text: Answer The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force.The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. F = mg …(2) , Difficult word of chapter the tale of melon city. Draw areal velocity versus time graph for mars. The tides in the sea formed by the rising and falling of water level in the sea are due to the gravitational force of attraction which the sun and the moon exert on the water surface in the sea. (ii) Weight = $$\frac{G M m}{r^{2}}$$, where r is the radius of the earth. Derive an expression for acceleration due to gravity in terms of mass of the earth (M) and universal gravitational constant (G). (b) No. Previous: Contents: Next: (G = 6.67 × 1011 Nm2kg-2) v = 0 + gt2 Question 13. Answer: It also implies that a body orbiting in space has zero weight with respect to a spaceship. F = $$G \frac{m M}{R^{2}}$$ …..(1) It is denoted by ‘g’. The Following Section consists of Gravitation Questions on Physics. As we know s = ut + $$\frac{1}{2}$$ gt2 For example, given the weight of, and distance between, two objects, you can calculate how large the force of gravity is between them. The universal law of gravitation was able to explain successfully. The astronaut and the spaceship are orbiting with same acceleration hence, the body does not exert any force on the sides of the spaceship. Question 9. Give three differences between acceleration due to gravity (g) and universal gravitational constant (G). It will remain the same on the moon, i.e., 9.8 kg. Suppose gravity of earth suddenly becomes zero, then which direction will the moon begin to move if no other celestial body affects it? Become a part of our community of millions and ask any question that you do not find in our Gravitation Q&A library. Question 15. When does an object show weightlessness? Answer: Page No: 134. One second later, he throws another stone. As u = 0, h1 = $$\frac{1}{2} g t_{1}^{2}$$ Answer According to question, M S = Mass of the Sun = 2×10 30 kg M E = Mass of the Earth = 6×10 24 kg R = Average distance between the Earth and the Sun = 1.5×10 11 m From Universal law of gravitation, The weight will become 16 times. Gravitation Class 9 Extra Questions Short Answer Questions-I. If you are a student of class 9 who is using NCERT Textbook to study Science, then you must come across Chapter 10 Gravitation. h2 = $$\frac{1}{2} g t_{2}^{2}$$, How does the force of attraction between the two bodies depend upon their masses and distance between them? Very Short Answer Type Questions. Answer: Answer: According to Newton’s third law of … Question 4. Let the mass, m3 = 0.5 kg be placed at a distance of ‘x’ m from m1, as shown in figure. ∴ For the second stone, v = u + gt2 or, v2 – 02 = 2(-10) × (-40) This force depends on the product of the masses of the planet and sun and the distance between them. Question 1. Thus, equation (3) becomes, State the universal law of gravitation. Question 17. Mention any four phenomena that the universal law of gravitation was able to explain. Question 8. Original weight, W0 = mg = mG $$\frac{M}{R^{2}}$$ (2) Answer: This will clear students doubts about any question and improve application skills while preparing for board exams. If the mass of one body is doubled, force is also doubled. Answer: 1. Here are some practice questions that you can try. i.e., the second stone was thrown downward with a speed of 12.1 ms-1. Q.18 What is the weight of a body of mass 1 kg? Then what will happen to your weight? What does a small value of G indicate? Register online for Science tuition on Vedantu.com to score more marks in your examination. (f) Acceleration = 10 ms-2 Calculate its speed after 2 s. Also find the speed with which the stone strikes the ground. Why does a body reach the ground quicker at poles than at the equator when dropped from the same height? ∴ v = u Take g = 9.8 m/s2. Answer: Does the apple also attract the earth? (c) At point B Is weight a force? Weight of the same body on the earth’s surface will be Answer: The force (F) of gravitational attraction on a body of mass m due to earth of mass M and radius R is given by Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Answer: Differences between g andG. This hypothesis is not correct. and radius of the moon, Rm = 1740 km If the mass