text: Solution: Download NCERT Class 11 Maths Solutions of Permutations and Combinations PDF through the direct links available. NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1. (b) ∠r + 120° + 80° + 60° = 360° [Angle sum property of a quadrilateral] Check NCERT Solutions now. These solutions explains the accurate method of solving problems. NCERT textbook of class 8 maths consist of 16 chapters having multiple exercises. We hope the NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Ex 14.3 help you. Students can access the NCERT MCQ Questions for Class 8 Maths Chapter 13 Direct and Indirect Proportions with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. ⇒ ∠A + ∠C + ∠D + ∠B = 360° What is a regular polygon? (ii) Square NCERT Solutions for Class 8 Maths Chapter 3 UNDERSTANDING QUADRILATERALS in English and Hindi Medium updated for new academic session 2020-2021 based on latest NCERT Books for current year, useful to all students. The subject experts at BYJU’S have solved each question of NCERT exercises with utmost care, in order to help the students in solving any question from the NCERT textbook. z + 60° = 180° (Linear pair) Follow the chapter link to NCERT solutions and download FREE NCERT solutions in PDF format. NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4 Ex 3.4 Class 8 Maths Question 1. NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities: In our daily life, we compare many things like his bottle is filled up to two-thirds of its height, his mark is 66% of my mark, my marks increased by 15% compared to the last test, etc. Solution: NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Class 8 Maths Understanding Quadrilaterals Ex 3.1 Class 8 Maths Understanding Quadrilaterals Ex 3.2 Class 8 Maths Understanding Quadrilaterals Ex ⇒ x = 108°. ⇒ x = 180° – 90° = 90° Classify each of the above figure on the basis of the following: ⇒ x = 360° – 300° = 60° (iii) ABC is a triangle which has no diagonal. NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals In class 8 the Maths started to get difficult and the fear of students started to increase. Resource: National Council of Educational Research and Training (NCERT) Solutions NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12, NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1. ⇒ x + 70° + 60° + 90° = 360° [∵ 180° – 90° = 90°] NCERT Solutions for Class 8 Maths Chapter 3 - Understanding Quadrilaterals:. ⇒ 30° + x + 110° + 120° + x = 540° [∵ 180° – 70° = 110°; 180° – 60° = 120°] Solution: Solutions are prepared by the Maths experts, keeping the latest CBSE Syllabus in mind. A polygon with equal sides and equal angles is called a regular polygon. ⇒ 2x = 280° NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.5 in English medium as well as Hindi Medium use online or download in PDF form free for academic session 2020-21. (c) A triangle NCERT Solutions For Class 8 Maths Chapter 3 – Understanding Quadrilaterals: In this article, we will provide you with detailed NCERT Solutions for Math Class 8 Chapter 3. (a) Cost of apples (b) Distance travelled by a car Register and get all exercise solutions in your emails. Find the angle measure x in the following figures: NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 8 so that you can refer them as and when required. in free pdf for Maths for Class 8 … These ncert book chapter wise … Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. The lesson uses various word problems and questions to explain kids addition and subtraction in a faster way. (a) Find x + y + z Free PDF download of NCERT Solutions for Class 8 Maths Chapter 3 Exercise 3.3 (EX 3.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. NCERT Solutions for Class 8 Maths Exercise 3.1 5. ⇒ y = 120° Reduce the following equations into slope-intercept form and find their slopes and the y intercepts. Angle sum = (8 – 2) × 180° = 6 × 180° = 1080° • NCERT Solutions for Class 8 Maths are given for the students so that they can get to know the answers to the questions in case they are not able to find it.It is important for all the students who are in Class 8 currently. Ex 15.3 Class 8 Maths Question 1. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals, drop a comment below and we will get back to you at the earliest. and, z + 30° = 180° [Linear pair] NCERT Solutions For Class 6 Maths Chapter 3 Playing with Numbers,These numbers can be written in single digits, double digits, three-digits in the generalized form. NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.3 Factorization in English Medium and Hindi Medium simplified form free to use it online. ⇒ ∠a + 120° = 180° Students can take the benefits of Hindi Medium and Here, in this article, you can download Ch 3 Maths Class 8 solutions in PDF format for free. (d) Number of sides = n NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals, are provided here, which can be downloaded for free in PDF format. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. (c) Number of sides = 10 Angle sum = (10 – 2) × 180° = 8 × 180° = 1440° Cloudflare Ray ID: 61841ff60d06f460 (ii) ABCDEF is a regular hexagon which has nine diagonals AE, AD, AC, BF, BE, BD, CF, CE and DF. EXERCISE 3.2. NCERT Solutions for Class 8 Maths Chapter 8 COMPARING QUANTITIES Exercise 8.1, Exercise 8.2 and Exercise 8.3 in English Medium as well as Hindi Medium updated for new academic session 2020-21 based on new NCERT Books. View NCERT solutions of all the questions of NCERT, including Examples, Exercises, Miscellaneous and Supplementary Questions for Class 6 to 12 free at teachoo. ⇒ z = 180° – 60° = 120° NCERT Solutions for Class 8 Social Science History Chapter 3 We have provided here the best and accurate answers to all the questions given in chapter 3 of the latest NCERT History Book of Class 8 . • What is the sum of the measures of the angles of a convex quadrilateral? Join AC diagonal which divides the quadrilateral into two triangles ABC and ADC. In this post, we published NCERT Solutions for class 10 Maths chapter 3. Ex 8.3 Class 8 Maths Question 1. We know that the sum of the exterior angles formed by NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.3 Ex 15.3 Class 8 Maths Question 1. Classify each of them on the basis of the following. The sum of the interior angles in a quadrilateral is always 360 degrees. Solution. Calculate the amount and compound interest on (a) ₹ 10,800 for 3 years at 12$$\frac { 1 }{ 2 }$$ % per annum compounded annually. Also after the chapter you can get links to Class 8 Maths Notes, NCERT Solutions, Important Question, Practice Papers, etc. Solutions are prepared by the Maths experts, keeping the latest CBSE Syllabus in mind.. NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.3 (a) In Fig. The NCERT Solutions for Class 8 Maths Chapter 8 deals with the concept of ratios and percentages, along with other main topics and concepts. NCERT Solutions for Class 8 Maths – Free PDF Download Chapterwise Chapter 1 Rational Numbers. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. ⇒ ∠a = 180° – 120° = 60° CBSE Guide of All Chapters given in the Book. NCERT Solutions for Class 8 Maths Exercise 8.3 Class 8 Maths book solutions are available in PDF format for free download. State the name of a regular polygon of: (a) 3 sides (b) 4 sides (c) 6 sides Ans. These ncert book chapter wise questions and answers are very helpful for CBSE exam. Until now, the student must have dealt only with whole numbers. We hope the NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 help you. (a) 7 (b) ₹ 18,000 for 2$$\frac { 1 }{ 2 }$$ years at 10% per annum compounded annually. also, x + 90° = 180° [Linear pair] ⇒ 5x = 540° ⇒ z = 180° – 30° = 150° Toppr’s NCERT Solutions for Class 8 Maths Chapter 3 will help them in understanding this Chapter. Given here are some figures. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1, drop a comment below and we will get back to you at the earliest. Examine the table. NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3. (b) Angle sum of a quadrilateral = 360° What can you say about the angle sum of a convex polygon with number of sides? Ex 3.1 Class 8 Maths Question 1. Chapter 1: Rational Numbers Chapter 2: Linear Equations in One Variable Chapter 3: Understanding Quadrilaterals Chapter 4 The NCERT solutions for Class 8 maths Chapter 3- Understanding Quadrilaterals includes the solutions of all the questions present in the NCERT textbook. NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Free PDF available on Vedantu are solved by experts. Ex 3.1 Class 8 Maths Question 7. (b) Simple closed curve: (1), (2), (5), (6) and (7) Solutions of the questions are prepared considering the latest CBSE Curriculum 2020 – 2021.. NCERT Solutions for Class 8 Maths Chapter 10 Exercise 10.3 (b) Find x + y + z + w ∠1 + ∠3 + ∠2 + ∠4 + ∠5 + ∠6 = 180° + 180° Get here NCERT Solutions for Class 8 Maths Chapter 3.These NCERT Solutions for Class 8 of Maths subject includes detailed answers of all the questions in Chapter 3 – Understanding Quadrilaterals provided in NCERT Book which is prescribed for class 8 in schools. Knowledge of variables gained from our NCERT Class 8 Maths Chapter 2 solutions will benefit you a lot in higher classes such as Class 9, Class 10, Class 11 and Class 12. ⇒ x + 220° = 360° ⇒ 2x = 540° – 260° 1. ⇒ 2x + 260° = 540° (a) ∠a + 30° + 90° = 180° [Angle sum property] Register and get all exercise solutions in your emails. Students can access the NCERT MCQ Questions for Class 9 Maths Chapter 8 Quadrilaterals with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. ⇒ x + x + x + x + x = 540° [All angles of a regular pentagon are equal] y + 80° = 180° (Linear pair) NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.3. A regular polygon: A polygon having all sides of equal length and the interior angles of equal size is known as regular polygon. All the solutions provided here are solved by the top academic experts at Embibe. NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.3 Factorization in English Medium and Hindi Medium simplified form free to use it online. (a) Simple curve: (1), (2), (5), (6) and (7) Draw the graphs for the following tables of values, with suitable scales on the axes. smartur Learning is (Super) rewarding! ∠r + 260° = 360° ⇒ 50° + 130° + 120° + x = 360° Chapter 1 - Number Systems; Chapter 2 - Polynomials; Chapter 3 - Coordinate Geometry; Chapter 4 - Linear Equations in Two Variables w = 180° – ∠r = 180° – 100° = 80° (Linear pair) NCERT Class 9 Maths chapter list is given below. Use MCQ Questions for Class 9 Maths with Answers during preparation and score maximum marks in the exam. NCERT Solutions for Class 8 Maths Chapter 10 Exercise 10.3 Visualizing Solid Shapes in English Medium and Hindi Medium format free to study online. The NCERT Solutions for chapter Understanding Quadrilaterals are prepared by the mathematics experts at BYJU’S, keeping the examination point of view in mind. Video Solutions has been prepared strictly as per latest syllabus issued by CBSE for the students associated with Class X. Free Download NCERT Solutions for Class 3 Maths Chapter 8 Who Is Heavier with a complete solution in PDF Form solved by expert teachers as per CBSE NCERT … We at Entrancei have devised specific NCERT solutions for class 8 Maths for all students. Another way to prevent getting this page in the future is to use Privacy Pass. Here you can read Chapter 3 of Class 8 Maths NCERT Book. CBSE NCERT solutions for class 8 Maths. (ii) 4 sides CBSE Class 8 Maths All Exercise Questions with Solutions to help you to revise complete Syllabus (b) ₹ 18,000 for 2 years at 10% per annum compounded annually. x = 180° – 120° = 60° Let us draw a non-convex quadrilateral. Free NCERT Solutions for Class 3 Math Chapter 8 - Who Is Heavier. NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3. Download Free Study Material for Class 8 to score more marks. Solutions of the questions are prepared considering the latest CBSE Curriculum 2020 – 2021. NCERT Solutions for Class 8 Maths Chapter 12: The first step towards your preparation for tests and exams will be solving the NCERT exercise questions.Every time you finish a chapter, you need to solve the questions at the end of it. NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable . If you want to download, you can. (a) Number of sides = 7 Calculate the amount and compound interest on (a) 10,800 for 3 years at 12 % per annum compounded annually. The NCERT solutions for Class 8 maths Chapter 3- Understanding Quadrilaterals includes the solutions of all the questions present in the NCERT textbook. NCERT Solutions for Class 8 Maths Chapter 10 Exercise 10.3 Visualizing Solid Shapes in English Medium and Hindi Medium format free to study online. Videos related to each questions are given along with the solutions. All questions are discussed by the experts of maths teachers and according to the guidelines of NCERT (CBSE). Free PDF download of NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.3 (EX 8.