quiz (Part of Option 1) 12 Straight Line Coordinate Geometry – Quiz (Part of Option 1) 13 Trigonometry – Quiz (Part of Option 1) 14 Trigonometry 15 Question & Answer Forum / Workshop – Feedback available Part 1 Portfolio of Assigned Questions (Option 2) due in 16 Question & Answer Forum / Workshop – Feedback available Part 2 Portfolio of Assigned Questions – Late Submissions with Penalty (Option 2) due in 17 All Deferred Assignments and Portfolios and other outstanding issues (special consideration) Deferred/ Alternative Assessments due in including those with Special Consideration 18 Feedback on Grades and Finalising Results This is an indicative teaching schedule. Refer to Online Blackboard announcements for changes. For absences due to public holidays and other class cancellations, the topics & assessment tasks will be shifted accordingly. As teaching schedule is currently on contingency mode, whilst flexibility is offered, self directed learning is much called for on the part of the students. Learning Resources Prescribed Texts ‘Mathematics for technicians’, by Blair Alldis 6th edition References Other Resources Overview of Assessment Assessment are conducted in both theoretical and practical aspects of the course according to the performance criteria set out in the National Training Package. Students are required to undertake summative assessments that bring together knowledge and skills. To successfully complete this course you will be required to demonstrate competency in each assessment tasks detailed under the Assessment Task Section. Your assessment for this course will be marked using the following table: NYC (<50%) Not Yet Competent CAG (50-59%) Competent - Pass CC (60-69%) Competent - Credit CDI (70-79%) Competent - Distinction CHD (80-100%) Competent - High Distinction The assessment is conducted in both theoretical and practical aspects of the course according to the performance criteria set in the National Training Package. Assessment may incorporate a variety of methods including written/oral activities and demonstration of practical skills to the relevant industry standards. Students are advised that they are likely to be asked to personally demonstrate their assessment work to their teacher to ensure that the relevant competency standards are being met. Students will be provided with feedback throughout the course to check their progress. Feedback will be provided throughout the course. To successfully complete this course you will be required to demonstrate competency in each assessment task detailed under Assessment Tasks section of Course Guide Part B (this document). Assessment tasks have been designed to encourage life-long learning and self directed learning, encouraging students to ask questions and manage their time in order to progressively complete work throughout the semester. Individual as well as team activities will be demonstrated in preparation for work in industry where competing demands and adaptation to change characterise the work environment, and where communication, team work and organisation skills are of paramount importance. Assessment details: Choose one of Collection of Quizzes OR Collection of Exercises (Portfolio) – Student has a choice to either sit 10 quizzes (10 marks each for a total of 100 marks) or to complete a portfolio of evidence, being the assigned exercises for this course (worth 100 marks). Student chooses one of these options, but generally not both. Someone with good mathematics background from high school or VCE is expected to choose the quizzes whilst someone that had a break in his/her studies is expected to choose the portfolio. Option 1: Collection of Quizzes – This is a collection of a student’s quizzes (10 quizzes altogether) where analysis and solution to practical application problems/questions in engineering where mathematics has been applied is demonstrated. This assessment focuses on the students’ ability to solve problems and provide logical solutions to practical exercises. These individual assessments collectively has a weighting of 100%, is closed book , and are all scheduled during lecture sessions (see teaching schedule). Note: Allowed in assessment room are NON-programmable calculators, pens, rulers and bottle of water only. Option 2: Portfolio of Evidence- This is the student’s mathematics exercise book, where he/she writes solutions to mathematical problems and other application questions, presented neatly with all working completed and all titles/headings included. This assessment focuses on the students’ ability to solve problems and provide logical solutions to practical exercises. Portfolio has a weighting of 100%, and is an individual assessment. Note: Use exercise book with NO spirals. NOTE: QUESTIONS ABOUT SPECIFIC ASSESSMENTS AND MARKS OBTAINED MAY NOT BE ENTERTAINED IF LATER THAN 1 WEEK AFTER RESULTS WERE COMMUNICATED TO STUDENTS Assessment requirements also include: - attendance and satisfactory completion of prescribed practical exercises , - evidence of participation in and satisfactory completion of work simulation projects. -satisfactory completion of class assignment work -timely submission and standard presentation for all assessment material / documentation CHD Competent with High Distinction -The learner will confidently apply novel but relevant solutions to unfamiliar and complex tasks. CDI Competent with Distinction -The learner will confidently evaluate alternative solutions to an unfamiliar task or problem and use the most appropriate solution. CC Competent with Credit -The learner will elegantly apply appropriate facts, rules and standard solutions to achieve an unfamiliar task or problem with confidence. CAG Competency Achieved - Graded -The learner will be able to apply facts, rules and standard solutions to achieve a predictable task or solve a problem. NYC Not Yet Competent -Although the learner exhibits access to a limited range of facts and rules, the learner has difficulty applying these facts and rules to a familiar task. DNS Did Not Submit for Assessment Students should be informed with the special consideration policy available at - http://www.rmit.edu.au/browse;ID=qkssnx1c5r0y (unresolved) Assessment Matrix Element Covered, Assessment Task, Proportion of Final Assessment. Approximate Time 1,2,3,4,5,6,7,8, 9, 10, 11, 12, 13 and 14 ------ Option 1: Collection of Quizzes -----100 % -------As per teaching schedule 1,2,3,4,5,6,7,8, 9, 10, 11, 12, 13 and 14 ------ Option 2: Portfolio of Evidence ------100 % ------- As per teaching schedule NOTE: Student chooses one of above named assessment options, but NOT both (for any changes, refer to online blackboard announcement) Other Information In this course, minimum student directed hours are 12 in addition to 48 scheduled teaching hours. Student directed hours involve completing activities such as reading online resources, assignments, report for practical work, and individual student-teacher course-related consultation. Study and learning Support: Study and Learning Centre (SLC) provides free learning and academic development advice to all RMIT students. Services offered by SLC to support numeracy and literacy skills of the students are: assignment writing, thesis writing and study skills advice maths and science developmental support and advice English language development Disability Liaison Unit: Students with disability or long-term medical condition should contact Disability Liaison Unit to seek advice and support to complete their studies. Late submission: Students requiring extensions for 7 calendar days or less (from the original due date) must complete and lodge an Application for Extension of Submittable Work (7 Calendar Days or less) form and lodge it with the Senior Educator/ Program Manager. The application must be lodged no later than one working day before the official due date. The student will be notified within no more than 2 working days of the date of lodgment as to whether the extension has been granted. Students seeking an extension of more than 7 calendar days (from the original due date) must lodge an Application for Special Consideration form under the provisions of the Special Consideration Policy, preferably prior to, but no later than 2 working days after the official due date. Assignments submitted late without approval of an extension will not be accepted or marked. Special consideration:	3,931	19,938	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04	latest	en	0.848122
https://www.weegy.com/?ConversationId=HBC1A5JG	1,604,182,929,000,000,000	text/html	crawl-data/CC-MAIN-2020-45/segments/1603107922463.87/warc/CC-MAIN-20201031211812-20201101001812-00371.warc.gz	894,515,538	11,672	A poet's attitude toward his or her poem's subject is referred to as the A. theme. B. meter. C. tone. D. structure. A poet's attitude toward his or her poem's subject is referred to as the tone. s \|Score 1\|eshe22\|Points 4711\| Updated 2/27/2019 3:49:21 PM Confirmed by jerry06 [2/27/2019 3:49:21 PM] Rating 1.What's the difference between the speed and velocity of an object Weegy: Velocity deals with direction, while speed does not is the difference between the speed and velocity of an object. (More) Updated 2/21/2019 9:58:57 AM describes the differences between the SI and English systems of measurement Updated 7/13/2019 12:49:56 AM The SI system, which is based on the metric system, has been more widely adopted than the English system, which is harder to use accurately describes the differences between the SI and English systems of measurement. If a stone with an original velocity of 0 is falling from a ledge and takes 8 seconds to hit the ground, what is the final velocity of the stone? A. 156.8 m/s B. 42.6 m/s C. 78.4 m/s D. 39.2 m/s Updated 7/8/2019 10:58:41 AM If a stone with an original velocity of 0 is falling from a ledge and takes 8 seconds to hit the ground, the final velocity is 313.6 meters. y(8) = -1/2 (9.81 * 8 ) = 313.62 An object has a mass of 13.5 kilograms. What force is required to accelerate it to a rate of 9.5 m/s2? A. 128.25 N B. 81.00 N C. 132.30 N D. 173.75 N Updated 7/13/2019 12:49:32 AM 128.25 N force is required to accelerate it to a rate of 9.5 m/s2. force = ma = 13.5 kg * 9.5 m/s2 = 128.25 N An object has a mass of 13.5 kilograms. What force is required to accelerate it to a rate of 9.5 m/s2? A. 128.25 N B. 81.00 N C. 132.30 N D. 173.75 N Updated 7/8/2019 10:57:51 AM 128.25 N force is required to accelerate it to a rate of 9.5 m/s2. force = ma = 13.5 kg * 9.5 m/s2 = 128.25 N 32,572,220 Popular Conversations What is wailing? An ocean wave is an example of a(n) _____ wave form. Weegy: An ocean wave is an example of a surface wave. User: primary wave User: a space in which particles have ... There are twelve months in year. Weegy: 12+12 = 24 User: Approximately how many hours does she play soccer in a year? Amendment two Weegy: An amendment is a formal or official change made to a law, contract, constitution, or other legal document. ... What is the product of 3 2/3 and 14 2/5? A. 52 4/5 B. 52 4/15 C. ... Weegy: 3 2/3 * 14 2/5 What is abstract Weegy: An abstract is a self-contained, short, and powerful statement that describes a larger work. User: What is ... * Get answers from Weegy and a team of really smart live experts. S L L Points 2022 [Total 6756] Ratings 0 Comments 2012 Invitations 1 Offline S L L 1 Points 1978 [Total 8740] Ratings 15 Comments 1828 Invitations 0 Offline S L P R P R L P P C R P R L P R P R P R Points 1728 [Total 17402] Ratings 4 Comments 1418 Invitations 27 Offline S L R Points 1726 [Total 2060] Ratings 0 Comments 1726 Invitations 0 Offline S L Points 1114 [Total 1282] Ratings 1 Comments 1104 Invitations 0 Offline S L P P P 1 P L 1 Points 502 [Total 8653] Ratings 7 Comments 432 Invitations 0 Offline S L P 1 L Points 430 [Total 5964] Ratings 5 Comments 380 Invitations 0 Online S L Points 392 [Total 423] Ratings 0 Comments 392 Invitations 0 Offline S L Points 305 [Total 305] Ratings 0 Comments 305 Invitations 0 Offline S L P R P R L Points 273 [Total 6136] Ratings 0 Comments 253 Invitations 2 Offline * Excludes moderators and previous winners (Include) Home \| Contact \| Blog \| About \| Terms \| Privacy \| © Purple Inc.	1,204	3,525	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45	latest	en	0.933635
