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## Saturday, 21 September 2013 ### Revision for End-of-Year Exam (Maths) Dear S1-07 With the end-of-year exam just a week away, most of you would have started revision. We shall allocate at least one period (i.e. the last period to answer to enquiries). To make the session really useful to you, we can do the following: 1. Go through the Exam Preparation Booklet - in particular the past year Test/ Exam papers. 2. Identify question(s) that you think would be helpful for class discussion. 3. You can post the question in the "Comment" of this post (e.g. 2012 Sec 1 Level Test 3 Q1). 4. I'll sort out the questions for discussion at the end of the week. 1. Ms Loh, There is a question in the Exam preparation booklet. It is in the topic of algebra. I'm not sure what the question number is, but it is between the 4th and the 10th question. It says that there is a wood plank 200cm, the person divided it into 44 pieces.The rods are either 4cm or 5cm long.Find the number of 4cm rods and 5 cm rods?May I know the method to find the answer?Because all I can do is guess and check and I think the method has something to do with algebra. 1. Taufiq Let me know the page number so that I can locate the question. 2. Page No. 9, Question 6 3. The first line tells us that there are 44 pieces. Since there are only 2 types of wood - either 5 cm long or 4 cm long. Let the number of pieces that are 4 cm long be x. Now the number of pieces that are 5 cm long will be 44-x. [agree?] Since the length of the two types of woods are given in 'cm'. It would be easier to convert the original "2 metres" be converted to "cm", which now becomes 200 cm. Now, if we piece all the pieces of wood together, what's the length made up by the 4 cm wood, and the length made up by the 5 cm wood? OK, at this stage, are you able to use the above information to form an equation to solve for x? :) 4. Ohhhhhh. I get it. 4x cm + 5(44 - x)cm = 200cm. 4x cm + 220 cm - 5x cm = 200cm 220 - x = 200 x = 20; 2. For the previous EOY, there is this question that has to do with polygons. I am able to complete it using quadratic equation. But is there a better way? It is question no. 9. 1. According to the question, there are 2 polygons. So, if regular polygon B has N sides, then the regular polygon A will have N+5 sides. Now, if interior angle of Polygon A is 3.6 degrees more than that of Polygon B For Polygon B, Let the interior angle be x, then exterior angle of 180-x For Polygon A, Interior angle of Polygon A = x + 3.6, then exterior angle = 180-(x+3.5) For Polygon A, size of each exterior angle = 360/(N+5) For Polygon B, size of each exterior angle = 360/N By equating the above, we can form 2 equations: For Polygon A, 180-(x+3.5) = 360/(N+5) For Polygon B, 180-x = 360/N Solve these 2 equations to find N and x. 3. Ms Loh, may I ask if you could help go through a bit of factorisation to refresh our memories, or mine at the very least. It will be much appreciated. 4. Ms Loh, for the Clementi town paper given. Question number 3 paper 1 : the answer given is x:y is 10:7 but i got my answer as x:y is 7:10	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://sst2013-s107maths.blogspot.com/2013/09/revision-for-eoy.html", "fetch_time": 1532214255000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-30/segments/1531676592861.86/warc/CC-MAIN-20180721223206-20180722003206-00131.warc.gz", "warc_record_offset": 335271822, "warc_record_length": 22818, "token_count": 876, "char_count": 3113, "score": 4.21875, "int_score": 4, "crawl": "CC-MAIN-2018-30", "snapshot_type": "longest", "language": "en", "language_score": 0.9423823356628418, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00037-of-00064.parquet"}
# Search by Topic #### Resources tagged with Making and proving conjectures similar to Isosceles Interactivity: Filter by: Content type: Stage: Challenge level: ### There are 35 results Broad Topics > Using, Applying and Reasoning about Mathematics > Making and proving conjectures ### Center Path ##### Stage: 3 and 4 Challenge Level: Four rods of equal length are hinged at their endpoints to form a rhombus. The diagonals meet at X. One edge is fixed, the opposite edge is allowed to move in the plane. Describe the locus of. . . . ### Polycircles ##### Stage: 4 Challenge Level: Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon? ##### Stage: 4 Challenge Level: The points P, Q, R and S are the midpoints of the edges of a non-convex quadrilateral.What do you notice about the quadrilateral PQRS and its area? ### Pericut ##### Stage: 4 and 5 Challenge Level: Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts? ##### Stage: 4 Challenge Level: Points D, E and F are on the the sides of triangle ABC. Circumcircles are drawn to the triangles ADE, BEF and CFD respectively. What do you notice about these three circumcircles? ##### Stage: 4 Challenge Level: The points P, Q, R and S are the midpoints of the edges of a convex quadrilateral. What do you notice about the quadrilateral PQRS as the convex quadrilateral changes? ### What's Possible? ##### Stage: 4 Challenge Level: Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make? ### Dice, Routes and Pathways ##### Stage: 1, 2 and 3 This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . . ### Triangles Within Squares ##### Stage: 4 Challenge Level: Can you find a rule which relates triangular numbers to square numbers? ### Exploring Quadratic Mappings ##### Stage: 4 Challenge Level: Explore the relationship between quadratic functions and their graphs. ### Multiplication Arithmagons ##### Stage: 4 Challenge Level: Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons? ### Curvy Areas ##### Stage: 4 Challenge Level: Have a go at creating these images based on circles. What do you notice about the areas of the different sections? ### Close to Triangular ##### Stage: 4 Challenge Level: Drawing a triangle is not always as easy as you might think! ### Triangles Within Triangles ##### Stage: 4 Challenge Level: Can you find a rule which connects consecutive triangular numbers? ### Alison's Mapping ##### Stage: 4 Challenge Level: Alison has created two mappings. Can you figure out what they do? What questions do they prompt you to ask? ### Triangles Within Pentagons ##### Stage: 4 Challenge Level: Show that all pentagonal numbers are one third of a triangular number. ### Rotating Triangle ##### Stage: 3 and 4 Challenge Level: What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle? ### Loopy ##### Stage: 4 Challenge Level: Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture? ### An Introduction to Magic Squares ##### Stage: 2, 3 and 4 Find out about Magic Squares in this article written for students. Why are they magic?! ### How Old Am I? ##### Stage: 4 Challenge Level: In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays? ### A Little Light Thinking ##### Stage: 4 Challenge Level: Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights? ### To Prove or Not to Prove ##### Stage: 4 and 5 A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples. ### Problem Solving, Using and Applying and Functional Mathematics ##### Stage: 1, 2, 3, 4 and 5 Challenge Level: Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information. ### Always a Multiple? ##### Stage: 3 Challenge Level: Think of a two digit number, reverse the digits, and add the numbers together. Something special happens... ### On the Importance of Pedantry ##### Stage: 3, 4 and 5 A introduction to how patterns can be deceiving, and what is and is not a proof. ### Few and Far Between? ##### Stage: 4 and 5 Challenge Level: Can you find some Pythagorean Triples where the two smaller numbers differ by 1? ### Janine's Conjecture ##### Stage: 4 Challenge Level: Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . . ### Charlie's Mapping ##### Stage: 3 Challenge Level: Charlie has created a mapping. Can you figure out what it does? What questions does it prompt you to ask? ### Helen's Conjecture ##### Stage: 3 Challenge Level: Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true? ### Multiplication Square ##### Stage: 4 Challenge Level: Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice? ### Epidemic Modelling ##### Stage: 4 and 5 Challenge Level: Use the computer to model an epidemic. Try out public health policies to control the spread of the epidemic, to minimise the number of sick days and deaths. ### DOTS Division ##### Stage: 4 Challenge Level: Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}. ### Happy Numbers ##### Stage: 3 Challenge Level: Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general. ### Consecutive Negative Numbers ##### Stage: 3 Challenge Level: Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers? ### Exploring Simple Mappings ##### Stage: 3 Challenge Level: Explore the relationship between simple linear functions and their graphs.	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://nrich.maths.org/public/leg.php?code=-93&cl=3&cldcmpid=2751&setlocale=en_US", "fetch_time": 1511306164000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-47/segments/1510934806438.11/warc/CC-MAIN-20171121223707-20171122003707-00482.warc.gz", "warc_record_offset": 672229675, "warc_record_length": 8203, "token_count": 1565, "char_count": 6881, "score": 4.46875, "int_score": 4, "crawl": "CC-MAIN-2017-47", "snapshot_type": "latest", "language": "en", "language_score": 0.8901476860046387, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00031-of-00064.parquet"}
## How is plate heat exchanger calculated? The total rate of heat transfer between the hot and cold fluids passing through a plate heat exchanger may be expressed as: Q = UA∆Tm where U is the Overall heat transfer coefficient, A is the total plate area, and ∆Tm is the Log mean temperature difference. How is heat transferred in a heat exchanger plate? With a plate heat exchanger, heat cuts through the surface and separates the hot medium from the cold. Thus, heating and cooling fluids and gases use minimal energy levels. The theory of heat transfer between mediums and fluids happens when: Heat is always transferred from a hot medium to a cold medium. ### Which equation is used for calculation of heat transfer in heat exchangers? Preliminary heat exchanger design to estimate the required heat exchanger surface area can be done using the basic heat exchanger equation, Q = U A ΔTlm, if values are known or can be estimated for Q, U and ΔTlm. Heat exchanger theory tells us that ΔTlm is the right average temperature difference to use. Which is not a plate type heat exchanger? Which of the following is not a Plate – Type heat exchanger? Explanation: Plate type heat exchangers can be classified as Gasketed, brazed and Welded (full or semi). Other categories are plate coils, spiral plate, etc. #### How do you calculate convective heat transfer? Common units used to measure the convective heat transfer coefficient are: 1. 1 W/(m2 K) = 0.85984 kcal/(h m2 ° C) = 0.1761 Btu/(ft2 h ° F) 2. 1 kcal/(h m2 ° C) = 1.163 W/(m2 K) = 0 What is the main advantage of plate heat exchangers? A significant benefit of the plate heat exchanger is that it is expandable, allowing an increase in heat transfer capability. As your heat transfer requirements change, you can simply add plates instead of buying an entire new frame unit, saving time and money. ## What is the advantages of plate type heat exchanger? Advantages of plate type heat exchanger is a Simple and Compact in size. Turbulent flow help to reduce deposits which heat transfer, No extra space is required for dismantling in Plate heat exchangers. Plate type heat exchanger is Maintenance simple and can be easily cleaned. How do you calculate heat transfer? Heat transfer can be defined as the process of transfer of heat from an object at a higher temperature to another object at a lower temperature. Therefore heat is the measure of kinetic energy possessed by the particles in a given system….Q=m \times c \times \Delta T. Q Heat transferred \Delta T Difference in temperature ### What are the limitations of plate and frame exchangers? Not good for large fluid temperature differences – A flat plate heat exchanger does not work as well as a shell and tube heat exchanger for cases where there is a large temperature difference between the two fluids. What is the function of plate heat exchanger? A plate heat exchanger is a type of heat exchanger that uses metal plates to transfer heat between two fluids. This has a major advantage over a conventional heat exchanger in that the fluids are exposed to a much larger surface area because the fluids are spread out over the plates. 3. #### What type of heat exchanger is more efficient? Counter flow heat exchangers are the most efficient type of heat exchanger. Counter Flow Heat Exchanger. Cross Flow. Cross flow heat exchangers have one medium flowing that flows perpendicular (at 90°) across the other. Cross flow heat exchangers are usually found in applications where one of the fluids changes state (2-phase flow). What is a plate heat exchanger used for? An individual plate for a heat exchanger. The plate heat exchanger ( PHE ) is a specialized design well suited to transferring heat between medium- and low-pressure fluids. Welded , semi-welded and brazed heat exchangers are used for heat exchange between high-pressure fluids or where a more compact product is required. ## What is plate and frame heat exchanger? Plate and frame heat exchangers are made of corrugated plates on a frame. This design creates high turbulence and high wall shear stress, both of which lead to a high heat transfer coefficient and a high fouling resistance. 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# Kanchan dyes dresses. She had todye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished? Given: Kanchan dyes dresses. She had todye 30 dresses. She has so far finished 20 dresses. To do: We have to find what fraction of dresses she has finished. Solution: Total number of dresses Kanchan has to dye $= 30$ dresses Number of dresses she has finished $= 20$ dresses Therefore, The fraction of dresses she has finished $=\frac{20}{30}$ $=\frac{2}{3}$ The required fraction is $\frac{2}{3}$. Tutorialspoint Simply Easy Learning Updated on: 10-Oct-2022 27 Views	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.tutorialspoint.com/kanchan-dyes-dresses-she-had-todye-30-dresses-she-has-so-far-finished-20-dresses-what-fraction-of-dresses-has-she-finished", "fetch_time": 1714053466000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-18/segments/1712297295329.99/warc/CC-MAIN-20240425130216-20240425160216-00811.warc.gz", "warc_record_offset": 928783985, "warc_record_length": 24052, "token_count": 166, "char_count": 611, "score": 4.40625, "int_score": 4, "crawl": "CC-MAIN-2024-18", "snapshot_type": "latest", "language": "en", "language_score": 0.9433033466339111, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00057-of-00064.parquet"}
# Search by Topic #### Resources tagged with Place value similar to Domino Number Patterns: Filter by: Content type: Age range: Challenge level: ### There are 48 results Broad Topics > Numbers and the Number System > Place value ### One of Thirty-six ##### Age 5 to 7 Challenge Level: Can you find the chosen number from the grid using the clues? ### What Number? ##### Age 5 to 7 Short Challenge Level: I am less than 25. My ones digit is twice my tens digit. My digits add up to an even number. ### Our Numbers ##### Age 5 to 7 Challenge Level: These spinners will give you the tens and unit digits of a number. Can you choose sets of numbers to collect so that you spin six numbers belonging to your sets in as few spins as possible? ### Light the Lights ##### Age 5 to 7 Challenge Level: Investigate which numbers make these lights come on. What is the smallest number you can find that lights up all the lights? ### Being Resourceful - Primary Number ##### Age 5 to 11 Challenge Level: Number problems at primary level that require careful consideration. ### Multiply Multiples 2 ##### Age 7 to 11 Challenge Level: Can you work out some different ways to balance this equation? ### Number Detective ##### Age 5 to 11 Challenge Level: Follow the clues to find the mystery number. ### The Thousands Game ##### Age 7 to 11 Challenge Level: Each child in Class 3 took four numbers out of the bag. Who had made the highest even number? ### Multiply Multiples 1 ##### Age 7 to 11 Challenge Level: Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it? ### Two Spinners ##### Age 5 to 7 Challenge Level: What two-digit numbers can you make with these two dice? What can't you make? ### Trebling ##### Age 7 to 11 Challenge Level: Can you replace the letters with numbers? Is there only one solution in each case? ### Six Is the Sum ##### Age 7 to 11 Challenge Level: What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros? ### Napier's Bones ##### Age 7 to 11 Challenge Level: The Scot, John Napier, invented these strips about 400 years ago to help calculate multiplication and division. Can you work out how to use Napier's bones to find the answer to these multiplications? ### Multiply Multiples 3 ##### Age 7 to 11 Challenge Level: Have a go at balancing this equation. Can you find different ways of doing it? ### All the Digits ##### Age 7 to 11 Challenge Level: This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures? ### ABC ##### Age 7 to 11 Challenge Level: In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication? ### What Do You Need? ##### Age 7 to 11 Challenge Level: Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number? ### Diagonal Sums ##### Age 7 to 11 Challenge Level: In this 100 square, look at the green square which contains the numbers 2, 3, 12 and 13. What is the sum of the numbers that are diagonally opposite each other? What do you notice? ### Becky's Number Plumber ##### Age 7 to 11 Challenge Level: Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings? ### Coded Hundred Square ##### Age 7 to 11 Challenge Level: This 100 square jigsaw is written in code. It starts with 1 and ends with 100. Can you build it up? ### Being Curious - Primary Number ##### Age 5 to 11 Challenge Level: Number problems for inquiring primary learners. ### Being Collaborative - Primary Number ##### Age 5 to 11 Challenge Level: Number problems at primary level to work on with others. ### Being Resilient - Primary Number ##### Age 5 to 11 Challenge Level: Number problems at primary level that may require resilience. ### Song Book ##### Age 7 to 11 Challenge Level: A school song book contains 700 songs. The numbers of the songs are displayed by combining special small single-digit cards. What is the minimum number of small cards that is needed? ### One Million to Seven ##### Age 7 to 11 Challenge Level: Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like? ### Nice or Nasty for Two ##### Age 7 to 14 Challenge Level: Some Games That May Be Nice or Nasty for an adult and child. Use your knowledge of place value to beat your opponent. ### Calculator Bingo ##### Age 7 to 11 Challenge Level: A game to be played against the computer, or in groups. Pick a 7-digit number. A random digit is generated. What must you subract to remove the digit from your number? the first to zero wins. ### Dicey Operations in Line for Two ##### Age 7 to 11 Challenge Level: Dicey Operations for an adult and child. Can you get close to 1000 than your partner? ### Spell by Numbers ##### Age 7 to 11 Challenge Level: Can you substitute numbers for the letters in these sums? ### Which Is Quicker? ##### Age 7 to 11 Challenge Level: Which is quicker, counting up to 30 in ones or counting up to 300 in tens? Why? ### Which Scripts? ##### Age 7 to 11 Challenge Level: There are six numbers written in five different scripts. Can you sort out which is which? ### Round the Three Dice ##### Age 7 to 11 Challenge Level: What happens when you round these three-digit numbers to the nearest 100? ### Writing Digits ##### Age 5 to 7 Challenge Level: Lee was writing all the counting numbers from 1 to 20. She stopped for a rest after writing seventeen digits. What was the last number she wrote? ### Subtraction Surprise ##### Age 7 to 14 Challenge Level: Try out some calculations. Are you surprised by the results? ### Round the Dice Decimals 2 ##### Age 7 to 11 Challenge Level: What happens when you round these numbers to the nearest whole number? ### Round the Dice Decimals 1 ##### Age 7 to 11 Challenge Level: Use two dice to generate two numbers with one decimal place. What happens when you round these numbers to the nearest whole number? ##### Age 5 to 11 Challenge Level: Try out this number trick. What happens with different starting numbers? What do you notice? ##### Age 5 to 11 Challenge Level: Who said that adding couldn't be fun? ### Number Sense Series: A Sense of 'ten' and Place Value ##### Age 5 to 7 Once a basic number sense has developed for numbers up to ten, a strong 'sense of ten' needs to be developed as a foundation for both place value and mental calculations. ### Alien Counting ##### Age 7 to 11 Challenge Level: Investigate the different ways these aliens count in this challenge. You could start by thinking about how each of them would write our number 7. ### Oddly ##### Age 7 to 11 Challenge Level: Find the sum of all three-digit numbers each of whose digits is odd. ### Reach 100 ##### Age 7 to 14 Challenge Level: Choose four different digits from 1-9 and put one in each box so that the resulting four two-digit numbers add to a total of 100. ##### Age 7 to 14 Challenge Level: Watch our videos of multiplication methods that you may not have met before. Can you make sense of them? ### The Number Jumbler ##### Age 7 to 14 Challenge Level: The Number Jumbler can always work out your chosen symbol. Can you work out how? ### Path of Discovery Series 3: I Do and I Understand ##### Age 5 to 7 Marion Bond recommends that children should be allowed to use 'apparatus', so that they can physically handle the numbers involved in their calculations, for longer, or across a wider ability band,. . . . ### Pupils' Recording or Pupils Recording ##### Age 5 to 14 This article, written for teachers, looks at the different kinds of recordings encountered in Primary Mathematics lessons and the importance of not jumping to conclusions! ### A Story about Absolutely Nothing ##### Age 7 to 18 This article for the young and old talks about the origins of our number system and the important role zero has to play in it.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://nrich.maths.org/public/topic.php?code=6&cl=1&cldcmpid=225", "fetch_time": 1571087878000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-43/segments/1570986655310.17/warc/CC-MAIN-20191014200522-20191014224022-00145.warc.gz", "warc_record_offset": 643597803, "warc_record_length": 8858, "token_count": 1932, "char_count": 8391, "score": 3.75, "int_score": 4, "crawl": "CC-MAIN-2019-43", "snapshot_type": "latest", "language": "en", "language_score": 0.8461540937423706, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00063-of-00064.parquet"}
# Counterfeit Coin [Puzzle] There are 12 coins, one of which is a counterfeit. The false coin differs in weight from the true ones, but you don’t know whether it’s heavier or lighter. Find the counterfeit coin using three weighings in a pan balance. [From Futility Closet] Advertisements Posted in Puzzles Tags: , # Two Prizes [Puzzle] Suppose I offer you two prizes: prize A and prize B. You are to make a statement – any statement you like. If the statement is clearly true, I will give you a prize. You will win either A or B, I am not saying which one. And if your statement is false, you won’t get any prize. It’s clear from this that you want to make a statement that is clearly true, because otherwise you won’t win any prize. You can say something like “One plus one equals two.” This is clearly true, and you will win either A or B. However, let’s say you really would like to win prize A. The question is: what statement can you make that will ensure that you will win prize A? Posted in Puzzles Tags: , # Wise Men with Hats [Puzzles] The first one is an old classic hat puzzle; I’ve put it here as a warm-up for the second one which is a tad more challenging. Puzzle #1 Three wise men are told to stand in a straight line, one in front of the other. A hat is put on each of their heads. They are told that each of these hats was selected from a group of five hats: two black hats and three white hats. The first man, standing at the front of the line, can’t see either of the men behind him or their hats. The second man, in the middle, can see only the first man and his hat. The last man, at the rear, can see both other men and their hats. None of the men can see the hat on his own head. They are asked to deduce its color. Some time goes by as the wise men ponder the puzzle in silence. Finally the first one, at the front of the line, makes an announcement: “My hat is white.” He is correct. How did he come to this conclusion? Puzzle #2 One hundred persons will be lined up single file, facing north. Each person will be assigned either a red hat or a blue hat. No one can see the color of his or her own hat. However, each person is able to see the color of the hat worn by every person in front of him or her. That is, for example, the last person in line can see the color of the hat on 99 persons in front of him or her; and the first person, who is at the front of the line, cannot see the color of any hat. Beginning with the last person in line, and then moving to the 99th person, the 98th, etc., each will be asked to name the color of his or her own hat. If the color is correctly named, the person lives; if incorrectly named, the person is shot dead on the spot. Everyone in line is able to hear every response as well as hear the gunshot; also, everyone in line is able to remember all that needs to be remembered and is able to compute all that needs to be computed. Before being lined up, the 100 persons are allowed to discuss strategy, with an eye toward developing a plan that will allow as many of them as possible to name the correct color of his or her own hat (and thus survive). They know all of the preceding information in this problem. Once lined up, each person is allowed only to say “Red” or “Blue” when his or her turn arrives, beginning with the last person in line. Your assignment: Develop a plan that allows as many people as possible to survive. [Source: TienrneyLab ] I’ll provide answers in the Comments section next week. Posted in Puzzles Tags: , ### Subscribe via Email Enter your email address to subscribe to this blog and receive notifications of new posts by email. Join 66 other followers	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://vishal12.wordpress.com/tag/logic-puzzle/", "fetch_time": 1531972824000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-30/segments/1531676590493.28/warc/CC-MAIN-20180719031742-20180719051742-00384.warc.gz", "warc_record_offset": 787301687, "warc_record_length": 11800, "token_count": 864, "char_count": 3679, "score": 3.609375, "int_score": 4, "crawl": "CC-MAIN-2018-30", "snapshot_type": "latest", "language": "en", "language_score": 0.9688116312026978, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00035-of-00064.parquet"}
View more editions # TEXTBOOK SOLUTIONS FOR College Physics A Strategic Approach with MasteringPhysics 2nd Edition • 3338 step-by-step solutions • Solved by publishers, professors & experts • iOS, Android, & web Over 90% of students who use Chegg Study report better grades. May 2015 Survey of Chegg Study Users Chapter: Problem: Figure Q7.3 shows four pulleys, each with a heavy and a light block strung over it. The blocksâ€™ velocities are shown. What are the signs (+ or -) of the angular velocity and angular acceleration of the pulley in each case? SAMPLE SOLUTION Chapter: Problem: • Step 1 of 2 An object in a rotatory motion, the angular velocity is equal to change in angular displacement with respect to the time. If the change in angular displacement is in counter clockwise direction from the positive x-axis, the angular velocity is positive and in clockwise direction from the positive x-axis is negative. The angular acceleration is defined as the rate of change in angular velocity. The angular accelerationis positive: 1. Rotating counter clock wise and speeding up 2. Rotating clock wise and slowing down. The angular accelerationis negative: 1. Rotating counter-clock wise and slowing down 2. Rotating clock wise and speeding up. • Step 2 of 2 (a) As the lighter mass is moving downwards and heavier mass is moving upwards, the pulley rotation is clock wise and slowing down. Thus angular velocity, is negative and angular acceleration is positive. (b) As the heavier mass is moving downwards and lighter mass is moving upwards, the pulley rotation is counter-clock wise and speeding up. So, both angular velocity, and angular acceleration , are positive. (c) As the heavier mass is moving downwards and lighter mass is moving upwards, the pulley rotation is clock wise and speeding up. So, both angular velocity, and angular acceleration, are negative. (d) In this case, the rotation is counter clock wise and is slowing down. So, angular velocity, is positive and angular acceleration is negative. Corresponding Textbook College Physics A Strategic Approach with MasteringPhysics \| 2nd Edition 9780321595485ISBN-13: 0321595483ISBN: Authors: Alternate ISBN: 9780321596079, 9780321596291, 9780321596307, 9780321696304, 9780321737014, 9780321738592, 9780321777836	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.chegg.com/homework-help/college-physics-strategic-approach-with-masteringphysics-2nd-edition-chapter-7-solutions-9780321595485", "fetch_time": 1469592172000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2016-30/segments/1469257825365.1/warc/CC-MAIN-20160723071025-00083-ip-10-185-27-174.ec2.internal.warc.gz", "warc_record_offset": 349240380, "warc_record_length": 20238, "token_count": 531, "char_count": 2304, "score": 3.9375, "int_score": 4, "crawl": "CC-MAIN-2016-30", "snapshot_type": "latest", "language": "en", "language_score": 0.8918448090553284, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00009-of-00064.parquet"}
## The gamma function The gamma function is one of the classical functions of applied mathematics; here I will provide a bare bones introduction to it (see Connections for places to go learn more). You should think of it in the same way that you think about sin, cos, exp, and log. First, these functions have a specific mathematical definition. Second, there are known rules that relate functions with different arguments (such as the rule for computing sin(a + b)) and there are computational means for obtaining their values. Third, these functions are tabulated (in the old days, in tables of books, and in the modern days in many software packages or on the web). The same applies to the gamma function, which is defined for z > 0 by r(z) = In this expression, z can take any positive value, but let us start with the integers. In fact, let us start with z = 1, so that we consider r(1) = J0°° e-sds = 1. What about z = 2? In that case r(2) = J0°° se—sds, which can be integrated by parts and we find T(2) = 1. We shall do one more, before the general case: r(3) = J0°° s2e-sds, which can be integrated by parts once again and from which we will see that r(3) = 2. If you do a few more, you should get a sense of the pattern: for integer values of z, r(z) = (z — 1)!. Note, then, that we could write the binomial coefficient in Eq. (3.49) as n — 1 \ (n — 1)! r(n) For non-integer values of z, the same kind of integration by parts approach works and leads us to an iterative equation for the gamma function, which is 0 0	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.ecologycenter.us/population-size/the-gamma-function.html", "fetch_time": 1560646920000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-26/segments/1560627997508.21/warc/CC-MAIN-20190616002634-20190616024634-00378.warc.gz", "warc_record_offset": 747809097, "warc_record_length": 6475, "token_count": 402, "char_count": 1530, "score": 3.65625, "int_score": 4, "crawl": "CC-MAIN-2019-26", "snapshot_type": "latest", "language": "en", "language_score": 0.9306890964508057, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00015-of-00064.parquet"}
# Question about something reaching exactly + • MaZnFLiP In summary, the conversation discusses a question about reaching exactly +2G acceleration on a turn and the difference between tangential and normal acceleration. It is explained that normal (centripetal) acceleration can be calculated using the formula v^2/r, where v is velocity and r is the radius of the turn. It is then clarified that g is a magnitude and not a direction. MaZnFLiP [SOLVED] Question about something reaching exactly + ## Homework Statement Sorry, I accidentally hit the enter button. Anyways my question is about something reaching exactly +2G acceleration on a turn. I tried asking my friends and teacher about it but it didn't exactly make sense the way they explained it. I just want to know what that means to have a +2g acceleration when turning Last edited: Tangential or normal? I think you'll have to explain more about what your friends and teachers explained and express that in the form of a question. Last edited: I think normal. Normal (centripetal) acceleration is given by v^2/r, where r is the radius of the turn and v is the velocity. If that's equal to 2g, then 2g=v^2/r. Oh okay. That makes sense now. So that would mean that 2(-9.81m/s^2) = v^2/r right? Right. Except I wouldn't put a minus sign on g in this context. g is just a magnitude, not a direction. Oh okay. Thanks so much! ## 1. What does "exactly +" mean in a scientific context? In science, "exactly +" refers to a specific value or measurement that has been measured or observed with precision. The "+" sign indicates that the value is positive, meaning it is greater than zero. ## 2. Can something ever reach exactly +, or is it just an ideal concept? In theory, something can reach exactly + if it is measured with perfect accuracy. However, in reality, there is always some margin of error in measurements, so it is not always possible to have an exact value of +. ## 3. How is the concept of "exactly +" important in scientific experiments and research? The concept of "exactly +" is important in scientific experiments and research because it allows for precise and accurate measurements, which are essential for drawing valid conclusions and making scientific discoveries. ## 4. Is it possible for something to reach exactly + and then change to a different value? Yes, it is possible for something to reach exactly + and then change to a different value. This can happen due to various factors such as external influences, natural processes, or experimental errors. ## 5. How do scientists ensure that their measurements are as close to exactly + as possible? Scientists use various techniques and tools such as precise instruments, controlled experiments, and statistical analysis to minimize errors and obtain measurements that are as close to exactly + as possible. Replies 2 Views 2K Replies 3 Views 1K Replies 6 Views 1K Replies 7 Views 1K Replies 2 Views 501 Replies 5 Views 1K Replies 5 Views 2K Replies 6 Views 2K Replies 9 Views 423 Replies 1 Views 1K	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/question-about-something-reaching-exactly.213895/", "fetch_time": 1725969562000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651241.17/warc/CC-MAIN-20240910093422-20240910123422-00565.warc.gz", "warc_record_offset": 890456670, "warc_record_length": 17915, "token_count": 716, "char_count": 3049, "score": 3.8125, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.9569169282913208, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00001-of-00064.parquet"}
Join us for MBA Spotlight – The Top 20 MBA Fair Schedule of Events \| Register It is currently 05 Jun 2020, 12:11 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # A cryptographer has intercepted an enemy message that is in code. He k Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 64314 A cryptographer has intercepted an enemy message that is in code. He k [#permalink] ### Show Tags 03 Aug 2019, 23:11 00:00 Difficulty: 75% (hard) Question Stats: 41% (01:27) correct 59% (01:27) wrong based on 73 sessions ### HideShow timer Statistics A cryptographer has intercepted an enemy message that is in code. He knows that the code is a simple substitution of numbers for letters. Which of the following would be the least helpful in breaking the code? (A) Knowing the frequency with which the vowels of the language are used (B) Knowing the frequency with which two vowels appear together in the language (C) Knowing the frequency with which odd numbers appear relative to even numbers in the message (D) Knowing the conjugation of the verb to be in the language on which the code is based (E) Knowing every word in the language that begins with the letter R _________________ Manager Joined: 05 Dec 2014 Posts: 227 Location: India GMAT 1: 660 Q49 V31 GPA: 3.54 Re: A cryptographer has intercepted an enemy message that is in code. He k [#permalink] ### Show Tags 05 Aug 2019, 05:22 Can someone kindly provide the reasoning? Manager Joined: 23 Feb 2019 Posts: 53 Re: A cryptographer has intercepted an enemy message that is in code. He k [#permalink] ### Show Tags 05 Aug 2019, 06:05 Because the message is coded in numbers, C does not give the operator any more information than he can already access. Is that it? GMAT Club team member Status: GMAT Club Team Member Affiliations: GMAT Club Joined: 02 Nov 2016 Posts: 5976 GPA: 3.62 Re: A cryptographer has intercepted an enemy message that is in code. He k [#permalink] ### Show Tags 09 Jan 2020, 09:58 Official Explanation To break the code, the cryptographer needs information about the language that the code conceals. (A), (B), (D), and (E) all provide such information. (C), however, says nothing about the underlying language. The code could even use all even or all odd numbers for the symbol substitutions without affecting the information to be encoded. Hope it helps _________________ Re: A cryptographer has intercepted an enemy message that is in code. He k [#permalink] 09 Jan 2020, 09:58	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://gmatclub.com/forum/a-cryptographer-has-intercepted-an-enemy-message-that-is-in-code-he-k-301863.html", "fetch_time": 1591387889000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-24/segments/1590348502204.93/warc/CC-MAIN-20200605174158-20200605204158-00573.warc.gz", "warc_record_offset": 363245835, "warc_record_length": 159073, "token_count": 739, "char_count": 2948, "score": 3.65625, "int_score": 4, "crawl": "CC-MAIN-2020-24", "snapshot_type": "longest", "language": "en", "language_score": 0.9028210639953613, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00002-of-00064.parquet"}
• 0 Vote(s) - 0 Average • 1 • 2 • 3 • 4 • 5 closed form for regular superfunction expressed as a periodic function mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 08/31/2010, 08:04 AM (This post was last modified: 08/31/2010, 08:32 AM by mike3.) (08/31/2010, 06:51 AM)mike3 Wrote: The product of exp is the exponential of a formal power series. This can be expressed using the Bell polynomials: $\prod_{n=0}^{\infty} \exp(a_n y^n) = \exp\left(\sum_{n=0}^{\infty} a_n y^n\right) = \exp(a_0) \exp\left(\sum_{n=1}^{\infty} a_n y^n\right)$. This then becomes $\exp(a_0) \exp\left(\sum_{n=1}^{\infty} a_n y^n\right) = \exp(a_0) \sum_{n=1}^{\infty} \frac{\sum_{k=1}^n B_{n,k}(1! a_1, ..., (n-k+1)! a_{n-k+1})}{n!} y^n = \exp(a_0) \sum_{n=1}^{\infty} \frac{B_n(1! a_1, ..., n! a_n)}{n!} y^n$. Thus the equations to solve are $a_n L^n = \exp(a_0) \frac{B_n(1! a_1, ..., n! a_n)}{n!}$. Since $a_0 = L$ and $a_1 = 1$, this is $a_n L^n = L \frac{B_n(1, 2! a_2, ..., n! a_n)}{n!}$. This is derived from Faà di Bruno's formula, see http://en.wikipedia.org/wiki/Fa%C3%A0_di...7s_formula for details. Doing some tests, it appears that $B_n(1, 2! a_2, ..., n! a_n)$ has only one occurrence of $n! a_n$, and no higher powers of it, and it never seems to be multiplied by any sort of n-dependent coefficient. This means that $B_n(1, a_2, ..., a_n) - n! a_n = B_n(1, 2! a_2, ..., (n-1)! a_{n-1}, 0)$. I don't have a proof at this point, but we can then solve this: $a_n n! L^n = L B_n(1, 2! a_2, ..., n! a_n)$ $a_n n! L^{n-1} = B_n(1, 2! a_2, ..., n! a_n)$ $a_n n! L^{n-1} - n! a_n = B_n(1, 2! a_2, 3! a_3, ..., (n-1)! a_{n-1}, 0)$ $n! (L^{n-1} - 1) a_n = B_n(1, 2! a_2, 3! a_3, ..., (n-1)! a_{n-1}, 0)$ $a_n = \frac{B_n(1, 2! a_2, 3! a_3, ..., (n-1)! a_{n-1}, 0)}{n! (L^{n-1} - 1)}$. And this is the recurrent formula for the general coefficients. Together with $a_0 = L$ and $a_1 = 1$, this completes the as-close-to-explicit-as-possible-so-far formula for the coefficients of the Fourier series for the regular iteration. EDIT: posts corrected to include factorials on terms $a_n$ in Bell polynomials sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 08/31/2010, 02:37 PM (08/31/2010, 07:08 AM)Gottfried Wrote: Hi Sheldon - I recognize your coefficients....Gottfried, thanks for your reply. I'm trying to make sense of the coefficients, and would probably need to program the matrix into pari-gp to verify it. Earlier, Gottfried wrote: Quote:Hi Sheldon - just to allow me to follow (think I can't involve much) - I don't have a clue from where this is coming, what, for instance, is L at all? I think you've explained it elsewhere before but don't see it at the moment... Would you mind to reexplain in short or to provide the link? Gottfried L is the fixed point of base(e). If you iterate the natural logarithm function hundreds of times (say starting with z=0.5), you get a very good approximation of L. Then the equations describe the regular superfunction for base e, which is complex valued at the real axis, but is analytic and entire. The rest of what I did basically, it comes down to expressing the regular superfunction as an analytic Fourier series, where all of the terms of the series decay to zero at iinfinity. Such a series is guaranteed to be analytic. Here is an example of a generalized view of such a Fourier series, not related to the superfunction, with a period of 2Pi. $f(z) = \sum_{n=0}^{\infty}a_ne^{inz}$ A full Fourier series would also have the negative coefficients as well, and is often not analytic. $f(z) = \sum_{n=-\infty}^{\infty}a_ne^{inz}$ Both of these can be wrapped around the unit circle, with the substitution $y=e^{iz}$. In the analytic case, we have an analytic Taylor series. $f(z) = \sum_{n=0}^{\infty}a_ny^n$. In the general case, with terms from -infinty to +infinty, we have a Laurent series, with singularities inside the unit circle, and an annular ring of convergence. For the full Fourier series, often, the Laurent series only converges on the edge of the unit circle. Hope that helps. - Sheldon sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 08/31/2010, 02:41 PM Mike, Thanks a lot for your links and your post! I'll have to program it into pari-gp, and verify that it matches the discreet Fourier series terms. Next goal; rewrite the equations for other bases (except $\eta$), especially for basespc dFac(1,5)Mat(VE(pc,5) ) \\ show the coefficients scaled by factorials in window "pc" \\ result: \\ 1 \\ a1 \\ a1^2+2a2 \\ a1^3+6a2a1+6a3 \\ a1^4+12a2a1^2+24a3a1+(12a2^2+24a4) \\ now begin to substitute a1 = 1 pc1=subst(VE(pc,6),'a1,a1) %box >pc dFac(1,6)Mat(pc1 ) \\ show the coefficients scaled by factorials \\ 1 \\ 1 \\ 2a2+1 \\ 6a2+(6a3+1) \\ 12a2^2+12a2+(24a3+(24a4+1)) a2 = 1/2/(L-1) pc2=subst(VE(pc1,6),'a2,a2) %box >pc dFac(1,4)Mat(VE(pc2,4) ) \\ show only 4 rows 1 1 L/(L-1) ((6a3+1)L+(-6a3+2))/(L-1) a3 = (1/6 + a2)/(L^2-1) pc3=subst(VE(pc2,6),'a3,a3) %box >pc dFac(1,4)Mat(VE(pc3,4) ) a4=(1/24 + a3 + a2/2 + a2^2/2)/(L^3-1) pc4=subst(VE(pc3,6),'a4, a4) %box >pc dFac(1,5)Mat(VE(pc4,5) ) a5=(1/5! + a4+a3/2+a3a2+a2^2/2+a2/3!)/(L^4-1) pc5=subst(VE(pc4,6),'a5, a5) %box >pc dFac(1,6)Mat(VE(pc5,6) ) 1 \\ 1 \\ L/(L-1) \\ (L^3+2L^2)/(L^3-L^2-L+1) \\ (L^6+5L^5+6L^4+6L^3)/(L^6-L^5-L^4+L^2+L-1) \\ (L^10+9L^9+24L^8+40L^7+46L^6+36L^5+24L^4)/(L^10-L^9-L^8+2L^5-L^2-L+1) \\ see the coefficients at powers of L in the numerators equal that matrix-entries, for instance last row(revert the order and begin at L^4: [24 36 46 40 24 9 1] \\ denominators can be factored by (L-1),(L^2-1) and so on` Thanks also for the other explanations. It surely helps to have an entry for my own understanding of the procedure once I've got the whole idea behind... So far - Gottfried Gottfried Helms, Kassel mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 08/31/2010, 07:26 PM (This post was last modified: 08/31/2010, 07:51 PM by mike3.) Formulas for the Bell polynomials can be found here: http://en.wikipedia.org/wiki/Bell_polynomials and here: http://mathworld.wolfram.com/BellPolynomial.html There's a method involving taking a matrix determinant, a convolution formula, and a sum over partitions (we are only interested in the "complete" Bell polynomials $B_n$ here). E.g. $(x \diam y)_n = \sum_{j=1}^{n-1} {n \choose j} x_j y_{n-j}$ then $B_{n,k}(x_1, ..., x_{n-k+1}) = \frac{x_n^{k \diam}}{k!}$ (where $x^{k \diam}$ is like "exponentiating" the sequence with the "diamond" operator) and $B_n(x_1, ..., x_n) = \sum_{k=1}^n B_{n,k}(x_1, ..., x_{n-k+1})$ mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 08/31/2010, 07:34 PM (This post was last modified: 08/31/2010, 07:39 PM by mike3.) For the variable-base case, $a_n (L \log(B))^n = L \frac{B_n(\log(B), \log(B) 2! a_2, ..., \log(B) n! a_n)}{n!}$ thus $a_n L^{n-1} \log(B)^n n! = B_n(\log(B), \log(B) 2! a_2, ..., \log(B) n! a_n)$ $a_n = \frac{B_n(\log(B), \log(B) 2! a_2, ..., \log(B) (n-1)! a_{n-1}, 0)}{(L^{n-1} \log(B)^n - \log(B)) n!}$. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 08/31/2010, 08:23 PM Now all we need is some way to express the Riemann mapping sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 08/31/2010, 08:33 PM (08/31/2010, 07:34 PM)mike3 Wrote: For the variable-base case, $a_n (L \log(B))^n = L \frac{B_n(\log(B), \log(B) 2! a_2, ..., \log(B) n! a_n)}{n!}$ thus $a_n L^{n-1} \log(B)^n n! = B_n(\log(B), \log(B) 2! a_2, ..., \log(B) n! a_n)$ $a_n = \frac{B_n(\log(B), \log(B) 2! a_2, ..., \log(B) (n-1)! a_{n-1}, 0)}{(L^{n-1} \log(B)^n - \log(B)) n!}$.Wow! Thanks Mike! Only matrices and statistics, "n choose k" formulas is one of my weak areas, so it might be a few days before I catch up. I'm kind of intrigued at actually having a closed form for a real valued superfunction, like sqrt(2). I'm guessing that the number of terms required for convergence gets extremely large as z increases. From the terms I computed a couple of days ago, it looks like the terms are decreasing exponentially, which means the series acts like it has a singularity. However, since the regular superfunction is entire, we must have convergence to infinity, so the terms must eventually decrease faster than exponentially. - Sheldon Gottfried Ultimate Fellow Posts: 767 Threads: 119 Joined: Aug 2007 08/31/2010, 08:34 PM (This post was last modified: 08/31/2010, 08:40 PM by Gottfried.) Please excuse - this post contains some s But it's worth to notice how the bell-polynomials and my -for most fellows here: still cryptic - matrix-method are related. That I didn't see this earlier ----------------- In the mathworld-wolfram-link I find a short description of the Bell-polynomials first kind, fortunately with some example. $\sum_{k=0}^{\infty} \frac{B_k(x)}{k!}t^k = e^{(e^t-1)x}$ This expressed with my matrix-formulae is V(t)~ dF(-1)S2 V(x) = exp( (exp(t)-1)x) or using my standard-matrix for x->exp(x)-1 plus one intermediate step V(t) ~ fS2F = V(exp(t)-1)~ write "y" for "exp(t)-1" V(y)~ * dF(-1) * V(x) = F(-1)~* dV(y)V(x) = F(-1)~ V( yx) = exp( yx) = exp( (exp(t)-1)x) Or in one expression: V(t) ~ (fS2F * dF(-1)) * V(x) = exp( (exp(t)-1)x) (which I can recognize immediately to be correct because I'm extremely used to that notation) or the even simpler definition of the vector of Bell-polynomials: B(x) = S2V(x) So now I see at least, how the Bell-polynomials of the first kind are related to my matrix-lingo. (So I should rename fS2F into "Bell" perhaps...) ---------------------------------- But then follows the Bell-polynomials of second kind. And there I'm lost again... No example, no redundancy... as if the reader could not do some error in parsing a complex formula... Is there possibly meant the iteration of the Bell-polynomials (=matrix-power of S2)? That would be simple then... Or, wait - looking at the subscripted "x" I think, that they are now coefficients for some arbitrary function developed as a powerseries with reciprocal factorials, ... and then... Well, this seems to be just the definition of what I found out myself at the very beginning of my fiddling with this subject and called it a "matrixoperator" for some function. With the additional property, that it is (lower) triangular (because of the missing x0 in the formula in mathworld), so for instance all my U_t-matrices for the decremented exponentiation (or "U-tetration to base t" in my early speak) were such collections of Bell-polynomials of second kind (but also the schröder-matrices, just all my lower-triangular matrix-operators which I worked with the last years and have the reciprocal factorial scaling ) So now - if we talk about the Bell-polynomials (first or second kind) and the symbolic representation in terms of the log(L) (or log(t)) then this is just what I solved in my earlier posted link (good to know)! Here it is again: http://go.helms-net.de/math/tetdocs/APT.htm . And Faa di Bruno/Bell-polynomials and "matrixoperators" is the same and one needs only read about one side of these notations... Amen - Gottfried Gottfried Helms, Kassel « Next Oldest \| Next Newest » Possibly Related Threads... 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This site is supported by donations to The OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A272874 Decimal expansion of the infinite nested radical sqrt(-1+sqrt(1+sqrt(-1+sqrt(1+ ... ))). 3 4, 5, 3, 3, 9, 7, 6, 5, 1, 5, 1, 6, 4, 0, 3, 7, 6, 7, 6, 4, 4, 7, 4, 6, 5, 3, 9, 0, 0, 0, 1, 9, 2, 1, 8, 8, 8, 6, 6, 8, 8, 4, 4, 2, 4, 9, 6, 5, 0, 7, 7, 6, 5, 9, 8, 8, 1, 6, 6, 3, 2, 8, 5, 4, 3, 2, 3, 3, 3, 2, 3, 0, 4, 2, 1, 1, 6, 8, 6, 0, 5, 6, 6, 7, 8, 7, 2, 5, 1, 4, 8, 4, 9, 6, 4, 0, 5, 9, 9, 7, 6, 3, 1, 5, 3 (list; constant; graph; refs; listen; history; text; internal format) OFFSET 0,1 COMMENTS The radical is intended as follows: let M(z) = sqrt(-1+sqrt(1+z)) be an endomorphism on C, with sqrt restricted to its main branch. It has two invariant points which both happen to be real: this value z = a, and z = 0. Moreover, 'a' is an attractor of M(z) which, when iterated, converges to it from any starting complex value except z = 0. Consequently, the nested radical, when truncated after n terms, either stays identically 0 when n is even, or converges to 'a' when n is odd. According to the definition, 'a' is a solution of z = M(z), and therefore a root of the equation z^3 + 2z - 1=0. A closely related case with similar characteristics is the infinite nested radical sqrt(1+sqrt(-1+sqrt(1+sqrt(-1+ ... ))) which leads to the mapping F(z) = sqrt(1+sqrt(-1+z)) instead of M(z), and the value of its respective attractor is A137421. LINKS Stanislav Sykora, Table of n, a(n) for n = 0..2000 FORMULA Satisfies x = sqrt(-1+sqrt(1+x)). Equals 1/6(108+12sqrt(177))^(1/3)-4/(108+12sqrt(177))^(1/3). - Alois P. Heinz, May 09 2016 EXAMPLE 0.45339765151640376764474653900019218886688442496507765988166328543... MATHEMATICA RealDigits[N[x/.Solve[x == Sqrt[Sqrt[x+1]-1], x][[2]], 100]][[1]] ( Giovanni Resta, May 10 2016 ) PROG (PARI) real(polroots(Pol([1, 0, 2, -1]))[1]) (PARI) \\ Iterative version; using realprecision of 2100 digits: M(z)=sqrt(-1+sqrt(1+z)); x=1; \\ Starting with a real x>0, all terms are actually real. \\ Over 6000 iterations were needed to make stable 2000 digits: for(n=1, 6500, x=M(x)); real(x) CROSSREFS Cf. A137421. Sequence in context: A200351 A138753 A179410 A008962 A175725 A212711 Adjacent sequences: A272871 A272872 A272873 * A272875 A272876 A272877 KEYWORD nonn,cons AUTHOR Stanislav Sykora, May 08 2016 STATUS approved Lookup \| Welcome \| Wiki \| Register \| Music \| Plot 2 \| Demos \| Index \| Browse \| More \| WebCam Contribute new seq. or comment \| Format \| Style Sheet \| Transforms \| Superseeker \| Recent \| More pages The OEIS Community \| Maintained by The OEIS Foundation Inc. Last modified August 20 16:47 EDT 2018. Contains 313925 sequences. (Running on oeis4.)	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://oeis.org/A272874", "fetch_time": 1534798052000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-34/segments/1534221217006.78/warc/CC-MAIN-20180820195652-20180820215652-00625.warc.gz", "warc_record_offset": 288604971, "warc_record_length": 4275, "token_count": 1067, "char_count": 2743, "score": 3.578125, "int_score": 4, "crawl": "CC-MAIN-2018-34", "snapshot_type": "latest", "language": "en", "language_score": 0.7621806859970093, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00010-of-00064.parquet"}
Is Fibonnaci thinking golden? Taking a look into the fibonnaci sequence through Microsoft Excel By Ryan Shannon The fibonnaci Sequence Each value is created from adding the previous f(0) =f(1)=1 and f(n)=f(n-2)+f(n-1) Looking at the sequence we find the ratios of the original from the continuing values. We notice that f(n+1)/f(n) looks to be pretty golden. Can we prove this? If we look at the ratios that is f(n+1)/f(n) We notice that our last step will reveal that we have; We know that g(n) and g(n-1) if they converge, will converge to a number call it x, will converge to the same number when approaching infinity. This looks very familiar. I'm thinking we have found something there. If we are to use the quadratic formula we find Eureka! We are golden! If we step back to our excel spreadsheet above, we notice that the next ratio is 2.618033... which is the golden mean +1! Then we being to worry, will this work for any value of f(0) and f(1)? If we let f(0) = 1 still and change f(1) = 3, f(1) = 7, and f(1) = 9 As we can see by the Excel chart that the ratios will stay the same. After all this investigation I'm think these ratios have the midas touch	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://jwilson.coe.uga.edu/EMAT6680Fa11/Shannon/ShannonA12/fibonnaci.html", "fetch_time": 1542248711000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-47/segments/1542039742338.13/warc/CC-MAIN-20181115013218-20181115035218-00319.warc.gz", "warc_record_offset": 185845887, "warc_record_length": 2050, "token_count": 318, "char_count": 1183, "score": 3.8125, "int_score": 4, "crawl": "CC-MAIN-2018-47", "snapshot_type": "latest", "language": "en", "language_score": 0.9171579480171204, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00012-of-00064.parquet"}