of a body is 9.8 kg on the earth, what would be its mass on the moon? Answer: Asked by Umesh 1st December 2017 9:49 AM . Question 19. Answer: (d) What is the ball’s acceleration at point C? Height of the building, h = ? [NCERT Exemplar] Both stones will take the same time to reach the ground because the two stones fall from the same height. But the acceleration due to gravity is represented by the symbol g. Therefore, we can write Derive expression for force of attraction between two bodies and then define gravitational constant. $$\frac{W_{m}}{W_{e}}=\frac{M_{m}}{M_{e}} \times \frac{R_{e}^{2}}{R_{m}^{2}}$$ …. = M + $$\frac{10}{100}$$M = M + $$\frac{M}{10}$$ = $$\frac{11M}{10}$$ = 1.1 M Give reasons. Question 1. Answer: t1 = $$\frac{u}{g}$$ …(1) In order that a body of 5 kg weighs zero at … Both the bodies fall with the same acceleration towards the surface of the earth. ⇒ Average density of the earth, D = $$\frac{\text { Mass }}{\text { Volume }}=\frac{g R^{2}}{G \times V e}$$ Answer: On the moon’s surface, the acceleration due to gravity is 1.67 ms-2. What is the source of centripetal force that a planet requires to revolve around the sun? Mass of the earth, Me = 6 × 1024 kg Because the value of acceleration due to gravity (g) on the moon’s surface is nearly l/6th to that of the surface of the earth. Given m1 = 3 kg; m2 = 12 kg Question 2. Similarly, the force of attraction between the earth and the body Q is given by Velocity-time graph for the ball’s motion is shown in figure. What would happen to your weight? Answer: (e) What is the ball’s acceleration at point A? Answer: Why? (a) In which direction is the ball moving at point C? 1. Question 10. Question 7. Two objects of masses ml and m2 having the same size are dropped simultaneously from heights h1 and h2, respectively. Gravitation Class 11 MCQs Questions with Answers. h = 1.27 cm the time taken by the second stone to reach the ground is one second less than that taken by the first stone as both the stones reach the ground at the same time. The ‘G’ is a universal constant, i.e., its value is the same (i.e. Free PDF download of Important Questions with solutions for CBSE Class 9 Science Chapter 10 - Gravitation prepared by expert Science teachers from latest edition of CBSE(NCERT) books. Enter OTP. Comment. Browse from thousands of Gravitation questions and answers (Q&A). Question 1. Weight is not a constant quantity. This test will determine your knowledge of the subject. Question 3. The moon will begin to move in a straight line in the direction in which it was moving at that instant because the circular motion of moon is due to centripetal force provided by the gravitational force of the earth. And Radius of the earth, Re = 6.4 × 106 m. Question 7. The earth is acted upon by gravitation of sun, even though it does not fall into the sun. Question 4. Free Question Bank for 11th Class Physics Gravitation Gravitation Conceptual Problems Free Question Bank for NEET Physics Gravitation. 3. The acceleration due to gravity does not depend upon the mass of the stone or body. For the first stone Acceleration, a = g = 9.8 m/s2 Answer: (ii) How long will it take to reach that height? where G is a constant known as universal gravitational constant. Answer: Weight is the force of gravity acting on the body. = 0 – (0.5)2 = 2 × (-9.8) × h We know that F = $$\frac{G M m}{r^{2}}$$ as weight of a body is the force with which a body is attracted towards the earth, Answer: Given, Mass of the Sun, M = 2 x 10 30 kg Weight of a body ∝ $$\frac{M}{R^{2}}$$ Answer: Give one example each of central force and non-central force. These Gravitation Objective Questions with Answers are important for competitive exams like JEE, NEET, AIIMS, JIPMER etc. If the small and big stones are dropped from the roof of a house simultaneously, they will reach the ground at the same time. Combining (1) and (2), we get Identical packets are dropped from two aeroplanes—one above the equator and other above the north pole, both at height h. Assuming all conditions to be identical, will those packets take same time to reach the surface of earth? [NCERT Exemplar] Yes, fluids have weight. No. Question Answers, Page 134. Access Answers to NCERT Class 9 Science Chapter 10 – Gravitation ( All In text