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# 4 cities 8 km apart
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Explanation: Paul) is the major administrative, economic, and cultural hub of Minnesota. Step out into the night-time streets of many Madrid neighbourhoods and you’ll find yourself swept along on a tide of people, accompanied by a happy crowd intent on dancing until dawn.”, “More than a decade after I fell for Madrid and decided to call it home, the life that courses relentlessly through the streets here still excites me. The distance is calculated in kilometers, miles and nautical miles, and the initial compass bearing/heading from the origin to the destination. Answer Save. Answer: (b) 2 Join. (c) x1/x2 = y2/y1 Contributed by US 1 ; Regional Information . (c) 5 When they met , let the distance covered by automobile 1 = D1 . x = 20 and y = 40 km = cm * 0.00001. km = cm / 100000. I will be trying out other ones in other cities. This is where you’ll find New Nordic pioneer Noma, (once again) voted the world’s best restaurant in 2014, and one of 15 Michelin-starred restaurants in town – not bad for a city of 1.2 million. (c) Neither directly nor inversely proportional Nov 21, 2013 . Question 31. This answer has been … × 10 ⇒ ? $$\frac{300}{6000}$$ = $$\frac{120}{? Question 6. An eclectic mix of academic essays, adventures abroad, projects, personal perspectives, news, and the best of the Web, Field Notes is the 4CITIES online journal. Explanation: City A to City B is a downstream journey on a stream which flows at a speed of 5km/hr. Find the speed of vehicle A if it travels 6.8 km/h faster than vehicle B and if they pass in 17.3 minutes. If two cities are 4.8 cm apart on the map, what is the actual distance between them?. Explanation: y = 1/2x Simply use our calculator above, or apply the formula to change the length 8.4 mi to km. Lv 4. He travels 20 km more due to an increase of speed of 4 km/hr. Explanation: Answer Save. (b) Rs 2000 Find the actual distance between them. When will the two boats meet for the first time? How many cows will graze the same field in 10 days ? \(\frac{3}{63}$$ = $$\frac{17}{? (a) 11.5 kg Explanation: 32 95448 Bayreuth. 36 km/hour C. 960 km/hour D. 4 km/hour (b) x – y = constant WE, DGGF, 4F65, Pleinlaan 2 Answer: (c) 13.5 kg = \(\frac{40×40}{25}$$ = 64. x1/y1 = x2/y2 (d) 10 days. }\) ⇒ ? Show more Show less. A map has a scale of 1 cm : 11.5 km. (d) 90 minutes. About the Multiple Marathons Program. (a) 6 days Organic art nouveau facades face off against 1960s concrete developments, and regal 19th-century mansions contrast with the brutal glass of the EU’s Gotham City. If you have any queries regarding Direct and Inverse Proportions CBSE Class 8 Maths MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon. We have provided Direct and Inverse Proportions Class 8 Maths MCQs Questions with Answers to help students understand the concept very well. (d) 40 metric tonnes. (a) Rs 400 1 Centimeter = 0.00001 Kilometer. Check the below NCERT MCQ Questions for Class 8 Maths Chapter 13 Direct and Inverse Proportions with Answers Pdf free download. (c) Rs 3000 = 8. $$\frac{8}{10}$$ = $$\frac{10}{? How much does he eat in 9 days ? (a) 1.2 kg 3 × 8 = 2 × ? Explanation: The ratio between the speeds of two trains is 7:8. (c) Neither directly nor inversely proportional The first map was published on 23 May 2015, last map added 2060 days ago. The actual distance from Detroit to Cleveland is 170 miles. The two cities are 240 kilometers apart. (b) 4 The power loss in the wire is, A car travels from Boston to Hartford in 4 hours. (c) 13.5 kg (b) 3 hours (c) 3.8 kg (a) 16 hectares (b) Rs 36 960 km/hour B. x and y vary inversely with each other. London is the Capital and largest city in England covered an area of 1,570 sq km with 8,294,058 inhabitants. (a) x + y = constant Download any 4 CiTieS song for FREE at http://www.4cities.co.uk Now taking bookings 4citiesuk@gmail.com (a) 10 Question 1094870: Two cities are 73.5km apart. Lv 7. If x and y are directly proportional, then which of the following is correct? Question 8. y = 4 x 300 = 1200. Answer: (b) 2.5 seconds y2 = 84, Question 36. what 2 cities are 1,000 kilometers apart. Definition of mile. So the scale is … The Links Are Full Duplex With Frame Size Of 1000 Bits. Victoria if they are 9 cm apart on a map with a scale of 1 cm : 18 km. (c) 50 km Relevance. Question 14. Das CityEL ist ein dreirädriges Leichtfahrzeug mit Elektroantrieb für eine Person, das auch als Cabrio gefahren werden kann. x = (12×50)/40 = 15. The actual distance between them is: Avg. Explanation: x = ky It will also display local time in each of the locations. 6 pipes are required to fill a tank in 1 hour 20 minutes. What is the rent of 16 hectares ? 1.The actual distance between two cities is 135 km the map scale is 1 cm:100 km. Students in the Classical Track go to Copenhagen for the third semester and Madrid for the fourth. Similar Questions. The rent of 7 hectares is Rs 875. To the nearest tenth of a kilometer, what is the actual distance corresponding to the map distance? What is the cost of 50 sticks at Rs 24 per score ? This is a proportion word problem. 0 . (d) 1300 km, Answer: (c) 1200 km “Belgium’s fascinating capital, and the administrative capital of the EU, Brussels is historic yet hip, bureaucratic yet bizarre, self confident yet unshowy, and multicultural to its roots. g. Expert answered|Score 1|emdjay23|Points 156670| Log in for more information. The average speed of the car was 60 km/hour. Erlebe die Rennsporttechnik des Abarth 595 Turismo 70th Anniversary Edition mit der neusten Generation des 7” Navigationssystems. Please tell us about your experience. = 8. Question 25. (d) Cannot be determined. 5,639,632. Both Links Are Fibre. Respond to this Question. These cities were chosen as representatives of geographic regions within Europe as well as of particular approaches to governance, culture, planning, and society-making. Algebra. On a certain map, Detroit and Cleveland are 3" apart. (d) x/y = constant. “One constant is the enviable quality of everyday life, with a café/bar scene that never gets old. The constant of variation, if x ∝ y, from the following table is 0 . Question 22. (c) Rs 537 City Maps for Minecraft Versions: The 4CITIES Master Course in Urban Studies. (b) 84 Explanation: (d) 8 hours. \(\frac{15}{40}$$ = $$\frac{6}{? If two cities are 8 centimeters apart on the map, what is the actual distance between the cities, to the nearest tenth of a kilometer? ⇒ ? III mmi. The power loss in the wire is A.On Carrie's map, Greenville and North Valley are 4.5 inches apart.The scale on her . \(\frac{7}{875}$$ = $$\frac{16}{? If two cities are 4.25 cm apart on the map, what is the actual distance between the cities? Contributed by US 1 ; Felt Report - Tell Us! Question 13. (d) 4. What are the currents supplied by the two stations if the train is at the distance of minimum potential? Respond to this Question. Answer: (a) Rs 400 math. \(\frac{10}{1000}$$ = $$\frac{4}{? A map has a scale of 1 cm : 5 km. You would have to figure out the exact distance between these two cities and add that distance further away either east or west from these two and find a small city close to either on the opposite side to get exactly 2000. Question. Click hereto get an answer to your question ️ Two cities are 150 km apart. = 50. If two cities are 4 cm apart on this map, how many km apart are they really? 1 decade ago. Explanation: (d) 3.5 seconds. (b) Rs 1500 Radio direction finders : 2014-01-09: Nathan pose la question : Radio direction finders set up two points A and B, which are 2.5 miles apart on an east-west line. A mile is a unit of length in a number of systems of measurement, including in the US Customary Units and British Imperial Units. (a) 32 km About 4CITIES. (a) x + y = constant = 25 mi? = 6. Stefan De Corte, Vrije Universiteit Brussel 92.5 km/hr: C. 85 km/hr: D. 87.5 km/hr : View Answer. Birmingham Shopping District, Birmingham Offering a free fitness centre, Aparthotel Adagio Birmingham City Centre is located 200 metres from the Bullring Shopping Centre. All this plays out in a cityscape that swings from majestic to quirky to rundown and back again. 0 Answers/Comments. \(\frac{5}{600}$$ = $$\frac{2}{? The two cities are 240 kilometers apart. Just wondering whether you're eligible or how to get your visa? You can share or return to this by using the link below. 5 (1) Experienced Teacher & Tutor in Round Rock, TX. The fare for a journey of 40 km is Rs 25. What you do … A train runs between San Diego, and Kansas City, which is 4 km apart and maintains voltages of 450 volts and 400 volts respectively. Driving distance: 0.0 miles , 0.0 kilometers (km) , 0 feet , 0 meters. Monrovia, Liberia to Natal, Brazil: 2700.7 km (1678.1 mi) Cape Town, South Africa to Boston: 12416.3 km (7715.2 mi) All distances are based on great circle distances between latitudinal and longitudinal coordinates. (b) 60 days If an amount of food last for 40 days for 120 men, how long will it last for 80 men at the same rate ? = 6. 1050 Brussel, Belgium, Information relating to individuals (personal data) is collected and used in accordance Citizen Scientist Contributions ; Did You Feel It? 8 = 4k ⇒ k = 2. 840060 840060 Answer: 8cm. 1K likes. To get rid of units, convert to the same units. the scale on the map is 1 inch = 200 miles. Answer: (d) Rs 50. 12/40 = x/2500 How long will it take to go 150 km ? Question 2. Apart-Hotel FirstBoarding Bayreuth. Answered in 1 hour by: 5/4/2005. See tutors like this. Answer: (c) 3 A TV is provided. Due to the COVID-19 pandemic, the 2020 student cohort will not begin until 2021. Enjoy your game. This is inverse proportion case: 1/200:4.5/m; Follow • 3. This website uses cookies to improve your experience while you navigate through the website. = \(\frac{40×6}{15}$$ = 16. Discuss: answer with explanation. MCQ Questions for Class 8 Maths with Answers, MCQ Questions for Class 12 History Chapter 10 Colonialism and the Countryside: Exploring Official Archives with Answers, MCQ Questions for Class 12 History Chapter 9 Kings and Chronicles: The Mughal Courts with Answers, Geography Class 12 Important Questions Chapter 4 Human Settlements, MCQ Questions for Class 12 History Chapter 8 Peasants, Zamindars and the State: Agrarian Society and the Mughal Empire with Answers, Geography Class 12 Important Questions Chapter 3 Human Development, MCQ Questions for Class 12 History Chapter 7 An Imperial Capital: Vijayanagara with Answers, Class 12 History Important Questions Chapter 11 Rebels and the Raj: The Revolt of 1857 and its Representations, MCQ Questions for Class 12 History Chapter 6 Bhakti-Sufi Traditions: Changes in Religious Beliefs and Devotional Texts with Answers, Geography Class 12 Important Questions Chapter 2 Migration: Types, Causes and Consequences, MCQ Questions for Class 12 History Chapter 5 Through the Eyes of Travellers: Perceptions of Society with Answers, Concise Mathematics Class 10 ICSE Solutions. Research and Practice Track students are expected to find their own internship, but 4CITIES maintains an ever-expanding network of partner institutions, organizations, and companies in cities across Europe upon which students are welcome to build. First Name. The distance between two cities A and B is 110 km. By: Bill F. answered • 12/19/12. But Brussels doesn’t go out of its way to impress. (a) Rs 357 How to convert 8.4 to miles. (c) 80 A map scale indicates that one centimeter equals 10 km two cities are 8 cm apart on the map how far apart are they in real life . If x ∝ y and x1= 5, y1 = 210 and x2 = 2, then find y2 ? The 4CITIES online journal . (d) 60. Question 20. $$\frac{12}{18}$$ = $$\frac{9}{? Question 21. 240/4=60. A. Answer: (a) Directly proportional ⇒ ? x ∝ y Two planes travel toward each other from cities that are about 2310 km apart at rates of 400 km/hr and 370 km/hr. Carrie and Krystal are taking a road trip from Greenville to North Valley . Solution: Since 1 cm on the map corresponds to an actual distance of 40 km However, coursework can also be completed at one of our academic partner universities in Aachen, Aalto, Amsterdam, Barcelona, Bern, Bordeaux, Budapest, Esch-sur-Alzette, Frankfurt am Main, Gdánsk, Gothenburg, Hamburg, Lausanne, Lille, Lund, Malmö, Milan, Oslo, Paris, Prague, Stockholm, Tartu, Trondheim, Utrecht, Warsaw, and Weimar (list subject to change). what 2 cities are 1,000 kilometers apart. A train starts from A at 8 a.m. and travels towards B at 60 km /hr. 5 Answers. Citizen Scientist Contributions ; Did You Feel It? Distance Between Cities. 15 men can mow 40 hectares of land in 1 day. 1/x=4/5 B. Two cities are 4 km apart on the map. (d) 36 days. Responses. Explanation: Required fields are marked *, Concise Mathematics Class 10 ICSE Solutions 2018. (a) Directly proportional On a map the scale means that the 1 represents 25,000 cm in reality (in this case - other scales, say 1:50000 means 1cm represents 50,000 cm) If 1cm represents 25,000cm or 250m then 1km is 1000 / 250 or 4 cm (1:50000 and 1cm is … Explanation: }$$ ⇒ ? See the latest article. Very clean and spacious property, close to local shops and the main centre. Question 11. x = k (1/y) Answer: (d) 6 What was the average speed of the car during the trip? (c) 3 }\) ⇒ ? Two cities are 3.8 cm apart. A mile is a unit of length in a number of systems of measurement, including in the US Customary Units and British Imperial Units. Fully renovated house close to the city centre is set in Belfast, 2.2 km from The Waterfront Hall, 2.8 km from SSE Arena, as well as 3.4 km from The Belfast Empire Music Hall. Apala types 200 words in half an hour. 9,2 (43 Bewertungen) WLAN Parkplatz am Hotel Entfernung (Luftlinie) 1,2 km 9,0 km 1,8 km ab 83,00 € Info & Buchen . Thank you. (c) Rs 48 12 ft. A 15 ft tall statue standing next to an adult elephant casts a 18 ft shadow. Contributed by US 1 ; Felt Report - Tell Us! On a map of New Mexico, 1 cm represents 45 km. (d) Rs 4800. Explanation: Converting 8.4 mi to km is easy. (b) 96 minutes Nov 20, 2013 . If the weight of 12 sheets of thick paper is 40 grams, how many sheets of the same paper would weigh 2500 grams? 2 × 3 = 6, 4 × 3 = 12, 8.4 Very good 4,887 reviews Check availability. Area 86,935 square miles (225,161 square km). 55°40′N 12°34′E > 590K residents > 86.4 km2 > since 1000s CE Edgier than Stockholm and worldlier than Oslo, the Danish capital gives Scandinavia the X factor. = $$\frac{875×16}{7}$$ = 2000. their net speed w/respect to each other is 240 km/h. (c) Rs 750 (b) 64 Similar Questions. Explanation: Ready to apply? However, generally you want to just have the scale without units. $$\frac{25}{40}$$ = $$\frac{40}{? User: A car travels from Boston to Hartford in 4 hours. (d) 240. Solvers Solvers. 1 : 300 = x : y }$$ ⇒ ? What is … Add to this its compact size, and you have what is possibly Europe’s most seamless urban experience.” more from Lonely Planet, 40°23′N 3°43′E > 3.1M residents > 604.3 km2 > since 800s CE The city has over 1,000 years of history and heritage under its belt, meaning there are some truly inspiring landmarks and architecture to explore. If we use 5 such types of pipes, how much time it will take to fill the tank? (a) 75 minutes $$\frac{1000}{10}$$ = $$\frac{600}{? Jan 1, 2018 - Caprice Suites Jersey City Located in Jersey City, Caprice Suites is 4.8 km from the Liberty State Park. Question 5. (d) 14.5 kg. }$$ ⇒ ? Question 24. = 400. Log On Ad: Over 600 Algebra Word Problems at edhelper.com: Word Problems: Unit Conversion Word. Answer: (d) 8 hours. The city is the most populous city in the United Kingdom with a population of 8.6 million inhabitants who speak over 300 languages. ⇒ ? Free WiFi is featured throughout the property. Hal offers several plans designed for runners intending to do two or more marathons with minimum rest between, including interactive programs on TrainingPeaks for those whose marathons are 2-8 weeks between. Kilometers to Miles conversion. Answer: (b) 72 minutes (a) Directly proportional ⇒ ? Answer: (b) 64 Nürnberger Str. x = ky Question 38. (a) 200 (a) Rs 1000 (b) 13 3 knives cost Rs 63. Answer: (b) 60 days Answer: (b) Rs 2400 Here is a place where the passions of Europe’s most passionate country are the fabric of daily life, a city with music in its soul and an unshakeable spring in its step. (d) 75. Answer: (d) 12 Emily. Using direct proportion concept: (c) Rs 200 = 72 minutes. Answer: (d) 240 Share this conversation. (a) 1 6 × 60 = 5 × ? Hence, y ∝ 1/x, where 1/2 is the proportionality constant. Explanation: = 2.5. The distance between two cities A and B is 330 km. How far apart are the real cities? Answer: (c) 1200 km Explanation: 1 : 300 = x : y 1/300 = 4/y y = 4 x 300 = 1200. 15 books weigh 6 kg. Report 2 Answers By Expert Tutors Best Newest Oldest. If x = 20 and y = 40, then x and y are: Explanation: = 357. 36 km/hour C. 60 km/hour D. 4 km/hour A car travels from Boston to Hartford in 4 hours. (d) x1.x2 = y1.y2, Answer: (c) x1/x2 = y2/y1 The scale of a map is 1:250000.The actual distance between two cities is 80 km. Weegy: A car travels from Boston to Hartford in 4 hours. x ∝ 1/y RD Sharma Solutions , RS Aggarwal Solutions and NCERT Solutions, October 1, 2020 by Veerendra Leave a Comment. Question 10. Distances are measured using a direct path, as the crow flies (c) 150 xarqi, Generalist. km (kilometers) miles; The output is via a measurement of the distance and also a map that shows that two locations and the path between them as the crow flies and the route by land transport. (a) 1000 km 0 . 1 2. (b) 8 (b) 2.4 kg Contribute to citizen science. ⇒ ? If x – 15 when y = 6, then the value of x when y = 15 is How far apart would the cities be on a map with a scale of 1 in. The resistance of go and return is 0.05 ohm per km. A motor cycle rider starts from A towards B at 7 a.m. with a speed of 20 Km/hr. (c) 4 minutes Answer. (d) 6 minutes. Martin. if two cities are 14 cm apart on the map, what is the distance between the cities? $$\frac{90}{1}$$ = $$\frac{225}{? (b) 4 hours Responses. How long would he take in walking 32 km ? If Union and Clayton are 8 km apart, then they are how far apart on the map? What will 10 g cost ? “Yet Copenhagen is more than just seasonal cocktails and geometric threads. It is mandatory to procure user consent prior to running these cookies on your website. (a) Rs 1200 Tech Support Specialist: xarqi, Generalist replied 15 years ago. (d) 70, Answer: (b) 84 (a) 120 (c) 100 0 . }$$ ⇒ ? = 2.5. Found 2 solutions by KMST, MathTherapy: Answer by KMST(5289) (Show Source): You can put this solution on YOUR website! (c) 6 hours 5.5 cm * 11.5 = 63.3 km. You can use the controls on the map to: Pan the map; Zoom the Map; Switch between Map and Satellite views; Some Examples of Distances. Answer: (b) 21 days Determine how far apart two cities are if they are 4 cm apart on the map? Another train starts from B at 9 a.m. and travels towards A at 75 km / hr. Necessary cookies are absolutely essential for the website to function properly. Explanation: If 18 women can reap a field in 7 days, in what time can 6 women reap the same field ? The two cities are 240 kilometers apart. 10 metres of cloth cost Rs 1000. (d) 4 hours. The scale of 1 cm : 12 km. y ∝ x, where 2 is the proportionality constant. math. Edgier than Stockholm and worldlier than Oslo, the Danish capital gives Scandinavia the X factor. }\) ⇒ ? (b) Inversely proportional (d) Rs 573. MCQ Questions for Class 8 Maths with Answers were prepared based on the latest exam pattern. “Copenhagen is the coolest kid on the Nordic block. Boats P and Q run a shuttle service between the two cities that are 300 kms apart. 6 pipes are required to fill a tank in 1 hour. x = k/y Expert Answer . To convert cm to km, multiply the cm value by 0.00001 or divide by 100000. The two cities are 240 kilometers apart. Show transcribed image text . You also have the option to opt-out of these cookies. Hence, the distance between the two cities is 1200 km. (b) Rs 2400 Explanation: $90$ km/hr: B. If x = 15 and y = 1/30, then x and y are: Answer: (a) 80 If 20 cows eat as much as 15 oxen, how many cows will eat at much as 36 oxen ? 8/3 = x/4 x = 32/3 = 10.7 km Steve. Below you will find a brief introduction to each. (d) Rs 50. (a) 2 Land Relief. Featuring projects, research, travels, photography, and editorials on the 4CITIES experience, cities, and all things urban, the blog also includes news and events and selected articles of relevance. x = 480/5 = 96, Question 40. Explanation: (a) 120 minutes Question 4. (b) 800 4 cities, 2 years, 1 master degree. 12 mi. Let number of sheets for 2500 is x. Answer: (a) Rs 1000 Take Dublin Castle, for instance. (a) 6 }\) ⇒ ? = 50. 50°51′N 4°21′E > 1.2M residents > 161.38 km2 > since 979 CE Nov 20, 2013 . 7 years ago . Asked 3/12/2014 11:25:11 AM . Answer: (c) xy = constant Seit Anfang 2009 ist auch ein Lithium-Ionen-Akku mit Batteriemanagementsystem erhäl… These cookies will be stored in your browser only with your consent. 40 cows can graze a field in 16 days. St. Paul is the state capital, and the Twin Cities region (Minneapolis–St. (c) 80 minutes 5 × 3 = 15, 7 × 3 = 21. Nov 21, 2013 . If two cities are 8 centimeters apart on the map, what is the actual distance between the cities, to the nearest tenth of a kilometer? Just ask style bibles Monocle and Wallpaper magazines, which fawn over its industrial-chic bar, design and fashion scenes, and culinary revolution. Distance Between United States cities, distance of cities listed on map, mileage distances are in kilometers and miles. If 3 quintals of coal cost Rs 6000, what is the cost of 120 kg ? 2 . 19. Je nach Streckenverbrauch und Zustand der Blei-Vlies-Batterien kann der CityEL bis 50 km Reichweite mit einer Batterieladung erzielen. This website uses cookies to improve your experience. (d) Rs 1200. Other tools to help with distance questions. (a) Rs 1000 1 decade ago. = 2400, Question 16. The Distance Calculator can find distance between any two cities or locations available in The World Clock. = 60. A man walks 20 km in 5 hours. To connect with 4 CiTieS, join Facebook today. “Madrid nights are the stuff of legend, and the perfect complement to the more sedate charms of fine arts and fine dining. or See distance between cities in kilometers (km), miles and nautical miles and their local time. England is a home to more than 10 ethnic groups including White British (85%), White (5%), Indian (1.8%), Pakistani (1.6%), Irish (1.2%), Black Caribbean (1%), Black African (0.8%), Chinese (0.4%), and Bangladeshi (0.5%). Verified. Answer: (a) Directly proportional (a) 2 seconds Explanation: Explanation: Explanation: Want to know how the process works and what you'll need to complete your application? How far apart are two cities that are 10" apart on this map? 18 × 7 = 6 × ? (c) 2.5 hours Question 12. Question 7. (c) 80 days Explanation: If 8 men can do a piece of work in 20 days, in how many days could 20 men do the same work ? In addition to this tool we also offer a couple other tools that can help find the distance on a map. Question 19. math. Question 3. (a) 3 hours }\) ⇒ ? (a) 40 (b) 1100 km 0 . Answer: (b) 90 metric tonnes The hardest part is setting it up. 15 × 6 = ? (d) Rs 60. 4 km * 1000 m/km * 100 cm/m = 400,000 cm. (d) 15, Answer: (d) 15 Explanation: What is the distance between two cities that are 4.2 cm apart on the map? (c) xy = constant $$\frac{20}{5}$$ = $$\frac{32}{? Question 30. Dear student, let speed of 1st automobile = v , the speed of 2 nd automobile = 8v/9 . Question 34. k = x/y, where k is a constant. A labourer is paid Rs 400 for 2 days work. The fall of potential per km is 8 V and the average resistance per km is 8 V and the average resistance per km is . A machine in a soft drink factory fills 600 bottles in 5 hours. 4 years ago. A. How to convert 8.4 to miles. is largely self-selected. (d) Rs 100. 55°40′N 12°34′E > 590K residents > 86.4 km 2 > since 1000s CE “Copenhagen is the coolest kid on the Nordic block. A boy runs 1 km in 10 minutes. We've got you covered. Contributed by US 1 ; Regional Information . Explanation: Answer: (d) Rs 60. The two-year journey of 4CITIES is divided into four semesters. Category: Computer. But opting out of some of these cookies may affect your browsing experience. We'll assume you're ok with this, but you can opt-out if you wish. (a) x1/y1 = x2/y2 If x and y are inversely proportional, then which one is true? Two cities are 5.5 cm apart on the map. 120 copies of a book cost Rs 600. 2021-01-10 03:23:28 (UTC) 23.721°N 121.552°E; 11.4 km depth; Interactive Map. This whole maelstrom swirls out from Brussels’ medieval core, where the Grand Place is surely one of the world’s most beautiful squares. (c) 99 = 60. 1 See answer wakeup2saysrry is waiting for your help. The application window for 2022 will open on 01 November, 2021. 4 Answers. Relevance. }$$ ⇒ ? Explanation: Category: Computer. 4 CiTieS. If necessary, round to the nearest hundredth. So the scale is: 8 cm : 32 km = 1 cm : 4 km. Round to the nearest tenth of a mile. 1 decade ago. (c) 3 (d) Rs 900. (c) 10 Put all files combined, it's 46 mb of Minecraft maps! Question 29. (a) 2 hours Answer: (d) 6 minutes. 12 × 50 = x × 40 Answer: (b) 2.4 kg (a) 15 days = 240. $$\frac{20}{150}$$ = $$\frac{12}{? Solvers Solvers. }$$ ⇒ ? (b) 90 metric tonnes Since they are going the same speed, they will meet in the middle. (c) 20 hectares Question: Question 4 (8 Points) Nodes A And B Are 3000 Km Apart And Nodes B And C Are 400 Km Apart. $$\frac{15}{20}$$ = $$\frac{36}{? Question 27. (c) 60 metric tonnes Jessica. Explanation: A. The city may have more bars than any other city on earth – a collection of storied cocktail bars and nightclubs that combine a hint of glamour with non-stop marcha (action). (c) 850 What I love most about the city is that not only does it hold on to its traditions, it incorporates them in everything from high-fashion Dirndls with pop-art motifs or punk conical studs to handmade Sacher Torte–flavoured doughnuts and inspired neo-retro cafes. 8.4 miles equal 13.5184896 kilometers (8.4mi = 13.5184896km). Distance Calculator now supports locations worldwide! 4.8 miles equal 7.7248512 kilometers (4.8mi = 7.7248512km). The scale of a map is given as 1 : 300. Answer: (b) Rs 2000 Free WiFi access is provided, and all the apartments are air-conditioned. ⇒ ? (b) Inversely proportional Vienna’s past is alive in its present, and, by extension, its future.” more from Lonely Planet. “With its rambling palaces, winding cobbled lanes, elegant Kaffeehäuser (coffee houses) and cosy wood-panelled Beisln (bistro pubs), Vienna is steeped in history. x = 15, y = 1/30 They started at the same time. }$$ ⇒ ? A shot travels 90 m in 1 second. x ∝ y (d) x/y = constant. This distance measure can be converted to U.S. miles and any map will have the distance measures between each city. The citizens’ humorous, deadpan outlook on life is often just as surreal as the canvases of one-time resident Magritte.” more from Lonely Planet, 48°12′N 16°22′E > 1.9M residents > 414.65 km2 > since 15 BCE Große und sc... - Februar 2020. so. Satisfied Customers: 596. Explanation: The scale of a map is 1:250000.The actual distance between two cities is 80 km. that is the actual distance between two cities that are 4.5 inches apart on the map. (a) 50 days 4 CiTieS is on Facebook. Lake Itasca, Itasca State Park, northwestern Minnesota. Boat P, which starts from City A has a still-water speed of 25km/hr, while boat Q, which starts from city B at the same time has a still-water speed of 15km/hr. (b) x1/x2 = y1/y2 electric power is sent from one city to another city through copper wires. Lv 7. If x and y are inversely proportional, then: If x = ky and when y = 4, x = 8 then k = How to Apply. 7 years ago. 1 25000 Scale Map. x = 750, Question 37. 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He works for 5 days, how many km apart on this map pass each other from that. = 200 miles 40×6 } { 15 } \ ) = 48: Here list of world! Is paid Rs 400 for 2 days 8 hours of design, architecture contemporary. 55°40′N 12°34′E > 590K residents > 86.4 km2 > since 1000s CE Copenhagen...
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475 views
What does the following code do?
var a, b: integer;
begin
a:=a+b;
b:=a-b;
a:a-b;
end;
1. exchanges $a$ and $b$
2. doubles $a$ and stores in $b$
3. doubles $b$ and stores in $a$
4. leaves $a$ and $b$ unchanged
5. none of the above
edited | 475 views
0
typo mistake in a:a-b; it should be a:=a-b;
Answer is simply $A$ i.e. it swaps the values of the two.. Take any two values for $A$ and $B$. and perform the given operations over them.
edited
(a)
initially a = 10 b = 5
a = 15
b = 10
a = 5
+1 vote
1. a= a+b
2. b= a-b= (a+b)-b= a (from 1)
3. a= a-b= (a+b) - (a) = b (from 1 and 2)
Hence the values are getting exchanged
1
2
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https://netlib.org/lapack/explore-html-3.6.1/d5/dfb/group__complex16_g_bcomputational_gabed606962bd46ec1794178450b3b12e1.html
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LAPACK 3.6.1 LAPACK: Linear Algebra PACKage
subroutine zgbcon ( character NORM, integer N, integer KL, integer KU, complex*16, dimension( ldab, * ) AB, integer LDAB, integer, dimension( * ) IPIV, double precision ANORM, double precision RCOND, complex*16, dimension( * ) WORK, double precision, dimension( * ) RWORK, integer INFO )
ZGBCON
Purpose:
``` ZGBCON estimates the reciprocal of the condition number of a complex
general band matrix A, in either the 1-norm or the infinity-norm,
using the LU factorization computed by ZGBTRF.
An estimate is obtained for norm(inv(A)), and the reciprocal of the
condition number is computed as
RCOND = 1 / ( norm(A) * norm(inv(A)) ).```
Parameters
[in] NORM ``` NORM is CHARACTER*1 Specifies whether the 1-norm condition number or the infinity-norm condition number is required: = '1' or 'O': 1-norm; = 'I': Infinity-norm.``` [in] N ``` N is INTEGER The order of the matrix A. N >= 0.``` [in] KL ``` KL is INTEGER The number of subdiagonals within the band of A. KL >= 0.``` [in] KU ``` KU is INTEGER The number of superdiagonals within the band of A. KU >= 0.``` [in] AB ``` AB is COMPLEX*16 array, dimension (LDAB,N) Details of the LU factorization of the band matrix A, as computed by ZGBTRF. U is stored as an upper triangular band matrix with KL+KU superdiagonals in rows 1 to KL+KU+1, and the multipliers used during the factorization are stored in rows KL+KU+2 to 2*KL+KU+1.``` [in] LDAB ``` LDAB is INTEGER The leading dimension of the array AB. LDAB >= 2*KL+KU+1.``` [in] IPIV ``` IPIV is INTEGER array, dimension (N) The pivot indices; for 1 <= i <= N, row i of the matrix was interchanged with row IPIV(i).``` [in] ANORM ``` ANORM is DOUBLE PRECISION If NORM = '1' or 'O', the 1-norm of the original matrix A. If NORM = 'I', the infinity-norm of the original matrix A.``` [out] RCOND ``` RCOND is DOUBLE PRECISION The reciprocal of the condition number of the matrix A, computed as RCOND = 1/(norm(A) * norm(inv(A))).``` [out] WORK ` WORK is COMPLEX*16 array, dimension (2*N)` [out] RWORK ` RWORK is DOUBLE PRECISION array, dimension (N)` [out] INFO ``` INFO is INTEGER = 0: successful exit < 0: if INFO = -i, the i-th argument had an illegal value```
Date
November 2011
Definition at line 149 of file zgbcon.f.
149 *
150 * -- LAPACK computational routine (version 3.4.0) --
151 * -- LAPACK is a software package provided by Univ. of Tennessee, --
152 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
153 * November 2011
154 *
155 * .. Scalar Arguments ..
156 CHARACTER norm
157 INTEGER info, kl, ku, ldab, n
158 DOUBLE PRECISION anorm, rcond
159 * ..
160 * .. Array Arguments ..
161 INTEGER ipiv( * )
162 DOUBLE PRECISION rwork( * )
163 COMPLEX*16 ab( ldab, * ), work( * )
164 * ..
165 *
166 * =====================================================================
167 *
168 * .. Parameters ..
169 DOUBLE PRECISION one, zero
170 parameter ( one = 1.0d+0, zero = 0.0d+0 )
171 * ..
172 * .. Local Scalars ..
173 LOGICAL lnoti, onenrm
174 CHARACTER normin
175 INTEGER ix, j, jp, kase, kase1, kd, lm
176 DOUBLE PRECISION ainvnm, scale, smlnum
177 COMPLEX*16 t, zdum
178 * ..
179 * .. Local Arrays ..
180 INTEGER isave( 3 )
181 * ..
182 * .. External Functions ..
183 LOGICAL lsame
184 INTEGER izamax
185 DOUBLE PRECISION dlamch
186 COMPLEX*16 zdotc
187 EXTERNAL lsame, izamax, dlamch, zdotc
188 * ..
189 * .. External Subroutines ..
190 EXTERNAL xerbla, zaxpy, zdrscl, zlacn2, zlatbs
191 * ..
192 * .. Intrinsic Functions ..
193 INTRINSIC abs, dble, dimag, min
194 * ..
195 * .. Statement Functions ..
196 DOUBLE PRECISION cabs1
197 * ..
198 * .. Statement Function definitions ..
199 cabs1( zdum ) = abs( dble( zdum ) ) + abs( dimag( zdum ) )
200 * ..
201 * .. Executable Statements ..
202 *
203 * Test the input parameters.
204 *
205 info = 0
206 onenrm = norm.EQ.'1' .OR. lsame( norm, 'O' )
207 IF( .NOT.onenrm .AND. .NOT.lsame( norm, 'I' ) ) THEN
208 info = -1
209 ELSE IF( n.LT.0 ) THEN
210 info = -2
211 ELSE IF( kl.LT.0 ) THEN
212 info = -3
213 ELSE IF( ku.LT.0 ) THEN
214 info = -4
215 ELSE IF( ldab.LT.2*kl+ku+1 ) THEN
216 info = -6
217 ELSE IF( anorm.LT.zero ) THEN
218 info = -8
219 END IF
220 IF( info.NE.0 ) THEN
221 CALL xerbla( 'ZGBCON', -info )
222 RETURN
223 END IF
224 *
225 * Quick return if possible
226 *
227 rcond = zero
228 IF( n.EQ.0 ) THEN
229 rcond = one
230 RETURN
231 ELSE IF( anorm.EQ.zero ) THEN
232 RETURN
233 END IF
234 *
235 smlnum = dlamch( 'Safe minimum' )
236 *
237 * Estimate the norm of inv(A).
238 *
239 ainvnm = zero
240 normin = 'N'
241 IF( onenrm ) THEN
242 kase1 = 1
243 ELSE
244 kase1 = 2
245 END IF
246 kd = kl + ku + 1
247 lnoti = kl.GT.0
248 kase = 0
249 10 CONTINUE
250 CALL zlacn2( n, work( n+1 ), work, ainvnm, kase, isave )
251 IF( kase.NE.0 ) THEN
252 IF( kase.EQ.kase1 ) THEN
253 *
254 * Multiply by inv(L).
255 *
256 IF( lnoti ) THEN
257 DO 20 j = 1, n - 1
258 lm = min( kl, n-j )
259 jp = ipiv( j )
260 t = work( jp )
261 IF( jp.NE.j ) THEN
262 work( jp ) = work( j )
263 work( j ) = t
264 END IF
265 CALL zaxpy( lm, -t, ab( kd+1, j ), 1, work( j+1 ), 1 )
266 20 CONTINUE
267 END IF
268 *
269 * Multiply by inv(U).
270 *
271 CALL zlatbs( 'Upper', 'No transpose', 'Non-unit', normin, n,
272 \$ kl+ku, ab, ldab, work, scale, rwork, info )
273 ELSE
274 *
275 * Multiply by inv(U**H).
276 *
277 CALL zlatbs( 'Upper', 'Conjugate transpose', 'Non-unit',
278 \$ normin, n, kl+ku, ab, ldab, work, scale, rwork,
279 \$ info )
280 *
281 * Multiply by inv(L**H).
282 *
283 IF( lnoti ) THEN
284 DO 30 j = n - 1, 1, -1
285 lm = min( kl, n-j )
286 work( j ) = work( j ) - zdotc( lm, ab( kd+1, j ), 1,
287 \$ work( j+1 ), 1 )
288 jp = ipiv( j )
289 IF( jp.NE.j ) THEN
290 t = work( jp )
291 work( jp ) = work( j )
292 work( j ) = t
293 END IF
294 30 CONTINUE
295 END IF
296 END IF
297 *
298 * Divide X by 1/SCALE if doing so will not cause overflow.
299 *
300 normin = 'Y'
301 IF( scale.NE.one ) THEN
302 ix = izamax( n, work, 1 )
303 IF( scale.LT.cabs1( work( ix ) )*smlnum .OR. scale.EQ.zero )
304 \$ GO TO 40
305 CALL zdrscl( n, scale, work, 1 )
306 END IF
307 GO TO 10
308 END IF
309 *
310 * Compute the estimate of the reciprocal condition number.
311 *
312 IF( ainvnm.NE.zero )
313 \$ rcond = ( one / ainvnm ) / anorm
314 *
315 40 CONTINUE
316 RETURN
317 *
318 * End of ZGBCON
319 *
subroutine zdrscl(N, SA, SX, INCX)
ZDRSCL multiplies a vector by the reciprocal of a real scalar.
Definition: zdrscl.f:86
double precision function dlamch(CMACH)
DLAMCH
Definition: dlamch.f:65
subroutine xerbla(SRNAME, INFO)
XERBLA
Definition: xerbla.f:62
complex *16 function zdotc(N, ZX, INCX, ZY, INCY)
ZDOTC
Definition: zdotc.f:54
subroutine zlacn2(N, V, X, EST, KASE, ISAVE)
ZLACN2 estimates the 1-norm of a square matrix, using reverse communication for evaluating matrix-vec...
Definition: zlacn2.f:135
subroutine zlatbs(UPLO, TRANS, DIAG, NORMIN, N, KD, AB, LDAB, X, SCALE, CNORM, INFO)
ZLATBS solves a triangular banded system of equations.
Definition: zlatbs.f:245
integer function izamax(N, ZX, INCX)
IZAMAX
Definition: izamax.f:53
logical function lsame(CA, CB)
LSAME
Definition: lsame.f:55
subroutine zaxpy(N, ZA, ZX, INCX, ZY, INCY)
ZAXPY
Definition: zaxpy.f:53
Here is the call graph for this function:
Here is the caller graph for this function:
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# physics
posted by .
A capacitor consists of two concentric spherical shells. The outer radius of the inner shell is a=0.76 mm and the inner radius of the outer shell is b=3.31 mm .
(a) What is the capacitance C of this capacitor? Express your answer in Farads.
(b) Suppose the Maximum possible electric field at the outer surface of the inner shell before the air starts to ionize is Emax(a)=3.0×106V⋅m-1 . What is the maximum possible charge on the inner sphere? Express your answer in Coulombs.
(c) What is the maximum amount of energy stored in the capacitor? Express your answer in Joules.
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https://www.convergencetraining.com/cooling-theory.html?src=cartMessage
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# Cooling Theory
SKU: RVI-11476Duration: 20 Minutes
How do you properly cool a building? Cooling systems are used to cool and condition the air in rooms and building. To do this, they must remove heat from the air. How much heat needs to be removed to reach the desired temperature? What about humidity - the amount of water vapor in the air? Humidity affects how much heat we feel so it must also be addressed. This interactive online course will address these and other cooling system-related topics and issues.
### Course Details
#### Specs
Training Time: 20 minutes
Compatibility: Desktop, Tablet, Phone
Based on: Industry Standards and Best Practices
Languages: English
### Learning Objectives
• State the First Law of Thermodynamics
• Understand that Total Cooling must take into account Sensible Heat and Latent Heat
• Identify that a thermometer measures sensible heat
• Identify that a sling psychrometer measures relative humidity
• Define "Enthalpy" and identify the properties of air that influence enthalpy
### Key Questions
The following key questions are answered in this module:
What is the definition of "heat"?
Heat is a measure of the warmth or coldness of a substance, usually in reference to a standard. Heat is also a form of energy. Heat energy, or thermal energy must be added or removed to change the temperature of the air in a building
What is the First Law of Thermodynamics?
The First Law of Thermodynamics, also called the Law of Conservation of Energy, states that energy cannot be created or destroyed; it can only change form.
What is Sensible Heat?
Heat that results in a temperature change only, with no change in phase. It is the heat in a substance, such as air, that can be "sensed" by you and a thermometer.
What is Latent Heat?
Heat that is used to change the phase of a substance - for example, from a solid to a liquid, or a liquid to a gas. The water vapor in air contains latent heat.
How do you measure the capacity of a heating or cooling system?
The capacity of a heating or cooling system is commonly reported in BTUs. One BTU, or one British thermal unit, is defined as the amount of heat required to heat one pound of liquid water by one degree Fahrenheit.
### Sample Video Transcript
Below is a transcript of the video sample provided for this module:
Air is a mixture of gases and water vapor that contains two types of heat: sensible heat and latent heat. “Sensible heat” is defined as heat that results in a temperature change only, with no change in phase. It is the heat in a substance, such as air, that can be “sensed” by you and a thermometer. When a solid melt or a liquid boil, it absorbs heat, but that heat is used to change the phase of the substance; the temperature of the substance does not change. “Latent heat” is heat that is used to change the phase of a substance – for example, from a solid to a liquid, or a liquid to a gas. The water vapor in air contains latent heat. Latent heat is not sensed by a thermometer, but you feel it because people rely upon latent heat for cooling. When we are hot, we perspire and the sweat on our skin evaporates into the air around us. The heat required for this phase change, from liquid to gas, cools us down. When the air around us is humid (when it contains a lot of water vapor), our sweat evaporates at a slower rate. This is why hot humid air “feels” hotter than hot dry air. It is also why HVAC systems must account for sensible heat and latent heat, and control both temperature and humidity. The Second Law governs how air is cooled in an AHU; heat is removed from the air by passing it through a coil that is cooler than the entering air.
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EMPLOYMENT VERIFICATION FOR FORMER ZOLIO INTERNS, CALL 917-909-4188
Select Page
Weighted average cost of capital (WACC) is a basic financial concept and formula that is frequently used in corporate finance and sometimes used in equity and debt valuation, although understanding what it is and its limitations is usually more beneficial to the external analyst trying to understand how a company thinks about itself and its future.
To understand this, let’s do a quick rundown of WACC. This formula tells you how much a firm must pay for debt and equity, with common stock, preferred stock, bonds, and other debt weighted accordingly to reach their total cost.
The formula is as follows: WACC = E/V * Re + D/V * Rd * (1-Tc)
E is the market value of the company’s equity
D is the market value of the company’s debt
V = E + D
Re = Cost of equity
Rd = Cost of debt
Tc = Tax rate for the company
To clarify, let’s use an example. Imagine a company with a market cap of \$1 billion in common stock that owes \$50 million in debt. Already, we know three variables here:
E = \$1 billion
D = \$50 million
V = \$1,050 million
Now Rd is easy: we just look at the company’s interest expense (listed on their 10-Qs and 10-Ks) and divide it by the total debt load. If the cost of debt is \$2 million per year for this company, then Rd = 4%.
Tc is even easier: most companies will tell you their tax rate in their SEC filings. If they don’t, just look at the company’s total taxes paid and divide that by pre-tax income. For our example, let’s say the Tc is 10%.
That just leaves Re, which is a bit harder to calculate. To do so, analysts typically (but not always!) refer to the capital asset pricing model (CAPM), in which the risk-free rate is added to the investment’s beta multiplied by expected market return minute market risk premium, or:
Re = Rf + B (MR – RP)
Rf = risk-free rate
B = investment beta
MR = expected market return
RP = risk premium
This will tell us the expected return on equity for this particular investment.
For the risk-free rate, let’s use the 10-year U.S. Treasury which is now at about 2.7%. Let’s say that the investment we are looking at is expected to be 10% riskier than the market, meaning the B = 1.1 (since 1 = same risk as the market). The expected market return for common stock is usually 7%, so let’s go with that for MR. Finally, let’s say our risk premium will be 1% to compensate for the greater risk of B = 1.1.
Now we’re ready to do some simple math:
Re = 2.7% + 1.1(7%-1%) = 2.7% = 9.3%.
And now, with everything in place, we can finally do our WACC calculation:
WACC = 1,000/1050 * .093 + 50/1050 * .04 * (1-.10)
or
WACC = 9.03%
(Note that the formula above simplifies billions and millions into thousands, but the end result is the same).
Doing this calculation on the corporate finance side, the analyst can then tell his boss that the company should not consider any investment that will not yield a 9.03% return on investment. On the external analyst side, however, one would run this analysis and (more importantly) look at the corporate finance analysis of WACC with some critical questions, mostly going back to the CAPM model.
For starters, is the risk-free rate assumption correct? Using a 10-year risk-free rate assumption for a 20-year investment, for instance, doesn’t make sense, but some analysts will try to fudge the RFR to get the result they want. Similarly, is the investment beta calculated honestly? There is a lot of leeway here, as companies can weight the volatility of debt more than equity in a CAPM model to get a more preferable result, and external analysts should be on the lookout for this. Similarly, is the expected market return and risk premium honest?
Making these numbers more generous can change things a lot. For instance, if our MR goes from 7% (the S&P 500 historical norm) to 3%, our WACC goes from 9.03% to 4.4%! All of a sudden, a slew of investment opportunities look more attractive, and can justify good investments that really are not that good.
This brings us to the primary lesson of the WACC formula: because it has numbers, it looks empirical and objective, but the assumptions in Re bring a lot of subjectivity into the equation (this is why the CAPM model is rejected by some analysts in some situations). Understanding how the WACC formula is used in a firm can teach you a lot about how healthy its financial future will be and how honest management is, while running these analyses for yourself to assess the success of a company’s past investment activity can also tell you whether management has deployed capital wisely or made major missteps.
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You are Here: Home >< Maths
C4 Edexcel - Coordinate Geometry; help! Watch
1. I've come across a question and i'm really not sure where to start.
Part a) of the question has given me a curve which looks pretty much like this. It has the parametric equations:
x=cost, y=sin2t, 0 <(or equal to) t < 2
I've been asked to find an expression for dy/dx in terms of the parameter t. I have done something (which I need for part b)) but I don't know if it's correct; I got -2cos2tcosect by doing dy/dt x dt/dx.
Part b) is where i'm having problems. It says 'Find the values of the parameter t at the points where dy/dx=0'. I think this is where y=1 and y=-1, but as i've said, i'm having difficulties tackling this question.
I would really appreciate if someone helped me out here.
2. First of all, your answer to part (a) is correct. It is simply an application of the Chain Rule.
For part (b) you have to set the expression that you get in (a) equal to zero, then solve for t. Personally I would draw a graph of the relevant function with the appropriate domain (the inequality that you have quoted). It might also help you to write 2t = theta or some other variable.
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# Choosing resistor and capacitor for active filter
I am new in designing the active filters though I designed Active low, high and bandpass and notch filter successfully (using Opamp, R and C). But these filters are not actually designed by me :-( I am using the online tool to design them.
When initially I designed a normal active low pass filter by seeing its general diagram and then calculating the value of R and C as per my cut off frequency. This filter works fine when I am seeing its result by using function generator as input. But I wanted to design it for an audio signal so when I gave an audio signal as input then my whole signal is vanished out. Because of this, I moved to the online tool.
So I wanted to know the impact of the resistance and capacitor on the input/output signal. What are the criteria for choosing the values of the components? For Eg. A cutoff frequency of 500Hz has many pairs of R and C which give 500Hz.
It will be great to hear your thoughts on this. As I wanted to get deep knowledge in the filter design as these are one of the important circuits in the electronics field.
I am attaching this schematic of low pass filter for 500Hz cutoff frequency which I designed by the online tool. But when I take some other random values of R and C then this circuit not works.
• Where is your schematic? Commented Oct 17, 2017 at 5:45
• Actually, I want the general idea behind selecting the R and C values. So that I can apply it in general for any filter. So I did not add the schematic but now I added it. So please share your thoughts about it. Thank you. Commented Oct 17, 2017 at 6:05
• Note that this is not an active filter. The filtering function does not rely on the opamp making it a passive filter (R1 and C2). The opamp, R3 and R5 form a non-inverting amplifier with a voltage gain of around 10. Commented Oct 17, 2017 at 7:30
• Ok I also tried the sallen key filter for band pass filtering which is purely active filter, I am just wanted to know how to calculate the values of componets. What are the parameters that I have to consider before selecting any value. Commented Oct 17, 2017 at 7:38
• Why do you want a bandpass filter? To cut out all lows so that men talk like chipmunksand music sounds tinny? Then you cut out all important high frequency consonant sounds in speech and make music sound like an old AM radio? For music I boost the lows a little because I do not use a sub-woofer and I leave the highs flat with no cut. My hifi speakers all have passive LC crossover filters, no active filters. Commented Oct 27, 2021 at 21:19
"I am attaching this schematic of low pass filter for 500Hz cutoff frequency which I designed by the online tool. But when I take some other random values of R and C then this circuit not works. "
Increasing the value of R and decreasing the capacitor C by the same factor (or vice versa) should have no effect on the filter as long as you are not approaching "exotic values" like 1 GOhm or 2 pF.
"As I wanted to get deep knowledge in the filter design as these are one of the important circuits in the electronics field."
Some general remarks:
It is not easy to gain "deep knowledge" in filter design because there are many, many alternative ciruits and many design strategies. It is a very challenging task to find the "best" (appropriate) circuit for a specific application.
In most cases, filters of higher orders (n>3) are composed of a series combination of filter stages n=2 (and n=1). However, this is not an absolute requirement. There are other design strategies based on passive RLC structures, which then are transferred into active realizations.
But - as said - in most cases the series approach is used and the following steps are performed:
1.) Filter specification based on typical requirements (path region, damping region and damping requirements),
2.) Selecting a corresponding transfer function (order, suitable approximation - Butterworth, Chebyschev, Besssel,...),
3.) Selecting one of the many available filter topologies (Sallen-Key, MFB, Integrator-stages, GIC-blocks,...),
4.) Using filter tables for finding the pole data for each second-order stage (pole frequency and pole-Q),
5) Using design formulas (available for the various topologies) for calculating the parts values,
Fazit: I think, it is clear now why it may be advantageous to use filter design programs. Otherewise you MUST consult a good text book on active filter design.
As you should know phase and impedance ratios of each ratio serve to negate or amplify the input. Since Zc(f) changes 20dB +/- 1 decade relative to R, sensitivity and group delay is low. The power of many C’s combined with optimal parameters and configurations trade off sensitivity with Q of each stage with steepness of skirts (Chebychev) or group delay (Bessel) or low inter-symbol-interference ,ISI Raised-Cosine) or high sensitivity with high Q. www.ti.com tools allow choice of tolerances to best understand. Try an 8th order filter in a quad Op Amp
• Thanks @tony Stewart for your help. Actually I tried the tool of ti.com webench design only. In that I have to select the BW and cutoff freq.They show different circuits for same cut off frequency with different Q value and unity gain BW of OPamp. Phase and impedance ration of what !! please clarify me, I am not getting it o_0 Commented Oct 17, 2017 at 7:43
• Atop filters must be defined this way with bandpass f and gain, then Bandstop f, dB. But it does not choose optimum R for low current, so scale every value to >>10k and reduce all C by same then add R for balanced input voltage or null DC offset. Commented Oct 17, 2017 at 7:51
• That low current point should be followed in designing all the filter!! And every time I take R > 10k, that's an efficient way to design a filter?? Commented Oct 17, 2017 at 7:54
• It depends. Sometimes I have used 100k range as optimal. But large signals or even DC with Rail-Rail output must use this. Commented Oct 17, 2017 at 8:04
• Alright, that means it is better to take R in kilos for low current. I will try it out and do the calculation then revert back to you Commented Oct 17, 2017 at 8:06
It’s perhaps useful to consider the energy involved in different RC values - for low voltages, capacitances Ike’s than about 1uF are convenient and for higher voltages somewhat smaller values are desirable if you want a low-cost and small-footprint device. For high frequencies you’ll want much smaller capacitances otherwise the power requirements will be high. As with all circuit designs there are trade offs between power consumption, parts cost, physical size and performance (noise immunity and output impedance particularly). So it’s a difficult question to answer without knowing all the parameters.
Lot of good suggestions here. Please also keep in mind that the output impedance of the driving source affects some filters more than others. For example, in your circuit, the output resistance of the source adds to R1, thereby lowering the corner frequency.
The same circuit has a nice, low output impedance because it's taken directly from the opamp. Though if the next stage has a low impedance input, you may have unexpected results.
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Homework Help: Elliptic functions proof f(z)-c has N zeros, N the order
1. Jun 28, 2017
binbagsss
1. The problem statement, all variables and given/known data
proof of theorem
2. Relevant equations
3. The attempt at a solution
Hi,
I have a couple of questions on the attached proof and theorem
1) On the last line, how is it we go from the order of the zeros = the number of zeros, or is it's meaning the number of zeros counted with multiplicity. - Am I correct in thinking that:
order of pole/zero = number of poles/zeros * multiplicity at each point ?
2) The zero expansion considered is fine since we are considering zeros of $f(z)-c$. But for expansion about a pole of $f(z)$, near that pole, how is it we can say that that same expansion holds for $f(z)-c$? Is this not making some assumptions on $c$ being small?
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# Free fall question
1. Jun 12, 2007
### hiddenlife5009
1. The problem statement, all variables and given/known data
In part (a) of this question suppose a stone is thrown verically upward with a speed of 12.9 m/s from the edge of a cliff and that h = 82 m. At what speed (in m/s) does the stone hit the ground?
2. Relevant equations
T = v/g
Free fall equation
V = (V^2 - 2gvt . sin(feta) + (g^2) . (t^2))^0.5
3. The attempt at a solution
By using the first equation, I have worked out how long it takes to reach max. Then by using the answer for that, and using the known values, I have used the free fall equation to find out how long it takes to reach the bottom of the cliff. Then by using the last equation I ended up with 52.956 m/s as the answer.
I was told this was not the answer and was wondering what steps I am missing or completely doing wrong.
The question may sound silly but I guess what its trying to ask is 'what is the velocity of the stone right before it hits the ground?'.
Any help would be most appreciated.
2. Jun 12, 2007
### prasannapakkiam
wow! You do not need to use such a complicated equation!
The rock is thrown up. It would soon change direction to start coming down. Calculate this height by using v^2=u^2-2as (by solving for s, also v=0 obviously).
Then add this calculated s to 85. Then use the same equation to figure out v. If you need further help ask away...:)
3. Jun 12, 2007
### hiddenlife5009
Sorry, but where are you getting 85 from?
4. Jun 12, 2007
### prasannapakkiam
Okay, there is a misunderstanding, may you please rephrase this question? I thought, h was the height of the cliff
5. Jun 12, 2007
### hiddenlife5009
Ah, height is part of the question, but I guess you meant 82, you wrote down 85. And what do you mean by add s (seconds I presume) to 82 (height)?
6. Jun 12, 2007
### prasannapakkiam
Oh sorry, Anyway, can you see how it works?
7. Jun 12, 2007
### hiddenlife5009
Not really. For starters, u = height?, and could you please refer to my previous post for the second bit I don't understand.
8. Jun 12, 2007
### prasannapakkiam
Okay, one of the 4 basic equations of motion is:
v^2=u^2+2as
v=final velocity
u=initial velocity
a=acceleration
s=distance
Now we want distance.
so we solve for s.
therefore: s=(v^2-u^2)/(2g)
Okay. When the ball reaches its maximal height, its velocity is 0 (Once you can accept that any projectile question becomes a poece of cake). So that is why we set:
v=0
g=-9.8 (the object is DECELERATING due to the effect of gravity...)
u=12.9
Okay do you get it so far? So work out s and find the distance from the ground to the maximal height of the rock (not from the cliff).
9. Jun 12, 2007
### prasannapakkiam
I see that you have used time. Try my method using distance.
10. Jun 12, 2007
### hiddenlife5009
Ok, so using that equation, I have gotten distance = 8.49m
Is that the maximum height the stone reaches? If so, where do I go from here to obtain the speed of the stone just before it hits the ground?
11. Jun 12, 2007
### prasannapakkiam
Good. So the distance from the ground (not cliff) is 8.49+h = 90.49
Okay. Now use v^2=u^2+2as
Here v is what we want to find.
Now is there an initial velocity? No. It is 0.
a=+9.8 (the stone/rock is ACCELERATING)
u=0
s=90.49
well I think the answer can be found now.
So remember acceleration whether it is a minus or plus. And try to find the easier equation to use in the context.
12. Jun 12, 2007
### hiddenlife5009
So therefore,
v = (2x9.8x90.49)^0.5
= 42.11 m/s
I'm fairly certain that is the answer. Thankyou very much for being so patient with me and helping me out, its my first time doing physics.
On a side note, there is another I need a tiny bit of help on. Got the first part of this question, but the second part I just can't grasp.
Part A
A cheetah, the fastest of all land animals over a short distance, can accelerate from zero to 18.0 m/s in three strides and to a full speed of 31.3 m/s in seconds. Assuming the first three strides are each 4.6 m long and that acceleration is constant until the cheetah reaches full speed, what is the cheetah's acceleration (in m/s2)?
Part B
If in part (a) of this question the cheetah's first three strides were each 3.7m long, how many seconds would it take to reach its full speed?
I keep saying to myself the equation I should be using is S = D/t
S = velocity/speed
D = distance/displacement
t = time
When it says the strides are now 3.7m long, I'm guessing a new acceleration is required, so I've come up with 14.77 m/s2. The answer is 2.14 seconds, but I just can't work out the solution. (Answer can be within .1).
13. Jun 12, 2007
### prasannapakkiam
That is wierd. Show how you got the answer to part a
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# Moment of Inertia of Door
## Homework Statement
A 32kg door that is 107 cm wide. What is the moment of inertia 10 cm inside the door with rotation around a vertical axis?
L=Iω
## The Attempt at a Solution
I=1/3mL^2
I= Icm+ md^2
1/3(32)(107)^2
Im not sure what to do after I find the I center of mass
## The Attempt at a Solution
Your question is confusing.Give more details.Whether the door is rotating in vertical plane or horizontal plane??
It is rotating around a vertical axis
It is rotating around a vertical axis
But in which plane???
I mean that whether the door is on floor and rotating about vertical axis or in an erect position rotating about vertical axis.Moment of inertia in both cases is different.
It seems fair to assume from the question that the door is mounted to a vertical pole 10cm from its longest edge and parallel to that edge.
JMUkid you should try doing a search for the "parallel axis theorem" :)
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# Thinking out loud about forks
#### DEGBERT
Member
I’m just trying to figure out the theory behind my forks operation, that’s it, I’m not looking for stack recommendations or anything. It’s driving me crazy, It’s like a song you can’t get out of your head. Any questions, comments, funny stories would be appreciated.
Anyway, while dissecting my 01 WP forks I made the following observations:
1. “Mid-valve” or active valving
Mid-Valve 24 mm OD * 8mm ID * .10mm (4X)
Very light mid-valve spring
Shims open approximately 5 mm
Correct me if I’m wrong, but it seems to me that the “mid-valve” or active piston has nothing or little to do with how much oil is pushed into the base valve ports. It would seem that would be a function of the ratio between the rod area vs. cartridge area.
So, as the rod moves its causes a differential pressure between the top of the active piston and the bottom of the active piston.
Example:
Rod OD= 14mm Area =154 mm2
Cartridge ID= 28 mm Area =616 mm2
So the area above the piston would be 462 mm2 vs. the area below the piston 616 mm2
So it would seem that any change in the rod dia. would have a substantial effect on the resulting pressure downstream with the same force. (Assuming the cartridge ID stays the same)
So I would guess your base valve stack would have to be based on the ratio between the rod and cartridge diameter.
Again, correct me if I’m wrong. A base valve compression stack that works so good on your friends Yamaha might not work so well on your KTM or Honda unless the rod/cartridge ratio is the same.
So this brings me to the question. Why the need for a mid-valve? The area of the active valve ports is almost as big as the area of the cartridge itself (I would guess 75%) and the shims are so easily pushed of the face of the piston (light spring) that it seems like there is no way there could be any kind of flow restriction (or very little) except at very, very, very slow rod speeds.
Would it not be more beneficial to use a substantially higher rate mid valve spring?
Wouldn’t this give a more progressive damping curve during very slow rod speeds?
It seems that if you could “shift” more of the dampening to the mid-valve for low speed stuff (landing jumps, G-outs) you could then substantially reduce the “stiffness” of your base valve stack, theoretically reducing harshness on roots, rocks, breaking bumps, etc.
OK, I know I’m starting to ramble and I’m probably way off in left field, but I’m almost done, I think.
2.“Base Valve” or passive valving
Everything I’ve read has said your valving is speed sensitive and not position sensitive, after looking at the stack I believe this to be true. Pressure can be traded off against velocity, by placing a different effective area at each side of the piston (the top shim / piston interface). The same pressure on a smaller area will move the piston at a higher speed but lower force for a given rate of fluid delivery. Translation: The more shim deflection the less resistance on the cartridge rod, thus less deflection at higher shaft speeds. So maybe this explains why if your high-speed stack is to “stiff” your forks will feel harsh and deflective in the choppy stuff.