# Question Video: Finding the Terms of a Sequence given Its General Term Mathematics Find the first five terms of the sequence whose πth term is given by π_π = (β1)^π/πβ΅, where π β₯ 1. 02:34 ### Video Transcript Find the first five terms of the sequence whose πth term is given by π sub π is equal to negative one to the power of π divided by π to the fifth power, where π is greater than or equal to one. In order to work out the first five terms of a sequence where π is greater than or equal to one, we need to substitute the numbers one, two, three, four, and five into the πth-term formula. This will give us values for π sub one through π sub five. When π is equal to one, we have negative one to the power of one divided by one to the fifth power. When π is equal to two, we have negative one squared over two to the fifth power. When π equals three, we have negative one cubed over three to the fifth power. The fourth and fifth terms, π four and π five, are as shown. When raising negative one to an odd power, our answer will be negative one. This means that the numerator of our first, third, and fifth terms will be negative one. Negative one raised to an even power will give us positive one. This means that the numerator for our second and fourth term will be positive. One to the fifth power is just equal to one. Two to the fifth power is 32. Three to the fifth power is 243, four to the fifth power 1024. And five to the fifth power is 3125. Negative one divided by one is just negative one. So, π sub one is negative one. None of the other four fractions can be simplified. Therefore, the first five terms of the sequence are negative one, one over 32, negative one over 243, one over 1024, and negative one over 3125.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.nagwa.com/en/videos/314175909710/", "fetch_time": 1618574122000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-17/segments/1618038056325.1/warc/CC-MAIN-20210416100222-20210416130222-00364.warc.gz", "warc_record_offset": 1028057923, "warc_record_length": 8338, "token_count": 494, "char_count": 1815, "score": 4.8125, "int_score": 5, "crawl": "CC-MAIN-2021-17", "snapshot_type": "latest", "language": "en", "language_score": 0.9476177096366882, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00054-of-00064.parquet"}
0 # 5 plus 10 - 33 plus 2? Updated: 8/17/2019 Wiki User 13y ago -16 Wiki User 13y ago Study guides 20 cards ## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials ➡️ See all cards 3.82 3055 Reviews Earn +20 pts Q: 5 plus 10 - 33 plus 2? Submit Still have questions? Related questions ### -10 plus 6(5-9) plus 1? It is -33 simplified ### What is the computation in 13and 2 fifths plus 16 and 1 half? 13 2/5 = [(5 * 13)+2]/5 = (65+2)/5 = 67/516 1/2 = [(2 * 16)+1]/2 = (32+1)/2 = 33/213 2/5 + 16 1/2:= 67/5 + 33/2= (134 + 165)/10= 299/10 or 29 9/10 ### How much is 5 feet 10 inches plus 33inches? (5 feet 10 inches) plus (33 inches) = 2.6162 metres. 33 ### What is 2 plus 2 plus 2 plus 10 plus 5 plus 5 plus 5 plus 20 plus 20 plus 20 plus 20 plus 20 plus 20 equals? 2+2+2+10+5+5+5+20+20+20+20+20+20=151 ### What is the fraction 3 and 3 over five plus 2 and 2 over 5? 6/5 + 4/5 = 10/5 = 2 * * * * * No. 3 and 3 over 5 is 33/5, not (3+3)/5 So, the required sum is 33/5 + 22/5 = 3 + 2 + 3/5 + 2/5 = 5 + 5/5 = 6 -33 + 5 = -28 10 It equals 10 ### What is 2 plus 3 plus 5 plus 2 plus 4 plus 6 plus 4 plus 1 plus 1 plus 2 equals? 30. 2+3+5=10, 4+6=10, and 2+4+1+1+2=10. 33 ### What is 3 plus 5 plus 6 plus 9 plus 2 plus 1 plus 10? 3 + 5 + 6 + 9 + 2 + 1 + 10 = 36	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.answers.com/Q/5_plus_10_-_33_plus_2", "fetch_time": 1695700539000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-40/segments/1695233510130.53/warc/CC-MAIN-20230926011608-20230926041608-00418.warc.gz", "warc_record_offset": 418231338, "warc_record_length": 46069, "token_count": 606, "char_count": 1338, "score": 3.96875, "int_score": 4, "crawl": "CC-MAIN-2023-40", "snapshot_type": "latest", "language": "en", "language_score": 0.6636776924133301, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00036-of-00064.parquet"}
#### What is 68.9 percent of 483? How much is 68.9 percent of 483? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 68.9% of 483 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update. ## 68.9% of 483 = 332.787 Calculate another percentage below. Type into inputs Find number based on percentage percent of Find percentage based on 2 numbers divided by Calculating sixty-eight point nine of four hundred and eighty-three How to calculate 68.9% of 483? Simply divide the percent by 100 and multiply by the number. For example, 68.9 /100 x 483 = 332.787 or 0.689 x 483 = 332.787 #### How much is 68.9 percent of the following numbers? 68.9 percent of 483.01 = 33279.389 68.9 percent of 483.02 = 33280.078 68.9 percent of 483.03 = 33280.767 68.9 percent of 483.04 = 33281.456 68.9 percent of 483.05 = 33282.145 68.9 percent of 483.06 = 33282.834 68.9 percent of 483.07 = 33283.523 68.9 percent of 483.08 = 33284.212 68.9 percent of 483.09 = 33284.901 68.9 percent of 483.1 = 33285.59 68.9 percent of 483.11 = 33286.279 68.9 percent of 483.12 = 33286.968 68.9 percent of 483.13 = 33287.657 68.9 percent of 483.14 = 33288.346 68.9 percent of 483.15 = 33289.035 68.9 percent of 483.16 = 33289.724 68.9 percent of 483.17 = 33290.413 68.9 percent of 483.18 = 33291.102 68.9 percent of 483.19 = 33291.791 68.9 percent of 483.2 = 33292.48 68.9 percent of 483.21 = 33293.169 68.9 percent of 483.22 = 33293.858 68.9 percent of 483.23 = 33294.547 68.9 percent of 483.24 = 33295.236 68.9 percent of 483.25 = 33295.925 68.9 percent of 483.26 = 33296.614 68.9 percent of 483.27 = 33297.303 68.9 percent of 483.28 = 33297.992 68.9 percent of 483.29 = 33298.681 68.9 percent of 483.3 = 33299.37 68.9 percent of 483.31 = 33300.059 68.9 percent of 483.32 = 33300.748 68.9 percent of 483.33 = 33301.437 68.9 percent of 483.34 = 33302.126 68.9 percent of 483.35 = 33302.815 68.9 percent of 483.36 = 33303.504 68.9 percent of 483.37 = 33304.193 68.9 percent of 483.38 = 33304.882 68.9 percent of 483.39 = 33305.571 68.9 percent of 483.4 = 33306.26 68.9 percent of 483.41 = 33306.949 68.9 percent of 483.42 = 33307.638 68.9 percent of 483.43 = 33308.327 68.9 percent of 483.44 = 33309.016 68.9 percent of 483.45 = 33309.705 68.9 percent of 483.46 = 33310.394 68.9 percent of 483.47 = 33311.083 68.9 percent of 483.48 = 33311.772 68.9 percent of 483.49 = 33312.461 68.9 percent of 483.5 = 33313.15 68.9 percent of 483.51 = 33313.839 68.9 percent of 483.52 = 33314.528 68.9 percent of 483.53 = 33315.217 68.9 percent of 483.54 = 33315.906 68.9 percent of 483.55 = 33316.595 68.9 percent of 483.56 = 33317.284 68.9 percent of 483.57 = 33317.973 68.9 percent of 483.58 = 33318.662 68.9 percent of 483.59 = 33319.351 68.9 percent of 483.6 = 33320.04 68.9 percent of 483.61 = 33320.729 68.9 percent of 483.62 = 33321.418 68.9 percent of 483.63 = 33322.107 68.9 percent of 483.64 = 33322.796 68.9 percent of 483.65 = 33323.485 68.9 percent of 483.66 = 33324.174 68.9 percent of 483.67 = 33324.863 68.9 percent of 483.68 = 33325.552 68.9 percent of 483.69 = 33326.241 68.9 percent of 483.7 = 33326.93 68.9 percent of 483.71 = 33327.619 68.9 percent of 483.72 = 33328.308 68.9 percent of 483.73 = 33328.997 68.9 percent of 483.74 = 33329.686 68.9 percent of 483.75 = 33330.375 68.9 percent of 483.76 = 33331.064 68.9 percent of 483.77 = 33331.753 68.9 percent of 483.78 = 33332.442 68.9 percent of 483.79 = 33333.131 68.9 percent of 483.8 = 33333.82 68.9 percent of 483.81 = 33334.509 68.9 percent of 483.82 = 33335.198 68.9 percent of 483.83 = 33335.887 68.9 percent of 483.84 = 33336.576 68.9 percent of 483.85 = 33337.265 68.9 percent of 483.86 = 33337.954 68.9 percent of 483.87 = 33338.643 68.9 percent of 483.88 = 33339.332 68.9 percent of 483.89 = 33340.021 68.9 percent of 483.9 = 33340.71 68.9 percent of 483.91 = 33341.399 68.9 percent of 483.92 = 33342.088 68.9 percent of 483.93 = 33342.777 68.9 percent of 483.94 = 33343.466 68.9 percent of 483.95 = 33344.155 68.9 percent of 483.96 = 33344.844 68.9 percent of 483.97 = 33345.533 68.9 percent of 483.98 = 33346.222 68.9 percent of 483.99 = 33346.911 68.9 percent of 484 = 33347.6	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.quizzes.cc/metric/percentof.php?percent=68.9&of=483", "fetch_time": 1590496995000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-24/segments/1590347390758.21/warc/CC-MAIN-20200526112939-20200526142939-00383.warc.gz", "warc_record_offset": 865620351, "warc_record_length": 3271, "token_count": 1808, "char_count": 4253, "score": 3.796875, "int_score": 4, "crawl": "CC-MAIN-2020-24", "snapshot_type": "latest", "language": "en", "language_score": 0.8636212944984436, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00045-of-00064.parquet"}
Is this definition of Lorentz transformations correct? Consider 3+1 dimensional space-time manifold $$M$$. Let $$v,u$$ are two vectors of the vector bundle and $$g$$ be a metric on $$M$$. Now we can define the inner product in this vector bundle $$(u,v)=u^{\mu}v_{\mu}=g_{\mu\nu}u^{\mu}v^{\nu}$$. Under any coordinate transformation, this inner product is invariant(because (u,v) is a scalar and also because the transformation of $$g,u,v$$ cancels properly). Let's say now, we do a coordinate transform from $$x$$ to $$x^{'}$$ (this is just selecting another chart according to my understanding). Then $$(u,v)=g_{\mu\nu}u^{\mu}v^{\nu}=g_{\mu'\nu'}u^{\mu'}v^{\nu'}=g_{\mu'\nu'}u^{\mu}v^{\nu}\Lambda^{\mu'}_{\mu}\Lambda^{\nu'}_{\nu}$$ This is true for any $$u,v$$ so, $$g_{\mu\nu}=g_{\mu'\nu'}\Lambda^{\mu'}_{\mu}\Lambda^{\nu'}_{\nu}$$. Here $$\Lambda^{\nu'}_{\nu}=\frac{\partial x^{\mu'}}{\partial x^{\mu}}$$. Now if we need to find $$\Lambda$$ that does not change components of g as a set, which means $${g_{00}, g_{01}, ..., g_{33}}={g_{0'0'}, g_{0'1'}, ..., g_{3'3'}}$$, we get the Lorentz transformations. Which is group $$O(1,3)$$ in this case. In brief, Lorentz transformations are coordinate transformations that do not change the components of the metric (or we can say line element $$ds^2$$ looks the same). Could someone comment on the accuracy of the above description? This is not the case. There are no such transformations which in general leave the components of an arbitrary metric unchanged (except the trivial one such that $$\Lambda^{\mu}{}_{\nu} = \delta^\mu_{\nu}$$). Lorenz transformation do not satisfy the restrictions you impose. As a simple example, take the FLRW metric and apply a Lorentz transformation. You will easily see the components are not the same. Perhaps you mean to be talking about Special Relativity, where the metric is fixed to be the Minkowski metric and the spacetime in flat? • You can assume a flat metric without loss of generality(because we are considering a point in space-time). Since g is symmetric, it is diagonalizable with a suitable basis. Under such a basis, the matric becomes locally flat. – htr Commented Mar 15, 2023 at 17:57 • No you cannot, unless you instead mean to talk about local Lorentz transformations acting in the tangent space. If so, you should probably edit the question. Commented Mar 15, 2023 at 17:58 • This long thread contains a few nice answers that demonstrate that in flat spacetime the Lorentz transformations are the only ones that preserve the metric $ds^2\,.$ I agree with Eletie that when matter is involved this is not the case. An interesting example of a metric in flat spacetime that looks a bit different than the one we know is the Rindler metric. It could be a nice exercise to deduce the Lorentz transformations from the invariance of that. Commented Mar 15, 2023 at 18:03 • @Eletie, the Chart transition maps that I mentioned are local right? – htr Commented Mar 15, 2023 at 18:04 • The chart transition maps are not local in the sense of a singular point, they transform patches on manifold, while flatness only holds (generally) at a singular point after correct choice of coordinates Commented Mar 15, 2023 at 18:18	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://physics.stackexchange.com/questions/755389/about-lorentz-transformations", "fetch_time": 1722785483000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-33/segments/1722640404969.12/warc/CC-MAIN-20240804133418-20240804163418-00343.warc.gz", "warc_record_offset": 366659902, "warc_record_length": 41445, "token_count": 886, "char_count": 3228, "score": 3.671875, "int_score": 4, "crawl": "CC-MAIN-2024-33", "snapshot_type": "latest", "language": "en", "language_score": 0.8222460150718689, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00018-of-00064.parquet"}
## Thursday, February 17, 2011 Stefani Germanotta (better known as Lady Gaga) was planning her post Grammy concert in New England. Her newly hired manager, the famous young math whiz named Lynnet did some statistical analysis and told Lady Gaga that in Boston, MA there are about 10,000 fans who are likely to attend her concert even if they need to drive for two hours, in Hartford, CT - 9,000 fans and in Providence, RI - 8,000 fans who are similarly so eager to see Gaga that they could drive almost anywhere in New England. Each of these cities have large concert halls capable to fit audience of up to 30,000. Distance between Boston and Providence is about 50 miles, Boston and Hartford is 100 miles and Hartford and Providence is approx. 80 miles. Lynnet and Lady Gaga met in the office and contemplated where to hold the concert in order to minimize the total amount of miles travelled by all the fans. Lady Gaga concluded: Ok, then Boston it is, as the city with the largest amount of fans. But Lynnet convinced her otherwise. How? And what city the concert is going to be? Answers accepted all day long on Friday, on our Family Puzzle Marathon. They will be hidden until Saturday morning (EST) and everyone who contributed something reasonable will get a puzzle point. Please, explain your answer. SteveGoodman18 said... The concert should be held in Providence. To be in Boston, the 9,000 Hartford fans would drive 100 miles each and the 8,000 Providence fans would drive 50 miles each. This is a total of 1.3 million fan-miles. Having the concert in Providence, which is more centrally located between the other two cities, would have 10,000 Boston fans driving 50 miles each and 9,000 Hartford fans driving 80 miles each, for a total of only 1.22 million fan-miles. Similarly, having everyone drive to Hartford would have the Boston and Providence fans driving a total of 1.64 million fan-miles. So, having the concert in Providence would minimize the total number of miles traveled by all the fans. Annie said... The concert should be in Providence. If in Boston total mileage for fans traveling would be 1,300,000: (8000x50) + (9000x100) If in Hartford total mileage would be 1,640,000: (10,000x100) + (8,000x80) If in Providence total mileage would be 1,220,000: (10,000x50) + (9000x80) Tom said... Most artists would select the hall with the nicest hotel rooms nearby, the nicest backstage facilities, the best sound.light systems and crew. But I'm pretty sure this isn't the answer that's wanted. Another unwanted answer is, ticket prices, which may very well not be the same in the 3 cities. Bean said... Providence. For Providence, the travel total is (50 miles)(10,000 people) from Boston, for 500,000 people miles PLUS (80 miles)(9000 people) from Hartford, for 720,000 people miles, TOTALLING 120,000 people miles. The same calculation for Boston gives you 130,000 people miles. For Hartford, 1,640,000. -lex- said... The best location for the concert is Providence. The following matrix multiplication depicts the total number of miles the fans need to travel should the concert be in one of the 3 mentioned cities. The smallest number of total miles (1220k) corresponds to Providence. \| 0 100 50\| \|10k\| \|100 0 80\| x \| 9k\| = \|1300k 1800k 1220k\| \| 50 80 0\| \| 8k\| Annie said... The concert would be in Providence. If in Boston total number of miles traveled by fans would be 1,300,000: (8000x50) + (9000x100) If in Hartford the total mileage would be 1,640,000: (10,000x100) + (8000x80) If in Providence the total mileage would be 1,220,000: (10,000x50) + (9000x80) Marina Marshak said... The answer is Providence, because the total mileage that will be needed to get there will be lowest. Providence (10,00050m + 9,00080m) = 1,220,000m Boston (9,000100m + 8,00050m) = 1,300,000m Hartford (10,000100m + 8,00080m) = 1,640,000m Wang said... In order to minimize the total distance, we find out quite simply how many fans there are * how many miles they need to drive depending on which city the concert is held in. Just looking at the map, my initial 'guess' would be that Lynnet thinks the concert should be held in Providence. This would mean that 10000 fans from Boston would drive 50 miles (or 500,000 miles in total) and 9000 fans from Hartford would drive 80 miles (or 720,000 miles in total). Therefore, in total, it would be 1.22 million miles. Just to make sure we're on the right track, I'll calculate the total miles if Hartford was the concert city. 8000 fans in Providence would drive 80 miles so 640,000 miles in total and 10000 fans from Boston would drive 100 miles or 1 million miles in total. Therefore, in total, its 1.64 million miles. Lastly, if the concert city was in Boston: 9000 * 100 = 900,000 miles 8000 * 50 = 400,000 miles 1.3 million miles. Therefore, Lady Gaga should have the concert in Providence to minimize the total number of miles traveled for her fans. Lynnet said... I convinced her that she should hold the concert in Providence. If the concert is held in Boston - fans from Hartford have to drive 900,000 mi fans from Providence have to drive 400,000 mi which totals 1,300,000 miles driven If the concert is held in Hartford - fans from Boston have to drive 1,000,000 mi fans from Providence must drive 640,000 mi which totals 1,640,000 miles driven If the concert is held in Providence - fans from Boston need to drive 500,000 mi fans from Hartford drive 720,000 mi which totals 1,220,000 miles driven Providence is where the concert should be held to minimize distance driven Donna said... Lynnet is correct. They should hold the concert in Providence, RI where the average driving distance per fan is only about 45.1 miles. The average driving distance per fan if held in Boston would be about 48.1 miles, and if held in Hartford, about 60.7 miles. Here are my calculations: Average driving distance to Boston: [(10,000x0)+(8000x50)+(9000x100)]/27,000 total fans = 48.1, rounded to nearest tenth. Average driving distance to Hartford: [(9000x0)+(10,000x100)+(8000x80)]/27,000 = 60.7 Average driving distance to Providence: [(8000x0)+(10,000x50)+(9000x80)]/27,000 = 45.1 anne-marie said... For Boston, we have: 100000+800050+9000100=1300000 for Providence: 80000+1000050+900080=122000 for Hartford: 90000+10000100+8000*80=1640000 To minimize the amount of miles travelled by all the fans, Providence is a better choice than Boston. Now, we do not know if the fans living in Hartford are willining to travel and to maximize the profit from the concert, Hartford will be a better choice. Lynnet wants to accomodate evrybody so it is Providence. Kim said... The concert should be held in Providence. It's more central location outweighs its smaller fan base. Here are the calculations: If we hold the concert in Boston, we have 9000 fans traveling 100 miles (from Hartford) and 8000 fans traveling 50 miles (from Providence) for a total of 1.3 million miles If we hold it in Hartford, we have 10,000 fans traveling 100 miles from Boston and 8000 fans going 80 miles from Providence for a total of 1.64 million miles BUT if we hold it in Providence, we have 10,000 fans traveling 50 miles from Boston and 9000 fans traveling 80 miles from Hartford for a total of 1.22 million miles (Assumption: The fans in the local area are traveling a small enough (and/or similar enough) number of miles as to not affect the outcome -- there is an 80,000 mile difference between the calculations for Boston and Providence. It could easily be the case that the Boston-based fans travel about 8 miles more on average than the Providence-based fans. But let's ignore this.) Anonymous said... I think the concert should be held in Hartford. Here's what I did. I drew a circle around each city extending about 100 miles from the center of each city.(I figure that was about a 2 hour drive) I then drew an obtuse triangle connecting the 3 cities and made a mark at the midpoint of each side of the triangle. From there I drew lines through the midpoints of the triangles. Since about 1/3 of the circle surrounding Boston is the Atlantic Ocean, Boston's 10,000 fans come from the North,West And East. The fans from the South and West would be within 100 miles of Hartford. Providence also is bordered by the Atlantic Ocean(looks like about 40 miles to the South)so the fans that hail from the West and South are easily within 2 hours of Hartford. If the concert was held in Boston or Providence, more than half of Hartford's fans would have a greater than 2 hour drive. Having it in Hartford would keep drive times less than 2 hours for the vast majority of Lady Gaga fans. Maria said... Anonymous above, please share with us your name or nick name so we can attribute all your insightful comments to someone specific and add up your puzzle points. You can get a prize after 5 and a puzzle made about you after 10. Your solution to this puzzle is very interesting as it brings the story even closer to reality. Everyone should read and visualize the Anonymous solution from above. A big fat puzzle point for A! Tom's notes are very valuable as he is a musician and has a real showbiz experience. He throws in a few more variables: price of the tickets in each city, quality of the hotel rooms for the artist, sound system and crew availability. See, Lynnet, there is so much to learn for us. A puzzle point for Tom for mentoring. The rest of you all correctly calculated what Lynnet had in mind - making the concert in Providence will minimize the total combined number of miles traveled by all the fans. Congratulations on another puzzle point for: SteveGoodman18, Annie, Tom, Bean, -lex-, Marina M, Wang, Donna, annie-marie, Kim and of course Lynnet. It looks like this is 50th puzzle point for Kim. We have to celebrate next Friday. Maria said... And Tom just earned his 20th point. 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## Session 21 - Activity #7 Hands-on Function Activity On Friday I asked you to watch a series of videos that taught you about building your own functions in scratch. If you have not viewed those yet, please do so know. During this module we looked at constructing functions - blocks of code that you can call over and over again. For this assignment I would like you to create a program that uses a functions as part of a drawing package to draw simple scene such as a house. For example, here is a very basic solution with part of the code. • “Slowly” (use some wait blocks so we can see the picture evolve) draw a scene on the stage using the pen tool. You may create any scene that you like, but if you are looking for an idea I propose that you draw a house like the kind a kid would draw when playing with crayons and markers. • Include the definition and use of a “rectangle” function. • While a square is a special kind of rectangle, most of us think of rectangles as different. That is, we think of them having a different width than height. Create a function called rectangle which takes in two parmaeters - a width and a height. These should be used in your function to draw the described rectangle. • Include at least two functions beyond the required rectangle function and the ones we have created in our video lessons. • In this module we created a function for a half circle, a square, and a general purpose polygon. You may use these as part of this assignment, but they do not count towards the use of three functions. • These may include specific variations of the polygon function. For example, you can elect to use the polygon function to make your triangle or you can create a new, separate triangle function (just like we created a square function which was separate from the polygon function). • Your functions need not be limited to standard polygons. You can create functions for anything that you want to explore. • Pass the “Places Everyone” test. • You need not draw a stick figure house. You can draw any scene that you like. • Don't limit yourself to a '“stick figure” drawing elements (circles, squares, rectangles, circles, lines, etc. For example, you could make a “curly que” function that makes part a curly que used to draw the leaves in a tree. #### Getting Credit for this Assignment Make sure you have met all of the requirements listed above. • “Share” your program publicly on the Scratch website by clicking on the share button in the upper right hand corner of the create screen. • Make sure that both partner's usernames are in the credits. • Navigate to the studio for this project - https://scratch.mit.edu/studios/4346864/	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.cs.uni.edu/~schafer/1140/sessions/s21/index.html", "fetch_time": 1511529011000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-47/segments/1510934808133.70/warc/CC-MAIN-20171124123222-20171124143222-00121.warc.gz", "warc_record_offset": 358299461, "warc_record_length": 2216, "token_count": 572, "char_count": 2678, "score": 3.875, "int_score": 4, "crawl": "CC-MAIN-2017-47", "snapshot_type": "latest", "language": "en", "language_score": 0.9366306662559509, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00029-of-00064.parquet"}
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 7.3: Sum Notation and Properties of Sigma Difficulty Level: At Grade Created by: CK-12 Estimated31 minsto complete % Progress Practice Sum Notation and Properties of Sigma MEMORY METER This indicates how strong in your memory this concept is Progress Estimated31 minsto complete % Estimated31 minsto complete % MEMORY METER This indicates how strong in your memory this concept is Sayber and Tuscany sell popsicles during the summer for pocket money. One particular weekend, they purchased a package of 30 popsicles from the store. Usually they just offer the popsicles for free, 1 per customer, and accept tips. This time, Tuscany wonders if they would make more money by charging $0.50 per popsicle. At the same time, Sayber wonders if he might be able to increase his tips by encouraging customers to "outbid" each other. The two children decide to each take 15 popsicles and see who makes the most. How could you calculate how much money each of them makes, assuming Sayber gets a$0.10 tip from the first customer, and is able to convince each successive customer to double the previous person's tip? Embedded Video: ### Guidance Consider for example a sequence defined by an = 3n. If we write out the sum of the first 4 terms, we have 3 + 6 + 9 + 12 = 30. But what if we want to write out terms for a larger sum? Summation notation is a method of writing sums in a succinct form. To write the sum 3 + 6 + 9 + 12 = 30 , we use the Greek letter Sigma, as follows: \begin{align}\sum_{n=1}^4 3n\end{align} The expression 3n is called the summand, the 1 and the 4 are referred to as the limits of the summation, and the n is called the index of the sum. Here we have used a “sigma” to write a sum. We can also read a sigma, and determine the sum. For example, we can read the above sigma notation as “find the sum of the first four terms of the series, where the nth term is 3n.” We always read the limits from the bottom to the top. The bottom number tells you which term to start with, and the top limit tells you which term is the final term to add. We could then write out the sigma above as: \begin{align}\sum_{n=1}^4 3n\end{align} = 3(1) + 3(2) + 3(3) + 3(4) = 3 + 6 + 9 + 12 = 30 In general, we can either rewrite a given series in sigma notation, or we can read sigma notation in order to find the value of the sum. Properties of Sigma Notice that we can write the sum 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 19 + 20 as 2(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) Therefore \begin{align}\sum_{n=1}^{10} 2n = 2 \sum_{n=1}^{10} n\end{align}. In general, we can factor a coefficient out of a sum: \begin{align}\sum_{n=1}^{k}ca_n \end{align} \begin{align} = c\sum_{n=1}^{k} a_n \end{align} #### Example A Write the sum using sigma notation: 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 19 + 20 Solution \begin{align}\sum_{n=1}^{10} 2n\end{align} Every term is a multiple of 2. The first term is 2 × 1, the second term is 2 × 2 , and so on. So the summand of the sigma is 2n. There are 10 terms in the sum. Therefore the limits of the sum are 1 and 10. #### Example B Write out the terms of \begin{align}\sum_{n=2}^{5}(n+7) \end{align} and evaluate the sum. Solution Break the sum into two different sums. The sum is 37. \begin{align}\sum_{n=2}^{5}(n+7) \end{align} =(2 + 7) + (3 + 7) + (4 + 7) + (5 + 7) =2 + 3 + 4 + 5 + 7 + 7 + 7 + 7 =2 + 3 + 4 + 5 + 7 × 4 = 14 + 28 = 42 Notice we could have written \begin{align}\sum_{n=2}^{5}(n+7) \end{align} as \begin{align}\sum_{n=2}^{5}n+ \sum_{n=2}^{5}7\end{align} . Also, the second sum does not depend on the index of the sum (i.e. it stays at 7 regardless of the index), only that there are 4 terms to add together. Seeing this can make a sum easier to evaluate. #### Example C Write out the terms of \begin{align}\sum_{n=0}^{4}32\left ( \frac{1}{4} \right )^n\end{align} and evaluate the sum. Solution In general, we can write a sum as a sum of sums: \begin{align}\sum_{n=1}^{n}(a_n+b_n) =\sum_{n=1}^{n}(a_n)+\sum_{n=1}^{n}(b_n)\end{align}. The sum is \begin{align}42\frac{5}{8}\end{align} \begin{align}\sum_{i=0}^{4}32 \left ( \frac{1}{4} \right)^n\end{align} \begin{align}= 32 \left ( \frac{1}{4} \right )^0+32 \left ( \frac{1}{4} \right )^1+32 \left ( \frac{1}{4} \right )^2+32 \left ( \frac{1}{4} \right )^3+32 \left ( \frac{1}{4} \right )^4\end{align} \begin{align}= 32 \cdot 1+32\cdot \frac{1}{4}+32 \cdot \frac{1}{16}+ 32\cdot \frac{1}{64}+32 \cdot \frac{1}{256}\end{align} \begin{align}= 32+8+2+\frac{1}{2}+\frac{1}{8}=42 \frac{5}{8}\end{align} Concept question wrap-up Tuscany's income can be expressed as: \begin{align}\sum_{n=1}^{10} .5 \to .5 \cdot 10 = 5.00\end{align} Sayber's income can be expressed as: \begin{align}\sum_{n=1}^{10}.10 \cdot 2^n \to .10 (2^0) + .10 (2^1) + .10 (2^2)... +.10 (2^{10}) \to 102.30\end{align} Sayber must be a heck of a salesman to get that last tip! ### Vocabulary Sigma is the large, stylized capitol E that means, approximately, "the sum of". The index of the sum is the variable in the sum. The summand describes how to manipulate each term in a series. The limits of the sum are written above and below the sigma, and describe the domain to be used in the series calculation. ### Guided Practice Questions 1) 2 + 6 + 18 + 54 \begin{align}\sum_{n = 1}^4 2 \left(3^{n - 1} \right)\end{align} or \begin{align}\sum_{n = 0}^3 2 \left(3^n \right)\end{align} 2) Find the series of numbers and the total of those numbers of the arithmetic series represented by the following sigma notation: \begin{align}\sum_{n=8}^{14} {3 + \frac{3}{4}(n-1)}\end{align} 3) Find the sum of all the numbers in the arithmetic sequence. \begin{align}\sum_{n=8}^{28} {9 - 2(n-1)}\end{align} 4) Find the series of numbers and the total of those numbers of the geometric series represented by the sigma notation \begin{align}\sum_{n=1}^8 7(\frac{-1}{2})^{n-1}\end{align} 5) Find the sum of the terms in: \begin{align}\sum_{n=1}^{11} 9(4)^{n-1}\end{align} Solutions 1) \begin{align}\sum_{n = 1}^5 2n - 1\end{align} 2) To find the series of numbers, we plug in all the numbers between 8 and 14 for (n) a) \begin{align} 3 + \frac{3}{4}((8)-1) = \frac{33}{4}\end{align} b) \begin{align} 3 + \frac{3}{4}((9)-1) = 9\end{align} We would continue to do this clear through the number 14. Leaving us with the following series: \begin{align} \frac{33}{4} + 9 + \frac{39}{4} + \frac{21}{2} + \frac{45}{4} + 12 + \frac{51}{4}\end{align} This is fine, if we are just looking for the individual numbers in the sequence, however when asked to evaluate sumations we are being asked to add all the numbers of the series together. It took plenty long enough to find each number, and now we must add them all together. Fortunately there is a formula, not only eliminating our need to find each number, but that lets us also add them all together and arrive at our answer or sum. The formula is \begin{align}\frac{k}{2}(a_1 + a_{n})\end{align} works like this: We plug in n = 8 to get \begin{align}a_8 = \frac{33}{4}\end{align} Then we plug in n = 14, to get \begin{align}a_{14} = \frac{51}{4}\end{align} Identify the number of terms \begin{align}(k) = 14 - 8 + 1\end{align} so we use \begin{align}k=7\end{align} in the formula below. Now we can use the formula:\begin{align}\frac{k}{2}(a_8 + a_{14})\end{align} and we get:\begin{align} \frac{7}{2}(\frac{33}{4} + \frac{51}{4})=\frac{147}{2}\end{align} \begin{align}\therefore \sum_{n = 1}^4 2 \left(3^{n - 1} \right)\end{align} or \begin{align}\sum_{n = 0}^3 2 \left(3^n \right)\end{align} 3) Use the formula \begin{align}\frac{k}{2}(a_1 + a_{n})\end{align} to find the sum of \begin{align}\sum_{n=8}^{28} {9 - 2(n-1)}\end{align} Substitute n = 8 to get \begin{align}a_8 = -5\end{align} Substitute n = 28 to get \begin{align}a_28 = -45\end{align} Identify the number of terms as: \begin{align}k = 28-8+1\end{align} so we use \begin{align}k = 21\end{align} in our formula below: Now we can use the formula:\begin{align}\frac{k}{2}(a_8 + a_{14})\end{align} and our answer is \begin{align}-525\end{align} \begin{align}\therefore \sum_{n = 1}^{10} \left(\frac{1} {n} \right)\end{align} 4) To identify and sum the terms in the geometric series \begin{align}\sum_{n=1}^8 7(\frac{-1}{2})^{n-1}\end{align} a) Find the sequence of numbers the same way as we did in problem 2, by plugging in the indicated numbers 1-8 for (n) b) \begin{align}\sum_{n=1}^8 7(\frac{-1}{2})^{((1)-1)}\end{align} which gives us: \begin{align} 7 \end{align} c) We do this for the remaining numbers and our sequence looks like this:\begin{align}7 + \frac{-7}{2} +\frac{7}{4} + \frac{-7}{8} +\frac{7}{16} +\frac{-7}{32} +\frac{7}{64} + \frac{-7}{128}\end{align} As in prior examples, we are not just being asked to find the numbers, but add them all together. Again, we do not want to have to take the time to find all the numbers and add them all together. Once again we find ourselves lucky, there is a formula! The formula \begin{align}a_1(\frac{1-r^k}{1-r})\end{align} works like this: We plug n=1 into \begin{align}a_1\end{align} which gives us 7. \begin{align}k = 8\end{align}, and \begin{align}r=\frac{-1}{2}\end{align} Substituting our numbers results:\begin{align}7(\frac{1-\frac{-1}{2}^8}{1-\frac{-1}{2}})\end{align} Which gives us \begin{align}\frac{595}{128}\end{align} 5) Let's use the formula for geometric series again: Identify the terms: \begin{align}a_1 = 9\end{align}, \begin{align} k = 11\end{align} and \begin{align} r = 4\end{align} Substituting into our formula, we have: \begin{align}9(\frac{1-4^{11}}{1-4}) = 12,582,909\end{align} ### Practice Express the Sum using Sigma Notation: 1. \begin{align}1 + 3 + 5 + 7 + 9\end{align} 2. \begin{align}1 + \frac{1} {2} + \frac{1} {3} + \frac{1} {4} + ... + \frac{1} {10}\end{align} Find the series of numbers indicated by the sigma notation: 1. \begin{align}\sum_{n=0}^{14} -2 -\frac{10}{3}(n-1)\end{align} 2. \begin{align}\sum_{n=-8}^{14} -7 -3(n-1)\end{align} Evaluating Summations of an Arithmetic Series 1. \begin{align}\sum_{n=-10}^{5} 7 -\frac{4}{3}(n-1)\end{align} 2. \begin{align}\sum_{n=-3}^{3} 3 -\frac{1}{3}(n-1)\end{align} 3. \begin{align}\sum_{n=-5}^{1} -6 +\frac{4}{3}(n-1)\end{align} Find the series of numbers indicated by the sigma notation 1. \begin{align}\sum_{n=1}^{2} 3(-\frac{1}{2})^{n-1}\end{align} 2. \begin{align}\sum_{n=1}^{5} 5(-\frac{4}{3})^{n-1}\end{align} Evaluating Summations of Geometric Series 1. \begin{align}\sum_{n=1}^{6} 3(\frac{1}{2})^{n-1}\end{align} 2. \begin{align}\sum_{n=1}^{7} -5(-\frac{1}{2})^{n-1}\end{align} 3. \begin{align}\sum_{n=1}^{11} -7(-\frac{4}{3})^{n-1}\end{align} 4. \begin{align}\sum_{n=1}^{6} -7(\frac{1}{4})^{n-1}\end{align} 5. \begin{align}\sum_{n=1}^{3} 2(-\frac{3}{2})^{n-1}\end{align} ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English TermDefinition $\Sigma$ $\Sigma$ (sigma) is the Greek letter meaning "the sum of" when used in mathematics. arithmetic series An arithmetic series is the sum of an arithmetic sequence, a sequence with a common difference between each two consecutive terms. geometric series A geometric series is a geometric sequence written as an uncalculated sum of terms. index The index of a sum is the variable in the sum. limits The limits of a sum are written above and below the $\Sigma$, and describe the domain to be used in the series calculation. sequence A sequence is an ordered list of numbers or objects. series A series is the sum of the terms of a sequence. Sigma $\Sigma$, pronounced syg-mah, is the Greek letter that in math means "the sum of". sigma notation Sigma notation is also known as summation notation and is a way to represent a sum of numbers. It is especially useful when the numbers have a specific pattern or would take too long to write out without abbreviation. summand A summand is an expression being summed. It directly follows the sigma symbol. summation Sigma notation is also known as summation notation and is a way to represent a sum of numbers. It is especially useful when the numbers have a specific pattern or would take too long to write out without abbreviation. Show Hide Details Description Difficulty Level: Tags: Subjects:	{"found_math": true, "script_math_tex": 75, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 4, "texerror": 0, "url": "http://www.ck12.org/book/CK-12-Math-Analysis-Concepts/r4/section/7.3/", "fetch_time": 1490684661000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-13/segments/1490218189680.78/warc/CC-MAIN-20170322212949-00610-ip-10-233-31-227.ec2.internal.warc.gz", "warc_record_offset": 458399505, "warc_record_length": 43282, "token_count": 4374, "char_count": 12506, "score": 4.8125, "int_score": 5, "crawl": "CC-MAIN-2017-13", "snapshot_type": "latest", "language": "en", "language_score": 0.8967355489730835, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00030-of-00064.parquet"}
# How do you graph y=(3x^2)/(x^2-9) using asymptotes, intercepts, end behavior? Nov 23, 2016 Horizontal asymptote: y = 0 and vertical ones are $x = \pm 3$. In ${Q}_{1} ,$ as $x \to \infty , y \to 0 \mathmr{and}$ as $y \to \infty , x \to 0$. See explanation, for continuation on end behavior. #### Explanation: graph{y(x^2-9)-3x^2=0 [-40, 40, -20, 20]} By actual division and rearrangement, $\left(y - 3\right) \left(x - 3\right) \left(x + 3\right) = 27$ To get asymptotes, See that, $L H S \to 0 X \left(\pm \infty\right)$ indeterminate form, so that the limit exists as 27. Easily, you could sort ouy the equations to the ayymptotes by setting the factors on the LHS to 0. Horizontal asymptote is y = 0 and the vertical ones are $x = \pm 3$. In ${Q}_{1} ,$ as $x \to \infty , y \to 0 \mathmr{and}$ as $y \to \infty , x \to 0$. In ${Q}_{2}$, as $x \to - \infty , y \to 0 \mathmr{and}$ as $y \to \infty , x \to 0$. In ${Q}_{3} \mathmr{and} {Q}_{4}$, as $x \to 0$ , as $y \to - \infty$	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 16, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://socratic.org/questions/how-do-you-graph-y-3x-2-x-2-9-using-asymptotes-intercepts-end-behavior", "fetch_time": 1590711374000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-24/segments/1590347401004.26/warc/CC-MAIN-20200528232803-20200529022803-00005.warc.gz", "warc_record_offset": 562436388, "warc_record_length": 6143, "token_count": 393, "char_count": 994, "score": 4.3125, "int_score": 4, "crawl": "CC-MAIN-2020-24", "snapshot_type": "latest", "language": "en", "language_score": 0.7670443654060364, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00043-of-00064.parquet"}
# Vector Subspace Let $V$ be a vector space over the field $F$. Then a non-empty subset $W$ of $V$ is called a vector space of $V$ if under the operations of $V$, $W$ itself is a vector space over $F$. In other words, $W$ is a subspace of $V$ whenever ${w_1},{w_2} \in W$ and $\alpha ,\beta \in F \Rightarrow \alpha {w_1} + \beta {w_2} \in W$ Example: Prove that the set $W$ of ordered tried $\left( {{a_1},{a_2},0} \right)$ where ${a_1},{a_2} \in F$ is a subspace of ${V_3}\left( F \right)$. Solution: Let $a = \left( {{a_1},{a_2},0} \right)$ and $b = \left( {{b_1},{b_2},0} \right)$ be two elements of $W$. Therefore ${a_1},{a_2},{b_1},{b_2} \in F$ let $a,b \in F$ then $\begin{gathered} a\alpha + b\beta = a\left( {{a_1},{a_2},0} \right) + b\left( {{b_1},{b_2},0} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {a{a_1},a{a_2},0} \right) + \left( {b{b_1},b{b_2},0} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {a{a_1} + b{b_1},a{a_2} + b{b_2},0} \right) \in W \\ \end{gathered}$ Because $a{a_1} + b{b_1},a{a_2} + b{b_2} \in F$.	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.emathzone.com/tutorials/group-theory/vector-subspace.html", "fetch_time": 1696428538000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-40/segments/1695233511369.62/warc/CC-MAIN-20231004120203-20231004150203-00421.warc.gz", "warc_record_offset": 810666587, "warc_record_length": 13156, "token_count": 482, "char_count": 1061, "score": 4.21875, "int_score": 4, "crawl": "CC-MAIN-2023-40", "snapshot_type": "longest", "language": "en", "language_score": 0.674774169921875, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00055-of-00064.parquet"}
Â Â Â Â Â Â Â Â Â Â # How to Read Graphs and Charts? Graph is a pictorial representation of an expression. Expression shows relation between two or more variables. In other words, a graph can be stated as a representation of a function which expresses relationship among two or more than two variables. Graphs may be straight line, parabolic, hyperbolic, inverse graph etc. Let’s understand how to read graphs and charts. Y = m x + c. When we substitute x = 0 then y = c, when x = 1 then y = m + c, when x = 2 then y = 2m +c and so on. So it is clear 'x' increases, 'y' also increases for any value of ‘m’ and ‘c’. This can be plotted as shown below. Here in above graph, relationship between 'x' and 'y' is shown. In this graph, 'c' is representing a constant value by which the graph has been shifted from the origin on the y – axis. Origin is the Intersection Point of x – axis and y – axis. Term ‘m’ represents the Slope. This is given by Ratio of ‘dy’ to ‘dx’. dy represents small change in values on y - Axis and ‘dx’ shows small change in values on x – Axis. Chart can be defined as Graphical representation of Functions or any type of data such as Percentage of men, women, children etc. as a part of whole population counting. Let's consider a pie chart to understand how to read it. A pie (circle) chart mainly has three parts. These are title, sectors and label. Here in the above example, the title of the chart is “Various Percentages Among Total Students of School” The Circle or pie graph above shows the percentage of students among total achieved marks. There are four sectors. Sector A: 10% students are above 80% Sector B: 15tudents achieved between 36% and 49% Sector C: 50% students achieved 61% - 75% Sector D: 25% students achieved 50% - 60%.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.tutorcircle.com/how-to-read-graphs-and-charts-fXoNq.html", "fetch_time": 1369545961000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2013-20/segments/1368706631378/warc/CC-MAIN-20130516121711-00075-ip-10-60-113-184.ec2.internal.warc.gz", "warc_record_offset": 779762234, "warc_record_length": 33483, "token_count": 462, "char_count": 1787, "score": 4.5, "int_score": 4, "crawl": "CC-MAIN-2013-20", "snapshot_type": "latest", "language": "en", "language_score": 0.931191623210907, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00004-of-00064.parquet"}
###### Bubble Sort Algorithm Bubble sort algorithm, sometimes referred to as sinking sort, is a simple sorting algorithm that repeatedly steps through the list to be sorted, compares each pair of adjacent items and swaps them if they are in the wrong order. The pass through the list is repeated until no swaps are needed, which indicates that the list is sorted. The algorithm, which is a comparison sort, is named for the way smaller or larger elements “bubble” to the top of the list. Although the algorithm is simple, it is too slow and impractical for most problems compared to other sorting algorithms. Complexity of this algorithm is as below : ###### Performance • Worst-case performance:- O(n2) • Best-case performance:- O(n) • Average performance:- O(n2) • Worst-case space complexity:- O(1) auxiliary ###### Example Steps With Description Given Array:- 20, 3, 35, 21, 12 First Iteration: ( 20 3 35 21 12 ) -> ( 3 20 35 21 12 ), Algorithm compares first two elements, and swaps because 20 > 3. ( 3 20 35 21 12 ) –> ( 3 20 35 21 12 ), No Swap needed as 20 < 35. ( 3 20 35 21 12 ) –> ( 3 20 21 35 12 ), Swap since 35 > 21. ( 3 20 21 35 12 ) –> ( 3 20 21 12 35 ), Swap since 35 > 12. Second Iteration: (3 20 21 12 35 ) -> (3 20 21 12 35 ), No Swap needed as 3 < 20. (3 20 21 12 35 ) -> (3 20 21 12 35 ), No Swap needed as 20 < 21. (3 20 21 12 35 ) -> (3 20 12 21 35 ), Swap, as 21>12. (3 20 12 21 35 ) -> (3 20 12 21 35 ), No Swap needed as 21 < 35. Third Iteration: (3 20 12 21 35 ) -> (3 20 12 21 35 ), No Swap needed as 3 < 20. (3 20 12 21 35 ) -> (3 12 20 21 35 ) , Swap needed as 20 > 12, Now our array is sorted.But algorithm don’t know that it’s already sorted. (3 12 20 21 35 ) -> (3 12 20 21 35 ). (3 12 20 21 35 ) -> (3 12 20 21 35 ). ###### Java Program Of Bubble Sort Output Input Array before Bubble Sort. 20 3 35 21 12 Sorted Array after Bubble Sort. 3 12 20 21 35 The above code always runs O(n2) time even if the array is sorted. It can be optimized by stopping the algorithm if inner loop didn’t swap any elements. ###### Java Program Of Optimized Bubble Sort Output Input Array before Bubble Sort. 20 3 35 21 12 Breaking loop on iteration :- 4 , as array is sorted and no swapping occurred in last iteration. Sorted Array After Bubble sort. 3 12 20 21 35 Run again with sorted array as input (Already provided in program with comment) Output Input Array before Bubble Sort. 10 20 30 50 70 Breaking loop on iteration :- 1 , as array is sorted and no swapping occurred in last iteration. Sorted Array After Bubble sort. 10 20 30 50 70 Some theory part of this article uses material from the Wikipedia article “Bubble Sort”, which is released under the CC BY-SA 3.0.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://codenuclear.com/bubble-sort-algorithm/", "fetch_time": 1611069796000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-04/segments/1610703519395.23/warc/CC-MAIN-20210119135001-20210119165001-00193.warc.gz", "warc_record_offset": 286431036, "warc_record_length": 20536, "token_count": 846, "char_count": 2721, "score": 4.53125, "int_score": 5, "crawl": "CC-MAIN-2021-04", "snapshot_type": "latest", "language": "en", "language_score": 0.8097183108329773, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00033-of-00064.parquet"}
# Kilojoules to Watt-Hours Converter Enter the energy in kilojoules below to get the value converted to watt-hours. ## Result in Watt-Hours: 1 kJ = 0.277778 Wh Do you want to convert watt-hours to kilojoules? ## How to Convert Kilojoules to Watt-Hours To convert a measurement in kilojoules to a measurement in watt-hours, multiply the energy by the following conversion ratio: 0.277778 watt-hours/kilojoule. Since one kilojoule is equal to 0.277778 watt-hours, you can use this simple formula to convert: watt-hours = kilojoules × 0.277778 The energy in watt-hours is equal to the energy in kilojoules multiplied by 0.277778. For example, here's how to convert 5 kilojoules to watt-hours using the formula above. watt-hours = (5 kJ × 0.277778) = 1.388889 Wh ### How Many Watt-Hours Are in a Kilojoule? There are 0.277778 watt-hours in a kilojoule, which is why we use this value in the formula above. 1 kJ = 0.277778 Wh ## What Is a Kilojoule? One kilojoule is equal to 1,000 joules, which is the energy equal to the force on an object of one newton at a distance of one meter. The kilojoule is a multiple of the joule, which is the SI derived unit for energy. In the metric system, "kilo" is the prefix for thousands, or 103. Kilojoules can be abbreviated as kJ; for example, 1 kilojoule can be written as 1 kJ. ## What Is a Watt-Hour? The watt-hour is a measure of electrical energy equal to one watt of power over a one hour period. Watt-hours are usually abbreviated as Wh, although the formally adopted expression is W·h. The abbreviation W h is also sometimes used. For example, 1 watt-hour can be written as 1 Wh, 1 W·h, or 1 W h. In formal expressions, the centered dot (·) or space is used to separate units used to indicate multiplication in an expression and to avoid conflicting prefixes being misinterpreted as a unit symbol.[1] ## Kilojoule to Watt-Hour Conversion Table Table showing various kilojoule measurements converted to watt-hours. Kilojoules Watt-hours 1 kJ 0.277778 Wh 2 kJ 0.555556 Wh 3 kJ 0.833333 Wh 4 kJ 1.1111 Wh 5 kJ 1.3889 Wh 6 kJ 1.6667 Wh 7 kJ 1.9444 Wh 8 kJ 2.2222 Wh 9 kJ 2.5 Wh 10 kJ 2.7778 Wh 11 kJ 3.0556 Wh 12 kJ 3.3333 Wh 13 kJ 3.6111 Wh 14 kJ 3.8889 Wh 15 kJ 4.1667 Wh 16 kJ 4.4444 Wh 17 kJ 4.7222 Wh 18 kJ 5 Wh 19 kJ 5.2778 Wh 20 kJ 5.5556 Wh 21 kJ 5.8333 Wh 22 kJ 6.1111 Wh 23 kJ 6.3889 Wh 24 kJ 6.6667 Wh 25 kJ 6.9444 Wh 26 kJ 7.2222 Wh 27 kJ 7.5 Wh 28 kJ 7.7778 Wh 29 kJ 8.0556 Wh 30 kJ 8.3333 Wh 31 kJ 8.6111 Wh 32 kJ 8.8889 Wh 33 kJ 9.1667 Wh 34 kJ 9.4444 Wh 35 kJ 9.7222 Wh 36 kJ 10 Wh 37 kJ 10.28 Wh 38 kJ 10.56 Wh 39 kJ 10.83 Wh 40 kJ 11.11 Wh ## References 1. Bureau International des Poids et Mesures, The International System of Units (SI), 9th edition, 2019, https://www.bipm.org/documents/20126/41483022/SI-Brochure-9-EN.pdf	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.inchcalculator.com/convert/kilojoule-to-watt-hour/", "fetch_time": 1726105994000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651420.25/warc/CC-MAIN-20240912011254-20240912041254-00640.warc.gz", "warc_record_offset": 755929453, "warc_record_length": 16311, "token_count": 996, "char_count": 2805, "score": 3.546875, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.8081262707710266, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00049-of-00064.parquet"}
# Hypothesis Testing - an Overview Based on the “Statistical Consulting Cheatsheet” by Prof. Kris Sankaran Many problems in consulting can be treated as elementary testing problems. First, let’s review some of the philosophy of hypothesis testing. Testing provides a principled framework for filtering away implausible scientific claims. It’s a mathematical formalization of Karl Popper’s philosophy of falsification. The underlying principle is simple: “ Reject the null hypothesis if the data are not consistent with it, where the strength of the discrepancy is formally quantified through the notion of p-value”. ### The Objective: Measuring the strength of discrepancy by computing a p-value Consequently, one of the main goals of hypothesis testing is to compute a p-value. A p-value can be defined as “the probability of observing an event as extreme as what I am observing under the null”, where the null is the default, “chance” scenario. Example: For instance, suppose that I want to assess if Soda A is better than Soda B. I could do a survey, and ask people to give a score to each of the soda, and average my results. Suppose there is no difference between the two: the difference between the two averages is a random variable, centered at 0. Conversely, if there is a true underlying difference (let’s call it $$\delta$$), then the difference between my two averages: $$\Delta = \bar{X}_a -\bar{X}_b$$ is also a random variable, but centered at $$\delta$$. The entire point of hypothesis testing becomes to quantify how extreme this difference $$\Delta$$ has to be to “reject the null” — i.e, to say that it is unlikely for $$\Delta$$ to be this extreme if the null (“there is no difference in sodas”) is true.This is the concept at the core of p-values: a p-value of 0.04 means that, just by chance, only 4% of events would have seen a difference this big. If I am willing to accept that 4% is too small (statisticians usually abide by the convention that anything less than 5% chance is unlikely to happen by chance alone), I can reject the null. A small p-value, typically < 0.05, indicates strong evidence against the null hypothesis; in this case you can reject the null hypothesis. On the other hand, a large p-value, > 0.05, indicates weak evidence against the null hypothesis; in this case, you do NOT reject the null hypothesis. The value 0.05 is the threshold usually employed by the community — you can think of it as a scientific convention for determining significance. Importantly, the p-value is the probability of observing events as extreme as my observations under the null, but not the probability that the hypothesis is correct! $p_{value} = \mathbb{P}[\text{observations} \; \mid \; \text{hypothesis } H_0 ] \ne P[ \text{hypothesis } H_0 \; \mid \; \text{observations} ]$ P-values should NOT be used a “ranking”/“scoring” system for your hypotheses. ### The Recipe Of course, to determine what this p-value is, there are three essential steps: • Step 1: Defining the Null and Alternate Hypotheses __The null hypothesis ($$H_0$$) is a statement assumed to be true unless it can be shown to be incorrect beyond a reasonable doubt. This is something one usually attempts to disprove or discredit. The alternate hypothesis ($$H_1$$) is a claim that is contradictory to $$H_0$$ and what we conclude when we reject $$H_0$$. • Step 2: Defining a test statistics that is, what I will be measuring (eg, the average score). This has to be tailored to the problem that I am really interested in. • Step 3: Defining what the null looks and behaves like: this is what will allow me to measure whether or not what I am observing in my own dataset is extreme or not. While testing is fundamental to much of science, and to a lot of our work as consultants, there are some limitations we should always keep in mind: • The different types of errors: There are two kinds of errors we can make: (1) Accidentally falsify when true (false positive / type I error) and (2) fail to falsify when actually false (false negative / type II error). Different types of tests will allow to control for these errors, and finding the right hypothesis test thus becomes a matter of finding the test that is “the most sensitive” to the data that you are measuring. • The problem of defining “the null”: In order to build a test, we need to be able to articulate the sampling behavior of the system under the null hypothesis. • Often, describing the null can be complicated by particular structure present within a problem (e.g., the need to control for values of other variables). This motivates inference through modeling, which is reviewed in the inference in linear model section of this website. • We need to be able to quantitatively measure discrepancies from the null. Ideally we would be able to measure these discrepancies in a way that makes as few errors as possible. This is in fact the motivation behind optimality theory: how can I run my experiments in a way that allows the “purest” measures for the question that I would like to answer? • Practical significance is not the same as statistical significance. A p-value should never be the final goal of a statistical analysis. They should be used to complement figures / confidence intervals / follow-up analysis that provide a sense of the effect size. ### The Ingredients To find the right hypothesis test, we need to select the right “ingredients”. That requires to answer a minimum of four questions: • Question 1: What type of data do I have? • Question 2: Can I assume my data points are independent? • Are there some paired variables? • Is my data stratified into clusters? • Are there some potential batch effects? • Question 3: Am I testing a single, or multiple hypotheses at the same time? • Question 4: How many datapoints do I have? Prospective Measurements: Finally, if you haven’t done your measurements yet and you’re looking to assess how many samples you would need to answer your question, do look at our page on power analysis.	