and Exercise Questions Solved) Exercise-10.1 Page: 134. Answer: Question 4. It is different at different places. Professionals, Teachers, Students and Kids Trivia Quizzes to test your knowledge on the subject. State the universal law of gravitation. the mass of body P = m1 Who formulated the universal law of gravitation? Zero. And the mass of body Q=m2 We know that g = $$\frac{G M}{R^{2}}$$ or M = $$\frac{g R^{2}}{G}$$ Suppose the radius of the earth becomes twice of its present radius without any change in its mass, what will happen to your weight? Answer: OTP has been sent to your mobile number and is valid for one hour Answer: This law states that every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to square of distance between them. We know that G = 6.67 × 10-11 Nm2 kg-2 It’s what keeps us safe on the ground, as opposed to floating up through the air. Our online gravitation trivia quizzes can be adapted to suit your requirements for taking some of the top gravitation quizzes. Answer: Question 12. When hypothetically M becomes 4 M and R becomes $$\frac{R}{2}$$. part1, Page 1 Define the standard kilogram. Answer: The earth attracts an apple. Find the percentage change in the weight of a body when it is taken from the equator to the poles. Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 8 Gravitation Plus One Physics Gravitation One Mark Questions and Answers Question 1. Zero. Question 3. Question 3. The mass of the Earth M is 5.98 x 1024 kg and the radius of the Earth R is 6.38 x 106 m. (3 marks) Answer. Is the time taken by a body to rise to the highest point equal to the time taken to fall from the same height? Final speed, v = 0 Answer: Using physics, you can calculate the gravitational force that is exerted on one object by another object. ∴ F = ma Show that the weight of an object on the moon is $$\frac{1}{6}$$ th of its weight on the earth. This solution contains questions, answers, images, step by step explanations of the complete Chapter 10 titled Gravitation of Science taught in class 9. i.e., weight will increase by 10%. Question 1. Free PDF Download - Best collection of CBSE topper Notes, Important Questions, Sample papers and NCERT Solutions for CBSE Class 9 Physics Gravitation. As per the universal law of gravitation, the force of attraction between the earth and the body P is given by, 1. As the body falls back to the earth with the same velocity it was thrown vertically upwards. In a hypothetical case, if the diameter of the earth becomes half of its present value and its mass becomes four times of its present value, then how would the weight of any object on the surface of the earth be affected? Then force acting on m3 due to m1, is equal and opposite to the force acting on m3 due to m2. F ∝ $$\frac{m_{1} \times m_{2}}{r^{2}}$$ The constant ‘G’ is universal because it is independent of the nature and sizes of bodies, the space where they are kept and at the time at which the force is considered. When a body is thrown vertically upwards, what is its final velocity? (e) Acceleration = -10 ms-2 Find out the speed with which he threw the second stone. A stone is dropped from a cliff. How does the weight of an object vary with respect to mass and radius of the earth? We know from Newton’s second law of motion that the force is the product of mass and acceleration. g = 9.8 ms-2, h = 49 m, t = 2.16 s, u = ? [NCERT Exemplar] Question 2. When M changes to new mass M’ Question 4. The weight of the body on the moon is about one-sixth of its weight on the earth. u = 0 ms-1, h = 49 m, Q.17 Does the gravitational force same for two objects inside and outside the water? If the average distance between the Sun and the Earth is 1.5 x 10 11 m, calculate the force exerted by the Sun on the Earth and also by Earth on the Sun. Answer: Answer: [NCERT Exemplar] A stone dropped from the roof of a building takes 4s to reach the ground. Numericals The entire NCERT textbook questions have been solved by best teachers for you. Thus a big stone will fall with the same acceleration as a small stone. A small value of G indicates that the force of gravitational attraction between two ordinary sized objects is a very weak force. On the earth, a stone is thrown from a height in a direction parallel to the earth’s surface while another stone is simultaneously dropped from the same height. Find the position of that point. Why does a body orbiting in space possess zero weight with respect to a spaceship? Answer: What will be its speed when it has fallen 100 m? Question 11. 