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. (c) Polygon: (1) and (2) Solution: NCERT Solutions Class 8 Maths PDF Download For Free. India’s First Rewards based Learning System for School Students. We hope the NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals, help you. (c) Angle sum of a pentagon = 540° Please enable Cookies and reload the page. Free PDF download of Class 8 Maths NCERT Solutions help students to understand and learn the chapters easily and score more marks in your exams. (b) In Fig. Calculate the amount and compound interest on (a) ₹ 10,800 for 3 years at 12 % per annum compounded annually. (c) Polygon Get NCERT Solutions of Chapter 3 Class 8 Understanding Quadrilaterals free at teachoo. Students can access the NCERT MCQ Questions for Class 8 Maths Chapter 7 Cubes and Cube Roots Roots with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. NCERT solutions for class 8 maths chapter 3 Understanding Quadrilaterals- A geometrical closed shape made by using 4 straight lines and having four connecting points is called Quadrilateral. (d) Angle sum of a regular pentagon = 540° Solution: The major concepts are listed below: 8.1 Recalling Ratios and Percentages 8.2 Finding the Increase or Decrease Per Cent 8.3 Finding Discounts 8.3.1 Estimation in percentages 8.4 Prices Related to Buying and Selling (Profit and Loss) 8.4.1 … Register and get all exercise solutions in your emails. Download Free NCERT Solutions for class 6th to 12th. ⇒ y = 180° – 60° From the above table, we conclude that the sum of all the angles of a polygon of side ‘n’ (i) 3 sides Given here are some figures. Chapter-wise NCERT Solutions for Class 8 Maths solved by Expert Teachers as per NCERT (CBSE) Book guidelines. (iii) Regular Hexagon, Ex 3.1 Class 8 Maths Question 6. Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. You may need to download version 2.0 now from the Chrome Web Store. (b) Simple closed curve (i) Equilateral triangle NCERT Solutions Class 8 Maths PDF Download For Free. Use MCQ Questions for Class 8 Maths with Answers during preparation and score maximum marks in the exam. What is a regular polygon? (b) 8 NCERT Book for Class 8 Maths Chapter 3 Understanding Quadrilaterals is available for reading or download on this page. Hence, the sum of all the angles of a convex quadrilateral = 360°. Use our NCERT solutions for Class 8 Maths Chapter 2 to learn concepts such as equations having the variable on both sides, equations reducible to the linear form etc. ⇒ y = 180° – 80° = 100° Solution: Find x in the following figures. CBSE Maths,NCERT Solutions,ncert Class 8 Maths,CBSE 8 Class Maths,e-learning, elearning,smartur.com. (Each figure is divided into triangles and the sum of the angles reduced from that). Get here NCERT Solutions for Class 8 Maths Chapter 3.These NCERT Solutions for Class 8 of Maths subject includes detailed answers of all the questions in Chapter 3 – Understanding Quadrilaterals provided in NCERT Book which is prescribed for class 8 in schools. ⇒ 300° + x = 360° NCERT Solutions for Class 3 Maths PDF Free Download October 17, 2018 by Kishen Here we are providing the most comprehensive NCERT Solutions for Class 3 Maths PDF Chapter-wise. NCERT Solutions for Class 8 Math Chapter 7 - Cubes and Cube Roots. (d) Convex polygon: (2) In ∆ADC, ∠1 + ∠2 + ∠5 = 180° …(ii) (angle sum property) NCERT Solutions for Class 8 Maths in PDF format provided by Vedantu are solved by subject expert teachers carefully from the latest edition books and as per NCERT (CBSE) guidelines. (e) Concave polygon: (1). (d) Convex polygon NCERT Solutions for Class 8 Maths Exercise 14.3 Class 8 Maths book solutions are available in PDF format for free download. Also, the solution is the breakdown of the content of the textbook. Given here are some figures. Maths is one such subject which requires considerable time and effort. (iii) 6 sides Access the Class 11 Maths NCERT Solutions for Chapter 7 Ex 7.1, Ex 7.2, Ex 7.3, Ex 7.4, and Miscellaneous Exercise provided in Hindi & … Students can access the NCERT MCQ Questions for Class 6 Maths Chapter 8 Decimals Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Maths NCERT Solutions Class 11 Maths Chapter 10 Exercise.10.3 Q1. (d) n (e) Concave polygon Ex 3.1 Class 8 Maths Question 3. NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals - Free PDF Students can easily download the free PDF available of NCERT Solutions for Class 8 Maths chapter 3 understanding quadrilaterals from the website. NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2 Ex 3.2 Class 8 Maths Question 1. NCERT Books chapter-wise Solutions (Text & Videos) are accurate, easy-to-understand and most helpful in Homework & Exam Preparations. (Make a non-convex quadrilateral and verify) In ∆ABC, ∠3 + ∠4 + ∠6 = 180°…(i) (angle sum property) RD Sharma Class 11 Solutions Free PDF Download, NCERT Solutions for Class 12 Computer Science (Python), NCERT Solutions for Class 12 Computer Science (C++), NCERT Solutions for Class 12 Business Studies, NCERT Solutions for Class 12 Micro Economics, NCERT Solutions for Class 12 Macro Economics, NCERT Solutions for Class 12 Entrepreneurship, NCERT Solutions for Class 12 Political Science, NCERT Solutions for Class 11 Computer Science (Python), NCERT Solutions for Class 11 Business Studies, NCERT Solutions for Class 11 Entrepreneurship, NCERT Solutions for Class 11 Political Science, NCERT Solutions for Class 11 Indian Economic Development, NCERT Solutions for Class 10 Social Science, NCERT Solutions For Class 10 Hindi Sanchayan, NCERT Solutions For Class 10 Hindi Sparsh, NCERT Solutions For Class 10 Hindi Kshitiz, NCERT Solutions For Class 10 Hindi Kritika, NCERT Solutions for Class 10 Foundation of Information Technology, NCERT Solutions for Class 9 Social Science, NCERT Solutions for Class 9 Foundation of IT, PS Verma and VK Agarwal Biology Class 9 Solutions, Class 8 Maths Understanding Quadrilaterals Exercise 3.1, Class 8 Maths Understanding Quadrilaterals Exercise 3.2, Class 8 Maths Understanding Quadrilaterals Exercise 3.3, Class 8 Maths Understanding Quadrilaterals Exercise 3.4, Understanding Quadrilaterals Class 8 Extra Questions, NCERT Solutions for Class 8 Social Science, NCERT Solutions for Class 8 English Honeydew, NCERT Solutions for Class 8 English It So Happened, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, Periodic Classification of Elements Class 10, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, CBSE Previous Year Question Papers Class 12, CBSE Previous Year Question Papers Class 10. (i) x + 6y = 0 (ii) 6x + 3y – 6 = 0 (iii) y = 1 Q2. Videos of questions and theory are available for your reference. Angle sum = (n – 2) × 180°, Ex 3.1 Class 8 Maths Question 5. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Ex 14.3, drop a comment below and we will get back to you at the earliest. How many diagonals does each of the following have? Ex 8.3 Class 8 Maths Question 1. Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Scroll down for Understanding Quadrilaterals from NCERT Book Class 8 Maths Book & important study material. NCERT Solutions for Class 12 Maths Ch 3 – Solved Exercises One of the best study materials to ace up your preparation is Class 12 Maths Ncert Solutions Chapter 3 Matrices. NCERT Solutions for Class 8 Maths Chapter 4 PRACTICAL GEOMETRY in English Medium as well as Hindi Medium updated for new academic session 2020-21 based on latest CBSE Curriculum 2020-2021 and updated Ex 3.1 Class 8 Maths Question 1. The NCERT solutions for Class 8 maths Chapter 3- Understanding Quadrilaterals contains solutions for all exercise questions. Apart from that, the solutions contain the details of the chapter and solved questions of the exercise. Free PDF download of NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.3 (EX 8.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. ∠r = 360° – 260° = 100° Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. ⇒ x = 140° facebook; ... Chapter 3 : Understanding Quadrilaterals 3.1 Introduction . Mathematics NCERT Grade 8, Chapter 7: Cube and Cube Roots- The chapter starts with an interesting story about India's great mathematician S.Ramanujan.In this chapter, students will learn about cube, cube roots and many interesting facts about them. Now, y = 180° – a (Linear pair) (c) In Fig. (a) Simple curve NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3 October 4, 2019 by Sastry CBSE Class 8 Maths Understanding Quadrilaterals Exercise 3.1 State the name of a regular polygon of Use MCQ Questions for Class 6 Maths with Answers during preparation and score maximum marks in the exam. (b) A regular hexagon Solution: Draw the graphs for the following tables of values, with suitable scales on the axes. Ex 3.1 Class 8 Maths Question 2. Yes, this property also holds true for a non-convex quadrilateral. NCERT Solutions for Class 8 Maths Chapter 3 – Understanding Quadrilaterals = (n – 2) × 180° The first section is about cubes.Perfect cubes or cube numbers are … Thus x + y + z = 90° + 120° + 150° = 360° ⇒ x = 360° – 220° = 140° (b) Number of sides = 8 In this chapter you will learn the concept of linear equation and methods to solve them, reducing equations to a simpler form and their applications. x + y + z + w = 60° + 100° + 120° + 80° = 360°. Now x + 120° = 180° (Linear pair) (c) 10 Your IP: 67.199.146.77 Chapter Wise NCERT Solutions for Class 9 Math. (a) Angle sum of a quadrilateral = 360° In the given figure, we have a quadrilateral ABCD. Ex 3.1 Class 8 Maths Question 4. NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.3 Linear equations in one variable. (i) ABCD is a convex quadrilateral which has two diagonals AC and BD. Students can access the NCERT MCQ Questions for Class 8 Maths Chapter 14 Factorisation with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. NCERT Solutions for Class 3 Maths Chapter 6 Fun with Give and Take This chapter elaborates the concept of addition and subtraction. CBSE Guide of All Chapters given in the Book. Angles sum = (7 – 2) × 180° = 5 × 180° = 900° Answers to all exercise questions and examples have been solved, with concepts of the chapter explained.In this chapter, we will learnWhat arecurves, open curves, closed curves, simple curvesWhat arepolygons, Diff (a) A convex quadrilateral Will this property hold if the quadrilateral is not convex? Adding, (i) and (ii) NCERT Class 10 maths Books are very important for students who want to get good marks in their exams. Performance & security by Cloudflare, Please complete the security check to access. Of Permutations and Combinations PDF through the direct links available the latest CBSE in. 8 to score More marks are discussed by the experts of Maths teachers and according the. Page in the ncert Solutions of Chapter 3 Understanding Quadrilaterals, are provided here solved. Are given along with the Solutions by cloudflare, Please complete the security check access. Divides the quadrilateral is always 360 degrees 61841ff60d06f460 • your IP: 67.199.146.77 Performance! All Chapters given in the Book 360 degrees, elearning, smartur.com through direct... 3 will help them in Understanding this Chapter marks in the future is use! Ex 3.1 help you to revise complete Syllabus and score maximum marks in the exam, Practice Papers,.. Students associated with Class X which has two diagonals AC and BD 8 Math Chapter 8 Quantities. To Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.3 help ncert solutions for class 8 maths chapter 3. 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text:
# Introduction