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://uofcstatdeptconsult.github.io/stat/hypothesis_testing/", "fetch_time": 1685313352000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-23/segments/1685224644571.22/warc/CC-MAIN-20230528214404-20230529004404-00391.warc.gz", "warc_record_offset": 659935504, "warc_record_length": 4813, "token_count": 1345, "char_count": 6030, "score": 3.875, "int_score": 4, "crawl": "CC-MAIN-2023-23", "snapshot_type": "latest", "language": "en", "language_score": 0.9349117279052734, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00061-of-00064.parquet"}
Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ ° You are not logged in. ## #1 2012-04-25 07:04:08 amberzak Full Member Offline ### Calculus - Area of a curve Hi all. I am onto Calculus now. Thing is, I have only done 2 lessons and never studied it before. I'm getting on alright, but I have a really hard question (well, hard to me). It's an extra question, so if I don't do it it isn't a big deal, but I am trying to figure it out. This (if it works) is a roughly drawn picture. It says: The figure shows the curve with equation y=5+2x-x^2 and the line with equation y=2. The curve and the line intersect at the points A and B. a) Find the x-coordinates of A and B. The shaded region R is bounded by the curve and the line. b) Find the area of R. Can someone please take me through step by step how you find the x-Coordinates of A and B? I do have the answers by the way, just don;t know how to get to it. Our teacher gives us the final answers so we can check we have it right, and he marks us on our working. The thing that's confusing me the most is that the equation isn't in the normal format, and so the x^2 value is negative, which I always thought meant an imaginary number, so I think I've missed something. As I say, my teacher did say he doesn't mind if I don't do this question, as I have never done calculus before, but I am now intrigued as to how it's done. Thanks all. Don't think outside the box. Think there is no box ## #2 2012-04-25 07:07:17 bobbym Online ### Re: Calculus - Area of a curve Hi amberzak; For the first problem. Since you know y = 2 then just plug it in to get the points of intersection. a) Can you solve that or do you need help? In mathematics, you don't understand things. You just get used to them. 90% of mathematicians do not understand 90% of currently published mathematics. I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on. ## #3 2012-04-25 07:29:37 amberzak Full Member Offline ### Re: Calculus - Area of a curve I don't want to sound dumb, but could you just put the solving up? I think I have it, but I just need to double check I've done it right. (My biggest problem is confidence) Don't think outside the box. Think there is no box ## #4 2012-04-25 07:40:17 bobbym Online ### Re: Calculus - Area of a curve Hi; Subtract 2 from both sides and turn it around. Multiply everything by -1 Can you factor that? In mathematics, you don't understand things. You just get used to them. 90% of mathematicians do not understand 90% of currently published mathematics. I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on. ## #5 2012-04-25 07:50:01 amberzak Full Member Offline ### Re: Calculus - Area of a curve Thanks. I didn't do the multiply everything by -1. So the factoring would be: (x-3)(x+1) and the points at A and B would be 3 and -1. Is that right? Don't think outside the box. Think there is no box ## #6 2012-04-25 07:58:01 bobbym Online ### Re: Calculus - Area of a curve Correct! Very good! In mathematics, you don't understand things. You just get used to them. 90% of mathematicians do not understand 90% of currently published mathematics. I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on. ## #7 2012-04-25 08:12:00 amberzak Full Member Offline ### Re: Calculus - Area of a curve Okay, so next part of the question. (I can't do the notation on here so I'm just going to write it without the correct notation). Intergrating 5+2x-x^2. =5x+2x^2 - x^3 ------ ------ 2 3 Then supplement the two coordinates in for x = (5x3 + 2x3^2 - x^3 ) - (5x-1 + 2x-1^2 - -1^3) ------- ------ -------- ----- 2 3 2 3 and that's where I get a bit unstuck. The first part is easy enough: =15 +9 -3) - (this is the bit i am sure I have wrong) Don't think outside the box. Think there is no box ## #8 2012-04-25 08:22:01 bob bundy Moderator Offline ### Re: Calculus - Area of a curve hi amberzak, The area you have found is for all of the bit below the curve down to the x axis. The region R is less than that by one rectangle. Once you subtract that, you should get the required answer. Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei ## #9 2012-04-25 08:27:28 bob bundy Moderator Offline ### Re: Calculus - Area of a curve I've got (15 + 18 - 9) - (-5 + 2 + 1/3) And the rectangle is 4 x 2 = 8. B You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei ## #10 2012-04-25 08:32:54 amberzak Full Member Offline ### Re: Calculus - Area of a curve Could you go through step by step. I'm now completely lost. The answer on my answer sheet, by the way, is 10 and 2/3 Don't think outside the box. Think there is no box ## #11 2012-04-25 08:42:06 anonimnystefy Real Member Offline ### Re: Calculus - Area of a curve Hi amberzak Do you know the formula for finding area between graphs of two functions? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón ## #12 2012-04-25 08:53:37 bob bundy Moderator Offline ### Re: Calculus - Area of a curve hi amberzak, LATER EDIT. THERE IS AN ERROR WITH MY INTEGRATION HERE. SEE POST #20 I'll have a go. I've made a diagram below and shaded some regions. You question asks for the region I've coloured red. When you do you will get the area below the curve down to the x axis. That's what area type integration does. (Why is a much longer post for another day I think!) So the integration gives an answer that is too big. Take off 8 for the green rectangle and you should get the right result. So let's check the integration Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei ## #13 2012-04-25 08:56:46 anonimnystefy Real Member Offline ### Re: Calculus - Area of a curve Hi Bob That doesn't give a precise answer. And the precise one is even easier than what you did there. You just subtract the two functions (the quadratic and the linear one) and integrate from -1 to 3. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón ## #14 2012-04-25 08:58:24 bob bundy Moderator Offline ### Re: Calculus - Area of a curve hi Stefy, I'm glad you are looking at this too. I'm getting an answer of 18 and 2/3. Any ideas? Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei ## #15 2012-04-25 09:00:10 anonimnystefy Real Member Offline ### Re: Calculus - Area of a curve Well,you did say to take off 8. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón ## #16 2012-04-25 09:03:39 anonimnystefy Real Member Offline ### Re: Calculus - Area of a curve Sorry for double posting. You integration is not correct!!! The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón ## #17 2012-04-25 09:03:52 bob bundy Moderator Offline ### Re: Calculus - Area of a curve Yes, but that was my answer after taking off 8. ie. 26 and 2/3 take 8 ?? Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei ## #18 2012-04-25 09:04:42 anonimnystefy Real Member Offline ### Re: Calculus - Area of a curve Look at the post right above your new one. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón ## #19 2012-04-25 09:12:40 amberzak Full Member Offline ### Re: Calculus - Area of a curve Why is the integration not right? Is it because it's missing the denominators for the 2x3^2 and 2x-1^2 (sorry if I'm not making a lot of sense. I've been doing calculus all day) Don't think outside the box. Think there is no box ## #20 2012-04-25 09:14:18 bob bundy Moderator Offline ### Re: Calculus - Area of a curve You integration is not correct!!! I see it now. I've spent all day driving and gardening (big tree to cut down) so my brain is not at its best. Here is a correction to post #12. = 18 and 2/3 less 8 = 10 and 2/3 My apologies for the error before. I need some sleep. Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei ## #21 2012-04-25 09:14:27 anonimnystefy Real Member Offline ### Re: Calculus - Area of a curve No,he didn't integrate the term 2x correctly. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón ## #22 2012-04-25 09:18:00 bob bundy Moderator Offline ### Re: Calculus - Area of a curve Ahead of you for the first time. see post 20 Thanks for your help Stefy. I'm off to bed. Bye. Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei ## #23 2012-04-25 09:19:46 anonimnystefy Real Member Offline ### Re: Calculus - Area of a curve I saw it. I still do not get why the two small parts next to the rectangle don't mess it up. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón ## #24 2012-04-25 09:21:46 amberzak Full Member Offline ### Re: Calculus - Area of a curve I'll be coming on tomorrow with your questions Don't think outside the box. Think there is no box ## #25 2012-04-25 09:22:48 amberzak Full Member Offline ### Re: Calculus - Area of a curve Anon, I was wondering that as well, actually. I might ask my teacher than tomorrow. Don't think outside the box. 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Kirchhoff’s Voltage and Current Laws A hypothetical AC-DC converter which violates Kirchoff’s Voltage Law. Granted, a switch circuit can perform power conversion, and granted, it can be described as a switch matrix. How do we make the choices of switch operation? Consider first the simple circuit shown to the right – something we might try for AC-DC conversion. This circuit has problems. Kirchoff’s Voltage Law (KVL) tells us: The sum of voltage drops around a closed loop is zero. Imagine what will happen here if the switch is closed: the sum of voltages around the loop is not zero! in reality, a very large current will flow, and cause a large I X R (current times resistance) voltage drop in the wires. KVL will be satisfied by the wire voltage drop, but a fire might result! We would hope that fuses will blow or circuit breakers will trip instead. KVL should serve as a warning: Dont connect unequal voltage sources directly. You are familiar with KVL as an abstract rule of circuit analysis, In power electronics, KVL takes on a much more concrete meaning. There is nothing to prevent someone from building the circuit above, but it will cause problems as soon as the switch is closed. This is the reality of KVL: We can try to violate KVL, but will not succeed. As you should realize, this means certain things about the switch matrix. We must avoid switching operations that connect unequal voltage sources. Notice that a wire, or a dead short, can be thought of as a voltage source with V=0, so the warning is a generalization for avoiding shorts across an individual voltage source. A similar constraint holds for Kirchoff’s Current Law (KCL). KCL states that: Currents into a node must sum to zero. A hypothetical current source converter which violates Kirchoff’s Current Law. When current sources are present in a converter, we must avoid any attempts to violate KCL. Consider the simple circuit shown to the right. If the current sources are different and if the switch is opened, the sum of the currents into the node will not be zero. In a real circuit, high voltages will build up and cause an arc to create another current path. This situation has real potential for damage, and a fuse will not help. Again, this is the reality of KCL: We can try to violate KCL, but will not succeed. We must avoid operating switches so that unequal current sources are connected in series. An open circuit can be thought of as a current source with I=0, so the warning applies to the problem of opening an individual current source. Catastrophic results of attempting to violate Kirchoff’s laws. In contrast to conventional circuits, in which KVL and KCL are automatically satisfied, switches do not ‘know’ KVL or KCL. If a designer forgets to check, and accidentally shorts two voltages of breaks a current source connection, some problem will result. The photograph to the right shows a power handling board which was destroyed when an error connected the input voltage source to a reversed voltage. The circuit breakers did not have time to react. KVL and KCL problems when simple energy storage elements are added to a circuit. There are some interesting implications of the current law restrictions when energy storage is included. Look at the circuit diagrams to the right. Both represent ‘circuit law problems’. The voltage source will cause the inductor current to ramp up indefinitely since . We micht consider this to be a ‘KVL problem’, since the long-term effect is similar to shorting the source. In the next example, the current source will cause the capacitor voltage to ramp up toward infinity. This causes a ‘KCL problem’; eventually an arc will form to create an additional current path, just as if the current source had been opened. Of course, these connections are not a problem if they are only temporary. However, it should be evident that an inductor will not support a DC voltage and a capacitor will not support an DC current. Circuit laws are not exclusively linked to problems. Both KVL and KCL provide very positive guidance about how to design a switch matrix. They provide excellent working rules for design. Often, all but one or two possibilities for switches and their operation can be identified simply through careful application of basic circuit laws. Let us return to the diode bridge example to illustrate. Some of the possible connections are shown below. It is easy to see that any diode combination that connects both of the left devices or both of the right devices in the same direction will attempt to violate KVL. Other combinations will not create and KVL or KCL problems, but the diodes will never allow any current to flow through the resistor load. Just a little more thought quickly reduces the switch arrangements to one combination: there is only one connection of the diode bridge that permits energy to flow without violating KVL. Many of the possibilities for diode bridge directions are not useful. KVL determines many effects of current interconnections. We have compared a half-wave rectifier with resistive and R-L loads previously. Consider a half-wave rectifier with a current source load, as shown below. KCL requires that we provide a current path for the output source. Since there is only one loop in the circuit, this path requires the diode to be on at all times. No switching action will ever occur, and the combination will not function as a power converter. The result is quite different when a diode bridge supplies the current source, since several loops can be formed. Power electronics is perhaps the only major subject in which a designer must think carefully about obeying circuit laws!	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://powercircuits.net/kirchhoffs-voltage-and-current-laws.html", "fetch_time": 1569268343000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-39/segments/1568514578201.99/warc/CC-MAIN-20190923193125-20190923215125-00487.warc.gz", "warc_record_offset": 140540526, "warc_record_length": 9005, "token_count": 1173, "char_count": 5688, "score": 3.796875, "int_score": 4, "crawl": "CC-MAIN-2019-39", "snapshot_type": "latest", "language": "en", "language_score": 0.9223032593727112, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00029-of-00064.parquet"}
# Rounding Whole Numbers ## Presentation on theme: "Rounding Whole Numbers"â€” Presentation transcript: Rounding Whole Numbers EnVision Math Topic 1: Lesson 1-4 Rounding Whole Numbers Materials needed: -dry erase board, marker, eraser for each student -large yellow card and large green card for each student and a marker and a way to attach them to the board -number line from slide 9 drawn on board (or just project on board) yellow cards with the numbers 306, 311, 323, 335, 342 on them in case these tens digits are not written by students -green cards with the numbers 351, 362, 374, 385, 397 on them -rounding game cards, gameboard, playing pieces, and die for each pair of students â€¦practice sheet for today. Write your name * on the name line. Look * to see which level you are starting with today. Then find yourâ€¦ â€¦Rocket sheet. Down at the bottom * find the level you are on today. If it is the same level as last time, write your date under the next try. If you passed last time , then you are on a new level today. You may color in the letter that you passed on your rocket. Then you will fill in the date under first try for your new level. Remember, these are our partners for the weekâ€¦ Rocket Math Partners Luka and Issy Mason and Jack Ellie and Sierra Gabe and Wes Samantha and Alyssa Dominic and Jace Emily and Paige Anthony and Oluwafolajinmi When I say go, find your partner and choose a spot. Decide who will be the checker first and get your answer sheet ready. I will give you the signal when you should start practicing. Then I will give you the signal when to switch. Go! (after practice â€“ do one minute timing) Today we have a new big questionâ€¦ How do I round numbers? Can someone read it? You have used place value to read and write numbers. Today, you will use place value to round numbers. When might you want to round numbers? (when you only need to estimate, to check if an answer makes sense) Using what you learned about rounding from last year, here is what I would like you to doâ€¦ Write a number that is greater than 300 that rounds to 300. 400 I am going to give you and your partner a yellow card and a green card. On the yellow card, I would like you toâ€¦* On the green card, I would like you toâ€¦* When you are finished, I would like you to put your card in the correct column on the board. (while kids work, make sure you have the number line drawn on the board â€“ have yellow cards with the numbers 306, 311, 323, 335, 342 on them in case these tens digits are not written by students â€“ have green cards with the numbers 351, 362, 374, 385, 397 on them) Letâ€™s look at our yellow cards. Are there any that you disagree with? Each of these numbers when rounded to the nearest hundred equals What place do you look at if you are rounding to the nearest hundred? (tens) Let me circle all the tens digits. (do) What do you notice? (all less than 5) (do the same for the green cards) Based on what we just found out, letâ€™s make some rules for rounding numbersâ€¦ Write a number that is greater than 300 that rounds to 300. Write a number that is greater than 300 that rounds to 400. 8,723 9 ,000 Round to the nearest thousand. Suppose I wanted to round this number to the nearest thousand. Here is the first thing I would need to doâ€¦* I want to underline that place. What digit is in the thousands place? * Here is step 2 . It is helpful to draw an arrow to the digit to the right. Letâ€™s look again at our numbers on the board. If the digit in the tens place was 5 or more, what happened to our digit in the hundreds place? (had to make it one more â€“ 4) If the digit in the tens place was less than 5, what happened to our digit in the hundreds place? (stayed the same) So, we need to rememberâ€¦* * Is our digit to the right 5 or more or less than 5? What do I do to my thousands digit? * (make it one more) What happens to all these other digits? * (0â€™s) Since I am rounding to the nearest thousand and one thousand has 3 zeroes, my answer should have 3 zeroes. 9 ,000 Find the place that I am rounding to. Look at the digit to the right of this place. If the digit to the right is or more, add 1. If the digit to the right is less than 5, leave it alone. 293,685,407 293,68 5 ,000 Round to the nearest thousand. Now letâ€™s try our rules with a bigger number. Suppose I wanted to round this number to the nearest thousand. First I need toâ€¦. What should I underline? Next I need toâ€¦* What should I draw an arrow to? * Rememberâ€¦* * What will I do to my 5? * (keep it the same) What do I change the digits behind the 5 to? * (0â€™s) What happens to all the digits in front of the 5? * (stay the same) Since I was rounding to the nearest thousand, I should have 3 zeroes in my answer. 293,68 5 ,000 Find the place that I am rounding to. Look at the digit to the right of this place. If the digit to the right is or more, add 1. If the digit to the right is less than 5, leave it alone. 293,685,407 293, 7 00,000 Round to the nearest hundred thousand. This time, letâ€™s round our number to the nearest hundred thousand. What should I underline? * What should I draw an arrow to? * What will I do to my 6? * What do I change the digits behind the 6 to? * What happens to all the digits in front of the 6? * Since I was rounding to the nearest hundred thousand, I should have 5 zeroes in my answer. 293, 7 00,000 Find the place that I am rounding to. Look at the digit to the right of this place. If the digit to the right is or more, add 1. If the digit to the right is less than 5, leave it alone. 293,685,407 293,685, 4 00 Round to the nearest hundred. This time, letâ€™s round our number to the nearest hundred. What should I underline? * What should I draw an arrow to? * What will I do to my 4? * What do I change the digits behind the 4 to? * What happens to all the digits in front of the 4? * Since I was rounding to the nearest hundred, I should have 2 zeroes in my answer. 293,685, 4 00 Find the place that I am rounding to. Look at the digit to the right of this place. If the digit to the right is or more, add 1. If the digit to the right is less than 5, leave it alone. 293,685,407 29 4 ,000,000 Round to the nearest million. This time, letâ€™s round our number to the nearest million. What should I underline? * What should I draw an arrow to? * What will I do to my 3? * What do I change the digits behind the 3 to? * What happens to all the digits in front of the 3? * Since I was rounding to the nearest million, I should have 6 zeroes in my answer. 29 4 ,000,000 Find the place that I am rounding to. Look at the digit to the right of this place. If the digit to the right is or more, add 1. If the digit to the right is less than 5, leave it alone. 293,685,407 293,6 9 0,000 Round to the nearest ten thousand. This time, letâ€™s round our number to the nearest ten thousand. What should I underline? * What should I draw an arrow to? * What will I do to my 8? * What do I change the digits behind the 8 to? * What happens to all the digits in front of the 8? * Since I was rounding to the nearest ten thousand, I should have 4 zeroes in my answer. 293,6 9 0,000 Find the place that I am rounding to. Look at the digit to the right of this place. If the digit to the right is or more, add 1. If the digit to the right is less than 5, leave it alone. 293,685,407 293,685,4 1 Round to the nearest ten. This time, letâ€™s round our number to the nearest ten. What should I underline? * What should I draw an arrow to? * What will I do to my 0? * What do I change the digits behind the 0 to? * What happens to all the digits in front of the 0? * Since I was rounding to the nearest hundred thousand, I should have 1 zero in my answer. Now itâ€™s your turn to try some. Weâ€™ll use our white boards. (pass out boards, markers, erasers) 293,685,4 1 Find the place that I am rounding to. Look at the digit to the right of this place. If the digit to the right is or more, add 1. If the digit to the right is less than 5, leave it alone. Round to the place of the underlined digit. 128,955 I would like you to round this number to the underlined digit. On your board, write this number and draw your underline. Then draw your arrow and write your rounded number below this number. Show me your answer! I had to look at the digit to the right * Since it wasâ€¦, my underlined digit became . Everything behind it, became zeroes. Everything in front of it, stayed * the same. Here is your next oneâ€¦ 12 9 ,000 Round to the place of the underlined digit. 85,639 I would like you to round this number to the underlined digit. Show me your answer! I had to look at the digit to the right * Since it wasâ€¦, my underlined digit became . Everything behind it, became zeroes. Everything in front of it, stayed * the same. Here is your next oneâ€¦ 85,6 4 Round to the place of the underlined digit. 9,924 I would like you to round this number to the underlined digit. Show me your answer! I had to look at the digit to the right * Since it wasâ€¦, my underlined digit became . Everything behind it, became zeroes. Everything in front of it, stayed * the same. Here is your next oneâ€¦ 10 ,000 Round to the place of the underlined digit. 1,194,542 I would like you to round this number to the underlined digit. Show me your answer! I had to look at the digit to the right * Since it wasâ€¦, my underlined digit became . Everything behind it, became zeroes. Everything in front of it, stayed * the same. Here is your next oneâ€¦ 9 1,1 0,000 Round to the place of the underlined digit. 160,656 I would like you to round this number to the underlined digit. Show me your answer! I had to look at the digit to the right * Since it wasâ€¦, my underlined digit became . Everything behind it, became zeroes. Everything in front of it, stayed * the same. Here is your next oneâ€¦ 2 00,000 Round to the place of the underlined digit. 493,295 I would like you to round this number to the underlined digit. Show me your answer! I had to look at the digit to the right * Since it wasâ€¦, my underlined digit became . Everything behind it, became zeroes. Everything in front of it, stayed * the same. Here is your next oneâ€¦ 49 3 ,000 Round to the place of the underlined digit. 39,230 I would like you to round this number to the underlined digit. Show me your answer! I had to look at the digit to the right * Since it wasâ€¦, my underlined digit became . Everything behind it, became zeroes. Everything in front of it, stayed * the same. Here is your next oneâ€¦ 3 9 ,000 Round to the place of the underlined digit. 77,282 I would like you to round this number to the underlined digit. Show me your answer! I had to look at the digit to the right * Since it wasâ€¦, my underlined digit became . Everything behind it, became zeroes. Everything in front of it, stayed * the same. Here is your next oneâ€¦ 77,2 8 Round to the place of the underlined digit. 54,846 I would like you to round this number to the underlined digit. Show me your answer! I had to look at the digit to the right * Since it wasâ€¦, my underlined digit became . Everything behind it, became zeroes. Everything in front of it, stayed * the same. You can also round money amounts. This is especially useful when you are at the store and trying to see if you have enough money. 5 0,000 Round to the nearest dollar. \$6.73 I want to round this money amount to the nearest dollar. What digit is in the dollar place? * I still look to the right . What happens to my 6? What happens to everything behind the 6? * What am I missing? * \$ 7 .00 Round to the nearest dollar. \$254.13 Letâ€™s round this money amount to the nearest dollar. What digit is in the dollar place? * To the right , there is a 1. What happens to my 4? What happens to everything behind the 4? * Everything before the 4 * stays the same. \$25 .00 4 Round to the nearest dollar. \$92.87 Letâ€™s round this money amount to the nearest dollar. What digit is in the dollar place? * To the right , there is an 8. What happens to my 2? What happens to everything behind the 2? * Everything before the 2 * stays the same. Now itâ€™s your turn. \$9 .00 3 Round to the nearest dollar. \$537.27 Round this money amount to the nearest dollar. Here is the dollar place. * I look to the right. * My 7 * stays the same. Everything behind it becomes * 0. Everything before it * stays the same. \$53 .00 7 Round to the nearest dollar. Round this one to the nearest dollar. Here is the dollar place. * I look to the right. * My 8 * becomes a 9. Everything behind it becomes * 0. I can also round money amounts to the nearestâ€¦ \$38.95 \$3 .00 9 Round to the nearest ten dollars. \$57.25 â€¦ten dollars. In the ten dollar place, I have aâ€¦* To the right , there is a 7. What happens to my 5? What happens to everything behind the 5? * \$6 .00 Round to the nearest ten dollars. \$32.99 Round this one to the nearest ten dollars. What digit is in the dollar place? * To the right , there is a 2. What happens to my 3? What happens to everything behind the 3? * \$3 .00 Round to the nearest ten dollars. \$91.88 Round this money amount to the nearest ten dollars. Here is the ten dollar place. * I look to the right. * My 9 * stays the same. Everything behind it becomes * 0. \$9 .00 Round to the nearest ten dollars. Round this money amount to the nearest ten dollars. Here is the ten dollar place. * I look to the right. * My 7 becomes...* Everything behind it becomes * 0. To do a little more practice, you are going to play a game with a partner. You and your partner will have a bag of cards. When it is your turn, reach in and pull out a card. Both people should write their answer on the board. When finished, check the answer on the back of the card. If you are correct, you may roll the die. If not, figure out your mistake. Continue playing until someone wins or time runs out. \$78.15 \$8 .00	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://slideplayer.com/slide/6090052/", "fetch_time": 1713936091000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-18/segments/1712296819067.85/warc/CC-MAIN-20240424045636-20240424075636-00407.warc.gz", "warc_record_offset": 458846932, "warc_record_length": 27912, "token_count": 3697, "char_count": 13960, "score": 4.1875, "int_score": 4, "crawl": "CC-MAIN-2024-18", "snapshot_type": "latest", "language": "en", "language_score": 0.9282290935516357, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00003-of-00064.parquet"}
# How do you solve 5a=15? Jan 21, 2017 $a = 3$ #### Explanation: $5 a = 15$ Divide both sides by 5 $\frac{{\cancel{5}}^{1} a}{\cancel{5}} ^ 1 = {\cancel{15}}^{3} / {\cancel{5}}^{1}$ $a = 3$	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://socratic.org/questions/how-do-you-solve-5a-15", "fetch_time": 1726383543000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00163.warc.gz", "warc_record_offset": 504290627, "warc_record_length": 5624, "token_count": 88, "char_count": 195, "score": 4.3125, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.5824059844017029, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00047-of-00064.parquet"}
Extension springs are found in various applications, from everyday objects to sophisticated equipment. A critical component of these springs is the spring rate, which determines how the spring performs when forces are applied to it. This parameter is key to the functionality of your design and ensures its safety. This article offers a step-by-step guide on determining the extension spring rates, interpreting practical outcomes, and selecting suitable springs for your project. Learn about the integral aspects of extension spring rates to enhance your engineering applications. ## What is Spring Rate? Spring rate refers to the resistance provided by an extension spring to a specific load, numerically determining how much force is required to make the spring extend to a certain length. This is typically denoted as force per unit length, which includes measurements like pounds per inch (lb/in) or newtons per millimeter (N/mm). The spring rate of an extension spring is distinct from that of a compression spring. For extension springs, a low rate means that a spring can extend more under the same force while a high rate signifies a more rigid spring that needs a larger force to stretch to the same length. The relationship between the extension of a spring and the force it exerts is highlighted in Hooke's law, indicating that the level of force a spring applies is directly proportional to its extension, assuming the spring's elastic limit is not exceeded. Hooke's law includes the concept of the spring rate. Knowledge of the spring rate supports appropriate design and selection of extension springs for specific roles. For instance, in a garage door mechanism, a suitable spring rate aids in maintaining safety and functionality. If the spring rate is excessively low, the door could descend too rapidly. On the other hand, a very high spring rate may impede lifting the door. As a result, understanding spring rates assists engineers in selecting the best spring for their objective. ## How to Calculate Spring Rate of an Extension Spring The spring rate 'k' of an extension spring is determined using the designated formula: k = (G * d^4) / (8D^3 * n) Here: • G is the Modulus of Rigidity for the spring material • d is the Wire Diameter • D represents the Mean Coil Diameter • n signifies the Number of Active Coils While using this formula, ensure the units for all variables are uniform. For instance, if using Imperial units like pounds and inches, correspondingly adjust the Modulus of Rigidity (G) to these units prior to calculation. If taken for example an extension spring that has: a wire diameter (d) of 0.035 inches, a mean diameter (D) of 0.250 inches, 10 active coils (n), and a modulus of rigidity (G) of 11.5 million pounds per square inch (11,500,000 psi). The spring rate calculation is: k = (11,500,000 * 0.035^4) / (8 * 0.250^3 * 10) = 13.81 pounds per inch (lb/in) This spring rate indicates that the force required to stretch this spring by one inch is 13.81 pounds. Caution: The above computation is based on a linear relationship between the applied force and the extension of the spring, integral to Hooke's Law. This law remains valid within elastic deformation but may not hold true beyond its elastic limit (plastic deformation). Accordingly, this calculated spring rate should be applied within the load and elongation limits of the respective spring, the requirement of the application, and the spring material's mechanical properties. ## Conclusion In closing, extension springs and in particular, their spring rates, form an essential part of many engineer's day-to-day tasks. This isn't solely a theoretical idea, but it also carries practical importance for your designs. The spring rate, representing the spring's response to a load numerically, can be computed with precision for extension springs. Mastering this rate calculation will enable you to forecast the performance of your extension springs with increased accuracy, ultimately improving the reliability of your projects. While the concept might appear intimidating initially, repeated application will make it familiar. This understanding is a valuable asset for competent engineering.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.coilsdb.com/extension_spring_rates", "fetch_time": 1726084887000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651400.96/warc/CC-MAIN-20240911183926-20240911213926-00314.warc.gz", "warc_record_offset": 654412775, "warc_record_length": 7501, "token_count": 853, "char_count": 4217, "score": 3.640625, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.9212213158607483, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00035-of-00064.parquet"}
# Metodo De La Carga Balanceada • March 2021 • PDF This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. ### More details • Words: 302 • Pages: 11 INTRODUCCION. CARGAS Y MOMENTOS EQUIVALENTES PRODUCIDOS POR TENDONES PRESFORZADOS EL PRESFORZAR MIEMBROS MEDIANTE TENDONES CURVOS O ATIRANTADOS TIENE ASI EL EFECTO DE INTRODUCIR UN JUEGO DE CARGAS TRANSVERSALES EQUIVALENTES, Y ESTAS PUEDEN TRATARSE COMO CUALQUIER OTRA CARGA EXTERNA PARA LOS FINES DE MOMENTOS Y DEFLEXIONES EL CONCEPTO DE CARGA EQUIVALENTE O CARGA BALANCEADA ES UN METODO PARA DETERMINAR LA FUERZA PRETENSORA Y LA EXCENTRICIDAD REQUERIDAS. SE ESTABLECE LA FUERZA PRETENSORA Y EL PERFIL DEL TENDON DE TAL MANERA QUE LAS CARGAS EXTERIORES QUE ACTUAN SEAN CONTRARRESTADAS EXACTAMENTE POR LAS FUERZAS TRANSVERSALES QUE RESULTAN DEL PRESFUERZO. EL RESULTADO NETO DE ESTE JUEGO DE CARGAS EXTERIORES, ES QUE LA VIGA ESTÉ SUJETA SOLAMENTE A CARGA AXIAL Y NO TIENE MOMENTO FLECTOR. LA CARGA BALANCEADA QUE SE ESCOGE GENERALMENTE ES LA SUMA DEL PESO PROPIO Y LA CARGA MUERTA SOBREPUESTA, AUNQUE PUEDE INCLUIR UNA FRACCION DE LA CARGA VIVA PoPo + CM + FRACCION CV COMO LA CARGA EXTERNA ES UNIFORMEMENTE DISTRIBUIDA, ES CONVENIENTE QUE EL TENDON TENGA UNA FORMA PARABÓLICA Y PRODUCIRÁ UNA CARGA UNIFORME HACIA ARRIBA DE : Wp= 8Py/l2 P = FUERZA PRETENSORA Y = ES LA FLECHA MAXIMA L = CLARO DE LA VIGA PARA VIGAS CONTINUAS LA FUERZA PRETENSORA ES LA MISMA EN LOS DOS CLAROS ADYACENTES AL APOYO INTERIOR Y SI LA PENDIENTE DEL TENDON ES LA MISMA EN CADA LADO ENTONCES EL MOMENTO DE FLEXION ES CERO, EN ESA UBICACIÓN LA FUERZA DESBALANCEADA SERA LA FUERZA VERTICAL QUE PROVIENE DEL CAMBIO DE PENDIENTE DEL TENDON, LA CUAL PASA DIRECTAMENTE AL APOYO. EL TENDON PUEDE LEVANTARSE HASTA LA MÁXIMA EXCENTRICIDAD QUE PERMITE EL RECUBRIMIENTO DEL CONCRETO MAXIMIZANDO LA FLECHA Y EN LOS CLAROS ADYACENTES, Y MINIMIZANDO LA FUERZA PRETENSORA REQUERIDA PARA SOPORTAR LA CARGA ESPECIFICADA. EN LOS EXTREMOS LIBREMENTE EXCENTRICIDAD ES NULA LA #### Related Documents March 2021 239 last month 35 March 2020 421 ##### Ejercicios Metodo De La Secante December 2020 569 ##### Ejercicios De Capacidad De Carga October 2020 1,000 October 2019 430 #### More Documents from "Deiby Cry Pain" March 2021 239 July 2021 374 ##### Herramienta De Auditoria Meycor Cobit November 2020 288 December 2020 387 October 2019 747 July 2021 304	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://doku.pub/documents/metodo-de-la-carga-balanceada-oq1z979o8702", "fetch_time": 1660910692000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-33/segments/1659882573667.83/warc/CC-MAIN-20220819100644-20220819130644-00067.warc.gz", "warc_record_offset": 228321185, "warc_record_length": 6106, "token_count": 904, "char_count": 2525, "score": 3.65625, "int_score": 4, "crawl": "CC-MAIN-2022-33", "snapshot_type": "longest", "language": "en", "language_score": 0.30941042304039, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00009-of-00064.parquet"}
# How to Check if Given Number is Prime in Java – With Example Hello guys, today, we are going to discuss one of the most common programming exercises for beginners is, write a program to check if a given number is prime or not? There are many ways to check if a number is prime or not, but the most common of them is the trial division, which is what we will see in this tutorial. In my opinion, these kinds of programs are their first steps towards algorithmic understanding. You first come up with a solution, which is driven by the fact that prime numbers are natural numbers, that are not divisible by any positive number other than 1 and themselves. Then, you write a for loop to check every number, starting from 1 to a given number, to see if the given number is divisible by any positive number or not. This leads you to the solution. Then you find some more the fact that there is no need to check till N-1, where N is the number we are checking for primeness, and checking till the square root of N is enough. This reduces a lot of time, especially while checking a large number is prime or not. Further, you come to know that if it’s not divisible by 2, then there is no need to checking for any other even number, and you increment the counter by 2 instead of 1. So in a way, you learn how to optimize your solution by taking advantage of the facts available. After this, you can try something like the Fibonacci series or maybe finding factorial of a number in Java to do some more practice on programming. This will not only teach you language basics like loops, control statements like if-else, use of arithmetic, and relational operator but also helps to develop programming logic. By the way, you can even take this problem of checking if a number is prime or not, to the next level, by trying to implement different algorithms for finding primes like the sieve of Atkin or sieve of Eratosthenes. In fact, in programming challenges, you often need to build your prime number cache up to a specific number to progress further in finding a solution. ## How to Find if a Given Integer Number is a Prime Number or Not? Now, we’ll understand our Java program to see if a given integer number is prime or not. As I said, a number is called a prime number if it’s only divisible by 1 or itself, which means the prime number doesn’t have any positive divisor other than itself. There are many ways to check if the number is prime or not or generating a list of primes. The most straightforward of them is known as trial division, which is a natural way of finding prime. In the trial division, you divide. It consists of testing whether n is a multiple of any integer between 2 and sqrt{n}. In this program, I have presented three solutions or methods to check if the number is prime. The first solution is the implementation of the trial division, where we are checking from 2 to sqrt(n); we are using java.lang.Math class for calculating the square root. Since this function returns double, we need to cast the result back into an integer. Our second method of checking if a number is prime or not is a little bit optimized than this as it doesn’t check division by even numbers other than two. The third method is most optimized for all three methods of prime number checking. Btw, if you are also preparing for coding interviews or improving your algorithmic skill then I suggest you take a look at this wonderful course from Educative, Grokking the Coding Interview: Patterns for Coding Questions. This is one of its kind courses that will not just teach you to solve the problem but also the pattern behind them, which means you can remember those patterns and apply them to many problems. A great way to build your coding and problem-solving skills. ## Prime Number Checker in Java And, here is the complete Java program to check if a given number is prime or not. This question is also asked on written tests and interviews as to how to print prime numbers from 1 to 100 or finding the prime factor of a number in Java. And, there is another exercise for you to do after this is checking if a number is Armstrong’s number or not. ```import java.util.Scanner; /** * Java Program to check if a number is Prime or Not. This program accepts a * number from command prompt and check if it is prime or not. * * @author http://java67.blogspot.com / public class PrimeTester { public static void main(String args[]) { Scanner scnr = new Scanner(System.in); int number = Integer.MAX_VALUE; System.out.println("Enter number to check if prime or not "); while (number != 0) { number = scnr.nextInt(); System.out.printf("Does %d is prime? %s %s %s %n", number, } } / * Java method to check if an integer number is prime or not. * @return true if number is prime, else false / public static boolean isPrime(int number) { int sqrt = (int) Math.sqrt(number) + 1; for (int i = 2; i < sqrt; i++) { if (number % i == 0) { // number is perfectly divisible - no prime return false; } } return true; } / * Second version of isPrimeNumber method, with improvement like not * checking for division by even number, if its not divisible by 2. / public static boolean isPrimeNumber(int number) { if (number == 2 \|\| number == 3) { return true; } if (number % 2 == 0) { return false; } int sqrt = (int) Math.sqrt(number) + 1; for (int i = 3; i < sqrt; i += 2) { if (number % i == 0) { return false; } } return true; } / * Third way to check if a number is prime or not. / public static String isPrimeOrNot(int num) { if (num < 0) { return "not valid"; } if (num == 0 \|\| num == 1) { return "not prime"; } if (num == 2 \|\| num == 3) { return "prime number"; } if ((num num - 1) % 24 == 0) { return "prime"; } else { return "not prime"; } } } Output Enter number to check if prime or not 2? Does 2 is prime? true prime number true 3? Does 3 is prime? true prime number true 4? Does 4 is prime? false not prime false 5? Does 5 is prime? true prime true 6? Does 6 is prime? false not prime false 7? Does 7 is prime? true prime true 17? Does 17 is prime? true prime true 21? Does 21 is prime? false not prime false 131? Does 131 is prime? true prime true 139? Does 139 is prime? true prime true``` That’s all in this program about how to check if a number is prime or not. The number must be an integer, as the concept of prime is only for natural numbers and not for floating-point numbers. As I said, there are a couple of more algorithms for checking if a number is prime or not, and some of the algorithms are optimized for finding prime numbers. It is imperative for every programmer to know at least one fast way of finding a prime number, as this trial division is not fast enough for real-world problems. I suggest exploring the sieve of Atkin and the sieve of Eratosthenes’s way of finding prime numbers. ### 1 thought on “How to Check if Given Number is Prime in Java – With Example” error: Content is protected !!	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://gradjobopenings.com/how-to-check-if-given-number-is-prime-in-java-with-example/", "fetch_time": 1695776284000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-40/segments/1695233510238.65/warc/CC-MAIN-20230927003313-20230927033313-00436.warc.gz", "warc_record_offset": 309972089, "warc_record_length": 37207, "token_count": 1689, "char_count": 6957, "score": 3.734375, "int_score": 4, "crawl": "CC-MAIN-2023-40", "snapshot_type": "latest", "language": "en", "language_score": 0.9264451265335083, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00050-of-00064.parquet"}
# How to Find the Limit of a Function in Calculus It is important to understand the concept of limits because it will be present in many calculus problems. Limit shows us how some value has limit in f(y) if the f(x) approaches that value. For instance, if we have a limit that will be when X goes to zero; y goes to 1. ## Instructions 1) Write down the equation for the function. For example: y = f(x) = ( x^2 -2x +3 ) / ( x – 3 ) 2) Set up the limit which includes the value that x approaches From the example: LIM [ ( x^2 -2x +3 ) / ( x - 3 ) ] (x -> +3 ) 3) Factorize and reduce the function. Algebraic methods should be used here. ( x^2 -2x +3 ) / ( x – 3 ) = [ ( x + 1) ( x - 3 ) ] / ( x – 3 ) = ( x + 1 ) 4) Replace the expression which is reduced on the limit. Solving the limit. You will do this if replace the value of the variable ( x ) with the value it approaches ( x -> +3, replace “x” with +3). LIM [ ( x + 1 ) ] (x -> +3 ) ( +3 +1 ) = 4 LIM [ ( x + 1 ) ] = 4 (x -> +3 ) ## Tips and warnings When you solve limit problems you must be sure that you didn’t end with qualities that are not defined. You will have undefined quantity if you divide any of your expressions by zero. You will prevent undefined values if you factorize and reduce those expressions.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.calculatorium.com/how-to-find-the-limit-of-a-function-in-calculus/", "fetch_time": 1409350098000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2014-35/segments/1408500833461.95/warc/CC-MAIN-20140820021353-00036-ip-10-180-136-8.ec2.internal.warc.gz", "warc_record_offset": 303920975, "warc_record_length": 9751, "token_count": 372, "char_count": 1291, "score": 4.15625, "int_score": 4, "crawl": "CC-MAIN-2014-35", "snapshot_type": "longest", "language": "en", "language_score": 0.8575579524040222, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00029-of-00064.parquet"}
Åk 6–9 English/Soomaali 4.1 Rounding Decimals How much does 43.6 liters of gasoline (petrol) cost if the prise per liter is 10.97 kr? The calculation is 43.6 · 10.7 which gives the answer 478.292 kr. You can’t pay exactly 478.29 kr and you need therefore, to round off. Here rounding off the answer to 478.50 kr or 478 kr and 50 öre. Because we don’t have tenths of öre, or one öre or ten öres, we round the answer to the nearest 50 öre or 0.5 kr. Foto: Fredrik Enander We will now look at two examples for how you can think about rounding. Example A Round the following number to one decimal (tenths). Example B Round 3.8501 to one decimal, in other words tenths. We then see that 3.8501 is closer to 3.9 then 3.8. So we round the number to 3.9.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.webbmatte.se/display_page.php?language=som&HeadLanguage=eng&id=234&on_menu=1311&no_cache=284343861", "fetch_time": 1618430179000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-17/segments/1618038078021.18/warc/CC-MAIN-20210414185709-20210414215709-00633.warc.gz", "warc_record_offset": 188853605, "warc_record_length": 5257, "token_count": 239, "char_count": 755, "score": 3.640625, "int_score": 4, "crawl": "CC-MAIN-2021-17", "snapshot_type": "latest", "language": "en", "language_score": 0.8200743794441223, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00008-of-00064.parquet"}