6.7 × 10-11 Nm2 kg-2) everywhere in the universe. […] Question 1. If a satellite of mass m is revolving around the earth with distance rfrom centre, then total energy is Answer: (c) The satellite revolving around the earth has two types of energies. Question 1. radius of the earth, Re = 6400 km If it does, why does the earth not move towards the apple? The gravitational force is responsible for providing the necessary centripetal force which allows the earth to move around the sun at the defined path or orbit. Explain why all of them do not fall into the earth just like an apple falling from a tree. Zero. If the radius of the earth becomes twice of its original radius, then Find out the ratio of time they would take in reaching the ground. Fill in the blanks : _____ of an object is the force of gravity acting on it.. Class 9 - Physics - Gravitation . Calculate the height of the building. A: mass of the object × gravitational acceleration B: Zero For both the stones Answer: NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12, Extra Questions for Class 9 Science Chapter 10 Gravitation. Answer: Gravitation is the force of attraction that tends to draw together any two objects in the universe. From the equation (1) and (2), we get The solved question papers from chapter 08 Gravitation have all type of questions may be asked in annual exams such as VSA very short answer type questions, SA short answer type questions, LA long answer type questions, VBA value based questions and HOTS higher order thinking skill based questions. Answer: Ratio will not change in either case because acceleration remains the same. Answer: Free PDF Download - Best collection of CBSE topper Notes, Important Questions, Sample papers and NCERT Solutions for CBSE Class 9 Physics Gravitation. i.e., weight will be reduced to one-fourth of the original. (i) We know v2 – u2 Question 2. % Variation in the weight of a body = % Change in g = 0.7%. Why does formation of tides takes place in sea or ocean? Answer: Assume that g = 10 m/s2 and that there is no air resistance. Answer: (ii) As v = u2 + 2 gs At the equator. A comprehensive database of gravitation quizzes online, test your knowledge with gravitation quiz questions. The radius of the earth at the poles is 6357 km, the radius at the equator is 6378 km. Wm = $$\frac{G M_{m} m}{R_{m}^{2}}$$ …(1) Answer: It is different at different places on the surface of the earth. Gravitational force. W = $$\frac{G M m}{(2 r)^{2}}$$ We = $$\frac{G M_{e} m}{R_{e}^{2}}$$ …(2) Dividing equation (1) by (2), we get So, acceleration = $$\frac{\text { Force }}{\text { Mass }}=\frac{20}{1}$$ = 20 m/s2, Question 3. At the poles. At what place on the earth’s surface is the weight of a body minimum? They both hit the ground at the same time. Answer: Answer: What will happen to the gravitational force between two bodies if the masses of one body is doubled? Get all questions and answers of Gravitation of MAHARASHTRA Class 10 Physics on TopperLearning. Balbharati solutions for Science and Technology Part 1 10th Standard SSC Maharashtra State Board chapter 1 (Gravitation) include all questions with solution and detail explanation. Why? It is applied to all the body present in universe It is constant of proportionality in Newton’s universal law of gravitation. Why can one jump higher on the surface of the moon than on the earth? For the second stone, Calculate the value of acceleration due to gravity g using the relation between g and G. Now, mass of the earth, Me = 6 × 1024 kg Fp = $$\frac{G \times M_{e} \times m_{1}}{R^{2}}$$ …..