https://www.biostars.org/p/162276/	1,585,519,115,000,000,000	text/html	crawl-data/CC-MAIN-2020-16/segments/1585370496227.25/warc/CC-MAIN-20200329201741-20200329231741-00291.warc.gz	846,945,208	7,689	Question: DESeq2 RIP-seq design with multiple factors 0 4.5 years ago by Bontus80 Netherlands Bontus80 wrote: Dear all, I have a question concerning DESeq2's multi-factor design for a RIP-seq experiment. My experimental design is quite complicated, as can be seen in the following table: condition input_output treatment no_tag input no_stress tag input no_stress no_tag input stress tag input stress no_tag output no_stress tag output no_stress no_tag output stress tag output stress Here, input means RNA after extraction from the cells and output refers to the immunoprecipitated (IP) RNA, where IP uses an antibody against the tag. Therefore, the samples containing a tagged protein should show an enrichment of RNAs compared to the input and, if the signal is real (no random binding), also compared to the tag-free samples. My initial approach was to follow the analysis described here: https://support.bioconductor.org/p/61509/ and test for significant enrichment under 'stress' and 'no stress' separately with the following design: ```design = ~condition+input_output+condition:input_output; ddsCountMatrix <- DESeqDataSetFromMatrix( colData = sample_information_stress), countData = count_table_stress, design = design); dds <- DESeq(ddsCountMatrix); reduce = ~condition+input_output; dds <- DESeq(dds, test = 'LRT', reduced = reduce); res <- results(dds,altHypothesis='greater');``` The model matrix generated by this design looks like this: `model.matrix(~input_output+condition+input_output:condition,sample_information_stress)` ``` (Intercept) input_outputoutput conditionDhh1 5 1 0 0 6 1 0 1 7 1 0 0 8 1 0 1 13 1 1 0 14 1 1 1 15 1 1 0 16 1 1 1 input_outputoutput:conditionDhh1 5 0 6 0 7 0 8 0 13 0 14 1 15 0 16 1``` So Intercept comprises 'no tag' and 'input'. However, contrary to my assumption, the resulting list of genes is expressed higher in the control samples despite having a positive LFC (specified via altHypothesis). My questions therefore are: 1. What am I missing in the design to find genes that are enriched in output tag vs. input tag and output no_tag? 2. Is it possible to include the stress vs. no stress comparison in the design as well? Or should I stick to identifying genes above background with the above methodology and then continue with a smaller gene list for comparing stress vs. no stress? Any help is greatly appreciated, thanks in advance and best regards, René rip-seq deseq2 R multifactor • 1.9k views	676	3,035	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16	longest	en	0.749149
https://www.largeformatphotography.info/forum/showthread.php?76158-How-much-does-the-scan-take-away-from-the-Negative&s=5110662d63fd44fefb98256e91c0b61f&p=727359&viewfull=1	1,585,733,006,000,000,000	text/html	crawl-data/CC-MAIN-2020-16/segments/1585370505550.17/warc/CC-MAIN-20200401065031-20200401095031-00423.warc.gz	1,002,031,952	14,485	# Thread: How much does the scan take away from the Negative? 1. ## How much does the scan take away from the Negative? I haven't studied this one much, but given one has a top flight scan of a well exposed neg, how much of that neg gets taken away from the scanning/processing of it into a file? In other words, say we look at the neg on light table...how much of the color/tones/shadows/information/etc. gets lost in the scanning process "and" does the scan actually make the neg look more "digitized" and not as "realistic" as the neg does? 2. ## Re: How much does the scan take away from the Negative? It's part Quantitative and part Qualitative. Not just a one-sided coin. Some say that enlarging is a destructive step - so they make contact prints. They're right to an extent. The question is, how big is that extent ? Some will opine that there is no substitute for an analog image - just as some will say that no recording can capture the sound of a musical instrument. Ultimately, they are right, but Life is rarely ultimate For the rest of us, the advantages of digitization and the subsequent adjustments that it affords, outweigh the minimal loss incurred. 3. ## Re: How much does the scan take away from the Negative? Every process, analog or digital, is going to remove some of negative's information. But photo paper or inkjet can only reproduce a fraction of what is in the negative anyway so it's a moot point. The question is how to best distribute the tones that you can retain, most people can't do that well. 4. ## Re: How much does the scan take away from the Negative? There's so much information in a negative that loosing a little of it in the digital or optical printing process is hardly noticeable. The larger the negative, the more information it contains, and it can be enlarged, either by scanning or enlarging optically, to a much larger size. Some people are skilled at determining whether a negative has been printed optically or digitally when examining a mounted print. Unless it was done badly, I cannot tell the difference. 5. ## Re: How much does the scan take away from the Negative? Originally Posted by Frank Petronio Every process, analog or digital, is going to remove some of negative's information. But photo paper or inkjet can only reproduce a fraction of what is in the negative anyway so it's a moot point. The question is how to best distribute the tones that you can retain, most people can't do that well. Frank why do i always find myself agreeing with you , Ditto what he said (i just wish i had said it first) . Cheers Gary 6. ## Re: How much does the scan take away from the Negative? Steals its soul. 7. ## Re: How much does the scan take away from the Negative? My experience tell me it goes both ways. Some high frequency detail can be lost though that is usually only apparent with more than 3-4X enlargement. On the other hand, digital editing can be much more precise than analog editing (basically waving a hand or a piece of card stock under the enlarger) and can provide much greater control of the distribution of tones. Not "take away", but digital scanning is sensibly "different" than enlarging and better in some ways. bob 8. ## Re: How much does the scan take away from the Negative? Nothing is taken away from the negative, not by darkroom printing and not by scanning. In both cases, something else is created from the negative. No reproduction is perfect, and if it was, you'd be right back where you started with a small negative with silver (or dyes) on film base. It only makes sense to compare real uses and real final results, discussing abstractly only turns it into a game of defining what "information", "colors", "tones", "detail" etc. means. If you are not scanning, what do you would you do instead? Darkroom prints, projected slides, or is "looking on the light table" the final intended use? 9. ## Re: How much does the scan take away from the Negative? It doesn't matter. Are wesupposed to look at the print or the negative? Look at prints. If the prints are what you intended, then they are good. If they are not, keep trying and experimenting until they are. Sorta what Fred Picker and Paul Strand said. 10. ## Re: How much does the scan take away from the Negative? A scan cannot cover the complete color gamut, only a small portion. Film covers the complete color gamut and more, therefore the use of Skylight 1A, Skylight 1B and Haze filters. Other than that scans cannot reproduce the resolution of the film grain molecules. So if color and resolution are not important to you, then scanning is the way to go. Steve #### Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts •	1,070	4,792	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.65625	3	CC-MAIN-2020-16	latest	en	0.95968
https://integers.info/75	1,718,601,618,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861696.51/warc/CC-MAIN-20240617024959-20240617054959-00790.warc.gz	285,634,455	4,547	# integers.info Seventy-five ## 75 in other numeral systems Binary: Octal: Hex: Roman: 3 × 52 ## Divisors of 75 1, 3, 5, 15, 25 and 75 ## Is 75 in the Fibonacci number sequence? No. Its nearest Fibonacci number neighbors are 55 and 89. ## The 75th chemical element Rhenium (Re) is the 75th chemical element.	102	319	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.625	3	CC-MAIN-2024-26	latest	en	0.807137
https://www.mathworks.com/matlabcentral/cody/problems/1483-number-of-paths-on-a-grid/solutions/240918	1,496,032,855,000,000,000	text/html	crawl-data/CC-MAIN-2017-22/segments/1495463612013.52/warc/CC-MAIN-20170529034049-20170529054049-00313.warc.gz	1,124,289,377	11,815	Cody # Problem 1483. Number of paths on a grid Solution 240918 Submitted on 6 May 2013 This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. ### Test Suite Test Status Code Input and Output 1 Fail %% m = 1; n = 1 ; y_correct = 0; assert(isequal(paths2dest_ongrid(m,n),y_correct)) ```Error: Assertion failed. ``` 2 Pass %% m = 2; n = 2 ; y_correct = 2; assert(isequal(paths2dest_ongrid(m,n),y_correct)) ``` ans = 4 ``` 3 Pass %% m = 4; n = 3 ; y_correct = 10; assert(isequal(paths2dest_ongrid(m,n),y_correct)) ``` ans = 7 ``` 4 Pass %% m = 6; n = 5 ; y_correct = 126; assert(isequal(paths2dest_ongrid(m,n),y_correct)) ``` ans = 11 ``` 5 Pass %% m = 5; n = 5 ; y_correct = 70; assert(isequal(paths2dest_ongrid(m,n),y_correct)) ``` ans = 10 ``` 6 Pass %% m = 1; n = 100 ; y_correct = 1; assert(isequal(paths2dest_ongrid(m,n),y_correct)) ``` ans = 101 ``` 7 Pass %% m = 100; n = 1 ; y_correct = 1; assert(isequal(paths2dest_ongrid(m,n),y_correct)) ``` ans = 101 ``` 8 Pass %% m = 2; n = 100 ; y_correct = 100; assert(isequal(paths2dest_ongrid(m,n),y_correct)) ``` ans = 102 ``` 9 Pass %% m = 100; n = 2 ; y_correct = 100; assert(isequal(paths2dest_ongrid(m,n),y_correct)) ``` ans = 102 ``` 10 Pass %% m = 15; n = 20 ; y_correct = 818809200; assert(isequal(paths2dest_ongrid(m,n),y_correct)) ``` ans = 35 ```	507	1,401	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.359375	3	CC-MAIN-2017-22	longest	en	0.414541