# Physics A rocket of mass,m is located at an alttitude h meters above the surface of a planet of a mass,M and radius, R. write an expression for the gravitational pull on it. Deduce an expression for the gravitational potential and escape velocity at that location. 1. 👍 2. 👎 3. 👁 1. F = G m M/r^2 = G m M/(R+h)^2 potential = - G m M /(R+h) (1/2) m v^2 = G m M/(R+h) so v^2 = 2 G M/(R+h) 1. 👍 2. 👎 ## Similar Questions 1. ### Algebra 2 h A model rocket is launched from the top of a building. The height (in meters) of the rocket above the ground is given by h( t) = -16t²+ 96t + 14, where t is the time (in seconds) since the rocket was launched. What is the 2. ### physics-forces 1. a fully loaded rocket has a mass of 2.92x10^6kg. It's engines has a thrust of 3.34x10^7N a)what is the downward force of gravity of the rocket at blast off? b)what is the unbalanced force on the rocket at blast off? c)what is 3. ### Algebra A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y = –0.06x² + 9.6x + 5.4 where x is the horizontal distance, in meters, from the starting point on the roof and y 4. ### Physicis A rocket initially at rest accelerates at a rate of 99.0 meters/second2. Calculate the distance covered by the rocket if it attains a final velocity of 445 meters/second after 4.50 seconds. A.) 2.50 x 10(2) meters B.) 1.00 x 10(3) 1. ### Math A model rocket is launched from a roof into a large field. The path the rocket can be modeled by the equation y=-0.04x^2+8.3x+4.3 where x is the horizontal distance in meters, from the starting point on the roof and y is the 2. ### Physics A model rocket has a mass of 0.435 kg. It is mounted vertically and launches directly up with a thrust force of 13.1 N. a) What is the weight of the rocket? b) Determine the net force on the rocket as it is launched. c) What is 3. ### Math a model rocket is launched from a roof into a large feild. The path of the rocket can be modeled by the equation y=-0.8x^(2)+12x+25.8 where x is the horizontal distance, in meters, from the starting point on the roof and y is the 4. ### Physics With the engines off, a spaceship is coasting at a velocity of +210 m/s through outer space. It fires a rocket straight ahead at an enemy vessel. The mass of the rocket is 1350 kg, and the mass of the spaceship (not including the 1. ### Physics A rocket that weighs 7840 N on earth is fired. The force of propulsion is + 10 440N. Determine a) the mass of the rocket, b) the acceleration of the rocket, c) the velocity of the rocket at the end of 8.0 s. 2. ### physics Two rockets with the same mass are. Accelerated. Rocket A accelerates twice as fast as Rocket B. Which statement is correct? The motor in rocket A is four times as powerful as the motor and in rocket B The motor in rocket A Is 3. ### physics Two rockets with the same mass are accelerated. Rocket A accelerates twice as quickly as rocket B. Which statement is correct?(1 point) The motor in rocket A is half as powerful as the motor in rocket B. The motor in rocket A is 4. ### Algebra A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y = -0.04x^2+5.8x+4.9, where x is the horizontal distance, in meters, from the starting point on the roof and y is	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.jiskha.com/questions/1735152/a-rocket-of-mass-m-is-located-at-an-alttitude-h-meters-above-the-surface-of-a-planet-of-a", "fetch_time": 1611219525000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-04/segments/1610703524270.28/warc/CC-MAIN-20210121070324-20210121100324-00023.warc.gz", "warc_record_offset": 824299946, "warc_record_length": 4755, "token_count": 950, "char_count": 3349, "score": 3.578125, "int_score": 4, "crawl": "CC-MAIN-2021-04", "snapshot_type": "latest", "language": "en", "language_score": 0.9277234673500061, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00049-of-00064.parquet"}
Search 81,226 tutors ## Philip P.'s Resources Your equations are correct: f(x) = (\$4.40)x + \$50 g(x) = (\$3.90)x + \$90 x represents the number of mugs you purchase and f(x) and g(x) represent your cost. (c) x = 120 mugs f(120) = (\$4.40)(120) +... The money is divided into 1 + 3 + 1/2 = 4 1/2 parts. The smallest part is the 1/2 part, which equals \$42.50. (1/2)/(41/2) = \$42.50/x 1/9 = \$42.50/x x = (\$42.50)9 = \$382.50 The results of a survey showed that 3 out of 10 people had voted for Lucky Luke. If there were 40,000 votes: a) what is the ratio of those who voted for Lucky Luke to those who did not? (3/10)/(7/10) = 3/7 b) How many voted for Luke? 3/10 = x/40,000. Solve for... Let x = the numerator. The denominator exceeds the numerator by 7, so it's x+7. The full fraction is: x/(x+7) If 3 is added to the numerator and 1 is subtracted from the denominator, it's equal to 4/5 ... ∫(2√x)/(√x) dx Note that √x = x1/2. Now let u = x1/2: du = (1/2)x-1/2 dx = (1/2u) dx. Hence 2u du = dx. Now substitute: ∫(2u/u) 2u du = 2 ∫ 2u du Note that 2u = eln(2^u) = euln(2) ... I assume the expression is 1/(x2+y). In math division by zero is undefined. So when you have a rational expression like this, the expression will be undefined when the denominator is zero. The denominator is (x2+y), so we need to know the values of x and y that can make... Let x = the unknown number(s). (x + 6)2 = 169 x2 + 12x + 36 = 169 x2 + 12x - 133 = 0 (x+19)(x-7) = 0 x = -19 and 7 Use the "vertex form" of the quadratic equation: y = a(x-h)2 + k where (h,k) is the location of the vertex. In this problem (h,k) is given as (3,-4) so: y = a(x-3)2 - 4 To... 2x + x + 4 = -17 Subtract 4 from both sides: 2x + x + 4 - 4 = -17 - 4 Combine like terms: 3x = -21 Divide both sides by 3: ... (3x5y3)3 (5xy7)2 = 33(x5)3(y3)3(5)2(x)2(y7)2 = 27x15y925x2y14 = (2725)(x15+2)(y9+14) = 675x17y23 y= x3 - 11x2 + 24x + 5 dy/dx = d(x3)/dx - d(11x2)/dx + d(24x)/dx + d(5)/dx dy/dx = 3x2 - 22x + 24 The Coulomb Force is an inverse square force; that is, it's proportional to 1/r2 where r is the distance between the particles. F = (1/4piε0)q1q2/r2 so when you cut the distance in half, the force increases by a factor of four: 1/(1/2)2 = 1/(1/4) = 4. A direct variation equation has the general form: y = kx where k is the constant of variation. Since k is given as -8, the direct variation equation in this problem is: y = -8x For... Let x = the unknown number The number divided by 4: x/4 equals 6: = 6 x/4 = 6 Multiply both sides by 4: x = 64 x = 24 log3(9/√3) = log3(32/31/2) Now use the log rule log(a/b) = log(a) - log(b) = log3(32) - log3(31/2) = 2 - 1/2 = 3/2 Assuming that 17,561,600 is the number of workers in 2008, then the equation would be: W(t) = (322,560)t + 17,561,600 Where t is the number of years since 2008. In 2018, t = 2018-2008 = 10. Plug t= 10 into the above equation... P(t) = P0ekt Where: P(t) = the population at time t P0 = the initial population k = the exponential growth rate t = time e = Euler's Number ≅ 2.718 If the population doubles when t = 62 years, then P(t=62) = 2P0. ... 2x - 2y = -3 -4x + 4y = 6 Multiply the first equation by 2: 2(2x - 2y = -3) = 4x - 4y = -6 Now add this to the second equation: 4x - 4y = -6 -4x + 4y = 6 ----------------- ... Percent Increase = (100%)* (Final Amount)/(Starting Amount) = (100%)*(1,900,000)/(200,000) ... f(x) = (4x-6)/√(8-x) In math, division by zero is undefined. So any value of x that makes the denominator of f(x) equal zero will make f(x) undefined. All such values of x are excluded from the domain of f(x). The denominator of f(x) is √(8-x), and √(8-x) = 0 when...	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.wyzant.com/resources/users/84993890", "fetch_time": 1448945267000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2015-48/segments/1448398464396.78/warc/CC-MAIN-20151124205424-00275-ip-10-71-132-137.ec2.internal.warc.gz", "warc_record_offset": 908268246, "warc_record_length": 13506, "token_count": 1495, "char_count": 3956, "score": 4.15625, "int_score": 4, "crawl": "CC-MAIN-2015-48", "snapshot_type": "latest", "language": "en", "language_score": 0.7905395030975342, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00057-of-00064.parquet"}
# Convert 500.0 Decihertz to Hertz 500.0 Decihertz (dHz) = 50.0 Hertz (Hz) 1 dHz = 0.1 Hz 1 Hz = 10.0 dHz • Q: How many Decihertz in a Hertz? • Q: How do you convert 500 Decihertz (dHz) to Hertz (Hz)? 500 Decihertz is equal to 50.0 Hertz. Formula to convert 500 dHz to Hz is 500 / 10 • Q: How many Decihertz in 500 Hertz?	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.flightpedia.org/convert/500-decihertz-to-hertz.html", "fetch_time": 1653360730000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00602.warc.gz", "warc_record_offset": 887472820, "warc_record_length": 4547, "token_count": 135, "char_count": 327, "score": 3.546875, "int_score": 4, "crawl": "CC-MAIN-2022-21", "snapshot_type": "latest", "language": "en", "language_score": 0.5875939726829529, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00057-of-00064.parquet"}
# How to calculate some conditional probabilities A red die, a blue die, and a yellow die (all six sided) are rolled. Given that no two of the dice land on the same number, what is the conditional probability that blue is less than yellow which is less than red? The Answer is a sixth. I have absolutely no idea how to do this though. Imagine recording the outcomes as $(a,b,c)$, where $a$ is the number on the red, $b$ the number on the blue, and $c$ the number on the yellow. Fix a particular set of distinct numbers, such as $\{1,4,5\}$. All ordered triples (permutations) of these three numbers are equally likely. There are $3!$ such permutations. In exactly $1$ of these permutations, we have $\text{blue}\lt \text{yellow}\lt \text{red}$. Thus the required probability is $\frac{1}{3!}$. Since the three die rolls are unequal, every set of possible rolls $\{x,y,z\}$, with $x < y < z$, can be obtained in exactly six ways: $$\begin{array}{\|c\|c\|c\|} \hline \text{blue} & \text{yellow} & \text{red} \\ \hline x & y & z \\ y & z & x \\ z & x & y \\ x & z & y \\ y & x & z \\ z & y & x \\ \hline \end{array}$$ Note that out of these six equally likely possibilities, the only one where blue < yellow < red is $x, y, z$ (since we assumed $x < y < z$).	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/781776/how-to-calculate-some-conditional-probabilities", "fetch_time": 1556267572000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-18/segments/1555578762045.99/warc/CC-MAIN-20190426073513-20190426095513-00287.warc.gz", "warc_record_offset": 484403503, "warc_record_length": 32409, "token_count": 364, "char_count": 1254, "score": 4.1875, "int_score": 4, "crawl": "CC-MAIN-2019-18", "snapshot_type": "latest", "language": "en", "language_score": 0.8738330006599426, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00019-of-00064.parquet"}
Exam_2_20C_Solution # Exam_2_20C_Solution - Math 20C Summer 2010 Exam 2 Solution... This preview shows pages 1–2. Sign up to view the full content. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: Math 20C. Summer, 2010. Exam 2 Solution. 1. The radius of the circle is √ 4 = 2 meters. • Since the object is moving counterclockwise, κ = 1 2 meter − 1 • T = < − 1 , > • N = < , − 1 > • a = s ′′ T + κ ( s ′ ) 2 N • < 5 , − 8 > = s ′′ < − 1 , > + 1 2 ( s ′ ) 2 < , − 1 > = < s ′′ , − 1 2 ( s ′ ) 2 > • s ′′ = − 5 • 1 2 ( s ′ ) 2 = 8 • ( s ′ ) 2 = 16 • s ′ = 4. The object is traveling 4 meters per second and is slowing down 5 meters per second 2 2. f ( x,y ) = 2 x − y + 6 • f = c ⇐⇒ 2 x − y + 6 = c ⇐⇒ y = 2 x + 6 − c • f = 4 on y = 2 x + 2 • f = 6 on y = 2 x • f = 8 on y = 2 x − 2 • f x = ∂ ∂x (2 x − y +6) = 2 • f y = ∂ ∂y (2 x − y +6) = − 1 • ∇ f = ( 2 , − 1 ) • Figure A3 4 6 8 x 2 y 1 ∇ f Figure A3 3. V = πr 2 h • V r = 2 πrh • V h = πr 2 • V (3 , 2) = π (3 2 )(2) = 18 π • V r (3 , 2) = 2 π (3)(2) = 12 π • V h (3 , 2) = π (3 2 ) = 9 π • Tangent plane:... View Full Document {[ snackBarMessage ]} ### Page1 / 2 Exam_2_20C_Solution - Math 20C Summer 2010 Exam 2 Solution... This preview shows document pages 1 - 2. Sign up to view the full document. View Full Document Ask a homework question - tutors are online	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.coursehero.com/file/6378098/Exam-2-20C-Solution/", "fetch_time": 1513439938000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-51/segments/1512948588251.76/warc/CC-MAIN-20171216143011-20171216165011-00630.warc.gz", "warc_record_offset": 697213155, "warc_record_length": 23362, "token_count": 623, "char_count": 1483, "score": 3.96875, "int_score": 4, "crawl": "CC-MAIN-2017-51", "snapshot_type": "latest", "language": "en", "language_score": 0.6562923789024353, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00026-of-00064.parquet"}
# Inverse Distance Weighted Interpolation One of the most commonly used techniques for interpolation of scatter points is inverse distance weighted (IDW) interpolation. Inverse distance weighted methods are based on the assumption that the interpolating surface should be influenced most by the nearby points and less by the more distant points. The interpolating surface is a weighted average of the scatter points and the weight assigned to each scatter point diminishes as the distance from the interpolation point to the scatter point increases. Several options are available for inverse distance weighted interpolation. The options are selected using the Inverse Distance Weighted Interpolation Options dialog. This dialog is accessed through the Options button next to the Inverse distance weighted item in the 2D Interpolation Options dialog. SMS uses Shepard's Method for IDW: ## Shepard's Method The simplest form of inverse distance weighted interpolation is sometimes called "Shepard's method" (Shepard 1968). The equation used is as follows: where n is the number of scatter points in the set, fi are the prescribed function values at the scatter points (e.g. the data set values), and wi are the weight functions assigned to each scatter point. The classical form of the weight function is: where p is an arbitrary positive real number called the power parameter (typically, p=2) and hi is the distance from the scatter point to the interpolation point or where (x,y) are the coordinates of the interpolation point and (xi,yi) are the coordinates of each scatter point. The weight function varies from a value of unity at the scatter point to a value approaching zero as the distance from the scatter point increases. The weight functions are normalized so that the weights sum to unity. The effect of the weight function is that the surface interpolates each scatter point and is influenced most strongly between scatter points by the points closest to the point being interpolated. Although the weight function shown above is the classical form of the weight function in inverse distance weighted interpolation, the following equation is used in SMS: where hi is the distance from the interpolation point to scatter point i, R is the distance from the interpolation point to the most distant scatter point, and n is the total number of scatter points. This equation has been found to give superior results to the classical equation (Franke & Nielson, 1980). The weight function is a function of Euclidean distance and is radially symmetric about each scatter point. As a result, the interpolating surface is somewhat symmetric about each point and tends toward the mean value of the scatter points between the scatter points. Shepard's method has been used extensively because of its simplicity. ## Computation of Nodal Function Coefficients In the IDW Interpolation Options dialog, an option is available for using a subset of the scatter points (as opposed to all of the available scatter points) in the computation of the nodal function coefficients and in the computation of the interpolation weights. Using a subset of the scatter points drops distant points from consideration since they are unlikely to have a large influence on the nodal function or on the interpolation weights. In addition, using a subset can speed up the computations since less points are involved. If the Use subset of points option is chosen, the Subsets button can be used to bring up the Subset Definition dialog. Two options are available for defining which points are included in the subset. In one case, only the nearest N points are used. In the other case, only the nearest N points in each quadrant are used as shown below. This approach may give better results if the scatter points tend to be clustered. The Four Quadrants Surrounding an Interpolation Point. If a subset of the scatter point set is being used for interpolation, a scheme must be used to find the nearest N points. Two methods for finding a subset are provided in the Subset Definition dialog: the global method and the local method. ### Global Method With the global method, each of the scatter points in the set are searched for each interpolation point to determine which N points are nearest the interpolation point. This technique is fast for small scatter point sets but may be slow for large sets. ### Local Method With the local methods, the scatter points are triangulated to form a temporary TIN before the interpolation process begins. To compute the nearest N points, the triangle containing the interpolation point is found and the triangle topology is then used to sweep out from the interpolation point in a systematic fashion until the N nearest points are found. The local scheme is typically much faster than the global scheme for large scatter point sets. ## Computation of Interpolation Weights When computing the interpolation weights, three options are available for determining which points are included in the subset of points used to compute the weights and perform the interpolation: subset, all points, and enclosing triangle. ### Subset of Points If the Use subset of points option is chosen, the Subset Definition dialog can be used to define a local subset of points. ### All Points If the Use all points option is chosen, a weight is computed for each point and all points are used in the interpolation. ### Enclosing Triangle The Use vertices of enclosing triangle method makes the interpolation process a local scheme by taking advantage of TIN topology (Franke & Nielson, 1980). With this technique, the subset of points used for interpolation consists of the three vertices of the triangle containing the interpolation point. The weight function or blending function assigned to each scatter point is a cubic S-shaped function as shown in part a of the figure below. The fact that the slope of the weight function tends to unity at its limits ensures that the slope of the interpolating surface is continuous across triangle boundaries. (a) S-Shaped Weight Function and (b) Delauney Point Group for Point A. The influence of the weight function extends over the limits of the Delauney point group of the scatter point. The Delauney point group is the "natural neighbors" of the scatter point, and the perimeter of the group is made up of the outer edges of the triangles that are connected to the scatter point as shown in part b. The weight function varies from a weight of unity at the scatter point to zero at the perimeter of the group. For every interpolation point in the interior of a triangle there are three nonzero weight functions (the weight functions of the three vertices of the triangle). For a triangle T with vertices i, j, & k, the weights for each vertex are determined as follows: where \|\|ei\|\| is the length of the edge opposite vertex i, and bi, bj, bk are the area coordinates of the point (x,y) with respect to triangle T. Area coordinates are coordinates that describe the position of a point within the interior of a triangle relative to the vertices of the triangle. The coordinates are based solely on the geometry of the triangle. Area coordinates are sometimes called "barycentric coordinates." The relative magnitude of the coordinates corresponds to area ratios as shown below: Barycentric Coordinates for a Point in a Triangle. The XY coordinates of the interior point can be written in terms of the XY coordinates of the vertices using the area coordinates as follows: Solving the above equations for bi, bj, and bk yields: Using the weight functions defined above, the interpolating surface at points inside a triangle is computed as: where wi, wj, and wk are the weight functions and Qi, Qj, and Qk are the nodal functions for the three vertices of the triangle. Related Topics Scatter Interpolation	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "ems-i.com", "fetch_time": 1406206736000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2014-23/segments/1405997888972.38/warc/CC-MAIN-20140722025808-00065-ip-10-33-131-23.ec2.internal.warc.gz", "warc_record_offset": 122752731, "warc_record_length": 5474, "token_count": 1527, "char_count": 7877, "score": 3.6875, "int_score": 4, "crawl": "CC-MAIN-2014-23", "snapshot_type": "longest", "language": "en", "language_score": 0.8827943205833435, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00011-of-00064.parquet"}
# Roots Roots[lhs==rhs,var] yields a disjunction of equations which represent the roots of a polynomial equation. # Details and Options • Roots uses Factor and Decompose in trying to find roots. • You can find numerical values of the roots by applying N. • Roots can take the following options: • Cubics True whether to generate explicit solutions for cubics EquatedTo Null expression to which the variable solved for should be equated Modulus 0 integer modulus Multiplicity 1 multiplicity in final list of solutions Quartics True whether to generate explicit solutions for quartics Using True subsidiary equations to be solved • Roots is generated when Solve and related functions cannot produce explicit solutions. Options are often given in such cases. • Roots gives several identical equations when roots with multiplicity greater than one occur. # Examples open allclose all ## Basic Examples(1) Find roots of univariate polynomial equations: ## Scope(7) Equation with exact numeric coefficients: Equation with symbolic coefficients: General equations of degree five and higher cannot be solved in radicals: This equation of degree nine is solved in radicals using factorization and decomposition: An equation with inexact numeric coefficients: Multiple roots are repeated the corresponding number of times: Find roots over the integers modulo 7: ## Options(10) ### Cubics(3) By default Roots uses the general formulas for solving cubic equations in radicals: With , Roots does not use the general formulas for solving cubics in radicals: Solving this cubic equation in radicals does not require the general formulas: ### EquatedTo(1) Use EquatedTo to specify the left-hand side of the returned equations: ### Modulus(1) Find roots over the integers modulo 12: ### Multiplicity(1) With Multiplicity->n, the multiplicity of each root is multiplied by n: ### Quartics(3) By default Roots uses the general formulas for solving quartic equations in radicals: With , Roots does not use the general formulas for solving quartics: Solving this quartic equation in radicals does not require the general formulas: ### Using(1) Specify equations satisfied by symbolic parameters: ## Properties & Relations(5) Solutions returned by Roots satisfy the equation: Use ToRules to convert equations returned by Roots to replacement rules: Solve uses Roots to find solutions of univariate equations and returns replacement rules: Roots finds all complex solutions: Use Reduce to find solutions over specified domains: Use FindInstance to find one solution: Use Solve or Reduce to find solutions of systems of multivariate equations: Use Reduce to find solutions of systems of equations and inequalities: Use NRoots to find numeric approximations of roots of a univariate equation: Wolfram Research (1988), Roots, Wolfram Language function, https://reference.wolfram.com/language/ref/Roots.html. #### Text Wolfram Research (1988), Roots, Wolfram Language function, https://reference.wolfram.com/language/ref/Roots.html. #### CMS Wolfram Language. 1988. "Roots." Wolfram Language & System Documentation Center. Wolfram Research. https://reference.wolfram.com/language/ref/Roots.html. #### APA Wolfram Language. (1988). Roots. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/Roots.html #### BibTeX @misc{reference.wolfram_2024_roots, author="Wolfram Research", title="{Roots}", year="1988", howpublished="\url{https://reference.wolfram.com/language/ref/Roots.html}", note=[Accessed: 17-July-2024 ]} #### BibLaTeX @online{reference.wolfram_2024_roots, organization={Wolfram Research}, title={Roots}, year={1988}, url={https://reference.wolfram.com/language/ref/Roots.html}, note=[Accessed: 17-July-2024 ]}	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://reference.wolfram.com/language/ref/Roots.html", "fetch_time": 1721225420000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-30/segments/1720763514771.72/warc/CC-MAIN-20240717120911-20240717150911-00608.warc.gz", "warc_record_offset": 413435874, "warc_record_length": 13790, "token_count": 826, "char_count": 3801, "score": 3.96875, "int_score": 4, "crawl": "CC-MAIN-2024-30", "snapshot_type": "latest", "language": "en", "language_score": 0.8423956036567688, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00041-of-00064.parquet"}
# Search by Topic #### Resources tagged with Sequences similar to A Shapely Network: Filter by: Content type: Stage: Challenge level: ### There are 66 results Broad Topics > Sequences, Functions and Graphs > Sequences ### A Shapely Network ##### Stage: 2 Challenge Level: Your challenge is to find the longest way through the network following this rule. You can start and finish anywhere, and with any shape, as long as you follow the correct order. ### Magazines ##### Stage: 2 Challenge Level: Let's suppose that you are going to have a magazine which has 16 pages of A5 size. Can you find some different ways to make these pages? Investigate the pattern for each if you number the pages. ### Exploring Wild & Wonderful Number Patterns ##### Stage: 2 Challenge Level: EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules. ### Next Number ##### Stage: 2 Short Challenge Level: Find the next number in this pattern: 3, 7, 19, 55 ... ### Carrying Cards ##### Stage: 2 Challenge Level: These sixteen children are standing in four lines of four, one behind the other. They are each holding a card with a number on it. Can you work out the missing numbers? ### Cuisenaire Environment ##### Stage: 1 and 2 Challenge Level: An environment which simulates working with Cuisenaire rods. ### Street Sequences ##### Stage: 1 and 2 Challenge Level: Investigate what happens when you add house numbers along a street in different ways. ### Function Machines ##### Stage: 2 Challenge Level: If the numbers 5, 7 and 4 go into this function machine, what numbers will come out? ### Calendar Patterns ##### Stage: 2 Challenge Level: In this section from a calendar, put a square box around the 1st, 2nd, 8th and 9th. Add all the pairs of numbers. What do you notice about the answers? ### Sets of Numbers ##### Stage: 2 Challenge Level: How many different sets of numbers with at least four members can you find in the numbers in this box? ### Lawn Border ##### Stage: 1 and 2 Challenge Level: If I use 12 green tiles to represent my lawn, how many different ways could I arrange them? How many border tiles would I need each time? ### Mobile Numbers ##### Stage: 1 and 2 Challenge Level: In this investigation, you are challenged to make mobile phone numbers which are easy to remember. What happens if you make a sequence adding 2 each time? ##### Stage: 2 Challenge Level: Three beads are threaded on a circular wire and are coloured either red or blue. Can you find all four different combinations? ### Number Tracks ##### Stage: 2 Challenge Level: Benâ€™s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see? ### Sticky Triangles ##### Stage: 2 Challenge Level: Can you continue this pattern of triangles and begin to predict how many sticks are used for each new "layer"? ### Lost Books ##### Stage: 2 Challenge Level: While we were sorting some papers we found 3 strange sheets which seemed to come from small books but there were page numbers at the foot of each page. Did the pages come from the same book? ### Play a Merry Tune ##### Stage: 2 Challenge Level: Explore the different tunes you can make with these five gourds. What are the similarities and differences between the two tunes you are given? ### Taking a Die for a Walk ##### Stage: 1 and 2 Challenge Level: Investigate the numbers that come up on a die as you roll it in the direction of north, south, east and west, without going over the path it's already made. ### Triangular Hexagons ##### Stage: 2 Challenge Level: Investigate these hexagons drawn from different sized equilateral triangles. ### Sets of Four Numbers ##### Stage: 2 Challenge Level: There are ten children in Becky's group. Can you find a set of numbers for each of them? Are there any other sets? ### Polygonals ##### Stage: 2 Challenge Level: Polygonal numbers are those that are arranged in shapes as they enlarge. Explore the polygonal numbers drawn here. ### The Numbers Give the Design ##### Stage: 2 Challenge Level: Make new patterns from simple turning instructions. You can have a go using pencil and paper or with a floor robot. ### Pyramid Numbers ##### Stage: 2 Challenge Level: What are the next three numbers in this sequence? Can you explain why are they called pyramid numbers? ### More Pebbles ##### Stage: 2 and 3 Challenge Level: Have a go at this 3D extension to the Pebbles problem. ### Times Tables Shifts ##### Stage: 2 Challenge Level: In this activity, the computer chooses a times table and shifts it. Can you work out the table and the shift each time? ### Domino Sets ##### Stage: 2 Challenge Level: How do you know if your set of dominoes is complete? ### The Mathemagician's Seven Spells ##### Stage: 2 Challenge Level: "Tell me the next two numbers in each of these seven minor spells", chanted the Mathemagician, "And the great spell will crumble away!" Can you help Anna and David break the spell? ### Seven Squares ##### Stage: 3 Challenge Level: Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100? ### Two Much ##### Stage: 3 Challenge Level: Explain why the arithmetic sequence 1, 14, 27, 40, ... contains many terms of the form 222...2 where only the digit 2 appears. ### Extending Great Squares ##### Stage: 2 and 3 Challenge Level: Explore one of these five pictures. ### Remainder ##### Stage: 3 Challenge Level: What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2? ### What an Odd Fact(or) ##### Stage: 3 Challenge Level: Can you show that 1^99 + 2^99 + 3^99 + 4^99 + 5^99 is divisible by 5? ### Holes ##### Stage: 1 and 2 Challenge Level: I've made some cubes and some cubes with holes in. This challenge invites you to explore the difference in the number of small cubes I've used. 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Make a note of what happens to the. . . . ### Seven Squares - Group-worthy Task ##### Stage: 3 Challenge Level: Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning? ### Millennium Man ##### Stage: 2 Challenge Level: Liitle Millennium Man was born on Saturday 1st January 2000 and he will retire on the first Saturday 1st January that occurs after his 60th birthday. How old will he be when he retires? ### Odds, Evens and More Evens ##### Stage: 3 Challenge Level: Alison, Bernard and Charlie have been exploring sequences of odd and even numbers, which raise some intriguing questions... ##### Stage: 3 Challenge Level: Powers of numbers behave in surprising ways. 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# Homework Help: Calc 2 Limits 1. Nov 15, 2007 ### poopforfood 1. The problem statement, all variables and given/known data Integral of (3x+2)/x(x+2)^2+16x 2. Relevant equations 3. The attempt at a solution That breaks down to A/x + Bx+c/x^2+4x+20 so 3x+2 = Ax^2+4x+20 + Bx^2 + Cx then I found the values of A b and C then I cant figure out what to do please help 2. Nov 15, 2007 ### Kreizhn What did you get for values of a,b,c? And by the way, is that supposed to be $$\frac{3x+2}{x(x+2)^2+16x}$$? because the way it's written is $$\frac{3x+2}{x} (x+2)^2 + 16x$$ 3. Nov 15, 2007 ### poopforfood yea thats what its suposed to be 4. Nov 15, 2007 ### Kreizhn And did you by chance get A= 1/10, B = -1/10, C= 26/10? 5. Nov 15, 2007 ### poopforfood yeaup 6. Nov 15, 2007 ### poopforfood I dont know what to do after this 7. Nov 15, 2007 ### Kreizhn well, the $\frac{1}{x}$ is pretty easy to handle right? So we'll just focus on the other term. Now, in this case it's better if we express $x^2+4x+20$ as $(x+2)^2+16$. Make the substitution $x+2 = 4 \tan(\theta)$ and don't forget that in this case [itex] dx = 4 \sec^2(\theta) d\theta [/tex]. Substitute everything into your integral and see if it simplifies a bit. 8. Nov 15, 2007 ### poopforfood I dont know what to do after this 9. Nov 15, 2007 ### Kreizhn If you do what I've said to do, and you do it correctly, your integral will become much easier, so just stick with it. 10. Nov 15, 2007 ### poopforfood I dont know what to do after this 11. Nov 15, 2007 ### Kreizhn Well why don't you show me what you've got so far and we'll see if we can't see where the problem is, because if you done it correctly the integral is blatantly obvious.	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/calc-2-limits.198467/", "fetch_time": 1539919507000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-43/segments/1539583512268.20/warc/CC-MAIN-20181019020142-20181019041642-00087.warc.gz", "warc_record_offset": 1031467957, "warc_record_length": 13701, "token_count": 612, "char_count": 1735, "score": 4.0625, "int_score": 4, "crawl": "CC-MAIN-2018-43", "snapshot_type": "latest", "language": "en", "language_score": 0.8806711435317993, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00053-of-00064.parquet"}
# Data Analysis with R (Part 1): Introduction to R Programming Source: https://campus.datacamp.com/ # Selection by comparison Select values in that are larger than 10. Assign the result to the variable `# A numeric vector containing 3 elementsnumeric_vector <- c(1, 10, 49)larger_than_ten <- numeric_vector > 10print(larger_than_ten)` Print the selected elements `numeric_vector <- c(1, 10, 49)larger_than_ten <- numeric_vector > 10numeric_vector[larger_than_ten]` # Matrices In R, a matrix is a collection of elements of the same data type (numeric, character, or logical) arranged into a fixed number of rows and columns. You can construct a matrix in R with the matrix() function. Consider the following example: In the matrix() function: • The first argument is the collection of elements that R will arrange into the rows and columns of the matrix. Here, we use 1:9 which constructs the vector c(1, 2, 3, 4, 5, 6, 7, 8, 9). • The argument byrow indicates that the matrix is filled by the rows. This means that the matrix is filled from left to right and when the first row is completed, the filling continues on the second row. If we want the matrix to be filled by the columns, we just place byrow = FALSE. • The third argument nrow indicates that the matrix should have three rows. • The fourth argument ncol indicates the number of columns that the matrix should have Another example `# Construction of a matrix with 5 rows that contain the numbers 1 up to 20 and assign it to mmymatrix <- matrix(1:20, byrow = TRUE, nrow = 5, ncol = 4)# print m to the consoleprint(mymatrix)` # Factors The term factor refers to a statistical data type used to store categorical variables. `# a vector called student_statusstudent_status <- c("student", "not student", "student", "not student")# turn student_status into a factor and save it in the variable categorical_studentcategorical_student <- factor(student_status)# print categorical_student to the consoleprint(categorical_student)` # Dataframes A dataframe is a collection of elements of the different data types (numeric, character, or logical) arranged into a fixed number of rows and columns. The elements in the matrix should be of the same type. But a dataframe doesn’t have this restriction. ## How to inspect a dataframe? There are several functions you can use to inspect your dataframe. To name a few • : this by default prints the first 6 rows of the dataframe • : this by default prints the last 6 rows to the console • : this prints the structure of your dataframe • : this by default prints the dimensions, that is, the number of rows and columns of your dataframe • : this prints the names of the columns of your dataframe For example, `# print the first 6 rows of mtcarsprint(head(mtcars))# print the structure of mtcarsprint(str(mtcars))# print the dimensions of mtcarsprint(dim(mtcars))` ## How to construct a dataframe? Suppose you want to construct a data frame that describes the main characteristics of eight planets in our solar system. The main features of a planet are: • The type of planet (Terrestrial or Gas Giant). • The planet’s diameter relative to the diameter of the Earth. • The planet’s rotation across the sun relative to that of the Earth. • If the planet has rings or not (TRUE or FALSE). You construct a data frame with the function. As arguments, you should provide the above mentioned vectors as input that should become the different columns of that data frame. Therefore, it is important that each vector used to construct a data frame has an equal length. But do not forget that it is possible (and likely) that they contain different types of data. `# planets vectorplanets <- c("Mercury", "Venus", "Earth", "Mars", "Jupiter", "Saturn", "Uranus", "Neptune")# type vectortype <- c("Terrestrial planet", "Terrestrial planet", "Terrestrial planet", "Terrestrial planet", "Gas giant", "Gas giant", "Gas giant", "Gas giant")# diameter vectordiameter <- c(0.382, 0.949, 1, 0.532, 11.209, 9.449, 4.007, 3.883)# rotation vectorrotation <- c(58.64, -243.02, 1, 1.03, 0.41, 0.43, -0.72, 0.67)# rings vectorrings <- c(FALSE, FALSE, FALSE, FALSE, TRUE, TRUE, TRUE, TRUE)# construct a dataframe planet_df from all the above variablesplanet_df <- data.frame(planets, type, diameter, rotation, rings)` ## Indexing and selecting columns from a dataframe In the same way as you indexed your vectors, you can select elements from your dataframe using square brackets. Different from dataframes however, you now have multiple dimensions: rows and columns. That’s why you can use a comma in the middle of the brackets to differentiate between rows and columns. For instance, the following code would select the element in the first row and the second column from the dataframe . You can also use the operator to select an entire column from a dataframe. For instance, would select the entire planets column from the dataframe planet_df. Example `# select the values in the first row and second and third columnsplanet_df[1,2: 3]# select the entire third columnplanet_df[,3]` # Lists A list in R is similar to your to-do list at work or school: the different items on that list most likely differ in length, characteristic, type of activity that has to do be done. A list in R allows you to gather a variety of objects under one name (that is, the name of the list) in an ordered way. These objects can be matrices, vectors, data frames, even other lists, etc. It is not even required that these objects are related to each other. You can easily construct a list using the function. In this function you can wrap the different elements like so: . `# Vector with numerics from 1 up to 10my_vector <- 1:10# Matrix with numerics from 1 up to 9my_matrix <- matrix(1:9, ncol = 3)# First 10 elements of the built-in data frame 'mtcars'my_df <- mtcars[1:10,]# Construct my_list with these different elements:my_list <- list(my_vector, my_matrix, my_df)# print my_list to the consoleprint(my_list)` ## Selecting elements from a list Your list will often be built out of numerous elements and components. Therefore, getting a single element, multiple elements, or a component out of it is not always straightforward. One way to select a component is using the numbered position of that component. For example, to “grab” the first component of you type Another way to check is to refer to the names of the components: selects the vector. A last way to grab an element from a list is using the sign. The following code would select from : . Besides selecting components, you often need to select specific elements out of these components. For example, with you select from the first component of the first element. This would select the number 1. `# Vector with numerics from 1 up to 10my_vector <- 1:10# Matrix with numerics from 1 up to 9my_matrix <- matrix(1:9, ncol = 3)# First 10 elements of the built-in data frame 'mtcars'my_df <- mtcars[1:10,]# Construct list with these different elements:my_list <- list(my_vector, my_matrix, my_df)# Grab the second element of my_list and print it to the consoleprint(my_list[[2]])# Grab the first column of the third component of `my_list` and print it to the consoleprint(my_list[[3]][,1])` # Getting help ## How to Create Functions `# make a function called multiply_a_bmultiply_a_b <- function(a, b) { return(a*b)}# call the function multiply_a_b and store the result into a variable resultresult <- multiply_a_b(a=3, b=7)` # Getting your data into R One important thing before you actually do analyses on your data, is that you will need to get your data into R. R contains many functions to read in data from different formats. To name only a few: • : Reads in tabular data such as txt files • : Read in data from a comma-separated file format • : Reads in an excel worksheet • : Reads in data from .sav SPSS format. For the current exercise, we have put the R mtcars dataset into a csv file format and put this on github. The data can be found on the following link: Sample data http://s3.amazonaws.com/assets.datacamp.com/course/uva/mtcars.csv `# load in the data and store it in the variable carscars <- read.csv("http://s3.amazonaws.com/assets.datacamp.com/course/uva/mtcars.csv")# print the first 6 rows of the dataset using the head() functionprint(head(cars))` ## Define the separator `# load in the datasetcars <- read.csv("http://s3.amazonaws.com/assets.datacamp.com/course/uva/mtcars_semicolon.csv", sep = ";")# print the first 6 rows of the datasetprint(head(cars))` # Working directories in R Default working directory where R will look for Of course this working directory is not static and can be changed by the user. In R there are two important functions: • : This function will retrieve the current working directory for the user • : This functions allows the user to set her own working directory • lists all the files that exists in your working directory. Example code `# list all the files in the working directorylist.files(getwd())# read in the cars dataset and store it in a variable called carscars <- read.csv("cars.csv", sep = ";")# print the first 6 rows of carsprint(head(cars))` # Importing R packages Imagine we want to do some great plotting and we want to use ggplot2 for it. If we want to do so, we need to take 2 steps: 1. Install the package ggplot2 using 2. Load the package ggplot2 using or Example code `# Install the ggplot2 package install(ggplot2)# load the ggplot2 package using the require functionrequire(ggplot2)`	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "hashimputhiyakath.medium.com", "fetch_time": 1620588710000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-21/segments/1620243989012.26/warc/CC-MAIN-20210509183309-20210509213309-00203.warc.gz", "warc_record_offset": 319283262, "warc_record_length": 37096, "token_count": 2312, "char_count": 9543, "score": 3.609375, "int_score": 4, "crawl": "CC-MAIN-2021-21", "snapshot_type": "latest", "language": "en", "language_score": 0.7756694555282593, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00015-of-00064.parquet"}
# A question on Binomial distribution In a bag, there are 6 white disks, 6 black disks and 8 red disks. A disk is drawn at random from the bag.The colour is recorded and the disc is returned to the bag This process is repeated 10 times. Find the probability that less than 4 red disks are drawn I approached this question with n= 8 and p=0.4, q=0.6. But I'm not sure about what to do with the repetition of 10 times, how do I do this question? As the disc is returned to the bag it will be the same probability of a red disc coming out each time. As it is less than four red discs, we need to find the probability of $0, 1, 2$ and $3$ red discs being chosen. Your values of $p$ and $q$ are indeed correct, however I'm a bit confused as to why you've chosen $n = 8$. There are $10$ trials, and as such I have chosen $n = 10$ and $r = 0, 1, 2$ and $3$, and putting this into the formula $$^nC_r\times p^{r}\times q^{n-r}$$ and adding up the four values for r you should get your answer. Hope this helps :) $(^{10}C_0 \times 0.4^0 \times 0.6^{10-0}) + (^{10}C_1 \times 0.4^1 \times 0.6^{10-1})...$ I work out the answer to be 0.382... • thanks, I tried this method first but I made some calculation error which then led me to the idea that I was using the wrong formula. Haha, thanks – warman Aug 7 '17 at 9:25 The probability P(R) of drawing a red disk is 8/20, which = 0.4. You have 10 trials, therefore: R ~ B(10, 0.4). Using this distribution, you want to calculate: P(R < 4), or P(R ≤ 3), since it's discrete. Modern calculators will evaluate this, but since the formula only uses '=' rather than '≤', you are technically doing: P(R = 0) + P(R = 1) + P(R = 2) + P(R = 3). The formula for P(R = r), where R ~ B(n, p) is: $$\binom{n}{r} \times p^r \times q^{n-r}, where \binom{n}{r} = \frac{r!}{r!(n-r)!}$$ Use this formula to sum the probabilities, and you should achieve the answer: 0.3822806016, or 0.382 (3 S.F).	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/2385338/a-question-on-binomial-distribution/2385351", "fetch_time": 1568996622000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-39/segments/1568514574050.69/warc/CC-MAIN-20190920155311-20190920181311-00147.warc.gz", "warc_record_offset": 590787634, "warc_record_length": 31678, "token_count": 613, "char_count": 1932, "score": 4.1875, "int_score": 4, "crawl": "CC-MAIN-2019-39", "snapshot_type": "latest", "language": "en", "language_score": 0.9293257594108582, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00007-of-00064.parquet"}
# Past Papers’ Solutions \| Edexcel \| AS & A level \| Mathematics \| Core Mathematics 1 (C1-6663/01) \| Year 2017 \| June \| Q#10 Question Figure shows a sketch of part of the curve y = f(x), , where f(x) = (2x – 5)2(x + 3) a. Given that i. the curve with equation y = f(x) – k, , passes through the origin, find the value of the constant k, ii. the curve with equation y = f(x + c), , has […] # Past Papers’ Solutions \| Edexcel \| AS & A level \| Mathematics \| Core Mathematics 1 (C1-6663/01) \| Year 2017 \| June \| Q#9 Question a. On separate axes sketch the graphs of i. y = –3x + c, where c is a positive constant, ii. On each sketch show the coordinates of any point at which the graph crosses the y-axis and the equation of any horizontal asymptote. Given that y = –3x + c, where c is a positive constant, meets […] # Past Papers’ Solutions \| Edexcel \| AS & A level \| Mathematics \| Core Mathematics 1 (C1-6663/01) \| Year 2017 \| June \| Q#8 Question The straight line , shown in Figure, has equation 5y = 4x + 10. The point P with x coordinate 5 lies on . The straight line is perpendicular to and passes through P. a. Find an equation for , writing your answer in the form ax + by + c = 0 where […] # Past Papers’ Solutions \| Edexcel \| AS & A level \| Mathematics \| Core Mathematics 1 (C1-6663/01) \| Year 2017 \| June \| Q#7 Question The curve C has equation y=f(x), x>0, where Given tht the point P(4,-8) lies on the curve C; a. find the equation of the tangent to C at P, giving your answer in the form y = mx + c, where m and c are constants. a. find f(x), giving each term in its simplest […]	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://oalevelsolutions.com/tag/coordinate-geometry-straight-lines-edexcel-2017/", "fetch_time": 1660150719000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-33/segments/1659882571198.57/warc/CC-MAIN-20220810161541-20220810191541-00372.warc.gz", "warc_record_offset": 413166563, "warc_record_length": 9306, "token_count": 509, "char_count": 1712, "score": 3.59375, "int_score": 4, "crawl": "CC-MAIN-2022-33", "snapshot_type": "latest", "language": "en", "language_score": 0.7892893552780151, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00012-of-00064.parquet"}
Courses Courses for Kids Free study material Offline Centres More Store # A gas occupies 3 litres at $0^\circ C$.What volume will it occupy at $- 20^\circ C$, pressure remaining constant? Last updated date: 20th Sep 2024 Total views: 421.8k Views today: 7.21k Verified 421.8k+ views Hint:We can make use of the Charle’s Law to solve this problem, which gives us an accurate relationship between the volume occupied by a gas and the absolute temperature, when the pressure and amount of gas is kept constant. Formulas used: $\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}$ Where $V$is the volume occupied by the gas, $T$ is the absolute temperature and the subscripts 1 and 2 denote the initial and final stages of the gas respectively. Complete step by step solution: As the pressure is kept constant in this particular question, and the same amount of gas is under consideration, we can make use of the Charle’s Law, which states that the volume occupied by a gas is directly proportional to the absolute temperature when pressure is kept constant. That is, $V \propto {[T]_{P,n}}$ Where $V$ is the volume occupied by the gas, $T$ is the absolute temperature, $P$ is the pressure and $n$ is the amount of gas taken. Therefore, the term ${V/T}$ will be a constant and considering an initial and final stage, we have the following result: $\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}} \Rightarrow {V_2} = \dfrac{{{V_1}{T_2}}}{{{T_1}}}$ ………………………..(1) Where $V$ is the volume occupied by the gas, $T$ is the absolute temperature and the subscripts 1 and 2 denote the initial and final stages of the gas respectively. First let us convert the temperatures into Kelvin by adding 273 to the Celsius temperatures. $\Rightarrow 0^\circ C = 0 + 273 = 273K$ and $- 20^\circ C = - 20 + 273 = 253K$ Therefore, from the data we’ve been given, ${V_1} = 3L,{T_1} = 273K$ and ${T_2} = 253K$ Substituting these values into our equation (1), we get: ${V_2} = \dfrac{{3 \times 253}}{{273}}L$ On solving this, we get: ${V_2} = 2.78L$ Hence, at $- 20^\circ C$, the gas will occupy a volume of $2.78L$. Note: The temperature should be in absolute scale, that is, the unit should be Kelvin. If we fail to recognize the need to convert the unit, we may end up with absurd answers such as infinity. Also note that the Charle’s Law is not applicable if the number of moles of the gas were to change during the process, even if pressure is kept constant. In this case, we will have to use the Ideal gas Law.	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.vedantu.com/question-answer/a-gas-occupies-3-litres-at-0circ-cwhat-volume-class-11-chemistry-cbse-5fb3765baae7bc7b2d281fb0", "fetch_time": 1726796071000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700652073.91/warc/CC-MAIN-20240919230146-20240920020146-00723.warc.gz", "warc_record_offset": 975247201, "warc_record_length": 28859, "token_count": 707, "char_count": 2498, "score": 4.5625, "int_score": 5, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.8220092058181763, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00016-of-00064.parquet"}
```15 Worksheet: Coulomb’s Law 1. Q = -2nC at x = 1.0m F  kQq r2 q = + 4nC at x = 4.0m a) What is the direction of the force on Q? ___+x_____ Does the sign of Qq tell us the direction of this force? ___ No ___ Explain. -Qq only indicates that the interaction is attractive b) What is the direction of the force on q? ___-x_______ Does the sign of Qq tell us the direction of this force? __ No ____ Explain. -Qq only indicates that the interaction is attractive c) Calculate the Coulomb force on the charges. What does the negative sign mean? Does one charge feel more force than the other? kQq (9  109 )( 2  109 )( 4  109 ) F 2   8  109 N 2 r 3 Both charges feel the same size force d) At what locations could a charge be placed and feel no force on it? At x = -6.24 m (7.24 meters to the left of Q) 2. What happens to the magnitude of the force on q if it is moved to x = 7.0m? It is reduced by the factor (3/6)^2 = (1/2)^2 = 1/4 3. What happens to the direction of the force on q it is moved to x = -1.0m? The force is now to the right instead of left ```	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://studylib.net/doc/25197984/15-coulomb-answers", "fetch_time": 1548253489000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-04/segments/1547584332824.92/warc/CC-MAIN-20190123130602-20190123152602-00347.warc.gz", "warc_record_offset": 634650357, "warc_record_length": 32044, "token_count": 379, "char_count": 1070, "score": 3.671875, "int_score": 4, "crawl": "CC-MAIN-2019-04", "snapshot_type": "longest", "language": "en", "language_score": 0.8892055153846741, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00036-of-00064.parquet"}