It's time to get your hands dirty with some OOP. For this project, you will be helping us write an adventure game where you create people, places, things, and then make them interact.

This project is designed to be done by two partners working in parallel, then combining your results into one finished product. Hereafter the two partners are called Person A and Person B. You will hand in one combined and finished project for your group.

The project begins with two exercises that both partners should do; these exercises do not require new programming, but rather familiarize you with the overall structure of the program as we've provided it. After that, each partner has separate exercises. There is one final exercise for everyone that requires the two partners' work to be combined. (Therefore, you should probably keep notes about all of the procedures that you've modified during the project, so you can notice the ones that both partners modified independently.)

This is a long project, and you should pace yourself accordingly. Once you finish all of the exercises for the entire project, your group should make one submission, including your modified adv.scm program with the modifications highlighted, and a transcript of the testing of your work. You MUST indicate at the top of your adv.scm file which of you is Partner A and which is Partner B. Make this is very prominent and easy to find!

## Scoring

Each partner works on nine problems. Three of these (numbers 1, 2, and 9) are common to the two partners; the others are separate. You hand in a single solution to each problem. Both partners get the points awarded to the group for problems 1, 2, and 9; each partner gets the points for his or her own problems 3 through 8. This means that your score for the project is mostly based on your individual work but also relies partly on the other member of your group. For the first two problems, you could get away with letting your partner do the work, but you shouldn't because those problems are necessary to help you understand the structure of the entire project. Problem 9 requires that both partners have already done their separate work, and meet together to understand each other's solutions, so neither partner will receive credit for it unless both have done their jobs.

If you can't find a partner and/or wish to work alone, please talk to a TA.

## Acknowledgement

This assignment is loosely based on an MIT homework assignment in their version of this course. But since this is Berkeley we've changed it to be politically correct; instead of killing each other, the characters go around eating gourmet food all the time. N.B.: Unless you are a diehard yuppie you may feel that eating gourmet food does not express appropriate sensitivity to the plight of the homeless. But it's a start.

In this laboratory assignment, we will be exploring two key ideas: the simulation of a world in which objects are characterized by a set of state variables, and the use of message passing as a programming technique for modularizing worlds in which objects interact.

## Notes on Working Together

• GitHub: Solid, but long set-up time. Try here for a taste.
• Dropbox: Easy to set-up. Be careful with overwriting your partner's files!
• Google Docs: Easy to set-up. Less likely to overwrite code.
• E-Mail: Nominate one partner. Send all of your code to that partner and they will be the one responsible for merging them together.
• Other: You may use whatever method you like, so long as it promotes academic honesty and successfully shares code among partners.

format: plain


text:
# Can sulfuric acid itself be electrolysed?

We all could know by now that water can be electrolysed by adding some sulfuric acid.:)

But is it possible that the sulfuric acid itself becomes electrolysed when the concentration of the acid in the water gets higher ?

Would it give sulfur dioxide or sulfur trioxide ?

• Rather hydrogen and peroxodisulphuric acid, with ozone as a byproduct. $$\ce{ 2 H2SO4 -> H2 + H2S2O8 }$$ en.wikipedia.org/wiki/Peroxydisulfuric_acid – Poutnik Aug 18 '19 at 17:08
• @Poutnik Thank you for the useful link. I wondered how the sulfuric acid clouds of Venus could be useful. – Cornelis Aug 18 '19 at 17:39

After some searching, I found this 1956 Thesis(Reference), which describes the production of peroxydisulfuric by electrolyzing concentrated sulfuric acid. The yield varied according to the concentration of the acid, temperature, current, and the nature of the electrode. In the Thesis, the chemist noted that discharged $$\ce{HSO4-}$$ united together to form peroxydisulfuric acid ($$\ce{H2S2O8}$$). It further stated that oxidation of sulfate ions by anodic oxygen was responsible for the peroxydisulfuric acid formation. Also, hydroxyl ion at anode played an important role in the acid formation.
The reaction mechanism was confirmed by electrolyzing sulfuric acid and ammonium sulfate at various concentration at $$\pu{16\!- \! 20 ^\circ C}$$ using platinum electrode and the reaction occurred in accordance to the theory. It was also confirmed that amount of peroxidic oxygen increased with increase in ammonium sulfate. The specific gravity of sulfuric acid and the current density was varied and accordingly the highest yield of 78.4% was obtained. It was also noted that increasing the current density increases the yield up to a certain point as peroxydisulfuric acid was found to convert into peroxymonosulfuric acid. At that moment, a small trace of hydrogen peroxide was noted.
So, the chemical species formed during electrolysis are $$\ce{HSO4-}$$, $$\ce{SO4^2-}$$, peroxide ion, protons, trace of hydrogen peroxide, and other side products.


text:
3 Corrected Math typo based on the reference provided

I am reading Huber's Robust Statistics (2nd). On page 2 and 3 he gave an example. The basic facts are summarized here. Let $$(X_n)$$ be a sequence of random variables and define two measures of spread as follows.

1. Mean Absolute Deviation: $$d_n := \frac{1}{n}\sum|x_i-\bar x|$$.
2. Standard Deviation: $$s_n := \sqrt{\frac{1}{n}\sum (x_i-\bar x)^2}$$.

Then he mentioned that Fisher claimed that for identically distributed normal observations $$s_n$$ is about 12% more efficient than $$d_n$$. In addition, $$s_n$$ converges to $$\sigma$$ while $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$$$\sigma\sqrt{2/\pi}\doteq 0.8\sigma$$. I have several questions about these statements.