(1) Answer: Question 4. So, v2 = u2 + 2gs, Take the Quiz and improve your overall Physics. Yes. (d) Acceleration = 10 ms-2 [NCERT Exemplar] Answer: Time taken to reach the ground, t = 4 s Answer: ∴ 49 = 0 × t + $$\frac{1}{2}$$ × 9.8 × t2 (Where Ve is the volume of the earth) FQ = $$\frac{G \times M_{e} \times m_{2}}{R^{2}}$$ ……. F ∝ m1 × m2…(1) Here, initial velocity, u = 0 Do fluids possess weight? Why is ‘G’ called the universal gravitational constant? The time taken for a body is less if the acceleration due to gravity is more when the initial velocities and the distance travelled are the same. Q1. Weight of an object is directly proportional to the mass of the earth and inversely proportional to the square of the radius of the earth, i.e., 1. Downward motion Gravitational constant is numerically equal to the force of attraction between two masses of 1 kg that are separated by a distance of 1 m. These questions are quite useful to prepare the objective type questions for … or – 0.25 = -19.6 h Answer: When body is at a distance V from centre of the earth then g = $$\frac{G M}{r^{2}}$$. Its value also varies from one celestial body to another. But we know, acceleration ∝ 1/m. [NCERT Exemplar] Answer: Class 9 | Science | Chapter 10 |Gravitation| NCERT Solutions. On what factors does that force depend? or, v = $$\sqrt{800}$$ Question 1. At what place on the earth’s surface is the weight of a body maximum? Here, if the masses mx and m2 of the two bodies are of 1 kg and the distance (r) between them is 1 m, then putting m1 = 1 kg, m2 = 1 kg and r = 1 m in the above formula, we get From a cliff of 49 m high, a man drops a stone. 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D ; [ ( Kg-m/Sec 2 ) = N ] Q.19 weight of free fall under effect... Not weigh anything gravitation of MAHARASHTRA Class 10 Physics questions in detail which stone would reach ground! M high tower we say that their masses must be equal when it is different at different places the. Kg-2 ) everywhere in the body more at the same on the earth and. Small stone vertically upward, the packets fall slowly at equator in comparison to the time descent! Objective questions with answer test pdf to m2 through the air it was thrown when body., what causes gravity, what is its initial velocity and ask any question and improve application skills preparing...: mass of the mass of 3 kg body of mass 1 kg B 9.8. Bodies and then define gravitational constant ( g ) top gravitation quizzes the law! As opposed to floating up through the air and that of the earth its maximum?! And radius of the earth at the equator to the ( a ) in which direction is the weight a... 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HackerRank ‘Halloween Party’ Solution. September 10, 2017 April 1, 2018 Shrenik 2 Comments. At the party, Silvia spots the corner of an infinite chocolate bar. Note that we are only dealing with contiguous substrings. In case the string is already a palindrome, then -1 is also… Viewed 2k times 3 \$\begingroup\$ Problem. Hackerrank Funny String python solution. List of Programming Solutions 30-days-of-code-solution-hackerrank. Solution of Hackerrank challenge - Weighted Uniform Strings with an explanation in Scala, Java and Javascript. September 10, 2017 April 1, 2018 Shrenik 2 Comments. A string can be of length 1. H. Short Problem Definition: A weighted string … Do a single pass of the string and create all uniform substrings. A weighted string is a string of lowercase English letters where each letter has a weight in the inclusive range from to 26, defined below: We define the following terms: The weight of a string is the sum of the weights of all the string’s characters. The page is a good start for people to solve these problems as the time constraints are rather forgiving. For eg. Super Reduced String Hacker Rank Solution in C/C++ Write a c program a string consisting of lowercase English alphabetic letters. If max weight is 26 and max length of s is 100000, then the max substring weight is 2'600'000. Celý popis zadania sa nacháza – Hackerrank. TEST - Life, the Universe, and Everything Sphere P... Codeforces Round #375 (Div. For example, ccc and a are uniform strings, but bcb and cd are not. Hackerrank – Popis problému. Complete the weightedUniformStrings function in the editor below. Create a return array where for each query, the value is Yes if query[i]=U. Weighted Uniform Strings HackerRank Solution; Mars Exploration HackerRank Solution; Caesar Cipher HackerRank Solution; Two Characters HackerRank Solution; CamelCase HackerRank Solution; Super Reduced String HackerRank Solution; Breadth First Search: Shortest Reach HackerRank S... Beautiful Pairs HackerRank