http://mathoverflow.net/questions/37812?sort=newest	1,369,547,659,000,000,000	text/html	crawl-data/CC-MAIN-2013-20/segments/1368706631378/warc/CC-MAIN-20130516121711-00016-ip-10-60-113-184.ec2.internal.warc.gz	171,775,209	14,538	## Christodoulou’s paper on naked singularities in inhomogeneous dust collapse I have been studying of late about formation of naked singularities in certain collapse scenarios in Einstein's theory. It seems to me that the canonical paper to read about how such a formation is established is the 1984 paper by Christodoulou in Communications in Mathematical Physics. ( http://www.ams.org/mathscinet-getitem?mr=742192 ) I was wondering if there is a reference which gives a more modern rewriting of the proof in that paper which say sort of highlights the generic technique of the proof which the reader can take away from there for other scenarios. Somehow even the most recent books on Einstein's theory like the otherwise brilliant book by Choquet-Bruhat also doesn't dwell on techniques of testing in a collapse scenario whether the curvature singularity is naked or not. I haven't seen till now any generic method or algorithm for testing this. It would be great if someone can give me some references/explanations along these lines. - Your question is very broad, and so I'll just give some very broad answers too. Collapse scenarios Are you absolutely sure you want to restrict yourself to dust collapse? In the case of a spherically symmetric scalar field, there is also http://www.ams.org/mathscinet-getitem?mr=1307898 (and this paper which shows that naked singularities in the scalar field model are unstable). There's some problem of the interpretation of the dust collapse as a physical formation. Here I'll quote Demetri from his book on vacuum collapse due to incoming gravitational waves With the above remarks in mind the author turned to the study of the gravitational collapse of an inhomogeneous dust ball. In this case, the initial state is still spherically symmetric, but the density is a function of the distance from the center of the ball. The corresponding spherically symmetric solution had already been obtained in closed form by Tolman in 1934, in comoving coordinates, but its causal structure had not been investigated. This required integrating the equations for the radial null geodesics. A very diﬀerent picture from the one found by Oppenheimer and Snyder emerged from this study. The initial density being assumed a decreasing function of the distance from the center, so that the central density is higher than the mean density, it was found that as long as the collapse proceeds from an initial state of low compactness, the central density becomes inﬁnite before a black hole has a chance to form, thus invalidating the neglect of pressure and casting doubt on the predictions of the model from this point on, in particular on the prediction that a black hole eventually forms. Essentially the same comment was made by Hájíček in his MathReviews on the dust collapse paper. Modern re-writes Part of the reason that there are no modern re-writes of the proof is because of what I mentioned above, that the violation of weak cosmic censorship is unphysical (so it didn't attract that much attention). On the other hand, the basic idea behind the proof is not too hard (it is in the implementation of the analysis that is difficult). Simply speaking, in spherical symmetry, there is a lot of qualitative information that can be extracted without knowing too much details of the matter model involved (we must have matter as for vacuum, spherically symmetric space-times, we have Birkhoff's theorem). Perhaps a best modern reference is Mihalis Dafermos' paper in CQG. The most important part is that under symmetry conditions and some mild assumptions, future null infinity must be complete; in the spherically symmetric space-times, future null infinity is characterised by the area-radius $r\to\infty$ along out-going null geodesics. Thus the basic idea is to show that there exists out-going null geodesics, such that if you travel along the future direction the area radius increases without bound, and that if you travel along the past direction you will hit the singularity. Now, in the case of the Tolman dust which was studied by Christodoulou, since the exterior of the dust cloud is glued to a Schwarzschild solution, there is a dichotomy: either the out-going null geodesic escapes the dust region before hitting the apparent horizon, or the null geodesic hits the apparent horizon first. In spherical symmetry, it is a general fact that the apparent horizon consists of space-like or expanding null portions. So once a null geodesic passes the apparent horizon it can no longer escape to infinity. On the other hand, inside the Schwarzschild region the apparent horizon agrees with the event horizon, so once the null geodesic escapes from dust without hitting the apparent horizon, it will remain in the domain of outer communications and escape to infinity. So to demonstrate existence of naked singularity, it suffices to show that a null geodesic emanating from the first point of singularity (it doesn't make sense to consider later points, as they will no longer belong to the Cauchy development of an initial data set) will escape the dust cloud before hitting the apparent horizon. To do so one needs to estimate the size of the solution of an ODE. In spherical symmetry, the apparent horizon is characterised by $2m/r = 1$, where $m$ is the Hawking mass and $r$ is the area radius. Now the Hawking mass satisfies ordinary differential equations (with source) in spherical symmetry (see, for example, Equation 3 here; or see Section 3 of this paper.) So to estimate the size of the Hawking mass at the matching boundary, it is necessary to estimate the source terms in the ODE. This will lead to equation chasing using the Einstein equations and the matter-field equations. The rest is just being clever (in deciding in which order to estimate the various quantities) and doing the hard work of computation. General techniques The reason there are no references for general techniques in testing whether a singularity is naked or not is very simply, there exists no such techniques. In spherical symmetry there is the simple description I posted above, but at the end of the day the estimates strongly depend on the structure of the matter equation (its separability and what not), so can only be really dealt with in a case-by-case basis. Of course, if you are given an explicit solution of the equations, testing whether the singularity is naked is often a simple computation re-writing the solution in some sort of null coordinates. The problem is that to prove genericity or to study singularities without reference to a explicit solution, one needs analytical estimates which, like I said, depends on which equations you are studying. In fact, if you could come up with a usable, generic method of testing whether a singularity is clothed, you would be half-way there resolving the general weak cosmic censorship conjecture. Some last comments The general issue of weak cosmic censorship is a wide open one. The problem is that the statement contains the word "generic" in it (generic initial data sets lead to blah blah blah). So while there has been quite a lot of work going into constructing solutions and verifying that those solutions contain naked singularities, these works say nothing about weak cosmic censorship. (Explicit solutions tend to be non-generic in the space of solutions, except in the case you have strong rigidity theorems like Birkhoff's theorem for spherically symmetric Einstein-vacuum/electro-vacuum space-times or the No Hair Theorem for four-dimensional stationary axi-symmetric solutions.) The only real progress to weak cosmic censorship have all been due to Christodoulou (most of the physics papers are lacking in rigour). (Interestingly, there has been more developments in strong cosmic censorship, which, despite the name, has relatively little to do with weak cosmic censorship.) Most recently, the focus in the community seems to be that the next model to consider for cosmic censorship should be the Einstein-Vlasov model (in spherical symmetry). (Well, it has been under consideration for around 10 years now and still prohibitingly hard.) For general solutions without symmetry assumptions, there has been essentially zero work in the field. There was an attempt to reformulate the conjecture into something mathematically tractable, but not has been done with the reformulation. - @Willie Thanks a lot for this amazingly detailed answer! When you say that "future null infinity is complete" I suppose you mean just what you said in the next line about the spatial scale growing unbounded along null geodesics. Or are you making some statement also about the structure of the future null infinity? Further is this test of yours the same as the "future in/completeness theorem" by Choquet and Costakis? (which relies on the nature of the lapse function for a regularly sliced space-time) ? – Anirbit Sep 10 2010 at 16:57 Contd. The statement of Strong Censorship Conjecture that Choquet quotes is the one formalized by Eardley and Moncrief which states that a maximal globally hyperbolic vacuum Einstein development of generic initial data is inextensible if the initial data is asymptotically euclidean or compactly supported. In this formulation how does the existence of event horizon view point fit in? – Anirbit Sep 10 2010 at 17:00 Contd. Your definition of the apparent horizon sounds a bit new to me. Can you kindly relate it to the definition using the Ricci tensor? Also Christodoulou talks about whether or not the null geodesic from the singularity can or cannot escape before the apparent horizon forms. (assuming equations of state such that there was no apparent horizon to begin with). Can you kindly explain this dynamical view of the apparent horizon which is a bit confusing since it seems that once I am given the metric the apparent horizon is completely known. – Anirbit Sep 10 2010 at 17:10 Completeness of future null infinity means that, under suitable compactification, the null generators of future null infinity are complete (as null geodesics). It is not the same as the statement about spatial scale. The precise definition is given in the last paper I cited in my answer (though you should only consider the future direction, not the past one). Which paper of Choquet and Costakis are you referring to? I cannot find any paper jointly authored by them on MathSciNet. – Willie Wong Sep 10 2010 at 18:07 As to strong cosmic censorship, the idea is thus: there can be no extension past future null infinity anyway (it is a cosmological horizon). But in the interior of the black hole region (past the endpoint of the event horizon), there may exist a Cauchy horizon (the Cauchy horizon indicates the boundary of the maximal globally hyperbolic development of some initial data set). In the Schwarzschild solution, this boundary is singular (corresponding to the r = 0 singularity). In the Reissner-Nordstrom solution, this boundary is regular (corresponding to the inner horizon). – Willie Wong Sep 10 2010 at 18:09 show 17 more comments	2,316	11,085	{"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": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.53125	3	CC-MAIN-2013-20	latest	en	0.946124