1. How to prove that $$s_n$$ is 12% more efficient, please? As least where to find the proof, please?
2. How to prove that $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$$$\sigma\sqrt{2/\pi}\doteq 0.8\sigma$$, please? Again at least where to find the proof, please?
3. I did some simulation to test all the above statements. Here are the codes and outcome.
n <- 10000 # number of samples
x <- array(list(), n)

set.seed(2014)

for(i in 1:n){
x[[i]] <- rnorm(10000) # the 10000 here is the size of each sample
}

dn <- rep(0, n) # mad
sn <- rep(0, n) # sd

for(i in 1:1000){
sn[i] <- sqrt(var(x[[i]])*999/1000) # sd
}

mean(dn) # **0.07979068 check out**
mean(sn) # **0.09995901 check out**

var(dn)/var(sn) # **0.6371817**

As the above simulation shows, the 12% efficiency of $$s_n$$ does not check out. Why is this the case, please? Did I make errors in my simulation, please? Thank you!

I am reading Huber's Robust Statistics (2nd). On page 2 and 3 he gave an example. The basic facts are summarized here. Let $$(X_n)$$ be a sequence of random variables and define two measures of spread as follows.

1. Mean Absolute Deviation: $$d_n := \frac{1}{n}\sum|x_i-\bar x|$$.
2. Standard Deviation: $$s_n := \sqrt{\frac{1}{n}\sum (x_i-\bar x)^2}$$.

Then he mentioned that Fisher claimed that for identically distributed normal observations $$s_n$$ is about 12% more efficient than $$d_n$$. In addition, $$s_n$$ converges to $$\sigma$$ while $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$. I have several questions about these statements.

1. How to prove that $$s_n$$ is 12% more efficient, please? As least where to find the proof, please?
2. How to prove that $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$, please? Again at least where to find the proof, please?
3. I did some simulation to test all the above statements. Here are the codes and outcome.
n <- 10000 # number of samples
x <- array(list(), n)

set.seed(2014)

for(i in 1:n){
x[[i]] <- rnorm(10000) # the 10000 here is the size of each sample
}

dn <- rep(0, n) # mad
sn <- rep(0, n) # sd

for(i in 1:1000){
sn[i] <- sqrt(var(x[[i]])*999/1000) # sd
}

mean(dn) # **0.07979068 check out**
mean(sn) # **0.09995901 check out**

var(dn)/var(sn) # **0.6371817**

As the above simulation shows, the 12% efficiency of $$s_n$$ does not check out. Why is this the case, please? Did I make errors in my simulation, please? Thank you!

I am reading Huber's Robust Statistics (2nd). On page 2 and 3 he gave an example. The basic facts are summarized here. Let $$(X_n)$$ be a sequence of random variables and define two measures of spread as follows.

1. Mean Absolute Deviation: $$d_n := \frac{1}{n}\sum|x_i-\bar x|$$.
2. Standard Deviation: $$s_n := \sqrt{\frac{1}{n}\sum (x_i-\bar x)^2}$$.

Then he mentioned that Fisher claimed that for identically distributed normal observations $$s_n$$ is about 12% more efficient than $$d_n$$. In addition, $$s_n$$ converges to $$\sigma$$ while $$d_n$$ converges to $$\sigma\sqrt{2/\pi}\doteq 0.8\sigma$$. I have several questions about these statements.

1. How to prove that $$s_n$$ is 12% more efficient, please? As least where to find the proof, please?
2. How to prove that $$d_n$$ converges to $$\sigma\sqrt{2/\pi}\doteq 0.8\sigma$$, please? Again at least where to find the proof, please?
3. I did some simulation to test all the above statements. Here are the codes and outcome.
n <- 10000 # number of samples
x <- array(list(), n)

set.seed(2014)

for(i in 1:n){
x[[i]] <- rnorm(10000) # the 10000 here is the size of each sample
}

dn <- rep(0, n) # mad
sn <- rep(0, n) # sd

for(i in 1:1000){
sn[i] <- sqrt(var(x[[i]])*999/1000) # sd
}

mean(dn) # **0.07979068 check out**
mean(sn) # **0.09995901 check out**

var(dn)/var(sn) # **0.6371817**

As the above simulation shows, the 12% efficiency of $$s_n$$ does not check out. Why is this the case, please? Did I make errors in my simulation, please? Thank you!

2 fixed code format

I am reading Huber's Robust Statistics (2nd). On page 2 and 3 he gave an example. The basic facts are summarized here. Let $$(X_n)$$ be a sequence of random variables and define two measures of spread as follows.

1. Mean Absolute Deviation: $$d_n := \frac{1}{n}\sum|x_i-\bar x|$$.
2. Standard Deviation: $$s_n := \sqrt{\frac{1}{n}\sum (x_i-\bar x)^2}$$.

Then he mentioned that Fisher claimed that for identically distributed normal observations $$s_n$$ is about 12% more efficient than $$d_n$$. In addition, $$s_n$$ converges to $$\sigma$$ while $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$. I have several questions about these statements.

1. How to prove that $$s_n$$ is 12% more efficient, please? As least where to find the proof, please?
2. How to prove that $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$, please? Again at least where to find the proof, please?
3. I did some simulation to test all the above statements. Here are the codes and outcome.

n <- 10000 # number of samples x <- array(list(), n)

set.seed(2014)

for(i in 1:n){ x[[i]] <- rnorm(10000) # the 10000 here is the size of each sample }

dn <- rep(0, n) # mad sn <- rep(0, n) # sd

for(i in 1:1000){ dn[i] <- mean(abs(x[[i]]-mean(x[[i]]))) # mad sn[i] <- sqrt(var(x[[i]])*999/1000) # sd }

mean(dn) # 0.07979068 check out mean(sn) # 0.09995901 check out

var(dn)/var(sn) # 0.6371817

I did some simulation to test all the above statements. Here are the codes and outcome.
n <- 10000 # number of samples
x <- array(list(), n)

set.seed(2014)

for(i in 1:n){
x[[i]] <- rnorm(10000) # the 10000 here is the size of each sample
}

dn <- rep(0, n) # mad
sn <- rep(0, n) # sd

for(i in 1:1000){
sn[i] <- sqrt(var(x[[i]])*999/1000) # sd
}

mean(dn) # **0.07979068 check out**
mean(sn) # **0.09995901 check out**

var(dn)/var(sn) # **0.6371817**

As the above simulation shows, the 12% efficiency of $$s_n$$ does not check out. Why is this the case, please? Did I make errors in my simulation, please? Thank you!

I am reading Huber's Robust Statistics (2nd). On page 2 and 3 he gave an example. The basic facts are summarized here. Let $$(X_n)$$ be a sequence of random variables and define two measures of spread as follows.

1. Mean Absolute Deviation: $$d_n := \frac{1}{n}\sum|x_i-\bar x|$$.
2. Standard Deviation: $$s_n := \sqrt{\frac{1}{n}\sum (x_i-\bar x)^2}$$.

Then he mentioned that Fisher claimed that for identically distributed normal observations $$s_n$$ is about 12% more efficient than $$d_n$$. In addition, $$s_n$$ converges to $$\sigma$$ while $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$. I have several questions about these statements.

1. How to prove that $$s_n$$ is 12% more efficient, please? As least where to find the proof, please?
2. How to prove that $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$, please? Again at least where to find the proof, please?
3. I did some simulation to test all the above statements. Here are the codes and outcome.

n <- 10000 # number of samples x <- array(list(), n)

set.seed(2014)

for(i in 1:n){ x[[i]] <- rnorm(10000) # the 10000 here is the size of each sample }

dn <- rep(0, n) # mad sn <- rep(0, n) # sd

for(i in 1:1000){ dn[i] <- mean(abs(x[[i]]-mean(x[[i]]))) # mad sn[i] <- sqrt(var(x[[i]])*999/1000) # sd }

mean(dn) # 0.07979068 check out mean(sn) # 0.09995901 check out

var(dn)/var(sn) # 0.6371817

As the above simulation shows, the 12% efficiency of $$s_n$$ does not check out. Why is this the case, please? Did I make errors in my simulation, please? Thank you!

I am reading Huber's Robust Statistics (2nd). On page 2 and 3 he gave an example. The basic facts are summarized here. Let $$(X_n)$$ be a sequence of random variables and define two measures of spread as follows.

1. Mean Absolute Deviation: $$d_n := \frac{1}{n}\sum|x_i-\bar x|$$.
2. Standard Deviation: $$s_n := \sqrt{\frac{1}{n}\sum (x_i-\bar x)^2}$$.

Then he mentioned that Fisher claimed that for identically distributed normal observations $$s_n$$ is about 12% more efficient than $$d_n$$. In addition, $$s_n$$ converges to $$\sigma$$ while $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$. I have several questions about these statements.

1. How to prove that $$s_n$$ is 12% more efficient, please? As least where to find the proof, please?
2. How to prove that $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$, please? Again at least where to find the proof, please?
3. I did some simulation to test all the above statements. Here are the codes and outcome.
n <- 10000 # number of samples
x <- array(list(), n)

set.seed(2014)

for(i in 1:n){
x[[i]] <- rnorm(10000) # the 10000 here is the size of each sample
}

dn <- rep(0, n) # mad
sn <- rep(0, n) # sd

for(i in 1:1000){
sn[i] <- sqrt(var(x[[i]])*999/1000) # sd
}

mean(dn) # **0.07979068 check out**
mean(sn) # **0.09995901 check out**

var(dn)/var(sn) # **0.6371817**

As the above simulation shows, the 12% efficiency of $$s_n$$ does not check out. Why is this the case, please? Did I make errors in my simulation, please? Thank you!

1

# Comparison between MAD and SD

I am reading Huber's Robust Statistics (2nd). On page 2 and 3 he gave an example. The basic facts are summarized here. Let $$(X_n)$$ be a sequence of random variables and define two measures of spread as follows.

1. Mean Absolute Deviation: $$d_n := \frac{1}{n}\sum|x_i-\bar x|$$.
2. Standard Deviation: $$s_n := \sqrt{\frac{1}{n}\sum (x_i-\bar x)^2}$$.