Again correct me if I’m wrong. Now I think I understand why the race tech gold valves I used on my last bike did not work so well. All the dampening was placed on the base valve compression shim stack. The forks worked very well on the low speed stuff (jump landings, G-outs) but on the fast choppy stuff (roots, rocks, etc.) it would beat my arms to a pulp. When I reduced the HS stack so it was acceptable in the choppy stuff, I would loose low speed control. I could never find the right combination that gave me the best of both worlds.
OK, I think I’m done for now, my head is starting to hurt.
#### marcusgunby
Hmmm you have been busy.IMO most of what you say makes sense my one comment would be on the midvalve you say it only causes a restriction at very slow movements but i think its the other way round-it causes a restriction at high piston speeds.The later spec 01 forks do have a stiffer midvalve spring but drewarm says it appears to make no difference.Could you go over the base valve part again as you talk about pistons moving at high speed but the piston is fixed.
I think you are correct about the gold valves.Hard to get suppleness and control at the same time.Also it would be interesting to try a Cr valve stack in the 01 fork-i think it would feel like concrete so yes the cartridge/piston ratio is important.IMO this is why most tuners find the Wp fork hard to get right.
[This message has been edited by marcusgunby (edited 04-05-2001).]
#### drehwurm
Member
Servus Degbert,
Correct me if I’m wrong, but it seems to me that the “mid-valve” or active piston has nothing or little to do with how much oil is pushed into the base valve ports.
Yes I'm of the same oppinion! Only oil which is displaced by the rod will go through the base valve. Any additional oil would have to come from outside the cartrige, i.e. passing by the piston rod seal which should not happen. And yes, the differential pressure is created behind the active piston thus working AGAINST the direction of flow! Also, did you notice that the active and passive piston valve are identical in port configuration and size! The mid valve is just a 'shimmed' version of the base rebound check valve!
A base valve compression stack that works so good on your friends Yamaha might not work so well on your KTM or Honda unless the rod/cartridge ratio is the same.
Theoretically yes, but practically I'm a little confused. The base valve shim stack of my 2001 KTM 520 14mm rod with mid valve WP fork is very, very similar to my 12mm Gold Valve without mid valve 1998 Yamaha WR stack. I like the KTM now and I loved the Yamaha back then. I'm faster now so the KTM has to be firmer, but the stacks are compareable! So what - I honestly don't know, it is just working for me.
Why the need for a mid-valve? The area of the active valve ports is almost as big as the area of the cartridge itself (I would guess 75%) and the shims are so easily pushed of the face of the piston (light spring) that it seems like there is no way there could be any kind of flow restriction (or very little) except at very, very, very slow rod speeds.
I replaced the spring with one at least two to three times stronger and didn't feel a difference. Maybe the WP mid valve configuration in general is just too weak, leaving it almost useless. But then I've seen severely bent Yamaha mid valves, so there must be something about it. The amount of oil which passes through the mid valve is 4 times bigger then the amount which goes through the base valve; ports on the base valve are substantially smaller than on the mid valve; valving is way stiffer on the base valve => what is the formula to convert mid valve damping to base valve damping to get the same overall damping? BTW, if you remove the base valve and seal the cartrige what have you got: right, a primitive shock! So, I'm still of the oppinion that active and passive valving are basically identical - you can theoretically replace each with the other to get the same overall effect. Practically though, two valves with different flow rates probably give you more possibilities of setup - I think I can accept that fact.
So maybe this explains why if your high-speed stack is to “stiff” your forks will feel harsh and deflective in the choppy stuff.
Hmmm, I don't know. Think of a stack #1 with 10 shims gradually rising OD from 10 to 20mm and stack #2 with 10 shims of 20mm OD. I think we agree that stack #2 is stiffer than #1 - BUT where, LSC or HSC? Does #2 has a HSC part and where would it start? Every stack has a CONTINUOUS DAMPING CURVE and not segments like LSC and HSC. My point is, that if the lSC part is not in the ballpark you can not blame such problems on HSC. It is like with jetting, if your main jet is too lean no matter how 'rich' your needle is, it won't solve your problems. Hope I could make my ideas clear somehow!
Michael
#### MACE
I'm the village idjut and all I've worked on are my poor KYBs and I use Visegrips on everyting but I love this kind of theory thread so here's my \$.02 (\$.03 CDN).
I think that the effect of the midvalve is felt when the flow through the rebound piston is sufficient to be restricted by the flow area between the bottom mv shim and the cartridge wall. When that area starts to restrict flow the restriction is modulated by the mv stack deflecting which results in a smaller shim O.D. and a larger flow area. All that coil spring does is shut the valve on rebound.
Now if this theory is true, replacing the mv with a rigid check plate (like Race Tech demands) would in fact increase the compression damping.
Ah the mysteries.....
------------------
MACE
One night I was layin' down,
I heard mama 'n papa talkin'
I heard papa tell mama, "you let that boy MOTO,
it's in him, and it got to come out..."
#### JTT
I too have pondered this. I think Mace is on the right track from what I have accumulated so far...
In order for the spring to have any significant effect, it would have to be extremely stiff (the pressures acting on it are considerable), but along this same thinking I am wondering why WP would upgrade to a stiffer spring (with no significant effect, as drehwurm has told us).
------------------
JTT
Logic Over Hype Coalition
#### drehwurm
Member
Servus Mace,
flow area between the bottom mv shim and the cartridge wall
No, I don't think so. Before the oil passes through the midvalve it already passed by a 24mm shim and the cartridge wall - the closed rebound valve!!! IMHO the MV damping is only dependent by the distance the shims can move away from the valve and the shim stack itself.
Now if this theory is true, replacing the mv with a rigid check plate (like Race Tech demands) would in fact increase the compression damping.
But then the rebound check plate on the base valve would also create a significant amount of (unwanted) rebound damping - wouldn't it?
Michael
------------------
#### drehwurm
Member
Servus JTT,
I am wondering why WP would upgrade to a stiffer spring
Me too, which leads to a mechanical question: If the spring is so stiff that the shims 'hover' between the valve and stop, isn't there a chance that the rather thin (very little axial guidance) shim stack could tilt?
Michael
------------------
#### MACE
Originally posted by drehwurm:
No, I don't think so. Before the oil passes through the midvalve it already passed by a 24mm shim and the cartridge wall - the closed rebound valve!!! IMHO the MV damping is only dependent by the distance the shims can move away from the valve and the shim stack itself.
Good point about the upstream restriction. There is either more area upstream or the MV does nothing. Next time I have parts in hand, I'll try to measure them.
I guarantee you that that wimpy MV spring is not holding the MV stack closed. The MV slams into the "hat" washer and (if there is sufficient flow) bends around the "hat" washer. Any other scenario and the MV stack is nothing but a rigid check valve.
Originally posted by drehwurm:
But then the rebound check plate on the base valve would also create a significant amount of (unwanted) rebound damping - wouldn't it?
Not necessarily "unwanted". It just must work within the overall rebound requirement.
#### drehwurm
Member
Servus MACE,
There is either more area upstream or the MV does nothing
Maybe it is like with electrical current: if you put two resistors in line, you get the added amount of each as overall resistance!
BTW: Did you ever notice that when bleeding your forks it is much harder to pull the rod out than to push it down! Is the rebound damping so much stronger than the comp damping?
Michael
------------------
#### Jeremy Wilkey
##### Owner, MX-Tech
Bravo... You guys got it!!!
------------------
"Danger is one thing but danger combined with long periods of suffering is quite another." Sir E. Hilary
wilkey@mx-tech.com
#### Murf
Member
drehwurm,
The reason (my guess) that it is so much easier to compress the fork than to extend it by hand when bleeding. The damping during extension (rebound) has to resist the spring. The damping while compressing the fork has the spring working in its favor to slow compression.
Am I right?
------------------
Murf
Y2K 300EXC
#### John Curea
The one nagging question that gets stuck in my mind relates to the different midvalve builds in the 28mm and 32mm cartridges.
I am trying to compare midvalve build designs on my 99KX250 with 28mm cartridge and 12mm piston rods vs. a late model YZ with 32mm cartridges with 12mm piston rods I dont want to throw the late model WP into the equation just yet because it has 28mm cartridge with 14mm piston rod. I want to keep it "apples to apples" so to speak.
Anyways, I am trying to determine the performance difference between the KX and YZ midvalves. I realize the only fluid displaced through the base valve eqauls the area of the piston rod in the cartridge at any given time. Reguardless if the piston rod lives in a 32mm or 28mm cartridge, the piston rod will still only displace X amount of fluid through the base valve.
The YZ midvalve is built extremely light, including float dimensions compared to the KX midvalve.
Other than factoring in the physical size difference (the YZ being 3mm larger) of the midvalve pistons, is there any other factor that influences the design/performance of the midvalve? (I have a feeling there is, I just cant get it nailed down).
Take Care, John
------------------
99 KX250
98 KTM50
88 LT250R
86 TRX70
#### JTT
Originally posted by KXVET#207:
Other than factoring in the physical size difference (the YZ being 3mm larger) of the midvalve pistons, is there any other factor that influences the design/performance of the midvalve?
Is there a variance in actual port size? How about the difference in leverage created by the dimension from the "pivot" location of the shims?
...just shooting ideas...
John, you also mention "lighter float dimensions"...do you mean there is a difference in how far the shim stack displaces (by compressing the spring)? I haven't seem a KX apart recently to take notice.
------------------
JTT
No matter where you go...there you are
#### Jeremy Wilkey
##### Owner, MX-Tech
Ok...
Man this kills me...
-The factors are Midvalve Float..
This impacts when the midvalve actually starts to work.... (In terms of speed.)
-Then the midvalve stiffness or stack build..
These two are directly realted to the swept volume of the clyinder. Volume is related to pie r2 x h.. So a larger cartrige is massively impacting to the total swept volume.
So lets look at to cases..
KX 28mm cartridge... Thge midvalve runs no float but does have .1mm bleed/float.. This stack is also only moderatly stiff however, because it pivots of a very small shim.. SO although in terms of speed the midvalve impacts lower speed ranges but the stack is not very progressive and does not impact high speeds as much as might be imagined..
Case 2...
YZ fork with much more float...
This set-up does not work until a slighlty faster speed, however the volume crossing the valve is much larger so apples to apples it's almost the same thing... The valving is also more progressive as it's stack is built..
A couple of more observations..
The basevalve stacks also vary greatly on these two bikes.. Why..Very simply one is stiffer one is softer. Like the circuts of a carborator the speed ranges overlap. The mid-valve of the KX has a slighly higher overall damping coeifecent and hence requires a slighly lighter valve stack and vise versa. Also the YZ has the CV which is bleeding fluid as well, also nesciating a stiffer build at the basevalve to acheive the ideal overall coifecent. Which works better? It depens on the aplication. I'll give the KX a better stadium build, and the YZ a better outdoor build.
The 28mm Midvalve trades plushness for instant damping, and hence more precise handling when riden by an expert. The YZ on the other hand is more suple at very low speeds. However in big hits is often more harsh.. (Volume of mid-valve and a very stiff midvalve just stacked up on you!)
Ok a couple of more issues I have talked about forever which no ever seems to remmber.. A KYB fork does not work effeceintly until the fork has compressed several inches.. That is often why pro's preffer the Showa forks.. Which work all the time.
This is the problem with backyard tunners.. They really don't understand how the circuits overalap..
I get imesly frustrated when I read this topic.. I feel like I've been beating this drum for literally years and noone ever gets it.... Sorry for the rant..
Also the reason compression feels so much softer is related to a few simple things.. Compression has a much smaller volume going through base-valve. (Only the rod.) If the basevalve restricted flow on return you would have chronic fork isseues such as zero damping..
The midvalve does not impact very low speeds so you don't feel it..
Rebound very stiff active valving, with a large volume.. Ofcourse you will feel it more.. And I ask you do you think you want as much compression as you have rebound?
Regards,
JEr
I know I can't spell and I rant but these are the issues. There are still some small fine points but this is good for now..
------------------
"Danger is one thing but danger combined with long periods of suffering is quite another." Sir E. Hilary
wilkey@mx-tech.com
#### John Curea
As the mighty light bulb above my head just illuminated (temporarily cancelling out the "blond moment")
The words "Swept Volume" gave me an Ah Ha moment!!
Take care, Grasshopper John
------------------
99 KX250
98 KTM50
88 LT250R
86 TRX70
#### drehwurm
Member
Servus Jer,
And I ask you do you think you want as much compression as you have rebound?
??? I'd say yes! I'd imagine that the force applied by compressed fork springs is less than the force applied by a bike landing after a big jump or even g-out. Am I looking at the wrong spot here?
Michael
------------------
#### JTT
Thanks Jer!! That clears up a lot of questions all at once. Good to have you back!
------------------
JTT
No matter where you go...there you are
#### Jeremy Wilkey
##### Owner, MX-Tech
As for the question Micheal asked,
Think that the rebound must slowly return the energy that the compression forces traped in the spring... If the rebound was as soft as the compression you would get launched into orbit upon landing..
Actually try it sometime.... :) I have.. By acident of course... :eek::
Reagards,
Jer
#### MACE
Originally posted by Jeremy Wilkey:
......
Ok a couple of more issues I have talked about forever which no ever seems to remmber.. A KYB fork does not work effeceintly until the fork has compressed several inches..
Say what? First time I've heard you mention that. Why is this true?
I get imesly frustrated when I read this topic.. I feel like I've been beating this drum for literally years and noone ever gets it.... Sorry for the rant..
[/QUOTE]
Actually, it seems the good discussions on midvalves all seem to have gone to the great server in the sky. I was hoping to find "The Truth About Midvalves" and some other classics and they seem to be lost. A shame. As for the ranting, all my best teachers used to throw chalk.... You have tenure.
#### John Curea
MACE
I used to have a history teacher who used to jump and down on the desks!! This guy was nuts!!
I will never forget "Law of the Retarding Lead"
John
------------------
99 KX250
98 KTM50
88 LT250R
86 TRX70
#### P_Taylor
##### Uhhh...
if I understand things right the KYB fork can cavitate some until the fork is compressed enough to create sufficient air pressure on the oil to prevent cavitation , the twin chamber showa on the other hand uses a second chamber with a spring loaded piston to keep pressure on the oil and to prevent cavitating
------------------
Patrick Taylor
#### Jeremy Wilkey
##### Owner, MX-Tech
Patrick,
You are right on! That is often why the expert level riders like the CR250 fork..
Regards,
Jer
When I was teaching HS I "Made an impaact" :eek:
#### John Curea
Jer
Is that why you see a remote resevoir on the forks of RC and Wey?
I thought at one time it might be oil, but I guess it makes more sense to have the air in the forks pressurized at full extension. It would prevent cavitation of the oil in the cartridge as the fork is on the last part of the rebound stroke.
Take Care,John
------------------
99 KX250
98 KTM50
88 LT250R
86 TRX70
#### motopuffs
Member
wouldn't be surprised if that is Nitrogen rather than air!
------------------
#### P_Taylor
##### Uhhh...
Has anyone seen the inside of Pro-Circuits new fork? \$4500 for a shock and set of forks
They claim the internal cartridge system is sealed and pressurized in the fork. I guess they have figured a way to do it without the external reservoirs.
The fork is made by Showa and has 49mm tubes. The shock uses a 50mm piston.
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Patrick Taylor
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## The Tridecagon, Hyperbola and Lill’s Method
The regular tridecagon is another regular polygon that cannot be constructed using a compass and straightedge. In this post I want to show how the tridecagon can be constructed using the intersection of a circle and a hyperbola. In my previous posts “The Heptagon, Hyperbola and Lill’s Circle” and “The Nonagon, Hyperbola and Lill’s Method”… Continue reading The Tridecagon, Hyperbola and Lill’s Method
## Lill’s Method, Prime Numbers and Tangent of Sum of Angles
In this post I want to explore again the property discussed in my paper “Lill’s Method and the Sum of Arctangents”. I’ll apply the property to this question: If tan(θ1)=2, tan(θ2)=3,tan(θ3)=5,…,tan(θn)=n-th prime number, then what is tan(θ1 + θ2 +… θn)? The question can be easily solved with a calculator. We’ll see that the answer… Continue reading Lill’s Method, Prime Numbers and Tangent of Sum of Angles
## The Nonagon, Hyperbola and Lill’s Method
The nonagon is another polygon that cannot be constructed with ruler and the compass (see OEIS sequence A004169). However, the nonagon can be constructed using conics (see OEIS sequence A051913). In this post I want to show how we can use the intersection of the Lill circle of the polynomial x3 – 0.75x + 0.125… Continue reading The Nonagon, Hyperbola and Lill’s Method
## The Heptagon, Hyperbola and Lill’s Circle
The heptagon cannot be constructed with just a ruler and a compass. However, in this post I’ll show how you can construct the heptagon using a hyperbola and the Lill’s circle of a third degree polynomial. The heptagon construction is very similar in nature to my trisection construction that I presented in my “Trisection Hyperbolas… Continue reading The Heptagon, Hyperbola and Lill’s Circle
## Trisection Hyperbolas and Lill’s Circle
In my previous blog post about trisection “Angle Trisection: a Neusis Construction using Lill’s Method and Lill’s Circle”, I showed a “mechanical” method for trisecting an angle smaller than 90 degrees. The neusis construction involved the Lill’s method representation of cubic equations of the form x3 -3tan(θ)x2 -3x + tan(θ). These cubic equations have 3 real… Continue reading Trisection Hyperbolas and Lill’s Circle
## Angle Trisection: a Neusis Construction using Lill’s Method and Lill’s Circle
The problem of trisecting an angle using just a compass and a straightedge is an impossible problem. However, the problem of angle trisection can be solved if we are allowed the use of a marked ruler. Geometric constructions that use a marked ruler are called neusis constructions. You can learn more about the topic of… Continue reading Angle Trisection: a Neusis Construction using Lill’s Method and Lill’s Circle
## Lill’s method and the Philo Line for Right Angles
(Note: This is an article from 2016 that I posted on Hubpages. In this article I used Riaz’s way of illustrating Lill’s method. To become more familiar with Lill’s method I recommend going through the links provided in my Lill’s Method page. Also see my Papers page and my Lill’s method articles/posts. ) In this… Continue reading Lill’s method and the Philo Line for Right Angles
## Lill’s Method and Geometric Solutions to Quadratic Equations With Complex Roots
I already wrote a paper called “Lill’s Method and Graphical Solutions to Quadratic Equations” that shows how to solve quadratic equations using Lill’s method. In this post I want to show an alternative way of solving quadratic equations that have complex or imaginary roots. This method was briefly mentioned in the paper “Geometric Solution of… Continue reading Lill’s Method and Geometric Solutions to Quadratic Equations With Complex Roots
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Question Paper: CAD / CAM / CIM : Question Paper May 2015 - Mechanical Engineering (Semester 7) | Mumbai University (MU)
0
## CAD / CAM / CIM - May 2015
### Mechanical Engineering (Semester 7)
TOTAL MARKS: 80
TOTAL TIME: 3 HOURS
(1) Question 1 is compulsory.
(2) Attempt any three from the remaining questions.
(3) Assume data if required.
(4) Figures to the right indicate full marks.
1 (a) Write short note on concurrent engineering in product desig. (6 marks)
1 (b) Write short note on computer aided process planing (CAPP). (7 marks)
1 (c) Plot the Bezier curve having endpoints P0(1, 3) and P3(7, 2). The other control points are P1(5, 6) and P2 (6, 0). Plot for values for u=0, 0.1, 0.2, ... 1, if the characteristic polygon is drawn in the sequence P0 - P1 - P2, P3. (7 marks)
2 (a) Write a part program using G' and M codes for finishing a forged component as shown in Figure 1. Assume the speed and feed on the turning cente as 500 rpm and 0.3 mm/rev respectively. Assume suitable data if necessary. (10 marks)
2 (b) Consider a triangle ABC having coordinates A(5,5) B(8,5) and C(5,10). Determine the new vertex position if.
i) The triangle is rotated by 60°, anticlockwise about the vertex A.
ii) The triangle is scale by 2 times in X direction and 3 times in Y direction about vertex A.
(10 marks)
### Write short note on following:
3 (a) Adaptive control in manufacturing. (8 marks)
3 (b) Artificial Intelligence in Computer Aided Process Planning (CAPP). (6 marks)
3 (c) Socio-techno-economic aspects with respect to Computer Integrated Manufacturing (CIM). (6 marks)
### Write short note on:
4 (a) (i) Essential of Computer Aided Design workstations and its functions. (5 marks)
4 (a) (ii) Visual realism. (5 marks)
4 (b) Write a complete APT program (geometric and motion commands) to machine the outline of the geometry as shown in Figure 2. The component is 5 mm tick. The milling tool used is 5 mm in diameter. Consider spindle speed as 1000 rpm and feed as 0.3 mm/rev. Assume suitable data if necessary. (5 marks)
5 (a) Write a program in object oriented language for 2D transformation which include functions for the following operations: (i) rotation @y-axis (ii) translation in x-direction. (10 marks)
### Write short note on:
5 (b) (i) Reverse engineering and data capture techniques. (5 marks)
5 (b) (ii) Green Manufacturing. (5 marks)
6 (a) Explain with block diagram Computer Aided Quality Control (CAQC). (10 marks)
### Write short note on:
6 (b) (i) Light and shade ray tracing. (5 marks)
6 (b) (ii) Design for assembly and disassembly. (5 marks)
### Write short note on:
7 (a) Automated material handling and storage system (6 marks)
7 (b) Flexible manufacturing systems (FMS). (6 marks)
7 (c) Feature recognition and design by feature. (8 marks)
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# Bell Curve – Definition
Cite this article as:"Bell Curve – Definition," in The Business Professor, updated March 4, 2019, last accessed December 4, 2020, https://thebusinessprofessor.com/lesson/bell-curve-definition/.
### Bell Curve Definition
A bell curve is referred to as a normal distribution because it is the most common type of distribution for a variable. The term ‘bell curve’ is derived from the graph depicting a normal distribution because it consists of a bell-shaped line. The highest point in the curve represents the most probable events in a series of data. All the other possible occurrences are distributed equally around the most probable event and end up forming a downward sloping line on either side of the peak.
### A Little More on What is the Bell Curve
A bell curve refers to the graphical representation of normal probability distribution. The underlying standard deviations of this distribution from the median or the highest point of the curve give it the shape of a curved bell. The standard deviation is a measurement that is utilized in the quantification of the variability of data dispersion in a given set of values, while the mean gives the average of the total data points in the data set.
The calculation of standard deviation is done after that of the mean, and it represents a percentage of the total data. For example, if a series of 100 test scores are collected and used in a normal probability distribution, 68% of the entire test scores are expected to fall below the mean or within one standard deviation.
When two standard deviations are moved away from the mean, 95% of the test scores are included. 99.7% of the test scores are represented when three standard deviations are moved away from the mean. Text scores which are extreme outliers are considered as long tail data points they lie outside the range of the three standard deviations. Such outliers include 100 or 0.
### Bell Curves in Finance
When analyzing the returns of the overall market or a security sensitivity, financial analysts and investors usually use a normal probability distribution. In finance, volatility is used to refer to standard deviations that show the returns of security. For example, blue chip stocks are the ones usually depicting a bell curve, and they have predictable low volatility.
By using the normal probability distribution of the past returns of stock, investors can make assumptions with regards to the expected future returns. In some cases though, stocks and other securities display distributions that are non-normal. Fatter tails than normal ones characterize these non-normal distributions. When the fatter tail is skewed negative, it signals to the investors that there is a higher probability of negative returns. The converse is also true.
### Academic Research on Bell Curve Portfolio
• A multifractal walk down Wall Street, Mandelbrot, B. B. (1999). Scientific American, 280(2), 70-73. This paper explains that individual investors and professional stock traders are aware that the prices usually quoted in any financial market can change very quickly.
• A focus on the exceptions that prove the rule, Mandelbrot, B., & Taleb, N. (2006). Financial Times, 23. This article presents how the current studies of uncertainty, such as economics and statistics, have been staying close to the bell curve which represents a probability distribution.
• Six ways companies mismanage risk, Stulz, R. M. (2009). Harvard Business Review, 87(3), 86-94. This paper presents six ways through which companies mismanage risk and also suggests solutions to these problems.
• Power laws and the new science of complexity management, Buchanan, M. (2004). Strategy+ Business, 34(Spring), 70-79. This is research that is aimed at gathering attention in the world of business since executives and scholars alike are starting to understand that the conventional theories of management forged in the industrialization era cannot cope anymore with the highly complex organizations that have emerged in the last two decades of increasing globalization and decentralization.
• Implementation issues in project web sites: a practioner’s viewpoint, O’Brien, W. J. (2000). Journal of management in engineering, 16(3), 34-39. This study uses the observations of a practitioner involved in the development and use of the first generation of project web sites to summarize the critical issues in implementing the sites on projects.
• On modified Black–Scholes equation, Ahmed, E., & Abdusalam, H. A. (2004). Chaos, Solitons & Fractals, 22(3), 583-587. This paper explains that since it is argued that from several points of view the telegraph equation is more suitable. The Black-Scholes equation is modified and then proposed.
• How Fractals Can Explain What’s Wrong with Wall Street, Mandelbrot, B. B. (2008). Sci. Am, 15. This paper presents the modern portfolio theory as a cornerstone of finance that tries to maximize returns given a specific level of risk.
• The Bell curve is wrong: so what?, Embrechts, P. (2000). Extremes and Integrated Risk Management, xxv-xxviii. This paper explains that the bell curve is considered as erroneous since various studies in different fields have shown that apparently, rare events are more common than predicted by the curve
• Rethinking modern portfolio theory, Warner, J. (2010). Bank Investment Consulting. This study shows how different investors are trying to decide whether to repair or abandon the theoretical foundations on which their portfolios are developed.
• A rational approach to pricing of catastrophe insurance, Dong, W., Shah, H., & Wong, F. (1996). This paper describes a methodology for rational pricing which contains solvency and a stability-based framework as well as a formula to quantify the loss variability driving solvency and stability.
• Portfolio crash testing: making sense of extreme event exposures, Novosyolov, A., & Satchkov, D. (2010). The Journal of Risk Model Validation, 4(3), 53. This article addresses various misunderstandings about stress testing which are common and then provides for its inclusion into the risk process as a supplement to risk measures like value-at-risk and tracking error.
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Mathematic
# Pi Day math celebration with Eugenia Cheng : Quick Wave : NPR
Eugenia Cheng is a mathematician and writer of the guide Easy methods to Bake Pi: An Edible Exploration of the Arithmetic of Arithmetic. All through the guide, she makes use of baking as a automobile for higher understanding arithmetic ideas.
Primary Books
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Primary Books
This March 14, Quick Wave is celebrating π … and pie!
We do this with the assistance of mathematician Eugenia Cheng, Scientist In Residence on the Faculty of the Artwork Institute of Chicago and writer of the guide Easy methods to Bake Pi.
This episode: The mathematical magic embedded in baking — and we do not imply the numbers or the measurements.
Making math enjoyable and relatable to the plenty is a lifelong ambition for Eugenia. Up to now, she’s the writer of seven in style math books, together with two youngsters’s books on math.
“I feel that we should always take any alternative we are able to to painting math in a means that’s enjoyable to individuals,” she says. “Simply associating math with enjoyable as an alternative of with trauma is an effective begin.”
Eugenia’s begin with arithmetic was joyful and got here early in life. See, since she was a toddler, baking and math had been linked. As a bit Eugenia realized from her mom that baking, like pure math, is not simply in regards to the substances — it is about the method.
Take for instance the primary recipe in Easy methods to Bake Pi. It is a information for making clotted cream, the fluffy staple of British tea time.
### Clotted Cream
Substances
Cream
Technique
1. Pour the cream right into a rice cooker.
2. Depart it on the “maintain heat” setting with the lid barely open, for about 8 hours.
3. Cool it within the fridge for about 8 hours.
4. Scoop the highest half off: that is the clotted cream.
That is it. One ingredient, 4 steps. It is the strategy, or recipe directions, that turns simply cream into unctuous clotted cream.
Eugenia loves this recipe as a result of, sure, it has an ingredient — however actually, it is all in regards to the course of. She loves pure math for a similar purpose. “And simply as a recipe could be a bit ineffective if it omitted the strategy, we won’t perceive what math is except we discuss the way in which it’s performed, not simply the issues it research,” she writes in her guide.
Throughout every chapter, Eugenia makes use of a recipe as a launching level for math. Take chapter two.
### Mayonnaise or Hollandaise Sauce
Substances
2 egg yolks
1.5 cups olive oil
seasoning
Technique
1. Whisk the egg yolks and seasoning utilizing a hand whisk or immersion blender.
2. Drip the olive oil in very slowly whereas persevering with to whisk.
For Hollandaise sauce, use 0.5 cup melted butter as an alternative of the olive oil.
The recipe is a playful place to begin for discussing the method of abstraction — in search of underlying similarities between gadgets like buildings, patterns or properties. Mayonnaise and hollandaise are two completely different sauces with barely completely different substances, however extra abstractly, the method of constructing them is similar.
By this and different playful examples, Eugenia brings arithmetic house and makes them shine with (principally) actual life use circumstances.
“I didn’t all the time love college math, however I used to be very fortunate as a result of my mom is mathematical. And so she confirmed me the true essence of math at house, in order that once I was bored by having to do timetables and reply questions at college, I knew that there was extra to math than that,” she remembers. “I held out that hope and that perception all over college till the very finish of faculty, when it acquired attention-grabbing once more — after which college when it lastly acquired actually attention-grabbing — after which lastly analysis.”
Eugenia’s been on Quick Wave earlier than! To listen to extra, try our episode, A Mathematician’s Manifesto For Rethinking Gender.
Interested by different math magic? Electronic mail us at [email protected].
Take heed to Quick Wave on Spotify, Apple Podcasts and Google Podcasts.
This episode was produced by Berly McCoy and edited by Rebecca Ramirez, with assist from Gisele Grayson. The audio engineer was Robert Rodriguez.
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# Calories (mean) to Kilotons (metric, explosive energy) conversion
Kilotons (metric, explosive energy) to Calories (mean) (Swap Units)
Format
Accuracy
Note: Fractional results are rounded to the nearest 1/64. For a more accurate answer please select 'decimal' from the options above the result.
Note: You can increase or decrease the accuracy of this answer by selecting the number of significant figures required from the options above the result.
Note: For a pure decimal result please select 'decimal' from the options above the result.
Show formula
## Calories (mean) to Kilotons (metric,explosive energy) formula
kt =
Calorie (mean) * 0.0000000000010014
Show working
Show result in exponential format
## Calories (mean)
The mean calorie is the total amount of energy needed to heat one gram of air-free water from 0 to 100°C at standard atmospheric pressure, divided by 100.It's value is 4.19002 J
## Calories (mean) to Kilotons (metric,explosive energy) formula
kt =
Calorie (mean) * 0.0000000000010014
## Kilotons (metric,explosive energy)
A kiloton of TNT is the amount of energy released in the detonation of 1,000 metric tons of TNT. It equates to 4.184 terajoules
## Calories (mean) to Kilotons (metric, explosive energy) table
Start
Increments
Accuracy
Format
Print table
< Smaller Values Larger Values >
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# Continuous partials implies differentiable
## Statement
Suppose $f$ is a real-valued function of $n$ variables $x_1,x_2,\dots,x_n$. Suppose that $(a_1,a_2,\dots,a_n)$ is a point in the domain of $f$ such that the partial derivatives $f_{x_1}, f_{x_2},\dots,f_{x_n}$ exist and are continuous at and around the point $(a_1,a_2,\dots,a_n)$ (i.e., they all exist and are continuous in an open ball containing $(a_1,a_2,\dots,a_n)$). Then, the gradient vector of $f$ exists at $(a_1,a_2,\dots,a_n)$ and is given by (as per relation between gradient vector and partial derivatives):
$(\nabla f)(a_1,a_2,\dots,a_n) = \langle f_{x_1}(a_1,a_2,\dots,a_n), f_{x_2}(a_1,a_2,\dots,a_n), \dots, f_{x_n}(a_1,a_2,\dots,a_n) \rangle$
Note that we say a function of multiple variables is differentiable if the gradient vector exists, hence this result can be restated as continuous partials implies differentiable.
## Significance
In general, computing partial derivatives is easy, but computing the gradient vector from first principles is hard. Fortunately, we have a result, called the relation between gradient vector and partial derivatives, that says that the coordinates of the gradient vector, wherever it does exist, are simply the partial derivatives. Explicitly, it says that:
$(\nabla f)(a_1,a_2,\dots,a_n) = \langle f_{x_1}(a_1,a_2,\dots,a_n), f_{x_2}(a_1,a_2,\dots,a_n), \dots, f_{x_n}(a_1,a_2,\dots,a_n) \rangle$
The problem with that result is that it holds conditional to the existence of the gradient vector. In fact, existence of partial derivatives not implies differentiable. Thus, after computing the partial derivatives, we know what the gradient vector should be if it exists, but we still aren't sure whether it exists.
That is where the result of this page comes in useful. It allows us to use the continuity of the partial derivatives, that we have already computed, to deduce the existence of the gradient vector, and hence also compute an expression for it.
## Related facts
### Converse
The converse is not true: the existence of the gradient vector does not imply the continuity of the partial derivatives, and it does not even imply the existence of partial derivatives around the point. In fact, we can construct counter-examples with functions of one variable. See differentiable not implies continuously differentiable.
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Balance Mesh Density for Smooth Geometry
Concept Procedure Quick Reference
When you render a model, the density of the mesh affects the smoothness of surfaces.
Mesh components are comprised of vertices, faces, polygons, and edges.
• A vertex is a point that forms the corner of a face or polygon.
• A face is a triangular portion of a surface object.
• A polygon is a quadrilateral portion of a surface object.
• An edge is the boundary of a face or polygon.
In a drawing, all faces have three vertices, except faces in polyface meshes, which are treated as adjoining triangles. For rendering purposes, each quadrilateral face is a pair of triangular faces that share one edge.
Smoothing of an object is handled automatically by the renderer. Two types of smoothing occur during the rendering process. One smoothing operation interpolates the face normals across a surface. The other operation takes into account the number of faces, the face count, that make up the geometry; greater face counts result in smoother surfaces but longer processing times.
While you cannot control the interpolation of face normals, you can control the display accuracy of curved objects by using the VIEWRES command and the FACETRES system variable.
Control Display of Circles and Arcs
The VIEWRES command controls the display accuracy of curved 2D linework like circles and arcs in the current viewport.
In the following example, line segments are more apparent as VIEWRES decreases - Upper left = 1000, Middle = 100, Lower right = 10.
These objects are drawn on the screen using many short straight line segments. Smoother arcs and circles display with higher VIEWRES settings, but they take longer to regenerate. To increase performance while you're drawing, set a low VIEWRES value.
Control Display of Curved Solids
FACETRES controls the mesh density and smoothness of shaded and rendered curved solids.
In the following example, facets display on curved geometry when FACETRES is low. FACETRES = .25.
When FACETRES is set to 1, there is a one-to-one correlation between the viewing resolution of circles and arcs and the tessellation, a means of subdividing the faces of solid objects. For example, when FACETRES is set to 2, the tessellation will be twice the tessellation set by VIEWRES. The default value of FACETRES is 0.5. The range of possible values is 0.01 to 10.
When you raise and lower the value of VIEWRES, objects controlled by both VIEWRES and FACETRES are affected. When you raise and lower the value of FACETRES, only solid objects are affected.
In the following example, smoother geometry is displayed when FACETRES is set to higher values. FACETRES = 5.
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https://www.gradesaver.com/textbooks/math/calculus/thomas-calculus-13th-edition/chapter-1-functions-section-1-1-functions-and-their-graphs-exercises-1-1-page-12/25
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## Thomas' Calculus 13th Edition
$f(0) = 0$ $f(1) = 1$ So, the first part of the graph will be a straight line from (0,0) to (1,1). $f(2) = 2-2$ $f(2) = 0$ So, the second part of the graph will be a straight line from (1,1) to (2,0).
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A117610 A Matrix Markov based on solved permutation Matrices Modulo 15 as 8 X 8 matrices extracted from the primes relative to the first set of eight primes free of {3,5}. 0
%I
%S 7,11,13,2,4,8,14,1,11,13,2,8,14,1,7,11,2,14,1,7,11,13,2,8,11,13,2,14,
%T 7,11,2,8,11,2,14,11,2,8,13,2,11,2,2,14,11,2,8,13,13,2,14,2,8,13,11,2,
%U 14,13,13,2,14,2,2,13,13,13,2,14,2,2,13,14
%N A Matrix Markov based on solved permutation Matrices Modulo 15 as 8 X 8 matrices extracted from the primes relative to the first set of eight primes free of {3,5}.
%C This method was a difficult model to program: it bifurcated at higher iterations to gives mostly {2,13,14} leaving out the other values. Observationally in terms of the modulo 10 endings {1,3,7,9} the modulo 15 ending pair as: 1 --> {1,11},3 --> {8,13},7 --> {2,7},9 --> {4,14} The idea is that an elliptically polarized partitioning of the primes should behave as permutation of these eight modulo 15 endings.
%F v[n]=vector v[n-1] permutated by Matrix M[n] a(n+m-1) =v[n][[m]]
%t (*a-> Prime[4]to Prime[12 modulo 15 as the reference sequence*) a = {7, 11, 13, 2, 4, 8, 14, 1}; (* finds permutations of the reference sequence to match the actual primes*) M = Table[Table[If[Mod[Prime[i + n], 15] - a[[m]] == 0, 1, 0], {n, 1, 8}, {m, 1, 8}], {i, 4, 68, 8}]; (* matrix Markov switches the permutation matrics in order*) v[0] = a; v[n_] := v[n] = M[[1 + Mod[n, 8]]].v[n - 1] a0 = Flatten[Table[v[n][[m]], {n, 0, 8}, {m, 1, 8}]]
%K nonn,uned
%O 0,1
%A _Roger L. Bagula_, Apr 06 2006
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# Readers ask: Suppose A Car Moves At Constant Speed Along A Hilly Road Where Does The Car Exert Quiz?
## When a car moves at constant speed on a straight road?
The acceleration of a car that travels in a straight line at a constant speed of 100 km/h is zero. Average acceleration = (change in velocity)/(time it takes). Since the car’s change in velocity is zero, its acceleration is zero.
## Is a car turning at a constant speed accelerating?
For example, if a car turns a corner at constant speed, it is accelerating because its direction is changing. The quicker you turn, the greater the acceleration. So there is an acceleration when velocity changes either in magnitude (an increase or decrease in speed) or in direction, or both.
You might be interested: What Happens If I Road Trip With My Car A Lot?
## Will the acceleration of a car be the same when a car travels around a sharp curve at a constant 60 km/h as when it travels around a gentle curve at the same speed?
6) Will the acceleration of a car be the same if it travels around a sharp curve at 60 km/hr as when it travels around a gentle curve at the same speed? Answer: No the acceleration will be greater for the sharp curve than for the gentle curve.
## What is true about a car with constant speed?
An object’s acceleration is the rate its velocity (speed and direction) changes. Therefore, an object can accelerate even if its speed is constant – if its direction changes. If an object’s velocity is constant, however, its acceleration will be zero. Since it travels in a straight line, its direction does not change.
## What forces act on a car at constant speed?
For example, when a car travels at a constant speed, the driving force from the engine is balanced by resistive forces such as air resistance and friction in the car’s moving parts. The resultant force on the car is zero.
## What is constant speed?
Definition: When the speed of an object remains the same – it does not increase or decrease – we say it is moving at a constant speed.
## Can u go around a curve with constant acceleration?
Can you go around a curve with constant acceleration? No, because the vector always points to the center of the circle and you are always changing the direction of that vector.
## Can a body have uniform speed but still have acceleration?
Since speed is equal to the magnitude of the velocity vector, it will also be constant. Again, since the acceleration is the rate of change of velocity of a body with time, if the velocity is constant, the acceleration of the body will be zero. Hence, a body can have a constant speed and still be accelerating.
You might be interested: What To Do If Your Car Slides Off The Road?
## What is tangential acceleration formula?
The tangential acceleration = radius of the rotation * its angular acceleration. It is always measured in radian per second square. Its dimensional formula is [T2]. When an object makes a circular motion, it experiences both tangential and centripetal acceleration.
## How many accelerators do you have in your car there are at least three controls?
Thus, your car actually has at least 3 accelerators: (1) the foot pedal called the “accelerator”, that changes the magnitude of the velocity, (2) the brake, which also changes the magnitude of the velocity, and (3) the steering wheel, which changes the direction of the velocity!
## At which point a B or C is the normal force acting on the car the largest?
The normal force on the car is largest at point C. In this case, the centripetal force keeping the car in a circular path of radius R is directed upward, so the normal force must be greater than the weight to provide this net upward force. (b) The normal force is smallest at point A, the crest of the hill.
## Which pulls harder the earth on the moon or the moon on the earth explain?
Which pulls harder gravitationally, the Earth on the moon or the moon on the Earth? The gravitational force on the moon due to the earth is only about half the force on the moon due to the sun.
## Does constant speed mean no acceleration?
The velocity vector is constant in magnitude but changing in direction. Because the speed is constant for such a motion, many students have the misconception that there is no acceleration. For this reason, it can be safely concluded that an object moving in a circle at constant speed is indeed accelerating.
You might be interested: Readers ask: How Long Can My Car Sit On The Side Of The Road?
## What is the acceleration of light at 300 000 m s?
Its direction does not shift because it moves in a straight line. Light’s velocity never varies since both its speed and direction are unchanged, so its acceleration is zero.
## What is a situation where you can accelerate even though your speed doesn’t change?
Acceleration therefore occurs whenever an object changes direction- for example, a car driving around a roundabout is constantly accelerating even if its speed does not change.
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https://nl.mathworks.com/matlabcentral/cody/problems/2020-area-of-an-isoceles-triangle/solutions/1836759
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Cody
# Problem 2020. Area of an Isoceles Triangle
Solution 1836759
Submitted on 3 Jun 2019 by MATTHEW RYCKMAN
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 5; y = 8; A_correct = 12; tolerance = 1e-12; assert(abs(isocelesArea(x,y)-A_correct)<tolerance)
A = 12
2 Pass
x = 2; y = 2; A_correct = sqrt(3); tolerance = 1e-12; assert(abs(isocelesArea(x,y)-A_correct)<tolerance)
A = 1.7321
3 Pass
x = 10; y = 2; A_correct = sqrt(99); tolerance = 1e-12; assert(abs(isocelesArea(x,y)-A_correct)<tolerance)
A = 9.9499
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https://wviechtb.github.io/metafor/reference/rma.mh.html
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Function to fit equal-effects models to $$2 \times 2$$ table and person-time data via the Mantel-Haenszel method. See below and the introduction to the metafor-package for more details on these models.
rma.mh(ai, bi, ci, di, n1i, n2i, x1i, x2i, t1i, t2i,
measure="OR", data, slab, subset,
correct=TRUE, level=95, verbose=FALSE, digits, ...)
## Arguments
ai
vector with the $$2 \times 2$$ table frequencies (upper left cell). See below and the documentation of the escalc function for more details.
bi
vector with the $$2 \times 2$$ table frequencies (upper right cell). See below and the documentation of the escalc function for more details.
ci
vector with the $$2 \times 2$$ table frequencies (lower left cell). See below and the documentation of the escalc function for more details.
di
vector with the $$2 \times 2$$ table frequencies (lower right cell). See below and the documentation of the escalc function for more details.
n1i
vector with the group sizes or row totals (first group). See below and the documentation of the escalc function for more details.
n2i
vector with the group sizes or row totals (second group). See below and the documentation of the escalc function for more details.
x1i
vector with the number of events (first group). See below and the documentation of the escalc function for more details.
x2i
vector with the number of events (second group). See below and the documentation of the escalc function for more details.
t1i
vector with the total person-times (first group). See below and the documentation of the escalc function for more details.
t2i
vector with the total person-times (second group). See below and the documentation of the escalc function for more details.
measure
character string to specify the outcome measure to use for the meta-analysis. Possible options are "RR" for the (log transformed) risk ratio, "OR" for the (log transformed) odds ratio, "RD" for the risk difference, "IRR" for the (log transformed) incidence rate ratio, or "IRD" for the incidence rate difference.
data
optional data frame containing the data supplied to the function.
slab
optional vector with labels for the $$k$$ studies.
subset
optional (logical or numeric) vector to specify the subset of studies that should be used for the analysis.
non-negative number to specify the amount to add to zero cells or even counts when calculating the observed effect sizes of the individual studies. Can also be a vector of two numbers, where the first number is used in the calculation of the observed effect sizes and the second number is used when applying the Mantel-Haenszel method. See below and the documentation of the escalc function for more details.
to
character string to specify when the values under add should be added (either "only0", "all", "if0all", or "none"). Can also be a character vector, where the first string again applies when calculating the observed effect sizes or outcomes and the second string when applying the Mantel-Haenszel method. See below and the documentation of the escalc function for more details.
drop00
logical to specify whether studies with no cases/events (or only cases) in both groups should be dropped when calculating the observed effect sizes or outcomes (the outcomes for such studies are set to NA). Can also be a vector of two logicals, where the first applies to the calculation of the observed effect sizes or outcomes and the second when applying the Mantel-Haenszel method. See below and the documentation of the escalc function for more details.
correct
logical to specify whether to apply a continuity correction when computing the Cochran-Mantel-Haenszel test statistic.
level
numeric value between 0 and 100 to specify the confidence interval level (the default is 95; see here for details).
verbose
logical to specify whether output should be generated on the progress of the model fitting (the default is FALSE).
digits
optional integer to specify the number of decimal places to which the printed results should be rounded. If unspecified, the default is 4. See also here for further details on how to control the number of digits in the output.
...
## Details
### Specifying the Data
When the outcome measure is either the risk ratio (measure="RR"), odds ratio (measure="OR"), or risk difference (measure="RD"), the studies are assumed to provide data in terms of $$2 \times 2$$ tables of the form:
outcome 1 outcome 2 total group 1 ai bi n1i group 2 ci di n2i
where ai, bi, ci, and di denote the cell frequencies and n1i and n2i the row totals. For example, in a set of randomized clinical trials (RCTs) or cohort studies, group 1 and group 2 may refer to the treatment/exposed and placebo/control/non-exposed group, respectively, with outcome 1 denoting some event of interest (e.g., death) and outcome 2 its complement. In a set of case-control studies, group 1 and group 2 may refer to the group of cases and the group of controls, with outcome 1 denoting, for example, exposure to some risk factor and outcome 2 non-exposure. For these outcome measures, one needs to specify the cell frequencies via the ai, bi, ci, and di arguments (or alternatively, one can use the ai, ci, n1i, and n2i arguments).
Alternatively, when the outcome measure is the incidence rate ratio (measure="IRR") or the incidence rate difference (measure="IRD"), the studies are assumed to provide data in terms of tables of the form:
events person-time group 1 x1i t1i group 2 x2i t2i
where x1i and x2i denote the number of events in the first and the second group, respectively, and t1i and t2i the corresponding total person-times at risk.
### Mantel-Haenszel Method
An approach for aggregating data of these types was suggested by Mantel and Haenszel (1959) and later extended by various authors (see references). The Mantel-Haenszel method provides a weighted estimate under an equal-effects model. The method is particularly advantageous when aggregating a large number of studies with small sample sizes (the so-called sparse data or increasing strata case).
When analyzing odds ratios, the Cochran-Mantel-Haenszel (CMH) test (Cochran, 1954; Mantel & Haenszel, 1959) and Tarone's test for heterogeneity (Tarone, 1985) are also provided (by default, the CMH test statistic is computed with the continuity correction; this can be switched off with correct=FALSE). When analyzing incidence rate ratios, the Mantel-Haenszel (MH) test (Rothman et al., 2008) for person-time data is also provided (again, the correct argument controls whether the continuity correction is applied). When analyzing risk ratios, odds ratios, or incidence rate ratios, the printed results are given both in terms of the log and the raw units (for easier interpretation).
### Observed Effect Sizes or Outcomes of the Individual Studies
The Mantel-Haenszel method itself does not require the calculation of the observed effect sizes of the individual studies (e.g., the observed log odds ratios of the $$k$$ studies) and directly makes use of the cell/event counts. Zero cells/events are not a problem (except in extreme cases, such as when one of the two outcomes never occurs in any of the $$2 \times 2$$ tables or when there are no events for one of the two groups in any of the tables). Therefore, it is unnecessary to add some constant to the cell/event counts when there are zero cells/events.
However, for plotting and various other functions, it is necessary to calculate the observed effect sizes for the $$k$$ studies. Here, zero cells/events can be problematic, so adding a constant value to the cell/event counts ensures that all $$k$$ values can be calculated. The add and to arguments are used to specify what value should be added to the cell/event counts and under what circumstances when calculating the observed effect sizes and when applying the Mantel-Haenszel method. Similarly, the drop00 argument is used to specify how studies with no cases/events (or only cases) in both groups should be handled. The documentation of the escalc function explains how the add, to, and drop00 arguments work. If only a single value for these arguments is specified (as per default), then these values are used when calculating the observed effect sizes and no adjustment to the cell/event counts is made when applying the Mantel-Haenszel method. Alternatively, when specifying two values for these arguments, the first value applies when calculating the observed effect sizes and the second value when applying the Mantel-Haenszel method.
Note that drop00 is set to TRUE by default. Therefore, the observed effect sizes for studies where ai=ci=0 or bi=di=0 or studies where x1i=x2i=0 are set to NA. When applying the Mantel-Haenszel method, such studies are not explicitly dropped (unless the second value of drop00 argument is also set to TRUE), but this is practically not necessary, as they do not actually influence the results (assuming no adjustment to the cell/event counts are made when applying the Mantel-Haenszel method).
## Value
An object of class c("rma.mh","rma"). The object is a list containing the following components:
beta
aggregated log risk ratio, log odds ratio, risk difference, log rate ratio, or rate difference.
se
standard error of the aggregated value.
zval
test statistics of the aggregated value.
pval
corresponding p-value.
ci.lb
lower bound of the confidence interval.
ci.ub
upper bound of the confidence interval.
QE
test statistic of the test for heterogeneity.
QEp
correspinding p-value.
MH
Cochran-Mantel-Haenszel test statistic (measure="OR") or Mantel-Haenszel test statistic (measure="IRR").
MHp
corresponding p-value.
TA
test statistic of Tarone's test for heterogeneity (only when measure="OR").
TAp
corresponding p-value (only when measure="OR").
k
number of studies included in the analysis.
yi, vi
the vector of outcomes and corresponding sampling variances.
fit.stats
a list with the log-likelihood, deviance, AIC, BIC, and AICc values under the unrestricted and restricted likelihood.
...
## Methods
The results of the fitted model are formatted and printed with the print function. If fit statistics should also be given, use summary (or use the fitstats function to extract them).
The residuals, rstandard, and rstudent functions extract raw and standardized residuals. Leave-one-out diagnostics can be obtained with leave1out.
Forest, funnel, radial, L'Abbé, and Baujat plots can be obtained with forest, funnel, radial, labbe, and baujat. The qqnorm function provides normal QQ plots of the standardized residuals. One can also just call plot on the fitted model object to obtain various plots at once.
A cumulative meta-analysis (i.e., adding one observation at a time) can be obtained with cumul.
Other extractor functions include coef, vcov, logLik, deviance, AIC, and BIC.
## Author
Wolfgang Viechtbauer (wvb@metafor-project.org, https://www.metafor-project.org).
## References
Cochran, W. G. (1954). Some methods for strengthening the common $$\chi^2$$ tests. Biometrics, 10(4), 417–451. https://doi.org/10.2307/3001616
Greenland, S., & Robins, J. M. (1985). Estimation of a common effect parameter from sparse follow-up data. Biometrics, 41(1), 55–68. https://doi.org/10.2307/2530643
Mantel, N., & Haenszel, W. (1959). Statistical aspects of the analysis of data from retrospective studies of disease. Journal of the National Cancer Institute, 22(4), 719–748. https://doi.org/10.1093/jnci/22.4.719
Nurminen, M. (1981). Asymptotic efficiency of general noniterative estimators of common relative risk. Biometrika, 68(2), 525–530. https://doi.org/10.1093/biomet/68.2.525
Robins, J., Breslow, N., & Greenland, S. (1986). Estimators of the Mantel-Haenszel variance consistent in both sparse data and large-strata limiting models. Biometrics, 42(2), 311–323. https://doi.org/10.2307/2531052
Rothman, K. J., Greenland, S., & Lash, T. L. (2008). Modern epidemiology (3rd ed.). Philadelphia: Lippincott Williams & Wilkins.
Sato, T., Greenland, S., & Robins, J. M. (1989). On the variance estimator for the Mantel-Haenszel risk difference. Biometrics, 45(4), 1323–1324. https://www.jstor.org/stable/2531784
Tarone, R. E. (1981). On summary estimators of relative risk. Journal of Chronic Diseases, 34(9-10), 463–468. https://doi.org/10.1016/0021-9681(81)90006-0
Tarone, R. E. (1985). On heterogeneity tests based on efficient scores. Biometrika, 72(1), 91–95. https://doi.org/10.1093/biomet/72.1.91
Viechtbauer, W. (2010). Conducting meta-analyses in R with the metafor package. Journal of Statistical Software, 36(3), 1–48. https://doi.org/10.18637/jss.v036.i03
rma.uni, rma.glmm, rma.peto, and rma.mv for other model fitting functions.
## Examples
### meta-analysis of the (log) odds ratios using the Mantel-Haenszel method
rma.mh(measure="OR", ai=tpos, bi=tneg, ci=cpos, di=cneg, data=dat.bcg)
#>
#> Equal-Effects Model (k = 13)
#>
#> I^2 (total heterogeneity / total variability): 92.68%
#> H^2 (total variability / sampling variability): 13.66
#>
#> Test for Heterogeneity:
#> Q(df = 12) = 163.9426, p-val < .0001
#>
#> Model Results (log scale):
#>
#> estimate se zval pval ci.lb ci.ub
#> -0.4734 0.0410 -11.5444 <.0001 -0.5538 -0.3930
#>
#> Model Results (OR scale):
#>
#> estimate ci.lb ci.ub
#> 0.6229 0.5748 0.6750
#>
#> Cochran-Mantel-Haenszel Test: CMH = 135.6889, df = 1, p-val < 0.0001
#> Tarone's Test for Heterogeneity: X^2 = 171.7567, df = 12, p-val < 0.0001
#>
### meta-analysis of the (log) risk ratios using the Mantel-Haenszel method
rma.mh(measure="RR", ai=tpos, bi=tneg, ci=cpos, di=cneg, data=dat.bcg)
#>
#> Equal-Effects Model (k = 13)
#>
#> I^2 (total heterogeneity / total variability): 92.13%
#> H^2 (total variability / sampling variability): 12.71
#>
#> Test for Heterogeneity:
#> Q(df = 12) = 152.5676, p-val < .0001
#>
#> Model Results (log scale):
#>
#> estimate se zval pval ci.lb ci.ub
#> -0.4537 0.0393 -11.5338 <.0001 -0.5308 -0.3766
#>
#> Model Results (RR scale):
#>
#> estimate ci.lb ci.ub
#> 0.6353 0.5881 0.6862
#>
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## Question 8.2
Expand Messages
• Question 8.2 asks Consider a knowledge base containing just two sentences: P(a) and P (b). Does this knowledge base entail For All x P(x)? Explain in terms of
Message 1 of 1 , Nov 13, 2003
"Consider a knowledge base containing just two sentences: P(a) and P
(b). Does this knowledge base entail For All x P(x)? Explain in terms
of models"
The answer, as noted the the authors answer sheets is No, the KB does
not entail this.
If the knowledge base only has those two facts, then that is the
agents universe. That is all it knows. So I would like to argue that
For All x P(x) is true. If the KB contained other facts, then this
might not be true.
If the agent in Wumpus world knew of two squares (only two squares,
no knowledge that any other squares exists) and each square had a pit
in it, can we not say that For every square in the Agents Wumpus
world, that sqaure has a pit?
Isn't each agents universe different? All kings are persons is true
in our universe, but certainly not in EVERY universe.
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When examining primary sources, it is important to evaluate the creator’s __________. purpose personality charisma effort
When examining primary sources, it is important to evaluate the creator's : PURPOSE.
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# How to find a rational number between 3 and 4?
• Last Updated : 09 Aug, 2021
In mathematics, a rational number is a kind of real number of the form p/q where q is not equal to 0. If the denominator and numerator are both integers and the denominator is not zero, we can categorize any fraction as a rational number. The outcome of splitting a rational number is a decimal number, which can be either a terminating or recurring decimal.
Examples of Rational Numbers
17 and -34 are rational numbers. It’s worth noting that the same rational number can be written in several ways as a ratio of integers.
7 and 21⁄3 are the same rational number.
### How to Find the Rational Numbers between Two Rational Numbers?
Between two rational numbers, there exist “n” numbers of rational numbers. Two alternative approaches can be used to find the rational numbers between two rational numbers. Let’s have a look at the two distinct approaches.
Approach 1:
Calculate the equivalent fractions of the given rational numbers and calculate the rational numbers in between them. Those figures should be the necessary reasonable figures.
Approach 2:
Calculate the mean of the two rational numbers supplied. The necessary rational number should be the mean value. Repeat the method with the old and newly obtained rational numbers to find more rational numbers.
### How to find a rational number between 3 and 4?
Solution:
Approach 1:
Let us follow the first approach to find out the rational number between 3 and 4.
The equivalent fraction for 3⁄1 can be 6⁄2 and for 4⁄1 can be 16⁄4.
Now, the numbers are 6⁄2 and 16⁄4, so the required rational number can be in between these numbers.
The numerator and denominator of the required number should be between the given number, i.e., numerator can be 10 and denominator can be 3.
Hence, the rational between 3 and 4 is 10⁄3.
Approach 2:
Let us follow the second approach to find out the rational number between 3 and 4.
The formula to calculate the mean is given as:
m = sum of the terms/number of the terms
Here, the given terms are 3 and 4, so the mean is:
m = (3 + 4)/2 = 7/2 = 3.5
Hence, the rational number between 3 and 4 is 7/2 or 3.5.
### Similar Questions
Problem 1: What is the rational number between 7 and 9?
Solution:
Here, the given terms are 7 and 9, so the mean is:
m = (7 + 9) / 2 = 16/2 = 8
Problem 2: What is the rational number between 1 and 4?
Solution:
Here, the given terms are 1 and 4, so the mean is:
m = (1 + 4)/2 = 5/2 = 2.5
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+ 1
# It's a integer or float?
My code works like this I input a random number program will divide the number and say wether the number is integer or float. I want to find the data type of number by using Boolean. I don't know who to do
18th Dec 2018, 12:14 PM
Shyamal Bhatt
+ 3
You could use modf from math.h library, in order to extract decimal part and check if it is zero. Example: https://code.sololearn.com/c29OHjXbSokC/?ref=app
18th Dec 2018, 4:09 PM
Javier Felipe Toribio
+ 3
TheZeroDoctor (result==(int)result) fails for big numbers. For example: 9999999999 / 3 = integer 3333333333. But your program shows result as float, because, result is 3333333248.0 !!! Using == with floats and double values which are results from a calculation is risky. modf detects that decimal is 0 in 9999999999 / 3 and you detect it's an integer
20th Dec 2018, 2:16 PM
Javier Felipe Toribio
18th Dec 2018, 12:51 PM
Lighton
+ 2
Javier Felipe Toribio you are right. You can do it by this way too but why? I think that it's make more complex code and it's not necessary.
20th Dec 2018, 1:03 PM
Lighton
+ 1
you can cast the number(result) to integer and check if it is equals to the number(result) or it is different. If it is the same thing that means that it was integer already. Otherwise it is a float
18th Dec 2018, 12:21 PM
Lighton
+ 1
Javier Felipe Toribio i agree, program wasn't work for big numbers
20th Dec 2018, 2:22 PM
Shyamal Bhatt
+ 1
With 99999999999 / 3 it still fails. Calculation with big numbers is inaccurate. Besides, types int, long int, are limited by its maximum value and you don't know how big the input number is. I think using casting is not a good way to check if something is integer or not.
20th Dec 2018, 5:22 PM
Javier Felipe Toribio
0
TheZeroDoctor can you write the code and post it , I will be thankful if you do that
18th Dec 2018, 12:24 PM
Shyamal Bhatt
0
Okay, wait a minute
18th Dec 2018, 12:31 PM
Lighton
0
TheZeroDoctor Thanks man
18th Dec 2018, 12:54 PM
Shyamal Bhatt
0
Javier Felipe Toribio can you explain again why it is not working with big numbers? And check again my code, I do some changes
20th Dec 2018, 4:52 PM
Lighton
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0
Q:
# The curved surface area and the volume of a cylinder are 264 sq.cm and 924 cub.cm, respectively. What is the ratio of its radius to height?
A) 4:3 B) 5:4 C) 7:6 D) 3:2
Explanation:
Q:
In triangle ABC, the length of BC is less than twice the length of AB by 2 cm. The length of AC exceeds the length of AB by 10 cm. The perimeter is 32 cm. The length (in cm)of the smallest side of the triangle is:
A) 10 B) 4 C) 6 D) 8
Explanation:
0 189
Q:
In triangle ABC, the length of BC is less than twice the length of AB by 3 cm. The length of AC exceeds the length of AB by 9 cm. The perimeter of triangle is 34 cm. The length (in cm) of the smallest side of the triangle is:
A) 9 B) 10 C) 8 D) 7
Explanation:
2 295
Q:
A circle is inscribed in a triangle ABC. It touches sides AB, BC and AC at the points P, Q and R respectively. If BP = 5.4 cm, CQ =7.3 cm and AR = 6.1 cm, then the perimeter (in cm) of the ABC is:
A) 37 B) 37.6 C) 37.25 D) 36
Explanation:
1 256
Q:
There are two circles of radius 5 cm and 3 cm respectively. The distance between their centres is 10 cm. The length (in cm) of a transverse common tangent is:
A) 9 B) 8 C) 10 D) 6
Explanation:
0 206
Q:
In triangle ABC, the length of BC is less than twice the length of AB by 3 cm. The length of the AC exceeds the length of AB by 1 cm. The perimeter of the triangle is 34 cm. The length (in cm) of the smallest side of the triangle is:
A) 10 B) 7 C) 9 D) 8
Explanation:
0 236
Q:
If each side of a rectangle is increased by 13%, then its area will increase by:
A) 27.69% B) 13% C) 26% D) 21.69%
Explanation:
0 5925
Q:
Chords AB and CD of a circle intersect externally at P. If AB = 6 cm, CD = 3 cm and PB = 4 cm, then the length (in cm) of PD is:
A) 5 B) 7 C) 6 D) 2
Explanation:
0 234
Q:
There is a polygon of 11 sides. How many triangles can be drawn by only using the vertices of the polygon?