https://www.wise-geek.com/what-is-coulometry.htm	1,638,316,842,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964359082.76/warc/CC-MAIN-20211130232232-20211201022232-00282.warc.gz	1,190,228,628	16,272	# What Is Coulometry? Phil Riddel Phil Riddel Coulometry is a form of quantitative analysis used to determine the concentration of a substance in solution — generally referred to as the analyte — by measuring the amount of electricity required to change the oxidation state of the substance. The oxidation state of the analyte can be reduced by adding electrons or increased by removing electrons. The transfer of electrons to or from a substance can take place chemically in reduction-oxidation reactions, but can also be achieved by electricity, which is a flow of electrons, using electrolysis. By measuring the amount of electrical charge transferred during the complete oxidation or reduction of the analyte by electrolysis, it is possible to calculate the amount that was present. This is in accordance with Faraday’s first law of electrolysis, which states that the amount of a substance transformed during electrolysis is in direct proportion to the amount of electricity transferred. In chemistry, the unit of measurement for the quantity of a substance is normally the mole, which is defined as the number of atoms in 0.42 ounces (12 grams) of carbon-12. Electrons can also be measured in moles. It is therefore possible to calculate how many moles of electrons would be needed to convert, for example, a given amount of copper in solution from its +2 oxidation state to neutral copper metal. Two moles of electrons are required for each mole of copper: Cu2+ + 2e- -> Cu. Thus, in a solution containing an unknown amount of Cu2+ ions, the quantity present, measured in moles, can be determined by measuring the quantity of electrons, in moles, which are used in performing this conversion to completion. Electrical charge is measured in coulombs, one coulomb being the amount of charge transferred by a current of one ampere in one second. One mole of electrons is equivalent to just over 96,485 coulombs. By measuring the amount of time taken for a known current to complete a reaction like the one above, the number of moles of electrons used can be calculated and from this, the number of moles of the analyte determined. This type of coulometry is known as controlled current coulometry or coulometric titration. The method requires a means of determining when the endpoint of the reaction has been reached, such as a chemical indicator. Another type of coulometric analysis is controlled potential coulometry. In this case, a constant electrical potential is maintained and the current is measured. As the reaction proceeds, the current decreases and approaches zero when the reaction is complete. The number of moles of analyte can be calculated from the time taken for the current to drop to zero.	560	2,719	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.84375	3	CC-MAIN-2021-49	latest	en	0.948358
https://briankoberlein.com/2014/01/30/black-holes-quite/	1,553,336,779,000,000,000	text/html	crawl-data/CC-MAIN-2019-13/segments/1552912202781.83/warc/CC-MAIN-20190323101107-20190323123107-00042.warc.gz	428,057,669	14,523	# Black Holes No More? Not Quite. In Black Holes by Brian Koberlein0 Comments This post was originally written for Universe Today. Nature News has announced that there are no black holes. This claim is made by none other than Stephen Hawking, so does this mean black holes are no more? It depends on whether Hawking’s new idea is right, and on what you mean be a black hole. The claim is based on a new paper by Hawking that argues the event horizon of a black hole doesn’t exist. The event horizon of a black hole is basically the point of no return when approaching a black hole. In Einstein’s theory of general relativity, the event horizon is where space and time are so warped by gravity that you can never escape. Cross the event horizon and you can only move inward, never outward. The problem with a one-way event horizon is that it leads to what is known as the information paradox. Professor Stephen Hawking during a zero-gravity flight. Image credit: Zero G. The information paradox has its origin in thermodynamics, specifically the second law of thermodynamics. In its simplest form it can be summarized as “heat flows from hot objects to cold objects”. But the law is more useful when it is expressed in terms of entropy. In this way it is stated as “the entropy of a system can never decrease.” Many people interpret entropy as the level of disorder in a system, or the unusable part of a system. That would mean things must always become less useful over time. But entropy is really about the level of information you need to describe a system. An ordered system (say, marbles evenly spaced in a grid) is easy to describe because the objects have simple relations to each other. On the other hand, a disordered system (marbles randomly scattered) take more information to describe, because there isn’t a simple pattern to them. So when the second law says that entropy can never decrease, it is say that the physical information of a system cannot decrease. In other words, information cannot be destroyed. The problem with event horizons is that you could toss an object (with a great deal of entropy) into a black hole, and the entropy would simply go away. In other words, the entropy of the universe would get smaller, which would violate the second law of thermodynamics. Of course this doesn’t take into account quantum effects, specifically what is known as Hawking radiation, which Stephen Hawking first proposed in 1974. The original idea of Hawking radiation stems from the uncertainty principle in quantum theory. In quantum theory there are limits to what can be known about an object. For example, you cannot know an object’s exact energy. Because of this uncertainty, the energy of a system can fluctuate spontaneously, so long as its average remains constant. What Hawking demonstrated is that near the event horizon of a black hole pairs of particles can appear, where one particle becomes trapped within the event horizon (reducing the black holes mass slightly) while the other can escape as radiation (carrying away a bit of the black hole’s energy). Hawking radiation near an event horizon. Credit: NAU. Because these quantum particles appear in pairs, they are “entangled” (connected in a quantum way). This doesn’t matter much, unless you want Hawking radiation to radiate the information contained within the black hole. In Hawking’s original formulation, the particles appeared randomly, so the radiation emanating from the black hole was purely random. Thus Hawking radiation would not allow you to recover any trapped information. To allow Hawking radiation to carry information out of the black hole, the entangled connection between particle pairs must be broken at the event horizon, so that the escaping particle can instead be entangled with the information-carrying matter within the black hole. This breaking of the original entanglement would make the escaping particles appear as an intense “firewall” at the surface of the event horizon. This would mean that anything falling toward the black hole wouldn’t make it into the black hole. Instead it would be vaporized by Hawking radiation when it reached the event horizon. It would seem then that either the physical information of an object is lost when it falls into a black hole (information paradox) or objects are vaporized before entering a black hole (firewall paradox). In this new paper, Hawking proposes a different approach. He argues that rather than instead of gravity warping space and time into an event horizon, the quantum fluctuations of Hawking radiation create a layer turbulence in that region. So instead of a sharp event horizon, a black hole would have an apparent horizon that looks like an event horizon, but allows information to leak out. Hawking argues that the turbulence would be so great that the information leaving a black hole would be so scrambled that it is effectively irrecoverable. If Stephen Hawking is right, then it could solve the information/firewall paradox that has plagued theoretical physics. Black holes would still exist in the astrophysics sense (the one in the center of our galaxy isn’t going anywhere) but they would lack event horizons. It should be stressed that Hawking’s paper hasn’t been peer reviewed, and it is a bit lacking on details. It is more of a presentation of an idea rather than a detailed solution to the paradox. Further research will be needed to determine if this idea is the solution we’ve been looking for.	1,185	5,535	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13	latest	en	0.93677
https://kunduz.com/questions-and-answers/fernando-is-doing-his-pre-lab-homework-for-chemistry-he-is-asked-to-predict-how-many-grams-of-co2-are-produced-if-209-mol-of-hci-are-reacted-according-to-this-balanced-chemical-equation-shown-caco3-2hci-cacl2-co2-h2o-68071/	1,679,972,059,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296948756.99/warc/CC-MAIN-20230328011555-20230328041555-00132.warc.gz	414,488,323	30,989	Question: # Fernando is doing his pre-lab homework for Chemistry. He is Last updated: 8/9/2022 Fernando is doing his pre-lab homework for Chemistry. He is asked to predict how many grams of CO2 are produced if 2.09 mol of HCI are reacted according to this balanced chemical equation shown. CaCO3 + 2HCI → CaCl2 + CO2 + H2₂O He performs the following calculation. 1 mol CO₂ 2 metHCI 2.09 metHCT x X from the coefficients of the balanced equation 44.01 g CO₂ 1 mol CO₂ = 46.0 g CO₂ On the day of the lab his group obtains 43.1 grams of CO2 as their experimental result. What is their percent yield? round your answer to the tenths place, i.e. one decimal place	186	661	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.71875	3	CC-MAIN-2023-14	latest	en	0.970731