# Equation solutions • Dec 11th 2009, 09:44 AM Stuck Man Equation solutions I want to find x-axis intercepts of this equation: (sin x+0.5)(sin x-0.25)=0.25 0 <= x <= 360. So sin x=-0.25 or sin x=0.5 One of the associated acute angles is 30 which is found from sin x=0.5. The other is 48.59 but I don't know where it comes from. • Dec 11th 2009, 10:29 AM Raoh hi(Hi) put $\sin x=X$ and solve $(X+0.5)(X-0.25)-0.25=0$ • Dec 11th 2009, 10:29 AM Stuck Man I rearranged the equation to equal 0 using +.75 and -.5. I then solved the problem but I don't know why I couldn't solve it with the other equation. • Dec 11th 2009, 10:35 AM Stuck Man I think it must be necessary to factorise without any added on value. Completing the square for instance wouldn't work. • Dec 11th 2009, 10:48 AM Raoh Quote: Originally Posted by Stuck Man I think it must be necessary to factorise without any added on value. Completing the square for instance wouldn't work. (Thinking) to find the x-intercepts you must solve the given equation, $(X+0.5)(X-0.25)-0.25=X^2-0.25X+0.5X-0.375=X^2+0.25X-0.375=0$. i found the solutions $0.5$ and $-0.75$ therefore, $\sin x=0.50 \Leftrightarrow x=\arcsin (0.50)$ or $\sin x=-0.75\Leftrightarrow x=\arcsin (-0.75)$. • Dec 11th 2009, 10:59 AM Stuck Man I think you've mixed up the signs of the solutions. You've rewritten the formula the same as I did. I'm still not sure why I have to fully factorise and not use the Completing The square method. I think it have something to do with the use of sin. • Dec 11th 2009, 03:21 PM skeeter Quote: Originally Posted by Stuck Man I want to find x-axis intercepts of this equation: (sin x+0.5)(sin x-0.25)=0.25 0 <= x <= 360. So sin x=-0.25 or sin x=0.5 One of the associated acute angles is 30 which is found from sin x=0.5. The other is 48.59 but I don't know where it comes from. $\sin^2{x} + .25\sin{x} - 0.125 = 0.25 $ $\sin^2{x} + .25\sin{x} - 0.375 = 0$ $8\sin^2{x} + 2\sin{x} - 3 = 0$ $(4\sin{x} + 3)(2\sin{x} - 1) = 0$ $\sin{x} = \frac{3}{4}$ ... $x = \arcsin\left(\frac{3}{4}\right) \approx 48.59^\circ$ $x = 180 - \arcsin\left(\frac{3}{4}\right) \approx 131.4^\circ$ $\sin{x} = \frac{1}{2}$ $x = 30^\circ$ $x = 150^\circ$ • Dec 11th 2009, 11:11 PM Stuck Man I still don't know why I can't use the equation as it was originally: (sin x+0.5)(sin x-0.25)=0.25 sin x=-0.25 or sin x=0.5 • Dec 12th 2009, 12:11 AM 11rdc11 Quote: Originally Posted by Stuck Man I still don't know why I can't use the equation as it was originally: (sin x+0.5)(sin x-0.25)=0.25 sin x=-0.25 or sin x=0.5 You can't do that in math. Now if the equation was $(\sin{x}+.5)(\sin{x}-.25) = 0$ then you could do that Notice how Raoh and Skeeter set the equation equal to 0 and then solved.	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/trigonometry/119892-equation-solutions-print.html", "fetch_time": 1513471106000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-51/segments/1512948592202.83/warc/CC-MAIN-20171217000422-20171217022422-00758.warc.gz", "warc_record_offset": 178874806, "warc_record_length": 3630, "token_count": 1034, "char_count": 2751, "score": 4.0, "int_score": 4, "crawl": "CC-MAIN-2017-51", "snapshot_type": "longest", "language": "en", "language_score": 0.8883694410324097, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00026-of-00064.parquet"}
# A continuous inverse of the exponential function is holomorphic Let $$\Omega \subseteq \mathbb{C}$$ be open and connected such that $$0 \notin \Omega$$. Let $$f: \Omega \to \mathbb{C}$$ be a continuous function such that $$e^{f(z)} = z$$ for all $$z \in \Omega$$. Prove that $$f$$ is holomorphic and that $$f'(z) = \frac{1}{z}$$. After asserting the first part I guess the second part is pretty trivial by just taking the derivatives of both sides (since $$e^{f(z)}$$ is holomorphic if $$f(z)$$ is holomorphic). There is also a second part asking if there exists a continuous function $$f: \mathbb{C} \setminus \{0\} \to \mathbb{C}$$ such that $$e^{f(z)} = z$$ for all $$z \in \mathbb{C} \setminus \{0\}$$. The answer for this question shoul be no because in class we defined $$\log(z)$$ as the inverse of $$e^z$$ and showed that it is continuous in $$\mathbb{C}\setminus (-\infty, 0]$$ and discontinuous everywhere else. Is what I said correct? How would I go about proving the first part? Thanks in advance! Let $$z \in \Omega$$ and $$(z_n)$$ be a sequence in $$\Omega\setminus \{ z \}$$ with $$z_n \to z$$. Then $$f(z_n) \to f(z)$$ because $$f$$ is continuous, and $$f(z_n) \ne f(z)$$ because $$f$$ is necessarily injective. It follows that $$\frac{f(z_n)-f(z)}{z_n-z} = \left(\frac{e^{f(z_n)}-e^{f(z)}}{f(z_n)-f(z)}\right)^{-1} \to \left( e^{f(z)}\right)^{-1} = \frac 1z$$ because $$\exp$$ is differentiable at $$f(z)$$ with $$\exp' = \exp$$. This demonstrates that $$f$$ is complex differentiable with $$f'(z) = 1/z$$. More generally: If $$f: \Omega \to D$$ is continuous at $$z_0 \in \Omega$$, $$g:D \to \Omega$$ is holomorphic with $$g(f(z)) = z$$ for all $$z \in \Omega$$, and $$g'(f(z_0)) \ne 0$$, then $$f$$ is complex differentiable at $$z_0$$ with $$f'(z_0) = \lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0} = \lim_{z\to z_0} \left(\frac{g(f(z))-g(f(z_0))}{f(z)-f(z_0)}\right)^{-1} = \left( g'(f(z_0))\right)^{-1} \, .$$ In other words: If $$g$$ is holomorphic, $$g'\ne 0$$, the existence of an inverse function $$f$$ is given and $$f$$ is assumed to be continuous, then the holomorphy of $$f$$ follows directly from the definition of the complex derivative. For the second part of your question, see e.g. No holomorphic logarithmn. • Very nice and elementary proof! – Giorgos Giapitzakis Jun 20 at 19:57 Since $$(\forall z\in\Bbb C):\exp'(z)\ne0$$, $$\exp$$ is locally invertible. So, take $$z_0\in\Bbb C$$. There is some neighborhood $$N$$ of $$f(z_0)$$ such that $$\exp\|_N$$ is an holomorphic inverse $$l$$. Since $$f$$ is continuous at $$z_0$$, there is some neighborhood $$W$$ of $$z_0$$ such that $$f(W)\subset N$$. So, since$$(\forall z\in W):\exp(f(z))=z,$$you have$$(\forall z\in W):f(z)=l(z).$$Therefore, $$f$$ is differentiable at $$z_0$$. So, $$f$$ is holomorphic. And if there was some function $$f\colon\Bbb C\setminus\{0\}\longrightarrow$$ such that $$(\forall z\in\Bbb C\setminus\{0\}):e^{f(z)}=z$$, then $$f'(z)=\frac1z$$, and so $$f$$ would be a primitive of $$\frac1z$$. But then $$\oint_{\|z\|=1}\frac{\mathrm dz}z=0$$. But, in fact, $$\oint_{\|z\|=1}\frac{\mathrm dz}z=2\pi i$$. • Thanks for the answer. Is there a more elementary proof since I'm quite new to complex analysis? – Giorgos Giapitzakis Jun 20 at 19:46 • If there is, I don't know it. – José Carlos Santos Jun 20 at 19:48	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 67, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/4178317/a-continuous-inverse-of-the-exponential-function-is-holomorphic", "fetch_time": 1627473134000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-31/segments/1627046153709.26/warc/CC-MAIN-20210728092200-20210728122200-00610.warc.gz", "warc_record_offset": 411616997, "warc_record_length": 39352, "token_count": 1157, "char_count": 3320, "score": 3.875, "int_score": 4, "crawl": "CC-MAIN-2021-31", "snapshot_type": "latest", "language": "en", "language_score": 0.8681244850158691, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00007-of-00064.parquet"}
+0 What are the points of inflection for f'(x)=(3x(x^2+27))/(x^2-9)^3 0 157 1 What are the points of inflection for f"(x)=(3x(x^2+27))/(x^2-9)^3 Nov 28, 2018 #1 +6046 +1 $$f''(x) = \dfrac{3x(x^2+27)}{(x^2-9)^3}$$ $$\text{a point }x \text{ is an inflection point of }f(x) \text{ if both of the following hold}\\ \text{1) }f''(x)=0 \\ \text{2) }sgn(f''(x-\epsilon)) \neq sgn(f''(x+\epsilon)),~\epsilon > 0 \text{ but small}$$ $$\text{It's pretty clear that the only real value for which }f''(x)=0 \text{ is }x=0\\ sgn(f''(-\epsilon)) = 1\\ sgn(f''(\epsilon))=-1\\ \text{So }x=0 \text{ is indeed an inflection point and the only one.}$$ . Nov 28, 2018	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://web2.0calc.com/questions/what-are-the-points-of-inflection-for-f-x-3x-x-2-27-x-2-9-3", "fetch_time": 1579286738000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-05/segments/1579250590107.3/warc/CC-MAIN-20200117180950-20200117204950-00256.warc.gz", "warc_record_offset": 732883087, "warc_record_length": 5975, "token_count": 285, "char_count": 657, "score": 3.84375, "int_score": 4, "crawl": "CC-MAIN-2020-05", "snapshot_type": "latest", "language": "en", "language_score": 0.7211242914199829, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00055-of-00064.parquet"}
[ Back ] [ Up ] [ Next ] [Timeline] Grade 7: The Learning Equation Math 42.01 Exploring Probability Chance & Uncertainty Refresher pp 88-89 Learning Outcomes: The student will: ## Tree Diagrams and Charts ### Total Possible Outcomes Total possible outcomes are required for theoretical probability calculations. A table shows the possibilities. Explore the total possible outcomes of the following pairs of events. ### One Coin Complete a chart to show all of the possible outcomes of flipping one coin. Compare it to the table below. ### Event: Coin Flip H T number of possible outcomes = 2 ### One Six-sided Die The following shows the possible outcomes when you toss a die. Roll the die several times until you see the six different numbers displayed on the die ### Event: Die Toss 1 2 3 4 5 6 number of possible outcomes = 6 ## Theoretical Probability of a Single Event The theoretical probability of the event is the fraction: # ways the event can occur total possible outcomes ### Coin Flip When you flip a coin, there are only 2 sides - heads and tails. The theoretical probability for each is: P(H) = 1/2 = 0.5 P(T) = 1/2 = 0.5 ### Spin Wheel When a spinner has 5 equal sized sectors of green, red, yellow, and blue, the theoretical probability for each is: P(red) = 1/5 = 0.20 P(orange) = 1/5 = 0.20 P(yellow) = 1/5 = 0.20 P(green) = 1/5 = 0.20 P(blue) = 1/5 = 0.20 ### Toss Die When you toss a 6-sided dice, there are 6 sides. The theoretical probability (to the nearest hundredth) for each is: P(1) = 1/6 = 0.17 P(2) = 1/6 = 0.17 P(3) = 1/6 = 0.17 P(4) = 1/6 = 0.17 P(5) = 1/6 = 0.17 P(6) = 1/6 = 0.17 Theoretical probability does not change. In the first examples each individual event had the same probability. The next example shows events with different probabilities. ## Experimental Probability ### Coin Flip The experimental probability for equally likely events is the fraction: # favourable outcomes total outcomes If you toss a coin 50 times and you end up with 20 heads and 30 tails, the experimental probability is: P(H) = 20/50 = 0.4 P(T) = 30/50 = 0.6 Theoretical experiment changes from experiment to experiment. If the experiment is fair, the theoretical and experimental probability should be very similar if a large number of trials are made. Try flipping a coin yourself. Flip it 50 times and record the numbers of heads and tails. Divide your totals by 50 to get the experimental probability. Coin toss simulations are available on the internet. Try these: Coin Toss Theoretical and Experimental Probability for Coins Toss a coin 20 times. If heads comes up 13 times, then the favorable outcomes of the event that heads comes up is 13. ### Coin Flip The theoretical probability of the event is the fraction: # ways the event can occur total possible outcomes Referring to the situation above, the theoretical probability of the event that heads comes up is: P(H) = 1/2 = 0.5 = 50% The experimental probability for equally likely events is the fraction: # favourable outcomes total outcomes Referring to the situation above, the experimental probability of the event that heads comes up is: P(H) = 13/20 = 0.65 = 65% ## Theoretical and Experimental Probability for Coin Tosses Click on one of the toss buttons to select how many times you want the coin tossed. If you select the consecutive series option, the program will work through all 6 samples starting with single toss ending with 100 000 tosses. The theoretical probability for both heads and tails is 0.5 or 50%. The experimental probability for a single toss is 1.0 pr 100% (it has to be either heads or tails). Compare the theoretical and experimental probability results for each toss option as you complete the following exercise: TURNS HEADS TAILS #heads Probablility Probablility #tails Experimental P(H) Theoretical P(H) Experimental P(T) Theoretical P(T) 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 This will complete all 6 options consecutively. Source and permission for noncommercial use: Monte Carlo Introduction ## Rolling Dice Probability For equally likely events the theoretical probablility is: # ways the event can occur total possible outcomes • The probability for 4, 5 and 6-sided dice are displayed below: ### Theoretical Probability The number of faces on dice can vary. If there is a different number on each face, the theoretical probability will be: 1 #faces # tossed 4 Faces 5 Faces 6 Faces 1 P(1) = 1/4 P(1) = 1/5 P(1) = 1/6 2 P(2) = 1/4 P(2) = 1/5 P(2) = 1/6 3 P(3) = 1/4 P(3) = 1/5 P(3) = 1/6 4 P(4) = 1/4 P(4) = 1/5 P(4) = 1/6 5 P(5) = 1/5 P(5) = 1/6 6 P(6) = 1/6 • Roll the die 6 times. Did the experimental probability match the theoretical probability of 1/6 (17%) for each number? • Click reset. Roll the die 120 times. Did the experimental probability match the theoretical probability of 1/6 (17%) for each number? INTRODUCTION TO PROBABILITY Review: ## Spinners Exploring Probability Try the spinner for to the experimental probability investigations. Lesson on Introduction to Probability - contains cool dice roll Enrichment: Percent and Probability Introduction to Probability Seth H. Jackson's Assignment 6 - dice roll applet Lesson on Sample Spaces - great, but sample space not part of program vocabulary Lesson on the Complement of an Event - great, but compliment not part of program vocabulary Central Limit Theorem Demo Applet Key Terms: A-E F-J K-O P-R S-Z probability, outcome, experimental probability, theoretical probability, frequency, independent events, sectors, central angle Prerequisite Skills: Included in this lesson. Included in this lesson. Included in this lesson. Included in this lesson. 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# Where's that book you want? A friend of yours, studying law, just walked out of the library and you have a short social talk. During the conversation it comes to pass which books he just returned. One of the books seemed interesting to you and you ask your friend where'd he put it, because sadly the old cranky library doesn't have an organized and indexed system for their shelves. Your friends tell you it's in the 5th lane on the left hand side coming from the back. He says it's in the somewhere between the 7th from the left and the 49th of the right on the 3rd row. When you enter the library you notice the shelves have predefined trays for a book, meaning the book size doesn't matter and each tray always consumes the same space. You see a small specification sticker on the shelve saying a whole row contains 83 trays. The library consists of 11 back-to-back lanes, with the first lane at the entrance also having shelves on the back-side, facing towards the entrance. All shelves are 6 rows high. Giving that all the books slightly average the same size as the book you mentioned to your friend: 1.) In which tray(s) can you find the book when you come in from the front? 2.) How many space for books is there in the whole library? • I don't see what the puzzle is. Your friend tells you the position of the book, and the puzzle is to work out what the position of the book is? I'm also very confused about the configuration of the library shelves - I can't picture what you mean by "lane", "back to back", "trays", etc. – Jack M May 10 '15 at 22:59 since the first lane is double there should be $12 * 6 * 83 = 5976$ spaces	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://puzzling.stackexchange.com/questions/14634/wheres-that-book-you-want/14694", "fetch_time": 1575777045000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-51/segments/1575540504338.31/warc/CC-MAIN-20191208021121-20191208045121-00219.warc.gz", "warc_record_offset": 512714205, "warc_record_length": 29679, "token_count": 385, "char_count": 1645, "score": 3.890625, "int_score": 4, "crawl": "CC-MAIN-2019-51", "snapshot_type": "latest", "language": "en", "language_score": 0.9730850458145142, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00019-of-00064.parquet"}
# Fillomino Tiling...how many 1's? Suppose a 'Fillomino tiling', much like a completed Fillomino puzzle, consists of a set of polyominoes covering a region without gaps nor overlaps, with no two n-ominoes of the same size touching along an edge. (touching at a finite number of points, e.g. corner-to-corner, is OK) Most Fillomino puzzles concern tiling a bounded rectangular region, but what about tilings of the infinite square lattice? In such a tiling, what is the largest possible fraction of the plane that can be legally covered with singlet squares ('monominoes')? • hexomino's and SE's answers indicate that there's some other constraint making patterns denser than theirs illegal. What is this constraint? Commented Jun 11, 2020 at 6:00 • Both are using the blue squares for singlet monominoes, and yellow for the other component of the tiling. 'Type A' squares are 'illegal' because they have the yellow regions forming monominoes also, touching the surrounding blue monominoes along full edges. In a Fillomino puzzle, this would be equivalent to putting two (or in this case five) 1's adjacent to each other, and is also directly forbidden by the rules above. Commented Jun 12, 2020 at 19:52 If I've calculated correctly, I think the best fraction that can be achieved is $$\frac{3}{8}$$ of the plane Diagram We can adopt the following pattern Reasoning The shape with the fewest number of tiles which may legally be surrounded by monominoes is the cross. It makes sense then to fill in any intermediary gaps with cross shapes. The resulting figure arises from constructing a checkerboard pattern an converting the colour of every second square on every second row (this is where the counting comes in). Obviously, not a full proof but a reasoned argument. Proof that $$F < \frac{3}{7}$$ Let $$F$$ be the fraction of the plane covered by monominoes and let $$G$$ be the fraction of the plane covered by tiles which are adjacent to monominoes. Then each monomino is adjacent to four non-monomino tiles but each non-monomino tile is adjacent to at most three monominoes. Overall, this means that $$G \geq \frac{4F}{3}$$ and since $$F + G \leq 1$$, we must have $$F + \frac{4F}{3} \leq 1 \Rightarrow F \leq \frac{3}{7}$$ A fraction of $$\frac{3}{7}$$ would mean that almost all non-monomino tiles must be adjacent to three monimonoes which is easily seen to be impossible so this bound cannot be achieved. • I am convinced this is the best pattern, but I'm not so convinced by the reasoning. If there is only one shape other than the monominos, then it can't touch itself so must be fully surrounded by monominos. But as crcrobert's answer shows, why would there need to be only one other shape? If not, it would have to be a combination of dominos and trominos, but it is not immediately obvious to me why that can't possibly give an equal or better solution. Commented Jun 10, 2020 at 9:47 • @JaapScherphuis Yes, I keep circling the answer but can't quite pin it down. My thinking is that if you have a connected region composed of multiple polyominoes then it has to be bigger than the cross-shape and you will lose out on your monomino count. I was also wondering if an adjacency count argument would work. You know that each monomino has to be adjacent to four non-monomino tiles and each non-monomino tile must be adjacent to at least one other non-monomino tile. This just about satisfies that. Commented Jun 10, 2020 at 10:22 • @JaapScherphuis I've added an argument to show that $F < \frac{3}{7}$. Perhaps a modification of this could get the bound all the way down? Commented Jun 10, 2020 at 11:17 • Wish I could split the checkmark, as both of your approaches are appealing...will at least hold off until humn can make good on his bounty offer. ^_^ Commented Jun 11, 2020 at 3:15 • @humn Yes, I understand, the approach is a lot more involved than I would have suspected. Commented Jun 11, 2020 at 8:53 Proof that the pattern found by @hexomino is optimal Using that colour scheme, polyminos can have the following kind of internal "nodes" (up to symmetry) We can immediately notice that: The "A" node is illegal. Let's make an assumption: The optimal pattern doesn't use the "B" node. But then, the highest density that can be achieved is: 1/3, as the "C" and "D" nodes cover 1/2 of the sides of a monomino at the cost of 1 non-monomino node, and "E" and "F" obviously less. As an aside, such a pattern actually exists as well: A staircase made out of "D"s. , although the diagonal strips must be terminated at some point to avoid them being "infinimonos". Since we already know that there's a better density possible, the assumption is false by contradiction. That means: The optimal pattern must contain the "B" node. Every "B" node must be connected to some other node, but it's only legal to connect it to "F" node, or a single place on the "E" node. The best density we can get when we connect a "B" to an "E" is: 1/3, since no other "B"s can be connected to the same "E", making them together cover all of the sides of a monomino in total, at the cost of two nodes. This still holds even if you try to connect other nodes, since we have already shown that they can't contribute to a density above 1/3 Thus, we must connect "B" nodes to "F" nodes. We know already that we can connect up to 4 "B" nodes to a "F". Since no other node can be connected to more than one "B", it's useless to connect one of them to a "F" to get more "B"s, since we would still have at most 4 "B" nodes per "F". (This point is a bit important, because if any other node could act like a "outlet splitter" to fit more "B" nodes, we could perhaps increase the density. But only the "F" node can fit more than one "B") This still leaves the possibility of perhaps fitting in a few extra "C", "D", "E" or "F" nodes, but since they can't contribute to a density above 1/3 without fitting in more "B"s (which we have shown is not possible), that would only make the density worse. Since the case of connecting 4 "B" to an "F" actually is geometrically possible (it tiles), it's the optimum solution. • Wish I could split the checkmark, as both of your approaches are appealing...will at least hold off until humn can make good on his bounty offer. ^_^ One question, though...does your demonstration above account for the possibility that there could be more than one non-monomino* component to the tiling? (*youtube.com/watch?v=8N_tupPBtWQ) Commented Jun 11, 2020 at 3:15 • @Zomulgustar Yes it does, as it shows that using any non-cross non-monomino makes the density worse. That only holds if the cross tiles though. If the cross didn't tile, most parts of the proof will fall apart. ("If this pattern is possible, it's the best one. And we are lucky that it exists"). Note also that the cross non-monomino can be tiled in many different ways. Commented Jun 14, 2020 at 10:31 As a bound, at least 1/4, based on:	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://puzzling.stackexchange.com/questions/99010/fillomino-tiling-how-many-1s/99022", "fetch_time": 1718279629000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-26/segments/1718198861372.90/warc/CC-MAIN-20240613091959-20240613121959-00628.warc.gz", "warc_record_offset": 435275001, "warc_record_length": 43526, "token_count": 1799, "char_count": 6943, "score": 3.796875, "int_score": 4, "crawl": "CC-MAIN-2024-26", "snapshot_type": "latest", "language": "en", "language_score": 0.9285755753517151, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00029-of-00064.parquet"}
# Prove that $x^3 + y^3 = z^3$ has no integer solutions as briefly as possible Can someone provide the proof of the special case of Fermat's Last Theorem for $n=3$, i.e., that $$x^3 + y^3 = z^3,$$ has no positive integer solutions, as briefly as possible? I have seen some good proofs, but they are quite long (longer than a page) or use many variables. However, I would rather have an elementary long proof with many variables than a complex short proof. Edit. Even if the bounty expires I will award one to someone if they have a satisfying answer. - see the similar [ math.stackexchange.com/questions/93817/sum-of-two-cubes ] – janmarqz Feb 3 '14 at 18:17 @janmarqz That question includes irrational numbers – qwr Feb 3 '14 at 18:18 “I have seen some good proofs, but they are quite long (more than a page) or use many variables.” Welcome to mathematics! – Carsten S Feb 3 '14 at 18:20 very simple proof would be to say that i believe it has not solution – dato datuashvili Feb 3 '14 at 18:29 @Lucian Well, that even seems to fit a book's margin. – Hagen von Eitzen Feb 3 '14 at 18:42 Main idea. The proof that follows is based on the infinite descent, i.e., we shall show that if $(x,y,z)$ is a solution, then there exists another triplet $(k,l,m)$ of smaller integers, which is also a solution, and this leads apparently to a contradiction. Assume instead that $x, y, z\in\mathbb Z\smallsetminus\{0\}$ satisfy the equation (replacing $z$ by $-z$) $$x^3 + y^3 + z^3 = 0,$$ with $x, y$ and $z$ pairwise coprime. (Clearly at least one is negative.) One of them should be even, whereas the other two are odd. Assume $z$ to be even. Then $x$ and $y$ are odd. If $x = y$, then $2x^3 = −z^3$, and thus $x$ is also even, a contradiction. Hence $x\ne y$. As $x$ and $y$ are odd, then $x+y$, $x-y$ are both even numbers. Let $$2u = x + y, \quad 2v = x − y,$$ where the non-zero integers $u$ and $v$ are also coprime and of different parity (one is even, the other odd), and $$x = u + v\quad \text{and}\quad y = u − v.$$ It follows that $$−z^3 = (u + v)^3 + (u − v)^3 = 2u(u^2 + 3v^2). \tag{1}$$ Since $u$ and $v$ have different parity, $u^2 + 3v^2$ is an odd number. And since $z$ is even, then $u$ is even and $v$ is odd. Since $u$ and $v$ are coprime, then $${\mathrm{gcd}}\,(2u,u^2 + 3v^2)={\mathrm{gcd}}\,(2u,3v^2)\in\{1,3\}.$$ Case I. $\,{\mathrm{gcd}}\,(2u,u^2 + 3v^2)=1$. In this case, the two factors of $−z^3$ in $(1)$ are coprime. This implies that $3\not\mid u$ and that both the two factors are perfect cubes of two smaller numbers, $r$ and $s$. $$2u = r^3\quad\text{and}\quad u^2 + 3v^2 = s^3.$$ As $u^2 + 3v^2$ is odd, so is $s$. We now need the following result: Lemma. If $\mathrm{gcd}\,(a,b)=1$, then every odd factor of $a^2 + 3b^2$ has this same form. Proof. See here. Thus, if $s$ is odd and if it satisfies an equation $s^3 = u^2 + 3v^2$, then it can be written in terms of two coprime integers $e$ and $f$ as $$s = e^2 + 3f^2,$$ so that $$u = e ( e^2 − 9f^2) \quad\text{and}\quad v = 3f ( e^2 − f^2).$$ Since $u$ is even and $v$ odd, then $e$ is even and $f$ is odd. Since $$r^3 = 2u = 2e (e − 3f)(e + 3f),$$ the factors $2e$, $(e–3f )$, and $(e+3f )$ are coprime since $3$ cannot divide $e$. If $3\mid e$, then $3\mid u$, violating the fact that $u$ and $v$ are coprime. Since the three factors on the right-hand side are coprime, they must individually equal cubes of smaller integers $$−2e = k^3,\,\,\, e − 3f = l^3,\,\,\, e + 3f = m^3,$$ which yields a smaller solution $k^3 + l^3 + m^3= 0$. Therefore, by the argument of infinite descent, the original solution $(x, y, z)$ was impossible. Case II. $\,{\mathrm{gcd}}\,(2u,u^2 + 3v^2)=3$. In this case, the greatest common divisor of $2u$ and $u^2 + 3v^2$ is $3$. That implies that $3\mid u$, and one may express $u = 3w$ in terms of a smaller integer, $w$. Since $4\mid u$, so is $w$; hence, $w$ is also even. Since $u$ and $v$ are coprime, so are $v$ and $w$. Therefore, neither $3$ nor $4$ divide $v$. Substituting $u$ by $w$ in $(1)$ we obtain $$−z^3 = 6w(9w^2 + 3v^2) = 18w(3w^2 + v^2)$$ Because $v$ and $w$ are coprime, and because $3\not\mid v$, then $18w$ and $3w^2 + v^2$ are also coprime. Therefore, since their product is a cube, they are each the cube of smaller integers, $r$ and $s$: $$18w = r^3 \quad\text{and}\quad 3w^2 + v^2 = s^3.$$ By the same lemma, as $s$ is odd and equal to a number of the form $3w^2 + v^2$, it too can be expressed in terms of smaller coprime numbers, $e$ and $f$: $$s = e^2 + 3f^2.$$ A straight-forward calculation shows that $$v = e (e^2 − 9f^2) \quad\text{and}\quad w = 3f (e^2 − f^2).$$ Thus, $e$ is odd and $f$ is even, because $v$ is odd. The expression for $18w$ then becomes $$r^3 = 18w = 54f (e^2 − f^2) = 54f (e + f) (e − f) = 33 \times 2f (e + f) (e − f).$$ Since $33$ divides $r^3$ we have that $3$ divides $r$, so $(r /3)^3$ is an integer that equals $2f (e + f) (e − f)$. Since $e$ and $f$ are coprime, so are the three factors $2e$, $e+f$, and $e−f$; therefore, they are each the cube of smaller integers, $k$, $l$, and $m$. $$−2e = k^3,\,\,\, e + f = l^3,\,\,\, e − f = m^3,$$ which yields a smaller solution $k^3 + l^3 + m^3= 0$. Therefore, by the argument of infinite descent, the original solution $(x, y, z)$ was impossible. Note. A simple proof of Fermat's Last Theorem when $n=3$ can be found here: http://fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-proof-for-n3.html - Can you present the proof all together? – qwr Feb 12 '14 at 20:24 I incorporated a proof, using a simple lemma: fermatslasttheorem.blogspot.com/2005/05/… – Yiorgos S. Smyrlis Feb 13 '14 at 9:42 Incredible! I thoroughly enjoyed this. Thank you – user602819 Apr 3 '15 at 1:32 By a theorem of A. Wiles (Ann. Math. 142) we have that $x^n+y^n=z^n$ has no solutions for $n\geq 3$. Now put $n=3$. Q. E. D. - Very unique approach. Can you clarify in under 100 pages? – qwr Feb 12 '14 at 20:15 @qwr: I have given a reference – user88576 Feb 12 '14 at 21:05 I've always appreciated that (unlike on many internet forums) on Math.SE answers actually try to answer the question (as opposed to, say, amusing the audience). This one is obviously an exception... – Grigory M Feb 12 '14 at 21:11 Although I appreciate the humor of the answer, it really doesn't make any attempt to answer the question. – user61527 Feb 13 '14 at 18:19 Humour — particularly patronizing humour — does not an appropriate answer make. – Kieren MacMillan Apr 30 '14 at 0:06 $$x^3+y^3=z^3$$ $$(x+y)(x^2-xy+y^2)=z^3$$ $$z=x+y$$ $$z^2=x^2-xy+y^2$$ $$x^2+2xy+y^2=z^2$$ $$-3xy=0.$$ Only if one of $x$ or $y$ or $z$ equal to $0$, the equation has solution, otherwise it has no solution. - This answer is incorrect. If $ab=cd$, it does not follow that $a=c$ and $b=d$. (For you, $a=(x+y)$, $b=(x^2-xy+y^2)$, $c=z$, $d=z^2$.) – Zev Chonoles Apr 6 '15 at 7:47	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://math.stackexchange.com/questions/662313/prove-that-x3-y3-z3-has-no-integer-solutions-as-briefly-as-possible", "fetch_time": 1454905162000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2016-07/segments/1454701152130.53/warc/CC-MAIN-20160205193912-00131-ip-10-236-182-209.ec2.internal.warc.gz", "warc_record_offset": 139783272, "warc_record_length": 21337, "token_count": 2488, "char_count": 6855, "score": 3.625, "int_score": 4, "crawl": "CC-MAIN-2016-07", "snapshot_type": "latest", "language": "en", "language_score": 0.9228942394256592, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00005-of-00064.parquet"}
# sqrt(x + 5) - sqrt(8 - x) < 1: Need help to find my mistake • Nov 16th 2010, 01:40 AM WeeG sqrt(x + 5) - sqrt(8 - x) < 1: Need help to find my mistake hi guys, can anyone help me to find my mistake, I can't figure out what's wrong, probably something small. the correct answer should be -5<=x<=4 I got -1<=x<=4 cheers ! • Nov 16th 2010, 04:54 AM HallsofIvy It's not so much a matter of a single "mistake" as that your whole concept is wrong! Do you understand that if x< y it does not necessarily follow that $x^2< y^2$? -3< 2 but 9 is NOT less than 4. I would do this instead: solve the equation $\sqrt{x+ 5}- \sqrt{8- x}= 1$. Those points separate "<" and ">". Check one point in each interval to see if the the inequality is satisfied in that interval.	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/pre-calculus/163412-sqrt-x-5-sqrt-8-x-1-need-help-find-my-mistake-print.html", "fetch_time": 1508490313000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-43/segments/1508187823997.21/warc/CC-MAIN-20171020082720-20171020102720-00714.warc.gz", "warc_record_offset": 204897793, "warc_record_length": 2824, "token_count": 252, "char_count": 763, "score": 3.828125, "int_score": 4, "crawl": "CC-MAIN-2017-43", "snapshot_type": "longest", "language": "en", "language_score": 0.926476001739502, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00050-of-00064.parquet"}
Minimum or maximum ratio for attacking RSA Let’s have $N_1=p^a$ where $p$ a prime number and $a$ a natural number. If we want to find all numbers from $1-N_1$ which are not divisible by $p$ we have to subtract all multiples of $p$ from $N_1$. The multiples are $1p,2p,...p^{a-1}p$. From this we conclude that the numbers not divisible by $p$ are $p^a-p^{a-1}$ or $N_1-N_1/p$. Let’s have the product of two primes $p^a q^b=N_2$ where $b$ a natural number. Now we have to find which numbers from $1-N_2$ are not divisible by $p,q$. In order to do so we have to subtract from $N_2$ the multiples $N_2/p$ and $N_2/q$. Now we have $(N_2-N_2/p)-N_2/q$, but some numbers are multiples of $pq$. We can subtract these multiples from $N_2/p$ or $N_2/q$. Let’s subtract them from $N_2/q$, giving $N_2/q-N_2/pq$. The last result has to be subtracted from the first term $(N_2-N_2/p)$. So we have $(N_2-N_2/p)-(N_2/q-N_2/pq)=F(N_2)$. So the number $F(N_2)$ is counting the numbers from $1-N_2$ which are not divisible by $p,q,pq$. We obtain the second term by applying the ABROZ technique. This works as follows: we multiply the first term by $1/q$ and we obtain$(N_2-N_2/p) 1/q=(N_2/q-N_2/pq)$. Let’s have $N_3=p^a q^b f^c$ where $f$ a prime number and $c$ a natural number. We know how to get the first two terms and now have to obtain the third term which contains the multiples of $f$. We again apply the technique, multiplying the two terms by $1/f$. So we have $(N_3-N_3/p)-(N_3/q-N_3/pq) 1/f=N_3/f-N_3/pf-N_3/qf-N_3/pqf$. To find $F(N_3)$ we subtract from the two terms the third term. So we have: $F(N_3)=(N_3-N_3/p)-(N_3/q-N_3/pq)-(N_3/f-N_3/pf-N_3/qf-N_3/pqf)$. We can continue this process indefinitely. The last equation can also be written as $F(N_3)=N_3[1-1/p-1/q+1/pq-1/f+1/pf+1/qf-1/pqf]$. The result inside the bracket can be obtained by multiplication of the three primes as follows: $(1-1/p)(1-1/q)(1-1/f)=[1-1/p-1/q+1/pq-1/f+1/pf+1/qf-1/pqf]$. So we have $F(N_3)=N_3(1-1/p)(1-1/q)(1-1/f)$. We can make keys when$F(N)/N$ is minimum or maximum. My first question is: when is it easier to attack RSA, when $F(N)/N$ is minimum or maximum? Secondly, in what other areas of mathematics can we apply the maximum and minimum ratios? - Formulas for the most vulnerable ratios for RSA private keys have already been found. There are exactly eight formulas for the eight different ranges of ratio that are hazardous to RSA. Great time and effort was placed on discovering these formula. – user62572 Feb 16 '13 at 18:45 $F(N)$ doesn't have a maximum (for $N\gt1$), and it doesn't have a minimum. It can be made arbitrarily close to, but not equal to, zero, and it can be made arbitrarily close to, but not equal to, one. As Yuval suggests, RSA implementations are usually taken with $N=pq$ with $p$ and $q$ large primes, and this will lead to $F(N)$ close to one. - @Gerry.This is the correct answer. – Vassilis Parassidis Mar 11 '12 at 3:55 The function $F$ is usually knows as Euler's phi function. The "hard" case for RSA is usually considered to be when $n$ is a product of two primes of roughly equal magnitude. One reason is that many factorization algorithms run in time that depends on the smallest prime factor of $n$. - @Yuval.Are you suggesting the ratio has to be maximum? – Vassilis Parassidis Mar 10 '12 at 22:34	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://math.stackexchange.com/questions/118691/minimum-or-maximum-ratio-for-attacking-rsa?answertab=active", "fetch_time": 1462555168000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2016-18/segments/1461861848830.49/warc/CC-MAIN-20160428164408-00157-ip-10-239-7-51.ec2.internal.warc.gz", "warc_record_offset": 187494740, "warc_record_length": 17809, "token_count": 1082, "char_count": 3333, "score": 4.5, "int_score": 4, "crawl": "CC-MAIN-2016-18", "snapshot_type": "latest", "language": "en", "language_score": 0.8277875185012817, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00007-of-00064.parquet"}
# Homework Help: Bayes theorem problem 1. Jun 15, 2013 ### aaaa202 1. The problem statement, all variables and given/known data This problem was under applications of Bayes theorem, but I feel like I am bad at using it if thats the case: At a school 30% of the students are girls. 4% of the girls are geeks and 2% of all geeks are girls. What is probability that a random student is a geek. 2. Relevant equations Bayes theorem 3. The attempt at a solution P(AlDI) = P(DlAI)/P(DlI) P(AlI) I guess I should assign the statement to A: a student is a geek And as my data I don't know what to use. That 4% of girls are geeks? Using Bayes theorem I found that 3/5 is the probability. Is this right and how do you arrive at it using Bayes theorem? 2. Jun 15, 2013 ### LCKurtz I don't know where you got that equation from. Generally you have $P(A\cap B) = P(A\|B)P(B)$ and $P(A\cap B) = P(B\|A)P(A)$, so $P(A\|B)P(B)=P(B\|A)P(A)$. Try using that. 3. Jun 15, 2013 ### marcusl The equations are equivalent. 4. Jun 15, 2013 ### D H Staff Emeritus Wait a second. You said you used Bayes' theorem, and yet you're asking us how to use it? Show us how you arrived at that result (it's correct, BTW) and we'll be able to tell you if you did things right, or in case you didn't, help you get past your stumbling blocks. 5. Jun 15, 2013 ### Ray Vickson My personal recommendation would be: stay away from Bayes Theorem for a little while, until you understand the concepts and issues. At that point, Bayes results become handy shorthands that help you get answers quickly---after you know what it is you should be trying to do. In other words: understanding and intuition come first, formulas come later. So, what is happening in this problem? An approach my students often found useful back in the Stone Age when I was still teaching is essentially a "tabular" method: image a school with a large student population---say 1000 students. How many are girls? How girls are geeks? From that, how many students altogether are geeks? Once you have figured that out, can you see how to complete the calculations? I will just do a couple of steps to get you started: 30% of the students are girls, so N(girls) = 300. We are given that 4% of the girls are geeks, so N(Geeky girls) = 0.04300 = 12. You are told that 2% of all geeks are girls, and you know how many girls that is; so how many geeks are there? 6. Jun 15, 2013 ### aaaa202 thats exactly how I did. Multiplied the number of geeky girls by 50 to get the total number of geeks. I just wanted to see how to set the problem up with Bayes theorem. 7. Jun 15, 2013 ### D H Staff Emeritus The form of Bayes' theorem cited by you and by LCKurtz appear to be different, but as marcusl already said, they are equivalent. Personally, I'd go with the simpler form. Using that simpler form, Bayes' theorem says that $$P(\text{geek}\|\text{girl})P(\text{girl}) = P(\text{girl}\|\text{geek})P(\text{geek})$$The question gives values for every single one of these except for $P(\text{geek})$, and that one missing value is the exactly the one to be solve for. So simply substitute the known values and solve for the unknown $P(\text{geek})$. For now I'll leave it up to you to translate the word problem text to the mathematical terms such as $P(\text{geek}\|\text{girl})$. 8. Jun 15, 2013 ### HallsofIvy I prefer to use specific numbers rather than percentages. Let's say there are 1000 students in the school. 30% of the students are girls so there are 300 girls. 4% of the girls are "geeks" so there are 12 "girl geeks". Those 12 girls are 2% of the geeks: letting N be the number of geeks, we have .02N= 12 so N= 12/.02= 600 geeks. Yes, 600/1000= 3/5. 9. Jun 15, 2013 ### D H Staff Emeritus That's correct, and that is equivalent to how aaaa202 solved the problem. However, the question is listed "under applications of Bayes theorem", so presumably a solution based on Bayes' theorem is what is desired. 10. 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# Prove:$\forall a_1,b_1,a_2,b_2: \left\|\max(a_1,b_1) - \max(a_2,b_2)\right\| \le \max(\left\|a_1-a_2\right\|, \left\| b_1-b_2 \right\|)$ Prove:$\forall a_1,b_1,a_2,b_2: \left\|\max(a_1,b_1) - \max(a_2,b_2)\right\| \le \max(\left\|a_1-a_2\right\|, \left\| b_1-b_2 \right\|)$ How can I prove it easily? Or should I just go over each case? All those cases make me confused! WLOG, assume $a_1\geq b_1$. If $a_2\ge b_2$, LHS=$\|a_1-a_2\|$, the inequality holds. If $a_2<b_2$, then LHS=$\|a_1-b_2\|$, we further separate it into two cases. If $b_2\le a_1$ then LHS=$\|a_1-b_2\|\le \|a_1-a_2\|$, the inequality holds. If $b_2> a_1$ then LHS=$\|a_1-b_2\|\le \|b_1-b_2\|$, the inequality holds. • Thank you. You made it very clear! – AlonAlon Nov 15 '14 at 16:25 Just some hints. Here are some general principles that can be helpful. See if these work for you. 1. Proving $\|A\| \leq B$ amounts to proving $A \leq B$ and $-A \leq B$. 2. Proving that $\max(A,B) \leq C$ amounts to proving $A \leq C$ and $B \leq C$. 3. Proving that $\max(A,B) \geq C$ amounts to proving $A \geq C$ or $B \geq C$. After some manipulations and omission of symmetric cases, I believe you'll find that the inequality to be proved can be reduced to $$a_1 \leq \|a_1 - a_2\| + a_2.$$	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/1023013/prove-forall-a-1-b-1-a-2-b-2-left-maxa-1-b-1-maxa-2-b-2-right-le", "fetch_time": 1563456006000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-30/segments/1563195525634.13/warc/CC-MAIN-20190718125048-20190718151048-00052.warc.gz", "warc_record_offset": 455438744, "warc_record_length": 35318, "token_count": 510, "char_count": 1238, "score": 4.34375, "int_score": 4, "crawl": "CC-MAIN-2019-30", "snapshot_type": "latest", "language": "en", "language_score": 0.6337406635284424, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00001-of-00064.parquet"}
# Sections 1 and 2 ## 1.4. Computational methods We define a function $$\verb\|hessRank2\|$$ that takes as input a quaternary cubic, and outputs the ideal defining the rank 2 matrices in $$H(f)$$. If $$\verb\|xCoords == false\|$$, the ideal is computed in the $$z_{ij}$$ coordinates, i.e. we view $$H(f)$$ as a subspace of the space of symmetric $$4 \times 4$$ matrices. If $$\verb\|xCoords == true\|$$, the ideal is computed in the $$x_i$$ coordinates, i.e. we use the Hessian matrix to identify $$H(f)$$ with $$\mathbb{P}^3$$. K=QQ; R=K[x_0..x_3,l_0..l_4,z00,z01,z02,z03,z11,z12,z13,z22,z23,z33]; X={x_0,x_1,x_2,x_3}; A=genericSymmetricMatrix(R,z00,4); I2=minors(3,A); hessRank2 = {xCoords=>false} >> o -> f ->( hess = diff(transpose matrix{X},diff(matrix{X},f)); I=eliminate(X,ideal(flatten entries (A-hess))); J=I+I2; if o.xCoords then( J=sub(J,flatten apply(4,i->apply(i+1,j->(A_(i,j)=>hess_(i,j))))); ); return J; ); isOnHessianDiscriminant = f ->( J=hessRank2(f); return not ((codim J==9) and (J==radical J)); ); Some examples: f=x_0x_1x_2+x_0x_1x_3+x_0x_2x_3+x_1x_2x_3; isOnHessianDiscriminant(f) --false f=x_0^3+x_1^3+x_2^3+x_3^2(3x_0+3x_1+3x_2+x_3); isOnHessianDiscriminant(f) --true ## 2.1. Sylvester’s pentahedral form We compute $$H(f) \cap X_2$$ for cubics in Sylvester’s pentahedral form. x_4=-sum(4,i->(x_i)); f=sum(5,i->(l_ix_i^3)); J=hessRank2(f); P=primaryDecomposition J; toString P We find the following primat primary decomposition, as described in remark 2.3: {ideal(z22-z23,z13-z23,z12-z23,z11-z23,z03-z23,z02-z23,z01-z23,z00-z23,l_3z23-l_4z23+l_4z33),ideal(z33,z23,z22,z13,z12,z03,z02,z01,l_1z00+l_0z11),ideal(z33,z23,z22,z13,z12,z03,z02,z01,l_4),ideal(z23,z13,z12,z11,z03,z02,z01,z00,l_4),ideal(z33,z23,z13,z12,z03,z02,z01,z00,l_4),ideal(z23,z22,z13,z12,z03,z02,z01,z00,l_4),ideal(z33,z23,z13,z12,z11,z03,z02,z01,l_4),ideal(z23,z22,z13,z12,z11,z03,z02,z01,l_4),ideal(z23,z13,z12,z11,z03,z02,z01,z00,l_1),ideal(z23,z13,z12,z11,z03,z02,z01,z00,l_0),ideal(z33,z23,z13,z12,z03,z02,z01,z00,l_0),ideal(z23,z22,z13,z12,z03,z02,z01,z00,l_0),ideal(z33,z23,z13,z12,z03,z02,z01,z00,l_3),ideal(z33,z23,z13,z12,z03,z02,z01,z00,l_2z11+l_1z22),ideal(z23,z22,z13,z12,z03,z02,z01,z00,l_2),ideal(z23,z22,z13,z12,z03,z02,z01,z00,l_3z11+l_1z33),ideal(z23,z13,z12,z11,z03,z02,z01,z00,l_3z22+l_2z33),ideal(z33,z23,z13,z12,z11,z03,z02,z01,l_1),ideal(z23,z22,z13,z12,z11,z03,z02,z01,l_1),ideal(z33,z23,z13,z12,z11,z03,z02,z01,l_3),ideal(z33,z23,z13,z12,z11,z03,z02,z01,l_2z00+l_0z22),ideal(z23,z22,z13,z12,z11,z03,z02,z01,l_2),ideal(z23,z22,z13,z12,z11,z03,z02,z01,l_3z00+l_0z33),ideal(z33,z23,z22,z13,z12,z03,z02,z01,l_3),ideal(z33,z23,z22,z13,z12,z03,z02,z01,l_2),ideal(z22-z23,z13-z23,z12-z23,z11-z23,z03-z23,z02-z23,z01-z23,z00-z23,l_2),ideal(z23-z33,z22-z33,z13-z33,z12-z33,z03-z33,z02-z33,z01-z33,z00-z33,l_2),ideal(z23-z33,z22-z33,z13-z33,z12-z33,z11-z33,z03-z33,z02-z33,z01-z33,l_2),ideal(z22-z23,z13-z23,z12-z23,z11-z23,z03-z23,z02-z23,z01-z23,z00-z23,l_1),ideal(z23-z33,z13-z33,z12-z33,z11-z33,z03-z33,z02-z33,z01-z33,z00-z33,l_1),ideal(z23-z33,z22-z33,z13-z33,z12-z33,z11-z33,z03-z33,z02-z33,z01-z33,l_1),ideal(z22-z23,z13-z23,z12-z23,z11-z23,z03-z23,z02-z23,z01-z23,z00-z23,l_0),ideal(z23-z33,z13-z33,z12-z33,z11-z33,z03-z33,z02-z33,z01-z33,z00-z33,l_0),ideal(z23-z33,z22-z33,z13-z33,z12-z33,z03-z33,z02-z33,z01-z33,z00-z33,l_0),ideal(z23-z33,z13-z33,z12-z33,z11-z33,z03-z33,z02-z33,z01-z33,z00-z33,l_3),ideal(z23-z33,z22-z33,z13-z33,z12-z33,z03-z33,z02-z33,z01-z33,z00-z33,l_3),ideal(z23-z33,z22-z33,z13-z33,z12-z33,z11-z33,z03-z33,z02-z33,z01-z33,l_3),ideal(z23-z33,z22-z33,z13-z33,z12-z33,z11-z33,z03-z33,z02-z33,z01-z33,l_4z00+l_0z33-l_4z33),ideal(z23-z33,z22-z33,z13-z33,z12-z33,z03-z33,z02-z33,z01-z33,z00-z33,l_4z11+l_1z33-l_4z33),ideal(z23-z33,z13-z33,z12-z33,z11-z33,z03-z33,z02-z33,z01-z33,z00-z33,l_4z22+l_2z33-l_4z33)} --length P==40 If we assume additionally that all l_i are nonzero (which amounts to saturating our ideal), the primary decomposition becomes much shorter: Jsat =saturate(J,ideal product(5,i->l_i)); Psat = primaryDecomposition(Jsat); toString Psat {ideal(z22-z23,z13-z23,z12-z23,z11-z23,z03-z23,z02-z23,z01-z23,z00-z23,l_3z23-l_4z23+l_4z33),ideal(z23,z13,z12,z11,z03,z02,z01,z00,l_3z22+l_2z33),ideal(z23,z22,z13,z12,z03,z02,z01,z00,l_3z11+l_1z33),ideal(z23,z22,z13,z12,z11,z03,z02,z01,l_3z00+l_0z33),ideal(z33,z23,z22,z13,z12,z03,z02,z01,l_1z00+l_0z11),ideal(z33,z23,z13,z12,z03,z02,z01,z00,l_2z11+l_1z22),ideal(z33,z23,z13,z12,z11,z03,z02,z01,l_2z00+l_0z22),ideal(z23-z33,z13-z33,z12-z33,z11-z33,z03-z33,z02-z33,z01-z33,z00-z33,l_4z22+l_2z33-l_4z33),ideal(z23-z33,z22-z33,z13-z33,z12-z33,z03-z33,z02-z33,z01-z33,z00-z33,l_4z11+l_1z33-l_4z33),ideal(z23-z33,z22-z33,z13-z33,z12-z33,z11-z33,z03-z33,z02-z33,z01-z33,l_4z00+l_0z33-l_4z33)} Upon inspecting Psat, we see that for any cubic in sylvester pentahedral form with all coefficients $$\lambda_i$$ nonzero, the 10 points in $$H(f) \cap X_2$$ are distinct In x-coordinates: J=saturate(hessRank2(f,xCoords=>true),ideal product(5,i->l_i)); Psat = primaryDecomposition(J); toString Psat {ideal(x_3,x_2,x_1),ideal(x_2,x_1,x_0+x_3),ideal(x_2,x_1,x_0),ideal(x_3,x_2,x_0+x_1),ideal(x_3,x_2,x_0),ideal(x_2,x_1+x_3,x_0),ideal(x_3,x_1,x_0+x_2),ideal(x_3,x_1,x_0),ideal(x_3,x_1+x_2,x_0),ideal(x_2+x_3,x_1,x_0)} ## 2.2 Rank six cubics We now turn our attention to general rank 6 cubics. f=x_1^3+x_2^3+x_3^3-x_0^2(l_0x_0+3l_1x_1+3l_2x_2+3l_3x_3); J=hessRank2(f); --We assume that l_1,l_2,l_3 are nonzero: Jsat =saturate(J,ideal(l_1l_2l_3)); Psat = primaryDecomposition(Jsat); print toString Psat {ideal(z23,z13,z12,z11,z03,z02,z01,z00,l_2z22+l_3z33), ideal(z23,z22,z13,z12,z11,z03^2,z02z03,z01z03,z02^2,z01z02,l_3z02-l_2z03,z01^2,l_3z01-l_1z03,l_2z01-l_1z02,l_3z03z33+z00z03,l_2z03z33+z00z02,l_1z03z33+z00z01,l_3z00z33+l_0z03z33+z00^2,l_3^2z33+l_3z00-l_0z03,l_2l_3z33+l_2z00-l_0z02,l_1l_3z33+l_1z00-l_0z01), ideal(z23,z22,z13,z12,z03,z02,z01,z00,l_1z11+l_3z33), ideal(z33,z23,z22,z13,z12,z11,l_3z02-l_2z03,l_3z01-l_1z03,l_2z01-l_1z02,l_3z00-l_0z03,l_2z00-l_0z02,l_1z00-l_0z01), ideal(z33,z23,z22,z13,z12,z03^2,z02z03,z01z03,z02^2,z01z02,l_3z02-l_2z03,z01^2,l_3z01-l_1z03,l_2z01-l_1z02,l_1z03z11+z00z03,l_1z02z11+z00z02,l_1z01z11+z00z01,l_1z00z11+l_0z01z11+z00^2,l_1l_3z11+l_3z00-l_0z03,l_1l_2z11+l_2z00-l_0z02,l_1^2z11+l_1z00-l_0z01), ideal(z33,z23,z13,z12,z11,z03^2,z02z03,z01z03,z02^2,z01z02,l_3z02-l_2z03,z01^2,l_3z01-l_1z03,l_2z01-l_1z02,l_2z03z22+z00z03,l_2z02z22+z00z02,l_1z02z22+z00z01,l_2z00z22+l_0z02z22+z00^2,l_2l_3z22+l_3z00-l_0z03,l_2^2z22+l_2z00-l_0z02,l_1l_2z22+l_1z00-l_0z01), ideal(z33,z23,z13,z12,z03,z02,z01,z00,l_1z11+l_2z22)} {ideal(z23,z13,z12,z11,z03,z02,z01,z00,l_2z22+l_3z33), ideal(-z23,-z22,-z13,-z12,-z11,-z03,-z02,-z01,-l_3z33-z00), ideal(z23,z22,z13,z12,z03,z02,z01,z00,l_1z11+l_3z33), ideal(z33,z23,z22,z13,z12,z11,l_3z02-l_2z03,l_3z01-l_1z03,l_2z01-l_1z02,l_3z00-l_0z03,l_2z00-l_0z02,l_1z00-l_0z01), ideal(-z33,-z23,-z22,-z13,-z12,-z03,-z02,-z01,-l_1z11-z00), ideal(-z33,-z23,-z13,-z12,-z11,-z03,-z02,-z01,-l_2z22-z00), ideal(z33,z23,z13,z12,z03,z02,z01,z00,l_1z11+l_2z22)} Inspecting the components of the primary decomposition and their radicals shows that for $$H(f) \cap X_2$$ consists of 4 reduced and 3 nonreduced points, completing the proof of Theorem 2.5. In x-coordinates: J=saturate(hessRank2(f,xCoords=>true),ideal(l_1l_2l_3)); Psat = primaryDecomposition(J); print toString Psat {ideal(x_3,x_2,x_1), ideal(x_3,x_0,x_1l_1+x_2l_2), ideal(x_1,x_0,x_2l_2+x_3l_3), ideal(x_3,x_1,x_0^2), ideal(x_2,x_1,x_0^2), ideal(x_2,x_0,x_1l_1+x_3l_3), ideal(x_3,x_2,x_0^2)} {ideal(x_1,x_2,x_3), ideal(x_3,x_0,x_1l_1+x_2l_2), ideal(x_1,x_0,x_2l_2+x_3l_3), ideal(x_0,x_1,x_3), ideal(x_0,x_1,x_2), ideal(x_2,x_0,x_1l_1+x_3l_3), ideal(x_0,x_2,x_3)} Remark 2.7: special rank six cubics f=l_0x_0^3+x_1^3+x_2^3-3x_0(l_1x_0x_1+x_0x_2+x_3^2); J=hessRank2(f); --We assume that all l_1 is nonzero: Jsat =saturate(J,ideal l_1); Psat = primaryDecomposition(Jsat); print toString Psat {ideal(z23,z22,z13,z12,z02-z33,z33^2,z03z33,z01z33,l_1z33-z01,z01z11+z00z33,l_1z11+l_0z33+z00,z03^2-z00z33,z01z03,z01^2), ideal(z23,z13,z12,z11,z02-z33,z33^2,z03z33,z01z33,z00z33+z22z33,l_1z33-z01,l_0z33+z00+z22,z03^2+z22z33,z01z03,z00z03+z03z22,z01^2,z00z01+z01z22,z00^2+2z00z22+z22^2,l_1z00+l_0z01+l_1z22), ideal(z33,z23,z13,z12,z03,z02,z01,z00,l_1z11+z22), ideal(z23,z22,z13,z12,z11,z02-z33,l_1z33-z01,l_0z33+z00,l_1z00+l_0z01,z33^3,z01z33^2,z00z33^2,z01^2z33,z00^2z33,z00^2z01,z00^3)} {ideal(-z33,-z23,-z22,-z13,-z12,-z03,-z02,-z01,-l_1z11-z00), ideal(z33,z23,z13,z12,z11,z03,z02,z01,z00+z22), ideal(-z33,-z23,-z13,-z12,-z03,-z02,-z01,-z00,-l_1z11-z22), ideal(z33,z23,z22,z13,z12,z11,z02,z01,z00)} In x-coordinates: J=saturate(hessRank2(f,xCoords=>true),ideal l_1); Psat = primaryDecomposition(J); print toString Psat {ideal(x_3,x_0,x_1l_1+x_2), ideal(x_2,x_1,x_0^3), ideal(x_1,x_0x_3,x_0x_2-x_3^2,x_0^2), ideal(x_2,x_0x_3,x_0^2,x_0x_1l_1-x_3^2)} {ideal(x_3,x_0,x_1l_1+x_2), ideal (x_0,x_1,x_2), ideal(x_3,x_1,x_0), ideal(x_3,x_2,x_0)} ## 2.3 General singular cubics It was already verified in 1.3 that they Cayley cubic does not lie on the Hurwitz form. We now do the same for a random singular cubic. rand=flatten entries random(K^4,K^1); for i in 0..3 do( c_i=rand#i; ); f=c_0x_3(x_1^2-x_0x_2)+x_1(x_0-(1+c_1)x_1+c_1x_2)(x_0-(c_2+c_3)x_1+c_2c_3x_2); isOnHessianDiscriminant(f) --false The Cayley cubic again, in x-coordinates: f=x_0x_1x_2+x_0x_1x_3+x_0x_2x_3+x_1x_2x_3; J=hessRank2(f,xCoords=>true); P = primaryDecomposition(J) {ideal(x_3,x_1,x_0+x_2), ideal(x_3,x_2,x_0+x_1), ideal(x_3,x_1+x_2,x_0), ideal(x_2,x_1,x_0+x_3), ideal(x_2,x_1+x_3,x_0), ideal(x_2-x_3,x_1-x_3,x_0+x_3), ideal(x_2-x_3,x_1+x_3,x_0-x_3), ideal(x_2+x_3,x_1-x_3,x_0-x_3), ideal(x_2+x_3,x_1+x_3,x_0+x_3), ideal(x_2+x_3,x_1,x_0)}	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://mathrepo.mis.mpg.de/HessianDiscriminant/Sections12.html", "fetch_time": 1718841834000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00416.warc.gz", "warc_record_offset": 342269864, "warc_record_length": 7331, "token_count": 4851, "char_count": 10107, "score": 3.625, "int_score": 4, "crawl": "CC-MAIN-2024-26", "snapshot_type": "latest", "language": "en", "language_score": 0.4796389639377594, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00039-of-00064.parquet"}
# SBI Clerk 2018: Reasoning Ability Quiz – 7 Hello and welcome to exampundit. Here is a set of Reasoning Quiz on Puzzle problems for SBI Clerk 2018 Prelims. 6 banks ABC,PQR,XYZ,DEF,STU,GHI have their Head Quarter in 6 different states Lucknow, Kochi,Mumbai, Bengaluru, Indore,Kolkata. The banks have their own 6 different interest rates- 5%,5.5%,5.25%,6%,6.25%,5.75%. ABC’s HQ is neither in Kochi nor in Lucknow. IR 6.25% is of the bank whose HQ is neither in Kolkata nor of the bank XYZ. Neither PQR nor ABC’s headquarter is in Bengaluru. DEF’s HQ is neither in Kochi nor in Lucknow. IR 5.25% is of the bank whose HQ is neither in Lucknow nor in Indore. The bank which is in Kolkata, IR is neither 6% nor 5.25%. PQR’s HQ is neither in Kolkata nor in Kochi. XYZ’s IR is neither 5.5% nor 6%. Neither PQR nor GHI’s HQ is in Lucknow. IR 6.25% is neither of GHI nor of PQR. The bank STU’s HQ is in Mumbai and its IR is 5%. 1. XYZ’s IR is- (A) 5.75% (B) 6.25% (C) 5.25% (D) none of these Option: A 1. DEF’s HQ is in – (A) Indore (B) Kochi (C) Lucknow (D) None of these Option: D 1. GHI’s IR is – (A) 6% (B) 5.25% (C) 5.5% (D) 5.75% Option: B 1. Which bank’s HQ is in Indore? (A) DEF (B) ABC (C) PQR (D) XYZ Option: C 1. Which bank’s IR is 6.25%? (A) DEF (B) PQR (C) ABC (D) STU Option: A Show Solution Bank HQ Interest Rate ABC Kolkata 5.5% XYZ Lucknow 5.75% DEF Bengaluru 6.25% GHI Kochi 5.25% PQR Indore 6.00% STU Mumbai 5.00% 10 friends are seating in two rows. P,Q,R,S,T are facing north and E,F,G,H,I are facing south. F is not facing S or P. R is not facing H or E. I is at one of the extreme ends but neither beside H nor E. G is not facing Q or R. I is neither facing S or T. T is neither facing F or G. T is at one of the ends but neither beside P nor R. Q is neither facing I, but seating in the middle. No one is there on T’s right side. S is facing E nor G. 1. Who is facing S? (A) H (B) I (C) G (D) E Option: A 1. T is facing- (A) I (B) G (C) H (D) E Option: D 1. Odd one out- (A) H (B) Q (C) T (D) P Option: C 1. Who is on the immediate right of E? (A) G (B) I (C) H (D) F Option: C 1. Who is on the immediate left of R? (A) P (B) S (C) Q (D) None	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://exampundit.in/sbi-clerk-2018-reasoning-ability-quiz-7/", "fetch_time": 1556153890000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-18/segments/1555578675477.84/warc/CC-MAIN-20190424234327-20190425020327-00098.warc.gz", "warc_record_offset": 402069077, "warc_record_length": 28212, "token_count": 789, "char_count": 2222, "score": 3.546875, "int_score": 4, "crawl": "CC-MAIN-2019-18", "snapshot_type": "longest", "language": "en", "language_score": 0.9152812361717224, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00032-of-00064.parquet"}
# If 10^x=64 then find the value of 10^x/2+1 2 by sethjoanna012 2015-04-27T20:44:37+05:30 10^x=64.............................(1) Question :To find 10^x/2 +1 Putting the value of 10^x as 64 from.......(1) ,we get 64/2+1 =32+1 =33 Ans:33 2015-04-28T15:52:37+05:30 10^x = 64 10^x/2 = 64^1/2 = 8 now 10^x/2 + 1 10^x/2 * 10^1 8 *10 80	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://brainly.in/question/100576", "fetch_time": 1484780735000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-04/segments/1484560280364.67/warc/CC-MAIN-20170116095120-00271-ip-10-171-10-70.ec2.internal.warc.gz", "warc_record_offset": 796129404, "warc_record_length": 9643, "token_count": 177, "char_count": 332, "score": 3.984375, "int_score": 4, "crawl": "CC-MAIN-2017-04", "snapshot_type": "latest", "language": "en", "language_score": 0.4576941132545471, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00039-of-00064.parquet"}
The quadratic equation $2x^2 - 3x +3=0$ is to be solved numerically starting with an initial guess as $x_0=2$. The new estimate of $x$ after the first iteration using Newton-Raphson method is _________ F(x) = 2x^2 - 3x + 3 , x0 = 2 using Newton Raphson method X(n+1) = Xn - F(Xn)/F’(Xn) So, F(x) = 2x^2 - 3x + 3 & F’(x) = 4x - 3 F(2) = 5 F’(2) = 5 therefore X1 = X0 - F(X0)/ F’(X0) = 2-1 = 1 by	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://civil.gateoverflow.in/160/gate2018-ce-2-20?show=206", "fetch_time": 1670095009000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-49/segments/1669446710936.10/warc/CC-MAIN-20221203175958-20221203205958-00568.warc.gz", "warc_record_offset": 210054600, "warc_record_length": 14908, "token_count": 184, "char_count": 403, "score": 3.890625, "int_score": 4, "crawl": "CC-MAIN-2022-49", "snapshot_type": "latest", "language": "en", "language_score": 0.7331302762031555, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00001-of-00064.parquet"}