Then he mentioned that Fisher claimed that for identically distributed normal observations $$s_n$$ is about 12% more efficient than $$d_n$$. In addition, $$s_n$$ converges to $$\sigma$$ while $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$. I have several questions about these statements.

1. How to prove that $$s_n$$ is 12% more efficient, please? As least where to find the proof, please?
2. How to prove that $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$, please? Again at least where to find the proof, please?
3. I did some simulation to test all the above statements. Here are the codes and outcome.

n <- 10000 # number of samples x <- array(list(), n)

set.seed(2014)

for(i in 1:n){ x[[i]] <- rnorm(10000) # the 10000 here is the size of each sample }

dn <- rep(0, n) # mad sn <- rep(0, n) # sd

for(i in 1:1000){ dn[i] <- mean(abs(x[[i]]-mean(x[[i]]))) # mad sn[i] <- sqrt(var(x[[i]])*999/1000) # sd }

mean(dn) # 0.07979068 check out mean(sn) # 0.09995901 check out

var(dn)/var(sn) # 0.6371817

As the above simulation shows, the 12% efficiency of $$s_n$$ does not check out. Why is this the case, please? Did I make errors in my simulation, please? Thank you!


text:
Comparison Testing

In the context of the Data Validation Library, a comparison test compares two catalogs. These tests fall under two key use cases:

• comparing two versions of the same catalog
• comparing two versions of two distinct catalogs from which contents can be extracted into shared data structures

The validation library's comparison API is exposed in three packages with different levels of abstraction:

• validation.core.comparison - provides an abstract recipe to implement comparisons for any kind of object using any kind of keys. You need to work directly with apache Spark's RDD objects.
• validation.core.comparison.metadiff - provides an abstract implementation of the base recipe using a combination of layer names and and partition identifiers as keys. This package checks for differences by comparing the metadata checksums. For grouping and actually creating the output data, you are still required to work with RDDs.
• validation.core.comparison.metadiff.grouped - provides two callbacks to work with pairs of differing catalog partition metadata; either layer-wise or partition-wise. Typically, this abstraction fulfills the majority of use cases.

The recommended process is to consider comparison.metadiff.grouped first, to see if it meets your requirements. Next, evaluate comparison.metadiff. If both of these packages do not meet your requirements, then consider the base level, comparison.

In turn, you can use comparison.metadiff as an example for how to use comparison.metadiff, and comparison.metadiff.grouped as an example for how to use comparison.metadiff.

To run a grouped comparison pipeline, refer to the quickstart-example in the SDK package.

In the sections below:

• reference refers to the baseline catalog version
• candidate refers to the catalog version being tested

The Comparison Package

The Comparator is the main class, which implements the inherited compile() function, that is called with the complete data of the reference.

You must provide access to the candidate's data. As shown in the code snippet below, the compile method extracts the candidate and reference data for comparison via the Joiner.join method, which you implement, and performs the actual comparison in the Comparison.compare method which you also implement.

scala
abstract class Comparator[K, C] (joiner: Joiner[K, C], comparison: Comparison[K, C])
extends NonIncrementalCompiler {

def queryReference(): InData

final override def compile(candidateData: InData, parallelism: Int)(
implicit logContext: LogContext): ToPublish = {
val referenceData = queryReference
val joinedData: JoinedData[K, C] = joiner.join(candidateData, referenceData)
val results: ToPublish = comparison.compare(joinedData)
results.partitionBy(outPartitioner(parallelism))
}
}

Since the following RDD declaration appears often, there is a type definition for convenience:

scala
type JoinedData[K, C] = RDD[(K, (Option[C], Option[C]))]

The Joiner trait joins both the reference and candidate data. It is your implementation's responsibility to do this in a way that is suitable for the corresponding comparison.

scala
trait Joiner[K, C] {
def join(referenceData: InData, candidateData: InData): JoinedData[K, C]
}

The Comparison trait does the actual comparison of the previously joined data and returns output data appropriately to the output layer configuration.

scala
trait Comparison[K, C] {
def compare(data: JoinedData[K, C]): ToPublish
}

The ContextHelper class queries the reference data and also provides Retrievers for the reference and the candidate catalog. These Retrievers are needed if you want to access the actual partition's content by retrieving the Payload for the given partition's metadata.

This package implements comparison by using a LayerKey as key to join the partitions' metadata.

scala
case class LayerKey(layer: Layer.Id, partition: Partition.Name)

The MetadataComparison class defines a retrieveResults() callback, that you need to implement to handle the metadata pairs that differ in their partitions' payload checksums. Since this data remains in an RDD, you can still group it according to your output needs.

scala
def retrieveResults(different: JoinedData[LayerKey, InMeta]): ToPublish

The Grouped Package

This package provides a quick and easy way for you to get a diff for a defined set of layers. As mentioned, this package offers two callbacks that you can use to work with pairs of differing catalog partition metadata: either layer-wise or partition-wise.

For layer-wise pairs:

scala
def handleDiff(layer: Layer.Id,
partitioned: Iterable[(Partition.Name, Option[InMeta], Option[InMeta])])
: Iterable[(OutKey, Option[Payload])]


text:
# Black Hole Rotation Speed

1. Jun 29, 2011

### mat1101

Is there a rotation speed for black holes, and if so, (either than infinity), what is the fastest speed one can travel at if it is about stellar mass black hole.

Also I would like a equation, but please it explain what it means because I am only 14.

2. Jun 29, 2011

### atyy

Last edited by a moderator: May 5, 2017
3. Jun 29, 2011

### bcrowell

Staff Emeritus
That's a great question. I have a PhD in physics, and I don't know the answer (although I can offer the extenuating circumstance that relativity isn't my specialty :-) ).

When a spinning figure skater pulls in her arms from r=1 unit to r =0.5 units, her velocity goes up. If you model her naively as two masses at the ends of a cord spinning end over end, then each mass's velocity doubles. You'd think that when r became zero, the velocity would reach infinity, but of course nothing can go faster than c. Relativistically, momentum doesn't depend on v, it depends on $v\gamma$, where $\gamma=1/\sqrt{1-v^2/c^2}$. This approaches infinity as v approaches c.

So I'm going to throw out a wild guess, which somebody like George Jones or Sam Gralla who is really knowledgeable about GR will probably shoot down. When a spinning black hole forms by gravitational collapse, the infalling matter has a radial velocity that approaches c as it reaches the event horizon. As it gets closer and closer to the singularity, its tangential velocity also approaches c.

The other thing that's not obvious to me is whether the angular momentum of a black hole is localized in its singularity, or whether the spacetime itself carries it. Or maybe it doesn't even make sense to ask where it's localized, since there isn't any really meaningful way in which we can say what's going on "now" inside the event horizon.

4. Jun 29, 2011

### xmangosaintx

Prof,
I have a few question related.. But as a pre-intro before an epic criticism, I'd like to say that I'm only reading physics as an enjoyable pass time, and I'm only a freshman in Bachelor of Psychology.

My question is:

1. The gravitational pull of a blackhole is very big, that even light couldn't escape. True?
2. If light, c, which is the cosmic speed limit, couldn't escape the pull, doesn't it mean that the black hole speed is larger?

These statements are contradicting, and I'm betting on my false information..

Hope to hear from yu,
-Hevind-

5. Jun 29, 2011

### mat1101

1. yes that is true
2. I would think so, but that would be something for a doctorate to answer

6. Jun 29, 2011

### mat1101

And thank you!
also would it be possible that if the black holes rotation was so fast, that not only space would warp around it, but stars, planets, etc. could too?

7. Jun 29, 2011

### bcrowell

Staff Emeritus
I'm pretty sure my answer is wrong, though :-) Let's wait for someone who is more of an expert to comment on your original question.

8. Jun 29, 2011

### mat1101

okay thanks!

9. Jun 29, 2011

### OnlyMe

Ben, it would seem from the results of the Gravity Probe B experiment that the frame dragging effect would have to apply equally to a black hole, thus the angular momentum of a rotating black hole should drag the surrounding space along with it. Essentially imparting some angular momentum to the surrounding space. While outside of the event horizon space could not be inertially locked to the rotation of the black hole, there should be a diminishing with distance "angular momentum"(?) of space itself???

10. Jun 29, 2011

### Pengwuino

Your first statement is true. The second statement is not true. Black holes can be stationary in fact, so that is not reason to assume there's a relationship between the angular velocity of a black hole and the speed of light.

Last edited: Jun 29, 2011
11. Jun 29, 2011

### mat1101

So is there an equation to solve the speed rotation of a black hole

Also, if i were to turn this into a science fair project, do you think it would have the chance to win a prize such as a scholarship.
I want to prove that, at such a rate of speed, not only space warps around the black hole, but stars, and planets too.

12. Jun 29, 2011

### Nabeshin

There are a few just general comments I want to make to clarify some of the stuff I'm reading here...

In a rotating black hole, it's good to note that nothing is actually moving. There's nothing to move, since a black hole is a solution which exists in vacuum. The event horizon, the only meaningful surface associated with a black hole, is not a physical object so it doesn't make any sense to talk about this spinning. Furthermore, the other obvious point, the singularity, is not described by GR, so while a quantum theory of gravity may shed light on some actual spinning taking place here, we certainly can say nothing within the context of GR.

Another point I want to make is that the particular quantity we call the spin of a black hole is just a parameter in the black hole solution. In this sense, it is on equal grounds with what we call the mass of the black hole. Identifying these two quantities with the familiar concepts of angular momentum and mass is a bit fuzzy. A nice analogy would be the spin of elementary particles, such as the electron. The electron, having no size, cannot be said to physically spin in any meaningful sense. Yet, we persist in calling a particular parameter associated with the properties of the electron 'spin'. Similarly with black holes, although with BH we are able to make the association much clearer since we understand about angular momentum and mass which may fall into the black hole, and we want to preserve a kind of continuity between the mass/angular momentum of infalling matter with the two quantities associated with the black hole.