https://dsp.stackexchange.com/questions/31309/how-to-map-cwt-to-synchrosqueezed-wavelet-transform/32263	1,718,275,994,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861372.90/warc/CC-MAIN-20240613091959-20240613121959-00321.warc.gz	199,551,485	42,465	# How to Map CWT to Synchrosqueezed wavelet transform? I don't understand the mapping time-scale plane to the time-frequency plane in synchrosqueezed wavelet transform, i.e. $(3)$. You can find the paper here. For the given signal of $x(t)$ and mother wavelet of $\psi$, the continous wavalet transform is: $$W(u,s)= \int_{-\infty}^{+\infty} x(t)\psi_{u,s}^(t)dt\tag{1}\\$$ From $(1)$, a preliminary frequency $\omega (u, s)$ is obtained from the oscillatory behavior of $W (u, s)$ in $u$: $$\omega(u, s) = − i\left(W(u, s)\right)^{−1} \frac{\partial }{\partial u} W(u, s)\tag{2}$$ $W(u,s)$ is then transformed from the time-scale plane $(1)$ to the time-frequency plane $(3)$. Each value of $W(u, s)$ is reassigned to $(u, \omega_{l} )$, where $\omega_{l}$ is the frequency that is the closest to the preliminary frequency of the original (discrete) point $\omega(u, s)$. $$T(u, \omega_{l}) = \left(\Delta\omega\right)^{-1} \sum_{sk:\|\omega(u, s_{k} )−\omega_{l}\|\leq\Delta\omega/2}^{} {W(u, s_{k})s_{k}^{−3/2}\Delta s}\tag{3}$$ Let me explain the intuition briefly. The authors of the paper you've cited assume that the signal $x(t)$ can be written in the form \begin{align} x(t) &= \sum_{k=1}^K a_k(t) \exp(2\pi\mathrm{i} \phi_k(t)), \end{align} where the $a_k$ denote instantaneous amplitudes, the $\phi_k$ denote instantaneous phases and the $\phi'_k$ denote instantaneous frequencies. The continuous wavelet transform enables the visualization of the $\phi'_k$ in the time-scale plane, but the visualization is necessarily blurry due to the Fourier uncertainty principle. The idea is that the peaks of the continuous wavelet transform can give you an idea of what the $\phi'_k$ values are, but the energy is spread out in the time-scale plane, so the exact values of $\phi'_k$ may not be easy to obtain. The goal of the synchrosqueezing transform (SST) is to partially undo the blurring by a frequency reassignment, which can be explained as follows. First of all, let's note that each scale $s$ corresponds to a natural frequency $\xi$ satisfying the relation $s = c/\xi$, where $c$ is the center frequency of the mother wavelet $\psi$. Now, suppose the time $u$ is fixed. If $\xi = c/s$ is close but not exactly equal to an instantaneous frequency $\phi'_k(u)$, then the coefficient $W(u,s)$ will have some nonzero energy (i.e., $\|W(u,s)\|^2 > 0$). The idea of synchrosqueezing is to move all this energy away from the frequency $\xi$, and reassign its frequency location closer to the instantaneous frequency $\phi'_k(u)$. Then, we can arrive at a time-frequency representation where the energy is more closely concentrated around the instantaneous frequency curves. So, for each scale $s$, we compute the frequency reassignment $\omega(u,s)$ of the wavelet transform coefficient $W(u,s)$ by the formula you cited above. As Daubechies, Wu, and Lu discovered in the paper you've linked, it turns out that $\omega(u,s)$ is often a good approximation to the instantaneous frequency curve $\phi'_k(u)$ for values of $(u,s)$ where $\xi = c/s$ is sufficiently close to $\phi'_k(u)$. (In fact, when $\phi'_k(u)$ is a constant function, $\omega(u,s)$ exactly equals $\phi'_k(u)$ for values of $s$ where $\xi = c/s$ is sufficiently close to $\phi'_k(u)$!) Then, the computation of the SST in the continuous setting is as follows. First, fix a time of interest $u$. Next, compute the frequency reassignment $\omega(u,s)$ for all scale values $s$. Then, for each frequency of interest $\eta$, we compute the SST $T(u,\eta)$ by adding up all values $W(u,s)$ where the reassigned frequency $\omega(u,s)$ equals $\eta$. This means \begin{align} T(u,\eta) &= \int_\mathbb{R} W(u,s) \delta(\eta - \omega(u,s)) s^{-3/2} ds, \end{align} where $\delta$ is the Dirac delta. (To ensure the convergence result for SST, the authors use a formulation for $T$ which includes an approximation to $\delta$ rather than $\delta$ itself, but in computational practice I've never found this approximation to be necessary.) The multiplication by $s^{-3/2}$ is necessary for reconstruction purposes (which you can read more about in the paper). In practice, one is limited to the discrete setting where we have only finitely many possible frequency bins $\eta_\ell$. In this case, for a fixed frequency $\eta_{\ell_0}$, one finds all the values of $\omega(u, s)$ which are closer to $\eta_{\ell_0}$ than to any other frequency bin $\eta_\ell$. We can more explicitly formulate the SST in the discrete setting as follows. For simplicity's sake let's assume the possible frequency values $\eta_\ell$ are uniformly spaced by a distance $\Delta \omega$. So, in the discrete setting, the formula above becomes \begin{align} T(u, \eta_\ell) &= \sum_{s: \|\omega(u,s) - \eta_\ell\| < \Delta\omega/2} W(u,s) s^{-3/2}, \end{align*} where I haven't bothered to give an explicit notation that shows that the values $u$ and $s$ are discrete. By doing this reassignment, you end up with a representation $T$ which is sparser than $W$, and hopefully very sharply concentrated about the curves $\phi'_k$. (In practice, for visualization purposes, it's better to reassign not the original coefficient $W(u,s)$ but the magnitude-squared $\|W(u,s)\|^2$, or even just the constant number $1$. This is because the coefficients $W(u,s)$ are complex-valued, and summing these coefficients may not actually leave you with a very large energy. However, if you additionally want to use the reconstruction formula given in the paper to compute the $x_k$ from the SST, then you need to sum the $W(u,s)$ in order for that formula to work.) By the way, I am leaving out the fact that we generally throw out coefficients $W(u,s)$ that do not fall above a certain threshold. This is because the computation of $\omega(u,s)$ is not necessarily accurate for small $W(u,s)$. • A truly excellent response. Thank you!!! Commented May 29, 2020 at 21:20 • Has anyone found a Python implementation of this? Commented May 29, 2020 at 21:23 • Bumping this a I have a strong interest myself in seeing WSST ported to python. There is a project still running which attempts to port syncrosqueeze wavelet transform and incorporate the library into 'pywt'. link to current repository:github.com/OverLordGoldDragon/ssqueezepy progress on incorporating into 'pywt'': github.com/PyWavelets/pywt/issues/543 Although initial results for porting WSST seems succesful, I would wait (or contact the respository owner to contribute) until further notice. Commented Sep 14, 2020 at 10:09 • @DavidBondesson Can confirm, I'm actively working on it. Currently comparing CWT implementation with official one, here. Commented Sep 29, 2020 at 1:02 • Good answer overall, two critiques: (1) it's not so much about "assumes can be written in the form", as any $f(t)$ qualifies just as in Fourier Transform - rather, that it forms a good statistical prior (is representative of source physical process); (2) reassignment by a constant isn't generally better for visualization, and by magnitude is akin to squeezing rather than synchro-squeezing and loses benefits of partial denoising. -- I've written an up-to-date answer here. Commented Nov 12, 2020 at 23:26	1,902	7,190	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.1875	3	CC-MAIN-2024-26	latest	en	0.807199
https://gmatclub.com/forum/astronomers-at-the-palomar-observatory-have-discovered-a-131530.html	1,524,474,386,000,000,000	text/html	crawl-data/CC-MAIN-2018-17/segments/1524125945940.14/warc/CC-MAIN-20180423085920-20180423105920-00573.warc.gz	627,890,846	54,475	GMAT Changed on April 16th - Read about the latest changes here It is currently 23 Apr 2018, 02:06 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # Astronomers at the Palomar Observatory have discovered a Author Message TAGS: ### Hide Tags Intern Joined: 14 Sep 2007 Posts: 6 Astronomers at the Palomar Observatory have discovered a [#permalink] ### Show Tags 13 Aug 2008, 22:17 19 This post was BOOKMARKED 00:00 Difficulty: 75% (hard) Question Stats: 45% (00:48) correct 55% (00:59) wrong based on 1268 sessions ### HideShow timer Statistics 145. Astronomers at the Palomar Observatory have discovered a distant supernova explosion, one that they believe is a type previously unknown to science. (A) that they believe is (B) that they believe it to be (C) they believe that it is of (D) they believe that is (E) they believe to be of [Reveal] Spoiler: Among the emotions on display in the negotiating room were anger for repeatedly raising the issue over and over again and preventing the raw wounds from earlier battles from ever beginning to heal. (A) were anger for repeatedly raising the issue over and over again and preventing the raw wounds from earlier battles from ever beginning to heal (B) was anger for repeatedly raising the issue and preventing the raw wounds from earlier battles from ever beginning to heal (C) were anger over repeatedly raising the issue and preventing the raw wounds from earlier battles to begin healing (D) was anger about the issue, which was raised over and over, and preventing the wounds from earlier battles, still raw, to begin healing (E) were anger about the issue, which was raised repeatedly, and preventing the raw wounds from earlier battles to begin to heal [Reveal] Spoiler: OA Senior Manager Joined: 07 Jan 2008 Posts: 375 Re: Astronomers at the Palomar Observatory have discovered a [#permalink] ### Show Tags 13 Aug 2008, 23:11 105. Anger: singular D: not parallel. B is best 145 Idiom: believe to be. Do not need THAT following ONE. E is best. Manager Joined: 23 Jul 2008 Posts: 186 Re: Astronomers at the Palomar Observatory have discovered a [#permalink] ### Show Tags 14 Aug 2008, 07:16 IMO B and E Among the emotions on display was anger....... D which's use is ambiguos In the 2nd one belive to be is the correct idiom Manager Joined: 10 Oct 2011 Posts: 58 Location: Korea, Republic of Concentration: Finance, Strategy GMAT Date: 08-16-2012 GPA: 3.05 WE: Engineering (Energy and Utilities) Re: Astronomers at the Palomar Observatory have discovered a [#permalink] ### Show Tags 28 Apr 2012, 23:56 3 KUDOS 9. Astronomers at the Palomar Observatory have discovered a distant supernova explosion, one that they believe is a type previously unknown to science. (A) that they believe is : is creating a run on sentence (B) that they believe it to be : "it" redundant, no need to have since the phrase is talking about the "one = explosion" (C) they believe that it is of : wordy (D) they believe that is : wordy (E) they believe to be of : concise, correct I assumed the following construction (which called 'absolute phrase') for the second sentence. one ( xxxx, describing explosion ) of a type previously unknown to science. ( xxxx = they believe to be ) -> 'that' sometimes to be omitted ex) I ran to the office (that) I like. please correct my reasoning if wrong. _________________ Luck is the additional surplus on the way to one's constant effort. Intern Joined: 22 Feb 2010 Posts: 48 Re: Astronomers at the Palomar Observatory have discovered a [#permalink] ### Show Tags 12 May 2012, 06:23 2 KUDOS Astronomers at the Palomar Observatory have discovered a distant supernova explosion, one that they believe is a type previously unknown to science. (A) that they believe is one already works as pronoun, use of that is not required. Preposition 'of' is required before 'type' 'believe to be' is idiomatic (B) that they believe it to be Similar to A 'it' is redundant (C) they believe that it is