A) 180 B) 150 C) 165 D) 175
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# Facility location problem
Authors: Liz Cantlebary, Lawrence Li (CHEME 6800 Fall 2020)
Stewards: Allen Yang, Fengqi You
## Introduction
The Facility Location Problem (FLP) is a classic optimization problem that determines the best location for a factory or warehouse to be placed based on geographical demands, facility costs, and transportation distances. These problems generally aim to maximize the supplier's profit based on the given customer demand and location. FLP can be further broken down into capacitated and uncapacitated problems, depending on whether the facilities in question have a maximum capacity or not.
## Theory and Formulation
### Weber Problem and Single Facility FLPs
The Weber Problem is a simple FLP that consists of locating the geometric median between three points with different weights. The geometric median is a point between three given points in space such that the sum of the distances between the median and the other three points is minimized. It is based on the premise of minimizing transportation costs from one point to various destinations, where each destination has a different associated cost per unit distance.
Given ${\displaystyle N}$ points ${\displaystyle (a_{1},b_{1})...(a_{N},b_{N})}$ on a plane with associated weights ${\displaystyle w_{1}...w_{N}}$, the 2-dimensional Weber problem to find the geometric median ${\displaystyle (x,y)}$ is formulated as(1)
{\displaystyle \min {\begin{aligned}W(x,y)=\sum _{i=1}^{N}w_{i}d_{i}(x,y,a_{i},b_{i})\\\end{aligned}}}
where
${\displaystyle d_{i}(x,y,a_{i},b_{i})={\sqrt {(x-a_{i})^{2}+(y-b_{i})^{2}}}}$
The above formulation serves as a foundation for many basic single facility FLPs. For example, the minisum problem aims to locate a facility at the point that minimizes the sum of the weighted distances to the given set of existing facilities, while the minimax problem consists of placing the facility at the point that minimizes the maximum weighted distance to the existing facilities(8). Additionally, in contrast to the minimax problem, the maximin facility problem maximizes the minimum weighted distance to the given facilities.
### Capacitated and Uncapacitated FLPs
FLPs can often be formulated as mixed-integer programs (MIPs), with a fixed set of facility and customer locations. Binary variables are used in these problems to represent whether a certain facility is open or closed and whether that facility can supply a certain customer. Capacitated and uncapacitated FLPs can be solved this way by defining them as integer programs.
A capacitated facility problem applies constraints to the production and transportation capacity of each facility. As a result, customers may not be supplied by the most immediate facility, since this facility may not be able to satisfy the given customer demand.
In a problem with ${\displaystyle N}$ facilities and ${\displaystyle M}$ customers, the capacitated formulation defines a binary variable ${\displaystyle x_{i}}$ and a variable ${\displaystyle y_{ij}}$ for each facility ${\displaystyle i}$ and each customer ${\displaystyle j}$. If facility ${\displaystyle i}$ is open, ${\displaystyle x_{i}=1}$; otherwise ${\displaystyle x_{i}=0}$. Open facilities have an associated fixed cost ${\displaystyle f_{i}}$ and a maximum capacity ${\displaystyle k_{i}}$. ${\displaystyle y_{ij}}$ is the fraction of the total demand ${\displaystyle d_{j}}$ of customer ${\displaystyle j}$ that facility ${\displaystyle i}$ has satisfied and the transportation cost between facility ${\displaystyle i}$ and customer ${\displaystyle j}$ is represented as ${\displaystyle t_{ij}}$. The capacitated FLP is therefore defined as(7)
${\displaystyle \min \ \sum _{i=1}^{N}\sum _{j=1}^{M}d_{j}t_{ij}y_{ij}+\sum _{i=1}^{N}f_{i}x_{i}}$
${\displaystyle s.t.\ \sum _{i=1}^{N}y_{ij}=1\ \ \forall \,j\in \{1,...,M\}}$
${\displaystyle \quad \quad \sum _{j=1}^{M}d_{j}y_{ij}\leq k_{i}x_{i}\ \ \forall \,i\in \{1,...,N\}}$
${\displaystyle \quad \quad y_{ij}\geq 0\ \ \forall \,i\in \{1,...,N\},\ \forall \,j\in \{1,...,M\}}$
${\displaystyle \quad \quad x_{i}\in \{0,1\}\ \ \forall \,i\in \{1,...,N\}}$
In an uncapacitated facility problem, the amount of product each facility can produce and transport is assumed to be unlimited, and the optimal solution results in customers being supplied by the lowest-cost, and usually the nearest, facility. Using the above formulation, the unlimited capacity means ${\displaystyle k_{i}}$ can be assumed to be a sufficiently large constant, while ${\displaystyle y_{ij}}$ is now a binary variable, because the demand of each customer can be fully met with the nearest facility(7). If facility ${\displaystyle i}$ supplies customer ${\displaystyle j}$, then ${\displaystyle y_{ij}=1}$; otherwise ${\displaystyle y_{ij}=0}$.
## Numerical Example
Suppose a paper products manufacturer has enough capital to build and manage an additional manufacturing plant in the United States in order to meet increased demand in three cities: New York City, NY, Los Angeles, CA, and Topeka, KS. The company already has distribution facilities in Denver, CO, Seattle, WA, and St. Louis, MO, and due to limited capital, cannot build an additional distribution facility. So, they must choose to build their new plant in one of these three locations. Due to geographic constraints, plants in Denver, Seattle, and St. Louis would have a maximum operating capacity of 400tons/day, 700 tons/day, and 600 tons/day, respectively. The cost of transporting the products from the plant to the city is directly proportional, and an outline of the supply, demand, and cost of transportation is shown in the figure below. Regardless of where the plant is built, the selling price of the product is $100/ton. To solve this problem, we will assign the following variables: ${\displaystyle i}$ is the factory location ${\displaystyle j}$ is the city destination ${\displaystyle C_{ij}}$ is the cost of transporting one ton of product from the factory to the city ${\displaystyle x_{ij}}$ is the amount of product transported from the factory to the city in tons ${\displaystyle A_{i}}$ is the maximum operating capacity at the factory ${\displaystyle D_{j}}$ is the amount of unmet demand in the city To determine where the company should build the factory, we will carry out the following optimization problem for each location to maximize the profit from each ton sold: max ${\displaystyle \sum _{j\epsilon J}x_{ij}(100-C_{ij})}$ subject to ${\displaystyle \sum _{j\epsilon J}x_{ij}\leq A_{i}}$ ${\displaystyle \forall i\epsilon I}$ ${\displaystyle \sum _{i\epsilon I}x_{ij}\leq D_{j}}$ ${\displaystyle \forall j\epsilon J}$ ${\displaystyle x_{ij}\geq 0}$ ${\displaystyle \forall i\epsilon I}$ ${\displaystyle \forall j\epsilon J}$ If the factory is built in Denver, 300tons/day of product go to Los Angeles and 100tons/day go to Topeka, for a total profit of$36,300/day.
If the factory is built in Seattle, 300tons/day of product go to Los Angela, 100tons/day of product go to Topeka, and 300tons/day go to New York City, for a total profit of $56,500/day. If the factory is built in St. Louis, 100tons/day of product go to Topeka and 500tons/day go to New York City, for a total profit of$55,200.
Therefore, to maximize profit, the factory should be built in Seattle.
## Applications
Facility location problems are utilized in many industries to find the optimal placement of various facilities, including warehouses, power plants, public transportation terminals, polling locations, and cell towers, to maximize efficiency, impact, and profit. In more unique applications, extensive research has been done in applying FLPs to humanitarian efforts, such as identifying disaster management sites to maximize accessibility to healthcare and treatment(4). A case study by Nigerian researchers discussed the application of mixed-integer FLPs in optimizing the locations of waste collection centers to provide sanitation services in crucial communities. More effective waste collection systems could combat unsanitary practices and environmental pollution, which are major concerns in many developing nations(2).
FLPs have also been used in clustering analysis, which involves partitioning a given set of elements (e.g. data points) into different groups based on the similarity of the elements. The elements can be placed into groups by identifying the locations of center points that effectively partition the set into clusters, based on the distances from the center points to each element(9).
## Conclusion
The facility location problems is an important application of computational optimization. The uses of this optimization technique are far-reaching, and can be used to determine anything from where a family should live based on the location of their workplaces and school to where a Fortune 500 company should put a new manufacturing plant or distribution facility to maximize their return on investment.
## References
1. http://www.pitt.edu/~lol11/ie1079/notes/ie2079-weber-slides.pdf
2. Adeleke, O. J.; Olukanni, D. O. (2020), Facility Location Problems: Models, Techniques, and Applications in Waste Management. Recycling, 5, 10.
3. Balcik, B.; Beamon, B. M. (2008), Facility Location in Humanitarian Relief. International Journal of Logistics Research and Applications, 11, 101-121.
4. Daskin, M. S.; Dean, L. K. (2004), Location of Health Care Facilities. Handbook of OR/MS in Health Care: A Handbook of Methods and Applications, 43-76.
5. Drezner, Z; Hamacher. H. W. (2004), Facility Location Applications and Theory. New York, NY: Springer.
6. Eiselt, H. A.; Marianov, V. (2019), Contributions to Location Analysis. Cham, Switzerland: Springer.
7. Francis, R. L.; Mirchandani, P. B. (1990), Discrete Location Theory. New York, NY: Wiley.
8. Hansen, P., et al. (1985), The Minisum and Minimax Location Problems Revisited. Operations Research, 33, 6, 1251–1265.
9. Meira, L. A. A.; Miyazawa, F. K.; Pedrosa, L. L. C. (2017), Clustering through Continuous Facility Location Problems. Theoretical Computer Science, 657, 137-145.
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# zbMATH — the first resource for mathematics
Hoops and fuzzy logic. (English) Zbl 1039.03016
BL-algebras are the algebras of Hájek’s basic logic [see P. Hájek, Metamathematics of fuzzy logic. Trends in Logic – Studia Logica Library 4, Kluwer Academic Publishers, Dordrecht (1998; Zbl 0937.03030)]. Working in the neighbourhood of BL-algebras and their $$0$$-free subreducts, the authors investigate the positive fragment of various logics related to Basic Logic. All the resulting structures are particular cases of Büchi-Owens hoops. Just as Basic Logic is the logic of continuous t-norms, MTL is the logic of left-continuous t-norms. The authors also investigate the weaker structures arising from MTL, called semi-hoops. They provide axiomatizations and prove completeness and conservativeness results for several such logics. Turning attention to predicate logics, the authors prove further completeness results with respect to safe interpretations over the corresponding totally ordered structures, generalizing previous results of Hájek for Basic Logic. Mutual interpretability issues are dealt with in a final section. Since all interpretations turn out to be computable in polynomial time, various co-NP-completeness results for the tautology problem of these logics are obtained from the well-known corresponding results about Łukasiewicz logic [D. Mundici, “Satisfiability in many-valued sentential logic is NP-complete”, Theor. Comput. Sci. 52, 145–153 (1987; Zbl 0639.03042)] and about Basic Logic etc. [M. Baaz, P. Hájek, F. Montagna, and H. Veith, “Complexity of t-tautologies”, Ann. Pure Appl. Logic 113, 3–11 (2002; Zbl 1006.03022)].
##### MSC:
03B52 Fuzzy logic; logic of vagueness 06D35 MV-algebras 03D15 Complexity of computation (including implicit computational complexity) 68Q17 Computational difficulty of problems (lower bounds, completeness, difficulty of approximation, etc.)
Full Text:
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12 questions linked to/from How are primes generated for RSA?
87k views
How can I generate large prime numbers for RSA?
What is the currently industry-standard algorithm used to generate large prime numbers to be used in RSA encryption? I'm aware that I can find any number of articles on the Internet that explain how ...
58k views
How does asymmetric encryption work?
I've always been interested in encryption but I have never found a good explanation (beginners explanation) of how encryption with public key and decryption with private key works. How does it ...
9k views
How are random numbers for RSA generated? [duplicate]
The RSA public key encryption requires two very large prime numbers as part of its encryption process that serve as secrets. These are typically generated with cryptographically secure random number ...
3k views
Can I select a large random prime using this procedure?
Say I want a random 1024-bit prime $p$. The obviously-correct way to do this is select a random 1024-bit number and test its primality with the usual well-known tests. But suppose instead that I do ...
6k views
Generating Random Primes
Although this has been extensively discussed around here, I'm curious whether my approach makes sense, or I should just stick to "the standard version". I'm implementing some homomorphic encryption ...
2k views
Example of cryptography random number
I read that random numbers are being used in cryptography and security. I think I have idea how to truly generate true random, non-deterministic number. But before continuing further I'd like to ask ...
892 views
Is it possible to correct the exponent of an RSA public key if it has been altered?
Is it possible to change the exponent of an RSA public key?
317 views
Does it necessarily mean that an RSA moduli generated with poor randomness is not random?
In 2012 a group of researchers collected a large amount of RSA moduli and calculated their greatest common divisor in order to find common factors between them. By finding a common factor they could ...
723 views
What is the danger if a non-prime is chosen for RSA? [duplicate]
I was reading this question about generating primes for RSA keys. The answers point out that most implementations of of the algorithm use probabilistic prime-ness checking algorithms. The answer by @...
554 views
Prime numbers in RSA encryption [duplicate]
I'm studying how the selection of prime numbers in RSA encryption may affect the security of the encryption in regards to the public key. Essentially, are there any specific types of prime numbers ...
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https://www.quantumstudy.com/an-astronomical-telescope-has-an-angular-magnification-of-magnitude-5-for-far-objects/
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# An astronomical telescope has an angular magnification of magnitude 5 for far objects…
Q: An astronomical telescope has an angular magnification of magnitude 5 for far objects. The separation between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length fo of the objective and the focal length fe of the eyepiece are
(A) fo = 45 cm and fe = – 9 cm
(B) fo = 50 cm and fe = 10 cm
(C) fo = 7.2 cm and fe = 5 cm
(D) fo = 30 cm and fe = 6 cm
Ans: (D)
Sol: $\large \frac{f_o}{f_e} = 5$ …(i)
$\large f_o + f_e = 36$ …(ii)
On solving , we get
fo = 30 cm and fe = 6 cm
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# Ultimate List of College Science Fair Projects
This article lists out the various science fair projects available for college students.The topics covered in this article include anatomy, biology, physics, chemistry and many more.
College science fairs have been a trend in many universities across the world as science students get to showcase their projects.
Students are expected to make innovative designs on at least one of their projects while having a topic related to their subject to be discussed in detail during the fair.
For college students, these fairs are an excellent opportunity to showcase their learning while organizing the required resources. We will guide you through the science project ideas that you can make use of during your college science fair.
Whatever you choose, make sure that your project is well-researched and that you have spent enough time pondering the questions that you want to answer. With a bit of hard work and a little luck, you can achieve some amazing results at your college science fair!
## College Science Fair Projects
### College-level science fair projects in Physics
1. How do you apply a force to another object at the home of gravity, and then how does it react? Use any type(compass) to determine which way an object would point after it was dropped from your height above the ground.
2. Experiment with speed dilation on our planet’s oceans. Explain why there is less light happening in front than behind/around them (Include sound).
3. Experiment with the solar wind. Explain what it is, determine the speed and strength of your planet’s magnetic field (Use magnetometer as necessary), its direction at noon/noon time in your place (+10 days) using maps, etc; compare this data to where you live on earth.
4. Measuring temperature differences between hot liquids that are next to a warm object such as a table or hand and more than one hot liquid next to a cooler object such as the base of an instrument; be creative and determine what temperature is most comfortable.
5. Using your knowledge of buoyancy, explain how you can influence your underwater exploration by adding weights (diving cowboy).
### Environmental science fair projects for college students
1. Using nature for power generation – You can use multiple sources such as water, wind, natural energy, etc.
2. Experiment to produce energy efficiently.
3. Recycling water waste into electricity (in residences, classrooms or offices)
4. Printer conserving materials – combine recycled objects with more durable material for use in the production of 3-dimensional printing.
### Psychology science fair projects for college students
1. Dissecting and identifying kinship, mating goals or a friendly relationship in an identification chart2. Doing statistics on which scent attracts the most flies when others sit down – behavioural biology is fast becoming integral to psychology; distinguish between empathy and sympathy (which are emotions); research effects of drugs like cocaine, caffeine etc increase blood flow, especially the prefrontal cortex (thinking part) necessary for critical thinking – link up with findings on behaviour.
3. The Stroop Effect – How thinking is affected by biased mental tasks.
### Biochemistry science fair projects for college
1. The five stages of protein degradation for experiments against E.coli
2. Testing the DNA in a stool sample (see “Scoop” below) to determine how long ago an infection occurred − 1st stage bacteria die, 2nd and 3rd inhibits their growth − can help isolate which type of bacteria cause diseases.
3. Test cellular behaviour involving cancer by observing the formation of cancer cells in certain organs continuously for 15 minutes thus obtaining a fluorescent stain that is produced after several days. The result will determine if there are any abnormalities with cell growth.
4. Assessing vitamin/mineral deficiency caused by food toxins as well as direct effects from foods. Vitamin A, C, E and K all decrease when given high doses but can be replenished again so serum levels must be monitored to test the effect.
5. Lab experiment of what foods will affect blood pressure (elevation), glucose, cholesterol and lipids in each dish as well as which should be watched because they can decrease harmful substances while also increasing the amounts needed for homeostasis
6. Test if antibiotics are causing oxidative stress by testing if effects occur under conditions where salt intake is very high.
7. Test if it is possible to measure uric acid levels using urine samples, do not want any technicians touching your sample during the storage/ analysis process and keeping results secret on the computer that backs up results of the experiment!
8. Detect viruses using an instrument that is able to bidirectionally transmit the virus but filters out all other non-viruses. Also, it needs to be able to show traps and blanks in the coverage method
9. Lab experiment where one food sample has 2 different species created as food and tested for one species over weeks to see what happens.
10. Lab experiment where you grow bacteria/ yeast on a glass slide using the same media used by other biology labs.
### Chemistry science fair project ideas for college
1. Smart bomb fruits course demonstration where product balls are inserted into fruit bombs resulting with implosions being formed.
2. Chemistry class experiment that measures pH using bicarbonate (HCO,) vs. calcium chloride titration methods or vinegar method.
3. Superconductivity project using soldered wire grid where 1% solution is placed on tensioned wires to measure the electrical resistance across each square inch of the surface.
4. Using salt and sugar water, conduct an experiment that shows a rate difference between adding large quantities slowly vs rapidly into test tube.
5. Drinking alcohol through wax paper membranes as used in forensic examiners.
Astronomy science fair projects for college
1. Using electronic cameras and sensors to remotely observe the universe
2. Making a ‘Virtual Star’ i.e card with velcro attached to the back that when held up in front of a light source on earth you can see it from far away and therefore you know where it is by its movement
3. Using a camera and website to show the ball of light created by the sun on earth during different times
4. Telescopes and telescopes’ parts can be used in 3 ways: 1) Antenna, 2) filters to let only specific wavelengths through depending upon wavelength etc. or for example making telescope lens blanks (which are very fragile let people use this as a science project because they will learn how frustrating it is when they cannot get the lens ‘right’ using their science project)
5. Designing and constructing a telescope with different lenses designs for example, making them bigger or smaller
6. Ways to find something you want – 1) magnifying glass – look where you think it is (possible sites in the universe), 2) some other kind of device that determines altitude by contact with ammonia otherwise known as Becquerel’s Blue Ray or through looking at stars and based upon quantities recorded on the ground which are used by early humans to predict where the sun and stars could be.
### Engineering science fair projects for college students
1. Using electrolysis techniques to study chlorine from water.
2. Sound level experiment where air compressor is used as an increasing distance between loudspeakers.
3. Doppler radar project analyzing a Skyping cell phone call.
4. Determine the velocity of an object dropped from a high point. Add 6-8 feet to gain science accuracy, or change distance between drop and subsequent measurement.
5. Doppler radar analysis of room speed vs occupant movement in a building recording via ammeter.
6. Estimate the roof speed at a building from dropped objects or meteors from space and can be published as an article.
7. Sound intensity level recordings of doors closing vs opening in test chambers.
8. Biscuit (cookie) experiment with 3 different colors of milk powder white, yellow & red.
9.Reflectivity samples measuring water on walls and floor surfaces where results are compared to known values taken.
10. Estimation of speed from soiled clothes.
### Forensic science fair projects for college students
1. Compare different kinds of paint used commonly on roadsides 13a Accurately register blood indicator from fingerprints onto a glass plate with a digital device, show students how this works.
2. Types of crime that could be solved with forensic science
3. Uses banana fibers to indicate saliva and chewing gum in dust.
4. Use colored powder on a piece of paper, place two cards facing each other, then use card as if it was an x-ray.
5. Identify soil samples from landfills using molecular biology
6. Forensic analysis of tire stains.
7. What is the DNA evidence left on, or next to a crime scene after someone has handled all their gloves and shoes?
8. Identify types and amount of chemicals in gas leaks using spectrometer – If you have dropped a coin into your coffee grounds , can you still tell if it was a quarter, dime or nickel based upon its size. Use telephone flashlights to determine whether a light is hot or not.
9. Consider how a forensic exam would change over time and allow an officer to determine when the below might be static in cars and trucks (tires, brakes, ghia , etc) Identify shapes of compound at different times without touching it!
Science experiments can help students learn and retain information in a number of ways. For one, they provide a hands-on way to explore scientific concepts. Experiments can also make learning more fun and engaging, which can lead to better retention rates. Finally, doing science experiments helps students develop problem-solving skills and critical thinking abilities.
In case you’re running out of time, try essay help for your science project. That might help you save time.
### Here are some key tips to winning a college science fair
1. Be creative! Don’t select something too difficult as this will take more time, and possibly even fewer observations with better quality repetitive results that will give you clues to others’ thoughts on measurement error etc.
2. If you are uncertain of your ability to successfully conduct a project, select an easier one first – perhaps something simpler and conceptually sound that involves collecting data every now and then – like in our class above for insect tracking or take this 15-question quiz on starting a research project with many questions focusing on the feasibility.
2. Consider your data! Choose to collect certain information in every collection, and then remember prior thoughts you had as it pertained to that particular item/event.
3. Organize your project in such a way that will help you keep track of your results Not only the date, time and names for each type of observation (e.g., take pictures in digital format) put them into categories such as ‘date’ or just simply write it down on a piece paper at first to help remember specifics – especially if something happens more than once.
The first thing students should know is that research can be easy or hard depending on how they approach things. You see non-scientists like us somehow get confused by what “the rules” of doing science really mean in practice –especially when we wish to achieve.
We hope that this article was helpful in finding the right project for you. By choosing one of the science fair projects listed out in this article, you’ll be well on your way to mastering the basics of that subject. Not only will you be able to develop your technical skills, but you’ll also gain an understanding of the scientific principles at play. If you’re still undecided on what science fair project to choose, explore more information on our website! We’re confident that you’ll find the perfect project for you!
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#### Data Units Calculator
###### Kilobit to Kibibit
Online data storage unit conversion calculator:
From:
To:
The smallest unit of measurement used for measuring data is a bit. A single bit can have a value of either zero(0) or one(1). It may contain a binary value (such as True/False or On/Off or 1/0) and nothing more. Therefore, a byte, or eight bits, is used as the fundamental unit of measurement for data storage. A byte can store 256 different values, which is sufficient to represent standard ASCII table, such as all numbers, letters and control symbols.
Since most files contain thousands of bytes, file sizes are often measured in kilobytes. Larger files, such as images, videos, and audio files, contain millions of bytes and therefore are measured in megabytes. Modern storage devices can store thousands of these files, which is why storage capacity is typically measured in gigabytes or even terabytes.
# 1 kbit to kibit result:
1 (one) kilobit(s) is equal 0.9765625 (zero point nine million seven hundred and sixty-five thousand six hundred and twenty-five) kibibit(s)
#### What is kilobit?
The kilobit is a multiple of the unit bit for digital information or computer storage. The prefix kilo- (symbol k) is defined in the International System of Units (SI) as a multiplier of 10^3 (1 thousand), and therefore, 1 kilobit = 10^3 bits = 1000 bits. The kilobit has the unit symbol kbit.
#### What is kibibit?
The kibibit is a multiple of the bit, a unit of digital information storage, using the standard binary prefix kibi, which has the symbol Ki, meaning 2^10. The unit symbol of the kibibit is Kibit. 1 kibibit = 2^10 bits = 1024 bits
#### How calculate kbit. to kibit.?
1 Kilobit is equal to 0.9765625 Kibibit (zero point nine million seven hundred and sixty-five thousand six hundred and twenty-five kibit)
1 Kibibit is equal to 1.024 Kilobit (one point zero × 1 twenty-four kbit)
1 Kilobit is equal to 1000.0000 bits (one thousand point zero × 4 zero bits)
1 Kibibit is equal to 1024 bits (one thousand and twenty-four bits)
1 Kilobit is equal to 1000 Bit (one thousand bit)
Kibibit is greater than Kilobit
Multiplication factor is 1.024.
1 / 1.024 = 0.9765625.
Maybe you mean Kibibit?
1 Kilobit is equal to 0.9765625 Kibibit (zero point nine million seven hundred and sixty-five thousand six hundred and twenty-five kibit) convert to kibit
### Powers of 2
kbit kibit (Kibibit) Description
1 kbit 0.9765625 kibit 1 kilobit (one) is equal to 0.9765625 kibibit (zero point nine million seven hundred and sixty-five thousand six hundred and twenty-five)
2 kbit 1.953125 kibit 2 kilobit (two) is equal to 1.953125 kibibit (one point nine hundred and fifty-three thousand one hundred and twenty-five)
4 kbit 3.90625 kibit 4 kilobit (four) is equal to 3.90625 kibibit (three point ninety thousand six hundred and twenty-five)
8 kbit 7.8125 kibit 8 kilobit (eight) is equal to 7.8125 kibibit (seven point eight thousand one hundred and twenty-five)
16 kbit 15.625 kibit 16 kilobit (sixteen) is equal to 15.625 kibibit (fifteen point six hundred and twenty-five)
32 kbit 31.25 kibit 32 kilobit (thirty-two) is equal to 31.25 kibibit (thirty-one point twenty-five)
64 kbit 62.5 kibit 64 kilobit (sixty-four) is equal to 62.5 kibibit (sixty-two point five)
128 kbit 125 kibit 128 kilobit (one hundred and twenty-eight) is equal to 125 kibibit (one hundred and twenty-five)
256 kbit 250 kibit 256 kilobit (two hundred and fifty-six) is equal to 250 kibibit (two hundred and fifty)
512 kbit 500 kibit 512 kilobit (five hundred and twelve) is equal to 500 kibibit (five hundred)
1024 kbit 1000 kibit 1024 kilobit (one thousand and twenty-four) is equal to 1000 kibibit (one thousand)
2048 kbit 2000 kibit 2048 kilobit (two thousand and forty-eight) is equal to 2000 kibibit (two thousand)
4096 kbit 4000 kibit 4096 kilobit (four thousand and ninety-six) is equal to 4000 kibibit (four thousand)
8192 kbit 8000 kibit 8192 kilobit (eight thousand one hundred and ninety-two) is equal to 8000 kibibit (eight thousand)
### Convert Kilobit to other units
kbit System Description
1 kbit 1000 bit 1 kilobit (one) is equal to 1000 bit (one thousand)
1 kbit 125 b 1 kilobit (one) is equal to 125 byte (one hundred and twenty-five)
1 kbit 0.125 kb 1 kilobit (one) is equal to 0.125 kilobyte (zero point one hundred and twenty-five)
1 kbit 0.000125 mb 1 kilobit (one) is equal to 0.000125 megabyte (zero point zero × 3 one hundred and twenty-five)
1 kbit 0.000000125 gb 1 kilobit (one) is equal to 0.000000125 gigabyte (zero point zero × 6 one hundred and twenty-five)
1 kbit 0.000000000125 tb 1 kilobit (one) is equal to 0.000000000125 terabyte (zero point zero × 9 one hundred and twenty-five)
1 kbit 0.1220703125 kib 1 kilobit (one) is equal to 0.1220703125 kibibyte (zero point one billion two hundred and twenty million seven hundred and three thousand one hundred and twenty-five)
1 kbit 0.00011920928955078125 mib 1 kilobit (one) is equal to 0.00011920928955078125 mebibyte (zero point zero × 3 eleven quadrillion nine hundred and twenty trillion nine hundred and twenty-eight billion nine hundred and fifty-five million seventy-eight thousand one hundred and twenty-five)
1 kbit 0.0000001164153218269348144531 gib 1 kilobit (one) is equal to 0.0000001164153218269348144531 gibibyte
1 kbit 0.0000000001136868377216160297 tib 1 kilobit (one) is equal to 0.0000000001136868377216160297 tebibyte (zero point zero × 9 one quintillion one hundred and thirty-six quadrillion eight hundred and sixty-eight trillion three hundred and seventy-seven billion two hundred and sixteen million one hundred and sixty thousand two hundred and ninety-seven)
1 kbit 0.001 mbit 1 kilobit (one) is equal to 0.001 megabit (zero point zero × 2 one)
1 kbit 0.000001 gbit 1 kilobit (one) is equal to 0.000001 gigabit (zero point zero × 5 one)
1 kbit 0.000000001 tbit 1 kilobit (one) is equal to 0.000000001 terabit (zero point zero × 8 one)
1 kbit 0.9765625 kibit 1 kilobit (one) is equal to 0.9765625 kibibit (zero point nine million seven hundred and sixty-five thousand six hundred and twenty-five)
1 kbit 0.00095367431640625 mibit 1 kilobit (one) is equal to 0.00095367431640625 mebibit (zero point zero × 3 ninety-five trillion three hundred and sixty-seven billion four hundred and thirty-one million six hundred and forty thousand six hundred and twenty-five)
1 kbit 0.000000931322574615478515625 gibit 1 kilobit (one) is equal to 0.000000931322574615478515625 gibibit
1 kbit 0.0000000009094947017729282379 tibit 1 kilobit (one) is equal to 0.0000000009094947017729282379 tebibit (zero point zero × 9 nine quintillion ninety-four quadrillion nine hundred and forty-seven trillion seventeen billion seven hundred and twenty-nine million two hundred and eighty-two thousand three hundred and seventy-nine)
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# The difference between Bayesian and Frequentist misspecified models
Hi all,
I’m currently comparing predictive projection, Lasso (and some other variable selection methods) on a logistic regression. For this comparison, I use simulated data (including collinear data) and for most simulated data the predictive performance of predictive projection and Lasso is similar.
I also use a misspecified model, where the logistic model and the data generating process have different relations. The logistic model has linear relations while the data generating process uses a step function and an exponential relation. In this case predictive projection performs better than Lasso.
I once heard that Bayesian models are relatively better than frequentist models under misspecification, however I could never find any evidence to support this claim. Does anyone know whether this is true? If yes, why is this the case?
Isn’t neither the Lasso nor predictive projection full-Bayesian?
Aki the expert here… but you are correct. The predictive projection is also not full-Bayesian, but arguable a lot closer than lasso as. One full Bayesian approach would be Bayesian model averaging, for example.
Presumably, there is already a “full Bayesian” model from which the projective approach is, say a sparse regression. But to me it is an unfair comparison: Lasso itself is not an inference, it is a model, which can also be Bayesianlised even though it is not the optimal model to use. Lasso with a student-t likelihood can also be very robust.
It is hard for me to believe there can be a universal conclusion on the relative robustness between Bayesian vs non-Bayesian. Bayesian and non-Bayesian tend to use different models. Are we really saying horseshoe is more robust than lasso, or are we saying Bayesian is more robust than non-Bayesian?
1 Like
Unfortunately I don’t have much time to contribute to the discussion, but I’ll comment the one thing for which I have an answer ready
The projection predictive (notice the order of the words) is beyond full-Bayesian. It starts from the fullest Bayesian model possible. Then we use decision theory to answer the question how to make optimal Bayesian inference in the future if we observe only some of the covariates. Using decision theory doesn’t make it less Bayesian (especially if we follow Bernardo & Smith, who start the axioms from preferences tying utilities to be part of the Bayesian theory). There are some computational approximations to make it faster, and we could then discuss how much these approximations are allowed to affect the result compared to theoretical exact computation. I think that usually the biggest weaknesses of the approach are 1) that the first model wasn’t full enough and 2) the cost of covariates is not often included explicitly (but that is a utility choice made by the the user).
To be honest, I always trip up on that word ordering! :)
Just to add to it: after projection onto a submodel, we still have all posterior samples available for that submodel, so any fully Bayesian post-estimation analyses can still be performed. Nothing much really gets lost (well, computational approximations apart).
This could be handled by adding penalties when the variable selection is performed, or at least that’s where I’d start.
Now I see that I did not yet mention that I use the horseshoe prior yet, but I see you already picked up on that.
The projection itself does not really change the predictive performance (by design). So I think it might have to do with something else.
If you take a Bayesian view of the lasso regression, then you could see it as the MAP estimate of the logistic regression with a Laplace prior on the regression coefficient. The Laplace prior has lighter tails than the horseshoe prior and this might make the latter method more robust against outliers (O’Hagan 1979). Could this also result in robustness against (certain types of) misspecification?
On the other hand, I could also imagine that, when using a posterior distribution to make a prediction, you get (slightly) different results to using a point estimate (as with Lasso).
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# Hands-On Standards, Common Core Edition, Grade 2, Teacher Resource Guide
Hands-On Standards teacher resource complete with 29 lessons built to the Common Core
\$49.99
\$49.99
Hands-On Standards Common Core Math Grade 2 Teacher Resource Guide is the perfect supplement for your core curriculum with targeted, hands-on instruction. Includes 29 lessons focusing on Addition and Subtraction, Comparing Three-Digit Numbers, Estimating and Measuring, Coins, Identify Plane Shapes, Recognizing Fractions, and much more. Each hands-on, standards-focused lesson uses one or more of the following most common manipulatives: Base Ten Blocks, Coin Tiles, Color Tiles, Cuisenaire® Rods, Geared Mini-Clocks, Geoboards, Inchworms™, Inchworms™ Rulers, Two-Color Counters, and 2-cm Color Cubes.
Hands-On Standards Common Core Math Grade 2 Teacher Resource Guide integrates mathematical practice and content standards for all domains into engaging hands-on activities. Reproducible student pages help students build understanding of math concepts. Full-color photographs walk you through implementing each lesson. Easy-to follow lesson plans include ideas to help you differentiate instruction and are aligned to the Common Core.
### TOPICS
Operations and Algebraic Thinking
• Writing number sentences
• Even and odd number patterns
Number and Operations in Base Ten
• Three-digit numbers
• Skip-counting by 5s
• Represent numbers
• Numbers in different forms
• Comparing three-digit numbers
• Adding and subtracting within 1,000
• Adding and subtracting 10 or 100
Measurement and Data
• Standard units
• Inches and feet
• Choosing a unit
• Estimating and measuring
• Comparing two numbers
• Whole numbers as lengths on a number line
• Time to 5 minutes
• Understanding coins
• Solve problems with coins
Geometry
• Identify plane shapes
• Building cubes and prisms
• Partitioning rectangles
• Recognizing fractions
• Identifying unit fractions
### What’s Included:
• 1 Hands-On Standards Math Teacher Resource Guide Grade 2 with 29 lessons
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https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_16&oldid=117690
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# 2015 AMC 8 Problems/Problem 16
## Problem
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If $\frac{1}{3}$ of all the ninth graders are paired with $\frac{2}{5}$ of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
$\textbf{(A) } \frac{2}{15} \qquad\textbf{(B) } \frac{4}{11} \qquad\textbf{(C) } \frac{11}{30} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{11}{15}$
## Solution
### Solution 1
Let the number of sixth graders be $s$, and the number of ninth graders be $n$. Thus, $\frac{n}{3}=\frac{2s}{5}$, which simplifies to $n=\frac{6s}{5}$. Since we are trying to find the value of $\frac{\frac{n}{3}+\frac{2s}{5}}{n+s}$, we can just substitute $\frac{6s}{5}$ for $n$ into the equation. We then get a value of $\frac{\frac{\frac{6s}{5}}{3}+\frac{2s}{5}}{\frac{6s}{5}+s} = \frac{\frac{6s+6s}{15}}{\frac{11s}{5}} = \frac{\frac{4s}{5}}{\frac{11s}{5}} = \boxed{\textbf{(B)}~\frac{4}{11}}$
### Solution 2
We see that the minimum number of ninth graders is $6$, because if there are $3$ then there is $1$ ninth-grader with a buddy, which would mean $2.5$ sixth graders with a buddy, and that's impossible. With $6$ ninth-graders, $2$ of them are in the buddy program, so there $\frac{2}{\tfrac{2}{5}}=5$ sixth-graders total, two of whom have a buddy. Thus, the desired fraction is $\frac{2+2}{5+6}=\boxed{\textbf{(B) }\frac{4}{11}}$.
### Solution 3
Let the number of sixth graders be $s$, and the number of ninth-graders be $n$. Then you get $\frac{n}{3}=\frac{2s}{5}$, which simplifies to $5n=6s$. We can figure out that $n=6$ and $s=5$ is a solution to the equation. Then you substitute and figure out that $\frac{5\cdot\frac{2}{5}+6\cdot\frac{1}{3}}{5+6}=\boxed{\textbf{(B) }\frac{4}{11}}$
-RedFireTruck
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https://timedate.org/if-today-is-tuesday-after-60-days
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# If Today Is Tuesday, what will be the day after 60 days
If today is tuesday,what day of the will it be 60 days from today ?
Saturday
## How do we solve it ?
A week has 7 days " Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday ".
60 / 7 = 8 ( Remainder = 4)
So you can say that after multiples of 7 days, the day will repeat itself as tuesday.
Hence after 56 days Tuesday will occur again.
So we calculate after the remainder (4) , the answer is Saturday.
Simple Logic = And whatever the remainder obtained is, add that to the day.
### Same Answer For Following Questions :
Or if tomorrow is Tuesday, what will be the day after 60 days
Tomorrow is Tuesday what is the day of the week after 60 days
### Weekdays
If 60 days for other weekdays.
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:: The Limit of a Real Function at Infinity. Halflines. Real Sequence Divergent to Infinity
:: by Jaros{\l}aw Kotowicz
::
:: Received August 20, 1990
:: Copyright (c) 1990-2019 Association of Mizar Users
Lm1: for r, r1, g being Real st 0 < g & r <= r1 holds
( r - g < r1 & r < r1 + g )
proof end;
Lm2: for seq being Real_Sequence
for f1, f2 being PartFunc of REAL,REAL st rng seq c= dom (f1 + f2) holds
( dom (f1 + f2) = (dom f1) /\ (dom f2) & rng seq c= dom f1 & rng seq c= dom f2 )
proof end;
Lm3: for seq being Real_Sequence
for f1, f2 being PartFunc of REAL,REAL st rng seq c= dom (f1 (#) f2) holds
( dom (f1 (#) f2) = (dom f1) /\ (dom f2) & rng seq c= dom f1 & rng seq c= dom f2 )
proof end;
notation
let r be Real;
synonym left_open_halfline r for halfline r;
end;
definition
let r be Real;
coherence
is Subset of REAL
proof end;
coherence
is Subset of REAL
proof end;
coherence
is Subset of REAL
proof end;
end;
:: deftheorem defines left_closed_halfline LIMFUNC1:def 1 :
for r being Real holds left_closed_halfline r = ;
:: deftheorem defines right_closed_halfline LIMFUNC1:def 2 :
for r being Real holds right_closed_halfline r = ;
:: deftheorem defines right_open_halfline LIMFUNC1:def 3 :
for r being Real holds right_open_halfline r = ;
theorem :: LIMFUNC1:1
for seq being Real_Sequence holds
( ( seq is non-decreasing implies seq is bounded_below ) & ( seq is non-increasing implies seq is bounded_above ) )
proof end;
theorem Th2: :: LIMFUNC1:2
for seq being Real_Sequence st seq is non-zero & seq is convergent & lim seq = 0 & seq is non-decreasing holds
for n being Nat holds seq . n < 0
proof end;
theorem Th3: :: LIMFUNC1:3
for seq being Real_Sequence st seq is non-zero & seq is convergent & lim seq = 0 & seq is non-increasing holds
for n being Nat holds 0 < seq . n
proof end;
theorem Th4: :: LIMFUNC1:4
for seq being Real_Sequence st seq is convergent & 0 < lim seq holds
ex n being Nat st
for m being Nat st n <= m holds
0 < seq . m
proof end;
theorem Th5: :: LIMFUNC1:5
for seq being Real_Sequence st seq is convergent & 0 < lim seq holds
ex n being Nat st
for m being Nat st n <= m holds
(lim seq) / 2 < seq . m
proof end;
definition
let seq be Real_Sequence;
attr seq is divergent_to+infty means :: LIMFUNC1:def 4
for r being Real ex n being Nat st
for m being Nat st n <= m holds
r < seq . m;
attr seq is divergent_to-infty means :: LIMFUNC1:def 5
for r being Real ex n being Nat st
for m being Nat st n <= m holds
seq . m < r;
end;
:: deftheorem defines divergent_to+infty LIMFUNC1:def 4 :
for seq being Real_Sequence holds
( seq is divergent_to+infty iff for r being Real ex n being Nat st
for m being Nat st n <= m holds
r < seq . m );
:: deftheorem defines divergent_to-infty LIMFUNC1:def 5 :
for seq being Real_Sequence holds
( seq is divergent_to-infty iff for r being Real ex n being Nat st
for m being Nat st n <= m holds
seq . m < r );
theorem :: LIMFUNC1:6
for seq being Real_Sequence st ( seq is divergent_to+infty or seq is divergent_to-infty ) holds
ex n being Nat st
for m being Nat st n <= m holds
seq ^\ m is non-zero
proof end;
theorem Th7: :: LIMFUNC1:7
for k being Nat
for seq being Real_Sequence holds
( ( seq ^\ k is divergent_to+infty implies seq is divergent_to+infty ) & ( seq ^\ k is divergent_to-infty implies seq is divergent_to-infty ) )
proof end;
theorem Th8: :: LIMFUNC1:8
for seq1, seq2 being Real_Sequence st seq1 is divergent_to+infty & seq2 is divergent_to+infty holds
seq1 + seq2 is divergent_to+infty
proof end;
theorem Th9: :: LIMFUNC1:9
for seq1, seq2 being Real_Sequence st seq1 is divergent_to+infty & seq2 is bounded_below holds
seq1 + seq2 is divergent_to+infty
proof end;
theorem Th10: :: LIMFUNC1:10
for seq1, seq2 being Real_Sequence st seq1 is divergent_to+infty & seq2 is divergent_to+infty holds
seq1 (#) seq2 is divergent_to+infty
proof end;
theorem Th11: :: LIMFUNC1:11
for seq1, seq2 being Real_Sequence st seq1 is divergent_to-infty & seq2 is divergent_to-infty holds
seq1 + seq2 is divergent_to-infty
proof end;
theorem Th12: :: LIMFUNC1:12
for seq1, seq2 being Real_Sequence st seq1 is divergent_to-infty & seq2 is bounded_above holds
seq1 + seq2 is divergent_to-infty
proof end;
Lm4:
by XREAL_0:def 1;
theorem Th13: :: LIMFUNC1:13
for seq being Real_Sequence
for r being Real holds
( ( seq is divergent_to+infty & r > 0 implies r (#) seq is divergent_to+infty ) & ( seq is divergent_to+infty & r < 0 implies r (#) seq is divergent_to-infty ) & ( r = 0 implies ( rng (r (#) seq) = & r (#) seq is constant ) ) )
proof end;
theorem Th14: :: LIMFUNC1:14
for seq being Real_Sequence
for r being Real holds
( ( seq is divergent_to-infty & r > 0 implies r (#) seq is divergent_to-infty ) & ( seq is divergent_to-infty & r < 0 implies r (#) seq is divergent_to+infty ) & ( r = 0 implies ( rng (r (#) seq) = & r (#) seq is constant ) ) )
proof end;
reconsider jj = 1 as Real ;
theorem :: LIMFUNC1:15
for seq being Real_Sequence holds
( ( seq is divergent_to+infty implies - seq is divergent_to-infty ) & ( seq is divergent_to-infty implies - seq is divergent_to+infty ) ) by ;
theorem :: LIMFUNC1:16
for seq, seq1 being Real_Sequence st seq is bounded_below & seq1 is divergent_to-infty holds
seq - seq1 is divergent_to+infty
proof end;
theorem :: LIMFUNC1:17
for seq, seq1 being Real_Sequence st seq is bounded_above & seq1 is divergent_to+infty holds
seq - seq1 is divergent_to-infty
proof end;
theorem :: LIMFUNC1:18
for seq, seq1 being Real_Sequence st seq is divergent_to+infty & seq1 is convergent holds
seq + seq1 is divergent_to+infty by Th9;
theorem :: LIMFUNC1:19
for seq, seq1 being Real_Sequence st seq is divergent_to-infty & seq1 is convergent holds
seq + seq1 is divergent_to-infty by Th12;
theorem Th20: :: LIMFUNC1:20
for seq being Real_Sequence st ( for n being Nat holds seq . n = n ) holds
seq is divergent_to+infty
proof end;
reconsider s1 = id NAT as Real_Sequence by ;
Lm5: for n being Nat holds s1 . n = n
by ;
theorem Th21: :: LIMFUNC1:21
for seq being Real_Sequence st ( for n being Nat holds seq . n = - n ) holds
seq is divergent_to-infty
proof end;
theorem Th22: :: LIMFUNC1:22
for seq1, seq2 being Real_Sequence st seq1 is divergent_to+infty & ex r being Real st
( r > 0 & ( for n being Nat holds seq2 . n >= r ) ) holds
seq1 (#) seq2 is divergent_to+infty
proof end;
theorem :: LIMFUNC1:23
for seq1, seq2 being Real_Sequence st seq1 is divergent_to-infty & ex r being Real st
( 0 < r & ( for n being Nat holds seq2 . n >= r ) ) holds
seq1 (#) seq2 is divergent_to-infty
proof end;
theorem Th24: :: LIMFUNC1:24
for seq1, seq2 being Real_Sequence st seq1 is divergent_to-infty & seq2 is divergent_to-infty holds
seq1 (#) seq2 is divergent_to+infty
proof end;
theorem Th25: :: LIMFUNC1:25
for seq being Real_Sequence st ( seq is divergent_to+infty or seq is divergent_to-infty ) holds
abs seq is divergent_to+infty
proof end;
theorem Th26: :: LIMFUNC1:26
for seq, seq1 being Real_Sequence st seq is divergent_to+infty & seq1 is subsequence of seq holds
seq1 is divergent_to+infty
proof end;
theorem Th27: :: LIMFUNC1:27
for seq, seq1 being Real_Sequence st seq is divergent_to-infty & seq1 is subsequence of seq holds
seq1 is divergent_to-infty
proof end;
theorem :: LIMFUNC1:28
for seq1, seq2 being Real_Sequence st seq1 is divergent_to+infty & seq2 is convergent & 0 < lim seq2 holds
seq1 (#) seq2 is divergent_to+infty
proof end;
theorem Th29: :: LIMFUNC1:29
for seq being Real_Sequence st seq is non-decreasing & not seq is bounded_above holds
seq is divergent_to+infty
proof end;
theorem Th30: :: LIMFUNC1:30
for seq being Real_Sequence st seq is non-increasing & not seq is bounded_below holds
seq is divergent_to-infty
proof end;
theorem :: LIMFUNC1:31
for seq being Real_Sequence st seq is increasing & not seq is bounded_above holds
seq is divergent_to+infty by Th29;
theorem :: LIMFUNC1:32
for seq being Real_Sequence st seq is decreasing & not seq is bounded_below holds
seq is divergent_to-infty by Th30;
theorem :: LIMFUNC1:33
for seq being Real_Sequence holds
( not seq is monotone or seq is convergent or seq is divergent_to+infty or seq is divergent_to-infty )
proof end;
theorem Th34: :: LIMFUNC1:34
for seq being Real_Sequence st ( seq is divergent_to+infty or seq is divergent_to-infty ) holds
( seq " is convergent & lim (seq ") = 0 )
proof end;
theorem Th35: :: LIMFUNC1:35
for seq being Real_Sequence st seq is convergent & lim seq = 0 & ex k being Nat st
for n being Nat st k <= n holds
0 < seq . n holds
seq " is divergent_to+infty
proof end;
theorem Th36: :: LIMFUNC1:36
for seq being Real_Sequence st seq is convergent & lim seq = 0 & ex k being Nat st
for n being Nat st k <= n holds
seq . n < 0 holds
seq " is divergent_to-infty
proof end;
theorem Th37: :: LIMFUNC1:37
for seq being Real_Sequence st seq is non-zero & seq is convergent & lim seq = 0 & seq is non-decreasing holds
seq " is divergent_to-infty
proof end;
theorem Th38: :: LIMFUNC1:38
for seq being Real_Sequence st seq is non-zero & seq is convergent & lim seq = 0 & seq is non-increasing holds
seq " is divergent_to+infty
proof end;
theorem :: LIMFUNC1:39
for seq being Real_Sequence st seq is non-zero & seq is convergent & lim seq = 0 & seq is increasing holds
seq " is divergent_to-infty by Th37;
theorem :: LIMFUNC1:40
for seq being Real_Sequence st seq is non-zero & seq is convergent & lim seq = 0 & seq is decreasing holds
seq " is divergent_to+infty by Th38;
theorem :: LIMFUNC1:41
for seq1, seq2 being Real_Sequence st seq1 is bounded & ( seq2 is divergent_to+infty or seq2 is divergent_to-infty ) holds
( seq1 /" seq2 is convergent & lim (seq1 /" seq2) = 0 )
proof end;
theorem Th42: :: LIMFUNC1:42
for seq, seq1 being Real_Sequence st seq is divergent_to+infty & ( for n being Nat holds seq . n <= seq1 . n ) holds
seq1 is divergent_to+infty
proof end;
theorem Th43: :: LIMFUNC1:43
for seq, seq1 being Real_Sequence st seq is divergent_to-infty & ( for n being Nat holds seq1 . n <= seq . n ) holds
seq1 is divergent_to-infty
proof end;
definition
let f be PartFunc of REAL,REAL;
attr f is convergent_in+infty means :: LIMFUNC1:def 6
( ( for r being Real ex g being Real st
( r < g & g in dom f ) ) & ex g being Real st
for seq being Real_Sequence st seq is divergent_to+infty & rng seq c= dom f holds
( f /* seq is convergent & lim (f /* seq) = g ) );
attr f is divergent_in+infty_to+infty means :: LIMFUNC1:def 7
( ( for r being Real ex g being Real st
( r < g & g in dom f ) ) & ( for seq being Real_Sequence st seq is divergent_to+infty & rng seq c= dom f holds
f /* seq is divergent_to+infty ) );
attr f is divergent_in+infty_to-infty means :: LIMFUNC1:def 8
( ( for r being Real ex g being Real st
( r < g & g in dom f ) ) & ( for seq being Real_Sequence st seq is divergent_to+infty & rng seq c= dom f holds
f /* seq is divergent_to-infty ) );
attr f is convergent_in-infty means :: LIMFUNC1:def 9
( ( for r being Real ex g being Real st
( g < r & g in dom f ) ) & ex g being Real st
for seq being Real_Sequence st seq is divergent_to-infty & rng seq c= dom f holds
( f /* seq is convergent & lim (f /* seq) = g ) );
attr f is divergent_in-infty_to+infty means :: LIMFUNC1:def 10
( ( for r being Real ex g being Real st
( g < r & g in dom f ) ) & ( for seq being Real_Sequence st seq is divergent_to-infty & rng seq c= dom f holds
f /* seq is divergent_to+infty ) );
attr f is divergent_in-infty_to-infty means :: LIMFUNC1:def 11
( ( for r being Real ex g being Real st
( g < r & g in dom f ) ) & ( for seq being Real_Sequence st seq is divergent_to-infty & rng seq c= dom f holds
f /* seq is divergent_to-infty ) );
end;
:: deftheorem defines convergent_in+infty LIMFUNC1:def 6 :
for f being PartFunc of REAL,REAL holds
( f is convergent_in+infty iff ( ( for r being Real ex g being Real st
( r < g & g in dom f ) ) & ex g being Real st
for seq being Real_Sequence st seq is divergent_to+infty & rng seq c= dom f holds
( f /* seq is convergent & lim (f /* seq) = g ) ) );
:: deftheorem defines divergent_in+infty_to+infty LIMFUNC1:def 7 :
for f being PartFunc of REAL,REAL holds
( f is divergent_in+infty_to+infty iff ( ( for r being Real ex g being Real st
( r < g & g in dom f ) ) & ( for seq being Real_Sequence st seq is divergent_to+infty & rng seq c= dom f holds
f /* seq is divergent_to+infty ) ) );
:: deftheorem defines divergent_in+infty_to-infty LIMFUNC1:def 8 :
for f being PartFunc of REAL,REAL holds
( f is divergent_in+infty_to-infty iff ( ( for r being Real ex g being Real st
( r < g & g in dom f ) ) & ( for seq being Real_Sequence st seq is divergent_to+infty & rng seq c= dom f holds
f /* seq is divergent_to-infty ) ) );
:: deftheorem defines convergent_in-infty LIMFUNC1:def 9 :
for f being PartFunc of REAL,REAL holds
( f is convergent_in-infty iff ( ( for r being Real ex g being Real st
( g < r & g in dom f ) ) & ex g being Real st
for seq being Real_Sequence st seq is divergent_to-infty & rng seq c= dom f holds
( f /* seq is convergent & lim (f /* seq) = g ) ) );
:: deftheorem defines divergent_in-infty_to+infty LIMFUNC1:def 10 :
for f being PartFunc of REAL,REAL holds
( f is divergent_in-infty_to+infty iff ( ( for r being Real ex g being Real st
( g < r & g in dom f ) ) & ( for seq being Real_Sequence st seq is divergent_to-infty & rng seq c= dom f holds
f /* seq is divergent_to+infty ) ) );
:: deftheorem defines divergent_in-infty_to-infty LIMFUNC1:def 11 :
for f being PartFunc of REAL,REAL holds
( f is divergent_in-infty_to-infty iff ( ( for r being Real ex g being Real st
( g < r & g in dom f ) ) & ( for seq being Real_Sequence st seq is divergent_to-infty & rng seq c= dom f holds
f /* seq is divergent_to-infty ) ) );
theorem :: LIMFUNC1:44
for f being PartFunc of REAL,REAL holds
( f is convergent_in+infty iff ( ( for r being Real ex g being Real st
( r < g & g in dom f ) ) & ex g being Real st
for g1 being Real st 0 < g1 holds
ex r being Real st
for r1 being Real st r < r1 & r1 in dom f holds
|.((f . r1) - g).| < g1 ) )
proof end;
theorem :: LIMFUNC1:45
for f being PartFunc of REAL,REAL holds
( f is convergent_in-infty iff ( ( for r being Real ex g being Real st
( g < r & g in dom f ) ) & ex g being Real st
for g1 being Real st 0 < g1 holds
ex r being Real st
for r1 being Real st r1 < r & r1 in dom f holds
|.((f . r1) - g).| < g1 ) )
proof end;
theorem :: LIMFUNC1:46
for f being PartFunc of REAL,REAL holds
( f is divergent_in+infty_to+infty iff ( ( for r being Real ex g being Real st
( r < g & g in dom f ) ) & ( for g being Real ex r being Real st
for r1 being Real st r < r1 & r1 in dom f holds
g < f . r1 ) ) )
proof end;
theorem :: LIMFUNC1:47
for f being PartFunc of REAL,REAL holds
( f is divergent_in+infty_to-infty iff ( ( for r being Real ex g being Real st
( r < g & g in dom f ) ) & ( for g being Real ex r being Real st
for r1 being Real st r < r1 & r1 in dom f holds
f . r1 < g ) ) )
proof end;
theorem :: LIMFUNC1:48
for f being PartFunc of REAL,REAL holds
( f is divergent_in-infty_to+infty iff ( ( for r being Real ex g being Real st
( g < r & g in dom f ) ) & ( for g being Real ex r being Real st
for r1 being Real st r1 < r & r1 in dom f holds
g < f . r1 ) ) )
proof end;
theorem :: LIMFUNC1:49
for f being PartFunc of REAL,REAL holds
( f is divergent_in-infty_to-infty iff ( ( for r being Real ex g being Real st
( g < r & g in dom f ) ) & ( for g being Real ex r being Real st
for r1 being Real st r1 < r & r1 in dom f holds
f . r1 < g ) ) )
proof end;
theorem :: LIMFUNC1:50
for f1, f2 being PartFunc of REAL,REAL st f1 is divergent_in+infty_to+infty & f2 is divergent_in+infty_to+infty & ( for r being Real ex g being Real st
( r < g & g in (dom f1) /\ (dom f2) ) ) holds
( f1 + f2 is divergent_in+infty_to+infty & f1 (#) f2 is divergent_in+infty_to+infty )
proof end;
theorem :: LIMFUNC1:51
for f1, f2 being PartFunc of REAL,REAL st f1 is divergent_in+infty_to-infty & f2 is divergent_in+infty_to-infty & ( for r being Real ex g being Real st
( r < g & g in (dom f1) /\ (dom f2) ) ) holds
( f1 + f2 is divergent_in+infty_to-infty & f1 (#) f2 is divergent_in+infty_to+infty )
proof end;
theorem :: LIMFUNC1:52
for f1, f2 being PartFunc of REAL,REAL st f1 is divergent_in-infty_to+infty & f2 is divergent_in-infty_to+infty & ( for r being Real ex g being Real st
( g < r & g in (dom f1) /\ (dom f2) ) ) holds
( f1 + f2 is divergent_in-infty_to+infty & f1 (#) f2 is divergent_in-infty_to+infty )
proof end;
theorem :: LIMFUNC1:53
for f1, f2 being PartFunc of REAL,REAL st f1 is divergent_in-infty_to-infty & f2 is divergent_in-infty_to-infty & ( for r being Real ex g being Real st
( g < r & g in (dom f1) /\ (dom f2) ) ) holds
( f1 + f2 is divergent_in-infty_to-infty & f1 (#) f2 is divergent_in-infty_to+infty )
proof end;
theorem :: LIMFUNC1:54
for f1, f2 being PartFunc of REAL,REAL st f1 is divergent_in+infty_to+infty & ( for r being Real ex g being Real st
( r < g & g in dom (f1 + f2) ) ) & ex r being Real st f2 | is bounded_below holds
f1 + f2 is divergent_in+infty_to+infty
proof end;
theorem :: LIMFUNC1:55
for f1, f2 being PartFunc of REAL,REAL st f1 is divergent_in+infty_to+infty & ( for r being Real ex g being Real st
( r < g & g in dom (f1 (#) f2) ) ) & ex r, r1 being Real st
( 0 < r & ( for g being Real st g in (dom f2) /\ () holds
r <= f2 . g ) ) holds
f1 (#) f2 is divergent_in+infty_to+infty
proof end;
theorem :: LIMFUNC1:56
for f1, f2 being PartFunc of REAL,REAL st f1 is divergent_in-infty_to+infty & ( for r being Real ex g being Real st
( g < r & g in dom (f1 + f2) ) ) & ex r being Real st f2 | is bounded_below holds
f1 + f2 is divergent_in-infty_to+infty
proof end;
theorem :: LIMFUNC1:57
for f1, f2 being PartFunc of REAL,REAL st f1 is divergent_in-infty_to+infty & ( for r being Real ex g being Real st
( g < r & g in dom (f1 (#) f2) ) ) & ex r, r1 being Real st
( 0 < r & ( for g being Real st g in (dom f2) /\ () holds
r <= f2 . g ) ) holds
f1 (#) f2 is divergent_in-infty_to+infty
proof end;
theorem :: LIMFUNC1:58
for f being PartFunc of REAL,REAL
for r being Real holds
( ( f is divergent_in+infty_to+infty & r > 0 implies r (#) f is divergent_in+infty_to+infty ) & ( f is divergent_in+infty_to+infty & r < 0 implies r (#) f is divergent_in+infty_to-infty ) & ( f is divergent_in+infty_to-infty & r > 0 implies r (#) f is divergent_in+infty_to-infty ) & ( f is divergent_in+infty_to-infty & r < 0 implies r (#) f is divergent_in+infty_to+infty ) )
proof end;
theorem :: LIMFUNC1:59
for f being PartFunc of REAL,REAL
for r being Real holds
( ( f is divergent_in-infty_to+infty & r > 0 implies r (#) f is divergent_in-infty_to+infty ) & ( f is divergent_in-infty_to+infty & r < 0 implies r (#) f is divergent_in-infty_to-infty ) & ( f is divergent_in-infty_to-infty & r > 0 implies r (#) f is divergent_in-infty_to-infty ) & ( f is divergent_in-infty_to-infty & r < 0 implies r (#) f is divergent_in-infty_to+infty ) )
proof end;
theorem :: LIMFUNC1:60
for f being PartFunc of REAL,REAL st ( f is divergent_in+infty_to+infty or f is divergent_in+infty_to-infty ) holds
abs f is divergent_in+infty_to+infty
proof end;
theorem :: LIMFUNC1:61
for f being PartFunc of REAL,REAL st ( f is divergent_in-infty_to+infty or f is divergent_in-infty_to-infty ) holds
abs f is divergent_in-infty_to+infty
proof end;
theorem Th62: :: LIMFUNC1:62
for f being PartFunc of REAL,REAL st ex r being Real st
( f | is non-decreasing & not f | is bounded_above ) & ( for r being Real ex g being Real st
( r < g & g in dom f ) ) holds
f is divergent_in+infty_to+infty
proof end;
theorem :: LIMFUNC1:63
for f being PartFunc of REAL,REAL st ex r being Real st
( f | is increasing & not f | is bounded_above ) & ( for r being Real ex g being Real st