I hope at least some of that made any sense to someone...

13. Jun 29, 2011

### Pengwuino

@Nabeshin: So what kind of inferences can we make when we talk about the Kerr solution in regards to the angular momentum? For example, the Schwarzschild solution doesn't just make sense for black holes; it works in general for spherically symmetric static situations. So I can look at the Sun and I really can identify to good approximation the mass of the sun with the Schwarzschild "mass". Going to the Kerr solution, does the "angular momentum" make a good approximation to what we traditionally know as angular momentum for something like a pulsar? I only use a pulsar as an example because of it's high rotation speed and I know it's closer to a black hole than something like our Sun; if there were a less massive object with high rotation speed that I knew of I would use it as my example.

14. Jun 29, 2011

### WannabeNewton

Are you asking if the black hole's rotation would cause other objects to rotate with it?

15. Jun 29, 2011

### Nabeshin

Yes, I think so.

I'm fairly sure you're familiar with things like the ADM mass? The general point is that notions such as mass, and also angular momentum, are not well defined in general relativity. However, if we take the schwarzschild solution and expand in a post-newtonian sort of formalism, we recover the normal Newtonian equations of motion. Only then do we observe the correspondence between the parameter "m" in the schwarzschild metric and the familiar "m" of Newtonian mechanics.

Without saying much more, I'll just state I think the situation is analogous as far as the angular momentum is concerned.

16. Jun 30, 2011

### bcrowell

Staff Emeritus
Hmm...this is just for the sake of argument, so feel free to shoot me down.

When we talk about a Schwarzschild black hole with a singularity at the center, we're talking about a spacetime that is in a sense unphysical, because a real, physical black hole probably has to form by gravitational collapse. An observer who falls through the event horizon is forced to believe in the singularity, because he will reach it in finite proper time. But an observer on the outside can say that infalling matter takes infinite time to pass through the event horizon, so no singularity exists. In this sense, can't we just take away the whole mystery of how the black hole can have mass if there is no mass at any point in the manifold? The mass is ordinary atoms and dust that are asymptotically approaching the horizon. And in this sense the mass of a black hole isn't just an arbitrary parameter, it's the mass of whatever collapsed.

So can't we do the same thing with the angular momentum as with the mass?

To me there's a problem with this analogy, since it's actually not possible to get a half-integral spin by having matter moving around in little circles, whereas it is possible to attribute the angular momentum of a black hole to the motion of the infalling matter.

17. Jun 30, 2011

### pervect

Staff Emeritus
From "Black holes and Time Warps",which I'd recommend reading, or even buying. The story below is fiction, of course, but it's based on physical calculations by the co-author of a very-well known GR textbook.

It's a fun read, and should answer this question and a lot of others in an entertaining manner.

18. Jun 30, 2011

### Nabeshin

This is an excellent point, and you're right, I was just thinking in terms of the mathematical schwarzschild solution rather than the more astrophysical context for a black hole. I actually think this argument makes a lot of sense, and yes to an observer who sees the black hole forming, it should be trivial to identify the black hole parameter mass with the infalling mass. To most observers though, it seems that any light from infalling matter will have long since redshifted so far down the spectrum as to be rendered invisible. So in this sense they will not be able to use the scheme you have described.

I don't immediately see why not... I'll just reiterate that 99% of the time I think of black holes as pre-existing objects in the universe which have (presumably) been there forever, so I'm not too versed on some of the finer points relating to measurements and whatnot while they form.

Again, this is the kind of thing that, like you say, depends on what kind of observer you are. If you're the (unfortunate) kind who is falling into the BH, and thus forced to believe in the singularity, as you put it, then you attribute the angular momentum to a point. Also, it doesn't have to be an electron if you don't want it to be. Pick a boson with spin 1 and that removes the issue of half-integer spins, and yet you still have a size zero particle possessing angular momentum.

19. Jun 30, 2011

### stevebd1

The quantity of spin for a black hole is normally expressed as a/M where a is the spin parameter in metres (a=J/mc) and M is the gravitational radius in metres (M=Gm/c2) where J is angular momentum, m is mass in kg, c is the speed of light in m/s and G is the gravitational constant. It's thought that a number of black holes have a spin of a/M=0.998 which is very fast (1 being the max).

a, the spin parameter, is normally established by looking at the marginally stable orbit (MSO) which normally coincides with the inner edge of the accretion disc, this is at 6M for a static black hole with no spin and at 1M for a black hole with maximum spin (i.e. a/M=1). Depending on where the MSO is tells us how much spin the black hole has. This is derived from the equations on page 258 of http://www.lsw.uni-heidelberg.de/users/mcamenzi/CObjects_06.pdf" [Broken] where, knowing where the MSO is, the equations can be rewritten relative to the spin parameter a.

Spinning black holes generate something called frame dragging where an object approaching the black hole with no angular momentum (ZAMO- Zero Angular Momentum Object) can begin to spin around the black hole. The angular velocity of this object when observed from infinity is expressed as-

$$\omega=\frac{2Mrac}{\Sigma^2}$$

where

$$\Sigma^2=(r^2+a^2)^2-a^2\Delta \sin^2\theta$$
$$\Delta= r^{2}+a^{2}-2Mr$$

The tangential velocity of the object as observed from infinity is expressed as-

$$v_t=\omega\,R$$

where R is the reduced circumference-

$$R=\frac{\Sigma}{\rho}\,\sin\theta$$
and $\rho=\sqrt{r^2+a^2 \cos^2\theta}$

For the local quantities of angular & tangential velocity, you would divide $\omega$ and vt by the red shift $(\alpha)$-

$$\alpha=\frac{\rho}{\Sigma}\sqrt{\Delta}$$

Last edited by a moderator: May 5, 2017
20. Jun 30, 2011

### mat1101

Thank you so much for explaining it really helped!!!


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# Limit Points in Excluded Point Space

## Theorem

Let $T = \left({S, \tau_{\bar p}}\right)$ be an excluded point space.

Let $x \in S$ such that $x \ne p$.

Then $p$ is the only limit point of $x$.

Similarly, let $U \subseteq S$.

Then $p$ is the only limit point of $U$.

## Proof

Let $U \subseteq S$, and let $x \in S, x \ne p$.

From:

Excluded Point Topology is Open Extension Topology of Discrete Topology
Limit Points in Open Extension Space

it follows that:

$p$ is a limit point of $U$
$p$ is a limit point of $x$.

Now suppose $y \in S, y \ne p$.

We have by definition of excluded point space that $\left\{{y}\right\}$ is open in $T$.

So there is no $z \in \left\{{y}\right\}: z \ne y, z \in U$.

Hence $y$ can not be a limit point of $U$.

Similarly $x \notin \left\{{y}\right\}$.

So $y$ can not be a limit point of $x$.

Hence the result.

$\blacksquare$


text:
Motion Under Central Forces (part 1)

This article covers an introduction to central forces , equation of motion under central forces and

Introduction

If the force $\mathop f\limits^ \to$ acting on a body has following characteristics then it is a central force (i) it depends on the distance between two particles (ii) it is always directed towards or away from a fixed point. Gravitational force is an example of central forces. Mathematically if we consider central point as origin

Matter Waves

This article is about matter waves and it contain brief introduction of the topic followed by few questions. Photoelectric effect and Compton scattering are evidences of the corpuscular nature of light. Interference , diffraction and polarisation reveal the wave nature of light. Louis de Brogli was the first one to suggest that matter may have …

Elementary Particles Revision Notes (B.Sc. and GATE)

(1) Introduction Elementary particles are those microscopic elementary constituents out of which all matter in this universe is made of. Bound neutron is stable but unbound neutron is unstable and it decays according to equation $n \to p + e + \overline {{\nu _e}}$ (anti-nutrino) Half life of free neutron is 14 min 49 …

Particle Properties of waves (Revision Notes)

This article is about the revision of Particle properties of wave. I’ll be doing detailed articles on each one of the sub-topics given in this article. Classical Mechanics do not put any restriction on values of a dynamical property of a particle, so small and large both values are permitted. Principle of superposition:- When two …

Total energy of earth in its circular orbit around the sun

Question : Find out the total energy of earth in its circular orbit around the sun in terms of gravitational constant Answer: Let R be the total distance between the earth and the sun. If


text:
Text editor capable of wrapping text at syllable boundaries

I am looking for a text editor that can break lines at syllable boundaries and not only at word boundaries, the latter being the default for products like Microsoft Word or Notepad++.

My purpose is to have nicely laid out texts, like in printed books, as I am not completely satisfied with the looks of Justified Text.

I am specifically searching for a solution for the Italian language.

Thanks

Latex is capable of doing this, if you are willing to go that route. An example:

\documentclass{article}
\usepackage[italian]{babel}

\begin{document}

Il contratto di trasporto concluso ai sensi delle Condizioni Generali,
di norma, è relativo all’utilizzo da parte del passeggero di un singolo
treno ed è rappresentato da un titolo di viaggio che legittima il
medesimo ad usufruire del servizio ferroviario prescelto.