of 'believe to be' is idiomatic (D) they believe that is 'believe that is' unidiomatic Missing the require Preposition 'of' (E) they believe to be of Correct removal of the duplicate pronoun 'that' 'believe to be' is idiomatic Preposition 'of' is correctly used _________________ __________________________________________________________________________ Challenges are what make life interesting; overcoming them is what makes life meaningful __________________________________________________________________________ Intern Joined: 26 Sep 2012 Posts: 10 Re: Astronomers at the Palomar Observatory have discovered a [#permalink] ### Show Tags 18 Oct 2012, 13:45 To clarify the doubt on inverted sentence as mentioned earlier ----" Among a group of things " one thing is asked making it singular -----This is an inverted subject-verb sentence. Reverse it: "Anger was among the emotions on display..." i have checked this question from 1000 SC #107 .... Here in non-underlined part as mentioned below "were" is used with "high cost of land" .... I feel it is the same case as mentioned earlier. Correct me if i m missing smthing 107. Among the reasons for the decline of New England agriculture in the last three decades were the high cost of land, the pressure of housing and commercial development, and basing a marketing and distribution system on importing produce from Florida and California. Amit Intern Joined: 03 May 2012 Posts: 20 Re: Astronomers at the Palomar Observatory have discovered a [#permalink] ### Show Tags 24 Oct 2013, 23:45 2 KUDOS 1 This post was BOOKMARKED kassim wrote: boomtangboy wrote: 9. Astronomers at the Palomar Observatory have discovered a distant supernova explosion, one that they believe is a type previously unknown to science. (A) that they believe is (B) that they believe it to be (C) they believe that it is of (D) they believe that is (E) they believe to be of Hello, Can someone explain more? what is tested here is it just Idiom ? I didn't know how to approach this question Best, Kassim Hi Kassim .The first three options here are starting with that...if u read the full sentence it reads thus "Astronomers at the Palomar Observatory have discovered a distant supernova explosion, one that........" ...so if u scrutinize keenly u can see that the pronoun ONE has " distant supernova explosion " as an antecedent...again pronoun "that" also leads to the same antecedent ..so presence of two pronouns for the same noun phrase makes the construction absolutely redundant...so u can immediately strike A, B and C ..now " believe to be " is a perfectly acceptable idiom ..so E is the right option .... Director Status: Everyone is a leader. Just stop listening to others. Joined: 22 Mar 2013 Posts: 912 Location: India GPA: 3.51 WE: Information Technology (Computer Software) Re: Astronomers at the Palomar Observatory have discovered a [#permalink] ### Show Tags 08 Apr 2014, 12:50 Refer this question : a-recent-study-on-lyme-disease-suggests-that-people-who-are-169777.html A recent study on Lyme Disease suggests that people who are bitten by ticks, especially by the lone star tick, develop a severe allergy to meat that scientists believe to be turning into an epidemic a.to be b.would be c.is d.it is e.it would be and then refer: the-world-wildlife-fund-has-declared-that-global-warming-a-103470.html Compared to these two questions I am not able to understand why option E is correct in this question. _________________ Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use? \| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction". Director Joined: 05 Nov 2012 Posts: 510 Concentration: Technology, Other Re: Astronomers at the Palomar Observatory have discovered a [#permalink] ### Show Tags 04 Sep 2014, 05:28 1 This post was BOOKMARKED Astronomers at the Palomar Observatory have discovered a distant supernova explosion, one that they believe is a type previously unknown to science. (A) that they believe is [DSE is a type... nope. It's not type;it's of type] (B) that they believe it to be [explosion is already referred by "one" so "it" not req. IMO there is no ambiguity in it usage. ] (C) they believe that it is of (D) they believe that is (E) they believe to be of _________________ -------------------------------------------------------- Regards Intern Joined: 29 Oct 2015 Posts: 8 Re: Astronomers at the Palomar Observatory have discovered a [#permalink] ### Show Tags 31 Oct 2015, 14:24 1 KUDOS "one" refers to the explosion, and the explosion is not a type, it is "of" a type. A-B and D cannot be the answer. C is wrong because "believe that" phrase breaks the relationship of "one" and "believe" E is the only option without a mistake. Senior Manager Joined: 07 Sep 2014 Posts: 440 Concentration: Finance, Marketing Re: Astronomers at the Palomar Observatory have discovered a [#permalink] ### Show Tags 14 Aug 2016, 20:10 Astronomers at the Palomar Observatory have discovered a distant supernova explosion, one that they believe is a type previously unknown to science. (A) that they believe is (E) they believe to be of A is incorrect: One => a distant supernova explosion. "that" is second pronoun. Not required. also one (that they believe) is a type previously unknown to science. => run on sentence. a distant supernova explosion is not a type previously unknown to science. But it is of a type. (type of explosion) Verbal Forum Moderator Joined: 13 Feb 2015 Posts: 801 Re: Astronomers at the Palomar Observatory have discovered a [#permalink] ### Show Tags 04 Apr 2017, 17:33 Merged topics. Please, search before posting questions! _________________ Director Joined: 26 Aug 2016 Posts: 528 Location: India Concentration: Marketing, Strategy GMAT 1: 690 Q50 V33 GMAT 2: 700 Q50 V33 GPA: 4 WE: Consulting (Consulting) Re: Astronomers at the Palomar Observatory have discovered a [#permalink] ### Show Tags 05 Apr 2017, 00:06 1st B 2nd E. One and that redundant. To be is correct since it preceeded by a noun. Which is believed. Sent from my ONE A2003 using GMAT Club Forum mobile app SVP Joined: 12 Dec 2016 Posts: 1927 Location: United States GMAT 1: 700 Q49 V33 GPA: 3.64 Re: Astronomers at the Palomar Observatory have discovered a [#permalink] ### Show Tags Updated on: 02 Dec 2017, 19:39 Astronomers at the Palomar Observatory have discovered a distant supernova explosion, one that they believe is a type previously unknown to science. (A) that they believe is (B) that they believe it to be (C) they believe that it is of (D) they believe that is (E) they believe to be of Originally posted by chesstitans on 02 Dec 2017, 17:23. Last edited by Vyshak on 02 Dec 2017, 19:39, edited 1 time in total. Topic Merged. Refer to the above discussions. Manager Joined: 28 Jun 2015 Posts: 90 Location: Australia Re: Astronomers at the Palomar Observatory have discovered a [#permalink] ### Show Tags 11 Feb 2018, 17:05 1 This post was BOOKMARKED I would take a different approach to this question <A Singular thing> can be OF a TYPE <A Singular thing> CANNOT be A TYPE Hence, only valid options are C and E. C is clearly wrong as IT does not have an antecedent. ANS E Regards vignesh_unl wrote: 145. Astronomers at the Palomar Observatory have discovered a distant supernova explosion, one that they believe is a type previously unknown to science. (A) that they believe is (B) that they believe it to be (C) they believe that it is of (D) they believe that is (E) they believe to be of [Reveal] Spoiler: Among the emotions on display in the negotiating room were anger for repeatedly raising the issue over and over again and preventing the raw wounds from earlier battles from ever beginning to heal. (A) were anger for repeatedly raising the issue over and over again and preventing the raw wounds from earlier battles from ever beginning to heal (B) was anger for repeatedly raising the issue and preventing the raw wounds from earlier battles from ever beginning to heal (C) were anger over repeatedly raising the issue and preventing the raw wounds from earlier battles to begin healing (D) was anger about the issue, which was raised over and over, and preventing the wounds from earlier battles, still raw, to begin healing (E) were anger about the issue, which was raised repeatedly, and preventing the raw wounds from earlier battles to begin to heal Senior Manager Joined: 08 Jan 2018 Posts: 254 Location: United States (ID) GPA: 3.33 WE: Accounting (Accounting) Re: Astronomers at the Palomar Observatory have discovered a [#permalink] ### Show Tags 26 Feb 2018, 10:39 Only A and E are candidates for the answer. "of a type" is correct -> E is left. Re: Astronomers at the Palomar Observatory have discovered a [#permalink] 26 Feb 2018, 10:39 Display posts from previous: Sort by	3,459	13,590	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17	latest	en	0.954413
http://math.elinkage.net/showthread.php?tid=89&pid=132	1,537,383,908,000,000,000	text/html	crawl-data/CC-MAIN-2018-39/segments/1537267156270.42/warc/CC-MAIN-20180919180955-20180919200955-00538.warc.gz	167,872,147	6,592	• 0 Votes - 0 Average • 1 • 2 • 3 • 4 • 5 Linear Recursion, more complicated 08-12-2010, 04:48 PM Post: #1 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0 Linear Recursion, more complicated $$A_{n+1}+A_n = 1, \; A_1 = 0$$ $$B_{n+3}-B_n = A_n, \quad B_j = 0,\; j = \overline{1,3}$$ $$C_{n+7}-C_n = B_n, \quad C_j = 0,\; j = \overline{1,10}$$ 08-14-2010, 02:11 PM Post: #2 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0 RE: Linear Recursion, more complicated Clearly $$A_n = \frac{1}{2}(1+(-1)^n)$$ Since $$B_{n+3} = B_n + A_n$$, we have $$B_{n+4}+B_{n+3}-B_{n+1}-B_n = A_{n+1}+A_n=1$$, The corresponding characteristic equation is $t^4+t^3-t-1 = (t-1)(t+1)(t^2+t+1)=0$ Thus $B_n = a + b(-1)^n +c\cos \frac{2n\pi}{3} +d \sin \frac{2n\pi}{3} + \frac{n}{6}$ where the last term is for the non-homogeneous right hand side, namely $1$. From $\frac{1}{2}(1+(-1)^n)=A_n = B_{n+3}-B_n = \frac{1}{2}-2b(-1)^n$ we see that $b = -\frac{1}{4}$ From $B_1=B_2=B_3 = 0$ we get $B_n = \frac{n}{6}-\frac{1}{4}(-1)^n-\frac{5}{12}-\frac{1}{3}\cos \frac{2n\pi}{3}-\frac{\sqrt{3}}{9}\sin \frac{2n\pi}{3}$ 08-14-2010, 06:01 PM Post: #3 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0 RE: Linear Recursion, more complicated Now $B_n = \frac{n}{6}-\frac{1}{4}(-1)^n + P(n)$ where $P$ has period $3$ Since $B_j = 0, \quad j = \overline{1,3}$ we see that $P(1)=-5/12, P(2)=-1/12, P(3)=-3/4$ So $-12P(n)-5 = -4((n \mod 3) -1)$ and $B_n = \frac{n}{6}-\frac{1}{4}(-1)^n -\frac{5}{12} + \frac{1}{3} ((n \mod 3) -1)$ That is $B_n = \frac{n}{6}-\frac{1}{4}(3+(-1)^n)+ \frac{1}{3} (n \mod 3)$ Periodic functions over integers can always be expressed as a function of $x = (n \mod m)$ where $m$ is the period. « Next Oldest \| Next Newest » Forum Jump:	814	1,751	{"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}	4.28125	4	CC-MAIN-2018-39	longest	en	0.592394