( r < g & g in dom f ) ) holds
f is divergent_in+infty_to+infty by Th62;
theorem Th64: :: LIMFUNC1:64
for f being PartFunc of REAL,REAL st ex r being Real st
( f | is non-increasing & not f | is bounded_below ) & ( for r being Real ex g being Real st
( r < g & g in dom f ) ) holds
f is divergent_in+infty_to-infty
proof end;
theorem :: LIMFUNC1:65
for f being PartFunc of REAL,REAL st ex r being Real st
( f | is decreasing & not f | is bounded_below ) & ( for r being Real ex g being Real st
( r < g & g in dom f ) ) holds
f is divergent_in+infty_to-infty by Th64;
theorem Th66: :: LIMFUNC1:66
for f being PartFunc of REAL,REAL st ex r being Real st
( f | is non-increasing & not f | is bounded_above ) & ( for r being Real ex g being Real st
( g < r & g in dom f ) ) holds
f is divergent_in-infty_to+infty
proof end;
theorem :: LIMFUNC1:67
for f being PartFunc of REAL,REAL st ex r being Real st
( f | is decreasing & not f | is bounded_above ) & ( for r being Real ex g being Real st
( g < r & g in dom f ) ) holds
f is divergent_in-infty_to+infty by Th66;
theorem Th68: :: LIMFUNC1:68
for f being PartFunc of REAL,REAL st ex r being Real st
( f | is non-decreasing & not f | is bounded_below ) & ( for r being Real ex g being Real st
( g < r & g in dom f ) ) holds
f is divergent_in-infty_to-infty
proof end;
theorem :: LIMFUNC1:69
for f being PartFunc of REAL,REAL st ex r being Real st
( f | is increasing & not f | is bounded_below ) & ( for r being Real ex g being Real st
( g < r & g in dom f ) ) holds
f is divergent_in-infty_to-infty by Th68;
theorem Th70: :: LIMFUNC1:70
for f, f1 being PartFunc of REAL,REAL st f1 is divergent_in+infty_to+infty & ( for r being Real ex g being Real st
( r < g & g in dom f ) ) & ex r being Real st
( (dom f) /\ c= (dom f1) /\ & ( for g being Real st g in (dom f) /\ holds
f1 . g <= f . g ) ) holds
f is divergent_in+infty_to+infty
proof end;
theorem Th71: :: LIMFUNC1:71
for f, f1 being PartFunc of REAL,REAL st f1 is divergent_in+infty_to-infty & ( for r being Real ex g being Real st
( r < g & g in dom f ) ) & ex r being Real st
( (dom f) /\ c= (dom f1) /\ & ( for g being Real st g in (dom f) /\ holds
f . g <= f1 . g ) ) holds
f is divergent_in+infty_to-infty
proof end;
theorem Th72: :: LIMFUNC1:72
for f, f1 being PartFunc of REAL,REAL st f1 is divergent_in-infty_to+infty & ( for r being Real ex g being Real st
( g < r & g in dom f ) ) & ex r being Real st
( (dom f) /\ c= (dom f1) /\ & ( for g being Real st g in (dom f) /\ holds
f1 . g <= f . g ) ) holds
f is divergent_in-infty_to+infty
proof end;
theorem Th73: :: LIMFUNC1:73
for f, f1 being PartFunc of REAL,REAL st f1 is divergent_in-infty_to-infty & ( for r being Real ex g being Real st
( g < r & g in dom f ) ) & ex r being Real st
( (dom f) /\ c= (dom f1) /\ & ( for g being Real st g in (dom f) /\ holds
f . g <= f1 . g ) ) holds
f is divergent_in-infty_to-infty
proof end;
theorem :: LIMFUNC1:74
for f, f1 being PartFunc of REAL,REAL st f1 is divergent_in+infty_to+infty & ex r being Real st
( right_open_halfline r c= (dom f) /\ (dom f1) & ( for g being Real st g in right_open_halfline r holds
f1 . g <= f . g ) ) holds
f is divergent_in+infty_to+infty
proof end;
theorem :: LIMFUNC1:75
for f, f1 being PartFunc of REAL,REAL st f1 is divergent_in+infty_to-infty & ex r being Real st
( right_open_halfline r c= (dom f) /\ (dom f1) & ( for g being Real st g in right_open_halfline r holds
f . g <= f1 . g ) ) holds
f is divergent_in+infty_to-infty
proof end;
theorem :: LIMFUNC1:76
for f, f1 being PartFunc of REAL,REAL st f1 is divergent_in-infty_to+infty & ex r being Real st
( left_open_halfline r c= (dom f) /\ (dom f1) & ( for g being Real st g in left_open_halfline r holds
f1 . g <= f . g ) ) holds
f is divergent_in-infty_to+infty
proof end;
theorem :: LIMFUNC1:77
for f, f1 being PartFunc of REAL,REAL st f1 is divergent_in-infty_to-infty & ex r being Real st
( left_open_halfline r c= (dom f) /\ (dom f1) & ( for g being Real st g in left_open_halfline r holds
f . g <= f1 . g ) ) holds
f is divergent_in-infty_to-infty
proof end;
definition
let f be PartFunc of REAL,REAL;
assume A1: f is convergent_in+infty ;
func lim_in+infty f -> Real means :Def12: :: LIMFUNC1:def 12
for seq being Real_Sequence st seq is divergent_to+infty & rng seq c= dom f holds
( f /* seq is convergent & lim (f /* seq) = it );
existence
ex b1 being Real st
for seq being Real_Sequence st seq is divergent_to+infty & rng seq c= dom f holds
( f /* seq is convergent & lim (f /* seq) = b1 )
by A1;
uniqueness
for b1, b2 being Real st ( for seq being Real_Sequence st seq is divergent_to+infty & rng seq c= dom f holds
( f /* seq is convergent & lim (f /* seq) = b1 ) ) & ( for seq being Real_Sequence st seq is divergent_to+infty & rng seq c= dom f holds
( f /* seq is convergent & lim (f /* seq) = b2 ) ) holds
b1 = b2
proof end;
end;
:: deftheorem Def12 defines lim_in+infty LIMFUNC1:def 12 :
for f being PartFunc of REAL,REAL st f is convergent_in+infty holds
for b2 being Real holds
( b2 = lim_in+infty f iff for seq being Real_Sequence st seq is divergent_to+infty & rng seq c= dom f holds
( f /* seq is convergent & lim (f /* seq) = b2 ) );
definition
let f be PartFunc of REAL,REAL;
assume A1: f is convergent_in-infty ;
func lim_in-infty f -> Real means :Def13: :: LIMFUNC1:def 13
for seq being Real_Sequence st seq is divergent_to-infty & rng seq c= dom f holds
( f /* seq is convergent & lim (f /* seq) = it );
existence
ex b1 being Real st
for seq being Real_Sequence st seq is divergent_to-infty & rng seq c= dom f holds
( f /* seq is convergent & lim (f /* seq) = b1 )
by A1;
uniqueness
for b1, b2 being Real st ( for seq being Real_Sequence st seq is divergent_to-infty & rng seq c= dom f holds
( f /* seq is convergent & lim (f /* seq) = b1 ) ) & ( for seq being Real_Sequence st seq is divergent_to-infty & rng seq c= dom f holds
( f /* seq is convergent & lim (f /* seq) = b2 ) ) holds
b1 = b2
proof end;
end;
:: deftheorem Def13 defines lim_in-infty LIMFUNC1:def 13 :
for f being PartFunc of REAL,REAL st f is convergent_in-infty holds
for b2 being Real holds
( b2 = lim_in-infty f iff for seq being Real_Sequence st seq is divergent_to-infty & rng seq c= dom f holds
( f /* seq is convergent & lim (f /* seq) = b2 ) );
theorem :: LIMFUNC1:78
for f being PartFunc of REAL,REAL
for g being Real st f is convergent_in-infty holds
( lim_in-infty f = g iff for g1 being Real st 0 < g1 holds
ex r being Real st
for r1 being Real st r1 < r & r1 in dom f holds
|.((f . r1) - g).| < g1 )
proof end;
theorem :: LIMFUNC1:79
for f being PartFunc of REAL,REAL
for g being Real st f is convergent_in+infty holds
( lim_in+infty f = g iff for g1 being Real st 0 < g1 holds
ex r being Real st
for r1 being Real st r < r1 & r1 in dom f holds
|.((f . r1) - g).| < g1 )
proof end;
theorem Th80: :: LIMFUNC1:80
for f being PartFunc of REAL,REAL
for r being Real st f is convergent_in+infty holds
( r (#) f is convergent_in+infty & lim_in+infty (r (#) f) = r * () )
proof end;
theorem Th81: :: LIMFUNC1:81
for f being PartFunc of REAL,REAL st f is convergent_in+infty holds
( - f is convergent_in+infty & lim_in+infty (- f) = - () )
proof end;
theorem Th82: :: LIMFUNC1:82
for f1, f2 being PartFunc of REAL,REAL st f1 is convergent_in+infty & f2 is convergent_in+infty & ( for r being Real ex g being Real st
( r < g & g in dom (f1 + f2) ) ) holds
( f1 + f2 is convergent_in+infty & lim_in+infty (f1 + f2) = () + () )
proof end;
theorem :: LIMFUNC1:83
for f1, f2 being PartFunc of REAL,REAL st f1 is convergent_in+infty & f2 is convergent_in+infty & ( for r being Real ex g being Real st
( r < g & g in dom (f1 - f2) ) ) holds
( f1 - f2 is convergent_in+infty & lim_in+infty (f1 - f2) = () - () )
proof end;
theorem :: LIMFUNC1:84
for f being PartFunc of REAL,REAL st f is convergent_in+infty & f " = {} & lim_in+infty f <> 0 holds
( f ^ is convergent_in+infty & lim_in+infty (f ^) = () " )
proof end;
theorem :: LIMFUNC1:85
for f being PartFunc of REAL,REAL st f is convergent_in+infty holds
( abs f is convergent_in+infty & lim_in+infty (abs f) = |.().| )
proof end;
theorem Th86: :: LIMFUNC1:86
for f being PartFunc of REAL,REAL st f is convergent_in+infty & lim_in+infty f <> 0 & ( for r being Real ex g being Real st
( r < g & g in dom f & f . g <> 0 ) ) holds
( f ^ is convergent_in+infty & lim_in+infty (f ^) = () " )
proof end;
theorem Th87: :: LIMFUNC1:87
for f1, f2 being PartFunc of REAL,REAL st f1 is convergent_in+infty & f2 is convergent_in+infty & ( for r being Real ex g being Real st
( r < g & g in dom (f1 (#) f2) ) ) holds
( f1 (#) f2 is convergent_in+infty & lim_in+infty (f1 (#) f2) = () * () )
proof end;
theorem :: LIMFUNC1:88
for f1, f2 being PartFunc of REAL,REAL st f1 is convergent_in+infty & f2 is convergent_in+infty & lim_in+infty f2 <> 0 & ( for r being Real ex g being Real st
( r < g & g in dom (f1 / f2) ) ) holds
( f1 / f2 is convergent_in+infty & lim_in+infty (f1 / f2) = () / () )
proof end;
theorem Th89: :: LIMFUNC1:89
for f being PartFunc of REAL,REAL
for r being Real st f is convergent_in-infty holds
( r (#) f is convergent_in-infty & lim_in-infty (r (#) f) = r * () )
proof end;
theorem Th90: :: LIMFUNC1:90
for f being PartFunc of REAL,REAL st f is convergent_in-infty holds
( - f is convergent_in-infty & lim_in-infty (- f) = - () )
proof end;
theorem Th91: :: LIMFUNC1:91
for f1, f2 being PartFunc of REAL,REAL st f1 is convergent_in-infty & f2 is convergent_in-infty & ( for r being Real ex g being Real st
( g < r & g in dom (f1 + f2) ) ) holds
( f1 + f2 is convergent_in-infty & lim_in-infty (f1 + f2) = () + () )
proof end;
theorem :: LIMFUNC1:92
for f1, f2 being PartFunc of REAL,REAL st f1 is convergent_in-infty & f2 is convergent_in-infty & ( for r being Real ex g being Real st
( g < r & g in dom (f1 - f2) ) ) holds
( f1 - f2 is convergent_in-infty & lim_in-infty (f1 - f2) = () - () )
proof end;
theorem :: LIMFUNC1:93
for f being PartFunc of REAL,REAL st f is convergent_in-infty & f " = {} & lim_in-infty f <> 0 holds
( f ^ is convergent_in-infty & lim_in-infty (f ^) = () " )
proof end;
theorem :: LIMFUNC1:94
for f being PartFunc of REAL,REAL st f is convergent_in-infty holds
( abs f is convergent_in-infty & lim_in-infty (abs f) = |.().| )
proof end;
theorem Th95: :: LIMFUNC1:95
for f being PartFunc of REAL,REAL st f is convergent_in-infty & lim_in-infty f <> 0 & ( for r being Real ex g being Real st
( g < r & g in dom f & f . g <> 0 ) ) holds
( f ^ is convergent_in-infty & lim_in-infty (f ^) = () " )
proof end;
theorem Th96: :: LIMFUNC1:96
for f1, f2 being PartFunc of REAL,REAL st f1 is convergent_in-infty & f2 is convergent_in-infty & ( for r being Real ex g being Real st
( g < r & g in dom (f1 (#) f2) ) ) holds
( f1 (#) f2 is convergent_in-infty & lim_in-infty (f1 (#) f2) = () * () )
proof end;
theorem :: LIMFUNC1:97
for f1, f2 being PartFunc of REAL,REAL st f1 is convergent_in-infty & f2 is convergent_in-infty & lim_in-infty f2 <> 0 & ( for r being Real ex g being Real st
( g < r & g in dom (f1 / f2) ) ) holds
( f1 / f2 is convergent_in-infty & lim_in-infty (f1 / f2) = () / () )
proof end;
theorem :: LIMFUNC1:98
for f1, f2 being PartFunc of REAL,REAL st f1 is convergent_in+infty & lim_in+infty f1 = 0 & ( for r being Real ex g being Real st
( r < g & g in dom (f1 (#) f2) ) ) & ex r being Real st f2 | is bounded holds
( f1 (#) f2 is convergent_in+infty & lim_in+infty (f1 (#) f2) = 0 )
proof end;
theorem :: LIMFUNC1:99
for f1, f2 being PartFunc of REAL,REAL st f1 is convergent_in-infty & lim_in-infty f1 = 0 & ( for r being Real ex g being Real st
( g < r & g in dom (f1 (#) f2) ) ) & ex r being Real st f2 | is bounded holds
( f1 (#) f2 is convergent_in-infty & lim_in-infty (f1 (#) f2) = 0 )
proof end;
theorem Th100: :: LIMFUNC1:100
for f, f1, f2 being PartFunc of REAL,REAL st f1 is convergent_in+infty & f2 is convergent_in+infty & lim_in+infty f1 = lim_in+infty f2 & ( for r being Real ex g being Real st
( r < g & g in dom f ) ) & ex r being Real st
( ( ( (dom f1) /\ c= (dom f2) /\ & (dom f) /\ c= (dom f1) /\ ) or ( (dom f2) /\ c= (dom f1) /\ & (dom f) /\ c= (dom f2) /\ ) ) & ( for g being Real st g in (dom f) /\ holds
( f1 . g <= f . g & f . g <= f2 . g ) ) ) holds
( f is convergent_in+infty & lim_in+infty f = lim_in+infty f1 )
proof end;
theorem :: LIMFUNC1:101
for f, f1, f2 being PartFunc of REAL,REAL st f1 is convergent_in+infty & f2 is convergent_in+infty & lim_in+infty f1 = lim_in+infty f2 & ex r being Real st
( right_open_halfline r c= ((dom f1) /\ (dom f2)) /\ (dom f) & ( for g being Real st g in right_open_halfline r holds
( f1 . g <= f . g & f . g <= f2 . g ) ) ) holds
( f is convergent_in+infty & lim_in+infty f = lim_in+infty f1 )
proof end;
theorem Th102: :: LIMFUNC1:102
for f, f1, f2 being PartFunc of REAL,REAL st f1 is convergent_in-infty & f2 is convergent_in-infty & lim_in-infty f1 = lim_in-infty f2 & ( for r being Real ex g being Real st
( g < r & g in dom f ) ) & ex r being Real st
( ( ( (dom f1) /\ c= (dom f2) /\ & (dom f) /\ c= (dom f1) /\ ) or ( (dom f2) /\ c= (dom f1) /\ & (dom f) /\ c= (dom f2) /\ ) ) & ( for g being Real st g in (dom f) /\ holds
( f1 . g <= f . g & f . g <= f2 . g ) ) ) holds
( f is convergent_in-infty & lim_in-infty f = lim_in-infty f1 )
proof end;
theorem :: LIMFUNC1:103
for f, f1, f2 being PartFunc of REAL,REAL st f1 is convergent_in-infty & f2 is convergent_in-infty & lim_in-infty f1 = lim_in-infty f2 & ex r being Real st
( left_open_halfline r c= ((dom f1) /\ (dom f2)) /\ (dom f) & ( for g being Real st g in left_open_halfline r holds
( f1 . g <= f . g & f . g <= f2 . g ) ) ) holds
( f is convergent_in-infty & lim_in-infty f = lim_in-infty f1 )
proof end;
theorem :: LIMFUNC1:104
for f1, f2 being PartFunc of REAL,REAL st f1 is convergent_in+infty & f2 is convergent_in+infty & ex r being Real st
( ( (dom f1) /\ c= (dom f2) /\ & ( for g being Real st g in (dom f1) /\ holds
f1 . g <= f2 . g ) ) or ( (dom f2) /\ c= (dom f1) /\ & ( for g being Real st g in (dom f2) /\ holds
f1 . g <= f2 . g ) ) ) holds
lim_in+infty f1 <= lim_in+infty f2
proof end;
theorem :: LIMFUNC1:105
for f1, f2 being PartFunc of REAL,REAL st f1 is convergent_in-infty & f2 is convergent_in-infty & ex r being Real st
( ( (dom f1) /\ c= (dom f2) /\ & ( for g being Real st g in (dom f1) /\ holds
f1 . g <= f2 . g ) ) or ( (dom f2) /\ c= (dom f1) /\ & ( for g being Real st g in (dom f2) /\ holds
f1 . g <= f2 . g ) ) ) holds
lim_in-infty f1 <= lim_in-infty f2
proof end;
theorem :: LIMFUNC1:106
for f being PartFunc of REAL,REAL st ( f is divergent_in+infty_to+infty or f is divergent_in+infty_to-infty ) & ( for r being Real ex g being Real st
( r < g & g in dom f & f . g <> 0 ) ) holds
( f ^ is convergent_in+infty & lim_in+infty (f ^) = 0 )
proof end;
theorem :: LIMFUNC1:107
for f being PartFunc of REAL,REAL st ( f is divergent_in-infty_to+infty or f is divergent_in-infty_to-infty ) & ( for r being Real ex g being Real st
( g < r & g in dom f & f . g <> 0 ) ) holds
( f ^ is convergent_in-infty & lim_in-infty (f ^) = 0 )
proof end;
theorem :: LIMFUNC1:108
for f being PartFunc of REAL,REAL st f is convergent_in+infty & lim_in+infty f = 0 & ( for r being Real ex g being Real st
( r < g & g in dom f & f . g <> 0 ) ) & ex r being Real st
for g being Real st g in (dom f) /\ holds
0 <= f . g holds
f ^ is divergent_in+infty_to+infty
proof end;
theorem :: LIMFUNC1:109
for f being PartFunc of REAL,REAL st f is convergent_in+infty & lim_in+infty f = 0 & ( for r being Real ex g being Real st
( r < g & g in dom f & f . g <> 0 ) ) & ex r being Real st
for g being Real st g in (dom f) /\ holds
f . g <= 0 holds
f ^ is divergent_in+infty_to-infty
proof end;
theorem :: LIMFUNC1:110
for f being PartFunc of REAL,REAL st f is convergent_in-infty & lim_in-infty f = 0 & ( for r being Real ex g being Real st
( g < r & g in dom f & f . g <> 0 ) ) & ex r being Real st
for g being Real st g in (dom f) /\ holds
0 <= f . g holds
f ^ is divergent_in-infty_to+infty
proof end;
theorem :: LIMFUNC1:111
for f being PartFunc of REAL,REAL st f is convergent_in-infty & lim_in-infty f = 0 & ( for r being Real ex g being Real st
( g < r & g in dom f & f . g <> 0 ) ) & ex r being Real st
for g being Real st g in (dom f) /\ holds
f . g <= 0 holds
f ^ is divergent_in-infty_to-infty
proof end;
theorem :: LIMFUNC1:112
for f being PartFunc of REAL,REAL st f is convergent_in+infty & lim_in+infty f = 0 & ex r being Real st
for g being Real st g in (dom f) /\ holds
0 < f . g holds
f ^ is divergent_in+infty_to+infty
proof end;
theorem :: LIMFUNC1:113
for f being PartFunc of REAL,REAL st f is convergent_in+infty & lim_in+infty f = 0 & ex r being Real st
for g being Real st g in (dom f) /\ holds
f . g < 0 holds
f ^ is divergent_in+infty_to-infty
proof end;
theorem :: LIMFUNC1:114
for f being PartFunc of REAL,REAL st f is convergent_in-infty & lim_in-infty f = 0 & ex r being Real st
for g being Real st g in (dom f) /\ holds
0 < f . g holds
f ^ is divergent_in-infty_to+infty
proof end;
theorem :: LIMFUNC1:115
for f being PartFunc of REAL,REAL st f is convergent_in-infty & lim_in-infty f = 0 & ex r being Real st
for g being Real st g in (dom f) /\ holds
f . g < 0 holds
f ^ is divergent_in-infty_to-infty
proof end;
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Community Tip - Did you get an answer that solved your problem? Please mark it as an Accepted Solution so others with the same problem can find the answer easily. X
20-Turquoise
Integral sinc(x)
Quora shows interesting math. Prime 7 can show third equation is also pi.
I want to calculate until
and
does not show pi/2
need very long time to solve.
ACCEPTED SOLUTION
Accepted Solutions
20-Turquoise
(To:-MFra-)
10 REPLIES 10
21-Topaz II
(To:ttokoro)
Please note that the definition of Mathcad's sinc function differs from that used elsewhere. The Mathcad definition is sinc(x) = sin(x)/x, while on certain texts it is defined like this: sinc(x) = sin(πx) / (πx) so in the comparison of the calculated values and those of the text there may be differences. I'm referring to Mathcad 15. Certain errors in MC15 have been fixed in Prime 7.
23-Emerald III
(To:-MFra-)
Maple says:
Luc
23-Emerald III
(To:ttokoro)
Luc
20-Turquoise
(To:LucMeekes)
This is limit of Prime 7.
Therefore, I can't get the results shown by Quora web site.
21-Topaz II
(To:ttokoro)
20-Turquoise
(To:-MFra-)
TOL=10^-3
TOL=10^-7
Therefore, Quora and Luc show the answer is pi/2 for n=0 to 6. But n>6 are not calculate symbolically when using 15 digit accuracy. May be n>6 also show pi/2 symbolically when calculate more digit accuracy. Or Quora is true mathematically.
21-Topaz II
(To:ttokoro)
I use the sinc function according to the definition of the text reported here below. Sin (x) / x is something else ...
20-Turquoise
(To:-MFra-)
23-Emerald III
(To:-MFra-)
The sinc function is not included in the ISO standard for mathematical functions, ISO 80000-2:2013, nor in its 2019 update.
The 'NIST Handbook of mathematical functions' defines the Sinc function:
(but that's not the sinc function...)
There's a difference between the sinc function and the normalised sinc function according to Wikipedia. Follow the link to Rayleigh's formula to find how it was originally defined...
Success!
Luc
24-Ruby II
(To:ttokoro)
When I try to solve it (Mathcad 15 F000, Windows 7 x64, 16 GB of RAM), I get an error - "Internal error [1321]":
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# A-level Biology Maths Q's help!Watch
#1
I'd appreciate it if someone could explain how you got to each answer. Thankyou
a) A scientist measured the rate of removal of amino acids from a polypeptide with and without an enzyme present. With the enzyme present, 578 amino acids were released per second. Without the enzyme, 3.0 × 10–9 amino acids were released per second.
Calculate by how many times the rate of reaction is greater with the enzyme present.
b) Lactulose can also be used to treat people who have too high a concentration of hydrogen ions (H+) in their blood.
The normal range for blood H+ concentration is 3.55 × 10–8 to 4.47 × 10–8 mol dm–3
A patient was found to have a blood H+ concentration of 2.82 × 10–7 mol dm–3
Calculate the minimum percentage decrease required to bring the patient’s blood H+ concentration into the normal range.
0
2 months ago
#2
Null
Last edited by UrbanIncentive; 2 months ago
0
#3
I NEED HELP ON THIS
Some
prokaryotic cells can divide every 30 minutes. A liquid culture contained a
starting population of 1.35 × 104
cells.
Assuming each
cell divides every 30 minutes, calculate how many cells there will be after 3
hours. Assume no cells die during this time.
0
#4
I hate MATHS! these questions stress me out!
0
2 months ago
#5
(Original post by HumbleBee_x)
I NEED HELP ON THIS
Some
prokaryotic cells can divide every 30 minutes. A liquid culture contained a
starting population of 1.35 × 104
cells.
Assuming each
cell divides every 30 minutes, calculate how many cells there will be after 3
hours. Assume no cells die during this time.
I'm not 100% sure on this but if it divides every 30 mins, that means the population doubles every 30 mins. 3 hours = 6 x 30 mins, so it doubles 6 times, or 2^6. Multiply the original population by 2^6, i believe that's the answer.
0
#6
markscheme:
8.64
× 105;;
Accept 864 000 /
however expressed, e.g. 864 × 103 for 2 marks
Allow one mark
for
26 = 64
OR
64 / 26 × (1.35
× 104)
(Original post by ephemeral02)
I'm not 100% sure on this but if it divides every 30 mins, that means the population doubles every 30 mins. 3 hours = 6 x 30 mins, so it doubles 6 times, or 2^6. Multiply the original population by 2^6, i believe that's the answer.
0
2 months ago
#7
(Original post by HumbleBee_x)
markscheme:
8.64
× 105;;
Accept 864 000 /
however expressed, e.g. 864 × 103 for 2 marks
Allow one mark
for
26 = 64
OR
64 / 26 × (1.35
× 104)
yeah so 1.35x10^4 x 2^6 = 864000 = 8.64x10^5
0
2 months ago
#8
(1.35 × 104)*2
then do "ans*2" on ur calculator 5 times. which part of the question trips u up the most?
Last edited by anonymoussse; 2 months ago
0
2 months ago
#9
b) = 84.1
how much have u downgraded compared to ur original value? that's what % decrease means. use that and apply it to all % decrease questions.
if ur old value was 12 and it decreased to 3, ur % decrease is 9/12=75% because u have downgraded by 9 right? compared to what u had before, (12), what is the %? u didn't downgrade 100%, u've only downgraded 9/12=75% basically u've taken away 75% off ur original (12) value.
if ur ex was a 10, how much have u downgraded if ur new partner is a 4? u've downgraded by 6. now compare how much that is. 6/10=60%. u've downgraded a whole 60%
Last edited by anonymoussse; 2 months ago
0
2 months ago
#10
a) = 578/(3.0*10^-9) then put that in standard form (2.72*10^1)
u might hate the maths in chemistry A2 , they are harder than these ones. which part do u find tricky
Last edited by anonymoussse; 2 months ago
0
#11
why do you do 2^6 why isn't it just 6?
thankyou so much!!
(Original post by ephemeral02)
yeah so 1.35x10^4 x 2^6 = 864000 = 8.64x10^5
0
#12
where did you get 104 from?
I understand standard form, but when it comes to q's with a biological context I mess up!
(Original post by anonymoussse)
(1.35 × 104)*2
then do "ans*2" on ur calculator 5 times. which part of the question trips u up the most?
0
#13
yeah I understand that, but thankyou
and love the humour haha
(Original post by anonymoussse)
b) = 84.1
how much have u downgraded compared to ur original value? that's what % decrease means. use that and apply it to all % decrease questions.
if ur old value was 12 and it decreased to 3, ur % decrease is 9/12=75% because u have downgraded by 9 right? compared to what u had before, (12), what is the %? u didn't downgrade 100%, u've only downgraded 9/12=75% basically u've taken away 75% off ur original (12) value.
if ur ex was a 10, how much have u downgraded if ur new partner is a 4? u've downgraded by 6. now compare how much that is. 6/10=60%. u've downgraded a whole 60%
0
2 months ago
#14
(Original post by HumbleBee_x)
why do you do 2^6 why isn't it just 6?
thankyou so much!!
because whenever bacteria divides, it will double in size. say you start off with one bacteria and it divides once, that leaves you with two (1x2=2). then, the two each divide again, leaving you with four total (2x2=4). this means that every time it divides, you are multiplying the original number by two. because it will divide 6 times, you multiply it by two 6 times, which is 2x2x2x2x2x2 or 2^6. hope that makes more sense
0
#15
Thankyou so much!!! I understand.
(Original post by ephemeral02)
because whenever bacteria divides, it will double in size. say you start off with one bacteria and it divides once, that leaves you with two (1x2=2). then, the two each divide again, leaving you with four total (2x2=4). this means that every time it divides, you are multiplying the original number by two. because it will divide 6 times, you multiply it by two 6 times, which is 2x2x2x2x2x2 or 2^6. hope that makes more sense
0
2 months ago
#16
that happens a lot with a level. the actual question is easy but theyll scare u with context.
just ignore everything apart from the numbers and what u need to do with them, then treat the daunting numbers like more familiar ones....even in timed exams, u don't have enough time to understand everything. but it's okay because u don't need to. the context isn't always important. skim read everything and treat daunting questions as easy ones, because they are. theyre just disguised as hard
(Original post by HumbleBee_x)
yeah I understand that, but thankyou
and love the humour haha
0
#17
True indeed, the context does get me a bit stressed out
Timing is an issue for me too, I panic and just leave the maths Q's blank
0
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# Traveling faster than sound shockwaves
Contra Todd Gardiner's answer, it's not actually definitional of a shock wave that it travels faster than the usual speed of sound. It's normally produced by.
If the source is traveling faster than the waves, the waves never catch up to the If a jet exceeds the speed of sound, it produces a loud sonic boom, which can.
You may have heard this word already used in the context of supersonic traveling . That is exactly it. Shockwave is the event, whether it is visible...
## Traveling faster than sound shockwaves traveling Seoul
For the sake of other readers of this answer who might not have any idea what. The Pantheon of Derivatives — Part III. Understanding reference frames allows us to calculate a more difficult problem: what is the Doppler shift if both the source and the observer are moving? Although you are just standing by the side of the road, you are also standing on the Earth, which is spinning. Of course when considering a plane, all of the shock waves are oblique which complicates things a little bit, but that doesn't really change their propagation characteristics. The shock wave travels at the speed of sound relative to a weighted average of medium velocity, and is thus not an exception to the rule that wave travel at the speed of sound relative to the average velocity of the medium. Where to find me.
This is now a. When an object or disturbance moves faster than the information about it can propagate into the surrounding fluid, fluid near the disturbance cannot react or "get out of the way" before the disturbance arrives. In the frame of reference where the shock wave is stationary, entering medium travels towards the shock wave at super sonic speeds, and exiting medium travels away from the shock wave at sub sonic speeds. User Name Remember Me? I just heard a brief sound, the whole apartment building shook, and then a second or two later, I heard what sounded like an explosion.
The Speed of Light and Sound: Science Fun with Balloons for Kids of All Ages
### Traveling faster than sound shockwaves flying
When the object is moving faster than sound, the resulting sounds travel behind the object, creating a sonic boom. However, the beat frequency is something easily detected. Rather than try to sense the small shift in frequency, the bat focuses on the beat frequency between the sound that it produced and the reflected sound. It turns out that there is no difference between a moving source and a moving observer when it comes to the Doppler shift of light. Jokes, jokes and more jokes. This moving source of sound causes the Doppler Effect. Discuss the workings and policies of this site.
### Traveling: Traveling faster than sound shockwaves
Traveling faster than sound shockwaves Visualizing traveling salesman problem using matplotlib python Traveling faster than sound shockwaves 694 TRIPS WNWQ NORWAY HIKING TOUR Traveling service jehovah witnesses case Travel guides united states guide Trips from york city
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Stat 200 - Assignment #1_ Descriptive Statistics Data Analysis Plan.rtf - University of Maryland University College STAT200 Assignment#1 Descriptive
# Stat 200 - Assignment #1_ Descriptive Statistics Data Analysis Plan.rtf
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University of Maryland University College STAT200 - Assignment #1: Descriptive Statistics Data Analysis Plan Identifying Information Student (Full Name): Class: STAT 200 Instructor: Date: 9/26/2018 Scenario: Please write a few lines describing your scenario and the four variables (in addition to income) you have selected. I am a 35 year old, married man with two children. My income is \$50,000 per year. In addition to income, four variables include: SE-Marital status, SE-Family size, USD-Food, and USD-Meat. Use Table 1 to report the variables selected for this assignment. Note: The information for the required variable, “Income,” has already been completed and can be used as a guide for completing information on the remaining variables. Table 1. Variables Selected for the Analysis Variable Name in the Data Set Description (See the data dictionary for describing the variables.) Type of Variable (Qualitative or Quantitative) Variable 1: Income Annual household income in USD. Quantitative Variable 2: Marital Status Marital status of Head of Household Qualitative Variable 3: Family Size Total number of people in family. Quantitative Variable 4: Food Total amount of annual expenditures on food. Quantitative Variable 5: Meat Total amount of annual expenditures on meat. Quantitative Reason(s) for Selecting the Variables and Expected Outcome(s): 1. Variable 1: “Income” - The income I chose is the average income that an E5 makes in the Navy.
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13-11-2009, 02:46 PM #1 iamcreasy Registered User Join Date: Jun 2008 Location: Dhaka, Bangladesh Posts: 168 Thanks: 6 Thanked 8 Times in 8 Posts Edges... 1. why soft edge and hard edge is imp?Is it because, I wanna show this edge more prominently comparing to others?Or it also has some other job to do?? Something like, a pipe will have soft edge in the main body but the the outline of a groove, the edge should be hard. The edge that pass through are white are soft edges and the ones indicated by red lines are hard edge. EDGES: RESULT: and what is the meaning of edges having angle between 0 to 180?is it like soft hard edge? 2. Normal map don't work correctly if the hard edges are not kept separate.It is most likely those edges(!) has two normals of different angles. so, it there any method of making a hard edge smooth without costing extra polygon. Last edited by iamcreasy : 13-11-2009 at 02:48 PM.
13-11-2009, 08:42 PM #2 Joopson Registered User Join Date: Jul 2005 Location: Los Angeles, CA Posts: 2,313 Thanks: 1 Thanked 33 Times in 33 Posts the set normal angle button is what I think you are looking for. In Maya 2009, this option is in the modeling menu set in the normals drop-down menu, The number you set it too means that any angle which is less than that will be soft. Therefore, an angle of 90 degrees would be soft if I set it to 91. You can select an edge you want hard and set it to 0, and it will make that one edge hard. You can select a loop and do the same thing. This tool is your best friend :bandit: __________________ Environment Artist @ Plastic Piranha www.joopson.com
14-11-2009, 09:17 AM #3 iamcreasy Registered User Join Date: Jun 2008 Location: Dhaka, Bangladesh Posts: 168 Thanks: 6 Thanked 8 Times in 8 Posts Thank you very much. But I asked , what is the significant of hard and soft edge.Is it because to make something more visible and something less? What is the importance of angle between 0 to 90(Excluding this 2 values)
14-11-2009, 09:25 PM #4 stwert EduSciVis-er Join Date: Dec 2005 Location: Toronto Posts: 3,371 Thanks: 74 Thanked 709 Times in 642 Posts Just a wild guess, but I would say it determines how the renderer calculates the normal of that face. If it's soft, it takes into account adjacent normals, if hard, then it maybe doesn't... am I on the right track?? EDIT: Or are you just asking what it's used for? Like soft areas vs. sharp edges...? __________________ -stwert Website - Blog - YouTube - Vimeo - Facebook - Twitter -->How to Effectively Ask Questions on Forums<--
14-11-2009, 09:52 PM #5 Joopson Registered User Join Date: Jul 2005 Location: Los Angeles, CA Posts: 2,313 Thanks: 1 Thanked 33 Times in 33 Posts When you soften an edge, it makes it appear more rounded, and when you harden it, it looks sharp. I am a bit confused about your question, so I am sorry if I am not answering it. __________________ Environment Artist @ Plastic Piranha www.joopson.com
15-11-2009, 11:49 AM #6 iamcreasy Registered User Join Date: Jun 2008 Location: Dhaka, Bangladesh Posts: 168 Thanks: 6 Thanked 8 Times in 8 Posts Do i need to use hard edge when the corner of a box is beveled? What is the significance(is case of gaming) if the angle is 30(or any value except 0 and 90)
15-11-2009, 08:12 PM #7 Chirone Subscriber Join Date: Dec 2007 Location: NZ Posts: 3,124 Thanks: 11 Thanked 147 Times in 143 Posts usually the bevel operator will make the edges soft except for the ones in the corner (show hard/soft edges and you'll see what i'm talking about) as for your second question i think what stwert said holds for games as well. it all depends on whether or not the game engine will read that data i guess... __________________ that's a "Ch" pronounced as a "K" Computer skills I should have: Objective C, C#, Java, MEL. Python, C++, XML, JavaScript, XSLT, HTML, SQL, CSS, FXScript, Clips, SOAR, ActionScript, OpenGL, DirectX Maya, XSI, Photoshop, AfterEffects, Motion, Illustrator, Flash, Swift3D
16-11-2009, 04:39 AM #8
iamcreasy
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Join Date: Jun 2008
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Originally posted by Chirone usually the bevel operator will make the edges soft except for the ones in the corner (show hard/soft edges and you'll see what i'm talking about) as for your second question i think what stwert said holds for games as well. it all depends on whether or not the game engine will read that data i guess...
hmm, i checked, but all edges of a simple cube becomes soft.
then , what about normal(something else different from gaming) modeling, what is the significant of different angles?Why should someone use them?
Similar Threads Thread Thread Starter Forum Replies Last Post jerry can ctbram Work In Progress 29 26-11-2010 03:51 PM
A little bit about who we are
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# The candle lab: calculations Data: 0.95 g, 30.6 °C 1. q = heat/energy change, units: J or Joules c = specific heat, units: J/g°C m = mass, units: g or.
## Presentation on theme: "The candle lab: calculations Data: 0.95 g, 30.6 °C 1. q = heat/energy change, units: J or Joules c = specific heat, units: J/g°C m = mass, units: g or."— Presentation transcript:
The candle lab: calculations Data: 0.95 g, 30.6 °C 1. q = heat/energy change, units: J or Joules c = specific heat, units: J/g°C m = mass, units: g or grams T= temperature change, units: °C 2. 200 g 3. q = cm T = 4.18 J/g°C x 200 g x 30.6 °C = 25 582 J or 25.6 kJ
4. 25.581 kJ/0.95 g wax = 26.9 kJ = 27 kJ 5. First, assume the heat absorbed by the water is equal to the heat released by the candle 1 mol C 25 H 52 = 352 g or 352.52 #mol= 0.95 gx 1 mol / 353 g= 0.00269 mol kJ/mol = 25.581 kJ / 0.00269 mol = 9 510 kJ/mol = 9.5 x 10 6 J/mol 6. Calorimeter: An apparatus used in the determination of the heat of a reaction Calorimetry: The science of measuring the quantities of heat that are involved in a chemical or physical change.
7. The calorimeter is made up of more than just water (the metal from the can) Other sources of error include anything that interferes with the transfer of heat or the measurement of that heat. E.g. incomplete transfer of heat, loss of heat, evaporation of water, incomplete combustion. Not:experimental error, inaccurate thermometer For more lessons, visit www.chalkbored.com www.chalkbored.com
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Questions on Fugue No. 6
Well-Tempered Clavier, Book I
by Johann Sebastian Bach
Your instructor may have referred you to this page with instructions to forward your answers to the following questions on the D minor fugue (Flash or Shockwave). To do this you will need to enter YOUR NAME and INSTRUCTOR'S EMAIL in the spaces provided. When you have answered all of the questions, punch the "Send to Instructor" button.
``` YOUR NAME:
INSTRUCTOR'S EMAIL:
```
1. If the universe was once compressed into a space smaller than the head of a pin, the universe of this fugue is compressed into the space of its:
tail.
subject.
countersubject.
2. The following example reveals that the countersubject is generated from the subject. Which statement best reflects this relationship?
Figure (c) is generated from (b), and (d) from (b).
Figure (c) is generated from (a), and (d) from (a).
Figure (c) is generated from (b), and (d) from (a).
Figure (c) is generated from (a), and (d) from (b).
3. The subject has three falling thirds. Two of them are followed by a rising second. How are these two falling-third-rising-second figures related?
The 2nd is a melodic inversion of the 1st.
The 2nd is a retrograde inversion of the 1st.
The 2nd is a rhythmic augmentation of the 1st.
The 2nd is a rhythmic diminution of the 1st.
4. This fugue contains the first instance of a melodically inverted subject in the Well-Tempered Clavier. How many times is the subject inverted?
twice
three times
four times
five times
5. This fugue contains many stretti. Which of the following combinations are heard?
• subject in stretto with itself
• subject in stretto with its inversion
• inverted subject in stretto with itself
none of the above
one of the above
two of the above
all of the above
6. Aggregates of the subject are heard in mm. 12 and 33. Which of the following statements are true?
• In both instances the subject's head is in counterpoint with its tail.
• In m. 12 the head is inverted while the tail is upright.
• In m. 33 the head is upright while the tail is inverted.
one of the above
two of the above
all of the above
7. What is the figural source for the inside voices of the second-to-last measure. (You may wish to refer to the musical example of question no. 2.)
• The tenor is derived from figure (a) and the alto from (b).
• The alto and tenor are both derived from figure (a).
• The alto is the retrograde-inversion of the tenor.
none of the above
one of the above
two of the above
all of the above
8. The countersubject is to the subject as:
the earth is to the moon.
the moon is to Saturn.
Saturn is to the earth.
Titan is to Saturn.
9. That everything in this fugue emanates from its subject reinforces the principle of:
• Fortspinnung
• iron being iron no matter where it is found.
• good ideas tending to generate new ones.
• related structures indicating a common origin.
one of the above
two of the above
three of the above
all of the above
10. With which of the following statements would Bach have agreed?
• A good composer finds motives that can generate new ones.
• Many self-accompanying motives have yet to be discovered.
• Elaboratio is the precursor to inventio.
none of the above
one of the above
two of the above
all of the above
Don't forget to enter your name and instructor's email at the top of this page, then click the "Send to Instructor" button.
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johnjax - 3 months ago 14
Python Question
# Optimizing Python distance calculation while accounting for periodic boundary conditions
I have written a Python script to calculate the distance between two points in 3D space while accounting for periodic boundary conditions. The problem is that I need to do this calculation for many, many points and the calculation is quite slow. Here is my function.
``````def PBCdist(coord1,coord2,UC):
dx = coord1[0] - coord2[0]
if (abs(dx) > UC[0]*0.5):
dx = UC[0] - dx
dy = coord1[1] - coord2[1]
if (abs(dy) > UC[1]*0.5):
dy = UC[1] - dy
dz = coord1[2] - coord2[2]
if (abs(dz) > UC[2]*0.5):
dz = UC[2] - dz
dist = np.sqrt(dx**2 + dy**2 + dz**2)
return dist
``````
I then call the function as so
``````for i, coord2 in enumerate(coordlist):
do something with i
``````
Recently I read that I can greatly increase performance by using list comprehension. The following works for the non-PBC case, but not for the PBC case
``````coord_indices = [i for i, y in enumerate([np.sqrt(np.sum((coord2-coord1)**2)) for coord2 in coordlist]) if y < radius]
for i in coord_indices:
do something
``````
Is there some way to do the equivalent of this for the PBC case? Is there an alternative that would work better?
You should write your `distance()` function in a way that you can vectorise the loop over the 5711 points. The following implementation accepts an array of points as either the `x0` or `x1` parameter:
``````def distance(x0, x1, dimensions):
delta = numpy.abs(x0 - x1)
delta = numpy.where(delta > 0.5 * dimensions, delta - dimensions, delta)
return numpy.sqrt((delta ** 2).sum(axis=-1))
``````
Example:
``````>>> dimensions = numpy.array([3.0, 4.0, 5.0])
>>> points = numpy.array([[2.7, 1.5, 4.3], [1.2, 0.3, 4.2]])
>>> distance(points, [1.5, 2.0, 2.5], dimensions)
array([ 2.22036033, 2.42280829])
``````
The result is the array of distances between the points passed as second parameter to `distance()` and each point in `points`.
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##### Question Info
This question is public and is used in 138 tests or worksheets.
Type: Multiple-Choice
Category: Fractions and Ratios
Standards: 5.NBT.A.1
Author: ksilvers12
View all questions by ksilvers12.
# Fractions and Ratios Question
View this question.
Add this question to a group or test by clicking the appropriate button below.
## Grade 5 Fractions and Ratios CCSS: 5.NBT.A.1
The Harris family pays $200,000 for a new house. They make a down payment that is $1//10$ of the price of the house. How much is the down payment? 1.$20
2. $200 3.$2,000
4. \$20,000
You need to have at least 5 reputation to vote a question down. Learn How To Earn Badges.
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #1 2012-12-12 00:06:57
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Proof
Hi,
I really need help seeing if I answered the following correct:
I'll give you an assumption, and ask you to provide proof for the assumption. If there is no proof for the assumption, the answer is "unfounded."
I choose A
1. If I have two coplanar lines, I must have a plane.
A-unfounded
B-Definition of a point
C-Definition of a plane
D-Given
E -Definition of a line
Last edited by zee-f (2012-12-12 00:41:33)
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Offline
## #2 2012-12-12 04:43:32
bob bundy
Registered: 2010-06-20
Posts: 8,462
### Re: Proof
hi zee-f,
I think this can be proved. Isn't a plane defined by 3 (non-colinear ) points?
How could you find 3 such points on those two lines?
But I don't understand how you can prove that with just a letter. Have you got any examples of this sort of thing?
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy
Offline
## #3 2012-12-12 05:54:47
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,039
### Re: Proof
A is correct.
The statement isn't even true. Two lines that are actually the same line do not define a single plane, but are coplanar.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
Offline
## #4 2012-12-12 06:02:55
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
I answered A and it was incorrect .
Yeah this lesson is confusing but I did answer 14 correctly like this one :
(F ) was correct
6. In the figure above, line segment MC is equal to imaginary line segment MI.
A Given
Bunfounded
CDefinition of supplementary angles
D1267200 inches
E Definition of an octagon
F Definition of a circle: all points are equidistant from the center
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Offline
## #5 2012-12-12 06:04:12
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
I have to use the information I know to proof the statement correct
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Offline
## #6 2012-12-12 06:47:20
bob bundy
Registered: 2010-06-20
Posts: 8,462
### Re: Proof
stefy wrote:
Two lines that are actually the same line do not define a single plane, but are coplanar.
I did consider this, but rejected this interpretation on the grounds that 'two lines' should mean exactly two distinct lines not one line counted twice.
That's the trouble with using English to make mathematical statements. It sometimes isn't precise enough.
As zee-f has had A marked wrong, I think we have further evidence that the questioner was thinking that way too.
So let's assume the lines are either (i) distinct and parallel or (ii) they cross at a point
So we either have 4 distinct points or at least 3.
What do we need to define a plane?
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy
Offline
## #7 2012-12-12 07:29:35
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
Hi,
According to my online courses A plane is defined by any of the following:
three points not lying on a line
a line and a point not lying on the line
two lines which intersect in a single point or are parallel
So I think C would be a correct answer
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Offline
## #8 2012-12-12 08:11:35
bob bundy
Registered: 2010-06-20
Posts: 8,462
### Re: Proof
Yes, C sounds good. But I thought you had to supply the proof as well. Maybe not.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy
Offline
## #9 2012-12-12 09:30:23
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
I am still confused on what am really doing
I got 6 incorrect out of 20
I choose A and it was incorrect
7. In the figure above, line segment EJ is equal to line segment JM
Bunfounded
CDefinition of an octagon
D1267200 inches
E Given
F Definition of supplementary angles
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Offline
## #10 2012-12-12 09:43:58
bob bundy
Registered: 2010-06-20
Posts: 8,462
### Re: Proof
6/20 eeekkk!
Oh hang on. 6 wrong. Oh that's not so bad. 70% is a good score.
But we'll get them sorted. Don't worry.
I need to see the diagram for this one.
bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy
Offline
## #11 2012-12-12 12:36:58
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
yup 14/ 20
The lesson uses the same chart for all the questions that use the chart
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Offline
## #12 2012-12-12 20:25:12
bob bundy
Registered: 2010-06-20
Posts: 8,462
### Re: Proof
hi zee-f,
Oh, that diagram. I remember that from another set of questions.
So EJ = JM ? They're not telling that; they're asking is it true?
Take a look at the diagram. Is J half way along EM ?
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy
Offline
## #13 2012-12-13 05:15:48
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
no EJ ≠ Jm
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Offline
## #14 2012-12-13 05:16:55
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
So B would be a good answer ?
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Offline
## #15 2012-12-13 06:06:32
bob bundy
Registered: 2010-06-20
Posts: 8,462
### Re: Proof
Yes, that's what I would choose.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy
Offline
## #16 2012-12-13 07:10:51
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
I choose E and it was incorrect
14. If a central angle is 30 degrees, then the arc it defines is also 30 degrees.
A Given
BDefinition of an inscribed angle
Cunfounded
DProperties of a central angle
E Properties of an arc
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Offline
## #17 2012-12-13 07:21:16
bob bundy
Registered: 2010-06-20
Posts: 8,462
### Re: Proof
I see why you said E to start with. I would say that D and E are the same. But if E isn't acceptable, D seems good to me.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy
Offline
## #18 2012-12-13 07:33:10
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
yeah I was stuck on which one to choose to. lol
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Offline
## #19 2012-12-13 07:34:26
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
E was incorrect I still don't see why O_o
16. If a radius bisects a chord, then the lengths of the parts of the radius on either side of the chord are equal.
A Given
BDefinition of a chord
Cunfounded
DDefinition of supplementary angles
E Definition of a bisector
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Offline
## #20 2012-12-13 07:46:55
bob bundy
Registered: 2010-06-20
Posts: 8,462
### Re: Proof
I think my diagrams will show you what to do here.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy
Offline
## #21 2012-12-13 08:23:42
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
No the radius isn't cut into equal parts so that whole statement is incorrect So C would be a good answer
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Offline
## #22 2012-12-13 08:27:39
bob bundy
Registered: 2010-06-20
Posts: 8,462
### Re: Proof
Yes. That's what I think.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy
Offline
## #23 2012-12-13 11:16:00
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
2 more left
I answered C for both it was incorrect
19. The given points (4, -8), (4, -5), and (-2, -6) make a right triangle.
A Distance Formula
B Definition of a right triangle
CDefinition of a triangle
Dunfounded
E Pythagorean Theorem
20. The given points (2, -3), (-7, -7), (2, -7), and (-7, -2) make a square.
A Definition of coordinate
BPythagorean Theorem
CDefinition of a square
DDefinition of supplementary angles
E Distance Formula
F unfounded
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Offline
## #24 2012-12-13 11:38:25
bob bundy
Registered: 2010-06-20
Posts: 8,462
### Re: Proof
hi
Did you try plotting the points? Have a look at my diagram.
I think you'll see what to do then.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy
Offline
## #25 2012-12-13 11:56:09
zee-f
Member
Registered: 2011-05-12
Posts: 1,220
### Re: Proof
No I didn't plot them probably why I got the question wrong the don't make a square or a right triangle So both (unfounded)
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Offline
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A203912 Number of (n+1)X5 0..6 arrays with column and row pair sums b(i,j)=a(i,j)+a(i,j-1) and c(i,j)=a(i,j)+a(i-1,j) such that rows of b(i,j) and columns of c(i,j) are lexicographically nondecreasing 1
2788842, 11575205434, 63988637491851, 270356138202215532, 898462061528909496153, 2459051354235725726537096, 5732451730121715144022348644, 11655318352946527961268400859823 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Column 4 of A203913 LINKS R. H. Hardin, Table of n, a(n) for n = 1..9 EXAMPLE Some solutions for n=4 ..6..4..4..4..4....2..5..1..1..1....1..0..0..4..4....2..3..2..4..6 ..4..6..6..6..6....5..2..6..6..6....0..1..4..2..2....6..5..6..5..5 ..6..5..5..5..5....1..6..3..3..6....1..4..0..4..4....5..6..5..6..4 ..5..6..6..6..6....4..4..3..3..0....0..5..1..2..4....6..5..6..5..5 ..5..6..6..6..6....5..3..6..6..2....6..2..2..1..2....5..6..5..6..4 CROSSREFS Sequence in context: A295471 A316487 A250535 * A234189 A321056 A033534 Adjacent sequences: A203909 A203910 A203911 * A203913 A203914 A203915 KEYWORD nonn AUTHOR R. H. Hardin Jan 07 2012 STATUS approved
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Search for a tool
Base 26 Cipher
Tool to decrypt/encrypt in Base 26. Base 26 uses 26 symbols, by using the alphabet's letter, one can encrypt words with numbers and conversely.
Results
Base 26 Cipher -
Tag(s) : Cryptography, Mathematics
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# Base 26 Cipher
## Base 26 Decoder/Converter
### Parameters
Value for A A=0, AA=00, Z=25, BA=26 (Recommanded for maths) A=1, AA=27, Z=26, BA=53 (Recommanded for crypto)
Letters order Normal Reversed (Over-encryption)
## Base 26 Encoder
### Parameters
Value for A A=0, AA=00, Z=25, BA=26 (Recommanded for maths) A=1, AA=27, Z=26, BA=53 (Recommanded for crypto)
Letters order Normal Reversed (Over-encryption)
Tool to decrypt/encrypt in Base 26. Base 26 uses 26 symbols, by using the alphabet's letter, one can encrypt words with numbers and conversely.
### How to encrypt using Base 26 cipher
The encoding with hexavigesimal uses an arithmetic base change from base 26 to base 10. The words are considered as written in base 26 (with 26 symbols: the 26 letters of the alphabet ABCDEFGHIJKLMNOPQRSTUVWXYZ) and converted to base 10.
Example: To code DCODE, written in base 26, convert it to base 10: D=3, C=2, O=14, D=3, E=4 so $$3 \times 26^4 + 2 \times 26^3 + 14 \times 26^2 + 3 \times 26^1 + 4 \times 26^0 = 1415626$$
This method is the most rigorous mathematically, but can raise problems for encrypting words starting with A (which corresponds to the 0 symbol in base 10) and is thus generally ignored at the beginning of the number (001 = 1).
### How to decrypt Base 26 cipher
Hexavigesimal decryption consists of the conversion from the base 10 to the base 26 (using the words as hexavigesimal numbers with the 26 letters of the alphabet as base symbols).
Example: $$1415626 = 3 \times 26^4 + 2 \times 26^3 + 14 \times 26^2 + 3 \times 26^1 + 4 \times 26^0$$ so [3,2,14,3,4] in base 26 and 3=D, 2=C, 14=O, 3=D, 4=E. The plain message is DCODE.
### How to recognize a Base 26 ciphertext?
The ciphered message is made of numbers, relatively big (for long words)
Usual word can appears multiple times with the same value in a long text.
### What is the reverse order letters option?
Rather than converting normally, the reverse order of letters can be considered (or the word reversed):
Example: DCODE = $$3 \times 26^0 + 2 \times 26^1 + 14 \times 26^2 + 3 \times 26^3 + 4 \times 26^4 = 1890151$$ (this is equivalent to coding EDOCD).
### How to deal with messages starting with 'A'?
as A is encoded 0 in base 26, when encoding it is null and disappear when decoding.
Example: AB = 0*26^1+1*26^0 = 1 and 1 = B
Add a zero at the beginning of a number to indicate a A at the beginning of a word.
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http://www.cs.mcgill.ca/~cs202/ 1) a)c >= '0' && c <= '9' b)i % j == 0 || j % i == 0 c)s1.equals(s2) d)Math.abs(d1-d2) <= .0001 2) a)int 0 b)boolean false c)String fox d)boolean false (note that it's always false since dividing by 5 never leaves you with remainder of 7. If you noticed this you got a bonus point, if you proved you knew what modulus meant you got full credit) 3) a) 2,4,6,8, b) F,E,D,C,B,A c) xoF n*ilE d) ***** **** *** ** * 4) a) public String[] voteByRank(String[] names, int[] ranks) { String[] orderedNames = new String[names.length]; for (int i=0; i < orderedNames.length; i++) { orderedNames[ranks[i] - 1] = names[i]; } return orderedNames; } b) public String[] voteByIndex(String[] names, int[] indexes) { String[] orderedNames = new String[names.length]; for (int i=0; i < orderedNames.length; i++) { orderedNames[i] = names[indexes[i] - 1]; } return orderedNames; } c) public int[] convertFromIndexToRank(int[] indexes) { int[] ranks = new int[indexes.length]; for (int i=0; i < ranks.length; i++) { ranks[indexes[i] - 1] = i + 1; } return ranks; } 5) Point2D A = new Point2D(1.0,1.0); Point2D B = new Point2D(2.5,1.5); Point2D C = new Point2D(2.0, 2.0); Vector2D AB = new Vector2D(A,B); Vector2D AC = new Vector2D(A,C); Vector2D AH = AC.projectOnto(AB); Point2D H = A.moveBy(AH); Vector2D HC = new Vector2D(H,C); System.out.println("The area is " + HC.getLength() * AB.getLength() / 2); 6) import java.util.Scanner; public class FlintAnalyser { public static void main(String[] args) { System.out.println("Please enter the name of the file to be processed"); Scanner keyboardReader = new Scanner(System.in); String filename = keyboardReader.nextLine(); Scanner scan = new Scanner(new File(filename)); double height = scan.nextDouble(); int blades = 0; int bladeletts = 0; int elongatedFlakes = 0; int wideFlakes = 0; int numFlints = 0; while ( height != 0.0 ) { double width = scan.nextDouble(); numFlints++; if ( height > 2 * width ) { blades++; if ( height < 5 && width < 1.3) { bladeletts++; } } else if ( height > width) { elongatedFlakes++; } else { wideFlakes++; } height = scan.nextDouble(); } System.out.println("Total number of flints processed: " + numFlints); System.out.println(" - " + numBlades + " blades (" + blades * 100.0/ numFlints + " %)," + bladeletts + "(" + (bladeletts * 100.0) / (blades) + ") of which are bladeletts."); System.out.println(" - " + (elongatedFlakes + wideFlakes) + " flakes (" + ((elongatedFlakes + wideFlakes) * 100.0 / numFlints) + " %), of which " + elongatedFlakes + "(" + (elongatedFlakes * 100.0 / (elongatedFlakes + wideFlakes)) + "%) are elongated, and " + wideFlakes + "(" + (wideFlakes * 100.0 / (elongatedFlakes + wideFlakes)) + "%) are wide"); } } 7) int numStars = 0; int numOnes = 0; for (int i =0; i < number.length(); i++) { char digit = number.charAt(i); if ( digit != '0' && digit != '1' && digit != '*') { System.out.println("Invalid input"); return; } if ( digit == '*') { numStars++; if (numStars > 1) { System.out.println("Corrupted beyond repair!"); return; } } if (digit == '1') { numOnes++; } } if (numOnes % 2 == 0) { String finalNumber = number.replace('*', '0'); System.out.println(finalNubmer); } else { System.out.println(number.replace('*', '0')); } //String finalNumber = numOnes % 2 == 0 ? number.replace('*','0') : number.replace('*', '0'); //String finalNumber = number.replace('*' , '0' + numOnes % 2);
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#### 24.1.1 Matlab-compatible solvers
Octave also provides a set of solvers for initial value problems for Ordinary Differential Equations that have a MATLAB-compatible interface. The options for this class of methods are set using the functions.
• `odeset`
• `odeget`
Currently implemented solvers are:
• Runge-Kutta methods
• `ode45` Integrates a system of non–stiff ordinary differential equations (non–stiff ODEs and DAEs) using second order Dormand-Prince method. This is a fourth–order accurate integrator therefore the local error normally expected is O(h^5). This solver requires six function evaluations per integration step.
• `ode23` Integrates a system of non–stiff ordinary differential equations (non-stiff ODEs and DAEs) using second order Bogacki-Shampine method. This is a second-order accurate integrator therefore the local error normally expected is O(h^3). This solver requires three function evaluations per integration step.
: [t, y] = ode45 (fun, trange, init)
: [t, y] = ode45 (fun, trange, init, ode_opt)
: [t, y, te, ye, ie] = ode45 (…)
: solution = ode45 (…)
Solve a set of non-stiff Ordinary Differential Equations (non-stiff ODEs) with the well known explicit Dormand-Prince method of order 4.
fun is a function handle, inline function, or string containing the name of the function that defines the ODE: `y' = f(t,y)`. The function must accept two inputs where the first is time t and the second is a column vector of unknowns y.
trange specifies the time interval over which the ODE will be evaluated. Typically, it is a two-element vector specifying the initial and final times (`[tinit, tfinal]`). If there are more than two elements then the solution will also be evaluated at these intermediate time instances.
By default, `ode45` uses an adaptive timestep with the `integrate_adaptive` algorithm. The tolerance for the timestep computation may be changed by using the options `"RelTol"` and `"AbsTol"`.
init contains the initial value for the unknowns. If it is a row vector then the solution y will be a matrix in which each column is the solution for the corresponding initial value in init.
The optional fourth argument ode_opt specifies non-default options to the ODE solver. It is a structure generated by `odeset`.
The function typically returns two outputs. Variable t is a column vector and contains the times where the solution was found. The output y is a matrix in which each column refers to a different unknown of the problem and each row corresponds to a time in t.
The output can also be returned as a structure solution which has a field x containing a row vector of times where the solution was evaluated and a field y containing the solution matrix such that each column corresponds to a time in x. Use `fieldnames (solution)` to see the other fields and additional information returned.
If using the `"Events"` option then three additional outputs may be returned. te holds the time when an Event function returned a zero. ye holds the value of the solution at time te. ie contains an index indicating which Event function was triggered in the case of multiple Event functions.
Example: Solve the Van der Pol equation
```fvdp = @(t,y) [y(2); (1 - y(1)^2) * y(2) - y(1)];
[t,y] = ode45 (fvdp, [0, 20], [2, 0]);
```
: [t, y] = ode23 (fun, trange, init)
: [t, y] = ode23 (fun, trange, init, ode_opt)
: [t, y, te, ye, ie] = ode23 (…)
: solution = ode23 (…)
Solve a set of non-stiff Ordinary Differential Equations (non-stiff ODEs) with the well known explicit Bogacki-Shampine method of order 3. For the definition of this method see http://en.wikipedia.org/wiki/List_of_Runge%E2%80%93Kutta_methods.
fun is a function handle, inline function, or string containing the name of the function that defines the ODE: `y' = f(t,y)`. The function must accept two inputs where the first is time t and the second is a column vector of unknowns y.
trange specifies the time interval over which the ODE will be evaluated. Typically, it is a two-element vector specifying the initial and final times (`[tinit, tfinal]`). If there are more than two elements then the solution will also be evaluated at these intermediate time instances.