\end{document}

Result:

(text taken from travel conditions of trenitalia)

Note that of course the editor does not do the wrapping - you write whole words -, but it is done during compiling the resulting pdf.


text:
## Current Direct Neutrino Mass Experiments    [PDF]

G. Drexlin, V. Hannen, S. Mertens, C. Weinheimer
In this contribution we review the status and perspectives of direct neutrino mass experiments. These experiments investigate the kinematics of $\beta$-decays of specific isotopes ($^3$H, $^{187}$Re, $^{163}$Ho) to derive model-independent information on the averaged electron (anti-) neutrino mass, which is formed by the incoherent sum of the neutrino mass eigenstates contributing to the electron neutrino. We first review the kinematics of $\beta$-decay and the determination of the neutrino mass, before giving a brief overview of past neutrino mass measurements (SN1987a-ToF studies, Mainz and Troitsk experiments for $^3$H, cryo-bolometers for $^{187}$Re). We then describe the Karlsruhe Tritium Neutrino (KATRIN) experiment which is currently under construction at Karlsruhe Institute of Technology. The large-scale setup will use the MAC-E-Filter principle pioneered earlier to push the sensitivity down to a value of 200 meV(90% C.L.). KATRIN faces many technological challenges that have to be resolved with regard to source intensity and stability, as well as precision energy analysis and low background rate close to the kinematic endpoint of tritium $\beta$-decay at 18.6 keV. We then review new experimental approaches such as the MARE, ECHO and Project8 experiments, which offer the promise to perform an independent measurement of the neutrino mass in the sub-eV region. This variety of methods and the novel technologies developed in all present and future experiments demonstrate the great potential of direct neutrino mass experiments in providing vital information on the absolute mass scale of neutrinos.
View original: http://arxiv.org/abs/1307.0101


text:
# Arithmetic Squence Tutorial

## Sequence

A sequence is a collection of numbers and terms.Sequence may be finite and infinite.

## Finite sequence :

$$\{a_1,a_2,a_3,a_4,a_5\}$$

## Infinite sequence :

$$\{a_1,a_2,a_3,a_4,......a_n\}$$

## Arithmetic sequence

A pattern of terms or numbers that increase or decrease at a constant amount.

Example:

$$\{1,5,9,13,17,21,25,29,33\}$$

This sequence has a difference of 4 between each number.

We could write like that:

$$\{a,a+d,a+2d,a+3d,......\}$$

a is the first term, and

d is the difference between them.

Now

$$a = 1$$

$$d = 4$$

So,

{1,1+4,1+2*4,1+3*4,1+4*4,1+5*4,...}

$$\{1,5,9,13,17,21……….\}$$


text:
# Closed set in Banach?

• April 15th 2010, 01:08 AM
veljko
Closed set in Banach?
Is set
$L=\{f\in X | \int_{-\infty}^{+ \infty} f(x)dx =0\}$.
closed in Banach space $X$?

a) $X = L^1 (-\infty, +\infty)$, b) $X = L^2 (-\infty, +\infty)$
• April 15th 2010, 02:45 AM
Enrique2
a) Is true because $f\mapsto \int_{\mathbb{R}}f$ is a continuous linear functional on $L_1(\mathbb{R})$. If we denote by $T$ this continuous mapping the space $L$ is
$T^{-1}(0)$.

b) Is false. To prove it just observe that there are dense subspaces in $L_2(\mathbb{R})$ formed by integrable functions and with integral 0. Look at the Haar system Haar wavelet - Wikipedia, the free encyclopedia.

If the set of functions with integral 0 were closed, this would imply that the closure of the span of the Haar system, that is the whole $L_2(\mathbb{R})$, would be included in $L$, that certainly is not true.


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# Mercurial/TortoiseHg: Move Files or Directories

A move is called a rename in Mercurial. So, to move a tracked file/directory called Foobar into FooDir use:

$hg rename Foobar FooDir\Foobar  The UI of TortoiseHg does not seem to be as mature as TortoiseSVN right now (v0.9.3). One cannot move a file/directory by keeping the right mouse button pressed while moving the picked items across to the destination directory. To move a file or directory, right-click and choose TortoiseHgRename File. Advertisements # Mercurial: Import Subversion Project Mercurial is a distributed VCS (DVCS). If you are planning to migrate from Subversion to Mercurial, it might be better to do this on a per-project basis. Due to the nature of a DVCS, it is better to spin off each of your Subversion projects in the repository as separate Mercurial repositories. Here is how: 1. If you have not already done so, make a copy of the Mercurial configuration file (Mercurial.ini) and add your username. See this post on how to do this. 2. The command to convert repositories is hg convert. It will not work unless the convert extension is enabled. To do this, open your Mercurial.ini and in the [extensions] section uncomment the convert = line. It should read: [extensions] convert =  3. The default Windows binaries of Mercurial cannot import from Subversion. If you attempt a hg convert using this binary, you will see an error: $ hg convert file:///C:/SVNRep/FoobarProject
Subversion python bindings could not be loaded

You will need the Python bindings of Subversion and Mercurial installed on top of a standalone Python to be able to do this. That is too messy! But, there is an easier alternative to this. Install TortoiseHg, the Explorer extension to Hg. Its hg.exe has been built with all the necessary bindings. Issue the convert command using its hg.exe:

$"C:\Program Files\TortoiseHg\hg.exe" convert file:///C:/SVNRep/FoobarProject  4. This should work. If your Subversion project has been cleanly maintained with the canonical trunks, branches and tags structure hg convert can import all of that correctly into the Mercurial structure of tags. # Mercurial: No username supplied The first error you might face while using Mercurial might be this: $ hg commit
abort: no username supplied (see "hg help config")

Mercurial needs a username and email for almost all its operations. It looks for this in a Mercurial configuration file.

For Windows: The Mercurial configuration file is named Mercurial.ini. You will find a copy of this file in the Mercurial installation directory. Copy it to your %HOMEDRIVE%%HOMEPATH% directory (C:\Users\AverageJoe\ on Windows 7).

For Linux: The Mercurial configuration file is named .hgrc. Create a file with this name in the home directory of the user if not already there.

Add these two lines to the Mercurial configuration file:

[ui]

Mercurial commands should work fine after this.

Tried with: Windows 7 Enterprise, Ubuntu 12.04 and Mercurial 2.0.2

# Ipe: Import Ipe Figures

Ipe is the editor of choice to draw figures for Computational Geometry papers and presentations. It is common to create a Ipe figure composed of other Ipe figures (stored in different .ipe files). But, there is no straightforward way to accomplish this import operation.

One way to achieve this is:

1. Open the source and destination Ipe figures in separate Ipe instances.
2. In the source Ipe instance, do a Select All to select the complete source Ipe figure. Copy it.
3. Switch back to the destination Ipe and Paste.

# CDBurnerXP: Copy a Bootable Disc

A little care is needed when making a copy of OS installation discs. Making a normal data copy does not copy the boot data and renders the new disc un-bootable. To create a bootable disc copy of a bootable disc using CDBurnerXP:

1. Put the original bootable disc in the drive. Use Copy Data Disc with the destination as your harddisk. This creates a bootable ISO copy of the disc.
2. Put a blank disc in the drive. Choose the Burn ISO Image option and use the bootable ISO created in the above step.

# OpenVPN: Route Add Failure on Windows Vista

Problem

OpenVPN (and OpenVPN GUI) on Windows Vista fails silently if executed without Administrator privileges. The error is almost not noticeable since VPN succeeds in obtaining an IP address (see the green OpenVPN icon in the screenshot). But, the VPN is not established due to the error:

ERROR: Windows route add command failed: system() returned error code 1
The requested operation requires elevation.

Solution

Run OpenVPN as Administrator or if this is a frequent used application change its shortcut to always run as Administrator.

# Windows Vista: Service Pack 2 Cleanup

I am using a LG C1 Express Dual tablet PC for a few weeks. It came with a fresh Vista image, which on connecting to the Internet underwent long hours of Windows updates, pulling itself through Vista SP1 (Service Pack 1) and later Vista SP2.

After these updates, I saw that the free space on C: had reduced quite a bit and ran Treesize Free to check out the situation. I found the C:\Windows\WinSxS directory occupying a lot of space!

It turns out that the WinSxS directory keeps components from all old updates so that uninstall (and rollback) of updates can be supported. Since, SP2 is a pretty significant update, I am not interested in keeping the files from the updates older than SP2. If you are sure of this, you can recover a lot of space by running the Windows Component Clean Tool (compcln.exe).

compcln.exe resides in a super-long-name directory inside C:\Windows\WinSxS, but it is hardlinked to the C:\Windows\Sytem32 directory. So, to execute it, open cmd and type:

\$ C:\Windows\System32\compcln

Press Y when it asks for confirmation of cleanup. Running compcln.exe recovered ~1GB of disk space on this tablet PC.

# How to do image animation in Beamer

Beamer makes it very easy to include a series of images and display them in a serial animated style in separate frames (slides).

For example, to display the images foobar-0.png, foobar-1.png and foobar-2.png in a serial fashion use:

\usepackage{xmpmulti}
\begin{frame}
\multiinclude[format=png]{foobar}
\end{frame}

xmpmulti is a package that ships along with the Beamer package. \multiinclude is the command that takes care of putting the foobar-x.png into different frames. Make sure that the files are named as basename-number.format, i.e., the basename and the number are separated by a hyphen. The format option is the extension of the files.

\multiinclude starts from file number 0. To start from a different number, say 9:

\multiinclude[format=png,start=9]{foobar}

\multiinclude inserts all files that are in the above format. To make it stop at a certain number, say 7:

\multiinclude[format=png,end=7]{foobar}

Typically, images are inserted using the \includegraphics command. The settings passed to \includegraphics can be passed to \multiinclude using its graphics option. For example, to set the scale of the image to 0.3:

\multiinclude[format=png,graphics={scale=0.3}]{foobar}

By default, the images are placed one on top of another. To replace each image by the next image try:

\multiinclude[<+>][format=png]{foobar}

# IrfanView: Replicate Crop on Multiple Images

I had a series of images of the same dimensions and wanted to crop all of them with the same bounds. Eyeballing the original crop and approximating the same on the rest of the images was out of the question, since the crops had to match to the very last pixel.