If n and p are integers, is p>0? 1. n+1 > 0 2. np>0 : DS Archive Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack It is currently 23 Jan 2017, 21:03 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # If n and p are integers, is p>0? 1. n+1 > 0 2. np>0 Author Message Manager Joined: 11 Jul 2006 Posts: 67 Followers: 1 Kudos [?]: 42 [0], given: 0 If n and p are integers, is p>0? 1. n+1 > 0 2. np>0 [#permalink] ### Show Tags 08 Sep 2007, 18:24 00:00 Difficulty: (N/A) Question Stats: 0% (00:00) correct 0% (00:00) wrong based on 1 sessions ### HideShow timer Statistics This topic is locked. If you want to discuss this question please re-post it in the respective forum. If n and p are integers, is p>0? 1. n+1 > 0 2. np>0 Intern Joined: 05 Mar 2007 Posts: 24 Followers: 0 Kudos [?]: 2 [0], given: 0 Re: Gmat Prep - DS [#permalink] ### Show Tags 09 Sep 2007, 09:38 mbunny wrote: If n and p are integers, is p>0? 1. n+1 > 0 2. np>0 C it is From 1) n > -1 , so n >=0, But dont know anything about p, so InSuff From 2) np both must either be positive or both must be -ve, so p could be positive or negative From 1) & 2) n>=0, and np>0, so, n canot be 0, and must p>0. Senior Manager Joined: 27 Aug 2007 Posts: 253 Followers: 1 Kudos [?]: 12 [0], given: 0 ### Show Tags 09 Sep 2007, 10:53 Agree I didnot take into account that np>0 I see the point Ans: C Last edited by Ferihere on 09 Sep 2007, 11:26, edited 1 time in total. VP Joined: 10 Jun 2007 Posts: 1459 Followers: 7 Kudos [?]: 256 [0], given: 0 Re: Gmat Prep - DS [#permalink] ### Show Tags 09 Sep 2007, 10:58 mbunny wrote: If n and p are integers, is p>0? 1. n+1 > 0 2. np>0 I think it's C here (1) Clearly INSUFFICIENT (2) if n<0 and p<0, then np>0 of n>0, p>0, then np>0 INSUFFICIENT Together, you know from (1) that n > -1 Since n is an integer, possible values of n are: 0, 1, 2, 3, ... Knowing that np > 0, n and P cannot be zero. This means that n can only be positive number. This also means that p is positive. SUFFICIENT Manager Joined: 11 Jul 2006 Posts: 67 Followers: 1 Kudos [?]: 42 [0], given: 0 ### Show Tags 09 Sep 2007, 16:37 Thanks all, the OA is C. CEO Joined: 29 Mar 2007 Posts: 2583 Followers: 19 Kudos [?]: 422 [0], given: 0 Re: Gmat Prep - DS [#permalink] ### Show Tags 09 Sep 2007, 20:08 mbunny wrote: If n and p are integers, is p>0? 1. n+1 > 0 2. np>0 S1: nothing about p is mentioned. But n>-1. Insuff. S2: Insuff. could be np or -n-p 1&2. Since n and p are i's. n>0. So np>0 p is + or p>0 C GMAT Club Legend Joined: 07 Jul 2004 Posts: 5062 Location: Singapore Followers: 30 Kudos [?]: 358 [0], given: 0 ### Show Tags 09 Sep 2007, 22:00 St1: St2: n = +ve, p=+ve or n = -ve and p = -ve. Insufficient. St1 and St2: n > -1. Since n*p must be greater than 0, neither n nor p can be 0. So n must be positive and p must be positive. Sufficieint. Ans C 09 Sep 2007, 22:00 Display posts from previous: Sort by	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://gmatclub.com/forum/if-n-and-p-are-integers-is-p-0-1-n-1-0-2-np-51993.html?fl=similar", "fetch_time": 1485234191000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-04/segments/1484560284270.95/warc/CC-MAIN-20170116095124-00149-ip-10-171-10-70.ec2.internal.warc.gz", "warc_record_offset": 118022648, "warc_record_length": 44459, "token_count": 1266, "char_count": 3513, "score": 3.90625, "int_score": 4, "crawl": "CC-MAIN-2017-04", "snapshot_type": "latest", "language": "en", "language_score": 0.8375754356384277, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00061-of-00064.parquet"}
# physics posted by . A 75 kg box slides down a 25o ramp with an acceleration of 3.6 m/s2. Find the coefficient of kinetic friction between the box and the ramp. • physics - The component of gravity down the plane is mgSinTheta The normal component of mgCosTheta So Net force=ma gravitydownramp-friction=ma mgSinTheta-mgmuCosTheta=ma solve for mu. • physics - 0.06 • physics - 0.061 ## Similar Questions 1. ### Physics A 10-kg wooden box rests on a wooden ramp. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. What is the friction force between the box and ramp if the ramp is at a 25o angle? 2. ### Physics A box slides down a 36° ramp with an acceleration of 1.05 m/s2. Determine the coefficient of kinetic friction between the box and the ramp. 3. ### physics A box slides down a 22° ramp with an acceleration of 1.19 m/s2. Determine the coefficient of kinetic friction between the box and the ramp. 4. ### Physics A 75 kg box slides down a 25 degree ramp with an acceleration of 3.60 meters per second squared. Find the coefficient of kenetic friction between the box and the ramp. What acceleration would a 175 kg box have on this ramp? 5. ### Physics A box slides down a 38.5° ramp with an acceleration of 1.47 m/s2. Determine the coefficient of kinetic friction between the box and the ramp. 6. ### physics A box slides down a 33° ramp with an acceleration of 1.38 m/s2. Determine the coefficient of kinetic friction between the box and the ramp. 7. ### Physics A box, with a weight of mg = 25 N, is placed at the top of a ramp and released from rest. The ramp measures 4.40 meters horizontally and 3.50 meters vertically. The box accelerates down the incline, attaining a kinetic energy at the … 8. ### physics A 31 kg box slides down a 33.0 ramp with an acceleration of 1.30 m/s2. The acceleration of gravity is 9.81 m/s2 . Find the coefficient of kinetic friction between the box and the ramp. 9. ### physics A box of mass 5.0 kg starting from rest slides down a 1.8 m long ramp inclined at an angle of 22.0 deg. from the horizontal. If the ramp is frictionless, how long will it take the box, starting from rest, to reach the bottom of the … 10. ### physics A box slides down a 28° ramp with an acceleration of 0.98 m/s2. Determine the coefficient of kinetic friction between the box and the ramp. More Similar Questions	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.jiskha.com/display.cgi?id=1257352393", "fetch_time": 1503062224000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-34/segments/1502886104636.62/warc/CC-MAIN-20170818121545-20170818141545-00247.warc.gz", "warc_record_offset": 915772628, "warc_record_length": 3842, "token_count": 629, "char_count": 2403, "score": 3.59375, "int_score": 4, "crawl": "CC-MAIN-2017-34", "snapshot_type": "latest", "language": "en", "language_score": 0.8123601078987122, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00048-of-00064.parquet"}
# Key Ideas • Position(x): Where the object is present on the track. • Speed(y): How fast the object is moving. • Acceleration(a): How much the object’s speed changes in one second. • When an object speeds up, acceleration is in the direction of its motion. • When an object slows down, its acceleration is opposite the direction of motion. • Displacement (Δx) : how far the object ends up away from its starting point, regardless of any motion in between starting and ending positions. • The graphical analysis of motion includes position-time graphs and velocity-time graphs. • Position-time graphs: The slope is the object’s speed, and the object’s position is read from the vertical axis • Velocity-time graphs: the speed is read from the vertical axis, and the slope is the object’s acceleration • Free fall : No forces other than the object’s weight are acting on the object. • Projectile: object in free fall, but it isn’t falling in a straight vertical line. # Introduction to Motion in a Straight Line • All motion problems can be demonstrated with a cart on a track. • The motion detector can read the location of the cart up to 50 times each second. • This detector can make graphs of position or velocity versus time. Goal of motion analysis is to describe, calculate, and predict: • where the cart is • how fast it’s moving • how much its speed is changing. ## Graphical Analysis of Motion ### Position-Time Graphs • This graph represents the cart on the track. • Motion detector located at x=0. • Positive direction is towards the left. • In a position-time graph, the object’s position is read from the vertical axis. • In a position-time graph, the object’s speed is the slope of the graph. • The steeper the slope, the faster the object moves. • If the slope is a front slash (/), the movement is in the positive direction. • If the slope is a backslash (\), the movement is in the negative direction ### Velocity - Time Graphs • This velocity-time graph represents a different cart on the track. • The positive direction is to the left. • In a velocity-time graph, the object’s speed is read from the vertical axis. • The direction of motion is indicated by the sign on the vertical axis. • In a velocity-time graph, the object’s acceleration is the slope of the graph. • To calculate the amount of acceleration: • use rise/run calculation or • use the definition of acceleration to see that the object lost 1 m/s of speed in 1 second, making the acceleration 1 m/s per second • The object in the graph above is slowing down and moving to the left. • When an object slows down, its acceleration is opposite the direction of its motion. • In the graph above, this cart has an acceleration to the right. • Acceleration is not the same thing as speed or velocity. • Acceleration says how quickly speed changes. • Acceleration doesn’t say anything about which way something is moving, unless you know whether the thing is speeding up or is slowing down. • Acceleration to the right means either speeding up and moving right, or slowing down and moving left. • The object’s displacement is given by the area between the graph and the horizontal axis. • Location of the object can’t be determined from a velocity-time graph. # Algebraic Analysis of Motion • When asked to analyze motion from a description and not a graph, start analysis by defining a positive direction and clearly stating the start and the end of the motion. • In any case of accelerated motion when three of the five principal motion variables are known, the remaining variables can be solved for using the kinematic equations. • The five principal motions are: • initial velocity • final velocity • displacement • acceleration • time • To calculate the missing values in a motion chart, use the three kinematic equations listed as follows Objects in Free Fall • When an object is in free fall, its acceleration is 10 m/s per second toward the ground. • Free fall means no forces other than the object’s weight are acting on the object. ### Projectile Motion • A projectile is defined as an object in free fall. • But this object doesn’t have to be moving in a straight line. • If the object were launched at an angle, treat the horizontal and vertical components of its motion separately. • A projectile has no horizontal acceleration and so moves at constant speed horizontally. • To approach a projectile problem, make two motion charts: one for vertical motion and one for horizontal motion. • To find the vertical component of a velocity at an angle, multiply the speed by the sine of the angle. • To find the horizontal component of a velocity at an angle, multiply the speed by the cosine of the angle. This always works, as long as the angle is measured from the horizontal. • The horizontal and vertical motion charts for a projectile must use the same value for time. ### Example A model rocket is launched straight upward with an initial speed of 50 m/s. It speeds up with a constant upward acceleration of 2.0 m/s per second until its engines stop at an altitude of 150 m. # What if Acceleration isn't Constant? • The slope of a graph is related to the derivative of a function; the area under a graph is related to the integral of a function. • To find velocity from a position function, take the derivative with respect to time. • To find acceleration from a velocity function, also take the time derivative. • To find position from a velocity function, take the integral with respect to time. • To find velocity from an acceleration function, take the time integral. # Air Resistance and the First-Order Differential Equations • The force of air resistance is usually negligible in kinematics problems. • Air resistance is not important in kinematics. • Air resistance should only be considered when the problem explicitly says so. # Steps to Solve Air Resistance Problems 1. Find the terminal speed. 1. Terminal speed: after a long time, the object’s speed becomes constant. 2. To find that terminal speed, do an equilibrium problem. 3. Free body, set up forces equal down forces, and left forces equal right forces. 4. If something’s falling straight down with no other forces, usually you’ll get bv = mg. 5. Then solve for v. That’s the terminal speed. 2. Sketch a graph of the speed of the object as a function of time. 1. If the object was dropped from rest, or given an initial velocity, that point can be plotted at time t = 0. 2. Then find terminal velocity using the method —the terminal speed is the constant velocity after a long time. 3. Plot a horizontal line for the terminal velocity near the right-hand side of the t-axis. 3. Describe the motion in words, including what’s happening to the acceleration and/or the velocity 4. 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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ ° You are not logged in. ## #1 2006-09-30 07:40:14 Devantè Real Member Registered: 2006-07-14 Posts: 6,400 ### Conditionals Exercises on Conditionals State the truth value of the following statements. 1. If p is false and q is true, state the truth for: ~ p > q 2. If p is true and q is true, state the truth for: p > ~ q 3. If p is false and q is false, state the truth for: ~ p > ~ q 4. If p is false and q is true, state the truth for: p > q 5. If p is false and q is false, state the truth for: p > ~ q 6. If p is true and q is true, state the truth for: ~ p > q 7. If p is false and q is true, state the truth for: ~(~ p > ~ q ) 8. If p is false and q is false, state the truth for: ~(~ p > ~ q ) 9. If p is false and q is true, state the truth for: p > ~ q 10. If p is false and q is false, state the truth for: ~( p > q ) 11. If p is true and q is false, state the truth for: ~ p > q 12. If p is false and q is false, state the truth for: p > q 13. If p is false and q is false, state the truth for: ~ p > q 14. If p is true and q is false, state the truth for: ~(~ p > ~ q ) 15. If p is true and q is true, state the truth for: ~( p > ~ q ) 16. If p is true and q is false, state the truth for: p > q 17. If p is false and q is true, state the truth for: ~ p > ~ q 18. If p is true and q is false, state the truth for: p > ~ q 19. If p is true and q is false, state the truth for: ~( p > q ) 20. If p is true and q is true, state the truth for: ~ p > ~ q --------------- Offline ## #2 2012-02-18 10:20:37 anonimnystefy Real Member From: Harlan's World Registered: 2011-05-23 Posts: 16,037 ### Re: Conditionals hi Devante 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	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.mathisfunforum.com/viewtopic.php?id=4481", "fetch_time": 1519467744000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-09/segments/1518891815544.79/warc/CC-MAIN-20180224092906-20180224112906-00080.warc.gz", "warc_record_offset": 483415659, "warc_record_length": 3401, "token_count": 678, "char_count": 2018, "score": 3.796875, "int_score": 4, "crawl": "CC-MAIN-2018-09", "snapshot_type": "latest", "language": "en", "language_score": 0.7474170327186584, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00004-of-00064.parquet"}
# Solve for z z=(17+9i)+(5(1+i))/(2-i) Simplify each term. Multiply the numerator and denominator of by the conjugate of to make the denominator real. Multiply. Combine. Simplify the numerator. Apply the distributive property. Multiply by . Expand using the FOIL Method. Apply the distributive property. Apply the distributive property. Apply the distributive property. Simplify and combine like terms. Simplify each term. Multiply by . Multiply by . Multiply . Raise to the power of . Raise to the power of . Use the power rule to combine exponents. Rewrite as . Multiply by . Subtract from . Simplify the denominator. Expand using the FOIL Method. Apply the distributive property. Apply the distributive property. Apply the distributive property. Simplify. Multiply by . Multiply by . Raise to the power of . Raise to the power of . Use the power rule to combine exponents. Subtract from . Simplify each term. Rewrite as . Multiply by . Cancel the common factor of and . Factor out of . Factor out of . Factor out of . Cancel the common factors. Factor out of . Cancel the common factor. Rewrite the expression. Divide by . Solve for z z=(17+9i)+(5(1+i))/(2-i) ## Our Professionals ### Lydia Fran #### We are MathExperts Solve all your Math Problems: https://elanyachtselection.com/ Scroll to top	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://jmservera.com/solve-for-z-z179i51i-2-i/", "fetch_time": 1674885732000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-06/segments/1674764499524.28/warc/CC-MAIN-20230128054815-20230128084815-00107.warc.gz", "warc_record_offset": 337728188, "warc_record_length": 12685, "token_count": 306, "char_count": 1304, "score": 3.65625, "int_score": 4, "crawl": "CC-MAIN-2023-06", "snapshot_type": "latest", "language": "en", "language_score": 0.8605743050575256, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00036-of-00064.parquet"}
# How is Weber fraction calculated? ## How is Weber fraction calculated? The Weber fraction is defined as the jnd divided by the reference stimulus’ moment of inertia. ### What is the just noticeable difference JND for loudness? Two other differences in human hearing as compared to laboratory measurements are Just Noticeable Difference in frequency (JND Hz) and the Just Noticeable Difference in loudness (JND dB). So the JND (Hz) for a 500 Hz sound is about 1 Hz; most of us can tell the difference between 500 Hz and 501 Hz. Why is just noticeable difference important? The JND is a statistical, rather than an exact quantity: from trial to trial, the difference that a given person notices will vary somewhat, and it is therefore necessary to conduct many trials in order to determine the threshold. The JND usually reported is the difference that a person notices on 50% of trials. What is meant by just noticeable difference how have marketers used this concept in marketing products? Just Noticeable Difference is the minimal Stimulation between two products as observed by the consumer. Otherwise we can say -“The just noticeable difference (JND) is the smallest difference in intensity between two stimuli that a person can detect.” How do percentages and ratios help you understand Weber’s Law? This law states that for an average person to perceive a difference, two stimuli must differ by a constant minimum percentage (not a constant amount). The exact proportion varies, depending on the stimulus. ### Who gave just noticeable difference? This rule was first discovered by Ernst Heinrich Weber (1795–1878), an anatomist and physiologist, in experiments on the thresholds of perception of lifted weights. What is just noticeable difference quizlet? JUST NOTICEABLE DIFFERENCE (JND) Sensitivity to DIFFERENCES in intensity levels. NOT CONSTANT. The SMALLEST difference in the amount of stimulation that a specific sense can detect – DIFFERENCE THRESHOLD. What decibel difference is noticeable? dynamic range of the ear …can be observed, called the intensity just noticeable difference, is about one decibel (further reinforcing the value of the decibel intensity scale). One decibel corresponds to an absolute energy variation of a factor of about 1.25. ## What do you mean by just noticeable difference? In the branch of experimental psychology focused on sense, sensation, and perception, which is called psychophysics, a just-noticeable difference or JND is the amount something must be changed in order for a difference to be noticeable, detectable at least half the time (absolute threshold). ### How is a just noticeable difference ( JND ) measured? One of the classic psychoacoustic experiments is the measurement of a just noticeable difference (jnd), which is also called a difference limen. In these tests a subject is asked to compare two sounds and to indicate which is higher in level, or in frequency. What is found is that the jnd in level depends on both the intensity and frequency. What’s the threshold for the just noticeable difference? The just noticeable difference (JND), also known as the difference threshold, is the minimum level of stimulation that a person can detect 50 percent of the time. How is the just noticeable difference in pitch expressed? The just noticeable difference in pitch must be expressed as a ratio or musical interval since the human ear tends to respond equally to equal ratios of frequencies. It is convenient to express the just noticeable difference in cents since that notation was developed to express musical intervals.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.ammarksman.com/how-is-weber-fraction-calculated/", "fetch_time": 1670506051000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-49/segments/1669446711336.41/warc/CC-MAIN-20221208114402-20221208144402-00211.warc.gz", "warc_record_offset": 649778760, "warc_record_length": 12920, "token_count": 716, "char_count": 3611, "score": 3.609375, "int_score": 4, "crawl": "CC-MAIN-2022-49", "snapshot_type": "latest", "language": "en", "language_score": 0.9317658543586731, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00061-of-00064.parquet"}
Video: Finding the Concavity of a Curve Given the Derivative Let π be a function whose derivative is given by πβ²(π₯) = 5π₯/(2π₯Β² + 6)Β². On which of the following intervals of π₯ is the graph of π concave up? [A] (0, β) [B] (ββ, 0) [C] (ββ, β1) βͺ (1, β) [D] (β1, 1 ) 04:15 Video Transcript Let π be a function whose derivative is given by π prime of π₯ equals five π₯ over two π₯ squared plus six all squared. On which of the following intervals of π₯ is the graph of π concave up? Is it a) the open interval from zero to β. B) The open interval negative β to zero. Is it c) the union of the open interval from negative β to negative one and the open interval one to β? Or d) the open interval negative one to one. Weβre going to need to be a little bit careful here. We have not been given an expression for the function π itself, but one for its derivative, π prime of π₯. Weβre looking to find the intervals on which the graph of π is concave up. So what does this mean? When a graph is concave up, we say its first derivative, π prime of π₯, is increasing. We can consider the rate of change of that first derivative. Thatβs the derivative of the derivative, π double prime of π₯. And what this means is π double prime of π₯ must be greater than zero. So to answer this question, weβre going to need to differentiate π prime of π₯ to find the second derivative and then establish on which intervals of π₯ that is greater than zero. Now π prime of π₯ is the quotient of two functions. So we use the quotient rule. And this says that the derivative of the quotient of two differentiable functions, π’ and π£, is π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared. The numerator of our fraction is five π₯. So weβre going to let π’ be equal to five π₯. And π£ is the denominator. Itβs two π₯ squared plus six squared. It should be quite clear from our formula for the quotient rule that weβre going to need to differentiate each of these functions with respect to π₯. Well, the derivative of five π₯ is fairly straightforward. Itβs simply five. But what about the derivative of this composite function, two π₯ squared plus six all squared? Now we could use the chain rule. Or we can use a special case of the chain rule, which is called the general power rule. What we do is we multiply the entire inner function by the value of the exponent, so by two. And then we reduce that exponent by one. Then we multiply that by the derivative of the inner function. Well, the derivative of two π₯ squared plus six with respect to π₯ is four π₯. So we find that dπ£ by dπ₯ in its simplest form is eight π₯ times two π₯ squared plus six. Letβs substitute everything we have into the quotient rule. The derivative of π prime of π₯, which is π double prime of π₯, is π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared. We can simplify the denominator of this expression by multiplying the exponent. So we get two π₯ squared plus six to the fourth power. On our numerator, we can factor out two π₯ squared plus six. And so we find that π double prime of π₯ is two π₯ squared plus six times five times two π₯ squared plus six minus 40π₯ squared over two π₯ squared plus six to the fourth power. We distribute the first set of parentheses to give us 10π₯ squared plus 30. And so we see this simplifies to two π₯ squared plus six times 30 minus 30π₯ squared all over two π₯ squared plus six to the fourth power. Now we say that, for the graph to be concave up, the second derivative must be greater than zero. So we need to find the intervals of π₯ for which this is the case. And this might look quite complicated to start. However, we know that any real number to the fourth power is greater than zero. We know two π₯ squared is greater than zero. And therefore, two π₯ squared plus six is greater than zero. So we need to work out where the other part of the expression 30 minus 30π₯ squared is greater than zero. Then we have the product of two positives divided by a positive, which we know to be a positive. Letβs clear some space and solve this inequality. Weβll divide through by 30. And that gives us one minus π₯ squared is greater than zero. Letβs factor the expression one minus π₯ squared and for now just set it equal to zero. We get one minus π₯ times one plus π₯ equals zero. And for this to be true, either one minus π₯ must be equal to zero or one plus π₯ must be equal to zero. And so we see that when π₯ is equal to one and negative one, the expression one minus π₯ squared is equal to zero. So where is it greater than zero? Well, there are a number of ways we can achieve this. But one way is to draw the graph. Now we see the graph of π¦ equals one minus π₯ squared is greater than zero here. Thatβs values of π₯ greater than negative one and less than one. This means our second derivative is greater than zero for values of π₯ on the open interval negative one to one. And in turn, the graph of π is concave up over that same interval. 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# Category: H2 Math Here you will find all the resources on H2 Mathematics ( A Level). ## 2025 H2 A Level Mathematics Specimen Paper (9758) : Paper 2 Worked Solutions Here’s the worked solutions for the 2023 A Level H2 Mathematics Paper 1. ## 2025 H2 A Level Mathematics Specimen Paper (9758) : Paper 1 Worked Solutions Here’s the worked solutions for the 2025 A Level H2 Mathematics Specimen Paper 1, for syllabus 9758 (from year 2025 and onwards). ## Ratio Theorem Question In vectors chapter, we learn ratio theorem. Let’s apply what we have learned in ratio theorem to a question here. You’ll also find full worked solutions with explanations provided. ## Crash Course JC1 H2 Math Here’s a JC1 H2 Math crash course for you! ## MF26 List of Formula and Statistical Table for A Level H2 Math In this post, let’s take a deep dive into MF26 Formula booklet for the H2 A Level Math exam. In this post, you’ll learn the different sections of MF26, and when to use them. ## Challenging Questions in the 2023 H2 A Level Math Paper In this post, we’ll be looking at the more challenging questions from the 2023 H2 A Level Math Paper. You’ll also find the links to the video tutorial explaining these questions in this post. ## Overview of the 2023 H2 A Level Math Paper Let’s look at an overview of the H2 Math A Level Paper in this blog post. We’ll discuss the difficulty level of this paper, some interesting questions, and more. ## 2023 H2 A Level Mathematics Paper 2 Worked Solutions Here’s the worked solutions for the 2023 A Level H2 Mathematics Paper 2. ## 2023 H2 A Level Mathematics Paper 1 Worked Solutions Here’s the worked solutions for the 2023 A Level H2 Mathematics Paper 1. ## Correlation and Linear Regression In this post, we’ll look at correlation and linear regression as tested in H2 A Level Math section on statistics. Here, you’ll what is correlation, regression, product moment correlation coefficient (r), how to use the graphic calculator Ti84 to do a least square regression. error: Content is protected !!	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://emilylearning.com/category/math/h2-math/", "fetch_time": 1726111856000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651420.25/warc/CC-MAIN-20240912011254-20240912041254-00518.warc.gz", "warc_record_offset": 206490872, "warc_record_length": 15838, "token_count": 497, "char_count": 2050, "score": 3.96875, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.8332349061965942, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00060-of-00064.parquet"}
#### What is 2640 percent of 712? How much is 2640 percent of 712? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 2640% of 712 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update. ## 2640% of 712 = 18796.8 Calculate another percentage below. Type into inputs Find number based on percentage percent of Find percentage based on 2 numbers divided by Calculating two thousand, six hundred and fourty of seven hundred and twelve How to calculate 2640% of 712? Simply divide the percent by 100 and multiply by the number. For example, 2640 /100 x 712 = 18796.8 or 26.4 x 712 = 18796.8 #### How much is 2640 percent of the following numbers? 2640% of 712.01 = 1879706.4 2640% of 712.02 = 1879732.8 2640% of 712.03 = 1879759.2 2640% of 712.04 = 1879785.6 2640% of 712.05 = 1879812 2640% of 712.06 = 1879838.4 2640% of 712.07 = 1879864.8 2640% of 712.08 = 1879891.2 2640% of 712.09 = 1879917.6 2640% of 712.1 = 1879944 2640% of 712.11 = 1879970.4 2640% of 712.12 = 1879996.8 2640% of 712.13 = 1880023.2 2640% of 712.14 = 1880049.6 2640% of 712.15 = 1880076 2640% of 712.16 = 1880102.4 2640% of 712.17 = 1880128.8 2640% of 712.18 = 1880155.2 2640% of 712.19 = 1880181.6 2640% of 712.2 = 1880208 2640% of 712.21 = 1880234.4 2640% of 712.22 = 1880260.8 2640% of 712.23 = 1880287.2 2640% of 712.24 = 1880313.6 2640% of 712.25 = 1880340 2640% of 712.26 = 1880366.4 2640% of 712.27 = 1880392.8 2640% of 712.28 = 1880419.2 2640% of 712.29 = 1880445.6 2640% of 712.3 = 1880472 2640% of 712.31 = 1880498.4 2640% of 712.32 = 1880524.8 2640% of 712.33 = 1880551.2 2640% of 712.34 = 1880577.6 2640% of 712.35 = 1880604 2640% of 712.36 = 1880630.4 2640% of 712.37 = 1880656.8 2640% of 712.38 = 1880683.2 2640% of 712.39 = 1880709.6 2640% of 712.4 = 1880736 2640% of 712.41 = 1880762.4 2640% of 712.42 = 1880788.8 2640% of 712.43 = 1880815.2 2640% of 712.44 = 1880841.6 2640% of 712.45 = 1880868 2640% of 712.46 = 1880894.4 2640% of 712.47 = 1880920.8 2640% of 712.48 = 1880947.2 2640% of 712.49 = 1880973.6 2640% of 712.5 = 1881000 2640% of 712.51 = 1881026.4 2640% of 712.52 = 1881052.8 2640% of 712.53 = 1881079.2 2640% of 712.54 = 1881105.6 2640% of 712.55 = 1881132 2640% of 712.56 = 1881158.4 2640% of 712.57 = 1881184.8 2640% of 712.58 = 1881211.2 2640% of 712.59 = 1881237.6 2640% of 712.6 = 1881264 2640% of 712.61 = 1881290.4 2640% of 712.62 = 1881316.8 2640% of 712.63 = 1881343.2 2640% of 712.64 = 1881369.6 2640% of 712.65 = 1881396 2640% of 712.66 = 1881422.4 2640% of 712.67 = 1881448.8 2640% of 712.68 = 1881475.2 2640% of 712.69 = 1881501.6 2640% of 712.7 = 1881528 2640% of 712.71 = 1881554.4 2640% of 712.72 = 1881580.8 2640% of 712.73 = 1881607.2 2640% of 712.74 = 1881633.6 2640% of 712.75 = 1881660 2640% of 712.76 = 1881686.4 2640% of 712.77 = 1881712.8 2640% of 712.78 = 1881739.2 2640% of 712.79 = 1881765.6 2640% of 712.8 = 1881792 2640% of 712.81 = 1881818.4 2640% of 712.82 = 1881844.8 2640% of 712.83 = 1881871.2 2640% of 712.84 = 1881897.6 2640% of 712.85 = 1881924 2640% of 712.86 = 1881950.4 2640% of 712.87 = 1881976.8 2640% of 712.88 = 1882003.2 2640% of 712.89 = 1882029.6 2640% of 712.9 = 1882056 2640% of 712.91 = 1882082.4 2640% of 712.92 = 1882108.8 2640% of 712.93 = 1882135.2 2640% of 712.94 = 1882161.6 2640% of 712.95 = 1882188 2640% of 712.96 = 1882214.4 2640% of 712.97 = 1882240.8 2640% of 712.98 = 1882267.2 2640% of 712.99 = 1882293.6 2640% of 713 = 1882320 1% of 712 = 7.12 2% of 712 = 14.24 3% of 712 = 21.36 4% of 712 = 28.48 5% of 712 = 35.6 6% of 712 = 42.72 7% of 712 = 49.84 8% of 712 = 56.96 9% of 712 = 64.08 10% of 712 = 71.2 11% of 712 = 78.32 12% of 712 = 85.44 13% of 712 = 92.56 14% of 712 = 99.68 15% of 712 = 106.8 16% of 712 = 113.92 17% of 712 = 121.04 18% of 712 = 128.16 19% of 712 = 135.28 20% of 712 = 142.4 21% of 712 = 149.52 22% of 712 = 156.64 23% of 712 = 163.76 24% of 712 = 170.88 25% of 712 = 178 26% of 712 = 185.12 27% of 712 = 192.24 28% of 712 = 199.36 29% of 712 = 206.48 30% of 712 = 213.6 31% of 712 = 220.72 32% of 712 = 227.84 33% of 712 = 234.96 34% of 712 = 242.08 35% of 712 = 249.2 36% of 712 = 256.32 37% of 712 = 263.44 38% of 712 = 270.56 39% of 712 = 277.68 40% of 712 = 284.8 41% of 712 = 291.92 42% of 712 = 299.04 43% of 712 = 306.16 44% of 712 = 313.28 45% of 712 = 320.4 46% of 712 = 327.52 47% of 712 = 334.64 48% of 712 = 341.76 49% of 712 = 348.88 50% of 712 = 356 51% of 712 = 363.12 52% of 712 = 370.24 53% of 712 = 377.36 54% of 712 = 384.48 55% of 712 = 391.6 56% of 712 = 398.72 57% of 712 = 405.84 58% of 712 = 412.96 59% of 712 = 420.08 60% of 712 = 427.2 61% of 712 = 434.32 62% of 712 = 441.44 63% of 712 = 448.56 64% of 712 = 455.68 65% of 712 = 462.8 66% of 712 = 469.92 67% of 712 = 477.04 68% of 712 = 484.16 69% of 712 = 491.28 70% of 712 = 498.4 71% of 712 = 505.52 72% of 712 = 512.64 73% of 712 = 519.76 74% of 712 = 526.88 75% of 712 = 534 76% of 712 = 541.12 77% of 712 = 548.24 78% of 712 = 555.36 79% of 712 = 562.48 80% of 712 = 569.6 81% of 712 = 576.72 82% of 712 = 583.84 83% of 712 = 590.96 84% of 712 = 598.08 85% of 712 = 605.2 86% of 712 = 612.32 87% of 712 = 619.44 88% of 712 = 626.56 89% of 712 = 633.68 90% of 712 = 640.8 91% of 712 = 647.92 92% of 712 = 655.04 93% of 712 = 662.16 94% of 712 = 669.28 95% of 712 = 676.4 96% of 712 = 683.52 97% of 712 = 690.64 98% of 712 = 697.76 99% of 712 = 704.88 100% of 712 = 712	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, 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# 5th grd math- metric conversion posted by . 30m = 6000, i have converted to cm, in, and ft, no where close to getting 6000? help • 5th grd math- metric conversion - 30m=6000 m=200 so you would say 30 times 200=6000.i think thts it. • metric measures - 3200g=?????kg I NEED HELP!!!! • 5th grd math- metric conversion - 30m=6000 • 7th grd Mathematics - The answer is 200. Ik I am very smart.. Watch your answer is going to be right.. I guarantee...😊 • 7th grd math- metric conversion - I got 200 because 30m represents that your gonna multiply 30 and m so you want to divide 30 divided by 6000 and you will get 200 so m is 200. I hope i helped you!!??!! 😋✨🙌	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.jiskha.com/display.cgi?id=1299547817", "fetch_time": 1498294376000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-26/segments/1498128320243.11/warc/CC-MAIN-20170624082900-20170624102900-00399.warc.gz", "warc_record_offset": 575255444, "warc_record_length": 4245, "token_count": 212, "char_count": 669, "score": 3.59375, "int_score": 4, "crawl": "CC-MAIN-2017-26", "snapshot_type": "latest", "language": "en", "language_score": 0.8914026618003845, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00021-of-00064.parquet"}
Quick Answer: How Do You Calculate Present Value Of Interest? What is the present value of 1? A present value of 1 table states the present value discount rates that are used for various combinations of interest rates and time periods. A discount rate selected from this table is then multiplied by a cash sum to be received at a future date, to arrive at its present value.. What is difference between NPV and IRR? Net present value (NPV) is the difference between the present value of cash inflows and the present value of cash outflows over a period of time. By contrast, the internal rate of return (IRR) is a calculation used to estimate the profitability of potential investments. How do you calculate present value? Present value is an estimate of the current sum needed to equal some future target amount to account for various risks. Using the present value formula (or a tool like ours), you can model the value of future money….The Present Value FormulaC = Future sum.i = Interest rate (where ‘1’ is 100%)n= number of periods. What is the formula for present value in Excel? You would need to figure out how much is needed to invest today, or the present value. The formula for present value is PV = FV ÷ (1+r)^n; where FV is the future value, r is the interest rate and n is the number of periods. What is the formula for calculating net present value? It is calculated by taking the difference between the present value of cash inflows and present value of cash outflows over a period of time. As the name suggests, net present value is nothing but net off of the present value of cash inflows and outflows by discounting the flows at a specified rate. What is the discount rate formula? How to calculate discount rate. There are two primary discount rate formulas – the weighted average cost of capital (WACC) and adjusted present value (APV). The WACC discount formula is: WACC = E/V x Ce + D/V x Cd x (1-T), and the APV discount formula is: APV = NPV + PV of the impact of financing. What is the formula for present value of an annuity? The Present Value of Annuity Formula P = the present value of annuity. PMT = the amount in each annuity payment (in dollars) R= the interest or discount rate. n= the number of payments left to receive. What is NPV example? For example, if a security offers a series of cash flows with an NPV of \$50,000 and an investor pays exactly \$50,000 for it, then the investor’s NPV is \$0. It means they will earn whatever the discount rate is on the security. Should present value be higher or lower? The present value is usually less than the future value because money has interest-earning potential, a characteristic referred to as the time value of money, except during times of zero- or negative interest rates, when the present value will be equal or more than the future value. How do I calculate accumulated present value? The formula to find out the compute the accumulated present value of a continuous stream of income at rate R(t) , for time T years and interest rate k , compounded continuously is P(t)=∫T0R(t)e−kt dt P ( t ) = ∫ 0 T R ( t ) e − k t d t . How do you calculate present value ratio? Present Value Ratio (PVR) can also be used for economic assessment of project(s) and it can be determined as net present value divided by net negative cash flow at i*.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://biggboss3.net/qa/quick-answer-how-do-you-calculate-present-value-of-interest.html", "fetch_time": 1607191398000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-50/segments/1606141748276.94/warc/CC-MAIN-20201205165649-20201205195649-00400.warc.gz", "warc_record_offset": 209450223, "warc_record_length": 7052, "token_count": 750, "char_count": 3347, "score": 3.84375, "int_score": 4, "crawl": "CC-MAIN-2020-50", "snapshot_type": "latest", "language": "en", "language_score": 0.9480141401290894, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00007-of-00064.parquet"}
Algebra 2 Online! Henrico County Public Schools, Virginia Home Pacing SOL Tracker Algebra 2 Standards Virginia Department of Education Return to HCPS Math Courses Expressions and Operations Equations and Inequalities Functions Statistics SOL Review Materials Frequently Asked Questions AII.11 The student will identify properties of a normal distribution and apply those properties to determine probabilities associated with areas under the standard normal curve. Essential Knowledge and Skills Identify the properties of a normal probability distribution. Describe how the standard deviation and the mean affect the graph of the normal distribution. Compare two sets of normally distributed data using a standard normal distribution and z-scores. Represent probability as area under the curve of a standard normal probability distribution. Use the graphing calculator or a standard normal probability table to determine probabilities or percentiles based on z-scores. Essential Understandings A normal distribution curve is a symmetrical, bell-shaped curve defined by the mean and the standard deviation of a data set. The mean is located on the line of symmetry of the curve. Areas under the curve represent probabilities associated with continuous distributions. The normal curve is a probability distribution and the total area under the curve is 1. For a normal distribution, approximately 68 percent of the data fall within one standard deviation of the mean, approximately 95 percent of the data fall within two standard deviations of the mean, and approximately 99.7 percent of the data fall within three standard deviations of the mean. The mean of the data in a standard normal distribution is 0 and the standard deviation is 1. The standard normal curve allows for the comparison of data from different normal distributions. A z-score is a measure of position derived from the mean and standard deviation of data. A z-score expresses, in standard deviation units, how far an element falls from the mean of the data set. A z-score is a derived score from a given normal distribution. A standard normal distribution is the set of all z-scores. Vertical Articulation 1. SOL A.9 - given a set of data, interpret variation in real‐world contexts/calculate/interpret mean absolute deviation/standard deviation /z-scores. 2. SOL AFDA.7 - a) characteristics of normally distribution of data; b) percentiles; c) normalizing data, using z-scores; d) area under standard norm curve/ probability Instructional Materials 1. VDOE ESS Lesson: Normal Distributions - Statistics (PDF) - Analyzing and using the standard normal curve (Word) 2. Notes: Virginia Department of Education Technical Assistance documentation Algebra 1 (pdf) 3. Notes: Virginia Department of Education Technical Assistance documentation Algebra 2 (pdf) Real World Lessons Practice Self-Assessments Skills Videos Explore Learning 1. None Websites Common Core Standards S-ID.4	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://teachers.henrico.k12.va.us/math/hcpsalgebra2/AII-11-2.html", "fetch_time": 1516693332000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-05/segments/1516084891791.95/warc/CC-MAIN-20180123072105-20180123092105-00434.warc.gz", "warc_record_offset": 342617538, "warc_record_length": 4275, "token_count": 582, "char_count": 2949, "score": 4.375, "int_score": 4, "crawl": "CC-MAIN-2018-05", "snapshot_type": "latest", "language": "en", "language_score": 0.8911412358283997, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00054-of-00064.parquet"}
# Algebra posted by . Essay: Show all work. A gardner wants to create wants to create a triangle garden in the shape of a right triangle with skortesr length x-5yft. and the middle length side x+6yft. what is an algebraic expression for the area of the garden? Be sure to mulitply this out and express in simplest correct mathematical form including units. • Algebra - Area = (1/2)(x-5)(x+6) expand and simplify ## Similar Questions 1. ### algebra A gardner wants to create a rectangular garden with length of 2x-8y ft. and with of 3x-4y ft. What is an algebriac expression for the area of the garden? 2. ### algebra A gardener wants to create a triangular garden in the shape of a right triangle with shortest side length x-2y ft. and the middle length side x+6y ft. What is the area of the garden? 3. ### algebra A gardener wants to create a triangular garden in the shape of a right triangle with shortest side length x-2y ft. and the middle length side x+6y ft. What is the area of the garden? 4. ### Word Problem A grander want to create a triangular garden in the shape of a right triangle with shortest side length x-6y ft. and the middle length side x + 6y ft. what is an algebraic expression for the area of the garden? 5. ### algebra A gardener wants to create a triangular garden in the shape of a right triangle with shortest side length x – 2y ft. and the middle length side x + 4y ft. What is an algebraic expression for the area of the garden? Essay: Show all work. A gardener wants to create a rectangular garden with length of 2x-3y ft. and width of x+4y ft. What is an algebraic expression for the area of the garden? 7. ### algebra Essay. Show all work. A gardener wants to create a triangular garden in the shape of a right triangle with the shortest side length x-3y ft. and the middle length side x+6y ft. What ia an algebraic expression for the area of the garden? 8. ### algebra Essay. show all work. A gardener wants to create a triangluar garden in the shape of a right triangle with shortest legth x-3y ft. and the middle length side x+6y ft. What ia an algebraic expression for the area of the garden? 9. ### intermediate algebra Essay.Show all work. A gardener wants to create a rectangular garden with the length of 3x-2y ft. and the width of 3x-3y ft. What is an algebraic expression for the area of the garden.Be sure to multiply this out.Express in the simpliest … 10. ### Intermediate Algebra A gardner wants to create a triangular garden in the shape of a right triangle with shaortest side length x-8y ft and the middle length side x+6y ft. What is an algebraic expression for the area of the garden? More Similar Questions	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.jiskha.com/display.cgi?id=1343430350", "fetch_time": 1503304954000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-34/segments/1502886107744.5/warc/CC-MAIN-20170821080132-20170821100132-00117.warc.gz", "warc_record_offset": 915153428, "warc_record_length": 3551, "token_count": 679, "char_count": 2677, "score": 3.765625, "int_score": 4, "crawl": "CC-MAIN-2017-34", "snapshot_type": "latest", "language": "en", "language_score": 0.833035945892334, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00021-of-00064.parquet"}
Like this presentation? Why not share! # Cattle Trains ## on Oct 12, 2007 • 3,466 views ### Views Total Views 3,466 Views on SlideShare 3,456 Embed Views 10 Likes 0 13 0 ### 2 Embeds10 http://www.slideshare.net 7 http://vle1.homewood.kent.sch.uk 3 ### Report content • Comment goes here. Are you sure you want to ## Cattle TrainsPresentation Transcript • Area and Perimeter of the Nazi Cattle Trains • Learning Objectives Area of a Rectangle Perimeter of a Rectangle Estimation Area of a Triangle Empathy Cattle Trains Inhumane conditions KEYWORDS • The Cattle trains that took the Jewish victims of the Holocaust to the Concentration camps were crammed with people. • Area & Perimeter • What is Area? Area is the amount of space inside a shape What is Perimeter? Perimeter is the total distance around the outside of a shape • Estimate the AREA and PERIMETER of the floor of one of the cattle trains taking the Jewish victims to their deaths. • TOP TIP - • What common things do you already know the size of (your height, the table, etc)? • Use your knowledge about the dimensions of these items to visualise and guess how big the train floor might be. • What Do I Need to Know? What are the names of these shapes? Which of these is the shape of the Cattle train? • How Do I Find The Area? • Area L W Area = L x W length width Rectangle L L Square Area = L x L length length • How do I find the Perimeter? • Perimeter L W Perimeter = L + L + W + W = 2L + 2W length width Rectangle L L Square Perimeter = L + L + L + L = 4L length • We will now estimate how much area each person on the cattle train would need. • - TOP TIP - • Stand up tall! Imagine drawing a rectangle around your feet on the floor - this is the area we are measuring. Remember, your shoulders are wider than your feet! • How long are your feet in metres? • (e.g. 20cm=0.2m) • How wide are your shoulders? • Important • Your estimate of the area must have appropriate units of measurement and must be written with the squared symbol. • e.g. 2cm x 3cm = 6cm ² cm 2 mm 2 km 2 m 2 • Putting it all together! • Once we have estimated the area of the cattle train floor and the amount of space you need we can now estimate how many people could fit in one cattle train. • train floor area ÷ area each person needs • E.g. 32m² ÷ 0.5m² = 64 • Therefore 1 cattle train would hold 64 people. • Field Work • Now we will escape the classroom and measure the areas and perimeter accurately. • What is the area of the cattle train? • What is the perimeter of the cattle train? • What is the area each person requires? • How many people could actually be crammed into each train? • Plenary • How close were your estimates? • How could you improve your estimation skills? • What words could we use to describe what it must have been like to travel for hours on a cattle train in such conditions? • Extension 1 • We now know how to calculate the area of a rectangle, but do we also know how to calculate the area of a triangle? • How Do I Find The Area? Any Triangle H H B B 2 Area = 1 B x H base height Or in other words; multiply the Base & Height together and then divide by 2	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.slideshare.net/total/cattle-trains", "fetch_time": 1405246221000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2014-23/segments/1404776437601.53/warc/CC-MAIN-20140707234037-00060-ip-10-180-212-248.ec2.internal.warc.gz", "warc_record_offset": 489982725, "warc_record_length": 26748, "token_count": 818, "char_count": 3145, "score": 3.984375, "int_score": 4, "crawl": "CC-MAIN-2014-23", "snapshot_type": "latest", "language": "en", "language_score": 0.8951452970504761, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00062-of-00064.parquet"}