By default, `ode23` uses an adaptive timestep with the `integrate_adaptive` algorithm. The tolerance for the timestep computation may be changed by using the options `"RelTol"` and `"AbsTol"`.
init contains the initial value for the unknowns. If it is a row vector then the solution y will be a matrix in which each column is the solution for the corresponding initial value in init.
The optional fourth argument ode_opt specifies non-default options to the ODE solver. It is a structure generated by `odeset`.
The function typically returns two outputs. Variable t is a column vector and contains the times where the solution was found. The output y is a matrix in which each column refers to a different unknown of the problem and each row corresponds to a time in t.
The output can also be returned as a structure solution which has a field x containing a row vector of times where the solution was evaluated and a field y containing the solution matrix such that each column corresponds to a time in x. Use `fieldnames (solution)` to see the other fields and additional information returned.
If using the `"Events"` option then three additional outputs may be returned. te holds the time when an Event function returned a zero. ye holds the value of the solution at time te. ie contains an index indicating which Event function was triggered in the case of multiple Event functions.
This function can be called with two output arguments: t and y. Variable t is a column vector and contains the time stamps, instead y is a matrix in which each column refers to a different unknown of the problem and the rows number is the same of t rows number so that each row of y contains the values of all unknowns at the time value contained in the corresponding row in t.
Example: Solve the Van der Pol equation
```fvdp = @(t,y) [y(2); (1 - y(1)^2) * y(2) - y(1)];
[t,y] = ode23 (fvdp, [0, 20], [2, 0]);
```
: odestruct = odeset ()
: odestruct = odeset ("field1", value1, "field2", value2, …)
: odestruct = odeset (oldstruct, "field1", value1, "field2", value2, …)
: odestruct = odeset (oldstruct, newstruct)
: odeset ()
Create or modify an ODE options structure.
When called with no input argument and one output argument, return a new ODE options structure that contains all possible fields initialized to their default values. If no output argument is requested, display a list of the common ODE solver options along with their default value.
If called with name-value input argument pairs "field1", "value1", "field2", "value2", … return a new ODE options structure with all the most common option fields initialized, and set the values of the fields "field1", "field2", … to the values value1, value2, ….
If called with an input structure oldstruct then overwrite the values of the options "field1", "field2", … with new values value1, value2, … and return the modified structure.
When called with two input ODE options structures oldstruct and newstruct overwrite all values from the structure oldstruct with new values from the structure newstruct. Empty values in newstruct will not overwrite values in oldstruct.
The most commonly used ODE options, which are always assigned a value by `odeset`, are the following:
AbsTol
Absolute error tolerance.
BDF
Use BDF formulas in implicit multistep methods. Note: This option is not yet implemented.
Events
Event function. An event function must have the form `[value, isterminal, direction] = my_events_f (t, y)`
InitialSlope
Consistent initial slope vector for DAE solvers.
InitialStep
Initial time step size.
Jacobian
Jacobian matrix, specified as a constant matrix or a function of time and state.
JConstant
Specify whether the Jacobian is a constant matrix or depends on the state.
JPattern
If the Jacobian matrix is sparse and non-constant but maintains a constant sparsity pattern, specify the sparsity pattern.
Mass
Mass matrix, specified as a constant matrix or a function of time and state.
MassSingular
Specify whether the mass matrix is singular. Accepted values include `"yes"`, `"no"`, `"maybe"`.
MaxOrder
Maximum order of formula.
MaxStep
Maximum time step value.
MStateDependence
Specify whether the mass matrix depends on the state or only on time.
MvPattern
If the mass matrix is sparse and non-constant but maintains a constant sparsity pattern, specify the sparsity pattern. Note: This option is not yet implemented.
NonNegative
Specify elements of the state vector that are expected to remain nonnegative during the simulation.
NormControl
Control error relative to the 2-norm of the solution, rather than its absolute value.
OutputFcn
Function to monitor the state during the simulation. For the form of the function to use see `odeplot`.
OutputSel
Indices of elements of the state vector to be passed to the output monitoring function.
Refine
Specify whether output should be returned only at the end of each time step or also at intermediate time instances. The value should be a scalar indicating the number of equally spaced time points to use within each timestep at which to return output. Note: This option is not yet implemented.
RelTol
Relative error tolerance.
Stats
Print solver statistics after simulation.
Vectorized
Specify whether `odefun` can be passed multiple values of the state at once.
Field names that are not in the above list are also accepted and added to the result structure.
: val = odeget (ode_opt, field)
: val = odeget (ode_opt, field, default)
Query the value of the property field in the ODE options structure ode_opt.
If called with two input arguments and the first input argument ode_opt is an ODE option structure and the second input argument field is a string specifying an option name, then return the option value val corresponding to field from ode_opt.
If called with an optional third input argument, and field is not set in the structure ode_opt, then return the default value default instead.
: stop_solve = odeplot (t, y, flag)
Open a new figure window and plot the solution of an ode problem at each time step during the integration.
The types and values of the input parameters t and y depend on the input flag that is of type string. Valid values of flag are:
`"init"`
The input t must be a column vector of length 2 with the first and last time step (`[tfirst tlast]`. The input y contains the initial conditions for the ode problem (y0).
`""`
The input t must be a scalar double specifying the time for which the solution in input y was calculated.
`"done"`
The inputs should be empty, but are ignored if they are present.
`odeplot` always returns false, i.e., don’t stop the ode solver.
Example: solve an anonymous implementation of the `"Van der Pol"` equation and display the results while solving.
```fvdp = @(t,y) [y(2); (1 - y(1)^2) * y(2) - y(1)];
opt = odeset ("OutputFcn", @odeplot, "RelTol", 1e-6);
sol = ode45 (fvdp, [0 20], [2 0], opt);
```
Background Information: This function is called by an ode solver function if it was specified in the `"OutputFcn"` property of an options structure created with `odeset`. The ode solver will initially call the function with the syntax `odeplot ([tfirst, tlast], y0, "init")`. The function initializes internal variables, creates a new figure window, and sets the x limits of the plot. Subsequently, at each time step during the integration the ode solver calls `odeplot (t, y, [])`. At the end of the solution the ode solver calls `odeplot ([], [], "done")` so that odeplot can perform any clean-up actions required.
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# Can two distinct strings create the same parse tree for a given grammar?
More specifically: for any arbitrary CFG, is there guaranteed to exist two distinct strings such that a leftmost parse of both strings creates the same parse tree? If not, does there exist any CFG that has this property?
I know that given a string and a CFG, there may be multiple leftmost parse trees for the same string (if the grammar is ambiguous), but I am unsure if the converse is true.
I suspect the answer is no and I have not come across a case that seems to suggest otherwise, but I cannot think of a rigorous justification.
• Welcome to CS.SE! Please edit to clarify what you are asking. Are you asking, does there exist any CFG where some pair of distinct strings can create the same parse tree? Are you asking, is there always a pair of distinct strings that create the same parse tree, for all possible CFGs? Are you asking about a specific CFG? (If so, which one?) Are you asking for an algorithm that takes a CFG as input and outputs whether there exist a pair of distinct strings with the same parse tree? What counts as "the same parse tree"? This can be considered for re-opening if it's edited to clarify. – D.W. Aug 3 '16 at 20:38
• Also, what are your thoughts? What have you tried? What self-study have you done? Have you tried working through some examples? Have you tried proving this on your own? We want you to make your best effort towards solving it on your own before asking, and show us in the question what you've tried. Please edit the question to clarify these points. Thank you! – D.W. Aug 3 '16 at 20:39
• Hint: what happens when you traverse the roots of parse-trees in leftmost order? If two trees are the same, their roots are the same. What effect does this have on the word produced? – jmite Aug 3 '16 at 21:51
• Hint: The string can be read off the leaves of the parse tree (traversed left to right). – Yuval Filmus Aug 6 '16 at 12:30
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https://ch.mathworks.com/matlabcentral/cody/problems/1678-count-consecutive-0-s-in-between-values-of-1/solutions/272479
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Cody
# Problem 1678. Count consecutive 0's in between values of 1
Solution 272479
Submitted on 3 Jul 2013 by Paul Berglund
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 1; y_correct = 0; assert(isequal(zero_count(x),y_correct))
2 Pass
%% x = [0 1 0 0 0 1 0 1 1 1 0 0 0 1 0 1 1 0 0]; y_correct = [1 3 1 0 0 3 1 0 2]; assert(isequal(zero_count(x),y_correct))
3 Pass
%% x = [0 0 0 0 0 0]; y_correct = 6; assert(isequal(zero_count(x),y_correct))
4 Pass
%% x = [0 0 1 0 0 1]; y_correct = [2 2]; assert(isequal(zero_count(x),y_correct))
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HW12AA
# HW12AA - 12 A&A 2 H 1 and H 2 begin their flights at...
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Unformatted text preview: 12 A&A 2. H 1 and H 2 begin their flights at time t = 0. Using just the z co-ordinate of its trajectory, we see that the time until H 2 breaks down is 1- (- 3) = 4 hours. So H 1 changes its heading at t = 6 hours, by which point its co-ordinates are: (6 + 40 · 6 ,- 3 + 10 · 6 ,- 3 + 2 · 6) = (246 , 57 , 9) . The distance from here to H 1 ’s position at (446 , 13 , 0) is p (246- 446) 2 + (57- 13) 2 + (9- 0) 2 = √ 42017 ≈ 205 miles . At a speed of 150mph, the trip will therefore take (205 / 150) ≈ 1 . 37 hours. 12 A&A 12. (a) Choose any point Q ( x ,y ,z ) on the plane Ax + By + Cz- D = 0, and let n = A i + B j + C k . Then (as outlined in Example 12 . 5 . 11 in the text) the distance between P 1 and this plane is equal to the magnitude of the projection from the vector * QP 1 onto n . d = proj n * QP 1 = * QP 1 · n | n | = | Ax 1 + By 1 + Cz 1- Ax- By- Cz | √ A 2 + B 2 + C 2 = | Ax 1 + By 1 + Cz 1- D | √ A 2 + B 2 + C 2 ....
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https://hindimaintutorial.in/self-defence-training/
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Self Defence Training codechef solution- After the phenomenal success of the 36th Chamber of Shaolin, San Te has decided to start 37th Chamber of Shaolin. The aim this time is to equip women with shaolin self-defence techniques. The only condition for a woman to be eligible for the special training is that she must be between 1010 and 6060 years of age, inclusive of both 1010 and 6060. Given the ages of NN women in his village, please help San Te find out how many of them are eligible for the special training.
Self Defence Training solution codechef
After the phenomenal success of the 36th Chamber of Shaolin, San Te has decided to start 37th Chamber of Shaolin. The aim this time is to equip women with shaolin self-defence techniques.
The only condition for a woman to be eligible for the special training is that she must be between 1010 and 6060 years of age, inclusive of both 1010 and 6060.
Given the ages of NN women in his village, please help San Te find out how many of them are eligible for the special training.
Input Format Self Defence Training solution codechef
• The first line of input contains a single integer TT, denoting the number of test cases. The description of TT test cases follows.
• The first line of each test case contains a single integer NN, the number of women.
• The second line of each test case contains NN space-separated integers A1,A2,...,ANA1,A2,…,AN, the ages of the women.
Output Format
For each test case, output in a single line the number of women eligible for self-defence training.
Self Defence Training solution codechef Constraints
• 1T201≤T≤20
• 1N1001≤N≤100
• 1Ai1001≤Ai≤100
Sample Input 1
3
3
15 23 65
3
15 62 16
2
35 9
Sample Output 1 Self Defence Training solution codechef
2
2
1
Explanation
Test Case 11: Out of the women, only the 1st1st and 2nd2nd women are eligible for the training because their ages lie in the interval [10,60][10,60]. Hence the answer is 2.
Test Case 22: Only the 1st1st and 3rd3rd women are eligible for the training because their ages lie in the interval [10,60][10,60]. Hence the answer is again 2.
Test Case 33: Only the 1st1st woman with age 3535 is eligible for the training. Hence the answer is 11.
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https://www.coursehero.com/file/8720339/quizzez-7-9/
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# quizzez 7-9 - Score for this attempt 18 out of 20 Submitted...
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Score for this attempt: 18out of 20Submitted Sep 30 at 10:20amThis attempt took 13 minutes.Question 1Correct!2 / 2 ptsWhat proportion of a normal distribution corresponds to <i>z</i> scores greater than 1.04?<br>What proportion of a normal distribution corresponds to zscores greater than 1.04?
Question 2Correct!2 / 2 ptsThe critical value(s) associated with a <i>p</i> level of .05 for a one-tailed hypothesis test using the <i>z</i> statistic is (are):<br>The critical value(s) associated with a plevel of .05 for a one-tailed hypothesis test using the zstatistic is (are):
Question 3Correct!2 / 2 ptsA single observation can be expressed in a number of ways that all refer to that same observation and the exact same place within the normal curve. These expressions are:<br>A single observation can be expressed in a number of ways that all refer to that same observation and the exact same place within the normal curve. These expressions are:
Question 4Correct!2 / 2 pts
Skip to question text.Unnithan, Houser, and Fernhall (2006) were interested in whether playing the game Dance Dance Revolution (DDR) affected the heart rate of overweight adolescents differently than it affected nonoverweight adolescents. Twenty-two adolescents, 10 classified as overweight and 12 as not overweight, played DDR for 12 minutes, while the researchers measured each participant's heart rate. On average, the researchers found no difference between the heart rate of the two groups. Which of the following statements is the null hypothesis for this study?
Question 5Incorrect0 / 2 ptsThe statement “It is hypothesized that men and women will differ on reaction time measures” best illustrates a:<br>The statement “It is hypothesized that men and women will differ on reaction time measures” best illustrates a:
Question 6Correct!2 / 2 pts________ requires that all members of a population have an equal chance of being selected for a study.<br>________ requires that all members of a population have an equal chance of being selected for a study.
Question 7Correct!2 / 2 ptsA hypothesis test is said to be ________ when it produces fairly accurate results, even when some of the assumptions underlying the hypothesis test are violated.<br>A hypothesis test is said to be ________ when it produces fairly accurate results, even when some of the assumptions underlying the hypothesis test are violated.
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https://www.gradesaver.com/textbooks/math/calculus/calculus-8th-edition/chapter-16-vector-calculus-16-3-the-fundamental-theorem-for-line-integrals-16-3-exercises-page-1134/1
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## Calculus 8th Edition
$40$
Suppose $C$ is a smooth curve. Since, the gradient function is continuous we know that $f$ is differentiable on $C$. Thus, $\int_{C}∇ f.dr$ is the difference of the values of $f$ which is $50-10=40$
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http://www.lofoya.com/Solved/1580/how-much-did-dipan-get-in-english-paper-ii
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# Difficult Tabular Data Solved QuestionData Interpretation Discussion
Common Information
In a Class X Board examination, ten papers are distributed over five Groups – PCB, Mathematics, Social Science, Vernacular and English. Each of the ten papers is evaluated out of 100. The final score of a student is calculated in the following manner.
First, the Group Scores are obtained by averaging marks in the papers within the Group. The final score is the simple average of the Group Scores.
The data for the top ten students are presented below. (Dipan’s score in English Paper II has been intentionally removed in the table).
Note: B or G against the name of a student respectively indicates whether the student is a boy or a girl.
Table below can be scrolled horizontally
PCB Group Math'sGroup Social ScienceGroup VernacularGroup EnglishGroup Name of the student Phy. Chem. Bio. Hist. Geo. Paper I Paper II Paper I Paper II Final Score Ayesha(G) 98 96 97 98 95 93 94 96 96 98 96.2 Ram(B) 97 99 95 97 95 96 94 94 96 98 96.1 Dipan(B) 98 98 98 95 96 95 96 94 96 ?? 96.0 Sagnik(B) 97 98 99 96 96 98 94 97 92 94 95.9 Sanjiv(B) 95 96 97 98 97 96 92 93 95 96 95.7 Shreya(G) 96 89 85 100 97 98 94 95 96 95 95.5 Joseph(B) 90 94 98 100 94 97 90 92 94 95 95.0 Agni(B) 96 99 96 99 95 96 82 93 92 93 94.3 Pritam(B) 98 98 95 98 83 95 90 93 94 94 93.9 Tirna(G) 96 98 79 99 85 94 92 91 87 96 93.7
Q. Common Information Question: 1/5 How much did Dipan get in English Paper II?
✖ A. 94 ✖ B. 96 ✔ C. 97 ✖ D. 98 ✖ E. 99
Solution:
Option(C) is correct
Dipan’s Group Scores are as follows:
PCB Group:
$= 98 × \dfrac{3}{3}$
$= 98$
Mathematics Group:
$= 95$
Social Science Group:
$= \dfrac{96 + 95}{2}$
$= 95.5$
Vernacular Group:
$= \dfrac{96 + 94}{2}$
$= 95$
Dipan’s final score,
$= 96$
⇒Sum of Dipan’s Group Scores:
$= 96 × 5 = 480$
$⇒ 98 + 95 + 95.5 + 95 + 48 +x/2= 480$
$⇒ x/2 = 48.5$
$⇒ x = 97$
Dipan scored 97 marks in English Paper II.
Hence, option C.
## (0) Comment(s)
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https://schneide.blog/2016/08/22/generating-a-spherified-cube-in-c/
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# Generating a spherified cube in C++
In my last post, I showed how to generate an icosphere, a subdivided icosahedron, without any fancy data-structures like the half-edge data-structure. Someone in the reddit discussion on my post mentioned that a spherified cube is also nice, especially since it naturally lends itself to a relatively nice UV-map.
# The old algorithm
The exact same algorithm from my last post can easily be adapted to generate a spherified cube, just by starting on different data.
After 3 steps of subdivision with the old algorithm, that cube will be transformed into this:
If you look closely, you will see that the triangles in this mesh are a bit uneven. The vertical lines in the yellow-side seem to curve around a bit. This is because unlike in the icosahedron, the triangles in the initial box mesh are far from equilateral. The four-way split does not work very well with this.
One way to improve the situation is to use an adaptive two-way split instead:
Instead of splitting all three edges, we’ll only split one. The adaptive part here is that the edge we’ll split is always the longest that appears in the triangle, therefore avoiding very long edges.
Here’s the code for that. The only tricky part is the modulo-counting to get the indices right. The vertex_for_edge function does the same thing as last time: providing a vertex for subdivision while keeping the mesh connected in its index structure.
```TriangleList
subdivide_2(ColorVertexList& vertices,
TriangleList triangles)
{
Lookup lookup;
TriangleList result;
for (auto&& each:triangles)
{
auto edge=longest_edge(vertices, each);
Index mid=vertex_for_edge(lookup, vertices,
each.vertex[edge], each.vertex[(edge+1)%3]);
result.push_back({each.vertex[edge],
mid, each.vertex[(edge+2)%3]});
result.push_back({each.vertex[(edge+2)%3],
mid, each.vertex[(edge+1)%3]});
}
return result;
}
```
Now the result looks a lot more even:
Note that this algorithm only doubles the triangle count per iteration, so you might want to execute it twice as often as the four-way split.
# Alternatives
Instead of using this generic of triangle-based subdivision, it is also possible to generate the six sides as subdivided patches, as suggested in this article. This approach works naturally if you want to have seams between your six sides. However, that approach is more specialized towards this special geometry and will require extra “stitching” if you don’t want seams.
# Code
The code for both the icosphere and the spherified cube is now on github: github.com/softwareschneiderei/meshing-samples.
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# More trig differentials
• Oct 14th 2007, 07:04 PM
pseizure2000
More trig differentials
Yeah I missed this lesson in class because my alarm never went off. So i have plenty of question por vous. :D
f(x) = cos^9(x^4)
not a clue
• Oct 14th 2007, 07:07 PM
Krizalid
Chain Rule rules here!
$\displaystyle f'(x)=9\cos^8(x^4)\cdot(\cos x^4)'$
Can you take it from there?
• Oct 14th 2007, 07:16 PM
pseizure2000
got it! thanks!
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Find the dimensions of the rectangle that has a length 3 meters more that its width and a perimeter equal in value to its area?
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Start by prove to them the the number 0.57 deserve to be converted right into a portion . A totality number can"t exactly be placed in come a fraction. 0.57 have the right to be convert to 14.25/25. This information way that 0.57 IS a rational number. Hope this persuaded them!
0.57 = 57/100 this way it needs to be rational.
You are watching: How would you convince a fellow student that 0.57 is a rational number
Explanation: You have actually to define to them that any kind of number that can be created in fraction type is a rational number. This contains integers, terminating decimals, and repeating decimals as well as fractions. An integer deserve to be written as a portion simply by giving it a denominator of one, so any type of integer is a reasonable number.
4+y = -36
y = -40
The definition of a reasonable number is the proportion of two integers
Lets rewrite .57 as a fraction
.57 = 57/100
Both 57 and also 100 space integers and 57/100 is the ratio of integers for this reason .57 is rational.
Step-by-step explanation:
4+y = -36
Subtract 4 from every side
4+y-4 = -36-4
y = -40
The definition of a reasonable number is the ratio of 2 integers
Lets rewrite .57 together a fraction
.57 = 57/100
Both 57 and 100 space integers and 57/100 is the ratio of integers for this reason .57 is rational.
Irrational numbers cannot be written in the type of a portion in i beg your pardon both the numerator and also denominator are whole numbers. Therefore, you have the right to prove come your other student through using57/100 together a portion in i beg your pardon both the numerator and denominator are totality numbers and is tantamount to 0.57
Sorry however i only understand the answer come number 3A rational number is a number that deserve to be written as a fraction where numerator and also denominator are both integers (and the denominator is not zero).Since 0.57 = 57/100, so the number 0.57 deserve to be composed as a fraction where the molecule (57) and denominator (100) room both integers.
You have actually to define to lock that any number that deserve to be created in fraction kind is a rational number. This includes integers, end decimals, and repeating decimals as well as fractions. An integer have the right to be written as a fraction simply by offering it a denominator of one, so any integer is a rational number.
Which of the adhering to is the equation that a line that passes with the suggest (1,4) and also is parallel to the x-axis a. Y=1 b. Y=4 c. X=1 d. X=4
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## 4 Revolutionary Riddles Resolved!
visningar 2,926,924
Veritasium
3 år sedan
The solution to 4 rotation-related riddles, including the mystery cylinder, bike pedal pulling puzzle, track problem, and train part going backwards. Thank you to everyone who responded, liked, shared, or made a video response.
Please fill out this short survey for research: ve42.co/Rresearch
Special thanks to:
Mathematician George Hart: georgehart.com/
For allowing me to use excerpts from his pedal pulling puzzle solution: ve42.co/ppp
Petr Lebedev for combing through thousands of comments and providing the stats I gave in this video.
Video responses I used in this video (or watched):
everWonder? svfrom.info/history/video/ZbibZnvZydGhaa4
A Random Nerdy Channel svfrom.info/history/video/eZyLdnG42L-AqNQ
The Physics DoJo svfrom.info/history/video/gNGsZISnrbGuccw
Oblivious Jim svfrom.info/history/video/YZWQjoGxq7WiaaI
Armchair Explorers svfrom.info/history/video/g5SyjJewutatpdQ
MrEngineeringGuy svfrom.info/history/video/dLV_bGq23p6Ykao
Professor Cubers svfrom.info/history/video/obKdaafWrpKjnq4
Scoop Science svfrom.info/history/video/pd2Fg26r19KfoKo
A few notes on the puzzle:
1. A half-full container of honey does pretty well in reproducing the behaviour of the mystery cylinder. I wonder if the motion is a little smoother or more periodic with the ping-pong balls because they move as organized objects - also the delays between motion seemed to be longer with them than without ping pong balls.
2. For the average speed track problem, every time I said velocity I meant speed. Sorry to the pedants out there who are perhaps looking for some trick answer due to displacement being zero when you run around a track.
3. Although a lot of people identified it was something about a train's wheels that move backwards, fewer identified that specifically it was the part of the flange below the rail. Some simply said the bottom half of the wheel.
4. The bicycle question is perhaps the most complex of these riddles. If you tried it with a bike you likely found that it went backwards. But what happens if you sit on the bike and only push backwards on the bottom pedal. The answer might surprise you so give it a shot!
###### Kommentarer
Nathan Meyer 2 dagar sedan
Oh I get the track one now. You'd have to run the track 3 times in the same amount of time it took you to run the first lap to double your velocity. You can't inthe riddle because he limits you to 1 run around the track each time
This video just made me think of What if you ride a bike at a particular speed and run over a travelator (the kind you see in the airports) that has the same speed. will the bike appear stationary with its wheels negating the travelator??
Bernhard Diener 7 dagar sedan
You said "a part of the train that is going backwards with respect to the ground, that is NOT backwards with respect to the train". Maybe it's just my bad englisch that causes me to search for something obviously impossible. But what about some outer parts of the motor, which is spinning so fast, that these parts move backwards faster than the train is moving forward. They are also moving backwards with respect to the ground. That brings me to the Bike question. OK I missed the fact, that you can pull it backwards this way with the pedal going forward while pulling it backwards. That was confusin but is logical. but I said it depends ;-)
Porfirio Ruth 8 dagar sedan
I regret filling my bike tires with honey.
King Donut 10 dagar sedan
Thank god I thought you were gonna leave me hanging
William O'Hara 10 dagar sedan
What do riddles have to do with rolling bottles?
Luke D 14 dagar sedan
Attached it to my bike so I can hook random objects while cycling. That will give me free accelerating.
Peter Zhang 14 dagar sedan
Is there a ratio where the bike doesn't move?
Neelam Sachan 19 dagar sedan
I got all the answer right
GoodLifeMedicine 20 dagar sedan
I go around thinking I'm smart, but sadly I am too lazy to challenge myself to become smart. Therefore, I'm average. How depressing.
gitterrost4 21 dag sedan
This video (the average-velocity-riddle) just sparked a discussion about special relativity in my friend group...
Basim 21 dag sedan
im sorry man but your mountain bike should NEVER go backwards when pedaling backwards. It should only ever be one thing, pedaling backwards activates coaster which makes the pedals disconnect from the wheel and freely spin without turning the wheel. Rolling backwards on the other hand, should indeed rotate you pedals arms, but vis versa should never happen on an anywhere near decent bike as its extremely dangerous.
מעין ניר 22 dagar sedan
If im running at 2 kms at the first round I'll have to run 6 kms at the second round so that will give me a 4 kms average which is 2 times the first one
Suhas Gupta 22 dagar sedan
3:56 Rather than running twice the distance, the equation will hold true if same distance is covered in half the time; i.e. 2V = d/t/2
The RealButcher (Peter from Holland) 22 dagar sedan
Pfff, at least got the wheels right. Thanks for this!
cgfreeandeasy 22 dagar sedan
Ok, i was two times right and the third question maybe i went for coffee, so that i did not recognice this. The fourth question answer i do not credited. In real drive-situation, this phenomenon (backwards) did not take place. If the Bike has no freewheel (no idling), then there is no other direction, as forwards - independent of gear ratio or lever length. And if you look accurate on your Video-sequence, then you would see, that this happends: the force of backwards-pull pulls the boika backwards, but only so long, as far as the gear and the chain-drive traction begins, and then the pedal moves forwards, while the bika moves backwards. What means this? That there is an unrealistik Force closure (in german: Kraftschluß/Kraftübertragung), wich is feasible only in this special constellation.
Gary McLoud 23 dagar sedan
But wait! Didn’t Richard Feynman just tell us that the flanges on the train wheel do not keep it on the track. And that’s not what those flanges are for?
Gary McLoud 23 dagar sedan
svfrom.info/wiki/PL2D30B1DEFFDA0310.html
Jacek2048 24 dagar sedan
This and the other video were oddly some of the most exciting videos I've seen in a long time :D
kwzieleniewski 24 dagar sedan
Normally flanges are not needed for stability of rail wheels---this is achieved by making them a bit conical.
Meet Patel 25 dagar sedan
ZORK GAMER 25 dagar sedan
Riddle 2 seems a bit of a problem. for example distance for a lap = d and time taken for 1st lap = t if v1 is the speed for 1st lap v1=d/t now if v2 is the speed of 2nd lap then (v1+v2)/2=vag=2v1 substituting v1 = d/t (v2 + d/t)/2=2d/t :. v2=3d/t now since we cant travel 3d as we need to complete our task of v average = 2v1 in just a single lap we can say we need to complete 2nd lap in 1/3 of the time of what it took to do the first. :. v2 = 3d/t = d/(t/3) but now if we take total time i.e t + t/3 = 4t/3 and total distance = 2d we get vag=total distance /total time = 2d/(4t/3) = 3d/2t = 3v1/2 ???????????????????????????????????????????????????????????????
Duplicant Lives Matter 25 dagar sedan
Your race track math is off. You’re not changing the distance, fool. The track length remains constant. The time it takes to complete the revolution is precisely the thing that changes. If you complete the distance in one third the original time, the initial velocity is irrelevant. The result will be an average of twice the original velocity. Why on earth would you set time as a constant?
Duplicant Lives Matter 25 dagar sedan
Your confusion is perplexing to me because you even WROTE OUT the fact that distance was the constant, but then went on to go 2d instead of .5 t1
Duplicant Lives Matter 25 dagar sedan
Here’s the proof you’re looking for: Vave = (v1 + v2) /2 Two is the number of times around the track. The distance of the track doesn’t matter. The time it takes to get there on the first run doesn’t matter. The distance is the constant, not the time. 2(V1) = (V1 + V2)/2 4(V1) = V1 + V2 3(V1) = V2 Saying it can’t be done because 2v1 = 2d over t. You’re not running the same track. You’re running a track that is twice the length of the track of the first run.
Duplicant Lives Matter 25 dagar sedan
You’re basically saying that you can’t do this problem because in order to get twice the velocity, you need to run the same exact track twice on the second go, and do it in the same time. No wonder you can’t solve it.
Tristan Hameleers 26 dagar sedan
3:55 wrong because we don't what the one way of the speed of light is and you said in a recent video (I thought) that the one way speed of light may be unlimited or instant
Tristan Hameleers 25 dagar sedan
@Anonymous Secret ye that's also true
Anonymous Secret 25 dagar sedan
Anything with mass can't travel that fast. And presumably we are the target runners, so no speed of light :(.
Abhay Kumar 26 dagar sedan
give me only one answer of this question below of experiment, Based on real experiment if done before but not by calculations. If we take two clocks set time together at same place.and one clock is placed 10metre away from point M and another clock 1km away from point M . A light bulb is glowed at point M now would the reading of the clocks when light reaches them be different or same if path for light was through vacuum. Would they show different or same reading as when light reaches them the watch goes off. And saves that time when light reaches them through vaccum.!.!.!
Fin Has 29 dagar sedan
That running riddle is not impossible if the velocity need not be constant for each lap. Oh, 3 years ago video, anyone already explained this?
You pull the pedal of a bike backwards then the bike moves forward... that's the bottom line. Unless it's a bike with a gazillion gears and you set up the gear ratios so it actually moves backwards... f*ck off...
Them toes look tasty
I finally caught it 3 years later
Ok I know its 3 years late but I got the ping pong ball thing right with the honey
1=1/1,. ≈3=1/.33,. ≈3+1≈4,. ≈4/2≈2 aka double the average speed no? 🤷🏽♂️
ur fooling people
I tried to find The avg speed using V1 and V2 ..and the came up V1+V2=V2 and V1=0 ..and I was like WTF!!
Is there anyone with a more expansive explanation to number 3? I seem to weigh my average differently or something - I just can't quite wrap my head around why we don't try a different method instead of saying its impossible
The Train scenario reminded me of the time a Virginia Tech Mechanical Engineering professor told our class that car tires had to be incredibly tough because each part of the tire on a car going 50 mph (for example) cycles from 0 to 50 mph and back to zero, because the part of the tire in contact with the ground is going zero (because the ground is 'going' zero) and the top of the tire is going 50. This struck me as the stupidest thing I'd heard an educated person say, but I couldn't articulate a rebuttal. Can you provide a good explanation? P.S. The first time I saw an advertisement for a tire & rubber company seeking a Tribologist, I laughed my ass off.
I NEED A BETTER EXPLANATION FOR THE SECOND PROBLEM. I THINK THAT IT IS POSIBLE TO DOUBLE THE TOTAL AVARAGE !!!!!!! PLEASE MAKE A VIDEO EXPLAINING!!!!!!!
You helped me confirm that I am not as smart as I thought I was😂 I really love your channel!!
Jesus loves you brochacho
riddle number 3 got me good, thats such a cool riddle
I have Dutch pedals so the last riddle would be false for me
doesn't time stop if you travel at the speed of light?
What about in the running riddle 1?
Where is riddle #3? 1. cylinder 2. race lap 4. bike ???
All those train videos when I was 3 really paid off.
My bike has a function where the pedals move backwards but the wheels stay still (I tried this myself with my bike)
I was thinking instatanous velocity
Well... Mathematically you could run Vavr=2V1... If you run at speed infinity in both laps Then it mathematically it is possible, physically though, no it is not possible
If you use a fixed gear bike that won't work even with the same ratio
I still don't understand the track one (number 3?), not even a little bit Usually I understand things =p
Ping Pong Cup Shots Månad sedan
You have to run twice the distance in the time it takes you to run a single lap, but you already ran the first lap so the second lap is instant
To the 3rd riddle: It is possible if v1=v2
Actually, what am I know about physics..??? 😅 I'm a kid in physics. You're amazing!!!
You're assuming both laps cover equal distance. let's say you're running on a track with n lanes. Each lane out from the center will have a greater total length than the last, making a solution at least possible as the distance of the second lap could be greater than the first.
I ran the second lap. Jus that no one saw me.
The bike one depends on gearing and how long you pull cuz you could end up just pulling it back forever
Oooooohhhhhhh...... Revolutionary....... Ooooooohhhhhhhhh.......
I consider the multiple components of the cylinder to be a bit of a cheat. I was always intuitively suspicious about the circuit running problem but lack the maths to pin down the problem. I already knew about the train wheels My guess was the bike wouldn't move. I guess that all amounts to a fail... but I learned more than I would have had I known them all, so it's also a win, yay!
Manav Jain - CS Månad sedan
Ohhh man I got all 4 of them right... Although I'm an engineering student....😅😅🙄🙄
Where is the vortex spinning riddle :(
4:08 "This may seem like a trick question, but actually, it's a trick question"
Duplicant Lives Matter 25 dagar sedan
It’s not a trick question, it’s an inaccurate response to that question. He simply isn’t correct. Sorry
lol yeah
The track riddle can be done! You gave us an object to run around, not a distance. Assuming the track is a standard 400m, if you spend your second lap running in a 5.2m wide x 2m long zig-zag, you could effectively run 1200m on your second lap, adding making it take just as long to run thrice as fast, equating to a 2v average.
huh, I ...don't actually know if I can say I got 1 wrong or not since it works on the same principle: I was thinking an oblong on an axis but that actually reverses a little bit each "step" down the hill. thaaaat one was admitably not instantly obvious until I stopped and looked at that 2, I just went right into trying to find a ratio on autopilot. I mean, yeah ,wheels, I certainly couldn't think of anything else that would make sense but that was REALLY obvious so I kept thinking it couldn't count as a "riddle" so there might be something else. wooo! I am the mostest correct!
With the bike thing I mean if you pull hard enough no matter what its going backwards because the force would be so great it would overpower the force pushing it forward lol
its matematickly imposible ??? Well im whud point to the fact that you can run infinite speed lap..... since your speed is infinite your time is practicly 0 ....... Well eat that .....
Answer to everything: It depends... (auto correct)
why not 3x? You double the time but quadruple distance. 4d/2t=2d/t. There is no limitation on more time because the question is about average velocity of both trials and not about increasing the average velocity of the original trial.
Ping Pong Cup Shots Månad sedan
Why would you quadruple the distance when you only want to run an extra lap
4:02 , the avg velocity can be equated to 2V1 if you run the same distance d, in the time period of t/2.
i was way too confident on my 3(V1)
Apologies if this is a stupid question, but what is meant by "forward in respect to the ground"?
Just a guy who’s a frog for fun Månad sedan
It’s just said to make sure there’s no loopholes
Just a guy who’s a frog for fun Månad sedan
You can either be stationary with things moving around you or you moving with other things being stationary (forward in respect to the ground)
5:14 Watch closely! It's crazy!
I feel stupid now
I dont get the second riddle, cant you Just complete the second lap in half the time?
That would make the total time for both laps 1.5 times the time for the first lap thus lowering the total average speed
For a guy with a Ph.D. in Physics, it's astonishing you're confused about the very definition of "average velocity." By definition, Vavg = (V1 + V2)/2 unless, of course, every physics book I've ever owned has been lying to me. If Vavg = 2V1 = (V1 + V2)/2 4V1 = V1 + V2, solve for V2 V2 = 4V1- V1 = 3V1 ...it ain't rocket science. I think what you're getting balled up on is the difference between arithmetic and geometric mean...?
This one took a while for me to get, but you're incorrect. The factor you're neglecting is the time spend at each speed. For easy numbers, imagine you ran at 1m/s for 100 seconds, then ran at 2m/s for 1 second, for that speed and time, you would have ran 102 meters in 101 seconds. Now, by your definition, the average speed is (1+2)/2=1.5m/s. But that's obviously wrong. If you had been running at an average of 1.5m/s for 101 seconds, you'd have travelled 151.5meters, not 102. The correct equation for these numbers is ((1m/s*100s)+(2m/s*1s))/(100s+1s)= about 1.001m/s. As a sanity check, just do the normal velocity calculation. You know the distance you ran(102m), and the time you ran it in(101s). So, distance over time, 102/101= about 1.001m/s. I think Derek did a really bad job of explaining this, but hopefully this has helped.
Anyone notice how he didn't use the original video for the bike riddle? My guess is that he was unaware that the gear ratio affected the answer until the 5% who said it depends pointed it out.
Your bikes are bad they should allow for back pedaling what kind of bike allows you not to paddle backwards and doing nothing because it rachets
hold up, you have havaianas??? that's cool
For the velocity question could you not just run the same distance in half the time?
Ping Pong Cup Shots Månad sedan
Then the average speed would be 1.3333 (V1)
Ok my guesses was not bad: i guessed there is a ball in cylinder; 3v XD; train wheels (I 3D model a lot so yeah XD); move forwards
Wow. O-O
Who is here after watching 4 revolutionary riddles?
What if you walk sooooooooo slow on that first lap? Like a 30 minute lap
What happens when the pedal is almost halfway between bottom and top dead centre? It will no longer be able to turn, therefore not driving the chain. The string will not be able to lift the pedal to top dead centre and push it forwards. The freewheel would stop the cycle's momentum pulling the pedals with it. I'm willing to be proven wrong, if you're willing and able to try making the pedal go 720 degrees by pulling the string backwards only.
Your bicycle question/answer for some reason reminded me of an old WWII pilot's bet. He bet his fellow pilots that he could move his B-17 BACKWARDS with just the engines (which did NOT have a reverse prop setting back then.) He always won his bet. How did he do it?
Dp you think it's worth it to learn physics and math after 30 with no background in it?
run the second lap in 0 seconds
I was just thinking you run 1 mph the first lap then 3 mph the second wouldn’t that average to 2 mph which is double 1 mph
I don't get it: If I ran the first lap of say 10m in 4 secs (so v1 = 2.5m/s). The equation to run the second lap so average speed is 2*v1 (5m/s) = 5 = (10 + 10) / (4 +x) 20 + 5x = 20 x = 0 So I have to run the second lap in 0 seconds????? HOW???
Isn't 2V1=2d/t1 the same as 2V1=d/t1/2 Meaning to travel twice as fast you dont have to travel twice as far in the same time but rather the same distance in half the time? If so, then that would mean running at 3V1 will in fact be the correct answer.
Derek did a really bad job explaining this one, and I think that line in particular just makes it more confusing, so ignore the "You can't run twice the distance" explanation for now. Here's my suggestion: Go into excel and use this formula Vave=((V1*t1)+(V2*t2))/(t1+t2) No matter how slow you make V1, or how fast you make V2, the average only approach 2V1, and since there's a universal speed limit, even if you could make the second lap infinitely quickly... you can't. The important thing is that the average isn't just (v1+v2)/2. You can see this yourself easily. Imagine you ran at 10m/s for 100seconds, then 20m/s for 1second. Using that calculation the average speed for your total journey was 15m/s. But that makes no sense, you only ran at twice your speed for >1% of the journey. This is why you need to weight each velocity with the time spent moving at that velocity.
3:30 just don't move
software at 5:05?
I wasn't sure if there was a smaller cylinder or a ball bearing inside the cylinder.. Funny it's both.
Light and Time 2 månader sedan
What if you had extended the distance by first running as usual and the second time running zig zag extending the distance?
GOD GAMER - BRAWL STARS 2 månader sedan
The no2 riddle is always possible , I can prove that...Where can I prove it?
at the speed of light - wouldn't you have to be (or behave like) a photon? and therefore not experience any time - so from your frame of reference, t_total equals t1. so, to double your average speed, you'd "just" have to turn into light.
Correct me if I am wrong, but I think it is mathematically possible to solve the second riddle. The average speed equals 2d/(t1+t2). In other words, in order to achieve an average speed of 2V1, t1 must be much higher than t2, so instead of running at lightening speed at the second round, simply run with very low speed at the first round to maximize t1. Thankfully the distance (d) isn't specified, so you can reduce it to almost zero to complete the 2 rounds within your life span.
YOU GUYS ARE FOOL it's is a bicycle riding tutorial
At the speed of light there is no time.
L'internet Sans FIl 2 månader sedan
but 2d/t = d/0.5t so running faster should work ?
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## Riddles 25
Darn Socks
You have 14 brown socks, 14 blue socks and 14 black socks in your sock drawer. How many socks must you remove (without looking to be sure) to have a matched pair?
Four. You will have a pair of one color or another.
Cost of the Stopper
If a cork and a bottle cost \$2.10 and the bottle costs \$2.00 more than the stopper, then what does the stopper cost?
Five Cents. Bottle: \$2.05.
The Wise Daughter
Many years ago, a wealthy old man was near death. He wished to leave his fortune to one of his three children. The old man wanted to know that his fortune would be in wise hands. He
stipulated that his estate would be left to the child who would sing him half as many songs as days that he had left to live.The eldest son said he couldnāt comply because he didnāt know how many days his father had left to live, and besides he was too busy. The youngest son said the same thing. The man ended up leaving his money to his third child, a daughter. What did his daughter do?
Every other day, the daughter sang her father a song.
Who is the Murderer?
There are five acquaintances. One of them shot and killed one of the other five. Which man is the murderer?1. Dan ran in the NY City marathon yesterday with one of the innocent men.
2. Mike considered being a farmer before he moved to the city.
3. Jeff is a top-notch computer consultant and wants to install Ben?s new computer next week.
4. The murderer had his leg amputated last month.
5. Ben met Jack for the first time six months ago.
6. Jack has been in seclusion since the crime.
7. Dan used to drink heavily.
8. Ben and Jeff built their last computers together.
9. The murderer is Jackās brother; they grew up together in Seattle.
1. Jack is not the murderer, because he is the brother of the murderer.
2. Dan canāt be the murderer since he ran a marathon, and the murderer recently had his leg amputated, and wouldn?t be running a marathon of any magnitude that quickly.
3. Ben is not the murderer if he just met Jack, since Jack and the murderer grew up together.
4. This leaves Jeff and Mike. Since Jeff is still alive (he wants to install a new computer next week, present tense) he must be the murderer. Mike also didnāt grow up with Jack.It has been determined that Jack, Dan and Jeff are all alive. Ben must also be alive since Jeff plans to install Ben?s computer next week. This means that Jeff killed Mike.
Pumpkins or Apples?
Lynn likes grapes but not potatoes. She likes squash but not lettuce, and she likes peas but not onions. Following the same rule, will she like pumpkins or apples?
Pumpkins. Lynn only likes things that grow on vines.
Rounds of Golf
Robert and David played several golf matches against each other in a week. They played for a pizza at each match, but no pizzas were purchased until the end of the week. If at any time Robert and David had the same number of wins, those pizzas were canceled. Robert won four matches (but no pizzas), and David won three pizzas. How many rounds of golf were played?
Eleven, David won 7 matches, 4 to cancel out Robertās 4 wins, and 3 more to win the pizzas.
Magnetized Iron
There are two bars of iron. One bar is magnetized along its length, while the other is not. Using just the two bars, without any other items, how can you tell which bar is magnetized and which bar is not?
Take the two bars, and put them together like a T. (Put one bar horizontally across and center the vertical bar under it.) Bar #1 is the top of the āTā and bar #2 is the vertical bar.If they stick together, then bar #2 is the magnet. If they donāt, bar #1 is the magnet.
Bar magnets are ādeadā in their centers and there is no magnetic force, since the two poles cancel out. So, if bar #1 is the magnet, then bar #2 wonāt stick to its center.
However, bar magnets are quite āaliveā at their edges and the magnetic force is concentrated. If bar #2 is the magnet, bar #1 will stick to it.
Surprise Party
David is throwing Robert a surprise birthday party but he has to stay within his budget. He spent half of his money plus \$2.00 on the cake. Half of what he had left plus \$2.00 was spent on balloons and streamers. Then he spent half of what he had left plus \$1.00 on candy. Now he is out of money, how much did he start with?
This one is best solved working backwards, the last part David spent half of what was left plus \$1.00 on candy and then was out of money. That means he must have spent \$2.00 on Candy as \$1.00 was half of what he had using the same logic backwards:\$2.00 on candy
\$6.00 on Balloons and Streamers
\$12.00 on the cake
Total of \$20.00.
Name of the Engineer
A train goes between Chicago and New York. The brakeman, the fireman and the engineer are named Smith, Jones and Brown. (The names are not necessarily in order). There are also three passengers named Mr. Smith, Mr. Jones and Mr. Brown. Mr. Brown lives in New York. The brakeman lives halfway between New York and Chicago. Mr. Jones earns exactly \$20,000 per year. Smith beat the fireman at their last game of golf. The passenger who lives in Chicago has the same name as the brakeman. The brakemanās next door neighbor is a passenger on this train and earns exactly three times as much as the brakeman. What is the
name of the engineer?
Determine the known facts. Also notice that the passengers are noted with the title Mr., where as the brakeman, engineer and fireman are identified by their last names only.1. Mr Brown Lives in New York City
2. The brakeman lives midway between NY and Chicago
3. Mr. Jones earns exactly \$20K per year
4. Smith beat the fireman at their last game of golf.
5. The brakemanās next-door neighbor, who is a passenger, earns exactly three times the brakemanās salary.
6. The passenger who lives in Chicago has the same name as the brakeman.
According to #1 and #2, the brakemanās neighbor cannot be Mr. Brown. According to #5, the brakemanās neighbor also cannot be Mr. Jones, because \$20,000 is not evenly divisible by three. This leaves Mr. Smith as the next door neighbor to the brakeman. Mr. Smith lives halfway between New York and Chicago (#2) as does the brakeman. Since Mr. Brown lives in New York, by process of elimination, it is now known that Mr. Jones lives in Chicago.
According to statement #6, this means that the brakeman is named Jones.
According to statement #4, the fireman cannot be Smith, so the fireman must be must be Brown, which leaves Smith as the engineer.
Wedding Shower Present
Janie?s friends were chipping in to buy her a wedding shower present. At first, 10 friends chipped in, but 2 of them dropped out. Each of the 8 had to chip in another dollar to bring the amount back up. How much money did they plan to collect?
\$40 (10 at \$4, or 8 at \$5).
Detective Code
Detective Marty had been trailing three men suspected of committing several fur robberies.
He received a message in a sealed envelope delivered by Roscoe, an underworld contact and
decided to call Detective Sam, his partner. āI think I found out where the merchandise is
stored,ā said Detective Marty. Obviously Marty didnāt want Roscoe to know what was in the
envelope, so he read it to Sam in code. This is what Marty had said:āFlame mate weighty soak shave comedy debut stake scared.ā
What did the message say?
Am at Eight Oak Ave, come but take care. Read through it again to see!
Special Letters
A, E, H, I, K, L, What are the next letters and why?
M, N, O, P, U, W. Letters in the Hawaiian alphabet.
Cannot Find It
I went into the woods and got it.
I sat down to seek it.
I brought it home with me because I couldnāt find it.
What is it?
A splinter.
Repeat After Me
I say everything I hear to others around.
Iām not an animal, nor part of the human race.
I will, immediately, repeat after you;
But only if my tail is in place.
What am I?
A microphone.
Beating
I can flutter and take your breath away.
I can take a beating, but do not bruise.
If I stopped you would be sure to lose.
Everyday I am with you.
What am I?
Bees and Flowers
In a pond there are some flowers with some bees hovering over the flowers. How many flowers and bees are there if both the following statements are true:1. If each bee lands on a flower, one bee doesnāt get a flower.
2. If two bees share each flower, there is one flower left out.
4 bees and 3 flowers.
Stupid Rich Guy
A kind man comes across a dirty box, a voice says to him āCoconuts, \$5 a dozenā With his lightning quick arithmetic he calculates that if he sold those same coconuts to the coconut air assault team for the accepted rate of \$3 per dozen that in no time at all he would be a millionaire. Whatās with this guy, assuming his math is accurate?
He was a billionaire.
My Daughters
Iāve got ten or more daughters. Iāve got less than ten daughters.
Iāve got at least one daughter. If only one of these statements is
true, how many daughters have I got?
If I have any daughters, there will always be two statements which are true. Therefore, I have no daughters.
Orange, Silver and Purple
Other than being colors, what do the words orange, silver and purple have in common?
There is no word in the English language that rhymes with them.
As Three
Pronounced as one letter,
But look and youāll see,
That really Iām written with three.
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# How many kilograms are in 135000 milligrams?
Updated: 9/25/2023
Wiki User
8y ago
135,000 milligrams is 0.135 kilogram.
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Related questions
### What does 135000 mg in kg?
135,000 milligrams is 0.135 kilograms.
### How many milligrams in 5.2 kilograms?
52000000 milligrams in 5.2 kilograms
### How many milligrams are in 1.25 kilograms?
1.25 kilograms is 1,250,000 milligrams.
### How many milligrams in 280 kilograms?
280 kilograms = 280,000,000 milligrams
### How many kilograms are in 8675309 milligrams?
There are 8.675309 kilograms in 8675309 milligrams.
### How many milligrams is 08 kilograms?
8,000,000 milligrams = 8 kilograms
### How many kilograms in 6014 milligrams?
6014 milligrams = 0.006014 kilograms
### How many kilograms is 40000 milligrams?
40,000 milligrams is 0.04 kilograms.
### How many milligrams is 454 kilograms?
454 kilograms is 454 million milligrams.
### How many milligrams is 433 kilograms?
433 kilograms equals 443,000,000 milligrams.
### How many kilograms are in 3530000 milligrams?
3,530,000 milligrams is equal to 3.53 kilograms.
### How many milligrams are in 0.019 kilograms?
0.019 kilograms = 19 grams = 19,000 milligrams
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# Multiply Large Numbers represented as Strings
Given two positive numbers as strings. The numbers may be very large (may not fit in long long int), the task is to find product of these two numbers.
Examples:
```Input : num1 = 4154
num2 = 51454
Output : 213739916
Input : num1 = 654154154151454545415415454
num2 = 63516561563156316545145146514654
Output : 41549622603955309777243716069997997007620439937711509062916```
Recommended Practice
The idea is based on school mathematics.
We start from last digit of second number multiply it with first number. Then we multiply second digit of second number with first number, and so on. We add all these multiplications. While adding, we put i-th multiplication shifted.
The approach used in below solution is to keep only one array for result. We traverse all digits first and second numbers in a loop and add the result at appropriate position.
## C++
`// C++ program to multiply two numbers represented` `// as strings.` `#include` `using` `namespace` `std;` `// Multiplies str1 and str2, and prints result.` `string multiply(string num1, string num2)` `{` ` ``int` `len1 = num1.size();` ` ``int` `len2 = num2.size();` ` ``if` `(len1 == 0 || len2 == 0)` ` ``return` `"0"``;` ` ``// will keep the result number in vector` ` ``// in reverse order` ` ``vector<``int``> result(len1 + len2, 0);` ` ``// Below two indexes are used to find positions` ` ``// in result. ` ` ``int` `i_n1 = 0; ` ` ``int` `i_n2 = 0; ` ` ` ` ``// Go from right to left in num1` ` ``for` `(``int` `i=len1-1; i>=0; i--)` ` ``{` ` ``int` `carry = 0;` ` ``int` `n1 = num1[i] - ``'0'``;` ` ``// To shift position to left after every` ` ``// multiplication of a digit in num2` ` ``i_n2 = 0; ` ` ` ` ``// Go from right to left in num2 ` ` ``for` `(``int` `j=len2-1; j>=0; j--)` ` ``{` ` ``// Take current digit of second number` ` ``int` `n2 = num2[j] - ``'0'``;` ` ``// Multiply with current digit of first number` ` ``// and add result to previously stored result` ` ``// at current position. ` ` ``int` `sum = n1*n2 + result[i_n1 + i_n2] + carry;` ` ``// Carry for next iteration` ` ``carry = sum/10;` ` ``// Store result` ` ``result[i_n1 + i_n2] = sum % 10;` ` ``i_n2++;` ` ``}` ` ``// store carry in next cell` ` ``if` `(carry > 0)` ` ``result[i_n1 + i_n2] += carry;` ` ``// To shift position to left after every` ` ``// multiplication of a digit in num1.` ` ``i_n1++;` ` ``}` ` ``// ignore '0's from the right` ` ``int` `i = result.size() - 1;` ` ``while` `(i>=0 && result[i] == 0)` ` ``i--;` ` ``// If all were '0's - means either both or` ` ``// one of num1 or num2 were '0'` ` ``if` `(i == -1)` ` ``return` `"0"``;` ` ``// generate the result string` ` ``string s = ``""``;` ` ` ` ``while` `(i >= 0)` ` ``s += std::to_string(result[i--]);` ` ``return` `s;` `}` `// Driver code` `int` `main()` `{` ` ``string str1 = ``"1235421415454545454545454544"``;` ` ``string str2 = ``"1714546546546545454544548544544545"``;` ` ` ` ``if``((str1.at(0) == ``'-'` `|| str2.at(0) == ``'-'``) && ` ` ``(str1.at(0) != ``'-'` `|| str2.at(0) != ``'-'` `))` ` ``cout<<``"-"``;` ` ``if``(str1.at(0) == ``'-'``)` ` ``str1 = str1.substr(1);` ` ` ` ``if``(str2.at(0) == ``'-'``)` ` ``str2 = str2.substr(1);` ` ``cout << multiply(str1, str2);` ` ``return` `0;` `}`
## Java
`// Java program to multiply two numbers ` `// represented as Strings.` `import` `java.util.*;` `import` `java.io.*;` `class` `GFG` `{` ` ` `// Multiplies str1 and str2, and prints result.` `static` `String multiply(String num1, String num2)` `{` ` ``int` `len1 = num1.length();` ` ``int` `len2 = num2.length();` ` ``if` `(len1 == ``0` `|| len2 == ``0``)` ` ``return` `"0"``;` ` ``// will keep the result number in vector` ` ``// in reverse order` ` ``int` `result[] = ``new` `int``[len1 + len2];` ` ``// Below two indexes are used to ` ` ``// find positions in result. ` ` ``int` `i_n1 = ``0``; ` ` ``int` `i_n2 = ``0``; ` ` ` ` ``// Go from right to left in num1` ` ``for` `(``int` `i = len1 - ``1``; i >= ``0``; i--)` ` ``{` ` ``int` `carry = ``0``;` ` ``int` `n1 = num1.charAt(i) - ``'0'``;` ` ``// To shift position to left after every` ` ``// multipliccharAtion of a digit in num2` ` ``i_n2 = ``0``; ` ` ` ` ``// Go from right to left in num2 ` ` ``for` `(``int` `j = len2 - ``1``; j >= ``0``; j--)` ` ``{` ` ``// Take current digit of second number` ` ``int` `n2 = num2.charAt(j) - ``'0'``;` ` ``// Multiply with current digit of first number` ` ``// and add result to previously stored result` ` ``// charAt current position. ` ` ``int` `sum = n1 * n2 + result[i_n1 + i_n2] + carry;` ` ``// Carry for next itercharAtion` ` ``carry = sum / ``10``;` ` ``// Store result` ` ``result[i_n1 + i_n2] = sum % ``10``;` ` ``i_n2++;` ` ``}` ` ``// store carry in next cell` ` ``if` `(carry > ``0``)` ` ``result[i_n1 + i_n2] += carry;` ` ``// To shift position to left after every` ` ``// multipliccharAtion of a digit in num1.` ` ``i_n1++;` ` ``}` ` ``// ignore '0's from the right` ` ``int` `i = result.length - ``1``;` ` ``while` `(i >= ``0` `&& result[i] == ``0``)` ` ``i--;` ` ``// If all were '0's - means either both ` ` ``// or one of num1 or num2 were '0'` ` ``if` `(i == -``1``)` ` ``return` `"0"``;` ` ``// genercharAte the result String` ` ``String s = ``""``;` ` ` ` ``while` `(i >= ``0``)` ` ``s += (result[i--]);` ` ``return` `s;` `}` `// Driver code` `public` `static` `void` `main(String[] args) ` `{` ` ``String str1 = ``"1235421415454545454545454544"``;` ` ``String str2 = ``"1714546546546545454544548544544545"``;` ` ``if` `((str1.charAt(``0``) == ``'-'` `|| str2.charAt(``0``) == ``'-'``) && ` ` ``(str1.charAt(``0``) != ``'-'` `|| str2.charAt(``0``) != ``'-'``))` ` ``System.out.print(``"-"``);` ` ``if` `(str1.charAt(``0``) == ``'-'``)` ` ``str1 = str1.substring(``1``);` ` ` ` ``if` `(str2.charAt(``0``) == ``'-'``)` ` ``str2 = str2.substring(``1``);` ` ``System.out.println(multiply(str1, str2));` `}` `}` `// This code is contributed by ankush_953`
## Python3
`# Python3 program to multiply two numbers ` `# represented as strings.` `# Multiplies str1 and str2, and prints result.` `def` `multiply(num1, num2):` ` ``len1 ``=` `len``(num1)` ` ``len2 ``=` `len``(num2)` ` ``if` `len1 ``=``=` `0` `or` `len2 ``=``=` `0``:` ` ``return` `"0"` ` ``# will keep the result number in vector` ` ``# in reverse order` ` ``result ``=` `[``0``] ``*` `(len1 ``+` `len2)` ` ` ` ``# Below two indexes are used to ` ` ``# find positions in result.` ` ``i_n1 ``=` `0` ` ``i_n2 ``=` `0` ` ``# Go from right to left in num1` ` ``for` `i ``in` `range``(len1 ``-` `1``, ``-``1``, ``-``1``):` ` ``carry ``=` `0` ` ``n1 ``=` `ord``(num1[i]) ``-` `48` ` ``# To shift position to left after every` ` ``# multiplication of a digit in num2` ` ``i_n2 ``=` `0` ` ``# Go from right to left in num2` ` ``for` `j ``in` `range``(len2 ``-` `1``, ``-``1``, ``-``1``):` ` ` ` ``# Take current digit of second number` ` ``n2 ``=` `ord``(num2[j]) ``-` `48` ` ` ` ``# Multiply with current digit of first number` ` ``# and add result to previously stored result` ` ``# at current position.` ` ``summ ``=` `n1 ``*` `n2 ``+` `result[i_n1 ``+` `i_n2] ``+` `carry` ` ``# Carry for next iteration` ` ``carry ``=` `summ ``/``/` `10` ` ``# Store result` ` ``result[i_n1 ``+` `i_n2] ``=` `summ ``%` `10` ` ``i_n2 ``+``=` `1` ` ``# store carry in next cell` ` ``if` `(carry > ``0``):` ` ``result[i_n1 ``+` `i_n2] ``+``=` `carry` ` ``# To shift position to left after every` ` ``# multiplication of a digit in num1.` ` ``i_n1 ``+``=` `1` ` ` ` ``# print(result)` ` ``# ignore '0's from the right` ` ``i ``=` `len``(result) ``-` `1` ` ``while` `(i >``=` `0` `and` `result[i] ``=``=` `0``):` ` ``i ``-``=` `1` ` ``# If all were '0's - means either both or` ` ``# one of num1 or num2 were '0'` ` ``if` `(i ``=``=` `-``1``):` ` ``return` `"0"` ` ``# generate the result string` ` ``s ``=` `""` ` ``while` `(i >``=` `0``):` ` ``s ``+``=` `chr``(result[i] ``+` `48``)` ` ``i ``-``=` `1` ` ``return` `s` `# Driver code` `str1 ``=` `"1235421415454545454545454544"` `str2 ``=` `"1714546546546545454544548544544545"` `if``((str1[``0``] ``=``=` `'-'` `or` `str2[``0``] ``=``=` `'-'``) ``and` ` ``(str1[``0``] !``=` `'-'` `or` `str2[``0``] !``=` `'-'``)):` ` ``print``(``"-"``, end ``=` `'')` `if``(str1[``0``] ``=``=` `'-'` `and` `str2[``0``] !``=` `'-'``):` ` ``str1 ``=` `str1[``1``:]` `elif``(str1[``0``] !``=` `'-'` `and` `str2[``0``] ``=``=` `'-'``):` ` ``str2 ``=` `str2[``1``:]` `elif``(str1[``0``] ``=``=` `'-'` `and` `str2[``0``] ``=``=` `'-'``):` ` ``str1 ``=` `str1[``1``:]` ` ``str2 ``=` `str2[``1``:]` `print``(multiply(str1, str2))` `# This code is contributed by ankush_953`
## C#
`// C# program to multiply two numbers ` `// represented as Strings. ` `using` `System;` `class` `GFG ` `{ ` ` ` `// Multiplies str1 and str2, and prints result. ` `static` `String multiply(String num1, String num2) ` `{ ` ` ``int` `len1 = num1.Length; ` ` ``int` `len2 = num2.Length; ` ` ``if` `(len1 == 0 || len2 == 0) ` ` ``return` `"0"``; ` ` ``// will keep the result number in vector ` ` ``// in reverse order ` ` ``int` `[]result = ``new` `int``[len1 + len2]; ` ` ``// Below two indexes are used to ` ` ``// find positions in result. ` ` ``int` `i_n1 = 0; ` ` ``int` `i_n2 = 0; ` ` ``int` `i;` ` ` ` ``// Go from right to left in num1 ` ` ``for` `(i = len1 - 1; i >= 0; i--) ` ` ``{ ` ` ``int` `carry = 0; ` ` ``int` `n1 = num1[i] - ``'0'``; ` ` ``// To shift position to left after every ` ` ``// multipliccharAtion of a digit in num2 ` ` ``i_n2 = 0; ` ` ` ` ``// Go from right to left in num2 ` ` ``for` `(``int` `j = len2 - 1; j >= 0; j--) ` ` ``{ ` ` ``// Take current digit of second number ` ` ``int` `n2 = num2[j] - ``'0'``; ` ` ``// Multiply with current digit of first number ` ` ``// and add result to previously stored result ` ` ``// charAt current position. ` ` ``int` `sum = n1 * n2 + result[i_n1 + i_n2] + carry; ` ` ``// Carry for next itercharAtion ` ` ``carry = sum / 10; ` ` ``// Store result ` ` ``result[i_n1 + i_n2] = sum % 10; ` ` ``i_n2++; ` ` ``} ` ` ``// store carry in next cell ` ` ``if` `(carry > 0) ` ` ``result[i_n1 + i_n2] += carry; ` ` ``// To shift position to left after every ` ` ``// multipliccharAtion of a digit in num1. ` ` ``i_n1++; ` ` ``} ` ` ``// ignore '0's from the right ` ` ``i = result.Length - 1; ` ` ``while` `(i >= 0 && result[i] == 0) ` ` ``i--; ` ` ``// If all were '0's - means either both ` ` ``// or one of num1 or num2 were '0' ` ` ``if` `(i == -1) ` ` ``return` `"0"``; ` ` ``// genercharAte the result String ` ` ``String s = ``""``; ` ` ` ` ``while` `(i >= 0) ` ` ``s += (result[i--]); ` ` ``return` `s; ` `} ` `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` ``String str1 = ``"1235421415454545454545454544"``; ` ` ``String str2 = ``"1714546546546545454544548544544545"``; ` ` ``if` `((str1[0] == ``'-'` `|| str2[0] == ``'-'``) && ` ` ``(str1[0] != ``'-'` `|| str2[0] != ``'-'``)) ` ` ``Console.Write(``"-"``); ` ` ``if` `(str1[0] == ``'-'` `&& str2[0] != ``'-'``) ` ` ``{ ` ` ``str1 = str1.Substring(1); ` ` ``} ` ` ``else` `if` `(str1[0] != ``'-'` `&& str2[0] == ``'-'``) ` ` ``{ ` ` ``str2 = str2.Substring(1); ` ` ``} ` ` ``else` `if` `(str1[0] == ``'-'` `&& str2[0] == ``'-'``) ` ` ``{ ` ` ``str1 = str1.Substring(1); ` ` ``str2 = str2.Substring(1); ` ` ``} ` ` ``Console.WriteLine(multiply(str1, str2)); ` `} ` `}` `// This code is contributed by Rajput-Ji`
## Javascript
`// JavaScript program to multiply two numbers ` `// represented as strings.` `// Multiplies str1 and str2, and prints result.` `function` `multiply(num1, num2)` `{` ` ``let len1 = num1.length;` ` ``let len2 = num2.length;` ` ``if` `(len1 == 0 || len2 == 0)` ` ``return` `"0"` ` ``// will keep the result number in vector` ` ``// in reverse order` ` ``let result = ``new` `Array(len1 + len2).fill(0)` ` ` ` ``// Below two indexes are used to ` ` ``// find positions in result.` ` ``let i_n1 = 0` ` ``let i_n2 = 0` ` ``// Go from right to left in num1` ` ``for` `(``var` `i = len1 - 1; i > -1 ; i --)` ` ``{` ` ``let carry = 0` ` ``let n1 = (num1[i]).charCodeAt(0) - 48` ` ``// To shift position to left after every` ` ``// multiplication of a digit in num2` ` ``i_n2 = 0` ` ``// Go from right to left in num2` ` ``for` `(``var` `j = len2 - 1; j > -1; j--)` ` ``{` ` ``// Take current digit of second number` ` ``let n2 = (num2[j]).charCodeAt(0) - 48` ` ` ` ``// Multiply with current digit of first number` ` ``// and add result to previously stored result` ` ``// at current position.` ` ``let summ = n1 * n2 + result[i_n1 + i_n2] + carry` ` ``// Carry for next iteration` ` ``carry = Math.floor(summ / 10)` ` ``// Store result` ` ``result[i_n1 + i_n2] = summ % 10` ` ``i_n2 += 1` ` ``}` ` ``// store carry in next cell` ` ``if` `(carry > 0)` ` ``result[i_n1 + i_n2] += carry` ` ``// To shift position to left after every` ` ``// multiplication of a digit in num1.` ` ``i_n1 += 1` ` ` ` ``// print(result)` ` ``}` ` ``// ignore '0's from the right` ` ``i = result.length - 1` ` ``while` `(i >= 0 && result[i] == 0)` ` ``i -= 1` ` ``// If all were '0's - means either both or` ` ``// one of num1 or num2 were '0'` ` ``if` `(i == -1)` ` ``return` `"0"` ` ``// generate the result string` ` ``let s = ``""` ` ``while` `(i >= 0)` ` ``{` ` ``s += String.fromCharCode(result[i] + 48)` ` ``i -= 1` ` ``}` ` ``return` `s` ` ``}` `// Driver code` `let str1 = ``"1235421415454545454545454544"` `let str2 = ``"1714546546546545454544548544544545"` `if``((str1[0] == ``'-'` `|| str2[0] == ``'-'``) && ` ` ``(str1[0] != ``'-'` `|| str2[0] != ``'-'``))` ` ``process.stdout.write(``"-"``)` `if``(str1[0] == ``'-'` `&& str2[0] != ``'-'``)` ` ``str1.shift()` `else` `if``(str1[0] != ``'-'` `&& str2[0] == ``'-'``)` ` ``str2.shift()` `else` `if``(str1[0] == ``'-'` `&& str2[0] == ``'-'``)` `{` ` ``str1.shift()` ` ``str2.shift()` `}` `console.log(multiply(str1, str2))` `// This code is contributed by phasing17`
Output:
`2118187521397235888154583183918321221520083884298838480662480`
The above code is adapted from the code provided by Gaurav.