I tried to note down the crop parameters and redo the same on the rest of the images. When a crop rectangle is drawn on the image, its parameters can be seen on the Irfanview titlebar. The first 2 numbers are the coordinates of the origin (top left) and the next two numbers are the dimensions of the crop rectangle.

Redoing this crop manually on many images turned out bad since it was tiring to draw the rectangle with a mouse correctly to match the parameters.

Thankfully, the swiss army knife that is IrfanView has a better solution. Choose EditCreate custom crop selection. If you have chosen a crop rectangle its parameters will already be displayed. Click on Save current values, click on Apply to image and crop the image. Open the rest of the images and do the same.

# Beamer: Frame Title and Subtitle

The common method to provide the title and subtitle for a frame (slide) is:

\begin{frame}
\frametitle{News}
\framesubtitle{Technology}
\end{frame}

A shorter version which involves lesser typing is to provide the title and subtitle to \begin{frame} itself:

\begin{frame}{News}{Technology}
\end{frame}


text:
# Letters in LaTeX

My standard letter in LaTeX with the class scrlttr2 (created by a German to adhere to German letter guidelines):
 \documentclass[fromalign=location, fromphone=true, fromemail=true, locfield=wide]{scrlttr2} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \setkomavar{fromname}{Max Mustermann} \setkomavar{fromaddress}{Musterstr.\ 1\\12345 Musterstadt} \setkomavar{fromphone}{01234/56789000} \setkomavar{fromemail}{mustermx@provider.xy} \setkomavar{signature}{Max Mustermann} \setkomavar{subject}[]{} \setkomavar{date}{7.\ August 2013}

\begin{document} \begin{letter}{Jane Doe\\ Example street 2\\ 54321 Exampleville } \opening{Dear Mrs.\ X,} this is the letter I promised. \closing{Kind regards,} \end{letter} \end{document}

The variables ‘from…’ set the sender, the recipient is given right after begin letter. The sender information can be set at strange places, for a very simple letter I use ‘fromalign=location’ which results in the sender information somewhere at the top right, a bit higher than the recipient, but not in the headline. With the standard settings the e-mail address is too wide for the sender field, so I widen it with ‘locfield=wide’.

This entry was posted in LaTeX and tagged , , , , by swk. Bookmark the permalink.


text:
# 2.81: Align objects by specified point

I have several arbitrary objects. I need to align them along a line by specified points. I have tried to do it with Origin point, but no success.

Any suggestions?

EDIT 1

"Oscurart Tools" solves the problem but partly. It distributes objects along the straight line only. See the image below :|

• @JACHYM Thank you. But there are 2 points about Oscurart Tools: 1 - using this tool I need to align the first and the last object nevertheless, 2 - it doesn't work for curves (see my Edit 1 in the question). – VictorDDT Jan 7 at 14:07


text:
# The Last Group Theory Feedback

My fifth assignment has been returned! And with it, the usual amount of feedback!

In this assignment, I failed to realise that a subgroup is normal if it is a union of conjugacy classes. So rather than form subgroups and state they were a union of conjugacy classes, I went through the process of proving each group was a sub group.  Too long-winded, and frankly a waste of time. (for reference, this refers to Theorem 3.5 on properties of subgroups, p.74 in the handbook for M208).

There was another question where I had to show that a particular group had no subgroup of order 4. I took this rather literally, and found the only 4 possible combinations of elements that could be a subgroup of order 4, and then individually proved each one did not qualify either as a subgroup or as normal.  Again, although I received full marks for this as it was correct, it was long-winded. It seems that all I needed to notice that a group of order 4 is either a typical cyclic group of order 4 ( $C_{4}$ ), or a Klein group ( $K_{4}$ ). A property of $C_{4}$ is that it contains an element of order 4. But the question says the group has no subgroup of order 4! Therefore, my possible subgroups must be isomorphic to $K_{4}$. Property of $K_{4}$: it contains three elements of order 2. This cuts the number of possibilities in half. Amazing how mindful you have to be of the properties of everything...

When having to write down subgroups of a matrix group I thought I'd written two different subgroups, but they were actually the same. One of the entries in one subgroup was $p-\frac{1}{p}$, and the other was $\frac{1}{p}-p$. Now, these ARE both different. But they don't constitute as different general entries in my matrix, because they are both in $\mathbb{R}-\{0\}$. The correct answer had the second matrix with an entry that read $2\left(p-\frac{1}{p}\right)$, which of course has a different domain: $2\mathbb{R}-\{0\}$, so it forms a different sub group. My brain had clearly just seen two different formulas and had gone "there we go, they're different!". Not so. Domains for subgroups must be checked!

This is incredibly easy to forget: When trying to prove something is not true, don't leave it at the generalised proof that it's not true. If it's not true, provide a specific counter-example. With numbers! Remember those? The things that you count with that aren't letters? Duh! Yes, so I know to be vigilant now...

I had some trouble explaining a mapping of complex numbers in English. This is problematic. I could've programmed it and shown you the mapping and in an instant you would've thought "ah yeah!". But with lengthy explanations about how we get to the point where the real part gets mapped to the inverse of the imaginary etc etc... things can get a bit muddled. It seems I need to resort to explaining things mathematically more often, using the facts I already have.

I need to do more reading on the Isomorphism Theorem (specifically with regard to the domain of the Image that the quotient group is isomorphic to). Wow, that was a bit wordy... Ultimately I need to be more mindful about domains of Images...

So that's it! Again, happy with my high mark, but lots to remain mindful of...

# Group Theory Returns (with the Triforce)

I've just finished the first draft of my second Group Theory assignment. I am exhausted. I think of all my assignments, this is my most logic-dense. Just proof-reading the damn thing is causing brain burn. Though doing this kind of thing when you're tired never works, so I'll look at it over lunch tomorrow.

Some of the more complex concepts are so abstract and don't follow intuitively. Some do, but others don't. It's been fun, but I'm glad I don't have to dive any deeper into Group Theory right now.

What made this weekend's work more difficult was the recent release of Nintendo's Zelda: Breath of the Wild. Trying to concentrate on my study knowing that game was underneath my tv was tricky. What was particularly unfair was turning the page in my coursework to see they'd set a question about the Triforce.

UNFAIR. THANKS A LOT MATHEMATICS.

(also, #maybeTooMuchZelda)


text:
Prove that cosα+cos(α+β)+cos(α+2β)+…
Question:

Prove that $\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos [\alpha+(n-1) \beta]=\frac{\cos \left\{\alpha+\left(\frac{n-1}{2}\right) \beta\right\} \sin \left(\frac{n \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$ for all $n \in \mathbf{N}$.       [NCERT EXEMPLAR]

Solution:

Let $\mathrm{p}(n): \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos [\alpha+(n-1) \beta]=\frac{\cos \left\{\alpha+\left(\frac{n-1}{2}\right) \beta\right\} \sin \left(\frac{n \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)} \forall n$ $\in \mathbf{N}$.

Step I : For $n=1$,

LHS $=\cos [\alpha+(1-1) \beta]=\cos \alpha$

$\mathrm{RHS}=\frac{\cos \left\{\alpha+\left(\frac{1-1}{2}\right) \beta\right\} \sin \left(\frac{\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}=\cos \alpha$

As, LHS = RHS

So, it is true for $n=1$.

Step II : For $n=k$,

Let $\mathrm{p}(k): \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos [\alpha+(k-1) \beta]=\frac{\cos \left\{\alpha+\left(\frac{k-1}{2}\right) \beta\right\} \sin \left(\frac{k \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$ be true $\forall k \in \mathbf{N}$.

Step III : For $n=k+1$,

LHS $=\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos [\alpha+(k-1) \beta]+\cos [\alpha+(k+1-1) \beta]$

$=\frac{\cos \left\{\alpha+\left(\frac{k-1}{2}\right) \beta\right\} \sin \left(\frac{k \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}+\cos (\alpha+k \beta)$

$=\frac{\cos \left\{\alpha+\left(\frac{k-1}{2}\right) \beta\right\} \sin \left(\frac{k \beta}{2}\right)+\sin \left(\frac{\beta}{2}\right) \cos (\alpha+k \beta)}{\sin \left(\frac{\beta}{2}\right)}$

$=\frac{\sin \left(\alpha+k \beta-\frac{\beta}{2}\right)-\sin \left(\alpha-\frac{\beta}{2}\right)+\sin \left(\alpha+k \beta+\frac{\beta}{2}\right)-\sin \left(\alpha+k \beta-\frac{\beta}{2}\right)}{2 \sin \left(\frac{\beta}{2}\right)}$

$=\frac{-\sin \left(\alpha-\frac{\beta}{2}\right)+\sin \left(\alpha+k \beta+\frac{\beta}{2}\right)}{2 \sin \left(\frac{\beta}{2}\right)}$

$=\frac{2 \cos \left(\frac{2 \alpha+k \beta}{2}\right) \sin \left(\frac{k \beta+\beta}{2}\right)}{2 \sin \left(\frac{\beta}{2}\right)}$

$=\frac{\cos \left(\alpha+\frac{k \beta}{2}\right) \sin \left(\frac{(k+1) \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$

$\mathrm{RHS}=\frac{\cos \left\{\alpha+\left(\frac{k+1-1}{2}\right) \beta\right\} \sin \left(\frac{(k+1) \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$

$=\frac{\cos \left(\alpha+\frac{k \beta}{2}\right) \sin \left(\frac{(k+1) \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$

As, $\mathrm{LHS}=\mathrm{RHS}$

So, it is also true for $n=k+1$.

Hence, $\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos [\alpha+(n-1) \beta]=\frac{\cos \left\{\alpha+\left(\frac{n-1}{2}\right) \beta\right\} \sin \left(\frac{n \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$ for all $n$ $\in \mathbf{N}$