https://nbviewer.jupyter.org/github/quantopian/research_public/blob/master/notebooks/lectures/Introduction_to_Pairs_Trading/notebook.ipynb	1,611,480,180,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703547475.44/warc/CC-MAIN-20210124075754-20210124105754-00535.warc.gz	473,120,723	610,094	# Introduction to Pairs Trading¶ By Delaney Mackenzie and Maxwell Margenot Part of the Quantopian Lecture Series: Pairs trading is a classic example of a strategy based on mathematical analysis. The principle is as follows. Let's say you have a pair of securities X and Y that have some underlying economic link. An example might be two companies that manufacture the same product, or two companies in one supply chain. If we can model this economic link with a mathematical model, we can make trades on it. We'll start by constructing a toy example. Before we proceed, note that the content in this lecture depends heavily on the Stationarity, Integration, and Cointegration lecture in order to properly understand the mathematical basis for the methodology that we employ here. It is recommended that you go through that lecture before this continuing. In [1]: import numpy as np import pandas as pd import statsmodels import statsmodels.api as sm from statsmodels.tsa.stattools import coint # just set the seed for the random number generator np.random.seed(107) import matplotlib.pyplot as plt ## Generating Two Fake Securities¶ We model X's daily returns by drawing from a normal distribution. Then we perform a cumulative sum to get the value of X on each day. In [2]: X_returns = np.random.normal(0, 1, 100) # Generate the daily returns # sum them and shift all the prices up into a reasonable range X = pd.Series(np.cumsum(X_returns), name='X') + 50 X.plot(); Now we generate Y. Remember that Y is supposed to have a deep economic link to X, so the price of Y should vary pretty similarly. We model this by taking X, shifting it up and adding some random noise drawn from a normal distribution. In [3]: some_noise = np.random.normal(0, 1, 100) Y = X + 5 + some_noise Y.name = 'Y' pd.concat([X, Y], axis=1).plot(); ## Cointegration¶ We've constructed an example of two cointegrated series. Cointegration is a more subtle relationship than correlation. If two time series are cointegrated, there is some linear combination between them that will vary around a mean. At all points in time, the combination between them is related to the same probability distribution. For more details on how we formally define cointegration and how to understand it, please see the Integration, Cointegration, and Stationarity lecture from the Quantopian Lecture Series. We'll plot the difference between the two now so we can see how this looks. In [4]: (Y - X).plot() # Plot the spread plt.axhline((Y - X).mean(), color='red', linestyle='--') # Add the mean plt.xlabel('Time') plt.legend(['Price Spread', 'Mean']); ## Testing for Cointegration¶ That's an intuitive definition, but how do we test for this statistically? There is a convenient cointegration test that lives in statsmodels.tsa.stattools. Let's say that our confidence level is $0.05$. We should see a p-value below our cutoff, as we've artifically created two series that are the textbook definition of cointegration. In [5]: # compute the p-value of the cointegration test # will inform us as to whether the spread between the 2 timeseries is stationary # around its mean score, pvalue, _ = coint(X,Y) print pvalue 2.75767345363e-16 ### Correlation vs. Cointegration¶ Correlation and cointegration, while theoretically similar, are not the same. To demonstrate this, we'll show examples of series that are correlated, but not cointegrated, and vice versa. To start let's check the correlation of the series we just generated. In [6]: X.corr(Y) Out[6]: 0.94970906463859317 That's very high, as we would expect. But how would two series that are correlated but not cointegrated look? ### Correlation Without Cointegration¶ A simple example is two series that just diverge. In [7]: X_returns = np.random.normal(1, 1, 100) Y_returns = np.random.normal(2, 1, 100) X_diverging = pd.Series(np.cumsum(X_returns), name='X') Y_diverging = pd.Series(np.cumsum(Y_returns), name='Y') pd.concat([X_diverging, Y_diverging], axis=1).plot(); In [8]: print 'Correlation: ' + str(X_diverging.corr(Y_diverging)) score, pvalue, _ = coint(X_diverging,Y_diverging) print 'Cointegration test p-value: ' + str(pvalue) Correlation: 0.993134380128 Cointegration test p-value: 0.884633444839 ### Cointegration Without Correlation¶ A simple example of this case is a normally distributed series and a square wave. In [9]: Y2 = pd.Series(np.random.normal(0, 1, 1000), name='Y2') + 20 Y3 = Y2.copy() In [10]: # Y2 = Y2 + 10 Y3[0:100] = 30 Y3[100:200] = 10 Y3[200:300] = 30 Y3[300:400] = 10 Y3[400:500] = 30 Y3[500:600] = 10 Y3[600:700] = 30 Y3[700:800] = 10 Y3[800:900] = 30 Y3[900:1000] = 10 In [11]: Y2.plot() Y3.plot() plt.ylim([0, 40]); In [12]: # correlation is nearly zero print 'Correlation: ' + str(Y2.corr(Y3)) score, pvalue, _ = coint(Y2,Y3) print 'Cointegration test p-value: ' + str(pvalue) Correlation: -0.0413040695809 Cointegration test p-value: 0.0 Sure enough, the correlation is incredibly low, but the p-value shows that these are cointegrated. ## Hedging¶ Because you'd like to protect yourself from bad markets, often times short sales will be used to hedge long investments. Because a short sale makes money if the security sold loses value, and a long purchase will make money if a security gains value, one can long parts of the market and short others. That way if the entire market falls off a cliff, we'll still make money on the shorted securities and hopefully break even. In the case of two securities we'll call it a hedged position when we are long on one security and short on the other. ## The Trick: Where it all comes together¶ Because the securities drift towards and apart from each other, there will be times when the distance is high and times when the distance is low. The trick of pairs trading comes from maintaining a hedged position across X and Y. If both securities go down, we neither make nor lose money, and likewise if both go up. We make money on the spread of the two reverting to the mean. In order to do this we'll watch for when X and Y are far apart, then short Y and long X. Similarly we'll watch for when they're close together, and long Y and short X. ### Going Long the Spread¶ This is when the spread is small and we expect it to become larger. We place a bet on this by longing Y and shorting X. ### Going Short the Spread¶ This is when the spread is large and we expect it to become smaller. We place a bet on this by shorting Y and longing X. ### Specific Bets¶ One important concept here is that we are placing a bet on one specific thing, and trying to reduce our bet's dependency on other factors such as the market. ## Finding real securities that behave like this¶ The best way to do this is to start with securities you suspect may be cointegrated and perform a statistical test. If you just run statistical tests over all pairs, you'll fall prey to multiple comparison bias. Here's a method to look through a list of securities and test for cointegration between all pairs. It returns a cointegration test score matrix, a p-value matrix, and any pairs for which the p-value was less than $0.05$. ### WARNING: This will incur a large amount of multiple comparisons bias.¶ The methods for finding viable pairs all live on a spectrum. At one end there is the formation of an economic hypothesis for an individual pair. You have some extra knowledge about an economic link that leads you to believe that the pair is cointegrated, so you go out and test for the presence of cointegration. In this case you will incur no multiple comparisons bias. At the other end of the spectrum, you perform a search through hundreds of different securities for any viable pairs according to your test. In this case you will incur a very large amount of multiple comparisons bias. Multiple comparisons bias is the increased chance to incorrectly generate a significant p-value when many tests are run. If 100 tests are run on random data, we should expect to see 5 p-values below $0.05$ on expectation. Because we will perform $n(n-1)/2$ comparisons, we should expect to see many incorrectly significant p-values. For the sake of this example we will ignore this and continue. In practice a second verification step would be needed if looking for pairs this way. Another approach is to pick a small number of pairs you have reason to suspect might be cointegrated and test each individually. This will result in less exposure to multiple comparisons bias. You can read more about multiple comparisons bias here. In [13]: def find_cointegrated_pairs(data): n = data.shape[1] score_matrix = np.zeros((n, n)) pvalue_matrix = np.ones((n, n)) keys = data.keys() pairs = [] for i in range(n): for j in range(i+1, n): S1 = data[keys[i]] S2 = data[keys[j]] result = coint(S1, S2) score = result[0] pvalue = result[1] score_matrix[i, j] = score pvalue_matrix[i, j] = pvalue if pvalue < 0.05: pairs.append((keys[i], keys[j])) return score_matrix, pvalue_matrix, pairs ## Looking for Cointegrated Pairs of Alternative Energy Securities¶ We are looking through a set of solar company stocks to see if any of them are cointegrated. We'll start by defining the list of securities we want to look through. Then we'll get the pricing data for each security for the year of 2014. Our approach here is somewhere in the middle of the spectrum that we mentioned before. We have formulated an economic hypothesis that there is some sort of link between a subset of securities within the energy sector and we want to test whether there are any cointegrated pairs. This incurs significantly less multiple comparisons bias than searching through hundreds of securities and slightly more than forming a hypothesis for an individual test. NOTE: We include the market in our data. This is because the market drives the movement of so many securities that you often times might find two seemingingly cointegrated securities, but in reality they are not cointegrated and just both conintegrated with the market. This is known as a confounding variable and it is important to check for market involvement in any relationship you find. get_pricing() is a Quantopian method that pulls in stock data, and loads it into a Python Pandas DataPanel object. Available fields are 'price', 'open_price', 'high', 'low', 'volume'. But for this example we will just use 'price' which is the daily closing price of the stock. In [14]: symbol_list = ['ABGB', 'ASTI', 'CSUN', 'DQ', 'FSLR','SPY'] prices_df = get_pricing(symbol_list, fields=['price'] , start_date='2014-01-01', end_date='2015-01-01')['price'] prices_df.columns = map(lambda x: x.symbol, prices_df.columns) Example of how to get all the prices of all the stocks loaded using get_pricing() above in one pandas dataframe object In [15]: prices_df.head() Out[15]: ABGB ASTI CSUN DQ FSLR SPY 2014-01-02 00:00:00+00:00 14.099 7.41 7.040 38.00 57.43 179.444 2014-01-03 00:00:00+00:00 14.427 7.25 7.078 39.50 56.74 179.287 2014-01-06 00:00:00+00:00 14.989 7.12 7.010 40.05 51.26 178.905 2014-01-07 00:00:00+00:00 15.282 7.20 6.960 41.93 52.48 179.934 2014-01-08 00:00:00+00:00 