Time Complexity: O(m*n), where m and n are length of two number that need to be multiplied.
Auxiliary Space: O(m+n), where m and n are length of two number that need to be multiplied.
The Way to understand:
The Approach:
Compute the ones-digit, then the tens-digit, then the hundreds-digit, etc. For example when multiplying 1234 with 5678, the thousands-digit of the product is 4*5 + 3*6 + 2*7 + 1*8 (plus what got carried from the hundreds-digit).
## C++
`#include ` `#include` `using` `namespace` `std;` `int` `main() {` ` ``string a = ``"1235421415454545454545454544"``;` ` ``string b = ``"1714546546546545454544548544544545"``;` ` ``if` `(a==``"0"` `|| b==``"0"``){` ` ``cout<<0<
## Java
`import` `java.util.*;` `public` `class` `Main {` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``String a = ``"1235421415454545454545454544"``;` ` ``String b = ``"1714546546546545454544548544544545"``;` ` ``if` `(a.equals(``"0"``) || b.equals(``"0"``)) {` ` ``System.out.println(``"0"``);` ` ``return``;` ` ``}` ` ``int` `m = a.length() - ``1``, n = b.length() - ``1``,` ` ``carry = ``0``;` ` ``String product = ``""``;` ` ``for` `(``int` `i = ``0``; i <= m + n || carry != ``0``; ++i) {` ` ``for` `(``int` `j = Math.max(``0``, i - n);` ` ``j <= Math.min(i, m); ++j) {` ` ``carry += (a.charAt(m - j) - ``'0'``)` ` ``* (b.charAt(n - i + j) - ``'0'``);` ` ``}` ` ``product += (``char``)(carry % ``10` `+ ``'0'``);` ` ``carry /= ``10``;` ` ``}` ` ``product = ``new` `StringBuilder(product)` ` ``.reverse()` ` ``.toString();` ` ``System.out.println(``"The Product is: "` `+ product);` ` ``}` `}`
## Python3
`def` `multiply_strings(a, b):` ` ``if` `a ``=``=` `"0"` `or` `b ``=``=` `"0"``:` ` ``return` `"0"` ` ` ` ``m, n ``=` `len``(a) ``-` `1``, ``len``(b) ``-` `1` ` ``carry ``=` `0` ` ``product ``=` `""` ` ``for` `i ``in` `range``(``0``, m``+``n``+``1``):` ` ``for` `j ``in` `range``(``max``(``0``, i``-``n), ``min``(i, m)``+``1``):` ` ``carry ``+``=` `(``ord``(a[m``-``j]) ``-` `48``) ``*` `(``ord``(b[n``-``i``+``j]) ``-` `48``)` ` ``product ``+``=` `str``(carry ``%` `10``)` ` ``carry ``/``/``=` `10` ` ``return` `product[::``-``1``]` `if` `__name__ ``=``=` `'__main__'``:` ` ``a ``=` `"1235421415454545454545454544"` ` ``b ``=` `"1714546546546545454544548544544545"` ` ``product ``=` `multiply_strings(a, b)` ` ``print``(``"The Product is: "` `+` `product)`
## Javascript
`function` `multiplyStrings(a, b) {` ` ``if` `(a === ``"0"` `|| b === ``"0"``) {` ` ``return` `"0"``;` ` ``}` ` ``let m = a.length - 1;` ` ``let n = b.length - 1;` ` ``let carry = 0;` ` ``let product = ``""``;` ` ` ` ``for` `(let i = 0; i <= m + n; i++) {` ` ``for` `(let j = Math.max(0, i - n); j <= Math.min(i, m); j++) {` ` ``carry += (a[m - j].charCodeAt(0) - 48) * (b[n - i + j].charCodeAt(0) - 48);` ` ``}` ` ``product += (carry % 10).toString();` ` ``carry = Math.floor(carry / 10);` ` ``}` ` ``return` `product.split(``""``).reverse().join(``""``);` `}` `// Example usage` `const a = ``"1235421415454545454545454544"``;` `const b = ``"1714546546546545454544548544544545"``;` `const product = multiplyStrings(a, b);` `console.log(``"The Product is: "` `+ product);`
## C#
`using` `System;` `public` `class` `Mainn {` ` ``public` `static` `void` `Main(``string``[] args) {` ` ``string` `a = ``"1235421415454545454545454544"``;` ` ``string` `b = ``"1714546546546545454544548544544545"``;` ` ``if` `(a.Equals(``"0"``) || b.Equals(``"0"``)) {` ` ``Console.WriteLine(``"0"``);` ` ``return``;` ` ``}` ` ``int` `m = a.Length - 1, n = b.Length - 1,` ` ``carry = 0;` ` ``string` `product = ``""``;` ` ``for` `(``int` `i = 0; i <= m + n || carry != 0; ++i) {` ` ``for` `(``int` `j = Math.Max(0, i - n);` ` ``j <= Math.Min(i, m); ++j) {` ` ``carry += (a[m - j] - ``'0'``)` ` ``* (b[n - i + j] - ``'0'``);` ` ``}` ` ``product += (``char``)(carry % 10 + ``'0'``);` ` ``carry /= 10;` ` ``}` ` ``char``[] charArray = product.ToCharArray();` ` ``Array.Reverse(charArray);` ` ``product = ``new` `string``(charArray);` ` ``Console.WriteLine(``"The Product is: "` `+ product);` ` ``}` `}`
Time Complexity: O(m*n), where m and n are length of two number that need to be multiplied.
Auxiliary Space: O(m+n), where m and n are length of two number that need to be multiplied.
Method 2:
## C++
`// Include header file` `#include ` `using` `namespace` `std;` `int` `main()` `{` ` ``string num1 = ``"1235421415454545454545454544"``;` ` ``string tempnum1 = num1;` ` ``string num2 = ``"1714546546546545454544548544544545"``;` ` ``string tempnum2 = num2;` ` ``// Check condition if one string is negative` ` ``if` `(num1[0] == ``'-'` `&& num2[0] != ``'-'``) {` ` ``num1 = num1.substr(1);` ` ``}` ` ``else` `if` `(num1[0] != ``'-'` `&& num2[0] == ``'-'``) {` ` ``num2 = num2.substr(1);` ` ``}` ` ``else` `if` `(num1[0] == ``'-'` `&& num2[0] == ``'-'``) {` ` ``num1 = num1.substr(1);` ` ``num2 = num2.substr(1);` ` ``}` ` ``string s1 = num1;` ` ``string s2 = num2;` ` ``reverse(s1.begin(), s1.end());` ` ``reverse(s2.begin(), s2.end());` ` ``vector<``int``> m(s1.length() + s2.length());` ` ``// Go from right to left in num1` ` ``for` `(``int` `i = 0; i < s1.length(); i++) {` ` ``// Go from right to left in num2` ` ``for` `(``int` `j = 0; j < s2.length(); j++) {` ` ``m[i + j]` ` ``= m[i + j] + (s1[i] - ``'0'``) * (s2[j] - ``'0'``);` ` ``}` ` ``}` ` ``string product = ``""``;` ` ``// Multiply with current digit of first number` ` ``// and add result to previously stored product` ` ``// at current position.` ` ``for` `(``int` `i = 0; i < m.size(); i++) {` ` ``int` `digit = m[i] % 10;` ` ``int` `carry = m[i] / 10;` ` ``if` `(i + 1 < m.size()) {` ` ``m[i + 1] = m[i + 1] + carry;` ` ``}` ` ``product = to_string(digit) + product;` ` ``}` ` ``// ignore '0's from the right` ` ``while` `(product.length() > 1 && product[0] == ``'0'``) {` ` ``product = product.substr(1);` ` ``}` ` ``// Check condition if one string is negative` ` ``if` `(tempnum1[0] == ``'-'` `&& tempnum2[0] != ``'-'``) {` ` ``product = ``"-"` `+ product;` ` ``}` ` ``else` `if` `(tempnum1[0] != ``'-'` `&& tempnum2[0] == ``'-'``) {` ` ``product = ``"-"` `+ product;` ` ``}` ` ``cout << ``"Product of the two numbers is :"` ` ``<< ``"\n"` ` ``<< product << endl;` `}` `// This code is contributed by Aarti_Rathi`
## Java
`// Java program to multiply two numbers represented` `// as strings.` `import` `java.util.Scanner;` `public` `class` `StringMultiplication {` ` ``// Driver code` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``String num1 = ``"1235421415454545454545454544"``;` ` ``String tempnum1 = num1;` ` ``String num2 = ``"1714546546546545454544548544544545"``;` ` ``String tempnum2 = num2;` ` ``// Check condition if one string is negative` ` ``if` `(num1.charAt(``0``) == ``'-'` ` ``&& num2.charAt(``0``) != ``'-'``) {` ` ``num1 = num1.substring(``1``);` ` ``}` ` ``else` `if` `(num1.charAt(``0``) != ``'-'` ` ``&& num2.charAt(``0``) == ``'-'``) {` ` ``num2 = num2.substring(``1``);` ` ``}` ` ``else` `if` `(num1.charAt(``0``) == ``'-'` ` ``&& num2.charAt(``0``) == ``'-'``) {` ` ``num1 = num1.substring(``1``);` ` ``num2 = num2.substring(``1``);` ` ``}` ` ``String s1` ` ``= ``new` `StringBuffer(num1).reverse().toString();` ` ``String s2` ` ``= ``new` `StringBuffer(num2).reverse().toString();` ` ``int``[] m = ``new` `int``[s1.length() + s2.length()];` ` ``// Go from right to left in num1` ` ``for` `(``int` `i = ``0``; i < s1.length(); i++) {` ` ``// Go from right to left in num2` ` ``for` `(``int` `j = ``0``; j < s2.length(); j++) {` ` ``m[i + j] = m[i + j]` ` ``+ (s1.charAt(i) - ``'0'``)` ` ``* (s2.charAt(j) - ``'0'``);` ` ``}` ` ``}` ` ``String product = ``new` `String();` ` ``// Multiply with current digit of first number` ` ``// and add result to previously stored product` ` ``// at current position.` ` ``for` `(``int` `i = ``0``; i < m.length; i++) {` ` ``int` `digit = m[i] % ``10``;` ` ``int` `carry = m[i] / ``10``;` ` ``if` `(i + ``1` `< m.length) {` ` ``m[i + ``1``] = m[i + ``1``] + carry;` ` ``}` ` ``product = digit + product;` ` ``}` ` ``// ignore '0's from the right` ` ``while` `(product.length() > ``1` ` ``&& product.charAt(``0``) == ``'0'``) {` ` ``product = product.substring(``1``);` ` ``}` ` ``// Check condition if one string is negative` ` ``if` `(tempnum1.charAt(``0``) == ``'-'` ` ``&& tempnum2.charAt(``0``) != ``'-'``) {` ` ``product = ``new` `StringBuffer(product)` ` ``.insert(``0``, ``'-'``)` ` ``.toString();` ` ``}` ` ``else` `if` `(tempnum1.charAt(``0``) != ``'-'` ` ``&& tempnum2.charAt(``0``) == ``'-'``) {` ` ``product = ``new` `StringBuffer(product)` ` ``.insert(``0``, ``'-'``)` ` ``.toString();` ` ``}` ` ``else` `if` `(tempnum1.charAt(``0``) == ``'-'` ` ``&& tempnum2.charAt(``0``) == ``'-'``) {` ` ``product = product;` ` ``}` ` ``System.out.println(``"Product of the two numbers is :"` ` ``+ ``"\n"` `+ product);` ` ``}` `}`
## Python3
`# Python3 program to implement the above approach` `# function to reverse the string` `def` `reverse(s):` ` ``str` `=` `""` ` ``for` `i ``in` `s:` ` ``str` `=` `i ``+` `str` ` ``return` `str` `num1 ``=` `"1235421415454545454545454544"` `tempnum1 ``=` `num1` `num2 ``=` `"1714546546546545454544548544544545"` `tempnum2 ``=` `num2` `# Check condition if one string is negative` `if` `(num1[``0``] ``=``=` `'-'` `and` `num2[``0``] !``=` `'-'``):` ` ``num1 ``=` `num1[``1``:]` `elif` `(num1[``0``] !``=` `'-'` `and` `num2[``0``] ``=``=` `'-'``):` ` ``num2 ``=` `num2[``1``:]` `elif` `(num1[``0``] ``=``=` `'-'` `and` `num2[``0``] ``=``=` `'-'``):` ` ``num1 ``=` `num1[``1``:]` ` ``num2 ``=` `num2[``1``:]` ` ` `s1 ``=` `num1` `s2 ``=` `num2` `s1 ``=` `reverse(s1)` `s2 ``=` `reverse(s2)` `m ``=` `[``0``]``*``(``len``(s1)``+``len``(s2))` `# Go from right to left in num1` `for` `i ``in` `range``(``len``(s1)):` ` ` ` ``# Go from right to left in num2` ` ``for` `j ``in` `range``(``len``(s2)):` ` ``m[i ``+` `j] ``=` `m[i ``+` `j] ``+` `(``ord``(s1[i]) ``-` `48``) ``*` `(``ord``(s2[j]) ``-` `48``)` ` ` ` ` `product ``=` `""` `# Multiply with current digit of first number` `# and add result to previously stored product` `# at current position.` `for` `i ``in` `range``(``len``(m)):` ` ``digit ``=` `m[i] ``%` `10` ` ``carry ``=` `m[i] ``/``/` `10` ` ``if` `(i ``+` `1` `< ``len``(m)):` ` ``m[i ``+` `1``] ``=` `m[i ``+` `1``] ``+` `carry` ` ``product ``=` `str``(digit) ``+` `product` ` ` `# ignore '0's from the right` `while` `(``len``(product) > ``1` `and` `product[``0``] ``=``=` `'0'``): ` ` ``product ``=` `product[``1``:]` ` ` `#check condition if one string is negative` `if` `(tempnum1[``0``] ``=``=` `'-'` `and` `tempnum2[``0``] !``=` `'-'``):` ` ``product ``=` `"-"` `+` `product` ` ` `elif` `(tempnum1[``0``] !``=` `'-'` `and` `tempnum2[``0``] ``=``=` `'-'``):` ` ``product ``=` `"-"` `+` `product` ` ` `print``(``"Product of the two numbers is :"``)` `print``(product)` `# This code is contributed by Abhijeet Kumar(abhijeet19403)`
## C#
`// C# program to multiply two numbers represented` `// as strings.` `using` `System;` `using` `System.Text;` `using` `System.Linq;` `public` `class` `GFG{` ` ``// Driver Code` ` ``static` `public` `void` `Main (){` ` ``String num1 = ``"1235421415454545454545454544"``;` ` ``String tempnum1 = num1;` ` ``String num2 = ``"1714546546546545454544548544544545"``;` ` ``String tempnum2 = num2;` ` ``// Check condition if one string is negative` ` ``if` `(num1[0] == ``'-'` `&& num2[0] != ``'-'``) {` ` ``num1 = num1.Substring(1);` ` ``}``else``{` ` ``if` `(num1[0] != ``'-'` `&& num2[0] == ``'-'``) {` ` ``num2 = num2.Substring(1);` ` ``}``else``{` ` ``if` `(num1[0] == ``'-'` `&& num2[0] == ``'-'``) {` ` ``num1 = num1.Substring(1);` ` ``num2 = num2.Substring(1);` ` ``}` ` ``}` ` ``}` ` ``String s1 = ``new` `string``(num1.Reverse().ToArray());` ` ``String s2 = ``new` `string``(num2.Reverse().ToArray());` ` ``int``[] m = ``new` `int``[s1.Length + s2.Length];` ` ``// Go from right to left in num1` ` ``for` `(``int` `i = 0; i < s1.Length; i++) {` ` ``// Go from right to left in num2` ` ``for` `(``int` `j = 0; j < s2.Length; j++) {` ` ``int` `x = ``int``.Parse((s1[i]-``'0'``).ToString());` ` ``int` `y = ``int``.Parse((s2[j]-``'0'``).ToString());` ` ``m[i + j] += (x*y);` ` ``}` ` ``}` ` ``String product = ``""``;` ` ``// Multiply with current digit of first number` ` ``// and add result to previously stored product` ` ``// at current position.` ` ``for` `(``int` `i = 0; i < m.Length; i++) {` ` ``int` `digit = m[i] % 10;` ` ``int` `carry = m[i] / 10;` ` ``if` `(i + 1 < m.Length) {` ` ``m[i + 1] += carry;` ` ``}` ` ``product = digit.ToString() + product;` ` ``}` ` ``// ignore '0's from the right` ` ``while` `(product.Length > 1 && product[0] == ``'0'``) {` ` ``product = product.Substring(1);` ` ``}` ` ``// Check condition if one string is negative` ` ``if` `(tempnum1[0] == ``'-'` `&& tempnum2[0] != ``'-'``) {` ` ``product = ``new` `StringBuilder(product).Insert(0, ``'-'``).ToString();` ` ``}``else``{` ` ``if` `(tempnum1[0] != ``'-'` `&& tempnum2[0] == ``'-'``) {` ` ``product = ``new` `StringBuilder(product).Insert(0, ``'-'``).ToString();` ` ``}` ` ``} ` ` ``Console.Write(``"Product of the two numbers is :\n"` `+ product);` ` ``}` `}` `// This code is contributed by shruti456rawal`
## Javascript
`// Javascript program to implement the approach` `let num1 = ``"1235421415454545454545454544"``;` `let tempnum1 = num1;` `let num2 = ``"1714546546546545454544548544544545"``;` `let tempnum2 = num2;` `// Check condition if one string is negative` `if` `(num1[0] == ``'-'` `&& num2[0] != ``'-'``) {` ` ``num1 = num1.substring(1);` `}` `else` `if` `(num1[0] != ``'-'` `&& num2[0] == ``'-'``) {` ` ``num2 = num2.substring(1);` `}` `else` `if` `(num1[0] == ``'-'` `&& num2[0] == ``'-'``) {` ` ``num1 = num1.substring(1);` ` ``num2 = num2.substring(1);` `}` `let s1 = num1.split(``""``);` `let s2 = num2.split(``""``);` `s1.reverse();` `s2.reverse();` `let m = ``new` `Array(s1.length + s2.length).fill(0);` `// Go from right to left in num1` `for` `(``var` `i = 0; i < s1.length; i++)` `{` ` ``// Go from right to left in num2` ` ``for` `(``var` `j = 0; j < s2.length; j++) {` ` ``m[i + j] = m[i + j]` ` ``+ (parseInt(s1[i]) * (parseInt(s2[j])));` ` ``}` `}` `let product = ``""``;` `// Multiply with current digit of first number` `// and add result to previously stored product` `// at current position.` `for` `(``var` `i = 0; i < m.length; i++) {` ` ``let digit = m[i] % 10;` ` ``let carry = Math.floor(m[i] / 10);` ` ``if` `(i + 1 < m.length) {` ` ``m[i + 1] = m[i + 1] + carry;` ` ``}` ` ``product = digit.toString() + product;` `}` `// ignore '0's from the right` `while` `(product.length > 1 && product[0] == ``'0'``) {` ` ``product = product.substring(1);` `}` `// Check condition if one string is negative` `if` `(tempnum1[0] == ``'-'` `&& tempnum2[0] != ``'-'``) {` ` ``product = ``"-"` `+ product;` `}` `else` `if` `(tempnum1[0] != ``'-'` `&& tempnum2[0] == ``'-'``) {` ` ``product = ``"-"` `+ product;` `}` `console.log(``"Product of the two numbers is :"``);` `console.log(product);` `// This code is contributed by phasing17`
Output
```Product of the two numbers is :
2118187521397235888154583183918321221520083884298838480662480```
Time Complexity: O(m*n), where m and n are length of two number that need to be multiplied.
Auxiliary Space: O(m+n), where m and n are length of two number that need to be multiplied.
Method 3:
• Convert the two input numbers from strings to lists of integers.
• A list with zeros.
• Iterate over each digit in the second number (num2) from right to left.
• For each digit, multiply it with each digit in the first number num1, and add the result in the result list.
• Convert the result list to a string.
## C++
`#include ` `#include ` `#include ` `using` `namespace` `std;` `string multiply(string num1, string num2) {` ` ``// Convert the input numbers from strings to vectors of integers` ` ``vector<``int``> vec1(num1.size());` ` ``for` `(``int` `i = 0; i < num1.size(); i++) {` ` ``vec1[i] = num1[num1.size() - i - 1] - ``'0'``;` ` ``}` ` ``vector<``int``> vec2(num2.size());` ` ``for` `(``int` `i = 0; i < num2.size(); i++) {` ` ``vec2[i] = num2[num2.size() - i - 1] - ``'0'``;` ` ``}` ` ``// Initialize the result vector with zeros` ` ``vector<``int``> result(vec1.size() + vec2.size());` ` ``// Multiply each digit in vec2 with vec1 and add the result to the appropriate position in the result vector` ` ``for` `(``int` `i = 0; i < vec2.size(); i++) {` ` ``int` `carry = 0;` ` ``for` `(``int` `j = 0; j < vec1.size(); j++) {` ` ``int` `product = vec1[j] * vec2[i] + carry + result[i + j];` ` ``carry = product / 10;` ` ``result[i + j] = product % 10;` ` ``}` ` ``result[i + vec1.size()] = carry;` ` ``}` ` ``// Remove leading zeros from the result vector and convert it back to a string` ` ``while` `(result.size() > 1 && result.back() == 0) {` ` ``result.pop_back();` ` ``}` ` ``string str(result.size(), ``'0'``);` ` ``for` `(``int` `i = 0; i < result.size(); i++) {` ` ``str[result.size() - i - 1] = result[i] + ``'0'``;` ` ``}` ` ``return` `str;` `}` `int` `main() {` ` ``string num1 = ``"4154"``;` ` ``string num2 = ``"51454"``;` ` ``cout << multiply(num1, num2) << endl;` ` ``num1 = ``"654154154151454545415415454"``;` ` ``num2 = ``"63516561563156316545145146514654"``;` ` ``cout << multiply(num1, num2) << endl;` ` ``return` `0;` `}`
## Java
`public` `class` `MultiplyStrings {` ` ``public` `static` `String multiply(String num1, String num2) {` ` ``int``[] n1 = ``new` `int``[num1.length()];` ` ``int``[] n2 = ``new` `int``[num2.length()];` ` ``for` `(``int` `i = ``0``; i < num1.length(); i++) {` ` ``n1[i] = num1.charAt(num1.length() - ``1` `- i) - ``'0'``;` ` ``}` ` ``for` `(``int` `i = ``0``; i < num2.length(); i++) {` ` ``n2[i] = num2.charAt(num2.length() - ``1` `- i) - ``'0'``;` ` ``}` ` ``int``[] result = ``new` `int``[num1.length() + num2.length()];` ` ``for` `(``int` `i = ``0``; i < n2.length; i++) {` ` ``int` `carry = ``0``;` ` ``for` `(``int` `j = ``0``; j < n1.length; j++) {` ` ``int` `product = n1[j] * n2[i] + carry + result[i + j];` ` ``carry = product / ``10``;` ` ``result[i + j] = product % ``10``;` ` ``}` ` ``result[i + n1.length] = carry;` ` ``}` ` ``StringBuilder sb = ``new` `StringBuilder();` ` ``int` `i = result.length - ``1``;` ` ``while` `(i > ``0` `&& result[i] == ``0``) {` ` ``i--;` ` ``}` ` ``while` `(i >= ``0``) {` ` ``sb.append(result[i--]);` ` ``}` ` ``return` `sb.toString();` ` ``}` ` ``public` `static` `void` `main(String[] args) {` ` ``String num1 = ``"4154"``;` ` ``String num2 = ``"51454"``;` ` ``System.out.println(multiply(num1, num2)); ``// Output: 213739916` ` ``num1 = ``"654154154151454545415415454"``;` ` ``num2 = ``"63516561563156316545145146514654"``;` ` ``System.out.println(multiply(num1, num2)); ``// Output: 41549622603955309777243716069997997007620439937711509062916` ` ``}` `}`
## Python3
`def` `multiply(num1, num2):` ` ``# Convert the input numbers from strings to lists of integers` ` ``num1 ``=` `[``int``(digit) ``for` `digit ``in` `num1][::``-``1``]` ` ``num2 ``=` `[``int``(digit) ``for` `digit ``in` `num2][::``-``1``]` ` ``# Initialize the result list with zeros` ` ``result ``=` `[``0``] ``*` `(``len``(num1) ``+` `len``(num2))` ` ``# Multiply each digit in num2 with num1 and add the result to the appropriate position in the result list` ` ``for` `i, digit2 ``in` `enumerate``(num2):` ` ``carry ``=` `0` ` ``for` `j, digit1 ``in` `enumerate``(num1):` ` ``product ``=` `digit1 ``*` `digit2 ``+` `carry ``+` `result[i ``+` `j]` ` ``carry ``=` `product ``/``/` `10` ` ``result[i ``+` `j] ``=` `product ``%` `10` ` ``result[i ``+` `len``(num1)] ``=` `carry` ` ``# Remove leading zeros from the result list and convert it back to a string` ` ``result ``=` `result[::``-``1``]` ` ``while` `len``(result) > ``1` `and` `result[``-``1``] ``=``=` `0``:` ` ``result.pop()` ` ``return` `''.join(``str``(digit) ``for` `digit ``in` `result)` `# Example 1` `num1 ``=` `"4154"` `num2 ``=` `"51454"` `print``(multiply(num1, num2))` `# Output: "213739916"` `# Example 2` `num1 ``=` `"654154154151454545415415454"` `num2 ``=` `"63516561563156316545145146514654"` `print``(multiply(num1, num2))` `# Output: "41549622603955309777243716069997997007620439937711509062916"`
## Javascript
`function` `multiply(num1, num2) {` ` ``// Convert the input numbers from strings to arrays of integers` ` ``let vec1 = num1.split(``""``).reverse().map(Number);` ` ``let vec2 = num2.split(``""``).reverse().map(Number);` ` ``// Initialize the result array with zeros` ` ``let result = Array(vec1.length + vec2.length).fill(0);` ` ``// Multiply each digit in vec2 with vec1 and add the result to the appropriate position in the result array` ` ``for` `(let i = 0; i < vec2.length; i++) {` ` ``let carry = 0;` ` ``for` `(let j = 0; j < vec1.length; j++) {` ` ``let product = vec1[j] * vec2[i] + carry + result[i + j];` ` ``carry = Math.floor(product / 10);` ` ``result[i + j] = product % 10;` ` ``}` ` ``result[i + vec1.length] = carry;` ` ``}` ` ``// Remove leading zeros from the result array and convert it back to a string` ` ``while` `(result.length > 1 && result[result.length - 1] === 0) {` ` ``result.pop();` ` ``}` ` ``let str = result.reverse().join(``""``);` ` ``return` `str;` `}` ` ``let Num1 = ``"4154"``;` ` ``let Num2 = ``"51454"``;` ` ``console.log(multiply(Num1, Num2));` ` ``let num1 = ``"654154154151454545415415454"``;` ` ``let num2 = ``"63516561563156316545145146514654"``;` `console.log(multiply(num1, num2));`
## C#
`using` `System;` `using` `System.Linq;` `public` `class` `Program` `{` ` ``public` `static` `string` `Multiply(``string` `num1, ``string` `num2)` ` ``{` ` ``// Convert the input numbers from strings to arrays of integers` ` ``int``[] vec1 = num1.Reverse().Select(c => c - ``'0'``).ToArray();` ` ``int``[] vec2 = num2.Reverse().Select(c => c - ``'0'``).ToArray();` ` ``// Initialize the result array with zeros` ` ``int``[] result = ``new` `int``[vec1.Length + vec2.Length];` ` ``// Multiply each digit in vec2 with vec1 and add the result to the appropriate position in the result array` ` ``for` `(``int` `i = 0; i < vec2.Length; i++)` ` ``{` ` ``int` `carry = 0;` ` ``for` `(``int` `j = 0; j < vec1.Length; j++)` ` ``{` ` ``int` `product = vec1[j] * vec2[i] + carry + result[i + j];` ` ``carry = product / 10;` ` ``result[i + j] = product % 10;` ` ``}` ` ``result[i + vec1.Length] = carry;` ` ``}` ` ``// Remove leading zeros from the result array and convert it back to a string` ` ``while` `(result.Length > 1 && result[result.Length - 1] == 0)` ` ``{` ` ``Array.Resize(``ref` `result, result.Length - 1);` ` ``}` ` ``string` `str = ``new` `string``(result.Reverse().Select(d => (``char``)(d + ``'0'``)).ToArray());` ` ``return` `str;` ` ``}` ` ``public` `static` `void` `Main()` ` ``{` ` ``string` `num1 = ``"4154"``;` ` ``string` `num2 = ``"51454"``;` ` ``Console.WriteLine(Multiply(num1, num2));` ` ``string` `num3 = ``"654154154151454545415415454"``;` ` ``string` `num4 = ``"63516561563156316545145146514654"``;` ` ``Console.WriteLine(Multiply(num3, num4));` ` ``}` `}`
Output
```213739916
41549622603955309777243716069997997007620439937711509062916```
Time Complexity: O(m*n), where m and n are length of two number that need to be multiplied.
Auxiliary Space: O(m+n), where m and n are length of two number that need to be multiplied.
Related Article :
Karatsuba algorithm for fast multiplication
If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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Cody
# Problem 1511. Related Vectors
Solution 255396
Submitted on 5 Jun 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
%% A = [1 0 1 0 1 0]; B=ones(1,6); y_correct = [1 0 1 0 1 0]; assert(isequal(your_fcn_name(A,B),y_correct))
```Error: Output argument "y" (and maybe others) not assigned during call to "/users/msssystem5/your_fcn_name.m>your_fcn_name". ```
2 Fail
%% A = [-1 2 3 4 5 3]; B=zeros(1,6); y_correct = zeros(1,6); assert(isequal(your_fcn_name(A,B),y_correct))
```Error: Output argument "y" (and maybe others) not assigned during call to "/users/msssystem5/your_fcn_name.m>your_fcn_name". ```
3 Fail
%% A = [0 2 0 4 5 3]; B=[1 2 3 4 9 -6]; y_correct = [0 2 0 4 9 -6]; assert(isequal(your_fcn_name(A,B),y_correct))
```Error: Output argument "y" (and maybe others) not assigned during call to "/users/msssystem5/your_fcn_name.m>your_fcn_name". ```
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## Tuesday, March 12, 2013
### Histograms - 2 : Histogram Equalization
Hi friends,
In last article, we saw what is histogram and how to plot it. This time we can learn a method for image contrast adjustment called "Histogram Equalization".
So what is it ? Consider an image whose pixel values are confined to some specific range of values only. For eg, brighter image will have all pixels confined to high values. But a good image will have pixels from all regions of the image. So you need to stretch this histogram to either ends (as given in below image, from wikipedia) and that is what Histogram Equalization does (in simple words). This normally improves the contrast of the image.
Histogram Equalization
Again, I would recommend you to read the wikipedia page on Histogram Equalization for more details about it. It has a very good explanation with worked out examples, so that you would understand almost everything after reading that. And make sure you have checked the small example given in "examples" section before going on to next paragraph.
So, assuming you have checked the wiki page, I will demonstrate a simple implementation of Histogram Equalization with Numpy. After that, I will present you OpenCV function. ( If you are not interested in implementation, you can skip this and go to the end of article)
Numpy Implementation
We start with plotting histogram and its cdf (cumulative distribution function) of the image in Wikipedia page. All the functions are known to us except np.cumsum(). It is used to find the cumulative sum (cdf) of a numpy array.
import cv2
import numpy as np
from matplotlib import pyplot as plt
hist,bins = np.histogram(img.flatten(),256,[0,256])
cdf = hist.cumsum()
cdf_normalized = cdf *hist.max()/ cdf.max() # this line not necessary.
plt.plot(cdf_normalized, color = 'b')
plt.hist(img.flatten(),256,[0,256], color = 'r')
plt.xlim([0,256])
plt.legend(('cdf','histogram'), loc = 'upper left')
plt.show()
Input Image and its histogram
You can see histogram lies in brighter region. We need the full spectrum. For that, we need a transformation function which maps the input pixels in brighter region to output pixels in full region. That is what histogram equalization does.
Now we find the minimum histogram value (excluding 0) and apply the histogram equalization equation as given in wiki page. But I have used here, the masked array concept array from Numpy. For masked array, all operations are performed on non-masked elements. You can read more about it from Numpy docs on masked arrays
cdf_m = (cdf_m - cdf_m.min())*255/(cdf_m.max()-cdf_m.min())
cdf = np.ma.filled(cdf_m,0).astype('uint8')
Now we have the look-up table that gives us the information on what is the output pixel value for every input pixel value. So we just apply the transform.
img2 = cdf[img]
Now we calculate its histogram and cdf as before ( you do it) and result looks like below :
Histogram Equalized Image and its histogram
You can see a better contrast in the new image, and it is clear from the histogram also. Also compare the cdfs of two images. First one has a steep slope, while second one is almost a straight line showing all pixels are equi-probable.
Another important feature is that, even if the image was a darker image (instead of a brighter one we used), after equalization we will get almost the same image as we got. As a result, this is used as a "reference tool" (I don't get a more suitable than this) to make all images with same light conditions. This is useful in many cases, for eg, in face recognition, before training the face data, the images of faces are histogram equalized to make them all with same light conditions. It provides better accuracy.
OpenCV Implementation
If you are bored of everything I have written above, just leave them. You need to remember only one function to do this, cv2.calcHist(). Its input is just grayscale image and output is our image.
Below is a simple code snippet showing its usage for same image we used :
equ = cv2.equalizeHist(img)
res = np.hstack((img,equ)) #stacking images side-by-side
cv2.imwrite('res.png',res)
See the result :
OpenCV Histogram Equalization
So now you can take different images with different light conditions, equalize it and check the results.
Histogram equalization is good when histogram of the image is confined to a particular region. It won't work good in places where there is large intensity variations where histogram covers a large region, ie both bright and dark pixels are present. I would like to share to SOF questions with you. Please checkout the images in the questions, analyze their histograms, check resulting images after equalization :
How can I adjust contrast in OpenCV in C?
How do I equalize contrast & brightness of images using opencv?
So I would like to wind up this article here. In this article, we learned how to implement Histogram Equalization, how to use OpenCV for that etc. So take images, equalize it and have your own hack arounds.
See you next time !!!
Abid Rahman K.
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# Goodness of Fit from Gaussian Fit
27 views (last 30 days)
Benjamin Zaugg on 24 Apr 2013
Hi there, I have to analyze data (specificly area under curve) from an table with 2 coloumns and abou 1700 rows. I have used a a fit on the data:
fit_1 = fit(table([x0:x1],[1]),table([x0:x1],[2]),'gauss4');
this gives me the CFIT file named 'fit_1'
However, I would like to find out the Goodness of the fit (R squared) of my fit_1 compared to my table data, without using the cftool but integrate it into my existing function which I wrote.
Can somebody tell me if there is the possibility to do so outside of cftool and where I could find Information of how to do so?
Thanks a lot! Sorry for any typos & lack of knowledge (first post)
Greetings
Benjamin Zaugg on 24 Apr 2013
bump
Eric on 24 Apr 2013
Edited: Eric on 24 Apr 2013
I use the following:
ybar = mean(y); %Mean of the data
sst = sum( (y - ybar).^2 ); %Total sum of squares
gof.sse = sum( (y - fit_vec).^2 ); %Residual sum of squares
gof.rsquare = 1 - gof.sse/sst;
See http://en.wikipedia.org/wiki/Coefficient_of_determination as a reference for the equation for gof.rsquare.
In this code y is the data and fit_vec is the fitted data.
-Eric
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# Define * on Z by a * b = a – b + ab
Question:
Define $*$ on $Z$ by $a * b=a-b+a b$. Show that $*$ is a binary operation on $Z$ which is neither commutative nor associative.
Solution:
* is an operation as $a * b=a-b+a b$ where $a, b \in Z$. Let $a=\frac{1}{2}$ and $b=2$ two integers.
$\mathrm{a} * \mathrm{~b}=\frac{1}{2} * 2=\frac{1}{2}-2+\frac{1}{2} \cdot 2 \Rightarrow \frac{1-4}{2}+1=\frac{-3+2}{2} \Rightarrow \frac{-1}{2} \in \mathrm{Z}$
So, $*$ is a binary operation from $Z \times Z \rightarrow Z$.
For commutative,
$\mathrm{b}^{*} \mathrm{a}=2-\frac{1}{2}+2 \cdot \frac{1}{2}=\frac{4-1}{2}+1 \Rightarrow \frac{3+2}{2}=\frac{5}{2} \in \mathrm{Z}$
Since a*b ≠ b*a, hence * is not commutative operation.
Again for associative,
$a^{*}\left(b^{*} c\right)=a^{*}(b-c+b c)$
$=a-(b-c+b c)+a(b-c+b c)$
$=a-b+c-b c+a b-a c+a b c$
$(a * b) * c=(a-b+a b) * c$
$=a-b+a b-c+(a-b+a b) c$
$=a-b-c+a b+a c-b c+a b c$
As $a *(b * c) \neq(a * b) * c$, hence $*$ not an associative operation.
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welcome:
# Fastfood Optimality Check
## Problem Description
This example is based on the "Fast Food" problem, number 662 of volume VI of the ACM programming contests problem set archive (http://acm.uva.es/p/v6/662.html).
The fastfood chain McBurger owns several restaurants along a highway. Recently, they have decided to build several depots along the highway, each one located at a restaurant and supplying several of the q restaurants with the needed ingredients. Each of the restaurants will be supplied by the nearest depot. If two or more depots are equidistantly nearest to a restaurant, it is supplied by exactly one of these. Naturally, the depots should be placed so that the average distance between a restaurant and its assigned depot is minimized, or equivalently that the sum of supply distances is minimized.
We are to verify whether a given schema for constructing depots is optimal (among all construction schemas that construct the same number of depots as in the given schema). If it is, then there should not be any answer set; if it is not optimal, answer sets should encode a better solution.
Instance have been created based on data taken from the website of "Tank&Rast", the company which runs most of the German autobahn service stations.
## Input format
Instances are given as facts over two predicates, restaurant/2 and depot/2.
restaurant(N,K) represents that a restaurant named N (this is a constant consisting of lowercase ASCII characters with enclosed underscores) is located at kilometer K of the highway (K is a nonnegative integer less than or equal to 910). Example: restaurant(auetal_nord,270).
depot(N,K) represents the fact that a depot is to be constructed at the restaurant named N, located at kilometer K, in the construction scheme, the optimality of which is to be checked. Example: depot(breisgau_ost,761).
## Output format
Answer sets should contain a better construction scheme as atoms of the form altdepot(N,K), where N is the name of the associated restaurant and K the kilometer at which it is located. If the input construction scheme is optimal, no answer set should be printed.
NOTES:
1. Issues about directions on the highway (e.g. that some restaurants are reachable only going in one particular direction on the highway) are not considered, it is hence assumed that all restaurants are connected to both directions of the highway.
2. There can be more than one depot at a given kilometer. This occurs rather frequently in the data, when there are restaurants on both sides of the motorway.
## Author(s)
Author: Wolfgang Faber
Affiliation: University of Calabria, Italy
ASP Competition 2011: ProblemsDescription/FastfoodOptimalityCheck (last edited 2011-01-10 10:39:02 by CarmenSantoro)
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1. ## Stirling's formula manipulation
Hi, I'm stuck on part of this question involving Stirling's formula. I have to show that for $p+q=1$ (they're probabilities),
$\sum_{i=1}^\infty {2n \choose n}p^n q^n = \infty$ if and only if $p=q= \frac{1}{2}$
by using Stirling's approximation, which is $n! \approx \sqrt{2\pi n}\, \left(\frac{n}{e}\right)^{n}$, for large $n$.
I know that I should be aiming to show that $
{2n \choose n}p^n q^n \approx \frac{(4pq)^n}{\sqrt{\pi n}}$
I've tried rearranging it but I can't seem to get rid of the terms involving $e$ (among others).
Also, anyone who can solve this is a clever clogs, so maybe they can shed some light on this one, which seems to have defeated the SOS Math community. S.O.S. Mathematics CyberBoard :: View topic - Predictors after a simple linear regression is done
Thanks VERY much
2. Hello,
Originally Posted by mongrel73
Hi, I'm stuck on part of this question involving Stirling's formula. I have to show that for $p+q=1$ (they're probabilities),
$\sum_{i=1}^\infty {2n \choose n}p^n q^n = \infty$ if and only if $p=q= \frac{1}{2}$
by using Stirling's approximation, which is $n! \approx \sqrt{2\pi n}\, \left(\frac{n}{e}\right)^{n}$, for large $n$.
I know that I should be aiming to show that $
{2n \choose n}p^n q^n \approx \frac{(4pq)^n}{\sqrt{\pi n}}$
I've tried rearranging it but I can't seem to get rid of the terms involving $e$ (among others).
Also, anyone who can solve this is a clever clogs, so maybe they can shed some light on this one, which seems to have defeated the SOS Math community. S.O.S. Mathematics CyberBoard :: View topic - Predictors after a simple linear regression is done
Thanks VERY much
I don't understand where your problem is
${2n \choose n}=\frac{(2n)!}{n!n!}=\frac{(2n)!}{(n!)^2}$
Using Stirling's formula :
$(2n)! \sim \sqrt{4\pi n} \cdot \left(\frac{2n}{e}\right)^{2n}=2\sqrt{\pi n} \cdot 2^{2n} \cdot n^{2n} \cdot e^{-2n}$
$(n!)^2 \sim \left[\sqrt{2\pi n} \cdot \left(\frac ne\right)^n\right]^2=2 \pi n \cdot \left(\frac ne\right)^{2n}=2\pi n \cdot n^{2n} \cdot e^{-2n}$
Hence :
${2n \choose n}=\frac{{\color{red}2}\sqrt{\pi n} \cdot 2^{2n} \cdot {\color{red}n^{2n}} \cdot {\color{red}e^{-2n}}}{{\color{red}2}\pi n \cdot {\color{red}n^{2n}} \cdot {\color{red}e^{-2n}}}=\frac{2^{2n}}{\sqrt{\pi n}}$
And finally :
${2n \choose n} p^n (1-p)^n \sim \frac{(4p(1-p))^n}{\sqrt{\pi n}}$
Now, $p(1-p)=\frac 14 \Leftrightarrow p=\frac 12$. Otherwise, $p(1-p)<\frac 14$, for any $p \neq \frac 12$ (and in [0,1] of course). This is VERY easy to prove, by taking the derivative of $f ~:~ x \mapsto x(1-x)$.
If $p=\frac 12$, then the series would be equivalent to $\sum_{n\geq 1} \frac{1}{\sqrt{n}}$, which is a divergent Riemann series.
If $p \neq \frac 12$, then note that $\frac{1}{\sqrt{n}} \leq 1$, for any $n \geq 1$
Thus $\frac{(4p(1-p))^n}{\sqrt{\pi n}} \leq \frac{(4p(1-p))^n}{\sqrt{\pi}}$
Since $p(1-p)<\frac 14$, $4p(1-p)<1$ and therefore, $\sum_{n \geq 1} \frac{(4p(1-p))^n}{\sqrt{\pi n}} \leq \frac{1}{\sqrt{\pi}} \sum_{n \geq 1} (4p(1-p))^n$ is a convergent series.
3. The only tricky part of this was trying to figure out what that subscript i meant.
4. Originally Posted by matheagle
The only tricky part of this was trying to figure out what that subscript i meant.
I guess you replied to the wrong thread...
5. NOPE, look at the original sum, instead of using n, the OP used i.
6. Originally Posted by matheagle
NOPE, look at the original sum, instead of using n, the imposter used i.
Yes, oops. I also can't work out what my problem was. Thanks anyway though...
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# Do I rneed a voltage regulator?
#### silverfox
Mar 31, 2012
6
I am not very experienced at circuit design but I am designing a circuit that I hope will become part of a low-cost commercial product of my own design. Consequently, I am trying to keep component count to a bare minimum. My question is, if my gadget has its own 3V CR2032 coin battery which powers a microcontroller, EEPROM, RTC, and some optoisolators, is there any need for a voltage regulator? Can I get away just leaving it out?
#### davenn
Moderator
Sep 5, 2009
14,085
What are the voltage and current requirements of the circuit? That is, what voltage range will they work over. If say, it's 3 V plus or minus a volt then there shouldn't really be any problem
Have you worked out how long the circuit is going to last on a CR2032 ?
Dave
#### silverfox
Mar 31, 2012
6
How to best switch 5V with a 3V circuit?
A circuit for a commercial gadget I am designing only requires 3V but one of the functions of the circuit is to occasionally supply +5V to a jack into which an external device is plugged. When the user wants to use that device, they plug an external battery pack into a jack on my gadget and the application program in my microcontroller turns the device on and off as required by switching the 5V supply. I am thinking that one way to do this is with a solid state relay that can switch the 5V on and off but can be operated entirely with 3 volts. But I wonder if this is the best solution. I am trying to make this circuit as inexpensive as possible since the end product needs to be very low cost and hopefully manufactured in high volume. Is there a better way than the relay?
#### silverfox
Mar 31, 2012
6
What are the voltage and current requirements of the circuit? That is, what voltage range will they work over. If say, it's 3 V plus or minus a volt then there shouldn't really be any problem
Have you worked out how long the circuit is going to last on a CR2032 ?
Dave
I am not really sure about the range of voltage and current requirements. I guess I need to check that information on the datasheets of the individual components. And I don't know for sure how long the CR2032 will last but I am aware of a similar circuit for which the CR2032 is quite adequate. It seems like it would be tough calculation to make since there is a microcontroller involved and I think that the current draw depends on what it is doing, which is not entirely predictable.
#### (*steve*)
##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
I see several ways:
1) power the circuit from 5V, not 3V
2) employ a boost regulator to get 5V from your 3V
3) use another relay that will operate from 3V
4) use something other than a relay (SSR, optocoupler, triac, mosfet, etc -- possibly in combination)
#### (*steve*)
##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
Also look at the datasheet for the battery. It's not a particularly difficult problem once you have all the information.
#### silverfox
Mar 31, 2012
6
I see several ways:
1) power the circuit from 5V, not 3V
2) employ a boost regulator to get 5V from your 3V
3) use another relay that will operate from 3V
4) use something other than a relay (SSR, optocoupler, triac, mosfet, etc -- possibly in combination)
1) I am planning on using a CR2032 because size and weight are an issue here for the most common usage. If the user wants to attach a device that is powered from my device then he can attach a battery. But that is not the primary application and I want the unit to be as small and light as possible for the primary usage.
2) I don't think a boost regulator is the best idea because the external device could drain the little coin battery too fast. I want them to use their own battery if they want to power their devices off my gadget.
3) Another relay? Other than what?
4) So what are the advantages of any of those (optocoupler, triac, mosfet, etc ) over using a solid state relay as I suggested? That was my question. Are any of them cheaper, use less power, smaller? And are there 3V versions of all of them? I was thinking about an optocoupler but just wasn't sure if there would be any problems. I think that they have current limitations but probably a SSR does too. I am not sure which is more tolerant. Something to research.
#### (*steve*)
##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
If you don't need the output to be isolated, you can use your circuit to switch a mosfet. That will require almost no additional power from your battery. The SSR will require much more because it is essentially operated by you powering a LED.
I initially misinterpreted you as saying that you provide an external 5V to operate a relay. Since you've started a new thread for this I did not realise this was about the thing operated from a small cell.
#### silverfox
Mar 31, 2012
6
Thanks, that was the answer I was looking for. Now I will have to research MOSFETs. That is what I like about doing things like electronic projects on my own but with help from the kind people in forums. I learn a lot more this way.
#### silverfox
Mar 31, 2012
6
I am having a little trouble finding the correct MOSFET device to use. I found this site that is useful: http://brunningsoftware.co.uk/FET.htm. But they are using 5V to trigger the MOSFET it isn't clear (to me) if the same device will also work with 3V. I did a Digikey search but I didn't see any filters that specified the minimum gate voltage to trigger the MOSFET. Any ideas?
#### BobK
Jan 5, 2010
7,682
Look for a logic-level MOSFET. These have a lower gate voltage requirement.
Bob
#### (*steve*)
##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
I did a Digikey search but I didn't see any filters that specified the minimum gate voltage to trigger the MOSFET. Any ideas?
What you're looking for is threshold voltage. Digikey list it as "Vgs(th) (Max) @ Id". Some value around 1V would probably be good.
Then you need to look at the datasheet to determine the saturation current you're going to see at your minimum supply voltage (less however much your gate drive differs from that). to see what load the device can switch. Also relevant is Rds(on) for this gate voltage. Both of these can be used to calculate the max load depending on the amount of heatsinking your device has (which you then de-rate for safety)
edit: OK I did this and picked the first N channel mosfet I found. it's a NDS331NTR-ND, and it's datasheet tells me that it has a RdsOn of 0.21 ohms at Vgs of 2.1V (it also gas a BVdss of only 20V) It's saturation current will be something over 3A (it's only rated for 1.3A continuous)
At 1A, the power dissipation would be 0.21W, and the junction to ambient thermal resistance is 250 degC/W, so the junction would rise about 52 degrees over ambient. That would be acceptable, but I'd label the device as being able to switch 12V at 500mA just to be sure.
You could go searching for other devices (that may be more easily available or have different specs) if you have a need for them. I don't recommend this part, as I said it was the first one on the list.
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## Thursday, March 17, 2011
### A VIP ( Very Important Problem) ( for me anyway)
Consider a line of lockers, numbered consecutively, starting from 1 and continuing without bound.
(As you might have already noticed, this is not a real world problem)
Person 1 walks through and, starting with locker 1, makes sure all the lockers are closed.
Person 2 walks through, and, starting with locker 2, opens every other locker.
Person 3 walks through, starts at locker 3, and changes every third locker, opening the closed ones and closing the open ones.
Person 4 starts with locker 4 and changes the state of every fourth locker.
This continues in like manner, with person n, starting at locker n, and changing every nth locker.
(You might have noticed that this problem is not intended to practice necessary algebra skills)
This problem changed my life. I guess P.J. would call it a puzzle. I am sure most of the readers of this have heard the problem.
Oh, this is not exactly the puzzle that changed my life. That one also included a question. I think it is a better problem if the student has to decide what would be a good question. When I give this problem to students, they ask what the question is. My response is, "Oh no, did I forget to ask a question again? What do you think would be a good question? "
Someone will say something that is roughly equivalent to "Which lockers are open eventually?"
That is the question I was asked. However sometimes student's questions are better than the one I had in mind. Isn't it true that an important aspect of mathematics is deciding what questions to ask? Do we give our students a chance to do this?
But, as is my lifelong habit, I digress.
How did this problem change my life?
It was given to me in a course named "problem solving". I had been teaching for several years. I thought I was making progress but now know I had a lot to learn about how children learn. I now know I still do have a lot to learn about how children learn. I found the problem challenging and interesting, even though it has no basis in reality. I worked on it for a while, and an amazing thing happened. Once I had an answer, I became curious about why the answer ended up being so nice. I started thinking about it and without much effort I had a proof of a theorem about factors of integers, and I understood and therefore learned some interesting mathematics I did not know before.
I realized that what made this problem good was that I found it interesting, and that it created a situation where solving it helped me to understand something. It was not just an answer, but a revelation, that lead to a new understanding.
Was it important? Well, I have since used it to solve other problems, so I would say, yes.
But most importantly, I learned some significant mathematics by doing a problem. I was proud, and happy , and wanted to find other problems to work on. And I got this crazy idea that perhaps I could arrange for this sort of thing to happen in my classes for my students.
And so, for the next thirty five years, I tried to write problems that would lead my students to figuring out something for themselves.
It took a while. I tried and rejected the method of having them work a few examples in the hope they would notice the pattern I wanted them to discover. For one thing, it seemed phony but for another, they didn't learn any mathematics from working on the problem because they didn't really figure out any mathematics, they simply arrived at what I wanted them to arrive at.
I also tried worksheets. That was a total failure. For one thing, once give a worksheet, the objective of a student is to finish the worksheet. If another student has a question or observation, that interferes with the task at hand, which is to finish the worksheet. I also learned that students did not understand that they were supposed to learn something from finishing the worksheet. They were just supposed to finish it.
So, I eventually fell into a groove that I was happy with. One problem at a time. Everyone works on that problem. If you finish, compare your result with others. If you all agree, ask the next question. What if we changed the problem this way? Is there a generalization? Is there a special case? Meanwhile I walk around, rapidly, listening and looking at what is happening, waiting for the teachable moment, that moment when the entire class is ready to share their thoughts on the problem. I usually also share my thoughts at this point, especially with regard to connections and what might happen later.
Then I give them another problem to work on.
And learning happens, the students are actively engaged, and frequently I learn something I didn't know. I also learn a lot about how my students work.
Teaching by giving students a good problem is different than other ways of teaching. The teacher has to be ready for anything that comes up, and must have a plan if nothing comes up. The teacher must be willing to give students time to think of things , time to make up their mind, and time for ambiguity. The teacher must be prepared for long periods of time where everyone is wrong, and be willing to let that happen. It takes practice and a belief that eventually truth will win and incorrect reasoning will be exposed. It means giving up the role of the person who is the authority about truth. But in the end, it worked better for me than any alternative.
And, the best part is, I have a great time doing it.
And it all started with the famous locker problem, a puzzle that taught me how to become a better teacher.
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# Cumulative lift chart for CART® Classification
The procedure for the calculation of cumulative lift depends on the validation method. For a multinomial response variable, Minitab displays multiple charts that treat each class as the event in turn.
## Training data set or no validation
For the chart for a training data set, each point on the chart represents a terminal node from the tree. The terminal node with the highest event probability is the first point on the chart and appears leftmost. The other terminal nodes are in order of decreasing event probability.
Use the following process to find the x- and y-coordinates for the points.
1. Calculate the event probability of each terminal node: where
• n1,k is the number of cases in the event class in the kth node
• Nk is the number of cases in the kth node
2. Rank the terminal nodes from highest to lowest event probability.
3. Use every event probability as a threshold. For a specific threshold, cases with estimated event probability greater than or equal to the threshold get 1 as the predicted class, 0 otherwise. Then, you can form a 2x2 table for all cases with observed classes as rows and predicted classes as columns to calculate the true positive rate for each terminal node.
For example, suppose the following table summarizes a tree with 4 terminal nodes:
A: Terminal node B: Number of events C: Number of cases D: Threshold (B/C)
4 18 30 0.60
1 25 67 0.37
3 12 56 0.21
2 4 36 0.11
Totals 59 189
Then the following are the corresponding four tables with their respective true positive rates to 2 decimal places:
Table 1. Threshold = 0.60. True positive rate = 18 / 59 = 0.31
Predicted
event nonevent
Observed event 18 41
nonevent 12 118
Table 2. Threshold = 0.37. True positive rate = (18 + 25) / 59 = 0.73
Predicted
event nonevent
Observed event 43 16
nonevent 54 76
Table 3. Threshold = 0.21. True positive rate = (18 + 25 + 12) / 59 = 0.93
Predicted
event nonevent
Observed event 55 4
nonevent 98 32
Table 4. Threshold = 0.11. True positive rate = (18 + 25 + 12 + 4) / 59 = 1
Predicted
event nonevent
Observed event 59 0
nonevent 130 0
4. From the sorted terminal nodes, find the percentage of the population in the terminal nodes: where
• Nk is the number of cases in the kth node
• N is the number of cases in the training data set
5. From the sorted list, calculate the cumulative percentage of the data in each terminal node. These cumulative values are the x-coordinates on the chart.
For example, if the terminal node with the highest predicted probability contains 0.16 of the data and the terminal node with the second-highest event probability has 0.35 of the population, then the cumulative percentage of the data for the first terminal node is 0.16 and the cumulative percentage of the population for the second terminal node is 0.16 + 0.35 = 0.51.
6. To find the cumulative lift for the y-coordinate, divide the true positive rate and the cumulative percentage of the population:
The following table shows an example of the computations for a small tree. The values are to 2 decimal places.
A: Terminal node B: Number of events C: Number of cases D: Event probability for sorting (B/C) E: True positive rate F: Percent in data (C/ sum of C) G: Cumulative percent in data, x-coordinate H: Cumulative lift (E/G), y-coordinate
4 18 30 0.60 0.31 0.16 0.16 1.92
1 25 67 0.37 0.73 0.35 0.51 1.42
3 12 56 0.21 0.93 0.30 0.81 1.15
2 4 36 0.11 1 0.19 1.00 1
## Separate test data set
Use the same steps as the training data set case but calculate the event probability from the cases for the test data set.
## Test with k-fold cross-validation
The procedure to define the x- and y-coordinates on the cumulative lift chart with k-fold cross-validation has an additional step. This step creates many distinct event probabilities. For example, suppose the tree diagram contains 4 terminal nodes. We have 10-fold cross-validation. Then, for the ith fold, you use 9/10 portion of the data to estimate the event probabilities for cases in fold i. When this process repeats for each fold, the maximum number of distinct event probabilities is 4 *10 = 40. After that, sort all the distinct event probabilities in decreasing order. Use the event probabilities as each of the threshold values to assign predicted classes for cases in the entire data set. After this step, steps from 3 to the end for the training data set procedure apply to find the x- and y-coordinates.