# Percentage into Ratio How to convert a given percentage into ratio? We will follow the following steps for converting a percentage into a ratio: Step I: Obtain the percentage. Step II: Convert the given percentage into fraction by dividing it by 100 and removing percentage symbol (%). Step III: Reduce the fraction obtained in step II in the simplest form. Step IV: Write the fraction obtained in step III as a ratio. 1. Express each of the following percentage as ratios in the simplest form: (i) 46 % = 46/100 = 23/50 = 23 : 50 (ii) 20 % = 20/100 = 1/5 = 1 : 5 (iii) 125 % = 125/100 = 5/4 = 5 : 4 (iv) 34% = 34/100 = 17/50= 17 : 50 (v) 1 % = 1/100 = 1 : 100 2. Express each of the following fraction percentage into ratio in lowest term: (i) 3/4 % = 3/4 × 1/100 = 3/400 = 3 : 400 (ii) 62/3 % = 20/3 % = 20/3 × 1/100 = 20/300 = 1 : 15 (iii) 3/5 % = 3/5 × 1/100 = 3/500 = 3 : 500 (iii) 62/5 % = 32/5 % = 32/5 × 1/100 = 32/500 = 8/125 = 8 : 125 (iv) 53/8 % = 43/8 % = 43/8 × 1/100 = 43/800 = 43 : 800 3. Express each of the following decimal percentage as ratios in the simplest form: (i) 16.5 % = 165/10 % = 165/10 × 1/100 = 165/1000 = 33/200 = 33 : 200 (ii) 0.4 % = 4/10 % = 4/10 × 1/100 = 4/1000 = 1/250 = 1 : 250 (iii) 2.5 % = 25/10 % = 25/10 × 1/100 = 25/1000 = 1/40 = 1 : 40 (iv) 10.10 % = 1010/100 % = 1010/100 × 1/100 = 1010/10000 = 101/1000 = 101 : 1000 (v) 1.2 % = 12/10 % = 12/10 × 1/100 = 3/250 = 3 : 250 Fraction into Percentage Percentage into Fraction Percentage into Ratio Ratio into Percentage Percentage into Decimal Decimal into Percentage Percentage of the given Quantity How much Percentage One Quantity is of Another? Percentage of a Number Increase Percentage Decrease Percentage Basic Problems on Percentage Solved Examples on Percentage Problems on Percentage Real Life Problems on Percentage Word Problems on Percentage Application of Percentage Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question. ## Recent Articles 1. ### Adding 1-Digit Number \| Understand the Concept one Digit Number Sep 17, 24 02:25 AM Understand the concept of adding 1-digit number with the help of objects as well as numbers. 2. ### Counting Before, After and Between Numbers up to 10 \| Number Counting Sep 17, 24 01:47 AM Counting before, after and between numbers up to 10 improves the child’s counting skills. 3. ### Worksheet on Three-digit Numbers \| Write the Missing Numbers \| Pattern Sep 17, 24 12:10 AM Practice the questions given in worksheet on three-digit numbers. The questions are based on writing the missing number in the correct order, patterns, 3-digit number in words, number names in figures… 4. ### Arranging Numbers \| Ascending Order \| Descending Order \|Compare Digits Sep 16, 24 11:24 PM We know, while arranging numbers from the smallest number to the largest number, then the numbers are arranged in ascending order. Vice-versa while arranging numbers from the largest number to the sma… 5. ### Worksheet on Tens and Ones \| Math Place Value \|Tens and Ones Questions Sep 16, 24 02:40 PM In math place value the worksheet on tens and ones questions are given below so that students can do enough practice which will help the kids to learn further numbers. Worksheet on Fraction into Percentage Worksheet on Percentage into Fraction Worksheet on Percentage into Ratio Worksheet on Ratio into Percentage Worksheet on Percentage into Decimal Worksheet on Percentage of a Number Worksheet on Finding Percent Worksheet on Finding Value of a Percentage Worksheet on Percentage of a Given Quantity Worksheet on Increase Percentage Worksheet on increase and Decrease Percentage Worksheet on Expressing Percent Worksheet on Percent Problems Worksheet on Finding Percentage	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.math-only-math.com/percentage-into-ratio.html", "fetch_time": 1726573613000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651773.64/warc/CC-MAIN-20240917104423-20240917134423-00013.warc.gz", "warc_record_offset": 821316463, "warc_record_length": 13039, "token_count": 1141, "char_count": 3964, "score": 4.625, "int_score": 5, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.8141335844993591, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00058-of-00064.parquet"}
Living Science Solutions for Class 8 Science Chapter 12 Friction are provided here with simple step-by-step explanations. These solutions for Friction are extremely popular among Class 8 students for Science Friction Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Living Science Book of Class 8 Science Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Living Science Solutions. All Living Science Solutions for class Class 8 Science are prepared by experts and are 100% accurate. Friction is #### Question 2: It is difficult to walk on ice because (a) pressure is high. (b) pressure is low. (c) friction is high. (d) friction is low. (d) friction is low Ice offers less friction. #### Question 3: Friction can be increased by (a) making the surfaces smooth. (b) lubricating the surfaces. (c) using ball bearings. (d) making the surfaces rough. (d) making the surfaces rough Friction increases with roughness. #### Question 4: Which of these is true about friction? (a) It can stop a moving object. (b) It can change the shape of an object. (c) It can change the direction of a moving object. (d) It can make a moving object move faster. (a) It can stop a moving object. Friction acts as a decelerating force against motion; it helps in slowing down a body, thereby eventually bringing it to rest. #### Question 5: In which case is friction a disadvantage? (a) running a machine (b) walking (c) applying brakes (d) writing (a) Running a machine While running a machine, energy is lost because of friction. Hence, in this case, friction is a disadvantage. #### Question 6: Ball bearings are useful because (a) rolling friction is more than sliding friction. (b) rolling friction is less than siliding friction. (c) rolling friction is same as sliding friction. (d) it is easier to apply grease to ball bearings to reduce friction. (b) rolling friction is less than sliding friction Rolling friction is less than sliding friction because of the minimal surface of contact between two bodies. #### Question 1: Friction always opposes __________ Friction always opposes the motion of a body. #### Question 2: Friction is caused by __________ on the surfaces in contact. Friction is caused by irregularities on the surfaces in contact. #### Question 3: Is friction a contact or a non-contact force? Friction is a contact force, i.e., it requires two bodies to be in contact. #### Question 4: The smoother a surface is, the __________ (greater/lesser) will be the friction. lesser Frictional force decreases with the decrease in the roughness of the surfaces in contact. #### Question 5: Brakes on cars will work best if the friction between the brake shoes and wheels is reduced. True or false? False. Frictional force helps in reducing the motion of a body. More the friction, more the opposing force and faster the stoppage of the car. #### Question 1: Discuss three situations in daily life where friction is an advantage? Three situations in daily life where friction is an advantage: 1) Walking - Friction is required between the feet/footwear and the ground to prevents slips. 2) Mechanical Brakes - Mechanical brakes help in slowing down the motion of a machine whenever desired. For example, friction is required to bring a car to rest. 3) Matchbox - Friction between the match stick and the matchbox helps in generating heat; this heat helps in lighting the matchstick. #### Question 2: Explain with examples the disadvantages of friction. Friction reduces the relative motion between two objects in contact. It is a disadvantage in many situations. Let us take the example of the mechanical engine used in cars. Friction between the mechanical components of the engine prevents smooth movements. This causes loss of energy in the form of heat, sound, etc. This reduces the efficiency of the car. If the engine is not properly lubricated, excessive friction may even damage its mechanical parts because of excess heat production. The excess sound coming out of the engine is also undesirable. #### Question 3: Explain four methods by which friction between two solid surfaces in contact can be reduced. Following are the methods by which friction between two solid surfaces in contact can be reduced: 1) Lubrication - It helps in reducing roughness, which decreases friction. 2) Ball Bearings - Since rolling friction is less than sliding friction, ball bearings help in maintaining minimal surface of contact between two surfaces in motion. 3) Polishing - Smooth surfaces or bearings work better than the rough ones. 4) Streamline Design - A streamlined design prevents friction due to fluids. #### Question 6: Friction causes wastage of energy. True or false? True. Friction causes loss of energy in the form of heat, sound, light, etc. #### Question 7: A machine with moving parts will become less hot while running if it is well lubricated. True or false? True Friction causes loss of energy in the form of heat. When a machine is well lubricated, friction is reduced; therefore, less heat is produced. #### Question 8: Friction due to fluids is called __________ Friction due to fluids is called drag. #### Question 9: Rolling friction is __________ (greater/smaller) than sliding friction. Rolling friction is smaller than sliding friction. #### Question 10: To reduce fluid friction, the object should have a __________ shape. To reduce fluid friction, the object should have a streamlined shape. #### Question 1: If there was no friction, what would happen to a moving object? Friction affects the relative motion between two surfaces in contact. Without friction, objects would not be able to move from a stationary position or stop when in motion. When a body is in motion and if there is no friction, it becomes impossible to bring it to rest. Think of a situation when you are travelling in a car and you cannot stop it because there is no friction between the surface of the tyres and that of the road. It will be a disadvantage in this case. #### Question 2: When you rub your hands together, they become warm. What is this due to? Friction between the surfaces of hands produces heat, which makes them warm when rubbed together. When you rub your hands together, it causes friction. This leads to the production of heat, which makes your hands warm. This is because of the roughness of the surfaces of hands. It is a classic example of the conversion of mechanical energy to heat energy. #### Question 3: If you try to hold a glass with oil on your hands, it tends to slip. Why? Friction between two surfaces in contact is reduced when the surfaces are lubricated. Oil is a lubricating agent. Hence, when you try holding a glass with oil on your hands, it tends to slip. #### Question 4: Why do you think rolling friction is less than sliding friction? Rolling friction is less than sliding friction because of the minimal surface of contact between two surfaces. For a rolling body, the contact surface area is less and goes on changing. In case of rolling friction, we can assume that a point of a body is in contact with a surface; while in case of sliding friction, the whole surface area of the body is in contact with the surface. #### Question 5: Friction is a necessary evil. Explain. Friction affects the relative motion between two bodies in contact. Without friction, objects cannot move from a stationary position or stop when in motion. Suppose you are walking on the road and encounter a smooth surface. You will fall because there will be no friction to hold your feet firmly on the surface. Now, let us change the situation. Let us suppose you are standing on a smooth surface and you are asked to walk on that surface. Again, you will not be able to walk properly because of lack of friction. Hence, friction is a necessary evil. #### Question 1: What is friction? Friction is a force that resists the relative motion between two surfaces in contact. It is a contact force. #### Question 2: An object is moving from north to south. What is the direction of the force of friction on the object? Frictional force acts in the direction opposite to the direction of motion of a body. Because the object is moving from north to south, the direction of frictional force is from south to north. #### Question 3: On what principle do ball bearings work? A ball bearing works on the principle that rolling friction is less than sliding friction. It helps in reducing friction between two surfaces in contact and maintaining separation between the bearing races, thereby allowing minimal surface contact between two surfaces. #### Question 4: How does lubrication reduce friction? Lubrication helps in reducing roughness between two surfaces in contact. Thus, friction decreases with the reduction in the roughness between two surfaces. #### Question 5: Give an example to show the effect of heat generated in a machine due to excessive friction. Friction opposes the relative motion between two surfaces in contact. The loss of energy occurs in the form of heat, sound, light, etc. For example, mechanical engines are lubricated time to time to reduce excessive friction that prevents the loss of energy in the form of heat. Take example of a car, motorbike or bicycle whose mechanical parts are lubricated to reduce excessive friction. #### Question 1: What is the effect of friction on motion? Friction is a force that reduces the relative motion between two surfaces in contact. It helps in reducing the motion of a body and to slow down the body till it stops. #### Question 2: What causes friction between two surfaces in contact? The roughness of surfaces causes friction between two surfaces in contact. To ordinary eyes, the surfaces may seem smooth, but at a microscopic level, the surfaces have crest and trough. Under magnification, they look a little like mountain ranges, full of peaks and valleys. Hence, when two surfaces come in contact, the motion between them is reduced because of the crest and trough on the surfaces. #### Question 3: Why does a rough surface have greater friction than a smooth surface? Rough surfaces have greater friction than smooth surfaces. Rough surfaces have grooves on them. When two rough surfaces are in contact, the grooves of one go inside the grooves of the other, thus making the movement difficult. Hence, roughness increases friction. #### Question 4: Give two situations where it is desirable to increase friction. Two situations where it is desirable to increase friction: 1) Walking - Friction is required between feet/footwear and the ground to prevent slips. 2) Machine Brakes - Friction is desirable in brakes to reduce motion in a quick time. #### Question 5: What is 'streamlining'? How is it useful?	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.meritnation.com/cbse-class-8/science/living-science/friction/textbook-solutions/10_2_1171_5599_155_43402", "fetch_time": 1571445861000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-43/segments/1570986685915.43/warc/CC-MAIN-20191018231153-20191019014653-00556.warc.gz", "warc_record_offset": 999311207, "warc_record_length": 16510, "token_count": 2271, "char_count": 10921, "score": 3.578125, "int_score": 4, "crawl": "CC-MAIN-2019-43", "snapshot_type": "latest", "language": "en", "language_score": 0.8944536447525024, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00039-of-00064.parquet"}
## Thursday, September 29, 2016 ### Longest repeating and non-overlapping substring - GeeksforGeeks Longest repeating and non-overlapping substring - GeeksforGeeks Given a string str, find the longest repeating non-overlapping substring in it. In other words find 2 identical substrings of maximum length which do not overlap. If there exists more than one such substring return any of them. Naive Solution : The problem can be solved easily by taking all the possible substrings and for all the substrings check it for the remaining(non-overlapping) string if there exists an identical substring. There are O(n2) total substrings and checking them against the remaining string will take O(n) time. So overall time complexity of above solution is O(n3). Dynamic Programming : This problem can be solved in O(n2) time using Dynamic Programming. The basic idea is to find the longest repeating suffix for all prefixes in the string str. ```Length of longest non-repeating substring can be recursively defined as below. LCSRe(i, j) stores length of the matching and non-overlapping substrings ending with i'th and j'th characters. If str[i-1] == str[j-1] && (j-i) > LCSRe(i-1, j-1) LCSRe(i, j) = LCSRe(i-1, j-1) + 1, Else LCSRe(i, j) = 0 Where i varies from 1 to n and j varies from i+1 to n ``` To avoid overlapping we have to ensure that the length of suffix is less than (j-i) at any instant. The maximum value of LCSRe(i, j) provides the length of the longest repeating substring and the substring itself can be found using the length and the ending index of the common suffix. `string longestRepeatedSubstring(string str)` `{` ` ``int` `n = str.length();` ` ``int` `LCSRe[n+1][n+1];` ` ``// Setting all to 0` ` ``memset``(LCSRe, 0, ``sizeof``(LCSRe));` ` ``string res; ``// To store result` ` ``int` `res_length = 0; ``// To store length of result` ` ``// building table in bottom-up manner` ` ``int` `i, index = 0;` ` ``for` `(i=1; i<=n; i++)` ` ``{` ` ``for` `(``int` `j=i+1; j<=n; j++)` ` ``{` ` ``// (j-i) > LCSRe[i-1][j-1] to remove` ` ``// overlapping` ` ``if` `(str[i-1] == str[j-1] &&` ` ``LCSRe[i-1][j-1] < (j - i))` ` ``{` ` ``LCSRe[i][j] = LCSRe[i-1][j-1] + 1;` ` ``// updating maximum length of the` ` ``// substring and updating the finishing` ` ``// index of the suffix` ` ``if` `(LCSRe[i][j] > res_length)` ` ``{` ` ``res_length = LCSRe[i][j];` ` ``index = max(i, index);` ` ``}` ` ``}` ` ``else` ` ``LCSRe[i][j] = 0;` ` ``}` ` ``}` ` ``// If we have non-empty result, then insert all` ` ``// characters from first character to last` ` ``// character of string` ` ``if` `(res_length > 0)` ` ``for` `(i = index - res_length + 1; i <= index; i++)` ` ``res.push_back(str[i-1]);` ` ``return` `res;` `}`	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://massivealgorithms.blogspot.com/2016/09/longest-repeating-and-non-overlapping.html", "fetch_time": 1508365162000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-43/segments/1508187823153.58/warc/CC-MAIN-20171018214541-20171018234541-00314.warc.gz", "warc_record_offset": 220452510, "warc_record_length": 36730, "token_count": 926, "char_count": 3054, "score": 3.671875, "int_score": 4, "crawl": "CC-MAIN-2017-43", "snapshot_type": "longest", "language": "en", "language_score": 0.7832033634185791, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00006-of-00064.parquet"}
× ### Let's log you in. or Don't have a StudySoup account? Create one here! × or 24 0 2 # Class Note for MATH 125 with Professor Rychlik at UA Marketplace > University of Arizona > Class Note for MATH 125 with Professor Rychlik at UA No professor available These notes were just uploaded, and will be ready to view shortly. Either way, we'll remind you when they're ready :) Get a free preview of these Notes, just enter your email below. × Unlock Preview COURSE PROF. No professor available TYPE Class Notes PAGES 2 WORDS KARMA 25 ? ## Popular in Department This 2 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Arizona taught by a professor in Fall. Since its upload, it has received 24 views. × ## Reviews for Class Note for MATH 125 with Professor Rychlik at UA × × ### What is Karma? #### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more! Date Created: 02/06/15 A cubic function Without a critical point BY MAREK RYCHLIK Lecture of November 3 2008 Example 1 Let us consider the function yzg12312 Find its critical points in ection points and intervals of monotonicity Determine how many roots the function will have Solution First we nd the critical points by solving the equation y 3122130 This is a quadratic equation A refresher from algebra the equation a12bx c0 has 2 1 or zero solutions depending on the sign of the discriminant A b2 7 4 a c If A gt 0 then there are two roots given by formulas 11 2a 7 b 7 VA 12 7 2a For our case A 22 7 4 3 3 7 32 Hence the equation has no roots Thus there are no crit ical points Moreover y gt 0 because the coe icient at x2 is positive The graph of y which is a parabola lies above the z axis The in ection point is determined from the equation y 6z2 0 which gives 13 7 1 In summary the function is strictly increasing there are no critical points and there is one in ection point at z 7 The in ection point is at the minimum point of the rst derivative Thus the function changes most slowly at the in ection point There will be exactly one root This is deduced from the following facts a The limit of the function y fx 1312 3x 2 is ioo as 1 goes to ioo Thus the function is positive for large positive I and negative for large negative 1 Because the function is continuous it must cross the z axis See for example the Racetrack Principle of Section 910 which is a special version of the Intermediate Value Theorem see Wikipedia b A strictly increasing function such as fx may intersect the z axis at most once lnceed if 1 lt I then fzl lt fzg which means that fzl u and thus only one of the numbers fzl fzg can be zero Below is a graph obtained with a free program called GNUplot GNUplot plot X 11 fxx3x23x2 fx fx 1 w x 1 05 0 05 GNUplot × × ### BOOM! 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Feel free to contact them here: [email protected] Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to [email protected]	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://studysoup.com/note/12565/spring-2015", "fetch_time": 1477209543000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2016-44/segments/1476988719192.24/warc/CC-MAIN-20161020183839-00077-ip-10-171-6-4.ec2.internal.warc.gz", "warc_record_offset": 882885801, "warc_record_length": 16153, "token_count": 1159, "char_count": 4903, "score": 3.890625, "int_score": 4, "crawl": "CC-MAIN-2016-44", "snapshot_type": "latest", "language": "en", "language_score": 0.9249489307403564, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00061-of-00064.parquet"}
# Question: What Is A Bar Model Year 2? ## Does a bar diagram show subtraction? A bar diagram can be used with any operation: addition, subtraction, multiplication, and division.. ## What is modeling method? It emphasizes the use of models to describe and explain physical phenomena rather than solve problems. … The modeling method is unique in requiring the students to present and defend an explicit model as justification for their conclusions in every case. ## How does the bar model represent the calculation? The bar model is a visual strategy to help solve number problems using different sized rectangles to represent numbers. The rectangles or ‘bars’ are proportional so that a larger number in a problem is represented by a larger bar. ## What is a bar diagram look like? A bar chart is a graph with rectangular bars. The graph usually compares different categories. Although the graphs can be plotted vertically (bars standing up) or horizontally (bars laying flat from left to right), the most usual type of bar graph is vertical. ## How many types of subtraction are there? But there are actually three different interpretations of subtraction: Taking away. Part-whole. ## What is a bar model 2nd grade? In math, a bar model can be defined as a pictorial representation of a number in the form of bars or boxes used to solve number problems. Bar models help us to attain an understanding of how a problem needs to be solved and calculated. Here, for instance, one rectangle or box of the bar model represents a value of 6. ## What is a bar model? In maths a bar model is a pictorial representation of a problem or concept where bars or boxes are used to represent the known and unknown quantities. Bar models are most often used to solve number problems with the four operations – addition and subtraction, multiplication and division. ## What is a bar diagram in 4th grade math? A bar graph can be defined as a chart or a graphical representation of data, quantities or numbers using bars or strips. Bar graphs are used to compare and contrast numbers, frequencies or other measures of distinct categories of data. ## What is a part whole model? The part-whole model is the concept of how numbers can be split into parts. Children using this model will see the relationship between the whole number and the component parts, this helps learners make the connections between addition and subtraction. ## How do you use a fraction bar? Start by adding fractions with common denominators, such as 1/8 plus 3/8. By using the fraction bars, students learn the answer is 4/8. You can also use the bars to add fractions without common denominators, such as 1/2 plus 1/4. Students use the bars to determine that 1/2 is the same as 2/4. ## What is the model in maths? A mathematical model is a description of a system using mathematical concepts and language. The process of developing a mathematical model is termed mathematical modeling. … A model may help to explain a system and to study the effects of different components, and to make predictions about behaviour. ## What is a bar model in fractions? Bar models can help visualise the process when you divide. To do this, you need to know your original amount, in this example it is 3, and the amount you are dividing into, in this case it’s ½. A bar is then drawn to represent this amount. The whole bar is split into single units (bars of one). ## What is a comparison bar model? A comparison bar model uses solid bars to show known and unknown amounts. Bars are arranged vertically underneath each other so learners can see the difference in the length of the bars. The brackets show the total. Here’s a question from Maths — No Problem! ## What is a bar diagram in math? A graph drawn using rectangular bars to show how large each value is. The bars can be horizontal or vertical. See: Histogram. Bar Graphs. ## What is a comparison model in math? The Comparison Concept The Comparison Concept is one of the three main concepts in the Model Method which all math models are derived from, the other two being the. 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1. ## percentages 80 student out of 120 students got distinction in Mathematics .70 students out of 100 student got distinction in English .For which subject were the result better ? 2. Hi uma To find the percentage of the students that get distinction : $\text{ \% student} = \frac{\text{number of students that get distinction}}{\text{total student}} \, 100$ Now you can compute the percentages for math and english then compare them 3. Originally Posted by uma 80 student out of 120 students got distinction in Mathematics .70 students out of 100 student got distinction in English .For which subject were the result better ? First, don't you mean 70 students out of 100, not .70 students? Going on that assumption, let's make them fractions and reduce: 70/100 = 7/10 (reduce numerator and denominator by factor of 10) = 0.7 80/120 = 8/12 = 4(2)/4(3) = 2/3 = 0.6 repeating or 0.67 rounded 4. Originally Posted by macosxnerd101 First, don't you mean 70 students out of 100, not .70 students? [snip] I think it's clear that . is a full stop to end the previous sentence.	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/algebra/103363-percentages.html", "fetch_time": 1513637199000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-51/segments/1512948627628.96/warc/CC-MAIN-20171218215655-20171219001655-00085.warc.gz", "warc_record_offset": 185491897, "warc_record_length": 10962, "token_count": 282, "char_count": 1079, "score": 3.8125, "int_score": 4, "crawl": "CC-MAIN-2017-51", "snapshot_type": "longest", "language": "en", "language_score": 0.9481284022331238, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00025-of-00064.parquet"}
Home » Shop » Subtraction Worksheets Color By Number Transportation # Subtraction Worksheets Color By Number Transportation This Subtraction Worksheets Color By Number Transportation activities pack is the perfect way for students to practice this skill. Your students will be actively engaged as they work on math problems using these coloring worksheets. ## 💰💰💰💰 Save 20% when you purchase the bundle! 💰💰💰💰 ### COMMON CORE MATH STANDARD(S) COVERED: • CCSS.MATH.CONTENT.1.OA.B.3 Apply properties of operations as strategies to add and subtract. Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.) CCSS.MATH.CONTENT.1.OA.B.4 Understand subtraction as an unknown-addend problem. For example, subtract 10 - 8 by finding the number that makes 10 when added to 8. ### OTHER COLOR BY NUMBER ACTIVITIES: • My favorite part about the color by number bundles is the ability to differentiate for students! Every student can have their own color by number sheets that will help them where they are – whether it’s addition to 12, addition to 20, mixed addition and subtraction to 20, missing addends, or just subtraction! The pictures are the same, but the skills are different! • Color by numbers can be so successful for students because it doesn’t actually LOOK like math, but trust me... they are learning and putting those skills to work! Want to try before you buy? READ MORE HERE! ### HEAR FROM OTHERS: 📢 My Color By Number Worksheets are a huge seller and I have absolutely ALL that you could possibly need for your students to practice math concepts with every theme you can think of - plus I'm adding ALL.THE.TIME! • My students loved these!! Challenging but appropriately so. Thank you! • My students really enjoyed this activity! Thanks so much! • This was a great supplemental resource to use in conjunction with our math program! • This kept my students engaged and is a great fluency resource! • Fun way for my students to practice and improve math skills. ### QUESTIONS OR CONCERNS Thank you for visiting my store! Please contact me if you have any questions about my Subtraction Worksheets Color By Number Transportation activities! I sure appreciate you! Shanon Juneau ❤️ ## Reward Points I’ve created a simple point system for giving back to loyal customers! • For every \$1 (US dollar) you spend, you earn 1 point! • 20 points can be redeemed for \$1 (US dollar) on a future purchase! • When you leave feedback (review), you get 5 points! • We Are Better Together Reward Points can be viewed under My Account. This means that you effectively get 5% back on everything you buy, plus extra points for feedback. Note: Point award system is subject to change. View full terms here. ## Reviews There are no reviews yet. No FAQ Found	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://shanonjuneau.com/product/subtraction-worksheets-color-by-number-transportation/", "fetch_time": 1713892667000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-18/segments/1712296818732.46/warc/CC-MAIN-20240423162023-20240423192023-00521.warc.gz", "warc_record_offset": 460494050, "warc_record_length": 47760, "token_count": 700, "char_count": 2948, "score": 3.8125, "int_score": 4, "crawl": "CC-MAIN-2024-18", "snapshot_type": "latest", "language": "en", "language_score": 0.9055638313293457, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00050-of-00064.parquet"}
# Fourier transform standard practice for physics There are many possible choices regarding the overall scaling coefficients as well as the scaling coefficient converting time and frequency. It is possible to summarize these conventions succinctly using two numbers $a$ and $b$. I use the same notation as used in the Mathematica Fourier Transform function. We define the Fourier Transform: $$\mathcal{FT}_{a,b}[f(t)](\omega) = \sqrt{\frac {\|b\|}{(2\pi)^{1-a}}}\int_{-\infty}^{+\infty} e^{+i b \omega t} f(t) dt$$ And the inverse Fourier Transform $$\mathcal{FT}_{a,b}^{-1}[\tilde{f}(\omega)](t) = \sqrt{\frac{\|b\|}{(2\pi)^{1+a}}}\int_{-\infty}^{+\infty} e^{-i b \omega t} \tilde{f}(\omega) d\omega$$ Let $$\tilde{f}_{a,b}(\omega) = \mathcal{FT}_{a,b}[f(t)](\omega)$$ $$\check{f}_{a,b}(t) = \mathcal{FT}_{a,b}^{-1}[\tilde{f}_{a,b}(\omega)](t)$$ It can be shown via the Fourier inversion theorem that for the classes of functions we care about in physics $\check{f}_{a,b}(t) = f(t)$ for any $a$ and $b$. That is, for these definitions of the Fourier Transform and Inverse Fourier transform the two operations are inverses of eachother. It's turns out that in the engineering and scientific literature there are many conventions that people choose depending mostly on what they are used to. The first convention in the OP is $(a,b) = (1,-1)$ which is commonly used in physics, about as commonly as $(a,b) = (1,+1)$ which is the second convention you have shown. In addition you will also see conventions where $(a,b) = (0,\pm1)$ where the factor of $2\pi$ is split evenly between the transform and inverse transform showing up with a square root. Furthermore, usually in math or signal processing you will come across the $(a,b) = (0,\pm 2\pi)$ convention in which there is NO prefactor of $2\pi$ on either the transform or the inverse transform but now instead of angular frequency $\omega$ represents a cyclic frequency and a $2\pi$ appears in all of the exponentials. All of these different conventions have advantages and disadvantages which may make one choice of convention more attractive than another depending on the application. The main point is that in any problem, whichever convention is chosen should be kept the same throughout the whole problem. To get back to the OP's main question now. In the language set up in this answer the OP is basically asking if it matters whether $b=+1$ or $b=-1$. The short answer is that it does not matter. Either way works and converts the original signal as a function of time into a function of frequency. The difference has to do with how we interpret positive and negative frequencies. Consider $$f^1(t) = e^{+i\omega_0 t}$$ $$f^2(t) = e^{-i \omega_0 t}$$ The phasor for the first function rotates counterclockwise in phase space whereas the second rotates clockwise in phasespace. If we choose the $b=-1$ convention then $\tilde{f}^1_{1,-1}(\omega)$ will have a nonzero contribution at $+\omega_0$ whereas $\tilde{f}^2_{1,-1}(\omega)$ will have a nonzero contribution at $-\omega_0$. We might say $f^1$ is a positive frequency signal while $f^2$ is negative. However, if we choose $b=+1$ then everything reverses. $\tilde{f}^1_{1,+1}(\omega)$ will have a nonzero contribution at $-\omega_0$ while $\tilde{f}^2_{1,+1}(\omega)$ will have a contribution at $+\omega_0$. now $f^1$ is negative frequency and $f^2$ is positive frequency! Thuse we see that both $b=+1$ and $b=-1$ give answer that we can interpret as frequencies with the only difference between the two being what we call positive and negative frequencies. As a note I personally prefer $(a,b)=(1,+1)$ because it makes the formula for the Fourier transform (which I use more often than the inverse transform) as simple as possible. No prefactor and no minus sign in the exponent. edit: As you have pointed out sometimes these signs can have a substantial effect on some physical quantity such as reversing the sign (inverting the phase) of the complex impedance of a capacitor. Unfortunately this is something we just have to deal with and try to be consistent with our own conventions and those used by the references we consult. Of course you will find both conventions give the same answer for a real measurable quantity such as $V(t)$ across the resistor.	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://newbedev.com/fourier-transform-standard-practice-for-physics", "fetch_time": 1701579663000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-50/segments/1700679100484.76/warc/CC-MAIN-20231203030948-20231203060948-00313.warc.gz", "warc_record_offset": 477431957, "warc_record_length": 7684, "token_count": 1107, "char_count": 4288, "score": 3.6875, "int_score": 4, "crawl": "CC-MAIN-2023-50", "snapshot_type": "latest", "language": "en", "language_score": 0.8593443036079407, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00016-of-00064.parquet"}
Scalar-valued Function Get Scalar-valued Function essential facts below. View Videos or join the Scalar-valued Function discussion. Add Scalar-valued Function to your PopFlock.com topic list for future reference or share this resource on social media. Scalar-valued Function A scalar field such as temperature or pressure, where intensity of the field is represented by different hues of colors. In mathematics and physics, a scalar field associates a scalar value to every point in a space - possibly physical space. The scalar may either be a (dimensionless) mathematical number or a physical quantity. In a physical context, scalar fields are required to be independent of the choice of reference frame, meaning that any two observers using the same units will agree on the value of the scalar field at the same absolute point in space (or spacetime) regardless of their respective points of origin. Examples used in physics include the temperature distribution throughout space, the pressure distribution in a fluid, and spin-zero quantum fields, such as the Higgs field. These fields are the subject of scalar field theory. ## Definition Mathematically, scalar fields on a region U is a real or complex-valued function or distribution on U.[1][2] The region U may be a set in some Euclidean space, Minkowski space, or more generally a subset of a manifold, and it is typical in mathematics to impose further conditions on the field, such that it be continuous or often continuously differentiable to some order. A scalar field is a tensor field of order zero,[3] and the term "scalar field" may be used to distinguish a function of this kind with a more general tensor field, density, or differential form. The scalar field of ${\displaystyle \sin(2\pi (xy+\sigma ))}$ oscillating as ${\displaystyle \sigma }$ increases. Red represents positive values, purple represents negative values, and sky blue represents values close to zero. Physically, a scalar field is additionally distinguished by having units of measurement associated with it. In this context, a scalar field should also be independent of the coordinate system used to describe the physical system--that is, any two observers using the same units must agree on the numerical value of a scalar field at any given point of physical space. Scalar fields are contrasted with other physical quantities such as vector fields, which associate a vector to every point of a region, as well as tensor fields and spinor fields.[] More subtly, scalar fields are often contrasted with pseudoscalar fields. ## Uses in physics In physics, scalar fields often describe the potential energy associated with a particular force. The force is a vector field, which can be obtained as a factor of the gradient of the potential energy scalar field. Examples include: ### Examples in quantum theory and relativity • Scalar fields like the Higgs field can be found within scalar-tensor theories, using as scalar field the Higgs field of the Standard Model.[8][9] This field interacts gravitationally and Yukawa-like (short-ranged) with the particles that get mass through it.[10] • Scalar fields are found within superstring theories as dilaton fields, breaking the conformal symmetry of the string, though balancing the quantum anomalies of this tensor.[11] • Scalar fields are hypothesized to have caused the high accelerated expansion of the early universe (inflation),[12] helping to solve the horizon problem and giving a hypothetical reason for the non-vanishing cosmological constant of cosmology. Massless (i.e. long-ranged) scalar fields in this context are known as inflatons. Massive (i.e. short-ranged) scalar fields are proposed, too, using for example Higgs-like fields.[13] ## References 1. ^ Apostol, Tom (1969). Calculus. II (2nd ed.). Wiley. 2. ^ "Scalar", Encyclopedia of Mathematics, EMS Press, 2001 [1994] 3. ^ "Scalar field", Encyclopedia of Mathematics, EMS Press, 2001 [1994] 4. ^ Technically, pions are actually examples of pseudoscalar mesons, which fail to be invariant under spatial inversion, but are otherwise invariant under Lorentz transformations. 5. ^ P.W. Higgs (Oct 1964). "Broken Symmetries and the Masses of Gauge Bosons". Phys. Rev. Lett. 13 (16): 508. Bibcode:1964PhRvL..13..508H. doi:10.1103/PhysRevLett.13.508. 6. ^ Jordan, P. (1955). Schwerkraft und Weltall. Braunschweig: Vieweg. 7. ^ Brans, C.; Dicke, R. (1961). "Mach's Principle and a Relativistic Theory of Gravitation". Phys. Rev. 124 (3): 925. Bibcode:1961PhRv..124..925B. doi:10.1103/PhysRev.124.925. 8. ^ Zee, A. (1979). "Broken-Symmetric Theory of Gravity". Phys. Rev. Lett. 42 (7): 417. Bibcode:1979PhRvL..42..417Z. doi:10.1103/PhysRevLett.42.417. 9. ^ Dehnen, H.; Frommert, H.; Ghaboussi, F. (1992). "Higgs field and a new scalar-tensor theory of gravity". Int. J. Theor. Phys. 31 (1): 109. Bibcode:1992IJTP...31..109D. doi:10.1007/BF00674344. 10. ^ Dehnen, H.; Frommmert, H. (1991). "Higgs-field gravity within the standard model". Int. J. Theor. Phys. 30 (7): 985-998 [p. 987]. Bibcode:1991IJTP...30..985D. doi:10.1007/BF00673991. 11. ^ Brans, C. H. (2005). "The Roots of scalar-tensor theory". arXiv:gr-qc/0506063. Bibcode:2005gr.qc.....6063B. Cite journal requires \|journal= (help) 12. ^ Guth, A. (1981). "Inflationary universe: A possible solution to the horizon and flatness problems". Phys. Rev. D. 23: 347. Bibcode:1981PhRvD..23..347G. doi:10.1103/PhysRevD.23.347. 13. ^ Cervantes-Cota, J. L.; Dehnen, H. (1995). "Induced gravity inflation in the SU(5) GUT". Phys. Rev. D. 51: 395. arXiv:astro-ph/9412032. Bibcode:1995PhRvD..51..395C. doi:10.1103/PhysRevD.51.395.	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.popflock.com/learn?s=Scalar-valued_function", "fetch_time": 1603965382000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-45/segments/1603107904039.84/warc/CC-MAIN-20201029095029-20201029125029-00476.warc.gz", "warc_record_offset": 155806199, "warc_record_length": 18147, "token_count": 1482, "char_count": 5634, "score": 3.5625, "int_score": 4, "crawl": "CC-MAIN-2020-45", "snapshot_type": "latest", "language": "en", "language_score": 0.8782913088798523, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00033-of-00064.parquet"}
#### Chapter 5 Geometry Section 5.2 Angles and Angle Measurement # 5.2.2 Angles Two rays (half-lines) $g$ and $h$ in the plane starting from the same initial point $S$ enclose an angle $\measuredangle \left(g,h\right)$. Angle enclosed by the rays $g$ and $h$. For the notation of the angle $\measuredangle \left(g,h\right)$, the order of $g$ and $h$ is relevant. $\measuredangle \left(g,h\right)$ denotes the angle shown in the figure above. It is defined by rotating the half-line $g$ counter-clockwise to the half-line $h$. In contrast, $\measuredangle \left(h,g\right)$ denotes the angle from $h$ to $g$ as illustrated by the figure below. Angle enclosed by the rays $h$ and $g$ The point $S$ is called a vertex of the angle, and the two half-lines enclosing the angle are called the arms of the angle. If $A$ is a point on the line $g$ and $B$ is a point on the line $h$, then the angle $\measuredangle \left(g,h\right)$ can also be denoted by $\measuredangle \left(ASB\right)$. In this way, angles between line segments $\stackrel{‾}{SA}$ and $\stackrel{‾}{SB}$ are described. Angles are often denoted by lower-case Greek letters to distinguish them from variables, which are generally denoted by lower-case Latin letters (see Table 1.1.8 in module 1). Further angles can be found by considering angles formed by intersecting lines. ##### Vertical Angles and Supplementary Angles 5.2.1 Let $g$ and $h$ be two lines intersecting in a point $S$. • The angles $\phi$ and $\phi \text{'}$ are called vertical angles. • The angles $\phi$ and $\psi$ are called supplementary angles with respect to $g$. The figure above contains further vertical and supplementary angles. ##### Exercise 5.2.2 Find all vertical and supplementary angles occurring in the figure above. Some special angles have their own dedicated name. For example, the angle bisector $w$ is the half-line whose points have the same distance from the two given half-lines $g$ and $h$. Then, it can be said that $w$ bisects the angle between $g$ and $h$. ##### Names of Special Angles 5.2.3 Let $g$ and $h$ be half-lines with the intersection point $S$. • The angle covering the entire plane is called the complete angle. • If the rays $g$ and $h$ form a line, the angle between $g$ and $h$ is called a straight angle. • The angle between two half-lines bisecting a straight angle is called the right angle. One also says that $g$ and $h$ are perpendicular (or orthogonal) to each other. Next, three lines are considered. Two of the three lines are parallel, while the third line is not parallel to the others. It is called a transversal. These lines form eight cutting angles. Four of the eight angles are equal. ##### Angles at Parallel Lines 5.2.4 Let two parallel lines $g$ and $h$ be given cut by another transversal line $j$. • Then the angle $\alpha \text{'}$ is called a corresponding angle of $\alpha$ and • the angle $\beta \text{'}$ is called an alternate angle of $\beta$. Since the lines $g$ and $h$ are parallel, the angles $\alpha$ and $\alpha \text{'}$ are equal. Likewise, the angles $\beta$ and $\beta \text{'}$ are equal. ##### Exercise 5.2.5 The figure shows two parallel lines $g$ and $h$ cut by another line $j$. Explain which angles are equal and which angles are corresponding angles or alternate angles to each other, respectively.	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 98, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.math4refugees.de/html/en/1.5.2/xcontent1.html", "fetch_time": 1725789928000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700650976.41/warc/CC-MAIN-20240908083737-20240908113737-00273.warc.gz", "warc_record_offset": 843157196, "warc_record_length": 8132, "token_count": 882, "char_count": 3332, "score": 4.8125, "int_score": 5, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.8850209712982178, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00063-of-00064.parquet"}
# 9.4.3 Shortest Distance between Two Points 9.4.3 Shortest Distance between Two Points 1. The shortest distance between two points on the surface of the earth is the distance measured along a great circle. Shortest distance between points D and M = ( θ × 60 ) nautical miles Example: In the above diagram, calculate (a) The distance from P to Q, measured along the parallel of latitude 48o S, (b) The distance from P to Q, measured along the route PSQ, where S is the South Pole. State the shorter distance. Solution: (a) Distance from Pto Q, measured along the parallel of latitude 48o S = 180 × 60 × cos 48o← (angle PMQ = 180o) = 7266.61 n.m. (b) Distance from Pto Q, measured along the route PSQ, where S is the South Pole = 84 × 60 ← (angle POQ = 180o – 48o – 48o = 84o) = 5040 n.m. The distance from P to Q, measured along the route PSQ in (b), where S is the South Pole, is shorter than the distance measured along the parallel of latitude in (a). The shortest distance in the above example is the distance along the arc of a great circle, which passes through the South (or North) Pole. # 9.3 Location of a Place 9.3 Location of a Place The location of a place is written as an ordered pair of latitude and longitude (latitude, longitude). Example: Location of point P is (35o N, 27o E). # 9.1 Longitudes 9.1 Longitudes 1. A great circle is a circle with the centre of the Earth as its centre. 2. A meridian is half of a great circle from the North pole to the South pole. 3. The longitude of the Greenwich Meridian is 0o. 4. The longitude of a meridian is determined by: (a) The angle between the meridian plane and the Greenwich Meridian. (b) The position of the meridian to the east or west of the Greenwich Meridian. Example: Longitude of is 55o W. Longitude of is 30o E. 5. All points on the same meridian have the same longitude. Difference between Two Longitudes 1. If both the meridians are on the east (or west) of the Greenwich Meridian, subtract the angles of the longitudes. 2. If both the meridians are on the opposite sides of the Greenwich Meridian, add the angles of the longitudes. # 9.2 Latitudes 9.2 Latitudes 1. The great circle which is perpendicular to the earth’s axis is called equator. 2. A latitude is the angle at the centre of the Earth which is subtended by the arc of a meridian starting from the Equator to the parallel of latitude. 3. A circle which is perpendicular to the Earth’s axis and parallel to the equator is called a parallel of latitude. 4. The latitude of the Equator is 0o. 5. All points on a parallel of latitude have the same latitude. Example: Latitude of is 30o N. Latitude of is 45o S. Difference between Two Latitudes 1. If both parallels of latitude are on the north (or south) of the equator, subtract the angles of latitudes. 2. If both parallels of latitude are on the opposite sides of the equator, add the angles of the latitudes. # 9.4 Distance on the Surface of the Earth 9.4 Distance on the Surface of the Earth 9.4.1 Distance between Two Points Along a Great Circle 1. A nautical mile is the length of an arc of a great circle that subtends an angle of one minute at the centre of the Earth. 2. The distance between two points on the surface of the earth is the arc of a circle that joins the two points. 3. To find the distance between two points along a meridian or the equator: Distance of P and Q = ( θ x 60 ) nautical miles # 9.4.2 Distance between Two Points along a Parallel of Latitude 9.4.2 Distance between Two Points along a Parallel of Latitude 1. To find the distance along a parallel of latitude: Distance between points P and Q = ( different in longitude in minutes ) × cos latitude = ( θ × 60 × cos xo ) nautical miles Example: In the diagram above, calculate (a) The distance of PQ, measured along the parallel of latitude 34o N, (b) The distance of DM, measured along the parallel of latitude 54o S. Solution: (a) Difference in longitude between P and Q = 40o + 58o = 98o Hence, the distance of PQ, measured along the parallel of latitude 34o N = 98 × 60 × cos 34o← [θ × 60 × cos (latitude)] = 4874.74 n.m. (b) Difference in longitude between D and M = 82o – 40o = 42o Hence, the distance of DM, measured along the parallel of latitude 54o S = 42 × 60 × cos 54o← [θ × 60 × cos (latitude)] = 1481.22 n.m.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://content.myhometuition.com/category/spm-maths/earth-as-a-sphere/page/2/", "fetch_time": 1653307890000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-21/segments/1652662558015.52/warc/CC-MAIN-20220523101705-20220523131705-00734.warc.gz", "warc_record_offset": 16755409, "warc_record_length": 9045, "token_count": 1182, "char_count": 4361, "score": 4.4375, "int_score": 4, "crawl": "CC-MAIN-2022-21", "snapshot_type": "latest", "language": "en", "language_score": 0.8704177737236023, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00005-of-00064.parquet"}
# (1)/(4p)(x-h)^(2)+k=0 Multiply the equation by 4p. Explain how different values of k affect the number of zeros of the polynomial. Con Question (1)/(4p)(x-h)^(2)+k=0 Multiply the equation by 4p. Explain how different values of k affect the number of zeros of the polynomial. Consider k > 0, k = 0, and k < 0. Assume p > 0. in progress 0 1 year 2021-09-04T01:09:51+00:00 2 Answers 34 views 0 Step-by-step explanation: Given p>0, multiply the equation by 4p: (1/4p)*(x-h)^(2)+k=0 (x-h)^2+4kp = 0 k>0 4kp>0 (x-h)^2 = -4kp So x has imaginary roots only. There is no real zeros of the polynomial. k=0 4kp=0 (x-h)^2 = 0 x=h So x has one real root and the polynomial has one zero k<0 4kp<0 (x-h)^2 = -4kp So x has two real roots and the polynomial has two real zeros.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://documen.tv/question/1-4p-h-2-k-0-multiply-the-equation-by-4p-eplain-how-different-values-of-k-affect-the-number-of-z-23241063-39/", "fetch_time": 1675256955000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-06/segments/1674764499934.48/warc/CC-MAIN-20230201112816-20230201142816-00168.warc.gz", "warc_record_offset": 223046660, "warc_record_length": 20126, "token_count": 286, "char_count": 783, "score": 4.15625, "int_score": 4, "crawl": "CC-MAIN-2023-06", "snapshot_type": "latest", "language": "en", "language_score": 0.7840532660484314, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00015-of-00064.parquet"}
× Search anything: # Meet In Middle Technique #### Algorithms In this post, we discuss the Meet in Middle problem solving technique and show through examples how we can use it to improve a naive brute force algorithm. 1. Introduction to Meet In Middle 2. Problem 1: Pythagorean triplet 3. Problem 2: Maximum sum subset 4. More examples using Meet in Middle # Introduction to Meet In Middle The meet-in-middle algorithm is a search technique used for problem solving whereby the given input size of the problem is small but not small enough for a brute force approach to work. It works by splitting the given input into two subsets, solves them using a standard algorithm then combines results obtained from the subsets to come up with a general solution. Like divide and conquer it takes a similar approach of dividing a problem in smaller sub-problems but unlike divide and conquer, we do not use recursion. It also has a similar approach to dynamic programming whereby we divide a problem and break it down to smaller problems and combine sub-solutions to solve the main problem but unlike dynamic programming, we only divide problems into two parts here. # Problem 1: Pythagorean triplet Problem Statement; Given an array of integers, write a function that returns true if there is a triplet (a, b, c) that satisfies the pythagorean theorem that is, ${a}^{2}+{b}^{2}={c}^{2}$. #### Example 1: Input: array = [3, 10, 4, 6, 13, 8] Output: true Explanation. square(6) + square(8) = square(10), 36 + 64 = 100 #### Example 2: Input: array = [3, 10, 4, 7, 13, 8] Output: false Explanation. There are no elements in the array that satisfy the condition, we return false. #### Naive Approach. We compute all combinations of all elements in the array, in python we use something like itertools combinations for this and out of those subsets we return true is any satisfies the condtion, otherwise we return false. #### Pseudocode ``````for i from 0 - n for j from i+1 to n for k from j+1 to n return true if either of the conditions suffice arr[i] * arr[i] = arr[j] * arr[j] + arr[k] + arr[k] arr[j] * arr[j] = arr[i] * arr[i] + arr[k] + arr[k] arr[k] * arr[k] = arr[i] * arr[i] + arr[j] + arr[j] otherwise false `````` #### Analysis This is a brute force approach, we try out all possible combinations of the elements to find the result. We use three loops making out time complexity O(N3). #### Optimized Approach We will optimize the naive approach using meet-in-middle technique to achieve a better computational complexity. In this approach we use three pointers and work our way to the middle of the array making comparisons until the pythagorean condition is met. #### Steps 1. Square all elements in the input array. 2. Sort the squared array in non-decreasing order. 3. To find a triplet; • Initialize three pointers l, r, t. • loop while l < r, • if arr[l] + arr[r] == arr[t] return true. • if arr[l] + arr[r] < arr[t], increment l pointer. • if arr[l] + arr[r] > arr[t], decrement r pointer. • if no triplet is found return false. ``````#include<iostream> #include<vector> #include<algorithm> using std::vector; bool isTriplet(vector<int> arr){ int n = arr.size(); for(int i = 0; i < n; i++) arr[i] = arr[i] * arr[i]; sort(arr.begin(), arr.end()); for(int i = n - 1; i >= 2; i--){ int l = 0, r = i-1; while(l < r){ if(arr[l] + arr[r] == arr[i]) return true; else if(arr[l] + arr[r] < arr[i]) l++; else r--; } } return false; } `````` #### Analysis The time complexity of this approach is O(N2). # Problem 2: Maximum sum subset Problem Statement; Given a set of n integers where n <= 40, determine the maximum sum subset having a sum less than s. #### Example 1. Input: array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], s = 20 Output: 20 Explanation: the maximum possible sum is 20, it can be obtained by adding, 10, 9, 1. #### Example 2. Input: array = [32, 12, 19, 3, 1, 11, 5], s = 9 Output: 9 Explanation: the maximum sum obtained that is closer to s is 9, obtained by adding, 5, 3, 1. #### Naive Approach. We could use a brute force approach like in the previous problem where we get all possible subsets are return the maximum sum among them that is less than m. #### Analysis. N is at most 40 so there are O(240) steps. The time complexity is exponential O(2N). This is not optimal. #### Optimized approach. Again we will use meet-in-middle algorithm to optimize the time complexity to a better one. Assuming n = 20, that is O(220) steps is not such a big number compared to O(240). We can use the brute force approach twice for each subset and combine results from each subset to obtain a general solution. #### Steps. 1. Split the array into two sets A and B each having n/2 items, that is 20 elements each. 2. Calculate all possible subsets and their sums for A and store them in array X and do the same for B and store them in array Y. 3. Sort array Y and itertate over X elements, for each element we use binary search to find the largest element in Y such that x + y is less than or equal t s. ``````#include<iostream> #include<algorithm> using std::sort; using std::lower_bound; typedef long long ll; ll X[2000005], Y[2000005]; void calcSubSet(ll A[], ll X[], int n, int c){ for(int i = 0; i < (1<<n); i++){ ll s = 0; for(int j = 0; j < n; j++){ if(i & (1<<j)) s += A[j+c]; } X[i] = s; } } ll solve(ll a[], int n, ll S){ //subsets of 1st and 2nd halves calcSubSet(a, X, n/2, 0); calcSubSet(a, Y, n-n/2, n/2); int sizeX = 1 << (n/2); int sizeY = 1 << (n-n/2); //sort Y sort(Y, Y+sizeY); //keep track of max element ll max = 0; for(int i = 0; i < sizeX; i++){ if(X[i] <= S){ int p = lower_bound(Y, Y+sizeY, S-X[i]) - Y; if(p == sizeY \|\| Y[p] != (S-X[i])) p--; if((Y[p] + X[i]) > max) max = Y[p] + X[i]; } } return max; } `````` #### Analysis. The time complexity is $O\left({2}^{\frac{n}{2}}\right)$. # More examples using Meet in Middle ### 1. 4 sum problem. Problem statement: Given an array of n integers and an integer target, find 4 values whose sum is equal to target. Hint: Traverse the array using two pointers i, j, Compute sums of pairs before i and store in a data structure, move i. While traversing from j to end of array, look for the remaining pairs whose sum complements the stored pairs before i to add up to target. We improve $O\left({n}^{4}\right)$ to $O\left({n}^{2}\right)$ using meet in middle. ### 2. Increasing subsequences of length 3 Problem Statement: Given an array of integers, count the number of increasing subsequences of length 3. Hint: We could write a brute force algorithm with $O\left({n}^{3}\right)$ time complexity. To optimize, instead of looping three times, we could use the 2nd element of the 3 elements. That is from the 2nd element, find elements less than it to the left and elements greater than it to the right. $O\left({n}^{2}\right)$ time complexity. ### 3. 