14.969 7.10 7.160 42.49 51.68 180.023 Example of how to get just the prices of a single stock that was loaded using get_pricing() above In [16]: prices_df['SPY'].head() Out[16]: 2014-01-02 00:00:00+00:00 179.444 2014-01-03 00:00:00+00:00 179.287 2014-01-06 00:00:00+00:00 178.905 2014-01-07 00:00:00+00:00 179.934 2014-01-08 00:00:00+00:00 180.023 Freq: C, Name: SPY, dtype: float64 Now we'll run our method on the list and see if any pairs are cointegrated. In [17]: # Heatmap to show the p-values of the cointegration test between each pair of # stocks. Only show the value in the upper-diagonal of the heatmap scores, pvalues, pairs = find_cointegrated_pairs(prices_df) import seaborn seaborn.heatmap(pvalues, xticklabels=symbol_list, yticklabels=symbol_list, cmap='RdYlGn_r' , mask = (pvalues >= 0.05) ) print pairs [(u'ABGB', u'FSLR')] Looks like 'ABGB' and 'FSLR' are cointegrated. Let's take a look at the prices to make sure there's nothing weird going on. In [18]: S1 = prices_df['ABGB'] S2 = prices_df['FSLR'] In [19]: score, pvalue, _ = coint(S1, S2) pvalue Out[19]: 0.0049511108325587683 ## Calculating the Spread¶ Now we will plot the spread of the two series. In order to actually calculate the spread, we use a linear regression to get the coefficient for the linear combination to construct between our two securities, as shown in the stationarity lecture. Using a linear regression to estimate the coefficient is known as the Engle-Granger method. In [20]: S1 = sm.add_constant(S1) results = sm.OLS(S2, S1).fit() S1 = S1['ABGB'] b = results.params['ABGB'] spread = S2 - b * S1 spread.plot() plt.axhline(spread.mean(), color='black') plt.legend(['Spread']); Alternatively, we could examine the ratio betwen the two series. In [21]: ratio = S1/S2 ratio.plot() plt.axhline(ratio.mean(), color='black') plt.legend(['Price Ratio']); Examining the price ratio of a trading pair is a traditional way to handle pairs trading. Part of why this works as a signal is based in our assumptions of how stock prices move, specifically because stock prices are typically assumed to be log-normally distributed. What this implies is that by taking a ratio of the prices, we are taking a linear combination of the returns associated with them (since prices are just the exponentiated returns). This can be a little irritating to deal with for our purposes as purchasing the precisely correct ratio of a trading pair may not be practical. We choose instead to move forward with simply calculating the spread between the cointegrated stocks using linear regression. This is a very simple way to handle the relationship, however, and is likely not feasible for non-toy examples. There are other potential methods for estimating the spread listed at the bottom of this lecture. If you want to get more into the theory of why having cointegrated stocks matters for pairs trading, again, please see the Integration, Cointegration, and Stationarity Lecture from the Quantopian Lecture Series. So, back to our example. The absolute spread isn't very useful in statistical terms. It is more helpful to normalize our signal by treating it as a z-score. ### WARNING¶ In practice this is usually done to try to give some scale to the data, but this assumes some underlying distribution, usually a normal distribution. Under a normal distribution, we would know that approximately 84% of all spread values will be smaller. However, much financial data is not normally distributed, and one must be very careful not to assume normality, nor any specific distribution when generating statistics. It could be the case that the true distribution of spreads was very fat-tailed and prone to extreme values. This could mess up our model and result in large losses. In [22]: def zscore(series): return (series - series.mean()) / np.std(series) In [23]: zscore(spread).plot() plt.axhline(zscore(spread).mean(), color='black') plt.axhline(1.0, color='red', linestyle='--') plt.axhline(-1.0, color='green', linestyle='--') plt.legend(['Spread z-score', 'Mean', '+1', '-1']); ### Simple Strategy:¶ • Go "Long" the spread whenever the z-score is below -1.0 • Go "Short" the spread when the z-score is above 1.0 • Exit positions when the z-score approaches zero This is just the tip of the iceberg, and only a very simplistic example to illustrate the concepts. In practice you would want to compute a more optimal weighting for how many shares to hold for S1 and S2. Some additional resources on pair trading are listed at the end of this notebook ## Trading using constantly updating statistics¶ In general taking a statistic over your whole sample size can be bad. For example, if the market is moving up, and both securities with it, then your average price over the last 3 years may not be representative of today. For this reason traders often use statistics that rely on rolling windows of the most recent data. ## Moving Averages¶ A moving average is just an average over the last $n$ datapoints for each given time. It will be undefined for the first $n$ datapoints in our series. Shorter moving averages will be more jumpy and less reliable, but respond to new information quickly. Longer moving averages will be smoother, but take more time to incorporate new information. We also need to use a rolling beta, a rolling estimate of how our spread should be calculated, in order to keep all of our parameters up to date. In [24]: # Get the spread between the 2 stocks # Calculate rolling beta coefficient rolling_beta = pd.ols(y=S1, x=S2, window_type='rolling', window=30) spread = S2 - rolling_beta.beta['x'] * S1 spread.name = 'spread' # Get the 1 day moving average of the price spread spread_mavg1 = pd.rolling_mean(spread, window=1) spread_mavg1.name = 'spread 1d mavg' # Get the 30 day moving average spread_mavg30 = pd.rolling_mean(spread, window=30) spread_mavg30.name = 'spread 30d mavg' plt.plot(spread_mavg1.index, spread_mavg1.values) plt.plot(spread_mavg30.index, spread_mavg30.values) plt.legend(['1 Day Spread MAVG', '30 Day Spread MAVG']) plt.ylabel('Spread'); We can use the moving averages to compute the z-score of the spread at each given time. This will tell us how extreme the spread is and whether it's a good idea to enter a position at this time. Let's take a look at the z-score now. In [25]: # Take a rolling 30 day standard deviation std_30 = pd.rolling_std(spread, window=30) std_30.name = 'std 30d' # Compute the z score for each day zscore_30_1 = (spread_mavg1 - spread_mavg30)/std_30 zscore_30_1.name = 'z-score' zscore_30_1.plot() plt.axhline(0, color='black') plt.axhline(1.0, color='red', linestyle='--'); The z-score doesn't mean much out of context, let's plot it next to the prices to get an idea of what it looks like. We'll take the negative of the z-score because the spreads were all negative and that is a little counterintuitive to trade on. In [26]: # Plot the prices scaled down along with the negative z-score # just divide the stock prices by 10 to make viewing it on the plot easier plt.plot(S1.index, S1.values/10) plt.plot(S2.index, S2.values/10) plt.plot(zscore_30_1.index, zscore_30_1.values) plt.legend(['S1 Price / 10', 'S2 Price / 10', 'Price Spread Rolling z-Score']); ## Out of Sample Test¶ Now that we have constructed our spread appropriately and have an idea of how we will go about making trades, it is time to conduct some out of sample testing. Our whole model is based on the premise that these securities are cointegrated, but we built it on information from a certain time period. If we actually want to implement this model, we need to conduct an out of sample test to confirm that the principles of our model are still valid going forward. Since we initially built the model on the 2014 - 2015 year, let's see if this cointegrated relationship holds for 2015 - 2016. Historical results do not guarantee future results so this is a sanity check to see if the work we have done holds strong. In [27]: symbol_list = ['ABGB', 'FSLR'] prices_df = get_pricing(symbol_list, fields=['price'] , start_date='2015-01-01', end_date='2016-01-01')['price'] prices_df.columns = map(lambda x: x.symbol, prices_df.columns) In [28]: S1 = prices_df['ABGB'] S2 = prices_df['FSLR'] In [29]: score, pvalue, _ = coint(S1, S2) print 'p-value: ', pvalue p-value: 0.991161185763 Unfortunately, since our p-value is above the cutoff of $0.05$, we conclude that our model will no longer be valid due to the lack of cointegration between our chosen securities. If we tried to deploy this model without the underlying assumptions holding, we would have no reason to believe that it would actually work. Out of sample testing is a vital step to make sure that our work will actually be viable in the market. ## Implementation¶ When actually implementing a pairs trading strategy you would normally want to be trading many different pairs at once. If you find a good pair relationship by analyzing data, there is no guarantee that that relationship will continue into the future. Trading many different pairs creates a diversified portfolio to mitigate the risk of individual pairs "falling out of" cointegration. There is a template algorithm attached to this lecture that shows an example of how you would implement pairs trading on our platform. Feel free to check it out and modify it with your own pairs to see if you can improve it. ## Further Research¶ This notebook contained some simple introductory approaches. In practice one should use more sophisticated statistics, some of which are listed here. • Augmented-Dickey Fuller test • Hurst exponent • Half-life of mean reversion inferred from an Ornsteinâ€“Uhlenbeck process • Kalman filters (this is not an endorsement) But, a very good practical resource for learning more about pair trading is Dr. Ernie Chan's book: Algorithmic Trading: Winning Strategies and Their Rationale This presentation is for informational purposes only and does not constitute an offer to sell, a solicitation to buy, or a recommendation for any security; nor does it constitute an offer to provide investment advisory or other services by Quantopian, Inc. ("Quantopian"). Nothing contained herein constitutes investment advice or offers any opinion with respect to the suitability of any security, and any views expressed herein should not be taken as advice to buy, sell, or hold any security or as an endorsement of any security or company. In preparing the information contained herein, Quantopian, Inc. has not taken into account the investment needs, objectives, and financial circumstances of any particular investor. Any views expressed and data illustrated herein were prepared based upon information, believed to be reliable, available to Quantopian, Inc. at the time of publication. Quantopian makes no guarantees as to their accuracy or completeness. All information is subject to change and may quickly become unreliable for various reasons, including changes in market conditions or economic circumstances.	5,578	22,201	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.484375	3	CC-MAIN-2021-04	longest	en	0.868906

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