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CC-MAIN-2024-18
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http://howded.com/en/pages/307008
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you need
• tester table resistivity.
Guide
1
To reduce power current on the circuit and change the values on which it depends.To determine these values, use the equation, which is a type of write Ohm's law I = U • S / (ρ • l).Assemble the chain by connecting to the test site a rheostat.Connect it to a power current .After that, changing the setting of the rheostat, reduce tension in the area.In order to obtain figures voltage, connect in parallel to the tester and make a measurement.Then, by connecting the tester to the site and changing consistently, measure force current in the circuit.Lower voltage on the circuit by n times.By measuring the force current , make sure that it also decreased n times.
2
Change the resistance of the circuit.To do this, determine the resistivity of the conductor material at a special table.To reduce the strength of the current pick conductors of the same size, but with a high resistivity.How many times will increase the resistivity, so many times will decrease the strength of the current .
3
If you can not pick up the other conductors change their geometric dimensions.Reduce the cross-sectional area of the conductor.For example, if stranded wire, remove a few strands.How many times will decrease the cross sectional area, the same factor decreases current.The second way - to increase the overall length of the conductors.How many times will increase the length of the conductors on the circuit, so many times will decrease the strength of the current .
4
Another easy way - attach the chain to the source current less electromotive force.How many times will decrease the value of the EMF, so many times will decrease the strength of the current .These techniques can be combined to achieve the desired effect.For example, lowering the voltage 2 times, increasing the length of the conductors 3 times and reduce the cross-sectional area 4 times, you will get a reduction of force current 2 • 3 • 4 = 24 times.
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http://uk.ask.com/question/the-inversion-of-your-image-in-a-plane-mirror-is-actually-an-inversion-of
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What is the inversion in a planemirror called?
Answer
It is called "Lateral Inversion" Lateral means "sideways" The term is used because the image often seems left-right reversed. In fact the inversion is really. at right angles to the mirror surface, but it can appear to be laterally reversed.
Q&A Related to "What is the inversion in a plane mirror called..."
You are assuming that the image has a life of its own. You move your right hand and the hand's image, which is on your right opposite your hand, moves. You move your foot and the http://answers.yahoo.com/question/index?qid=201310...
Think it on. if two person stand in front of plane mirror, You will found that Left person is visible Right as the direction of mirror. THINK IT ON. THINK IT ON. http://answers.yahoo.com/question/index?qid=201401...
In theory, an infinite number of images is formed. Now, you would notice that the images keep becoming smaller and smaller. At a certain image, due to the inaccurate position of the http://wiki.answers.com/Q/When_two_plane_mirror_ar...
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Lateral inversion is simple to explain. It happens when a mirror produces virtual images. It is when right becomes left and appears right. If you write something ...
Lateral inversion in mirrors is the reversal of a picture or image when placed against a flat mirror. It may also be referred to as 'mirror image'. This means ...
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Pull up a chair!
Discussions is your place to get to know your peers, tackle the bigger challenges together, and have fun along the way.
• Looking for techniques improve your MATLAB or Simulink skills? Tips & Tricks has you covered!
• Sharing the perfect math joke, pun, or meme? Look no further than Fun!
• Think there's a channel we need? Tell us more in Ideas
Victoria
### How to Draw a Happy Sheep in MATLAB!
Let us consider how to draw a Happy Sheep. A Happy Sheep was introduced in the MATLAB Mini Hack contest: Happy Sheep!
In this contest there was the strict limitation on the code length. So the code of the Happy Sheep is very compact and is only 280 characters long. We will analyze the process of drawing the Happy Sheep in MATLAB step by step. The explanations of the even more compact version of the code of the same sheep are given below.
So, how to draw a sheep? It is very easy. We could notice that usually a sheep is covered by crimped wool. Therefore, a sheep could be painted using several geometrical curves of similar types. Of course, then it will be an abstract model of the real sheep. Let us select two mathematical curves, which are the most appropriate for our goal. They are an ellipse for smooth parts of the sheep and an ellipse combined with a rose for woolen parts of the sheep.
Let us recall the mathematical formulas of these curves. A parametric representation of the standard ellipse is the following:
Also we will use the following parametric representation of the rose (rhodonea) curve:
This curve was named by the mathematician Guido Grandi.
Let us combine them in one curve and add possible shifts:
Now if we would like to create an ellipse, we will set and . If we would like to create a rose, we will set and . If we would like to shift our curve, we will set and to the required values. Of course, we could set all non-zero parameters to combine both chosen curves and use the shifts.
Let us describe how to create these curves using the MATLAB code. To make the code more compact, it is possible to program both formulas for the combined curve in one line using the anonymous function. We could make the code more compact using the function handles for sine and cosine functions. Then the MATLAB code for an example of the ellipse curve will be the following.
% Handles
s=@sin;
c=@cos;
% Ellipse + Polar Rose
F=@(t,a,f) a(1)*f(t)+s(a(2)*t).*f(t)+a(3);
% Angles
t=0:.1:7;
% Parameters
E = [5 7;0 0;0 0];
% Painting
figure;
plot(F(t,E(:,1),c),F(t,E(:,2),s),'LineWidth',10);
axis equal
The parameter t varies from 0 to 7, which is the nearest integer greater than , with the step 0.1. The result of this code is the following ellipse curve with and .
This ellipse is described by the following parametric equations:
The MATLAB code for an example of the rose curve will be the following.
% Handles
s=@sin;
c=@cos;
% Ellipse + Polar Rose
F=@(t,a,f) a(1)*f(t)+s(a(2)*t).*f(t)+a(3);
% Angles
t=0:.1:7;
% Parameters
R = [0 0;4 4;0 0];
% Painting
figure;
plot(F(t,R(:,1),c),F(t,R(:,2),s),'LineWidth',10);
axis equal
The result of this code is the following rose curve with and .
This rose is described by the following parametric equations:
Obviously, now we are ready to draw main parts of our sheep! As we reproduce an abstract model of the sheep, let us select the following main parts for the representation: head, eyes, hoofs, body, crown, and tail. We will use ellipses for the first three parts in this list and ellipses combined with roses for the last three ones.
First let us describe drawing of each part independently.
The following MATLAB code will be used to do this.
% Handles
s=@sin;
c=@cos;
% Ellipse + Polar Rose
F=@(t,a,f) a(1)*f(t)+s(a(2)*t).*f(t)+a(3);
% Angles
t=0:.1:7;
% Parameters
Eyes = 2:3;
Hoofs = 4:7;
Body = 8;
Crown = 9;
Tail = 10;
G=-13;
P=[5 7 repmat([.1 .5],1,6) 6 4 14 9 3 3;zeros(1,14) 8 8 12 12 4 4;...
-15 2 G 3 -17 3 -3 G 0 G 9 G 12 G -15 12 4 3 20 7];
% Painting
figure;
hold;
plot(F(t,P(:,2*i-1),c),F(t,P(:,2*i),s),'LineWidth',10);
end
axis([-25 25 -15 20]);
figure;
hold;
for i=Eyes
plot(F(t,P(:,2*i-1),c),F(t,P(:,2*i),s),'LineWidth',10);
end
axis([-25 25 -15 20]);
figure;
hold;
for i=Hoofs
plot(F(t,P(:,2*i-1),c),F(t,P(:,2*i),s),'LineWidth',10);
end
axis([-25 25 -15 20]);
figure;
hold;
for i=Body
plot(F(t,P(:,2*i-1),c),F(t,P(:,2*i),s),'LineWidth',10);
end
axis([-25 25 -15 20]);
figure;
hold;
for i=Crown
plot(F(t,P(:,2*i-1),c),F(t,P(:,2*i),s),'LineWidth',10);
end
axis([-25 25 -15 20]);
figure;
hold;
for i=Tail
plot(F(t,P(:,2*i-1),c),F(t,P(:,2*i),s),'LineWidth',10);
end
axis([-25 25 -15 20]);
The parameters , , , , , and are written in the different submatrices of the matrix P. The code generates the following curves to illustrate the different parts of our sheep.
The following ellipse describes the head of the sheep.
The following submatrix of the matrix P represents its parameters.
The parametric equations of the head are the following:
The following ellipses describe the eyes of the sheep.
The following submatrices of the matrix P represent their parameters.
The parametric equations of the left and right eyes correspondingly are the following:
The following ellipses describe the hoofs of the sheep.
The following submatrices of the matrix P represent their parameters.
The parametric equations of the right front, left front, right hind, and left hind hoofs correspondingly are the following:
The following ellipse combined with the rose describes the crown of the sheep.
The following submatrix of the matrix P represents its parameters.
The parametric equations of the crown are the following:
The following ellipse combined with the rose describes the body of the sheep.
The following submatrix of the matrix P represents its parameters.
The parametric equations of the body are the following:
The following ellipse combined with the rose describes the tail of the sheep.
The following submatrix of the matrix P represents its parameters.
The parametric equations of the tail are the following:
Now all the parts of our sheep should be put together! It is very easy because all the parts are described by the same equations with different parameters.
The following code helps us to accomplish this goal and ultimately draw a Happy Sheep in MATLAB!
% Happy Sheep!
% By Victoria A. Sablina
% Handles
s=@sin;
c=@cos;
% Ellipse + Rose
F=@(t,a,f) a(1)*f(t)+s(a(2)*t).*f(t)+a(3);
% Angles
t=0:.1:7;
% Parameters
% Eyes (3:6)
% Hoofs (7:14)
% Crown (15:16)
% Body (17:18)
% Tail (19:20)
G=-13;
P=[5 7 repmat([.1 .5],1,6) 6 4 14 9 3 3;zeros(1,14) 8 8 12 12 4 4;...
-15 2 G 3 -17 3 -3 G 0 G 9 G 12 G -15 12 4 3 20 7];
% Painting
hold;
for i=1:10
plot(F(t,P(:,2*i-1),c),F(t,P(:,2*i),s),'LineWidth',10);
end
This code is even more compact than the original code from the contest. It is only 253 instead of 280 characters long and generates the same Happy Sheep!
Our sheep is happy, because of becoming famous in the MATLAB community, a star!
Congratulations! Now you know how to draw a Happy Sheep in MATLAB!
Chen Lin
### Enhancing GitHub and File Exchange connection: MATLAB and Simulink Integration for GitHub Unveiled
The File Exchange team is thrilled to introduce a more streamlined approach to working with GitHub and File Exchange - the MATLAB and Simulink Integration for GitHub!
Key Enhancements:
- Improves the existing connection between File Exchange and GitHub, ensuring quicker reflection of changes made in GitHub within File Exchange.
- Aligns with GitHub's standard and supported approach to building integrations.
Action Required for File Exchange Contributors!
If you are a File Exchange contributor and have linked any submissions to GitHub, it is essential to install the App.
Starting April 16, 2024, your File Exchange submissions will no longer update automatically unless you take the following steps:
1. Visit your My File Exchange page.
2. Follow the prompts on the page to install MATLAB and Simulink Integration for GitHub.
3. Complete the necessary steps in GitHub.
4. Return to the My File Exchange page and verify the installation.
A detailed description of the process is available here.
If you prefer your File Exchange submission not to update automatically from GitHub, no action is required. Users will still be able to find and download your submissions. However, to release a new version of your code, you must either install the GitHub App or disconnect from GitHub and manually upload new versions of your code.
Should you have any questions or encounter issues with the App, please feel free to comment on this post!
Emma Farnan
Posted on 28 Feb 2024 at 20:25
### Set colormap limits when creating m colors
Several of the colormaps are great for a 256 color surface plot, but aren't well optimized for extracting m colors for plotting several independent lines. The issue is that many colormaps have start/end colors that are too similar or are suboptimal colors for lines. There are certainly many workarounds for this, but it would be a great quality of life to adjust that directly when calling this.
Example:
x = linspace(0,2*pi,101)';
y = [1:6].*cos(x);
figure; plot(x,y,'LineWidth',2); grid on; axis tight;
And now if I wanted to color these lines, I could use something like turbo(6) or gray(6) and then apply it using colororder.
colororder(turbo(6))
But my issue is that the ends of the colormap are too similar. For other colormaps, you may get lines that are too light to be visible against the white background. There are plenty of workarounds, with my preference being to create extra colors and truncate that before using colororder.
cmap = turbo(8); cmap = cmap(2:end-1,:); % Truncate the end colors
figure; plot(x,y,'LineWidth',2); grid on; axis tight;
colororder(cmap)
I think it would be really awesome to add some name-argument input pair to these colormaps that can specify the range you want so this could even be done inside the colororder calling if desired. An example of my proposed solution would look something like this:
cmap = turbo(6,'Range',[0.1 0.8]); % Proposed idea to add functionality
Where in this scenario, the resulting colormap would be 6 equally spaced colors that range from 10% to 80% of the total color range. This would be especially nice because you could more quickly modify the range of colors, or you could set the limits regardless of whether you need to plot 3, 6, or 20 lines.
Nikolaos
Posted on 27 Feb 2024 at 14:55
### How to access English-speaking forum?
I asked my question in the general forum and a few minutes later it was deleted. Perhaps this is a better place?
Rather than using my German regional forum (as I do not speak German), I want to ask questions in an international English-speaking forum. Presumably there should be an international English forum for everyone around the world, as English is the first or second language of everyone who has gone to school. Where is it?
Chen Lin
Posted on 26 Feb 2024 at 21:58
### Celebrating @VBBV: Our Newest Member of the Editors' Club
Big congratulations to @VBBV for achieving the remarkable milestone of 3,000 reputation points, earning the prestigious title of Editor within our community.
This achievement is a testament to @VBBV's exceptional contributions and steadfast commitment to the community. These efforts have also been endorsed by fellow top contributors, underscoring the value and impact of @VBBV's expertise.
Welcome to the Editors' Club, @VBBV – we are excited to witness and support your continued journey and influence within our community!
goc3
Posted on 26 Feb 2024 at 14:19
### Which of the following will not produce a 3x3 array of zeros in MATLAB?
lazymatlab
Posted on 23 Feb 2024 at 13:24
### MATLAB O/X QUIZ
MATLAB O/X Quiz
1. An infinite loop can be made using "for".
2. "A == A" is always true.
3. "round(2.5)" is 3.
4. "round(-0.5)" is 0.
Ben Pasquariello
Posted on 22 Feb 2024 at 13:47
### New Quantum Computing Cheat Sheet!
MATLAB Support Package for Quantum Computing lets you build, simulate, and run quantum algorithms.
Check out the Cheat Sheet here!
lazymatlab
Posted on 17 Feb 2024 at 8:38
### 행렬식은 면적이다? Determinant is area?
2 x 2 행렬의 행렬식은
• 행렬의 두 row 벡터로 정의되는 평행사변형의 면적입니다.
• 물론 두 column 벡터로 정의되는 평행사변형의 면적이기도 합니다.
• 좀 더 정확히는 signed area입니다. 면적이 음수가 될 수도 있다는 뜻이죠.
• 행렬의 두 행(또는 두 열)을 맞바꾸면 행렬식의 부호도 바뀌고 면적의 부호도 바뀌어야합니다.
일반적으로 n x n 행렬의 행렬식은
• 각 row 벡터(또는 각 column 벡터)로 정의되는 N차원 공간의 평행면체(?)의 signed area입니다.
• 제대로 이해하려면 대수학의 개념을 많이 가지고 와야 하는데 자세한 설명은 생략합니다.(=저도 모른다는 뜻)
• 더 자세히 알고 싶으시면 수학하는 만화의 '넓이 이야기' 편을 추천합니다.
• 수학적인 정의를 알고 싶으시면 위키피디아를 보시면 됩니다.
• 이렇게 생겼습니다. 좀 무섭습니다.
아래 코드는...
• 2 x 2 행렬에 대해서 이것을 수식 없이 그림만으로 증명하는 과정입니다.
• gif 생성에는 ScreenToGif를 사용했습니다. (gif 만들기엔 이게 킹왕짱인듯)
Determinant of 2 x 2 matrix is...
• An area of a parallelogram defined by two row vectors.
• Of course, same one defined by two column vectors.
• Precisely, a signed area, which means area can be negative.
• If two rows (or columns) are swapped, both the sign of determinant and area change.
More generally, determinant of n x n matrix is...
• Signed area of parallelepiped defined by rows (or columns) of the matrix in n-dim space.
• For a full understanding, a lot of concepts from abstract algebra should be brought, which I will not write here. (Cuz I don't know them.)
• For a mathematical definition of determinant, visit wikipedia.
• A little scary, isn't it?
The code below is...
• A process to prove the equality of the determinant of 2 x 2 matrix and the area of parallelogram.
• ScreenToGif is used to generate gif animation (which is, to me, the easiest way to make gif).
% 두 점 (a, b), (c, d)의 좌표
a = 4;
b = 1;
c = 1;
d = 3;
% patch 색 pre-define
lightgreen = [144, 238, 144]/255;
lightblue = [169, 190, 228]/255;
lightorange = [247, 195, 160]/255;
% animation params.
anim_Nsteps = 30;
% create window
figure('WindowStyle','docked')
ax = axes;
ax.XAxisLocation = 'origin';
ax.YAxisLocation = 'origin';
ax.XTick = [];
ax.YTick = [];
hold on
ax.XLim = [-.4, a+c+1];
ax.YLim = [-.4, b+d+1];
area = patch([0, a, a+c, c], [0, b, b+d, d], lightgreen);
p_ab = plot(a, b, 'ko', 'MarkerFaceColor', 'k');
p_cd = plot(c, d, 'ko', 'MarkerFaceColor', 'k');
p_ab.UserData = text(a+0.1, b, '(a, b)', 'FontSize',16);
p_cd.UserData = text(c+0.1, d-0.2, '(c, d)', 'FontSize',16);
area.UserData = text((a+c)/2-0.5, (b+d)/2, 'ad-bc', 'FontSize', 18);
pause
pause
area.UserData.Visible = 'off';
pause
rect_ad = patch([0, a, a, 0], [0, 0, d, d], lightblue, 'EdgeAlpha', 0, 'FaceAlpha', 0);
pause
rect_bc = patch([0, c, c, 0], [0, 0, b, b], lightorange, 'EdgeAlpha', 0, 'FaceAlpha', 0);
draw_gridline(rect_bc, ["23", "34"])
rect_bc.UserData = text(b/2, c/2, 'bc', 'FontSize', 20, 'HorizontalAlignment', 'center');
pause
[0, 0, 0, 0], [0, b, b, 0], t_pause=0.004)
pause
[0, 0, d/(d/c-b/a), d/(d/c-b/a)],...
[0, 0, b/a*d/(d/c-b/a), b/a*d/(d/c-b/a)], t_pause=0.004)
pause
%% slide bc block
uistack(p_cd, 'top')
patch_slide(rect_bc, ...
[0, 0, 0, 0], [d, d, d, d], t_pause=0.004)
draw_gridline(rect_bc, "34")
pause
%% slide bc block
patch_slide(rect_bc, ...
[0, 0, a, a], [0, 0, 0, 0], t_pause=0.004)
draw_gridline(rect_bc, "23")
pause
%% slide bc block
patch_slide(rect_bc, ...
[d/(d/c-b/a), 0, 0, d/(d/c-b/a)], ...
[b/a*d/(d/c-b/a), 0, 0, b/a*d/(d/c-b/a)], t_pause=0.004)
pause
rect_bc.UserData.Visible = 'off';
area.UserData.Visible = 'on';
%% functions
arguments
objs
options.anim_Nsteps = 30
options.t_pause = 0.003
end
for alpha = linspace(0, 1, options.anim_Nsteps)
for i = 1:length(objs)
switch objs(i).Type
case 'patch'
objs(i).FaceAlpha = (inout(i)==1)*alpha + (inout(i)==0)*(1-alpha);
objs(i).EdgeAlpha = (inout(i)==1)*alpha + (inout(i)==0)*(1-alpha);
case 'constantline'
objs(i).Alpha = (inout(i)==1)*alpha + (inout(i)==0)*(1-alpha);
end
pause(options.t_pause)
end
end
end
function patch_slide(obj, x_dist, y_dist, options)
arguments
obj
x_dist
y_dist
options.anim_Nsteps = 30
options.t_pause = 0.003
end
dx = x_dist/options.anim_Nsteps;
dy = y_dist/options.anim_Nsteps;
for i=1:options.anim_Nsteps
obj.XData = obj.XData + dx(:);
obj.YData = obj.YData + dy(:);
obj.UserData.Position(1) = mean(obj.XData);
obj.UserData.Position(2) = mean(obj.YData);
pause(options.t_pause)
end
end
function draw_gridline(patch, where)
ax = patch.Parent;
for i=1:length(where)
v1 = str2double(where{i}(1));
v2 = str2double(where{i}(2));
x1 = patch.XData(v1);
x2 = patch.XData(v2);
y1 = patch.YData(v1);
y2 = patch.YData(v2);
if x1==x2
xline(x1, 'k--')
else
fplot(@(x) (y2-y1)/(x2-x1)*(x-x1)+y1, [ax.XLim(1), ax.XLim(2)], 'k--')
end
end
end
Wolfgang Schwanghart
Posted on 16 Feb 2024 at 12:47
### Render landscapes
I think that MATLAB's Flipbook Mini Hack had quite some inspiring entries. My work largely deals with digital elevation models (DEMs). Hence I really liked the random renderings of landscapes, in particular this one written by Tim which inspired me to adopt the code and apply to the example data that comes with my software TopoToolbox. The results and code are shown here.
Mike Croucher
Posted on 16 Feb 2024 at 12:07
### Producing animated gifs from MATLAB Flipbook Mini Hack entries
On Valentine's day, the MathWorks linkedIn channel posted this animated gif
and immeditaely everyone wanted the code! It turns out that this is the result of my remix of @Zhaoxu Liu / slandarer's entry on the MATLAB Flipbook Mini Hack.
I pointed people to the Flipbook entry but, of course, that just gave the code to render a single frame and people wanted the full code to render the animated gif. That way, they could make personalised versions
I just published a blog post that gives the code used by the team behind the Mini Hack to produce the animated .gifs https://blogs.mathworks.com/matlab/2024/02/16/producing-animated-gifs-from-matlab-flipbook-mini-hack-entries/
Thanks again to @Zhaoxu Liu / slandarer for a great entry that seems like it will live for a long time :)
Tim
Posted on 15 Feb 2024 at 23:19
### Tricks for Sphere Texturing
If you've dabbled in "procedural generation," (algorithmically generating natural features), you may have come across the problem of sphere texturing. How to seamlessly texture a sphere is not immediately obvious. Watch what happens, for example, if you try adding power law noise to an evenly sampled grid of spherical angle coordinates (i.e. a "UV sphere" in Blender-speak):
% Example: how [not] to texture a sphere:
rng(2, 'twister'); % Make what I have here repeatable for you
% Make our radial noise, mapped onto an equal spaced longitude and latitude
% grid.
N = 51;
b = linspace(-1, 1, N).^2;
r = abs(ifft2(exp(6i*rand(N))./(b'+b+1e-5))); % Power law noise
r = rescale(r, 0, 1) + 5;
[lon, lat] = meshgrid(linspace(0, 2*pi, N), linspace(-pi/2, pi/2, N));
[x2, y2, z2] = sph2cart(lon, lat, r);
r2d = @(x)x*180/pi;
subplot(1, 3, 1);
imagesc(r, 'Xdata', r2d(lon(1,:)), 'Ydata', r2d(lat(:, 1)));
xlabel('Longitude (Deg)');
ylabel('Latitude (Deg)');
% View from z axis
subplot(1, 3, 2);
surf(x2, y2, z2, r);
axis equal
view([0, 90]);
title('Top view');
% Side view
subplot(1, 3, 3);
surf(x2, y2, z2, r);
axis equal
view([-90, 0]);
title('Side view');
The created surface shows "pinching" at the poles due to different radial values mapping to the same location. Furthermore, the noise statistics change based on the density of the sampling on the surface.
How can this be avoided? One standard method is to create a textured volume and sample the volume at points on a sphere. Code for doing this is quite simple:
rng default % Make our noise realization repeatable
% Create our 3D power-law noise
N = 201;
b = linspace(-1, 1, N);
[x3, y3, z3] = meshgrid(b, b, b);
b3 = x3.^2 + y3.^2 + z3.^2;
r = abs(ifftn(ifftshift(exp(6i*randn(size(b3)))./(b3.^1.2 + 1e-6))));
% Modify it - make it more interesting
r = rescale(r);
r = r./(abs(r - 0.5) + .1);
% Sample on a sphere
[x, y, z] = sphere(500);
% Plot
ir = interp3(x3, y3, z3, r, x, y, z, 'linear', 0);
surf(x, y, z, ir);
axis equal off
set(gcf, 'color', 'k');
colormap(gray);
The result of evaluating this code is a seamless, textured sphere with no discontinuities at the poles or variation in the spatial statistics of the noise texture:
But what if you want to smooth it or perform some other local texture modification? Smoothing the volume and resampling is not equivalent to smoothing the surficial features shown on the map above.
A more flexible alternative is to treat the samples on the sphere surface as a set of interconnected nodes that are influenced by adjacent values. Using this approach we can start by defining the set of nodes on a sphere surface. These can be sampled almost arbitrarily, though the noise statistics will vary depending on the sampling strategy.
One noise realisation I find attractive can be had by randomly sampling a sphere. Normalizing a point in N-dimensional space by its 2-norm projects it to the surface of an N-dimensional unit sphere, so randomly sampling a sphere can be done very easily using randn() and vecnorm():
N = 5e3; % Number of nodes on our sphere
g=randn(3,N); % Random 3D points around origin
p=g./vecnorm(g); % Projected to unit sphere
The next step is to find each point's "neighbors." The first step is to find the convex hull. Since each point is on the sphere, the convex hull will include each point as a vertex in the triangulation:
k=convhull(p');
In the above, k is an N x 3 set of indices where each row represents a unique triangle formed by a triplicate of points on the sphere surface. The vertices of the full set of triangles containing a point describe the list of neighbors to that point.
What we want now is a large, sparse symmetric matrix where the indices of the columns & rows represent the indices of the points on the sphere and the nth row (and/or column) contains non-zero entries at the indices corresponding to the neighbors of the nth point.
How to do this? You could set up a tiresome nested for-loop searching for all rows (triangles) in k that contain some index n, or you could directly index via:
c=@(x)sparse(k(:,x)*[1,1,1],k,1,N,N);
t=c(1)|c(2)|c(3);
The result is the desired sparse connectivity matrix: a matrix with non-zero entries defining neighboring points.
So how do we create a textured sphere with this connectivity matrix? We will use it to form a set of equations that, when combined with the concept of "regularization," will allow us to determine the properties of the randomness on the surface. Our regularizer will penalize the difference of the radial distance of a point and the average of its neighbors. To do this we replace the main diagonal with the negative of the sum of the off-diagonal components so that the rows and columns are zero-mean. This can be done via:
w=spdiags(-sum(t,2)+1,0,double(t));
Now we invoke a bit of linear algebra. Pretend x is an N-length vector representing the radial distance of each point on our sphere with the noise realization we desire. Y will be an N-length vector of "observations" we are going to generate randomly, in this case using a uniform distribution (because it has a bias and we want a non-zero average radius, but you can play around with different distributions than uniform to get different effects):
Y=rand(N,1);
and A is going to be our "transformation" matrix mapping x to our noisy observations:
Ax = Y
In this case both x and Y are N length vectors and A is just the identity matrix:
A = speye(N);
Y, however, doesn't create the noise realization we want. So in the equation above, when solving for x we are going to introduce a regularizer which is going to penalize unwanted behavior of x by some amount. That behavior is defined by the point-neighbor radial differences represented in matrix w. Our estimate of x can then be found using one of my favorite Matlab assets, the "\" operator:
smoothness = 10; % Smoothness penalty: higher is smoother
x = (A+smoothness*w'*w)\Y; % Solving for radii
The vector x now contains the radii with the specified noise realization for the sphere which can be created simply by multiplying x by p and plotting using trisurf:
p2 = p.*x';
trisurf(k,p2(1,:),p2(2,:),p2(3,:),'FaceC', 'w', 'EdgeC', 'none','AmbientS',0,'DiffuseS',0.6,'SpecularS',1);
light;
set(gca, 'color', 'k');
axis equal
The following images show what happens as you change the smoothness parameter using values [.1, 1, 10, 100] (left to right):
Now you know a couple ways to make a textured sphere: that's the starting point for having a lot of fun with basic procedural planet, moon, or astroid generation! Here's some examples of things you can create based on these general ideas:
Posted on 15 Feb 2024 at 22:46
### Create hyperlinks in the command window
The MATLAB command window isn't just for commands and outputs—it can also host interactive hyperlinks. These can serve as powerful shortcuts, enhancing the feedback you provide during code execution. Here are some hyperlinks I frequently use in fprintf statements, warnings, or error messages.
1. Open a website.
url = "https://blogs.mathworks.com/graphics-and-apps/";
hypertext = "Go to website"
fprintf(1,'%s <a href="matlab: web(''%s'') ">%s</a>\n',msg,url,hypertext);
2. Open a folder in file explorer (Windows)
msg = "File saved to current directory.";
directory = cd();
hypertext = "[Open directory]";
fprintf(1,'%s <a href="matlab: winopen(''%s'') ">%s</a>\n',msg,directory,hypertext)
File saved to current directory. [Open directory]
3. Open a document (Windows)
msg = "Created database.csv.";
filepath = fullfile(cd,'database.csv');
hypertext = "[Open file]";
fprintf(1,'%s <a href="matlab: winopen(''%s'') ">%s</a>\n',msg,filepath,hypertext)
Created database.csv. [Open file]
4. Open an m-file and go to a specific line
msg = 'Go to';
file = 'streamline.m';
line = 51;
fprintf(1,'%s <a href="matlab: matlab.desktop.editor.openAndGoToLine(which(''%s''), %d); ">%s line %d</a>', msg, file, line, file, line);
5. Display more text
msg = 'Incomplete data detected.';
extendedInfo = '\tFilename: m32c4r28\n\tDate: 12/20/2014\n\tElectrode: (3,7)\n\tDepth: ???\n';
warning('%s <a href="matlab: fprintf(''%s'') ">%s</a>', msg,extendedInfo,hypertext);
<click>
• Filename: m32c4r28
• Date: 12/20/2014
• Electrode: (3,7)
• Depth: ???
6. Run a function
Chen Lin
Posted on 14 Feb 2024 at 15:39
### Happy Valentine's Day
See code here in our community contest area. (author: @Zhaoxu Liu / slandarer)
And what do you do for Valentine's Day?
K
Posted on 10 Feb 2024 at 13:59
### contact technical support
which technical support should I contact/ask for the published Simscape example?
Chen Lin
Posted on 10 Feb 2024 at 3:46
### Fun Spring Festival Couplets
Happy year of the dragon.
Image Analyst
Posted on 10 Feb 2024 at 2:15
### Trick to enlarge a matrix by padding it with zeros.
To enlarge an array with more rows and/or columns, you can set the lower right index to zero. This will pad the matrix with zeros.
m = rand(2, 3) % Initial matrix is 2 rows by 3 columns
mCopy = m;
% Now make it 2 rows by 5 columns
m(2, 5) = 0
m = mCopy; % Go back to original matrix.
% Now make it 3 rows by 3 columns
m(3, 3) = 0
m = mCopy; % Go back to original matrix.
% Now make it 3 rows by 7 columns
m(3, 7) = 0
Hannah
Posted on 8 Feb 2024 at 18:24
### Prize Wheel?
I was looking into the possibility of making a spin-to-win prize wheel in MATLAB. I was looking around, and if someone has made one before they haven't shared. A labeled colored spinning wheel, that would slow down and stop (or I would take just stopping) at a random spot each time. I would love any tips or links to helpful resources!
Mike Croucher
Posted on 7 Feb 2024 at 22:13
### [MATLAB Example] Increase Image Resolution using Deep Learning
Many of the examples in the MATLAB documentation are extremely high quality articles, often worthy of attention in their own right. Time to start celebrating them! Today's is how to increase Image Resolution using deep learning
Chen Lin
Posted on 7 Feb 2024 at 19:14
### How many circles do you find in this picture?
Can you see them?
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# npm
## @stdlib/stats-base-dists-discrete-uniform-mean
0.0.8 • Public • Published
# Mean
Discrete uniform distribution expected value.
The expected value for a discrete uniform random variable is
where `a` is the minimum support and `b` the maximum support of the distribution.
## Installation
`npm install @stdlib/stats-base-dists-discrete-uniform-mean`
## Usage
`var mean = require( '@stdlib/stats-base-dists-discrete-uniform-mean' );`
#### mean( a, b )
Returns the expected value of a discrete uniform distribution with parameters `a` (minimum support) and `b` (maximum support).
```var v = mean( 0, 1 );
// returns 0.5
v = mean( 4, 12 );
// returns 8.0
v = mean( 2, 8 );
// returns 5.0```
If `a` or `b` is not an integer value, the function returns `NaN`.
```var v = mean( 0.1, 2 );
// returns NaN
v = mean( 0, 2.2 );
// returns NaN
v = mean( NaN, 2 );
// returns NaN
v = mean( 2, NaN );
// returns NaN```
If provided `a > b`, the function returns `NaN`.
```var y = mean( 3, 2 );
// returns NaN
y = mean( -1, -2 );
// returns NaN```
## Examples
```var randu = require( '@stdlib/random-base-randu' );
var round = require( '@stdlib/math-base-special-round' );
var mean = require( '@stdlib/stats-base-dists-discrete-uniform-mean' );
var a;
var b;
var v;
var i;
for ( i = 0; i < 10; i++ ) {
a = round( randu()*10.0 );
b = round( ( randu()*10.0 ) + a );
v = mean( a, b );
console.log( 'a: %d, b: %d, E(X;a,b): %d', a.toFixed( 4 ), b.toFixed( 4 ), v.toFixed( 4 ) );
}```
## Notice
This package is part of stdlib, a standard library for JavaScript and Node.js, with an emphasis on numerical and scientific computing. The library provides a collection of robust, high performance libraries for mathematics, statistics, streams, utilities, and more.
For more information on the project, filing bug reports and feature requests, and guidance on how to develop stdlib, see the main project repository.
## Keywords
### Install
`npm i @stdlib/stats-base-dists-discrete-uniform-mean`
stdlib.io
69
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https://cs.stackexchange.com/questions/75648/all-nearest-neighbor-in-a-changing-2d-euclidean-space
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# All nearest neighbor in a changing 2d euclidean space
I am in need of an algorithm for a part of a game (a mod) I am making. I have abstracted the problem:
Given a 2D space with $N$ random points $p_1...p_n$, calculate the nearest neighbor of each of the points, where the distance is at most $C$.
Note that our list of points is unsorted, and that each point is an actual object so I could give it any property I want.
Now this can easily be done in $O(n^2)$ time, it is also the easiest and not really an issue. However, the time it takes when a new point is added or a point is removed is also $O(n^2)$, and I would like to get that down, because this is done during gameplay (while the initial step is only done at start).
Does anyone know a possible improvement over the simple 'bruteforce'? I tried sorting, twice, and keeping it up to date (using a 4x linked list (up, down, left, right)) but that did not seem to work.
• Welcome to CS.SE! Can you clarify what "where the distance is at most C" means? Does that mean that you are only interested in the nearest neighbor to a point $p_i$ if the distance is $\le C$, otherwise you want it to output "no neighbor near enough"? Are you most interested in practical algorithms, or in something with provable worst-case running time?
– D.W.
Commented May 19, 2017 at 18:07
• Thanks! I am only interested if there is a neighbor within distance C of a point. I want it practical. Commented May 19, 2017 at 18:46
• OK, cool. Then my answer should take care of that.
– D.W.
Commented May 19, 2017 at 18:49
• With your brute-force approach, adding a point takes $O(n)$ time, not $O(n^2)$ time, as you only need to compare the new point to all pre-existing points (there's no need to recompute the distance between two pre-existing points). Deleting a point also takes only $O(n)$ time, not $O(n^2)$ time: there is no need to recompute all pairs of distances, to delte point $x$ you only need to look up the nearest neighbor $y$ and update $y$'s nearest neighbor (by comparing $y$ to all other points).
– D.W.
Commented May 19, 2017 at 19:04
• With the bruteforce solution i had: storing only the nearest + distance, for all points the new nearest had to be found as well. So I guess what I now found is just a better bruteforce. Commented May 19, 2017 at 19:08
You want a data structure that supports nearest neighbor search in 2D. There are many options, but a simple one that is widely used for practical situations is a quadtree data structure. It supports both efficient lookup and efficient insert operations.
• The mod is in LUA and I think quadtrees might make the implementation much more complicated than necessary, so for my case I think my answer might be best. For others, use quad trees. (not sure what to mark as the answer) Commented May 19, 2017 at 18:57
• @Rahkiin, I wouldn't expect the programming language to make a major difference here; it's just a constant factor. Sophisticated data structures are harder to implement than simpler data structures, but if you want an efficient solution that's often necessary. If you want to optimize for ease of implementation, use a linked list and pairwise comparisons and accept that it won't be as fast as other options. Your question made it sound like you wanted to speed up running time. If so, quadtrees (or similar data structures) should provide substantial speedups, especially when $n$ is large.
– D.W.
Commented May 19, 2017 at 19:06
After I linked this to a friend, he found this solution:
Store for every point all other points that are in distance $C$ in a hashmap with the key as the node. (This is $O(n^2)$ with amortized $O(1)$ for hash put) When adding a node $x$, search all nodes and add every node with distance less than $C$ to the list of $x$. Add $x$ to all nodes in the list of $x$ ($O(n + some)$) When removing a node $y$, remove $y$ from the lists of the nodes in the list of $y$. $O(some)$
I tried this quickly in Processing (java-ish) to see if it works, and it seems to have a bit slower initial step, but removing and adding a node is almost no-time.
• This looks identical to the brute-force solution described in your question: you do $O(n^2)$ work to construct the data structure, and $O(n)$ work per insertion or deletion, same as the approach in your question. I don't see how the hashtable helps or is relevant here. This will likely be considerably slower than a quadtree or other nearest neighbor data structure, when $n$ is large.
– D.W.
Commented May 19, 2017 at 19:02
• True, but $n$ is max 8000. Commented May 19, 2017 at 19:06
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Students appreciate clear and concise ISC Maths Class 12 Solutions Chapter 5 Continuity and Differentiability Ex 5.10 that guide them through exercises.
ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.10
Find $$\frac{d y}{d x}$$ in the following (1 to 6) questions when :
Question 1.
(i) x = at2, y = 2 at (NCERT)
(ii) x = 2at2, y = at4 (NCERT)
Solution:
(i) Given x =at2 …………(1)
and y = 2at
Differentiate (1) and (2) w.r.t. ‘t’, we have
$$\frac{d x}{d t}$$ = 2at
and $$\frac{d y}{d t}$$ = 2a
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
= $$\frac{2 a}{2 a t}=\frac{1}{t}$$, t ≠ 0
(ii) Given, x = 2at2 ………….(1)
and y = at4 ………….(2)
Differentiate (1) and (2) w.r.t. ‘t’, we have
$$\frac{d x}{d t}$$ = 4at
and $$\frac{d y}{d t}$$ = 4at3
∴ $$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}$$
= $$\frac{4 a t^3}{4 a t}$$
= t2
Question 2.
(i) x = a cos θ, y = b sin θ (NCERT)
(ii) x = a cos θ, y = a sin θ (NCERT)
Solution:
(i) Given, x = a cos θ ;
y = b sin θ
Differentiating both sides w.r.t. θ ; we get
∴ $$\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}$$
= $$\frac{b \cos \theta}{-a \sin \theta}$$
= – $$\frac{b}{a}$$ cot θ
(ii) Given, x = a cos θ ………..(1)
and y = a sin θ ………….(2)
Diff. (1) and (2) w.r.t. ‘θ’, we have
$$\frac{d x}{d \theta}$$ = – a sin θ
and $$\frac{d y}{d \theta}$$ = a cos θ
∴ $$\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}$$
= $$\frac{a \cos \theta}{-a \sin \theta}$$
= – cot θ.
Question 3.
(i) x = a sec θ, y = b tan θ (NCERT)
(ii) x = sin t, y = cos 2t (NCERT)
Solution:
(i) Given x = a sec θ
and y = b tan θ
Diff. both given eqn. w.r.t. θ, we have
∴ $$\frac{d x}{d \theta}$$ = a sec θ tan θ
and $$\frac{d y}{d \theta}$$ = b sec2 θ
Thus $$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$$
= $$\frac{b \sec ^2 \theta}{a \sec \theta \tan \theta}$$
= $$\frac{b}{a}$$ cosec θ
(ii) Given x = sin t
and y = cos 2t ………..(2)
Diff. eqn. (1) and eqn. (2) w.r.t. t, we have
$$\frac{d x}{d t}$$ = cos t
$$\frac{d y}{d t}$$ = – sin 2t . 2
Thus, $$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
= $$\frac{-2 \sin 2 t}{\cos t}$$
= $$\frac{-4 \sin t \cos t}{\cos t}$$
= – 4 sin t
Question 4.
(i) x = 4t, y = (NCERT)
(ii) x = t + $$\frac{1}{t}$$, y = t – $$\frac{1}{t}$$ (NCERT Exampler)
(iii) x = a (t – sin t), y = a (1 + cos t)
Solution:
(i) Given x = 4t ……….(1)
and y = 4/t ………..(2)
Diff. (1) and (2) w.r.t. ‘t’, we have
$$\frac{d x}{d t}$$ = 4
and $$\frac{d y}{d t}$$ = – 4/t2
∴ $$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}$$
= $$\frac{-\frac{4}{t^2}}{4}$$
= – $$\frac{1}{t^2}$$
(ii) Given x = (t + $$\frac{1}{t}$$)
and y = (t – $$\frac{1}{t}$$) ……………(1)
Differentiating both sides of eqn. (1) w.r.t. x ; we have
$$\frac{d x}{d t}=\left(1-\frac{1}{t^2}\right)$$ ;
$$\frac{d y}{d t}=\left(1+\frac{1}{t^2}\right)$$
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
= $$\frac{\left(t^2+1\right)}{\left(t^2-1\right)}$$
= $$\frac{\left(t+\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)}=\frac{x}{y}$$ [using eqn. (1)]
(iii) x = a (t – sin t)
and y = a (1 + cos t)
∴ $$\frac{d x}{d t}$$ = a (1 – cos t)
and $$\frac{d y}{d t}$$ = – a sin t
Thus $$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}$$
= $$\frac{-a \sin t}{a(1-\cos t)}$$
= $$\frac{-2 a \sin \frac{t}{2} \cos \frac{t}{2}}{2 a \sin ^2 \frac{t}{2}}$$
= – cot $$\frac{t}{2}$$
Question 5.
(i) x = a cos3 t, y = b sin t
(ii) x = a (1 – sin t), y = a (1 + cos t)
Solution:
Given x = a cos3 t ………..(1)
and y = b sin3 t ………..(2)
Diff. eqn. (1) and eqn. (2) w.r.t. t, we have
$$\frac{d x}{d t}$$ = 3a cos2 t (- sin t)
and $$\frac{d y}{d t}$$ = 3b sin2 t (cos t)
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
= $$\frac{3 b \sin ^2 t \cos t}{3 a \cos ^2 t \sin t}$$
= – $$\frac{b}{a}$$ tan t
(ii) Given x = a (1 – sin t) ……….(1)
and y = a (1 + cos t) ………..(2)
Diff. both eqns. w.r.t. t, we have
$$\frac{d x}{d t}$$ = – a cos t ;
$$\frac{d y}{d t}$$ = – a sin t
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
= $$\frac{-a \sin t}{-a \cos t}$$
= tan t.
Question 5 (old).
(i) x = a sin3 t, y = a cos3 t (ISC 2004)
Solution:
Given x = a sin3 t
and y = a cos3 t
Diff. eqn. (1) and eqn. (2) w.r.t. ‘t’, we have
$$\frac{d x}{d t}$$ = 3a sin2 t cos t ;
$$\frac{d y}{d t}$$ = 3a cos2 t (- sin t)
Thus, $$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
= $$\frac{-3 a \cos ^2 t \sin t}{3 a \sin ^2 t \cos t}$$
= – cos t
Question 6.
(i) x = a (θ + sin θ), y = a (1 – cos θ)
(ii) x = a (1 – cos θ), y = (θ + sin θ)
Solution:
(i) Given x = a (θ + sin θ) ………..(1)
and y = a (1 – cos θ) …………..(2)
Diff. eqn. (1) and (2) w.r.t. θ, we have
$$\frac{d x}{d \theta}$$ = a (1 + cos θ)
$$\frac{d x}{d \theta}$$ = a sin θ
Thus, $$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$$
= $$\frac{a \sin \theta}{a(1+\cos \theta)}$$
= $$\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}}$$
= tan $$\frac{\theta}{2}$$
(ii) Given x = a (1 – cos θ) ;
y = a (θ + sin θ)
Diff. both sides w.r.t. θ ; we get
$$\frac{d x}{d \theta}$$ = a sin θ;
$$\frac{d x}{d \theta}$$ = a (1 + cos θ)
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$$
= $$\frac{a(1+\cos \theta)}{a \sin \theta}$$
= $$\frac{1+\cos \theta}{sin \theta}$$
= $$\frac{2 \cos ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}$$
= 2 cot $$\frac{\theta}{2}$$
Question 7.
(i) x = et (sin t + cos t), y = et (sin t – cos t)
(ii) x = cos θ – cos 2θ, y = sin θ – sin 2θ (NCERT)
Solution:
(i) Given x = et (sin t + cos t)
and y = et (sin t – cos t)
Diff. both eqn.’s w.r.t. t, we have
$$\frac{d x}{d t}$$ = et (cos t – sin t) + (sin t + cos t) et
= et (2 cos t)
and $$\frac{d y}{d t}$$ = et (cos t + sin t) + (sin t – cos t) et
= et (2 sin t)
∴ $$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}$$
= $$\frac{2 e^t \sin t}{2 e^t \cos t}$$
= tan t
(ii) Given x = cos θ – cos 2θ ……….(1)
and y = sin θ – sin 2θ ………..(2)
Diff. eqn. (1) and eqn. (2) w.r.t. θ, we have
$$\frac{d x}{d \theta}$$ = – sin θ + 2 sin 2θ
and $$\frac{d y}{d \theta}$$ = cos θ – 2 cos 2θ
∴ $$\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}$$
= $$\frac{\cos \theta-2 \cos 2 \theta}{-2 \sin \theta+2 \sin 2 \theta}$$
Question 8.
(i) x = a (cos θ + cos 2θ), y = b (sin θ + sin 2θ)
(ii) x = $$\frac{1+\log t}{t^2}$$, y = $$\frac{3+ 2\log t}{t^2}$$ (NCERT Exampler)
Solution:
Given x = a (cos θ + cos 2θ) …………(1)
and y = b (sin θ + sin 2θ) ………..(2)
Diff. eqn. (1) and eqn. (2) w.r.t. θ, we have
$$\frac{d x}{d \theta}$$ = a (- sin θ – 2 sin 2θ) ……….(3)
and $$\frac{d y}{d \theta}$$ = b (cos θ + 2 cos 2θ) …………..(4)
∴ $$\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}$$
= $$\frac{b(\cos \theta+2 \cos 2 \theta)}{a(-\sin \theta-2 \sin 2 \theta)}$$
= $$-\frac{b(\cos \theta+2 \cos 2 \theta)}{a(\sin \theta+2 \sin 2 \theta)}$$
[using eqn. (3) and eqn. (4)]
(iv) Given x = $$\frac{1+\log t}{t^2}$$ ……….(1)
and y = $$\frac{3+ 2\log t}{t^2}$$ …………..(2)
Differentiate eqn. (1) and (2) both sides w.r.t. ‘t’ ; we have
Question 9.
x = 3 cos θ – 2 cos3 θ, y = 3 sin θ – 2 sin3 θ
Solution:
Given x = 3 cos θ – 2 cos3 θ
and y = 3 sin θ – 2 sin3 θ
Diff. both given eqn’s w.r.t. θ, we have
$$\frac{d x}{d \theta}$$ = – 3 sin θ – 6 cos2 θ (- sin θ)
= – 3 sin θ + 6 cos2 θ sin θ ………..(1)
and $$\frac{d y}{d \theta}$$ = 3 cos θ – 6 sin2 θ cos 2 …………….(2)
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$$
= $$\frac{3 \cos \theta-6 \sin ^2 \theta \cos \theta}{-3 \sin \theta+6 \cos ^2 \theta \sin \theta}$$
[using (1) and (2)]
= $$\frac{3 \cos \theta\left[1-2 \sin ^2 \theta\right]}{-3 \sin \theta\left[1-2 \cos ^2 \theta\right]}$$
= $$\frac{3 \cos \theta\left(1-2 \sin ^2 \theta\right)}{-3 \sin \theta\left[1-2\left(1-\sin ^2 \theta\right)\right]}$$
= $$\frac{3 \cos \theta\left(1-2 \sin ^2 \theta\right)}{3 \sin \theta\left(1-2 \sin ^2 \theta\right)}$$
= cos θ.
Question 10.
(i) If x = 2 cos t – cos 2t and y = 2 sin t – sin 2t, find $$\frac{d y}{d x}$$ at t = $$\frac{\pi}{4}$$.
(ii) If x = 3 sin t – sin 3t, y = 3 cos t – cos 3t, find $$\frac{d y}{d x}$$ at t = $$\frac{\pi}{3}$$. (NCERT Exampler)
Solution:
(i) Given x = 2 cos t – cos 2t
and y = 2 sin t – sin 2t
Diff. both sides w.r.t. ‘t’ ; we have
$$\frac{d x}{d t}$$ = – 2 sin t + 2 sin 2t ;
$$\frac{d y}{d t}$$ = 2 cos t – 2 cos 2t
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
= $$\frac{2(\cos t-\cos 2 t)}{2(\sin 2 t-\sin t)}$$
at t = $$\frac{\pi}{4}$$;
$$\frac{d y}{d x}=\frac{\cos \left(\frac{\pi}{4}\right)-\cos \frac{\pi}{2}}{\sin \frac{\pi}{2}-\sin \frac{\pi}{4}}$$
= $$\frac{\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}}$$
= $$\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}$$
= √2 + 1
(ii) Given x = 3 sin t – sin 2t
= 3 sin t – (3 sin t – 4 sin3 t)
⇒ x = 4 sin3 t …….(1)
and y = 3 cos t – cos 3t ………..(2)
Diff. eqn. (1) and (2) w.r.t. t ; we get
$$\frac{d x}{d t}$$ = 12 sin2 t cos t
$$\frac{d y}{d t}$$ = – 3 sin t + 3 sin 3t
∴ $$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}$$
= $$\frac{-\sin t+\sin 3 t}{4 \sin ^2 t \cos t}$$
at t = $$\frac{\pi}{3}$$ ;
$$\frac{d y}{d x}=\frac{-\frac{\sqrt{3}}{2}+0}{4 \times \frac{3}{4} \times \frac{1}{2}}$$
= $$\frac{-\sqrt{3} \times 2}{2 \times 3}$$
= $$-\frac{1}{\sqrt{3}}$$
Question 11.
(i) If x = $$\frac{2 b t}{1+t^2}$$ and y = $$\frac{a\left(1-t^2\right)}{1+t^2}$$, find $$\frac{d y}{d x}$$ at t = 2.
(ii) If x = aeθ (sin θ – cos θ) and y = aeθ (sin θ + cos θ), find $$\frac{d y}{d x}$$ at θ = $$\frac{\pi}{4}$$.
Solution:
Given x = $$\frac{2 b t}{1+t^2}$$
and y = $$\frac{a\left(1-t^2\right)}{1+t^2}$$
Diff. both equations w.r.t. t ; we have
(ii) Given x = a eθ (sin θ – cos θ) ……………..(1)
and y = a eθ (sin θ + cos θ)
diff. eqn. (1) w.r.t. θ ; we have
$$\frac{d x}{d \theta}$$ = a eθ (sin θ – cos θ) + a eθ (cos θ + sin θ)
= a eθ (sin θ – cos θ + cos θ + sin θ)
= 2aeθ sin θ
and y = aeθ (sin θ + cos θ) ………….(2)
diff. eqn. (2) w.r.t. θ ; we have
$$\frac{d y}{d \theta}$$ = aeθ (sin θ + cos θ) + aeθ (cos θ – sin θ)
= aeθ (sin θ + cos θ + cos θ – sin θ)
= 2aeθ cos θ
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$$
= $$\frac{2 a e^\theta \cos \theta}{2 a e^\theta \sin \theta}$$
= cot θ
at θ = $$\frac{\pi}{4}$$ ;
$$\frac{d y}{d x}$$ = cot $$\frac{\pi}{4}$$ = 1.
Question 12.
(i) $$\frac{x^2}{1-x^2}$$ w.r.t. x2
(ii) sin x2 w.r.t. x3 (ISC 2009)
(iii) cot3 (2x + 1) w.r.t. x2 + 1
(iv) sin2 x w.r.t. ecos x
Solution:
(i) Let y = $$\frac{x^2}{1-x^2}$$
and z = x2
So we want to differentiate y w.r.t. i.e. we want to find $$\frac{d y}{d z}$$
Diff. given eqns w.r.t. x, we have
$$\frac{d y}{d x}=\frac{\left(1-x^2\right) 2 x-x^2(-2 x)}{\left(1-x^2\right)^2}$$
= $$\frac{2 x}{\left(1-x^2\right)^2}$$
and $$\frac{d z}{d x}$$ = 2x
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{2 x}{\frac{\left(1-x^2\right)^2}{2 x}}$$
= $$\frac{1}{\left(1-x^2\right)^2}$$
(ii) Let y = sin x2
and z = x3
So we want to diff. y w.r.t. z i.e. to find $$\frac{d y}{d z}$$
Diff. both eqn’s w.r.t. x : we have
$$\frac{d y}{d x}$$ = cos x2 . 2x ;
$$\frac{d z}{d x}$$ = 3x2
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{2 x \cos x^2}{3 x^2}$$
= $$\frac{2}{3 x}$$ cos x2
(iii) Let y = cot3 (2x + 1)
and z = x2 + 1
So, we want to diff. y w.r.t. z i.e. to find $$\frac{d y}{d z}$$.
Diff. given eqn’s both sides w.r.t. x, we have
$$\frac{d y}{d x}$$ = 3 cot2 (2x + 1) {- cosec2 (2x + 1)} . 2
and $$\frac{d z}{d x}$$ = 2x
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{-6 \cot ^2(2 x+1) \ {cosec}^2(2 x+1)}{2 x}$$
= $$\frac{-3 \cot ^2(2 x+1) \ {cosec}^2(2 x+1)}{x}$$
(iv) Let y = sin2 x ………….(1)
and z = ecos x
We want to diff. y w.r.t. z i.e. to find $$\frac{d y}{d z}$$
diff. eqn. (1) and eqn. (2) w.r.t. x ;
$$\frac{d y}{d x}$$ = 2 sin x cos x ;
$$\frac{d z}{d x}$$ = ecos x (- sin x)
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{2 \sin x \cos x}{e^{\cos x}(-\sin x)}$$
= – $$\frac{2 \cos x}{2^{\cos x}}$$
Question 8 (old).
(iv) log (sin x) w.r.t. $$\sqrt{cos x}$$
Solution:
Given, y = log (sin x)
and z = $$\sqrt{cos x}$$
i.e. we want diff. y w.r.t. z i.e. to find $$\frac{d y}{d z}$$
Now, diff. both eqn’s w.r.t. x, we have
$$\frac{d y}{d x}$$ = $$\frac{1}{sin x}$$ cos x
= cot x
and $$\frac{d z}{d x}$$ = $$\frac{1}{2}$$ (cos x)$$-\frac{1}{2}$$ .(- sin x)
= – $$\frac{\sin x}{2 \sqrt{\cos x}}$$
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{2 \cot x}{-\sin x}$$ $$\sqrt{cos x}$$
= – 2 cot x cosec x $$\sqrt{cos x}$$
Question 13.
(i) Differentiate sin-1 $$\left(\frac{2 x}{1+x^2}\right)$$ w.r.t. tan-1 x.
(ii) Differentiate tan-1 $$\left(\frac{2 x}{1-x^2}\right)$$ w.r.t. tan-1 x.
Solution:
(i) Let y = sin-1 $$\left(\frac{2 x}{1+x^2}\right)$$ ……….(1)
and z = tan-1 x
i.e. we want to find $$\frac{d y}{d z}$$
put x = tan θ in eqn. (1) ; we have
y = sin-1 $$\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)$$
= sin-1 (sin 2θ)
= 2θ
= 2 tan-1 x
∴ $$\frac{d y}{d x}$$ = $$\frac{2}{1+x^2}$$
and $$\frac{d z}{d x}$$ = $$\frac{1}{1+x^2}$$
Thus, $$\frac{d y}{d z}=\frac{d y / d x}{d z / d x}$$
= $$\frac{2 / 1+x^2}{1 / 1+x^2}$$
= 2
(ii) Let y = tan-1 $$\left(\frac{2 x}{1-x^2}\right)$$ …………….(1)
and z = tan-1 x ………(2)
i.e. we want to find $$\frac{d y}{d z}$$
put x = tan θ
⇒ θ = tan-1 x in eqn. (1) ; we have
∴ y = tan-1 $$\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)$$
⇒ y = tan-1 (tan 2θ)
= 2θ
= 2 tan-1 x
Thus $$\frac{d y}{d x}$$ = $$\frac{2}{1+x^2}$$
and $$\frac{d z}{d x}$$ = $$\frac{1}{1+x^2}$$
∴ $$\frac{d y}{d x}$$ = 2.
Question 14.
Differentiate cos-1 $$\left(\frac{1}{\sqrt{1+t^2}}\right)$$ w.r.t. sin-1 $$\left(\frac{t}{\sqrt{1+t^2}}\right)$$.
Solution:
Let y = cos-1 $$\left(\frac{1}{\sqrt{1+t^2}}\right)$$ ………..(1)
and z = sin-1 $$\left(\frac{t}{\sqrt{1+t^2}}\right)$$
putting t = tan θ
i.e. θ = tan-1 t in eqn (1), we have
y = cos-1 $$\left(\frac{1}{\sqrt{1+\tan ^2 \theta}}\right)$$
= cos-1 (cos θ)
⇒ y = θ = tan-1 t
Diff. w.r.t. t ; we have
∴ $$\frac{d y}{d t}=\frac{1}{1+t^2}$$ ………..(3)
putting t = tan Φ i.e. Φ = tan-1 t, in eqn. (2) ; we have
z = sin-1 $$\left(\frac{\tan \phi}{\sec \phi}\right)$$
= sin-1 (sin Φ) = Φ
⇒ z = tan-1 t
Diff. both sides w.r.t. t, we get
∴ $$\frac{d z}{d t}=\frac{1}{1+t^2}$$ ……………..(4)
Now, we want to diff. y w.r.t. z to find $$\frac{d y}{d z}$$.
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d t}}{\frac{d z}{d t}}$$
= $$\frac{\frac{1}{1+t^2}}{\frac{1}{1+t^2}}$$
= 1 [using (3) and (4)]
Question 15.
(i) Differentiate sin-1 $$\left(\frac{2 x}{1+x^2}\right)$$ w.r.t. cos-1 $$\left(\frac{1-x^2}{1+x^2}\right)$$.
(ii) Differentiate tan-1 $$\left(\frac{3 x-x^3}{1-3 x^2}\right)$$ w.r.t. tan-1 $$\frac{2 x}{1-x^2}$$.
Solution:
(i) Let y = sin-1 $$\left(\frac{2 x}{1+x^2}\right)$$ …………(1)
and z = cos-1 $$\left(\frac{1-x^2}{1+x^2}\right)$$ …………(2)
i.e. we want to diff. y w.r.t. z to find $$\frac{d y}{d z}$$
To simplify eqn. (1) ;
we put x = tan θ
i.e. θ = tan-1 x
y = sin-1 $$1\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)$$
= sin-1 (sin 2θ)
⇒ y = 2θ
= 2 tan-1 x
Diff. both sides w.r.t. x, we have
$$\frac{d y}{d x}=\frac{2}{1+x^2}$$ …………….(3)
To simplify eqn. (2)
we put x = tan Φ
i.e. Φ = tan-1 x
∴ z = cos-1 $$\left(\frac{1-\tan ^2 \phi}{1+\tan ^2 \phi}\right)$$
= cos-1 (cos 2Φ)
⇒ z = 2Φ = 2 tan-1 x
Diff. both sides w.r.t. x, we have
$$\frac{d z}{d x}=\frac{2}{1+x^2}$$ ………………(4)
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{\frac{2}{1+x^2}}{\frac{2}{1+x^2}}$$
= 1 [using eqn. (3) and eqn. (4)]
(ii) Let y = tan-1 $$\left(\frac{3 x-x^3}{1-3 x^2}\right)$$ …………….(1)
and z = tan-1 $$\frac{2 x}{1-x^2}$$ ……………(2)
i.e. we want to diff. y w.r.t. z i.e. to find $$\frac{d y}{d z}$$.
Putting x = tan θ
⇒ θ = tan-1 x in eqn. (1)
and x = tan Φ
⇒ Φ = tan-1 x in eqn. (2) ; we have
∴ y = tan-1 $$\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)$$
On differentiating both eqn.’s w.r.t. x, we have
∴ $$\frac{d y}{d x}=\frac{3}{1+x^2}$$
and $$\frac{d z}{d x}=\frac{2}{1+x^2}$$
Thus, $$\frac{d y}{d z}=\frac{d y / d x}{d z / d x}$$
= $$\frac{3}{1+x^2} \times \frac{1+x^2}{2}$$
= $$\frac{3}{2}$$
Question 16.
(i) Differentiate tan-1 $$\left(\frac{\sqrt{1+x^2}-1}{x}\right)$$ w.r.t. tan-1 x.
(ii) Differentiate tan-1 $$\left(\frac{\sqrt{1-x^2}}{x}\right)$$ w.r.t. cos-1 (2x $$\sqrt{1-x^2}$$).