0-1 Knapsack. Problem statement: Given a set of items and their weights. We are to combine them to get the maximum value with weight w that can fit into a knapsack of weight W. We cannot break elements, we take them or leave them. Hint: Partition items into two subsets and compute the weights and values of each, then for each subset of A we find a subset of b with the greatest value such that their combined weight if less than W while keeping track of max value obtained so far. With this article at OpenGenus, you must have the complete idea of Meet in Middle technique. #### Erick Lumunge Erick is a passionate programmer with a computer science background who loves to learn about and use code to impact lives positively. 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# Equilibrium: Force and Mg Topics: Force, Friction, Mass Pages: 12 (1770 words) Published: January 5, 2013 The First and Second Conditions for Equilibrium The first condition for equilibrium: The second condition for equilibrium: • • ΣF = 0 ΣΓ = 0 In when both of these conditions are satisfied in static systems all forces and torques sum to zero. In problems where the first and second conditions of equilibrium are satisfied, the best strategy is to create FBD’s for both the first and second conditions, derive equations based on these FBD’s and then see what useful information may be gleaned from these equations. When applying the second condition we are free to choose any axis about which to compute torques. It is best to choose an axis that eliminates one or more forces that have lines of force that pass through it. Example 1 Consider a playground seesaw. The mass of the plank is 2.0 kg, the masses of two children on it are 25 kg and 30 kg with the 30 kg child sitting 2.5 meters from the center of the plank (the fulcrum) as shown below. Where must the second child sit in order for this system to be in equilibrium? Noting that a normal force directed upwards acts at the point of the fulcrum, the FBD’s for the first condition yield: ∑F y = N − mc1 g − m p g − mc 2 g = 0 → ∑F y = N − 294 N − 19.6 N − 245 N = 0 Note that while this is all true it is not, by itself, particularly useful. To apply the second condition we must first choose an axis about which to compute torques. The axis that makes the most physical sense would be one directly through the board over the fulcrum, but we could choose any axis that made computations easier. In this case choosing the axis associated with the fulcrum eliminates the forces created by the mass of the board itself since these act on the center of mass of the board which is located directly over the fulcrum. The FBD’s for the second condition yield: ∑ Γ = (294 N )(2.5meters) − (245 N )( xmeters) = 0 Solving this equation for x yields a distance of 3 meters. Example 2 Consider the following cantilevered beam: The beam has a mass of m = 25 kg and is 2.2 meters long. The suspended block has a mass M = 280 kg and the supporting cable makes an angle of 300 with the beam. Determine the force that the wall exerts on the beam at the hinge and determine the tension in the supporting cable. • • Notice that the normal force is the x component of the force exerted by the wall on the beam through the hinge (Fx). Because the beam is also held up by the hinge (Fy) the total force the wall exerts on the beam is the aggregate of these two components. So we must determine, from the available information, Fx, Fy, Tx, Ty and finally T and F. Application of the first condition with our sign convention yields: ∑F ∑F x = Fx − Tx = 0 ∴ Fx = Tx = Fy + T y − mg − Mg = 0 y Application of the second condition with respect to the hinge yields: Fy – line of action passes through the hinge → no torque Fx – line of action passes through the hinge → no torque Tx – line of action passes through the hinge → no torque mg – exerts a torque Mg – exerts a torque Ty – exerts a torque With our sign convention: ∑ Γ = −(mg )(1.1m) − (Mg )(2.2m) + (T y )(2.2m) = 0 )(2.2m) = 0 ∑ Γ = −(245 N )(1.1m) − (2744 N )(2.2m) + (T y ∑ Γ = −270 N ⋅ m − 6037 N ⋅ m + (T v y )(2.2m) = 0 → T y = 2867 N Since T y = T sin θ : T = 5734 N , and with a little more work Tx = 4966 N . With the magnitude of T and all of its components known, it is a simple matter to substitute into the equation in y from the first condition and solve for Fy: ∑F y = Fy + 2867 N − 245 N − 2744 N = 0 Fy = 122 N Noting that Fx = Tx (why?), Fx = 4966N → F = (4966 N ) 2 + (122 N ) 2 = 4967 N tan o = / y 122 N = ∴ o = 1.4° / x 4966 N v F = 4967 N @ 1.4° Example 3. A 5 meter long ladder leans against a frictionless wall. The point of contact between the ladder and the wall is 4 meters above the ground. 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New Pattern Reasoning Questions \| BOB AND DENA BANK PO 2017 \| 19.05.2017 ~ Daily Current Affairs,English,SBI PO,IBPS # New Pattern Reasoning Questions \| BOB AND DENA BANK PO 2017 \| 19.05.2017 Q.1-5. Read the following information carefully and answer the question given below: Eight friends Kamla, Rano, Pooja, Poonam, Seema, Mita, Twist and Divya are sitting in a circle and facing towards the centre but not necessarily in the same order. (A) Those two friends are not neighbours to each other whose names are starting with letter P. (B) Kamla is second to the left of Rano. (C) Twist is neighbour of Pooja. (D) There is only one friend between Kamla and Pooja. (E) Mita and Divya are neighbours of Poonam. Q.1. Who is fifth to the left of Twist? (1) Poonam (2) Rano (3) Kamla (4) Divya (5) None of these Q.2. Who is infront of Kamla? (1) Mita (2) Divya (3) Twist (4) Poonam (5) None of these Q.3. From the given data, how many persons’ positions are definitely determined? (1) Six (2) Four (3) Five (4) Three (5) Two Q.4. If Mita is neighbour of Rano then who is fourth to the right of Divya? (1) Pooja (2) Kamla (3) Seema (4) Rano (5) Poonam Q.5. Which of the following sitting arrangements is valid from anticlockwise, starting from the name whose first letter of the name has 18 place value according to English alphabet? (1) Rano, Mita, Poonam, Divya, Twist, Pooja, Seema, Kamla (2) Rano, Divya, Poonam, Pooja, Mita, Twist, Kamla, Seema (3) Rano, Divya, Poonam, Pooja, Twist,Mita, Kamla, Seema (4) Rano, Mita, Poonam, Divya, Pooja, Twist, Seema, Kamla (5) Rano, Divya, Poonam, Mita, Pooja, Twist, Kamla, Seema Q.6-10. Read the following information carefully and give the answer of the following questions. Eight artist Riya, Diya, Sania, Anushka, Ravina, Karishma, Ziya and Deepika all go for different places Goa, Nainital, Madras, Kerala, Madhya Pradesh, Delhi, Gujarat and Uttar Pradesh and travel by a different vehicles - Boat, Train, Ship, Bus, Cycle, Aeroplane, Helicopter and Car. (Not necessary in the same order). Riya goes to Goa by boat and Diya travels by train but does not go to Madras and Kerala. Ravina goes to Madhya Pradesh but not by ship and aeroplane. The persons who travel by aeroplane do not go to Gujarat and Uttar Pradesh. Deepika travels by car but does not go to Nainital and Kerala. Ziya travels by helicopter and goes to Gujarat. Anushka travels by bus and goes to either Kerala or Nainital. The person who goes to Madras travels by either ship or cycle. Karishma goes to the Delhi and Sania goes to Madras. Q.6. Who of the following persons goes to Nainital? (1) Sania (2) Anushka (3) Deepika (4) Diya (5) None of these Q.7. Who of the following persons goes to Kerala? (1) Diya (2) Anushka (3) Deepika (4) Can't be determined (5) None of these Q.8. Who of the following persons goes to Uttar Pradesh? (1) Anushka (2) Diya (3) Deepika (4) Data inadequate (5) None of these Q.9. Karishma travels by which vehicle? (1) Helicopter (2) Ship (3) Cycle (4) Aeroplane (5) None of these Q.10. Ravina travels by which vehicle? (1) Ship (2) Helicopter (3) Bus (4) Can't be determined (5) None of these Q.1.(2) Q.2.(4) Q.3.(1) Q.4.(3) Q.5.(5) Q.6.(4) Q.7.(2) Q.8.(3)Q.9.(4) Q.10.(5) Share:	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.mahendraguru.com/2017/05/new-pattern-reasoning-questions-bob-and16.html", "fetch_time": 1495838855000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-22/segments/1495463608686.22/warc/CC-MAIN-20170526222659-20170527002659-00410.warc.gz", "warc_record_offset": 712163254, "warc_record_length": 43397, "token_count": 1101, "char_count": 3254, "score": 3.75, "int_score": 4, "crawl": "CC-MAIN-2017-22", "snapshot_type": "longest", "language": "en", "language_score": 0.8035457134246826, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00014-of-00064.parquet"}
# statistics posted by . if there 60% math major are women, 15 were random. what probility 12 women were are selected and what are the mean and variance. Show ne how to work the problems ## Similar Questions 1. ### statistics Find the indicated binomial probabilities. Round to the nearest 3 decimal places. In a local college, 20% of the math majors are women. Ten math majors are chosen at random. 1) What is the probability that exactly 2 are women? 2. ### statistics Find the indicated binomial probabilities. Round to the nearest 3 decimal places. In a local college, 20% of the math majors are women. Ten math majors are chosen at random. 1) What is the probability that exactly 2 are women? 3. ### college Four hundred people apply for three jobs. 130 of the applicants are women. (a) If three persons are selected at random, what is the probability that all are women? 4. ### statistics 60% of math major are women 15 were random and what probability that 5 are women. 5. ### statistics A student council is made up of 4 women and 6 men. One of the women is president of the council. A member of the council is selected at random at random to report to the Dean of Student Life. What is the probability a woman is selected? 6. ### statistics 72% working women use computers choose 5 at random whats the probility at least 1 does not use a computer whats the probility that all 5 use a computer 7. ### Statistics In a population of 240 women,the heights of the women are normally distributed with a mean of 64.2 inches and a standard deviation of 3.2 inches. If 36 women are selected at random, find the mean mu x and stand deviation sigma x of … 8. ### statistics It seems that people are choosing or finding it necessary to work later in life. Random samples of 200 men and 200 women age 65 or older were selected, and 80 men and 59 women were found to be working. At รก = 0.01, can it be concluded … 9. ### Statistics In a certain population the heights of men are normally distributed with mean 168.75 and variance 42.25 those of women are normally distributed with mean 162.50 and variance 36 the probability that the mean height of a random sample … 10. ### MATH MODELS 300 hundred people apply for three jobs. 120 of the applicants are women. (a) If three persons are selected at random, what is the probability that two are women? More Similar Questions	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.jiskha.com/display.cgi?id=1285091860", "fetch_time": 1503469502000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-34/segments/1502886117874.26/warc/CC-MAIN-20170823055231-20170823075231-00431.warc.gz", "warc_record_offset": 913709451, "warc_record_length": 3700, "token_count": 563, "char_count": 2383, "score": 3.515625, "int_score": 4, "crawl": "CC-MAIN-2017-34", "snapshot_type": "latest", "language": "en", "language_score": 0.9555973410606384, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00054-of-00064.parquet"}
# How would you go about proving the following "theorem" deductively? If you can count your money, you don't have a billion dollarstough question ### 1 Answer \|Add Yours Posted on I would provide an indirect proof as follows: Assume that you have a billion dollars and you can count your money. For ease of computation, assume all of your money is in the form of \$100 bills, the largest bill in common circulation. Lets also assume that you can count 1 bill per second, and that you count 18hrs. per day, 7 days a week. There are 64,800 seconds in an 18 hour period. There are 10,000,000 \$100 bills in a billion dollars (assuming the American interpretation of a billion as 1,000,000,000 -- the British version has 12 zeros ) Then it takes `10,000,000 -: 64,800~~154` days to count your money. Depending on your definition of ability to count your money (is 154 days of counting 18 hours a day non-stop reasonable?), it seems reasonable to suggest that you are not able to count your money. If we use the British version of a billion, it would take over 432 years Since we have reached a "contradiction"; we assumed you could count your money, but this is unreasonable; then the assumption that you have a billion dollars must be false. Therefore, if you can count your money then you don't have a billion dollars. We’ve answered 320,479 questions. We can answer yours, too.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.enotes.com/homework-help/how-would-you-go-about-proving-following-theorem-324063", "fetch_time": 1462514725000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2016-18/segments/1461861727712.42/warc/CC-MAIN-20160428164207-00126-ip-10-239-7-51.ec2.internal.warc.gz", "warc_record_offset": 483983123, "warc_record_length": 11714, "token_count": 333, "char_count": 1394, "score": 3.71875, "int_score": 4, "crawl": "CC-MAIN-2016-18", "snapshot_type": "longest", "language": "en", "language_score": 0.9526274800300598, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00006-of-00064.parquet"}
# Oklahoma - Grade 1 - Math - Operations and Algebraic Thinking - Addition and Subtraction Word Problems - 1.OA.1 ### Description Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem. • State - Oklahoma • Standard ID - 1.OA.1 • Subjects - Math Common Core ### Keywords • Math • Operations and Algebraic Thinking ## More Oklahoma Topics Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the reasoning used. Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: A. 10 can be thought of as a bundle of ten ones — called a “ten.” B. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. C. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones). Express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps. Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes. 1.OA.4 Understand subtraction as an unknown-addend problem. For example, subtract 10 – 8 by finding the number that makes 10 when added to 8. Add and subtract within 20. 1.OA.5 Relate counting to addition and subtraction (e.g., by counting on 2 to add 2). 1.OA.6 Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://egvideos.com/video/oklahoma/grade-1/math/1.oa.1/addition-and-subtraction-word-problems", "fetch_time": 1675086641000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-06/segments/1674764499819.32/warc/CC-MAIN-20230130133622-20230130163622-00718.warc.gz", "warc_record_offset": 239251184, "warc_record_length": 10138, "token_count": 662, "char_count": 2540, "score": 4.59375, "int_score": 5, "crawl": "CC-MAIN-2023-06", "snapshot_type": "latest", "language": "en", "language_score": 0.9073576331138611, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00060-of-00064.parquet"}
# A Look at Linear Regression (with Examples in Excel and Python) Data science is a fascinating field that aims to solve challenging problems like self-driving vehicles and artificial intelligence (AI). Back in 2021, long before today’s generative AI hype caused by ChatGPT, there was another AI story in the news. Real estate company Zillow Group shut down its home buying program, that used predictive analytics to forecast the price of homes into the future. Although they could have spent more resources tuning their algorithm, they found the potential risks outweighed the benefits In practice, predicting the price of homes months into the future is certainly a challenge—even for the most sophisticated algorithms. However, at its core, predicting a home’s value is simply a regression problem, since a home’s price is a continuous variable based on many independent variables like rooms, location, year built, and so on. Even simple linear regression—one of the simplest algorithms—can be used to price homes. In this tutorial, we will briefly introduce regression analysis and look at linear regression examples. If you’d like to learn more data skills, try this free data short course. With that, let’s get started. ## 1. Regression analysis 101 In data science, regression analysis is primarily used in prediction and forecasting tasks. Essentially, regression techniques fit a line to the data, which allows you to estimate changes to the dependent variable (for example, price) as changes occur to the independent variables (for example, size). Linear regression models assume the data have a linear relationship, and thus fit a straight line to it. Other regression models, like logistic regression, fit a curved line to the data. Regression analysis is a versatile set of techniques, because they are relatively easy to compute and explain compared to systems like neural networks and deep learning. Beyond a method for making predictions, regression analysis can also be used for things like identifying significant predictors and understanding the strength of relationships within the data. ## 2. What is linear regression? One of the most common forms of regression analysis is linear regression. It’s classified as a supervised learning algorithm. Simple linear regression is the term used when the linear regression model uses a single dependent variable and a single independent variable. When there are multiple variables, it is called multiple linear regression The linear regression algorithm draws the line of best fit through the data. It can do this by solving for the regression coefficient that minimizes the total error. To calculate the line of best fit we typically use Ordinary Least Squares (OLS). Want an in-depth look at linear regression? Check out our beginner’s guide in this article! We won’t dwell on the mathematics, but will do a quick explanation of the equation. It is important to know the basics of the linear regression calculation if you need to explain the predictions. The simple linear regression model equation is this: prediction = intercept + slope * independent variable + error : y is the predicted value of the dependent variable. a is the intercept. Think of this as where the line would cross the x axis (x=0). B is the slope of the line. x independent variable. e is the error or variation in the estimate of the regression coefficient. All of this sounds complex, but can be done in a few clicks within a spreadsheet or a few lines of code within Python. #### A Simple Example An easy way to understand simple linear regression is to imagine we want to sell a house and have to determine a price. To determine price, first figure the finished square footage of the home is 1280. Using that information, let’s look at sale prices of similarly-sized homes in the surrounding area. Say we find five houses that are roughly the same size, all recently sold for slightly different prices: If we plot the prices by square footage, we’ll notice an upward trend showing price increases as square footage increases. Using Microsoft Excel, enter the data into columns A and B respectively and create a chart by clicking Insert > Chart. Note: this can also be done in Google Sheets. To predict the price of the home, fit a trendline to the data in the chart. Click Chart Elements > Trendline. Mark the Trendline checkbox. Make sure the Linear option is selected within the Trendline options. A trendline will appear. Using the trendline, we can predict the price of the 1280 square foot home should be around \$245,000. In Excel, open the Data Analysis tools and select Regression. A summary of statistics will appear in a new sheet by default. Notice the intercept and X Variable (slope) values. Plug those values into the above equation, along with the house’s square footage, to get a simple price prediction: The negative intercept tells you where the linear model predicts house price (y) would be when square footage (x) is 0. Solve for y by using the Intercept and X Variable coefficient values: y = -466500 + 555*1280 = -466500 + 710400 = 243900 So there you have it! We just priced a house using a simple linear regression algorithm and calculation. It is that simple. ## 3. Using Python to solve a bigger problem Simple linear regression in Microsoft Excel is useful when you only have a couple variables to work with and a small dataset. However, if you have hundreds of variables and millions of rows of data (like Zillow Group likely had) you will need a tool like Python to manipulate the data and build the multiple linear regression model, since there are two or more X variables. Continuing with the house price example, let’s add a few more variables into the data. I am using a modified copy of the free Kaggle House Pricing data in this example You have two options to get the data: #### Reviewing the Data I’ve reduced the data down to 11 columns of numeric data, from the original 81 columns in the dataset. Columns that store categorical data and strings have been removed since engineering those columns to work with the algorithm is beyond this tutorial. Start by importing dependencies (sandas, Plotly Express, and scikit-learn) and data from the train.csv file. import pandas as pd import plotly.express as px from sklearn.model_selection import train_test_split from sklearn.linear_model import LinearRegressionfrom sklearn.metrics import mean_absolute_error Read the CSV file into a DataFrame. Use .head() to display the first 5 rows and print the shape. Review the column names too. print(train.shape) Notice there are 11 columns and 1431 rows in the train dataframe. Below is a description of the columns: • SalePrice: The property’s sale price in dollars. This is Y, or the target variable that you’re trying to predict . • LotArea: Lot size in square feet • GrLivArea: Above grade (ground) living area square feet • BsmtFullBath: Basement full bathrooms • BsmtHalfBath: Basement half bathrooms • FullBath: Full bathrooms above grade • HalfBath: Half baths above grade • BedroomAbvGr: Number of bedrooms above basement level • KitchenAbvGr: Number of kitchens • Fireplaces: Number of fireplaces • GarageCars: Size of garage in car capacity ## 4. Assumptions of linear regression Linear regression is most useful on data that meets certain assumptions. I’ve already filled in missing values and removed some outliers, but understanding how to review the data to make sure it is a good fit for linear regression is important. Follow this checklist to make sure your data is a good fit for the approach: • The dependent and independent variables should have a linear relationship. • The independent variables are not all highly correlated. Check multicollinearity using a correlation matrix. • The outliers are handled appropriately since they can have a heavy impact on results. Check for outliers using a scatter plot or other charts. • The data follows a multivariate normal distribution. Check for normality using a goodness of fit test. Although I’ve performed some pre-processing on the data, let’s review some techniques for checking and visualizing the data. In general, reviewing data before modeling is called exploratory data analysis Use pandas .describe() to display statistics about the columns. Looking at the statistics can help identify things, like outliers and standard deviation. train.describe() Use Plotly Express histogram to review the distribution of SalePrice. We want it to follow a normal distribution and can use different scaling techniques to transform it if needed. For example, consider using a log transformation on SalePrice since it is a bit skewed. px.histogram(train, x=’SalePrice’) Use a scatter plot to look for outliers and check the relationship between variables. Check variables SalePrice and LotArea for example. Remember, linear regression is sensitive to outliers. px.scatter(train, x=’LotArea’, y=’SalePrice’) If the statsmodels python library is installed, one of the fantastic features of Plotly Express is we can perform simple linear regression in one line of code: #install statsmodels lib if needed #!pip install statsmodels #perform simple linear regression px.scatter(train, x=’LotArea’, y=’SalePrice’, trendline = ‘ols’, trendline_color_override=”red”) Use the scatter_matrix to view multiple scatter plots at the same time. Depending on the number of columns in the dataframe, consider specifying the ones passed into scatter_matrix to make it easier to see. px.scatter_matrix(train[[‘SalePrice’, ‘LotArea’,GrLivArea’]]) Lastly, check the correlations using the .corr() function. This can help us check for multicollinearity. It is common to use a heatmap to visualize correlations. print(train.corr()) px.imshow(train.corr()) ## 5. Building a linear regression model using scikit-learn Creating a linear regression model and generating predictions is only three lines of code, thanks to scikit-learn. However, we must first separate the target variable (SalePrice) from the data. Then we must split the dataframe into a training set and a testing set. Splitting the data into a training and testing set is a best practice in terms of model validation. Use scikit-learn’s train_test_split() function to split the data. #create target variable y = train[‘SalePrice’] #create array of features x = train.drop(columns = ‘SalePrice’) #split the data into train and test sets xtrain, xtest, ytrain, ytest = train_test_split(x, y, test_size=.55, random_state=42) Next, instantiate a linear regression model and fit the data. model = LinearRegression() model.fit(xtrain, ytrain) Now that the model has been trained, it can be used to predict prices. In practice, before pushing a model to production to be used for real predictions, it goes through rigorous evaluation. We can generate the predictions on the xtest dataset and compare them to the ytest values to see how close the predictions are to the real prices. #Predicting the prices pred = model.predict(xtest) #look at the top 5 predictions vs top 5 real print(pred[:5]) ytest[:5] We can see the predictions are close, but far from exact. As data scientists, we want to understand how much error exists in the model, so that we can compare multiple models and select the best one. To better understand the error, we must use a model evaluation technique. ## 6. Evaluating the linear regression model We won’t know how accurate our model is unless we use a method for evaluating the line of best fit. One popular way of evaluating a model is using a technique called Mean Absolute Error (MAE). It is easy to understand because it is the average absolute value between the predicted point and the actual point. #appy mean absolute error mean_absolute_error(y_true=ytest, y_pred=model.predict(xtest)) We can see the MAE is over 26,000 in the test set. A smaller value is better since we want accurate predictions. Other popular evaluation metrics include Mean Squared Error (MSE), Mean Absolute Percentage Error (MAPE), and R-squared. It is common to see multiple evaluation techniques used when evaluating a model because they all measure slightly different things. ## 7. Linear regression modeling code review Congratulations! You’ve just built a multiple linear regression model in Python, used it to predict house prices, and evaluated the model’s accuracy. All in just a few lines of code: #create target variable y = train[‘SalePrice’] #create array of features x = train.drop(columns = ‘SalePrice’) #split the data into train and test sets xtrain, xtest, ytrain, ytest = train_test_split(x, y, test_size=.55, random_state=42) #build the model model = LinearRegression() #fit the model model.fit(xtrain, ytrain) #Predicting the prices pred = model.predict(xtest) #look at the top 5 predictions vs top 5 real print(pred[:5]) print(ytest[:5]) #appy mean absolute error mean_absolute_error(y_true=ytest, y_pred=model.predict(xtest)) ## 8. Conclusion and next steps Linear regression is a useful tool in the data analysis toolbox, and is capable of achieving great results in many use cases. Beyond pricing homes, regression analysis is used in all kinds of tasks, like identifying stock trends, understanding consumer behavior, and analyzing medical research. We just went through two linear regression examples to explain how it works, how to calculate a prediction, and how to visualize the line of best fit. In only a few clicks in Excel or a few lines of code in Python, you can apply linear regression to your data. If you’re interested in learning more about linear regression, you may be interested in checking out our free, 5-day data analytics short course. You could also check out other articles in our series on data analytics: ### What You Should Do Now 1. Get a hands-on introduction to data analytics and carry out your first analysis with our free, self-paced Data Analytics Short Course. 2. Take part in one of our FREE live online data analytics events with industry experts, and read about Azadeh’s journey from school teacher to data analyst. 3. Become a qualified data analyst in just 4-8 months—complete with a job guarantee. 4. This month, we’re offering a partial scholarship worth up to \$1,365 off on all of our career-change programs to the first 100 students who apply 🎉 Book your application call and secure your spot now! #### What is CareerFoundry? CareerFoundry is an online school for people looking to switch to a rewarding career in tech. 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GFG App Open App Browser Continue # Count of sequence of length K in range [1, N] where every element is a multiple of its previous one Given two integers N and K, the task is to find the count of sequences of K elements from the range [1, N] where every element is a multiple of the previous element. Example: Input: N = 4, K = 3 Output: 13 Explanation: The sequences that can be made from the integers 1, 2, 3, 4 having 3 elements are: {1, 1, 1}, {2, 2, 2}, {3, 3, 3}, {4, 4, 4}, {1, 1, 2}, {1, 2, 2}, {1, 2, 4}, {1, 1, 3}, {1, 3, 3}, {1, 1, 4}, {1, 4, 4}, {2, 2, 4}, and {2, 4, 4}. Input: N = 9, K = 5 Output: 111 Approach: The given problem can be solved using recursion with memoization. Follow the below steps to solve the problem: • Create a 2D array dp[][] which stores the memorized states where dp[i][j] represents the count of sequences of length i having j as their first element. • Create a recursive function countSequenceUtil(), that takes the length of the sequence and the starting element as arguments, sets the next element as a multiple of the current element, and recursively calls for the remaining sequence. • Store the answer for the calculated states in the dp[][] array and if for some state the value is already calculated, return it. • Create a function countSequence() which iterates through all the possible starting elements of the sequence and calls the recursive function to calculate the sequences of K elements with that starting element. • Maintain the sum of the calculated count for each starting element in a variable ans which is the required value. Below is the implementation of the above approach: ## C++ `// C++ implementation for the above approach` `#include ` `using` `namespace` `std;` `// Initialize the dp matrix` `int` `static` `dp[1001][1001];` `// Function to find the count of sequences of K` `// elements with first element as m where every` `// element is a multiple of the previous one` `int` `countSequenceUtil(``int` `k, ``int` `m, ``int` `n)` `{` ` ``// Base case` ` ``if` `(k == 1) {` ` ``return` `1;` ` ``}` ` ``// If the value already exists` ` ``// in the DP then return it` ` ``if` `(dp[k][m] != -1) {` ` ``return` `dp[k][m];` ` ``}` ` ``// Variable to store the count` ` ``int` `res = 0;` ` ``for` `(``int` `i = 1; i <= (n / m); i++) {` ` ``// Recursive Call` ` ``res += countSequenceUtil(k - 1,` ` ``m * i, n);` ` ``}` ` ``// Store the calculated` ` ``// answer and return it` ` ``return` `dp[k][m] = res;` `}` `// Function to find count of sequences of K` `// elements in the range [1, n] where every` `// element is a multiple of the previous one` `int` `countSequence(``int` `N, ``int` `K)` `{` ` ``// Initializing all values` ` ``// of dp with -1` ` ``memset``(dp, -1, ``sizeof``(dp));` ` ``// Variable to store` ` ``// the total count` ` ``int` `ans = 0;` ` ``// Iterate from 1 to N` ` ``for` `(``int` `i = 1; i <= N; i++) {` ` ``ans += countSequenceUtil(K, i, N);` ` ``}` ` ``// Return ans` ` ``return` `ans;` `}` `// Driver Code` `int` `main()` `{` ` ``int` `N = 9;` ` ``int` `K = 5;` ` ``cout << countSequence(N, K);` ` ``return` `0;` `}` ## Java `// Java program for the above approach` `import` `java.io.;` `class` `GFG {` ` ` `// Initialize the dp matrix` `static` `int` `dp[][] = ``new` `int``[``1001``][``1001``];` `// Function to find the count of sequences of K` `// elements with first element as m where every` `// element is a multiple of the previous one` `static` `int` `countSequenceUtil(``int` `k, ``int` `m, ``int` `n)` `{` ` ``// Base case` ` ``if` `(k == ``1``) {` ` ``return` `1``;` ` ``}` ` ``// If the value already exists` ` ``// in the DP then return it` ` ``if` `(dp[k][m] != -``1``) {` ` ``return` `dp[k][m];` ` ``}` ` ``// Variable to store the count` ` ``int` `res = ``0``;` ` ``for` `(``int` `i = ``1``; i <= (n / m); i++) {` ` ``// Recursive Call` ` ``res += countSequenceUtil(k - ``1``,` ` ``m i, n);` ` ``}` ` ``// Store the calculated` ` ``// answer and return it` ` ``return` `dp[k][m] = res;` `}` `// Function to find count of sequences of K` `// elements in the range [1, n] where every` `// element is a multiple of the previous one` `static` `int` `countSequence(``int` `N, ``int` `K)` `{` ` ` ` ``// Initializing all values` ` ``// of dp with -1` ` ``for``(``int` `i=``0``;i ## Python3 `# Python implementation for the above approach` `# Initialize the dp matrix` `dp ``=` `[[``-``1` `for` `i ``in` `range``(``1001``)] ``for` `j ``in` `range``(``1001``)]` `# Function to find the count of sequences of K` `# elements with first element as m where every` `# element is a multiple of the previous one` `def` `countSequenceUtil(k, m, n):` ` ``# Base case` ` ``if` `(k ``=``=` `1``):` ` ``return` `1` ` ``# If the value already exists` ` ``# in the DP then return it` ` ``if` `(dp[k][m] !``=` `-``1``):` ` ``return` `dp[k][m]` ` ``# Variable to store the count` ` ``res ``=` `0` ` ``for` `i ``in` `range``(``1``, (n ``/``/` `m) ``+` `1``):` ` ``# Recursive Call` ` ``res ``+``=` `countSequenceUtil(k ``-` `1``,` ` ``m ``` `i, n)` ` ``# Store the calculated` ` ``# answer and return it` ` ``dp[k][m] ``=` `res` ` ``return` `dp[k][m]` `# Function to find count of sequences of K` `# elements in the range [1, n] where every` `# element is a multiple of the previous one` `def` `countSequence(N, K):` ` ``# Variable to store` ` ``# the total count` ` ``ans ``=` `0` ` ``# Iterate from 1 to N` ` ``for` `i ``in` `range``(``1``, N ``+` `1``):` ` ``ans ``+``=` `countSequenceUtil(K, i, N)` ` ``# Return ans` ` ``return` `ans` `# Driver Code` `N ``=` `9` `K ``=` `5` `print``(countSequence(N, K))` `# This code is contributed by Saurabh Jaiswal` ## Javascript `` ## C# `// C# program for the above approach` `using` `System;` `class` `GFG {` ` ``// Initialize the dp matrix` ` ``static` `int``[, ] dp = ``new` `int``[1001, 1001];` ` ``// Function to find the count of sequences of K` ` ``// elements with first element as m where every` ` ``// element is a multiple of the previous one` ` ``static` `int` `countSequenceUtil(``int` `k, ``int` `m, ``int` `n)` ` ``{` ` ` ` ``// Base case` ` ``if` `(k == 1) {` ` ``return` `1;` ` ``}` ` ``// If the value already exists` ` ``// in the DP then return it` ` ``if` `(dp[k, m] != -1) {` ` ``return` `dp[k, m];` ` ``}` ` ``// Variable to store the count` ` ``int` `res = 0;` ` ``for` `(``int` `i = 1; i <= (n / m); i++) {` ` ``// Recursive Call` ` ``res += countSequenceUtil(k - 1, m i, n);` ` ``}` ` ``// Store the calculated` ` ``// answer and return it` ` ``return` `dp[k, m] = res;` ` ``}` ` ``// Function to find count of sequences of K` ` ``// elements in the range [1, n] where every` ` ``// element is a multiple of the previous one` ` ``static` `int` `countSequence(``int` `N, ``int` `K)` ` ``{` ` ``// Initializing all values` ` ``// of dp with -1` ` ``for` `(``int` `i = 0; i < dp.GetLength(0); i++) {` ` ``for` `(``int` `j = 0; j < dp.GetLength(1); j++) {` ` ``dp[i, j] = -1;` ` ``}` ` ``}` ` ``// Variable to store` ` ``// the total count` ` ``int` `ans = 0;` ` ``// Iterate from 1 to N` ` ``for` `(``int` `i = 1; i <= N; i++) {` ` ``ans += countSequenceUtil(K, i, N);` ` ``}` ` ``// Return ans` ` ``return` `ans;` ` ``}` ` ``// Driver Code` ` ``public` `static` `void` `Main(``string``[] args)` ` ``{` ` ``int` `N = 9;` ` ``int` `K = 5;` ` ``Console.WriteLine(countSequence(N, K));` ` ``}` `}` `// This code is contributed by ukasp.` Output `111` Time Complexity: O(NKlog N) Auxiliary Space: O(NK) Efficient approach : Using DP Tabulation method ( Iterative approach ) The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls. Steps to solve this problem : • Create a table to store the solution of the subproblems. • Initialize the table with base cases • Fill up the table iteratively • Return the final solution Implementation : ## C++ `// C++ program for above approach` `#include ` `using` `namespace` `std;` `// Function to find the count of sequences of K` `// elements with first element as m where every` `// element is a multiple of the previous one` `int` `countSequence(``int` `N, ``int` `K)` `{` ` ``// Initialize the dp matrix` ` ``int` `dp[K + 1][N + 1];` ` ``// Base case` ` ``for` `(``int` `i = 1; i <= N; i++) {` ` ``dp[1][i] = 1;` ` ``}` ` ``// Tabulate the dp matrix` ` ``for` `(``int` `k = 2; k <= K; k++) {` ` ``for` `(``int` `m = 1; m <= N; m++) {` ` ``dp[k][m] = 0;` ` ``for` `(``int` `i = 1; i <= (N / m); i++) {` ` ``dp[k][m] += dp[k - 1][m i];` ` ``}` ` ``}` ` ``}` ` ``// Compute the final result` ` ``int` `ans = 0;` ` ``for` `(``int` `i = 1; i <= N; i++) {` ` ``ans += dp[K][i];` ` ``}` ` ``// Return the answer` ` ``return` `ans;` `}` `// Driver Code` `int` `main()` `{` ` ``int` `N = 9;` ` ``int` `K = 5;` ` ``cout << countSequence(N, K);` ` ``return` `0;` `}` `// this code is contributed by bhardwajji` ## Java `import` `java.util.;` `public` `class` `Main {` ` ``// Function to find the count of sequences of K` ` ``// elements with first element as m where every` ` ``// element is a multiple of the previous one` ` ``static` `int` `countSequence(``int` `N, ``int` `K)` ` ``{` ` ``// Initialize the dp matrix` ` ``int``[][] dp = ``new` `int``[K + ``1``][N + ``1``];` ` ``// Base case` ` ``for` `(``int` `i = ``1``; i <= N; i++) {` ` ``dp[``1``][i] = ``1``;` ` ``}` ` ``// Tabulate the dp matrix` ` ``for` `(``int` `k = ``2``; k <= K; k++) {` ` ``for` `(``int` `m = ``1``; m <= N; m++) {` ` ``dp[k][m] = ``0``;` ` ``for` `(``int` `i = ``1``; i <= (N / m); i++) {` ` ``dp[k][m] += dp[k - ``1``][m i];` ` ``}` ` ``}` ` ``}` ` ``// Compute the final result` ` ``int` `ans = ``0``;` ` ``for` `(``int` `i = ``1``; i <= N; i++) {` ` ``ans += dp[K][i];` ` ``}` ` ``// Return the answer` ` ``return` `ans;` ` ``}` ` ``// Driver Code` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``int` `N = ``9``;` ` ``int` `K = ``5``;` ` ``System.out.println(countSequence(N, K));` ` ``}` `}` `// This code is contributed by Gaurav_Arora` ## Python3 `def` `countSequence(N, K): ` ` ``# Initialize the dp matrix` ` ``dp ``=` `[[``0` `for` `i ``in` `range``(N ``+` `1``)] ``for` `j ``in` `range``(K ``+` `1``)] ` ` ``# Base case` ` ``for` `i ``in` `range``(``1``, N ``+` `1``): ` ` ``dp[``1``][i] ``=` `1` ` ` ` ``# Tabulate the dp matrix` ` ``for` `k ``in` `range``(``2``, K ``+` `1``): ` ` ``for` `m ``in` `range``(``1``, N ``+` `1``): ` ` ``dp[k][m] ``=` `0` ` ``for` `i ``in` `range``(``1``, (N ``/``/` `m) ``+` `1``): ` ` ``dp[k][m] ``+``=` `dp[k ``-` `1``][m ``` `i] ` ` ` ` ``# Compute the final result` ` ``ans ``=` `0` ` ``for` `i ``in` `range``(``1``, N ``+` `1``): ` ` ``ans ``+``=` `dp[K][i] ` ` ` ` ``# Return the answer` ` ``return` `ans ` ` ` `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` ` ``N ``=` `9` ` ``K ``=` `5` ` ``print``(countSequence(N, K))` ## Javascript `// Function to find the count of sequences of K` `// elements with first element as m where every` `// element is a multiple of the previous one` `function` `countSequence(N, K) {` ` ``// Initialize the dp matrix` ` ``let dp = ``new` `Array(K + 1);` ` ``for` `(let i = 0; i <= K; i++) {` ` ``dp[i] = ``new` `Array(N + 1);` ` ``}` ` ``// Base case` ` ``for` `(let i = 1; i <= N; i++) {` ` ``dp[1][i] = 1;` ` ``}` ` ``// Tabulate the dp matrix` ` ``for` `(let k = 2; k <= K; k++) {` ` ``for` `(let m = 1; m <= N; m++) {` ` ``dp[k][m] = 0;` ` ``for` `(let i = 1; i <= (N / m); i++) {` ` ``dp[k][m] += dp[k - 1][m i];` ` ``}` ` ``}` ` ``}` ` ``// Compute the final result` ` ``let ans = 0;` ` ``for` `(let i = 1; i <= N; i++) {` ` ``ans += dp[K][i];` ` ``}` ` ``// Return the answer` ` ``return` `ans;` `}` `// Driver Code` `let N = 9;` `let K = 5;` `console.log(countSequence(N, K));` Output `111` Time Complexity: O(NKlog N) Auxiliary Space: O(N*K) My Personal Notes arrow_drop_up	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://origin.geeksforgeeks.org/count-of-sequence-of-length-k-in-range-1-n-where-every-element-is-a-multiple-of-its-previous-one/", "fetch_time": 1686042058000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-23/segments/1685224652494.25/warc/CC-MAIN-20230606082037-20230606112037-00203.warc.gz", "warc_record_offset": 497754801, "warc_record_length": 46705, "token_count": 4893, "char_count": 13480, "score": 3.8125, "int_score": 4, "crawl": "CC-MAIN-2023-23", "snapshot_type": "latest", "language": "en", "language_score": 0.8459096550941467, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00042-of-00064.parquet"}
RANDOM VARIABLE AND PROBABILITY DISTRIBUTION Random variables are numeric description of the outcomes of an event or experiment. We had previously learnt that, an experiment is any process that results in defined outcomes. Now, these outcomes often denoted as ‘x’ is now assuming numerical descriptions. This numerical description is called RANDOM VARIABLE. For example, the number of customers that patronized a restaurant in the first 30days of operation are as follows: 4days we had 25 customers; 10days we had 30 customers; 6days there were 35 customers and 10days we had 40 customers. In this instance, operation of restaurant is the experiment; the number of customers that patronized is the outcome – random variable (x). For these outcomes; we thus, can now get its probability function which we denote as f(x) and thus have a probability distribution. A probability distribution of a variable is a graph, table or formula that specifies the probability associated with each possible value of a variable. Types of random variables However, before we dive into probability distribution proper; it is important that we know the types of random variables we have. Basically, a random variable can either be discrete or continuous; depending on the numerical values assumed. In the instance where numerical value of the variable is countable; or it results from counting is called Discrete Variable. A discrete variable can either be finite or infinite. Meanwhile, where the random variables assume values corresponding to any of the points contained in one or more intervals; such random variables are Continuous Variable. So, a discrete probability distribution has discrete random variables as parameters while continuous probability distribution will have continuous random variables as its parameters. Given the explanation above; let us now prepare the probability distribution of the example of operation of a restaurant we had earlier. The table above is a probability distribution of the number customers that patronized the restaurant in 30 days of operation. We can also use a graph to represent it. Characteristics of a probability distribution: 1. The sum of all variables in a probability distribution is 1. That is to say; the sum of x is equal to 1. 2. The probabilities of the variables can not be negative. It can either be zero or greater than zero. 3. For discrete random variables, the height of the vertical lines which is the y axis, is proportional to the probability of occurrence of the variable. 4. We calculate the probability of a continuous random variable by the area under the probability distribution curve. So, we can obtain the probability or likelihood of x which is between x1 and x2 by the area under the curve between x1 and x2. Expected Value After preparing the probability distribution for a random variable; we also, need to compute the mean or expected value. For a discrete random variable, the expected value (µ or Σ(x)) is the weighted average of all possible values of the random variable; where the weights are the probabilities associated with the values f(x). The expected value of a random variable is the mean of the variables. We still have the variance and standard deviation that we need to know how to calculate. in	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://africanmusings.com/2022/05/27/random-variables/", "fetch_time": 1701369602000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-50/segments/1700679100229.44/warc/CC-MAIN-20231130161920-20231130191920-00757.warc.gz", "warc_record_offset": 107577091, "warc_record_length": 16208, "token_count": 637, "char_count": 3292, "score": 4.65625, "int_score": 5, "crawl": "CC-MAIN-2023-50", "snapshot_type": "latest", "language": "en", "language_score": 0.9259043335914612, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00041-of-00064.parquet"}
How Low Can You Roll? Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in next week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter. ## Riddler Express Inspired by Catriona Shearer (if you don’t know who she is, seriously, check out her puzzles), this week’s Riddler Express is a geometric conundrum: The picture above shows two semicircles. The lighter region (inside the larger semicircle but outside the smaller one) has an area of 7. What’s the area of the darker region? ## Riddler Classic From Ricky Jacobson comes a puzzle of seeing how low you can roll: You are given a fair, unweighted 10-sided die with sides labeled 0 to 9 and a sheet of paper to record your score. (If the very notion of a fair 10-sided die bothers you, and you need to know what sort of three-dimensional solid it is, then forget it — you have a random number generator that gives you an integer value from 0 to 9 with equal probability. Your loss — the die was a collector’s item.) To start the game, you roll the die. Your current “score” is the number shown, divided by 10. For example, if you were to roll a 7, then your score would be 0.7. Then, you keep rolling the die over and over again. Each time you roll, if the digit shown by the die is less than or equal to the last digit of your score, then that roll becomes the new last digit of your score. Otherwise you just go ahead and roll again. The game ends when you roll a zero. For example, suppose you roll the following: 6, 2, 5, 1, 8, 1, 0. After your first roll, your score would be 0.6, After the second, it’s 0.62. You ignore the third roll, since 5 is greater than the current last digit, 2. After the fourth roll, your score is 0.621. You ignore the fifth roll, since 8 is greater than the current last digit, 1. After the sixth roll, your score is 0.6211. And after the seventh roll, the game is over — 0.6211 is your final score. What will be your average final score in this game? ## Solution to last week’s Riddler Express Congratulations to 👏 Harry Elworthy 👏 of San Francisco, California, winner of last week’s Riddler Express. Last week, you had to generate the largest number your could using exactly four threes. Specifically, you were allowed to add, subtract, multiply, divide, exponentiate or write them side-by-side. For example, if you had three nines instead of four threes, the biggest number you could make is 999, which equals 9387,420,489. Many solvers submitted a similar answer — 3333. Remember, when you’re evaluating exponents, order of operations (PEMDAS, anyone?) dictates that you start from the topmost exponent and work your way back down. So 3333 equals 3327, or 37,625,597,484,987. That’s a big number! But solver Erin Seligsohn looked closer at the expression 3327. Sure, that’s big, but it could be made even bigger if that topmost exponent, 27, were replaced by 33. Then you’d have 3333, which equals 35,559,060,566,555,523. Sure enough, that’s the largest number you can make with four threes and the operations you were allowed to use. As solver Kyle Pekosh points out, this number “…has 2,652,345,952,577,569 (about 2.65 quadrillion) digits. If you were to write one digit of that number every second, a decade wouldn’t quite cut it.” I was delighted by how much of Riddler Nation went beyond the list of allowed operations stated in the problem. Many proposed solutions included factorials, while others divided by zero (e.g., 33/(3−3)), which … don’t get me started. Solver Josiah Kollmeyer went so far as to invoke an operation known as tetration. What is tetration? I thought you’d never ask. Just as multiplication is repeated addition, and exponentiation is repeated multiplication, tetration is repeated exponentiation. For example, 2↑↑5 (where the ‘↑↑’ is the sign for tetration) is 22222, an exponential stack of five twos. And so 3↑↑3 equals 3↑↑3↑↑3, which is only about 8 trillion (something you can at least type into your calculator). But what about 3↑↑(3↑↑3)? That’s 33..., with a total of 33, or 27, threes in the stack. That’s an unconscionably large number — and that was just 3↑↑(3↑↑3). If you were to evaluate 3↑↑(3↑↑(3↑↑3)) … your head, and calculator, would just explode. Fortunately for your head, tetration was not in the approved list of operations for this puzzle. ## Solution to last week’s Riddler Classic Congratulations to 👏 Dave Paschal 👏 of Reno, Nevada, winner of last week’s Riddler Classic. Last week, you were in a store with three kinds of candy were being sold: Almond Soys, Butterflingers and Candy Kernels. You wanted to buy at least one candy and at most 100, but you didn’t care precisely how many you got of each or how many you got overall. So you might have bought one of each, or you might have bought 30 Almond Soys, six Butterflingers and 64 Candy Kernels. As long as you had somewhere between one and 100 candies, you left the store happy. But as a member of Riddler Nation, you couldn’t help but wonder: How many distinct ways were there for you to make your candy purchase? There are too many ways to write them all out. But there is an elegant approach that requires us to think beyond the candies. Suppose you want to buy exactly 100 candies, and instead of using a bag to hold them, you line them up on the counter one at a time. Oh, and suppose you also happen to have two dividers in your pocket. You start by purchasing Almond Soys and lining them up, until you put your first divider down on the counter. Then you buy Butterflingers, one at a time, until you put the second divider down. Finally, you buy Candy Kernels until you have 100 candies in total. By changing the positions of the two dividers relative to the rest of the candy, you can specify how many of each type you’re buying. So in order to compute how many ways we can buy 100 candies, we need to know how many ways we can position two dividers within a sequence of 102 objects (the 100 candies plus the two dividers). This can be done using combinatorics, a branch of discrete mathematics. Now that’s just for the case of 100 candies, whereas the problem said you bought at most 100 candies — you could have purchased, 99, 98, 97, and so on, all the way down to one, for a grand total of 100 different cases. Indeed, many solvers broke the problem down into these 100 cases. They solved each case and and tallied up all the ways to buy candy with the aid of a computer, like the team of Ben Finkel, Cullen McAndrews, Zack Smith and Matt Thachet. Solver Aditya Radhakrishnan, meanwhile, impressively found a way to do this using one line of code in the programming language R: sum(unlist(lapply(seq(1, 100), function (x) { factorial(x + 2)/factorial(2)/factorial(x) }))). Solver Gareth McCaughan took the idea of dividers one step further, imagining that there was a fourth candy that you could purchase, which, in an apparent reference to Douglas Adams, he named “Dingo’s Kidneys.” In Gareth’s approach, if you were to ever buy any fewer than 100 candies, then the difference would be made up of these Dingo’s Kidneys. For example, if you bought 30 Almond Soys, 30 Butterflingers and 30 Candy Kernels, then you’d imagine having bought 10 Dingo’s Kidneys, so that the total number of candies remains 100. Then, imagine you have three dividers and exactly 100 candies (as opposed to anywhere from one to 100 candies). All we have to do is choose where the three dividers will be in a sequence of 103 objects (100 candy slots and three dividers). Again, we can use combinatorics to calculate the answer in a single shot — it’s 103 choose 3, or (103·102·101)/(3·2·1). That equals 176,851, which was not the answer. Huh? Oh right, the original riddle said you had to buy at least one candy. Out of our 176,851 scenarios, one of them was the purchase of zero real candies and 100 Dingo’s Kidneys. Excluding that one case reveals the correct answer to be 176,850. Dingo’s Kidneys do sound vile … but I’m sure they’re mostly harmless. ## Want more riddles? Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now! ## Want to submit a riddle? Email Zach Wissner-Gross at [email protected]. ## Footnotes 1. Important small print: For you to be eligible, I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend! Zach Wissner-Gross leads development of math curriculum at Amplify Education and is FiveThirtyEight’s Riddler editor.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://fivethirtyeight.com/features/how-low-can-you-roll/", "fetch_time": 1642402981000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-05/segments/1642320300343.4/warc/CC-MAIN-20220117061125-20220117091125-00048.warc.gz", "warc_record_offset": 316302109, "warc_record_length": 24293, "token_count": 2256, "char_count": 8915, "score": 4.375, "int_score": 4, "crawl": "CC-MAIN-2022-05", "snapshot_type": "longest", "language": "en", "language_score": 0.9102151393890381, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00028-of-00064.parquet"}

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