Solution:
(i) Let y = tan-1 $$\left(\frac{\sqrt{1+x^2}-1}{x}\right)$$ …………..(1)
and z = tan-1 x …………(2)
Now we want to find $$\frac{d y}{d z}$$
put x = tan θ in eqn. (1) ; we have
y = tan-1 $$\left(\frac{1-\cos \theta}{\sin \theta}\right)$$
= tan-1 $$\left(\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)$$
= tan-1 (tan $$\frac{\theta}{2}$$)
⇒ y = $$\frac{\theta}{2}$$
= $$\frac{1}{2}$$ tan-1 x
and z = tan-1 x
Diff. both given eqn’s w.r.t. x, we have
⇒ $$\frac{d y}{d x}=\frac{1}{2\left(1+x^2\right)}$$
and $$\frac{d z}{d x}=\frac{1}{1+x^2}$$
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{\frac{1}{2\left(1+x^2\right)}}{\frac{1}{1+x^2}}$$
= $$\frac{1}{2}$$
(ii) Let y = tan-1 $$\left(\frac{\sqrt{1-x^2}}{x}\right)$$ ………..(1)
and z = cos-1 (2x$$\sqrt{1-x^2}$$) ………………(2)
Put x =sin θ
i.e. θ = sin-1 x in eqn. (1) ; we have
y = tan-1 $$\left(\frac{\sqrt{1-\sin ^2 \theta}}{\sin \theta}\right)$$
= tan-1 (cot θ)
⇒ y = tan-1 (tan ($$\frac{\pi}{2}$$ – θ))
⇒ y = $$\frac{\pi}{2}$$ – θ
= $$\frac{\pi}{2}$$ – sin-1 x
Diff. both sides w.r.t. x, we have
$$\frac{d y}{d x}=-\frac{1}{\sqrt{1-x^2}}$$ ………..(3)
putting x = sin Φ i.e. Φ = sin-1 x in eqn. (2) ; we have
∴ z = cos-1 (2 sin Φ $$\sqrt{1-\sin ^2 \phi}$$)
= cos-1 (2 sin Φ cos Φ)
z = cos-1 (sin 2Φ)
= cos-1 (cos ($$\frac{\pi}{2}$$ – 2Φ))
z = $$\frac{\pi}{2}$$ – 2Φ
= $$\frac{\pi}{2}$$ – sin-1 x
Diff. both sides w.r.t. x ; we have
$$\frac{d z}{d x}=-\frac{2}{\sqrt{1-x^2}}$$ ……………..(4)
We want to diff. y w.r.t. z i.e. to find $$\frac{d y}{d z}$$
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{-\frac{1}{\sqrt{1-x^2}}}{\frac{-2}{\sqrt{1-x^2}}}$$
= $$\frac{1}{2}$$
Question 17.
Prove that the derivative of tan-1 $$\left(\frac{x}{1+\sqrt{1-x^2}}\right)$$ w.r.t. sin-1 x is independent of x.
Solution:
Let y = tan-1 $$\left(\frac{x}{1+\sqrt{1-x^2}}\right)$$ ………..(1)
and z = sin-1 x
Now we want to diff. y w.r.t. z
Put x = sin θ
∴ θ = sin-1 x in eqn. (1) ; we have
y = tan-1 $$\left(\frac{\sin \theta}{1+\cos \theta}\right)$$
= tan-1 $$\left(\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}}\right)$$
⇒ y = tan-1 (tan $$\frac{\theta}{2}$$)
= $$\frac{\theta}{2}$$
= $$\frac{1}{2}$$ sin-1 x
Diff. both sides w.r.t. x, we have
Thus $$\frac{d y}{d x}$$ = $$\frac{1}{2} \frac{1}{\sqrt{1-x^2}}$$
Diff. eqn. (2) w.r.t. x
$$\frac{d z}{d x}=\frac{1}{\sqrt{1-x^2}}$$
∴ $$\frac{d y}{d z}=\frac{d y / d x}{d z / d x}$$
= $$\frac{1}{2 \sqrt{1-x^2}} \times \sqrt{1-x^2}$$
= $$\frac{1}{}2$$
∴ $$\frac{d y}{d z}$$ is independent of x.
Question 18.
Differentiate xx w.r.t. x log x.
Solution:
Let y = xx ………..(1)
and z = x log x ………..(2)
We want to diff. y w.r.t. z i.e. to find $$\frac{d y}{d z}$$
Taking logarithm on both sides of eqn. (1) we have
log y = x log x ;
diff. both sides w.r.t. x ; we have
$$\frac{1}{y} \frac{d y}{d x}$$ = x × $$\frac{1}{x}$$ + log x . 1
⇒ $$\frac{d y}{d x}$$ = xx [1 + log x] ………..(3)
Diff. eqn. (2) w.r.t. x ; we have
$$\frac{d z}{d x}$$ = x × $$\frac{1}{x}$$ + log x . 1
= 1 + log x
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{x^x(1+\log x)}{1+\log x}$$
= xx.
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# Trouble with overheating LDOs
#### kramzar
Joined Mar 7, 2022
28
I am designing a custom pcb with a 3.3V and 5V LDO power regulator that converts 24V DC input to 3.3 and 5V to power a microcontroler , ethernet modbus transcievers, and some sensors. The whole system drawns maximum 200mA of current during operation
5V LDO = L78M05ABDT-TR
3.3V LDO= LM317AG-TN3-R
The board works well when I power it directly on the 3.3V rail via usb. But when I power it through the LDOs, they work for about 2 mins before they get very hot and the current starts droping (presumebly because the internal thermal shutdown kicks in).
My question is why is this happening? The LDOs are both rated for much higher currents and voltages, yet already start failing at 24V input.
The system consumes about 200mA of current during normal operation, which is also far below the rated limit.
I am using an external adjustable power supply to power it, which can supply up to 5A.
I also tried powering the system with 8V DC and it worked fine for an extended period, and while the LDOs did get hot, they didnt get too hot to touch or started droping current.
Do i need better cooling or is there an input/load resistor i am missing that i should add?
I am attaching a picture of the schematic and PCB layout of the power supply section.
#### ericgibbs
Joined Jan 29, 2010
17,410
hi k,
Have you calculated the power dissipation in the LDO's.
E
#### MrSalts
Joined Apr 2, 2020
2,767
0.2A * (24v - 3.3v) = 0.2A * 21.7v = 4.2 W. Oh, that poor LDO! Scalded to death!
#### Sensacell
Joined Jun 19, 2012
3,180
Better to use a switching regulator for the 5V, then use an LDO for the 3.3V, powered from the 5V.
Linear regulators suffer terrible power efficiency when dropping large voltages.
#### crutschow
Joined Mar 14, 2008
32,026
As noted, as well as a current rating, the devices have a power dissipation rating (as determined their heat -sink, if any).
That power equals (Vin-Vout) * Iout.
That power needs to be dissipated by an appropriate heat-sink.
Alternately, you can reduce the input voltage by putting a power resistor in series with the LDO input.
#### MrSalts
Joined Apr 2, 2020
2,767
Linear regulators suffer terrible power efficiency when dropping large voltages.
They work by turning the excess energy to heat. I'd call that inefficient unless the heat is needed somewhere.
#### dl324
Joined Mar 30, 2015
15,535
5V LDO = L78M05ABDT-TR
3.3V LDO= LM317AG-TN3-R
Why are you calling these regulators low drop out? They look like typical step down regulators to me. Both have a drop out voltage of about 2V.
At 200mA, the LM317 would be dissipating over 4W. Thermal resistance (junction to ambient) is 62 degrees C per W, so it wouldn't take long for thermal protection to kick in at half that current.
Do as recommended in post #4. Use a switching regulator for 5V and use an LDO regulator on the 5V output to generate 3.3V.
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# Zero-point energy, dark energy and space
1. Apr 11, 2010
### Jean Paul
Hello.
I've looked on the internet a lot about this: what is the zero-point energy and dark energy? From what I gathered, they both mean the same thing. Is that so?
Also these two energies represent minimal energy densities *in* space. But what about energy *of* space itself. I mean here the constitution of space itself. This is what physicists used to call the ether, I think. This idea has been abandonned. But don't you think that space has to be constituted of something, for it certainly cannot a void. If it was a void, then how could matter be 'floating' is a void?
I'd appreciate a lot if someone could clarify this for me. Thanks.
Jean
2. Apr 12, 2010
### Born2bwire
They are not the same thing.
In quantum mechanics, we often talk about the harmonic oscillator. The quantum oscillator is similar to the classical oscillator, like a pendulum or a mass on a spring. It turns out though that the quantum harmonic oscillator's lowest energy state is non-zero. This is also called the zero-point energy though I typically hate to use that term now that it has been used as a buzz word so much. This is different from the classical oscillator where the lowest energy state is zero.
It turns out that when we analyze the electromagnetic fields using quantum field theory that the fields behave similar to quantum oscillators. Long story short, and a lot of simplification, it means that the ground state for the vacuum for electromagnetic fields has a non-zero energy. In the vacuum state (not exactly the same as the vacuum of space, we mean just an absence of any photons), the fields fluctuate about a mean value of zero and have an energy greater than zero. So the energy density of the vacuum, if you do not allow for an upper limit of the frequency that electromagnetic waves can have, is infinite. However, this is not useful in anyway because we can only work with the change in energy, which gives rise to a force. In fact, we can renormalize the vacuum energy to be zero and the physics ends up being the same. In addition, this energy is the energy of the electromagnetic field states. In the vacuum state, there isn't any real photons so this energy does not represent any existing mass. You may hear about virtual photons popping out of the vacuum. This is different though. This is just a mathematical tool and it is based upon a situation where the energy of the system may have a large range of values such that the energy may be large enough to momentarily create and destroy a photon. But again this is viewed as a mathematical technique. So right now we do not see a physical consequence of the potentially infinite vacuum energy.
Dark energy/matter, which is out of my area of expertise, is a means by which astronomers account for the apparent deviation in the mass of the Universe. Dark matter, unlike the zero-point energy, is treated as a real thing though its properties are beyond my knowledge. One of the many astrophysics experts can expand on this in more detail.
However, I will comment on ether. Ether is a specific theory, the luminous aether, and it was the medium which mediated the propagation of light. However, the consequences of a luminous aether were not seen in experiments. This caused the theory to become more and more convoluted in attempt to keep current. In the end, it was dropped because it became overly complex and unsustainable. While you are not the first to use the term in this manner, you should understand the baggage that comes with the term and why it was rejected originally. Nothing in current physics has allowed the idea of a luminous aether to be supported.
3. Apr 12, 2010
### tedkelly
VERY well put. I just wish you could convince all of the perpetual motion enthusiasts out there.
Dark Energy is a term used to identify the unknown force that accounts for the constant acceleration of everything we can see in the Universe. The Doppler Effect has been used to calculate the speed at which everything in the Universe is moving away from us. From this calculation, they have determined that everything, (outside of our Galaxy), seems to be moving away from us - faster and faster every moment we observe those objects.
Dark Matter has been mostly narrowed down to a type of particle, called a WIMP.
Last edited: Apr 12, 2010
4. Apr 15, 2010
### Jean Paul
Thank you for the clarification. I didn't understand all the fine details, but I understand now that zero-point energy and dark energy aren't the same. And I suspected that much already before. I was seeking confirmation.
But you have not answered my other question: the constitution of space, or the fabric of space. What makes up space? I don't mean what is inside space, but what is space itself. The way I see it, there are two concepts: the energy *in* space, and then energy *of* space. Is dark energy the energy of space, the energy that space itself is made of? Think of it this way: after the Big Bang, space had to be created first, then second, energy went there. It's like if you wish to go across a river. You build first a bridge, then people can go across. Do you see what I'm trying to get across?
Looking forward for some more clarification. Thank you.
Jean
5. Apr 15, 2010
### bapowell
Dark energy and zero point energy are very similar. Dark energy behaves gravitationally just like quantum vacuum energy -- it leads to accelerated expansion of space. After the accelerated expansion of the universe was discovered, cosmologists thought that it was the result of the nonzero quantum vacuum energy! However, when they added up all those fluctuations, they got a number 120 orders of magnitude larger than the observed expansion rate! So, we're not sure where the dark energy is coming from, but it behaves just like quantum vacuum energy, and very well could be some form of it.
Space itself isn't made out of anything as far as we know, however, the geometry of spacetime is manifested as gravity. There is no energy *of* space -- if you remove all sources of energy density, then space becomes truly empty and has no dynamics (you recover special relativity). Dark energy would be energy that you put *in* space.
EDIT: Since writing this, it occurred to me that perhaps we should touch on gravity waves. These are quadrupolar waves in spacetime itself, just as light is a dipolar wave resulting from oscillating electric and magnetic fields. Gravitational radiation is an example of spacetime storing energy, and you can even think of gravity waves as spin 2 particles just as we now know that light is actually made up of photons. One can take this analogy further. Just as magnetic and electrostatic forces are often described as mediated by 'virtual photons', some have suggested that the manifold of spacetime actually be made up of virtual gravitons. However, these are by definition unobservable, and any energy the virtual gravitons have themselves won't gravitate, so it really becomes a question of interpretation. Sorry if I've terribly confused things...but it's an interesting question without an easy answer (that I can come up with anyway...)
Last edited: Apr 16, 2010
6. Apr 16, 2010
### Jean Paul
You explain:
" Space itself isn't made out of anything as far as we know, however, the geometry of spacetime is manifested as gravity. There is no energy *of* space -- if you remove all sources of energy density, then space becomes truly empty and has no dynamics (you recover special relativity). Dark energy would be energy that you put *in* space. "
If space isn't made of anything -- not even energy -- then space isn't anything at all. The reason is that deep down in the quantum world, everything comes down to energy, correct? One way to see that is that only energy came out of the Big Bang. So, if it doesn't possess energy, it's nothing--more precisely, it's nothingness. So then space isn't made of energy, it's nothingness. But then there is space-time that explains gravitation. Gravitation occurs in space. However since space is made of nothing, gravitation--and everything else--takes place in nothingness. Could space be an illusion?
Also the space-time model suggests that space-time has a 'shape'. But how can space which is nothingness have a shape? This is not only a philosophical impasse, but a physical one too. I cannot see how space cannot have a constitution made of energy.
Can someone clarify this for me, please? This impasse makes my head spin. My hunch is that dark energy is what the fabric of space is made of. I read today an article that says that dark energy was discovered around 1998, and that physicists admit of not knowning much about this energy. So then the question is still open to debate.
Jean
7. Apr 16, 2010
### bapowell
I'd like to address some of your questions but I don't have time now. But I'd like to comment on the above now. As I explained, dark energy is not what space is 'made out of'. We don't know where the dark energy comes from, but we understand how dark energy behaves -- how it gravitates and how it might behave quantum mechanically. This particular part of your question is not open for debate.
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# Events & Promotions
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Open Detailed Calendar
# HBS R2 Interview Invite
Author Message
TAGS:
### Hide Tags
Manager
Joined: 03 Nov 2006
Posts: 161
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Kudos [?]: 4 [0], given: 0
### Show Tags
31 Jan 2007, 11:05
I just got my invite!!!!! Just when I was expecting a Facebook message in my inbox, I see this. I'm on cloud 9!
GMAT Club Legend
Affiliations: HHonors Diamond, BGS Honor Society
Joined: 05 Apr 2006
Posts: 5926
Schools: Chicago (Booth) - Class of 2009
GMAT 1: 730 Q45 V45
Followers: 320
Kudos [?]: 2085 [0], given: 7
### Show Tags
31 Jan 2007, 11:09
Huge huge congrats!!!! Thats amazing stuff! Good luck!
Manager
Joined: 20 Feb 2005
Posts: 118
Followers: 2
Kudos [?]: 1 [0], given: 0
Re: HBS R2 Interview Invite [#permalink]
### Show Tags
31 Jan 2007, 12:03
hosam wrote:
I just got my invite!!!!! Just when I was expecting a Facebook message in my inbox, I see this. I'm on cloud 9!
What an achievement dude!! Super Congratulations!
Manager
Joined: 28 May 2006
Posts: 239
Followers: 1
Kudos [?]: 1 [0], given: 0
### Show Tags
31 Jan 2007, 12:08
Wow! Congrats and best of luck!
VP
Joined: 20 Sep 2005
Posts: 1018
Followers: 3
Kudos [?]: 36 [0], given: 0
Re: HBS R2 Interview Invite [#permalink]
### Show Tags
31 Jan 2007, 12:21
Awesome...go nail it and make us proud !
hosam wrote:
I just got my invite!!!!! Just when I was expecting a Facebook message in my inbox, I see this. I'm on cloud 9!
VP
Joined: 24 Sep 2006
Posts: 1359
Followers: 10
Kudos [?]: 199 [0], given: 0
### Show Tags
31 Jan 2007, 13:13
Extreme congrats! Make us proud.
Cheers. L.
VP
Joined: 15 Jun 2006
Posts: 1124
Schools: Chicago Booth
Followers: 5
Kudos [?]: 44 [0], given: 0
### Show Tags
31 Jan 2007, 13:27
Congrats!!! You got your foot in their door Good luck with the interview!
Manager
Joined: 03 Nov 2006
Posts: 161
Followers: 2
Kudos [?]: 4 [0], given: 0
### Show Tags
31 Jan 2007, 14:50
Thanks so much guys!
It's a phone interview on the 24th of February so its not for a while, but I think I'll start preparing tomorrow morning
VP
Joined: 24 Sep 2006
Posts: 1359
Followers: 10
Kudos [?]: 199 [0], given: 0
### Show Tags
01 Feb 2007, 07:08
Hosam et all,
I have been invited to interview by HBS, as well! I need to arrange the date with the alumn within 3 days.
I was assuming HBS would ding me, especially after MIT and Stanford did so for R1, but here I am, invited to interview by HBS (and I'm 30 years old!).
Anyway, let's keep our fingers crossed.
Cheers. L.
Manager
Joined: 03 Nov 2006
Posts: 161
Followers: 2
Kudos [?]: 4 [0], given: 0
### Show Tags
01 Feb 2007, 07:11
lepium, congratulations!!! Excellent news!!
Let me know when your interview date is.
VP
Joined: 15 Jun 2006
Posts: 1124
Schools: Chicago Booth
Followers: 5
Kudos [?]: 44 [0], given: 0
### Show Tags
01 Feb 2007, 07:14
lepium wrote:
Hosam et all,
I have been invited to interview by HBS, as well! I need to arrange the date with the alumn within 3 days.
I was assuming HBS would ding me, especially after MIT and Stanford did so for R1, but here I am, invited to interview by HBS (and I'm 30 years old!).
Anyway, let's keep our fingers crossed.
Cheers. L.
I am so happy for ya, Amigo! Great news. Good luck!
Current Student
Joined: 29 Jan 2005
Posts: 5222
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### Show Tags
01 Feb 2007, 07:15
Congrats to both hosam and lepium. HSB is the apotheosis of b-schools. Would love to see another few club members get in. Keep us updated.
Manager
Joined: 12 May 2006
Posts: 116
Followers: 1
Kudos [?]: 2 [0], given: 0
### Show Tags
01 Feb 2007, 11:23
Congratulations!!!!!
I can only imagine how you guys feel at the moment. I am confident that you guys will 'clean up' during the interview too!
Does anyone know when HBS is done granting interviews though? I have heard that they will continue to grant them up until March. However, I read from one forum chat that they will cease after the first week of Feb. . Anyone have any idea?
VP
Joined: 20 Sep 2005
Posts: 1018
Followers: 3
Kudos [?]: 36 [0], given: 0
### Show Tags
01 Feb 2007, 11:34
Congrats ! Ace it now...
lepium wrote:
Hosam et all,
I have been invited to interview by HBS, as well! I need to arrange the date with the alumn within 3 days.
I was assuming HBS would ding me, especially after MIT and Stanford did so for R1, but here I am, invited to interview by HBS (and I'm 30 years old!).
Anyway, let's keep our fingers crossed.
Cheers. L.
Manager
Joined: 15 Mar 2005
Posts: 230
Followers: 1
Kudos [?]: 0 [0], given: 0
### Show Tags
01 Feb 2007, 13:29
way to go!!....congrats guys...best of luck!
Manager
Joined: 04 Jan 2007
Posts: 53
Followers: 0
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### Show Tags
01 Feb 2007, 20:55
congrats guys!
SVP
Joined: 30 Mar 2006
Posts: 1730
Followers: 1
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### Show Tags
01 Feb 2007, 21:06
Great work Lepium
Make it count. All the best
Manager
Joined: 05 Jun 2006
Posts: 155
Followers: 1
Kudos [?]: 2 [0], given: 0
### Show Tags
01 Feb 2007, 21:13
congrats to Lepium and hosam and best wishes to increase the HBS count @ d club!!
VP
Joined: 24 Sep 2006
Posts: 1359
Followers: 10
Kudos [?]: 199 [0], given: 0
### Show Tags
01 Feb 2007, 22:06
Thanks a lot, guys! I'm just terrified of being grilled by HBS. Yet on the other hand, I'm thrilled to have reached this stage.
I haven't scheduled the interview yet. I'll let y'all know how it goes.
Cheers. L.
Manager
Joined: 03 Nov 2006
Posts: 161
Followers: 2
Kudos [?]: 4 [0], given: 0
### Show Tags
24 Feb 2007, 12:13
I just finished my phone interview and I really don't know what to feel. It was very focused on my work and leadership experiences., strenths, weaknesses...
Very little about anything else. No introduction and no curveballs. He asked me some pretty detailed questions about how I would differentiate the funds at my private equity firm (long-term goal). Not sure how much he liked my answer. All in all there was very little feedback since it was on the phone.
He ended by saying that he thought I was a strong team-player and have the potential for becoming a strong leader, which is hopefully a good sign.
He then asked whether there was anything else I would have liked to talk about. I said that I wanted to talk a bit about my hobbies and interests and at one point I became too phylosophical by comparing football teams with corporate teams and management influences yada yada, at which point he politely shut me up
34 days to go!!
24 Feb 2007, 12:13
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Distance between Manaus, Brazil and the Arctic Circle
7755 km = 4819 miles
During our calculation of the distance to the Arctic Circle we make two assumptions:
1. We assume a spherical Earth as a close approximation of the true shape of the Earth (an oblate spheroid). The distance is calculated as great-circle or orthodromic distance on the surface of a sphere.
2. We calculate the distance between a point on the Earth’s surface and the Arctic Circle as the length of the arc of the meridian passing through this point and crossing the Arctic Circle.
Find out the distance between Manaus and the North Pole, the South Pole, the Equator, the Tropic of Cancer, the Tropic of Capricorn, the Antarctic Circle
Find out the distance between Manaus and other cities
Manaus, Brazil
Country: Brazil
Manaus’ coordinates: 3°06′06″ S, 60°01′29″ W
Population: 1,598,210
Find out what time it is in Manaus right now
Wikipedia article: Manaus
The Arctic Circle
The Arctic Circle is a circle of latitude or parallel on the Earth's surface. The polar night and the midnight sun phenomena can be observed to the north of the Arctic Circle.
During the midnight sun there is no sunset and the Sun is over the horizon continuously for 24 hours or more.
During the polar night there is no sunrise and the Sun is below the horizon continuously for 24 hours or more.
For the points on the Earth's surface located at the Arctic Circle the polar night and the midnight sun coincide with the winter solstice and the summer solstice correspondingly and their duration equals 24 hours.
For the points on the Earth's surface located to the north of the Arctic Circle the duration of the polar night and the midnight sun grows with latitude.
The latitude of the Arctic Circle depends on the tilt of the Earth's axis which changes with time.
The current latitude of the Arctic Circle equals 66°33′43″ N.
Wikipedia article: the Arctic Circle
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# How to Calculate Frequency of a Wave
## Introduction
Understanding the frequency of a wave is essential in various fields, such as physics, engineering, and telecommunications. The frequency represents the number of complete cycles or oscillations a wave completes in a given time period. In this article, we will explore the fundamental principles behind calculating the frequency of a wave and provide you with a comprehensive guide to mastering this concept.
## How to Calculate Frequency of a Wave
To calculate the frequency of a wave, you need to consider two primary factors: the wavelength and the wave speed. The wavelength refers to the distance between two consecutive points on a wave that are in phase, while the wave speed is the rate at which the wave propagates through a medium. By utilizing the relationship between wavelength, wave speed, and frequency, we can determine the frequency of a wave accurately.
## The Relationship Between Wavelength, Wave Speed, and Frequency
The relationship between wavelength, wave speed, and frequency can be expressed through a simple formula:
Frequency (f) = Wave Speed (v) / Wavelength (λ)
This formula shows that the frequency of a wave is inversely proportional to its wavelength. In other words, as the wavelength decreases, the frequency increases, and vice versa. Similarly, the frequency is directly proportional to the wave speed. As the wave speed increases, the frequency also increases, and as the wave speed decreases, the frequency decreases.
## Step-by-Step Guide to Calculating Frequency
Now, let’s delve into the step-by-step process of calculating the frequency of a wave.
### Step 1: Determine the Wavelength
To calculate the frequency, you first need to measure or determine the wavelength of the wave. The wavelength can be measured by identifying two consecutive points on the wave that are in phase and measuring the distance between them.
### Step 2: Measure the Wave Speed
Next, you need to measure or determine the wave speed. The wave speed represents the rate at which the wave propagates through a medium. The speed can vary depending on the characteristics of the medium, such as its density and elasticity.
### Step 3: Apply the Formula
Once you have obtained the values for the wavelength and wave speed, you can apply the formula mentioned earlier:
Frequency (f) = Wave Speed (v) / Wavelength (λ)
Simply divide the wave speed by the wavelength to calculate the frequency of the wave.
### Step 4: Perform the Calculation
Now, perform the calculation using the obtained values:
Frequency = Wave Speed / Wavelength
Substitute the values you measured or determined in Steps 1 and 2 into the formula and calculate the frequency.
## Factors Affecting Wave Frequency
Several factors can affect the frequency of a wave. It is crucial to understand these factors as they play a significant role in real-life applications and scenarios involving waves. Let’s explore some of the key factors that influence wave frequency.
### 1. Wave Source
The source of the wave, such as an oscillating object or an electrical signal generator, determines the frequency of the wave. Different sources produce waves with varying frequencies.
### 2. Medium Properties
The properties of the medium through which the wave propagates can affect its frequency. Factors such as density, elasticity, and temperature of the medium can alter the speed of the wave and consequently impact its frequency.
### 3. Interference and Resonance
Interference occurs when two or more waves interact with each other. Depending on their relative frequencies, interference can result in constructive or destructive interference, altering the overall frequency. Resonance, on the other hand, is a phenomenon where the frequency of an external force matches the natural frequency of an object, resulting in amplified vibrations.
### 4. Doppler Effect
The Doppler effect is observed when there is relative motion between the source of the wave and the observer. It causes a change in the perceived frequency of the wave due to the compression or expansion of the wavefront.
## FAQs about Calculating the Frequency of a Wave
### Q: Can I calculate the frequency of any type of wave using the same formula?
Yes, the formula for calculating the frequency of a wave is applicable to all types of waves, including electromagnetic waves, sound waves, and water waves.
### Q: What are the units of frequency?
Frequency is typically measured in hertz (Hz), which represents the number of cycles or oscillations per second.
### Q: Is frequency the same as pitch?
No, frequency and pitch are related but not identical. Frequency refers to the objective measurement of the number of cycles per second, while pitch is a subjective perception of how high or low a sound appears to be.
### Q: How does frequency affect the energy of a wave?
The frequency of a wave is directly proportional to its energy. Higher frequency waves carry more energy compared to lower frequency waves.
### Q: Are there any practical applications of wave frequency calculations?
Yes, wave frequency calculations have various practical applications. They are used in telecommunications, radio and TV broadcasting, medical imaging, music, and many other fields.
### Q: Can I use frequency calculations to determine the speed of a wave?
No, frequency calculations alone cannot determine the speed of a wave. To determine the speed, you need to know the wavelength as well.
## Conclusion
Understanding how to calculate the frequency of a wave is crucial in numerous scientific and technological fields. By following the step-by-step guide outlined in this article, you can accurately determine the frequency using the wave’s wavelength and speed. Remember the fundamental formula: frequency equals wave speed divided by wavelength. Additionally, consider the various factors that can influence wave frequency, such as the wave source, medium properties, interference, resonance, and the Doppler effect. By mastering the concept of wave frequency, you can gain a deeper understanding of wave behavior and apply it to real-world scenarios.
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Algebra I: Finding the domain of function
This is a free lesson from our course in Algebra I
This lesson explains how to find a function's domain from its equation. If a function f is defined by an expression with variable x, then the domain is the set of all real numbers that can be substituted for x such that the resulting value of the function is a real number. E.g. if you've to find the domain of the function f(x) = x + 3, you can see that there is no restriction on x as all values of x as long as it is a real number, lead to a real number for f(x). Hence, the domain of this function is the set of all real numbers. Also, the set of all correspondence values of f(x) is called range.E.g. Find the domain of function f defined by f (x) = 1 / ( x - 1) Solution: x can take any real number except 1 since x = 1 would make the denominator equal to zero and the division by zero is not allowed. Hence the domain in interval notation is given by (- , 1) U (1 , +).
Other useful lessons:
Types of functions One to many relationship Finding the domain and range of the functions Identify if the graph is a function or not Finding the domain and range of an equation Determine domain and range from a finite set of ordered pairs
Winpossible's online math courses and tutorials have gained rapidly popularity since their launch in 2008. Over 100,000 students have benefited from Winpossible's courses... these courses in conjunction with free unlimited homework help serve as a very effective math-tutor for our students.
- All of the Winpossible math tutorials have been designed by top-notch instructors and offer a comprehensive and rigorous math review of that topic. - We guarantee that any student who studies with Winpossible, will get a firm grasp of the associated problem-solving techniques. Each course has our instructors providing step-by-step solutions to a wide variety of problems, completely demystifying the problem-solving process! - Winpossible courses have been used by students for help with homework and by homeschoolers. - Several teachers use Winpossible courses at schools as a supplement for in-class instruction. They also use our course structure to develop course worksheets.
Copyright © Winpossible, 2010 - 2011 Best viewed in 1024x768 & IE 5.0 or later version
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Search a number
1089 = 32112
BaseRepresentation
bin10001000001
31111100
4101001
513324
65013
73114
oct2101
91440
101089
11900
12769
1365a
1457b
154c9
hex441
1089 has 9 divisors (see below), whose sum is σ = 1729. Its totient is φ = 660.
The previous prime is 1087. The next prime is 1091. The reversal of 1089 is 9801.
Multipling 1089 by its reverse (9801), we get a square (10673289 = 32672).
It can be divided in two parts, 10 and 89, that added together give a palindrome (99).
1089 = T32 + T33.
The square root of 1089 is 33.
It is a perfect power (a square), and thus also a powerful number.
1089 is an esthetic number in base 8, because in such base its adjacent digits differ by 1.
It is an interprime number because it is at equal distance from previous prime (1087) and next prime (1091).
It is a tau number, because it is divible by the number of its divisors (9).
It is not a de Polignac number, because 1089 - 21 = 1087 is a prime.
It is a Duffinian number.
It is a Curzon number.
It is a plaindrome in base 14.
It is a nialpdrome in base 3, base 11 and base 16.
It is a zygodrome in base 3.
It is not an unprimeable number, because it can be changed into a prime (1087) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (3) of ones.
It is a polite number, since it can be written in 8 ways as a sum of consecutive naturals, for example, 94 + ... + 104.
It is a Proth number, since it is equal to 17 ⋅ 26 + 1 and 17 < 26.
1089 is the 33-rd square number and also the 18-th nonagonal number.
1089 is the 17-th centered octagonal number.
It is an amenable number.
1089 is a deficient number, since it is larger than the sum of its proper divisors (640).
1089 is a wasteful number, since it uses less digits than its factorization.
1089 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 28 (or 14 counting only the distinct ones).
The product of its (nonzero) digits is 72, while the sum is 18.
The cubic root of 1089 is about 10.2882764781.
The spelling of 1089 in words is "one thousand, eighty-nine".
Divisors: 1 3 9 11 33 99 121 363 1089
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0
# What are multiples of 300?
Updated: 4/28/2022
Wiki User
13y ago
The first twelve positive integer multiples of 300 are as follows:
1 x 300 = 300
2 x 300 = 600
3 x 300 = 900
4 x 300 = 1200
5 x 300 = 1500
6 x 300 = 1800
7 x 300 = 2100
8 x 300 = 2400
9 x 300 = 2700
10 x 300 = 3000
Wiki User
13y ago
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# My Weblog
## Learning Agda
Finally I am feeling bit comfortable with Agda and feeling great :). I started reading Ulf Norell‘s Dependently Typed Programming in Agda. Tutorials on Agda-wiki , AgdaTutorial , papers which mention Agda , stackoverflow post, Introduction to Agda and Programming in Martin-Löf’s Type Theory. I wrote some of the codes from the paper. You can also try a similar language Idris ( See the stackoverflow post about difference between these languages ).
```module Basic where
data Bool : Set where
true : Bool
false : Bool
not : Bool → Bool
not true = false
not false = true
or : Bool → Bool → Bool
or false false = false
or _ _ = true
and : Bool → Bool → Bool
and true true = true
and _ _ = false
data ℕ : Set where
zero : ℕ
suc : ℕ → ℕ
_+_ : ℕ → ℕ → ℕ
zero + m = m
suc n + m = suc ( n + m )
_⋆_ : ℕ → ℕ → ℕ
zero ⋆ m = zero
suc n ⋆ m = ( n ⋆ m ) + m
fold-ℕ : ℕ → ( ℕ → ℕ ) → ℕ → ℕ
fold-ℕ u _ zero = u
fold-ℕ u f ( suc n ) = f ( fold-ℕ u f n )
if_then_else_ : { A : Set } → Bool → A → A → A
if true then x else y = x
if false then x else y = y
data List ( A : Set ) : Set where
[] : List A
_::_ : A → List A -> List A
identity : ( A : Set ) → A → A
identity A x = x
zero′ : ℕ
zero′ = identity ℕ zero
apply : ( A : Set ) ( B : A → Set ) → ( ( x : A ) → B x ) → ( a : A ) → B a
apply A B f a = f a
identity₂ : ( A : Set ) → A → A
identity₂ = \A x → x
identity₃ : ( A : Set ) → A → A
identity₃ = \(A : Set ) ( x : A ) → x
identity₄ : ( A : Set ) → A → A
identity₄ = \ ( A : Set ) x → x
id : { A : Set } → A → A
id x = x
true′ : Bool
true′ = id true
one : ℕ
one = identity _ ( suc zero )
_∘_ : { A : Set } { B : A → Set } { C : ( x : A ) → B x → Set }
( f : { x : A } ( y : B x ) → C x y ) ( g : ( x : A ) → B x ) ( x : A )
→ C x ( g x )
( f ∘ g ) x = f ( g x )
plus-two = suc ∘ suc
plus-three = suc ∘ ( suc ∘ suc )
map : { A B : Set } → ( A → B ) → List A → List B
map f [] = []
map f ( x :: xs ) = f x :: map f xs
_++_ : { A : Set } → List A → List A → List A
[] ++ ys = ys
( x :: xs ) ++ ys = x :: ( xs ++ ys )
foldl : { A B : Set } → ( A → B → A ) → A → List B → A
foldl f val [] = val
foldl f val ( x :: xs ) = foldl f ( f val x ) xs
foldr : { A B : Set } → ( A → B → B ) → B → List A → B
foldr f val [] = val
foldr f val ( x :: xs ) = f x ( foldr f val xs )
--type of Vec A is ℕ → Set. This mean Vec A is family of types indexed by
-- natural numbers
data Vec ( A : Set ) : ℕ → Set where
[] : Vec A zero
_::_ : { n : ℕ } → A → Vec A n → Vec A ( suc n )
head : { A : Set } { n : ℕ } → Vec A ( suc n ) → A
head ( x :: xs ) = x
vmap : { A B : Set } { n : ℕ } → ( A → B ) → Vec A n → Vec B n
vmap f [] = []
vmap f ( x :: xs ) = f x :: vmap f xs
data Vec₂ ( A : Set ) : ℕ → Set where
nil : Vec₂ A zero
cons : ( n : ℕ ) → A → Vec₂ A n → Vec₂ A ( suc n )
{--
The rule for when an argument should be dotted is:
if there is a unique
type correct value for the argument it should be dotted
--}
vmap₂ : { A B : Set } ( n : ℕ ) → ( A → B ) → Vec₂ A n → Vec₂ B n
vmap₂ .zero f nil = nil
vmap₂ .( suc n ) f ( cons n x xs ) = cons n ( f x ) ( vmap₂ n f xs )
vmap₃ : { A B : Set } ( n : ℕ ) → ( A → B ) → Vec₂ A n → Vec₂ B n
vmap₃ zero f nil = nil
vmap₃ ( suc n ) f ( cons .n x xs ) = cons n ( f x ) ( vmap₃ n f xs )
pow : ℕ → ℕ → ℕ
pow _ zero = suc zero
pow a ( suc n ) = a ⋆ pow a n
t : ℕ
t = pow ( suc ( suc zero ) ) ( suc ( suc ( suc zero ) ) )
```
To evaluate the expression , type C-c C-n. Here is the output of t
suc (suc (suc (suc (suc (suc (suc (suc zero))))))).
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Helper I
## Passing filter in a weekly calculation measure
Good day
i am trying to create measure for the good sold in the current week using the below formula
Weekly_Sold = CALCULATE(COUNT(table[GoodSold]),FILTER(table,WEEKNUM(TODAY(),1)=WEEKNUM(MAX(table[SalesDate]),1)))
everytime it gives the total count of sold goods across the column, passing any filters
5 REPLIES 5
Community Support
You could refer to the following measure.
``````Measure =
CALCULATE (
COUNT ( 'Table'[GoodSold] ),
FILTER (
'Table',
WEEKNUM ( TODAY(), 1 )
= WEEKNUM (
CALCULATE (
MAX ( 'Table'[SalesDate] ),
ALLEXCEPT ( 'Table', 'Table'[GoodSold] )
),
1
)
)
)``````
If this post helps, then please consider Accept it as the solution to help the other members find it.
Helper I
@amitchandakthanks a million for the rich response, it would help me a lot in the future cases. i was just after the logic issue of my formula.
@v-eachen-msftthanks v, i tried your formula but it returned with the events of below dates
Date WeekNUM 6/9/2020 24 6/10/2020 24 6/11/2020 24 6/14/2020 25 6/15/2020 25 6/16/2020 25 6/17/2020 25 6/18/2020 25 6/20/2020 25 6/21/2020 26 6/22/2020 26
Community Support
Maybe I misunderstood your needs. You could share your expected output with files or screenshots here.
If this post helps, then please consider Accept it as the solution to help the other members find it.
Helper I
hello @v-eachen-msft please have a look below, i need to count goods solds based on this week and previous week for example, today the 23rd June in my region, i want to find W26 & W25 sales.
thanks
Good Sold ID Date 244128 4/7/2020 244133 4/16/2020 244140 4/16/2020 244143 4/16/2020 244146 4/16/2020 244149 6/14/2020 244153 4/9/2020 244155 4/14/2020 244158 4/16/2020 244159 6/14/2020 244179 4/16/2020 244190 4/15/2020 244211 4/18/2020 244214 4/14/2020 244216 4/14/2020 244229 6/4/2020 244234 5/7/2020 244240 6/10/2020 244241 6/22/2020 244242 4/18/2020 244251 4/29/2020 244255 4/2/2020 244258 4/9/2020 244271 4/15/2020 244272 4/18/2020 244273 4/18/2020 244280 6/18/2020 244281 4/15/2020 244321 4/18/2020 244326 4/15/2020 244331 4/18/2020 244336 4/18/2020 244341 4/18/2020 244355 6/4/2020 244362 4/7/2020 244393 6/22/2020 244406 5/7/2020 244420 6/15/2020 244424 4/18/2020 244425 4/18/2020 244430 4/18/2020 244435 4/18/2020 244455 5/5/2020 244467 4/18/2020 244473 6/18/2020 244476 4/13/2020 244477 4/18/2020 244481 4/5/2020 244482 4/18/2020 244485 5/31/2020 244492 4/18/2020 244504 6/15/2020 244514 5/31/2020 244519 4/18/2020 244527 4/29/2020 244569 4/18/2020 244571 5/10/2020 244581 4/18/2020 244587 6/18/2020 244588 4/27/2020 244594 4/9/2020 244600 5/20/2020 244626 5/21/2020 244628 6/7/2020 244639 5/5/2020 244654 5/31/2020 244670 6/18/2020 244673 6/8/2020 244698 6/16/2020 244816 6/15/2020 244822 4/18/2020 244851 6/21/2020 244853 5/31/2020 244889 6/21/2020 244902 4/13/2020 244914 6/21/2020 244916 4/1/2020 244926 5/12/2020 244937 6/21/2020 244951 6/18/2020 244953 4/18/2020 244995 4/18/2020 245026 5/5/2020 245031 6/4/2020 245033 4/18/2020 245063 5/19/2020 245078 4/18/2020 245147 6/21/2020 245159 4/18/2020 245191 4/18/2020 245225 4/18/2020 245747 4/9/2020 245812 4/9/2020 245819 6/17/2020 245828 4/1/2020 245846 5/10/2020 245850 6/16/2020 245853 4/20/2020 245874 4/25/2020 245895 4/8/2020 245915 6/2/2020 245938 5/11/2020 245981 4/21/2020 245993 5/16/2020 246139 4/29/2020 294221 5/21/2020 296089 4/18/2020 296092 4/14/2020 296099 6/14/2020 296103 4/14/2020 296136 5/20/2020 296162 4/14/2020 296362 6/4/2020 296392 6/9/2020 297101 4/29/2020 297121 4/1/2020 297131 4/25/2020 298877 4/18/2020 298909 4/18/2020 298916 4/18/2020 298937 4/18/2020 298938 6/8/2020 298944 5/11/2020 298950 5/10/2020 298958 5/5/2020 298959 6/4/2020 298961 4/18/2020 298965 6/15/2020 299183 4/1/2020 302243 4/18/2020 302247 5/10/2020 302250 4/18/2020 302252 4/18/2020 302255 5/17/2020 302259 6/14/2020 302265 4/18/2020 302293 4/7/2020 302298 4/14/2020 302301 4/9/2020 302305 4/2/2020 302307 4/18/2020 302323 4/15/2020 302340 4/18/2020 302344 5/3/2020 302351 4/7/2020 302353 4/18/2020 302370 4/18/2020 302379 5/10/2020 302386 4/18/2020 302390 5/3/2020 302393 4/18/2020 302395 4/18/2020 302407 4/18/2020 302408 6/8/2020 302410 6/8/2020 302430 6/15/2020 302442 4/18/2020 302447 5/10/2020 302455 5/19/2020 302465 4/18/2020 302468 4/18/2020 302471 4/18/2020 302475 4/9/2020 302484 6/18/2020 302493 4/2/2020 302510 5/5/2020 302520 5/20/2020 302524 4/18/2020 302529 5/10/2020 302540 4/29/2020 302544 5/10/2020 302545 4/18/2020 302553 4/18/2020 302562 4/29/2020 302569 5/31/2020 302586 5/10/2020 302594 5/3/2020 302619 5/5/2020 302622 6/11/2020 302650 4/29/2020 302652 4/18/2020 302671 4/18/2020 302672 4/18/2020 302676 5/10/2020 302686 5/10/2020 302696 4/18/2020 302705 4/18/2020 302720 6/21/2020 302721 6/22/2020 302735 5/20/2020 302736 4/18/2020 302745 5/17/2020 302761 5/3/2020 302787 5/5/2020 302790 5/31/2020 302791 6/9/2020 302799 4/18/2020 302815 4/5/2020 302817 4/5/2020 302839 4/18/2020 302846 4/18/2020 302847 4/18/2020 302889 4/18/2020 302938 4/5/2020 302941 6/22/2020 302954 4/18/2020 302967 6/18/2020 303011 5/5/2020 303060 4/18/2020 303061 6/21/2020 303096 4/18/2020 303097 4/18/2020 303161 4/18/2020 303215 6/10/2020 304094 4/22/2020 305231 5/21/2020 305300 5/31/2020 306168 5/30/2020 306173 6/20/2020 306180 5/11/2020 306182 4/18/2020 308711 4/18/2020 309025 5/17/2020 309029 6/2/2020 309522 4/21/2020 309523 4/18/2020 309524 4/5/2020 309556 5/19/2020 309557 5/20/2020 309558 5/13/2020 309726 4/18/2020 309727 5/12/2020 309728 4/2/2020 316390 6/4/2020 316391 6/4/2020 316392 5/31/2020 316393 6/4/2020 316511 6/10/2020 316670 6/11/2020 316671 6/14/2020 244291 6/22/2020 244611 4/18/2020 245996 4/29/2020 296145 6/3/2020 297119 5/13/2020 302309 4/13/2020 302478 6/15/2020 302479 6/16/2020 302716 6/10/2020 302731 5/21/2020 306178 6/18/2020 316097 6/11/2020 316098 6/15/2020
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# Thread: Exponential Growth and Decay Help?
1. ## Exponential Growth and Decay Help?
I need help with part c (mainly with how fast the salt is dissolving) and d please, any help would be appreciated. Thanks.
2. ## Re: Exponential Growth and Decay Help?
When you're finding the rate, or how fast or slow something is, you use the derivative expression.
so for c)
you find dS/dt and then sub in t = 5 and your value of k from part b), that's your rate.
d)sub S = 4 (as that's the amount that's left undissolved ) and your k from b)
Then you solve for t, assuming you have you the logarithm knowledge to work out the algebra.
3. ## Re: Exponential Growth and Decay Help?
Originally Posted by integral95
When you're finding the rate, or how fast or slow something is, you use the derivative expression.
so for c)
you find dS/dt and then sub in t = 5 and your value of k from part b), that's your rate.
d)sub S = 4 (as that's the amount that's left undissolved ) and your k from b)
Then you solve for t, assuming you have you the logarithm knowledge to work out the algebra.
Thanks.
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## August 03, 2005
### Laaast Guessss...
Continuing the story of Anthony's Guess-A-Number game. This was my six-year-old son's first try at programming a game together with me. You would play the little python program by trying to guess a random number fom 1 to 100. And most importantly, the opponent was a Very Scary Vampire.
I was surprised that Anthony was so enthusiastic about programming such a simple game. In 2005, we are decades beyond a time when Zork was the coolest game ever - today's run-of-the-mill free online Flash game "La Casa de Dora" for 3-year-olds is an animated multimedia extravaganza.
But Anthony was psyched. He had a lot of ideas on how to make his Guess A Number Vampire into a worthy opponent, and they were easy to program in python. The vampire would just give you just five guesses. If you did not get it in five, you would lose. And when you were about to run out of guesses, the vampire would taunt you.
Let us take a look at his program....
Last guess!
This is what Anthony's program looked like when it was finally "finished." At least, this is when I thought it was finished.
```
#!/usr/bin/env python
import time
import random
secret_number = random.randint(1, 100)
print "HAAARG! I've got the secret"
print "it's a number. 1-100"
guesses_left = 5
time.sleep(2)
print "GUESS!"
while guesses_left != 0:
guess = input()
if guess == secret_number:
print "you win!"
break
if guess > secret_number:
print "(TOO BIG!)"
else :
print "(too small!)"
guesses_left -= 1
if guesses_left == 0:
print "I WIN!"
print secret_number
elif guesses_left == 1:
print "LAST GUESS!"
else:
print "GUESS AGAIN!"
```
You can see that Anthony was able to understand all the basic programming concepts: variables, branches, loops. It seemed to come pretty naturally. Key enablers like importing and using a library were so easy in python that a six-year old could use it all without any trouble. Even the idosyncratic python syntax with indenting and the colons seemed okay.
There were only a couple mysteries, like the spelling of the strange word "elif". And there was the mixup between assignment "=" and equality "==". My son's intuition was that "=" should be used to test equality, and assignment shouldn't have any symbol at all. But at six, you are a sponge. He learned the actual syntax of Python quickly enough.
The game was very nicely balanced, beating you most of the time but letting you win quite often.
Here is a transcript of me battling it out with the vampire:
```
>./GuessANumber.py
HAAARG! I've got the secret
it's a number. 1-100
GUESS!
50
(TOO BIG!)
GUESS AGAIN!
25
(TOO BIG!)
GUESS AGAIN!
12
(too small!)
GUESS AGAIN!
19
(TOO BIG!)
LAST GUESS!
16
(too small!)
I WIN!
18
>
```
Anthony would shout out along with the terminal printout "LAST GUESS!" whenever the Vampire was about to defeat me.
One secret to a great game is having fun when you win or lose.
What Games Do in 2005
"I love my game!" Anthony said, "But, the vampire needs to say, 'Laaaast guuuesssss!'" He said the words using his best, most mocking sing-song nyah-nyah you-are-going-to-lose schoolyard voice.
"Sure, Anthony, it already does." And he used all capital letters, too. What a great game.
"No, Dad. I mean, the vampire has got to say it, you know? I mean SAY it, so that you can hear it! Laaaast guuuessssss!"
Ah hah.
Anthony was brought up in a world with Reader Rabbit, Talking Battleship and Nick Jr. Flash Games. Talking is a multimedia game feature that he took for granted. Of course the game should talk! What game doesn't?
But I had no idea how to make a python game talk. I did not even know if it would be possible.
"Let's look at it tomorrow, Anthony. The game is great right now. It is time to go to bed."
Pygame, Audacity, and The Secret to Being a Dad in 2005
6AM the next morning Anthony ran into my bedroom and shook me awake. There was only one thing on his mind.
"Can we make sounds for the vampire now?"
Have you noticed that Google can transform you from an Ordinary Witless Dope into the World's Best Dad? It happens all the time. Yesterday my kids asked me [how you fold a paper fortune teller] and I had no idea. All the elementary school girls had played with them growing up, but I had never made one. No matter, because, of course, Google pointed me to a website that shows how, so that meant I could show them. And soon our house filled up with wonderful little folded paper fortune tellers, all crayoned up with fortunes for every future possibility a child could imagine.
And overnight that previous evening Google had taught me how to make python talk. Here is what I found:
1. [sound in python games] brought me to pygame.org, a wonderful python game library by Pete Shinners that is layered on top of the SDL libraries by Sam Latinga. I soon learned how to use pygame.mixer.Sound to play WAV files.
2. [free sound recording on OS X] brought me straight to Audacity, a GPL sound recorder for Mac and Linux made freely available by Dominic Mazzoni and Vaughan Johnson and many other contributing programmers.
Pygame is easy to install on stock OS X, thanks to Bob Ippolito's pymac packages. If you are following along on your mac, you will want to install both pygame and pyObjC. Later I needed Numeric and py2app as well.
The open-source world is just great. Everything worked as advertised. So after running a couple simple tests I went to bed too, ready for the next morning.
Listening to Anthony's Inner Vampire
Here is what we did in the morning. Anthony and I counted the eight different prompts that would need voice clips - "too big," "too small," "i win," and so on. And then, using Audacity, we recorded Anthony reading each of the clips in his best vampire voice.
Here is an example - Anthony says Haaarg:
Anthony enjoyed recording the sound, but he was not happy with the playback.
"It sounds like a boy! It is not scary at all!" He was ready to give up. "Can you make the Vampire, Dad? You can sound like a vampire!" Of course, that would be no fun. It was Anthony's game, and it should be an Anthony vampire.
Audacity saved the day. The free program comes with dozens of great audio effects, and one of them was perfectly suited to what we needed: change pitch. To make a vampire, we just dropped Anthony's recording by a whole octave.
Meet our scary opponent - The Guess-A-Number Vampire:
Anthony was delighted. "That's me, that's my voice" he explained to his little sister who had come running when she heard the vampire.
Using pygame.mixer.Sound
Once we had the sounds recorded, we loaded them into the program and played them at the right spots. I typed in a few of the sounds, and Anthony followed the template. I am not sure if he understood what was going on - although the code was simple enough, we also had to insert time delays. It was important to wait for a sound to finish playing before starting the next sound.
But the beauty of it was it was simple. There was just not that much more code to write.
Here is the fully audio-enabled version of Anthony's vampire:
```
#!/usr/bin/env python
import random
import time
import pygame
pygame.mixer.init()
intro_sound = pygame.mixer.Sound("intro.wav")
guess_sound = pygame.mixer.Sound("guess.wav")
you_win_sound = pygame.mixer.Sound("you_win.wav")
big_sound = pygame.mixer.Sound("big.wav")
small_sound = pygame.mixer.Sound("small.wav")
i_win_sound = pygame.mixer.Sound("i_win.wav")
last_guess_sound = pygame.mixer.Sound("last_guess.wav")
guess_again_sound = pygame.mixer.Sound("guess_again.wav")
secret_number = random.randint(1, 100)
intro_sound.play()
print "HARRRG! I've got the secret"
print "it's a number. 1-100"
guesses_left = 5
time.sleep(7)
guess_sound.play()
print "GUESS!"
while guesses_left != 0:
guess = input()
if guess == secret_number:
you_win_sound.play()
print "you win!"
time.sleep(5)
break
if guess > secret_number:
big_sound.play()
print "(TOO BIG!)"
else :
small_sound.play()
print "(too small!)"
time.sleep(2)
guesses_left -= 1
if guesses_left == 0:
i_win_sound.play()
print "I WIN!"
print secret_number
time.sleep(5)
elif guesses_left == 1:
last_guess_sound.play()
print "LAST GUESS!"
else:
guess_again_sound.play()
print "GUESS AGAIN!"
```
At this point, the game was genuinely entertaining! The kids all wanted a turn playing, and I did too. Anthony was right. The sounds made a huge difference.
This is when I decided to write a few weblog entires about it.
The game was still not good enough for Anthony, however, because he wanted to see his vampire too. It took a couple weeks before we got back to the project, but with the help of some cutting and pasting, we finally finished it up.
I will write a bit more about it next time. We should have a download for you next time, too.
Posted by David at August 3, 2005 07:40 PM
Great parenting. As of now, my 23 month old is interested in cars, computers, and anything else electronic. He loves to 'type.' So much so that I let him pound away on an old keyboard with the cable removed.
Your story is an inspiration, and a proves just how much potential kids have if they're just given room to grow.
Can't wait to get the last bit of the story!
Posted by: Devin M at August 4, 2005 10:41 PM
Hi David, what a great story. Your son has quite an attention span. My son wants to make a game on the computer too, a haunted house with a skeleton that falls out of the ceiling. I haven't decided how we'll write that one :) I like the idea of using python.
My kids like to play some of the old BSD games; Rogue, the old dungeon crawling text game, and robot (Daleks).
Posted by: ScottLu at August 5, 2005 04:10 AM
Hi David!
Wow. I'm a bit behind. Your 6 year old knows how to program better than I do. He sounds like an awesome creative little guy. I hope someday I can be as cool a parent as you (with the help of Google of course!).
I'll probably start coding again soon if some changes occur at work, but I'll email you about it some time.
Posted by: mapgirl at August 14, 2005 09:39 PM
I'm just a random seo surfing with the last minutes of a friday afternoon, and your blog enties on your son's adventures with python are wonderful! I've really enjoyed reading, and you're doing a fantastic job - i'm so glad there are some kids out there really getting that kind of growth and learning. :)
Posted by: camilla todd at August 26, 2005 12:29 PM
I've found flash to be an excellent platform for easy coding (essentially javascript) and very simple to connect visual and audio assets to an event-based model - may be something to consider for some more in-depth forrays into the multimedia world :)
Posted by: Mike J at October 12, 2005 09:53 AM
Posted by: Travis Cripps at October 14, 2005 09:45 PM
I am having some trouble with PyObjC (pygame at the Instalation will not acknolige it). If you did too, how did you solve the problem, and if you didn't, which versions of python, pygame and pyobjc did you use and where did you get them from?
Posted by: Alexander Lupas at April 26, 2006 12:16 PM
I don't remember any specific problem. I got my PyObjC package and other python-mac packages from Bob Ippolito's website here:
http://pythonmac.org/packages/
Posted by: David at April 26, 2006 11:03 PM
thank you
Posted by: at April 27, 2006 07:58 AM
So when do you start programming with your daughter?
Posted by: Dorothea at May 26, 2006 10:34 AM
Using Audacity was a great idea. I'm one webmaster for my choir, and I use it all the time to cut sound clips to post on the site. It's very easy to use.
Posted by: Joanna at July 29, 2007 11:00 PM
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# What is electric car energy economy
O:Convert the unit to all units
T:Convert the unit to another unit
kilometer per kilowatt hourkm/kWhO:km/kWhT:km/kWh
kilowatt hour per 100 kilometerskWh/100kmO:kWh/100kmT:kWh/100km
kilowatt hour per 100 mileskWh/100miO:kWh/100miT:kWh/100mi
kilowatt hour per kilometerkWh/kmO:kWh/kmT:kWh/km
kilowatt hour per milekWh/miO:kWh/miT:kWh/mi
mile per kilowatt hourmi/kWhO:mi/kWhT:mi/kWh
mile per watt hourmi/WhO:mi/WhT:mi/Wh
miles per gallon gasoline equivalentMPGeO:MPGeT:MPGe
watt hour per kilometerWh/kmO:Wh/kmT:Wh/km
watt hour per mileWh/miO:Wh/miT:Wh/mi
#### Foods, Nutrients and Calories
NATURAL CABANA, 100% NATURAL LEMONADE, CHERRY, UPC: 856190003020 contain(s) 25 calories per 100 grams (≈3.53 ounces) [ price ]
6121 foods that contain Fatty acids, total trans. List of these foods starting with the highest contents of Fatty acids, total trans and the lowest contents of Fatty acids, total trans
#### Gravels, Substances and Oils
CaribSea, Freshwater, Instant Aquarium, Rio Grande weighs 1 601.85 kg/m³ (100.00023 lb/ft³) with specific gravity of 1.60185 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Magnesium perchlorate hexahydrate [Mg(ClO4)2 ⋅ 6H2O] weighs 1 980 kg/m³ (123.60736 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-434A, liquid (R434A) with temperature in the range of -40°C (-40°F) to 60°C (140°F)
#### Weights and Measurements
short ton per square meter (short tn/m²) is a non-metric measurement unit of surface or areal density
Radioactivity is the act of emitting radiation spontaneously by an atomic nucleus that, for some reason, is unstable.
lb/cm² to lb/nm² conversion table, lb/cm² to lb/nm² unit converter or convert between all units of surface density measurement.
#### Calculators
Humidity calculations using temperature, relative humidity, pressure
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Next: Generating code Up: Interval evaluation in ALIAS-Maple Previous: Interval valuation and the Contents
# Interval evaluation of an expression in Maple
Maple provides the package evalr that calculate the interval evaluation of an expression. An example follows:
```
X:=INTERVAL(0..1):
evalr(sin(X)*cos(X)+X);
```
But this package does not evaluate correctly some expressions (for example X-X will be evaluated to 0 whatever is the interval for X). ALIAS-Maple provides the procedure Interval that allows for the interval evaluation of an expression as shown in the following example:
```
Interval([x*cos(y)+sin(x*y),x^2-cosh(y)],[x,y],[[-2,2],[-2,2]]);
[[[-3, 3]], [[-7.3890560989307,3.8646647167634]]]
```
The first argument is a list of expression, the second argument is the name of the variables and the third a list of intervals for the variables. The procedure generates a C++ program and hence may take some seconds. But as soon as the executable has been created it is possible to re-use it with other intervals using the Restart procedure
```
Restart("Interval",[[-0.5,0.5],[-0.5,0.5]]);
```
Note however that as soon as you have defined the `ALIAS/ID` string before generating an executable it is necessary to reset this string to the same name before using the Restart procedure to re-run the same executable
An optional 4th argument allows to control the evaluation. Indeed by default the expression will be transformed into a compact form (as provided by the procedure MinimalCout) that may lead to a better evaluation. The fourth argument may be:
• "AsIt": the expressions are not transformed and are evaluated as they are provided
• "Gradient": the derivatives of the expressions are used to try to improve the evaluation
Here is an example:
```
Interval([x*sin(x)+cos(x)],[x],[[0,1]]);
[[[.54030230586814, 1.8414709848079]]]
[[[1, 1.381773290676]]]
```
in which the use of the "Gradient" option allows to get the exact interval evaluation. It must be reminded that the interval evaluation of an expression is very sensitive to the manner with which the expression is written. For example the expression for in the interval [-1,1] is evaluated as:
```
Interval([x^2+2*x+1],[x],[[-1,1]],"AsIt");
[[[-1, 4]]]
```
If the option AsIt is not used ALIAS will convert the expression into the Horner form :
```
Interval([x^2+2*x+1],[x],[[-1,1]]);
[[[-2, 4]]]
```
Here it may be noticed that the Horner form leads to a worst interval evaluation. But we may factor this expression as :
```
Interval([(x+1)^2],[x],[[-1,1]],"AsIt");
[[[0, 4]]]
```
Next: Generating code Up: Interval evaluation in ALIAS-Maple Previous: Interval valuation and the Contents
Jean-Pierre Merlet 2012-12-20
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# Fundamental, derived quantities and unit questions and answers
#### 1. Name three important fundamental quantities
Answer: The three important fundamental quantities are Mass, Length, and time
#### 2. Why is Mass, length and Time most important fundamental quantities?
Answer: They are most important fundamental quantities because, the units of these quantities form base unit upon which the units of other quantities depend on.
### 3. What is the full meaning of S.I. Units?
Answer: The S.I.Units stand for system international units that is the universal accepted measuring units.
#### 4. What are the units of Luminous intensity and Amount of substance respectively?
Answer: They are candela (cd) and mole (mol) respectively
### 5. What are derived quantities?
Answer: The derived quantities are those quantities obtained by combination of units of fundamental quantities. Examples, the unit of volume is obtained by multiplying the unit of length, that is m x m x m =m3, density is also derived quantity is give by the ratio of mass and volume that is Kg/m3
#### 6. List the example of fundamental quantities and their units.
Answer: The examples of fundamental quantities are as follow:
(a).Length , the unit is meter (m)
(b).Mass, the unit is Kilogram (KG)
(C).Time, the unit is Second (s)
(d).Temperature, the unit is Kelvin (k)
(e). Electric current, the unit is Ampere
(f).Amount of substance, the unit is b Mole (mol)
(g). Luminous intensity, the unit is Candela (cd)
### 8. List 12 derived quantities and their units
Answer: (i) Area, the units is m2
(ii) Volume, the unit is m3
(ii). Density, the unit is Kg/m3
(iv). Speed, the unit is m/s
(v). Velocity, the unit is m/s
(vi). Acceleration, the unit is m/s2
(vii). Force, the unit is Newton (N)
(viii). Weight, the unit is Newton (N)
(ix) Momentum, the unit is Newton second (Ns)
(x). Pressure, the unit is Pascal (Pa) or N/m2
(xi) Power, the unit is watts (w) or N/m 0s J/s
(xii) Work/Energy, the unit is Joules (J) or Nm
### Have any Question or Comment?
#### 2 comments on “Fundamental, derived quantities and unit questions and answers”
###### Maelle
I like this site because it is interesting ans it has answered my questions with well established explanations thank you vert much. I recommend it to every body.
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