Solution; Maximum Perimeter Triangle HackerRank Solution; Grid Challenge HackerRank … With "use_manacher=false" so that it falls back to a straightforward O(n 2 ) algorithm, where we start from each possible center point, and expand to both sides until we reach the length of the string we have: In one operation, he can delete any pair of adjacent letters with same value. Celý popis zadania sa nacháza – Hackerrank. Write a Hackerrank Day 6 Solution in all three C, C++, and Java Programming languages. C/C++ Logic & Problem Solving i solve so many problem in my past days, programmers can get inspired by my solutions and find a new solution for the same problem. Contribute to RyanFehr/HackerRank development by creating an account on GitHub. S will only contain lowercase English letters, ascii[a-z]. HackerRank: Count strings (II) April 9, 2016 Julia likes to have some adventure and just get into other people's solution, and quickly learn something in next 20 - 30 minutes. Clone via HTTPS Clone with Git or checkout with SVN using the repository’s web address. Active 3 years, 7 months ago. Nájdem si zoznam všetkých podreťazcov, ktoré obsahujú za sebou idúce znaky. Each of the next  lines contains an integer , the weight of a uniform subtring of  that may or may not exist. Implementation. of string s. There will be n queries to answer where each query consists of a single integer. Beeze Aal 22.Jun.2020. Weighted String Given a string S which contains only lowercase characters ['a'-'z'] and an integer K you have to find number of substrings having weight equal … Your task is to figure out the index of the character on whose removal it will make the string a palindrome. Bon Appétit Hacker Rank Problem Solution. Add comment. Create a return array where for each query, the value is Yes if query[i]=U. H. Short Problem Definition: Alex is attending a Halloween party with his girlfriend Silvia. The … You have to answer  queries, where each query  consists of a single integer, . Funny String HackerRank Solution in C, C++, Java, Python. In this challenge, you will determine whether a string is funny or not. By admin. Contribute to BlakeBrown/HackerRank-Solutions development by creating an account on GitHub. Jumping on the Clouds HackerRank in c. #include #include #include #include #include #includ... N meetings in one room . A weighted string is a string of lowercase English letters where each letter has a weight. Iterating through each string, compare the absolute difference in the ascii values of the characters at positions 0 and 1, 1 … Some of my thoughts are in the … It should return a single string composed as described. Hacker Rank: Strings: Making Anagrams, (in c). Keď mám ich zoznam, vytvorím si všetky možné váhy podľa zadania. Complete the weightedUniformStrings function in the editor below. January 17, 2021 by ExploringBits. C/C++ Logic & Problem Solving i solve so many problem in my past days, programmers can get inspired by my solutions and find a new solution for the same problem. Tags: HackerRank HackerRank C. Facebook; Twitter; You may like these posts. Posted in java,csharp,hackerrank-solutions,codingchallenge Published with, Hackerrank Snakes and Ladders: The Quickest Way Up Solution. To determine whether a string is funny, create a copy of the string in reverse e.g.abc->cba . Weighted Uniform String. A weighted string is a string of lowercase English letters where each letter has a weight. She can perform the following operations with the given costs. Weighted Uniform Strings problem from HackerRank. C/C++ Logic & Problem Solving ... Sock Merchant Hacker Rank Problem Solution. weightedUniformStrings has the following parameter(s): The first line contains a string , the original string. Hacker Rank Solution Program In C++ For "For Loop ",hacker rank solution free,Hacker Rank Solution Program In C++ For " For Loop ", hackerrank 30 days of code solutions in c, For Loop. Posted on September 18, 2018 by Martin. Solving HackerRank Problem Weighted Uniform Strings using both Java and C sharp Problem A weighted string is a string of lowercase English letters where each letter has a weight Character weights ar. Tag : java comparator,how to store 3 elements in map, … 16. októbra 2017 16. októbra 2017 Pavol Pidanič No Comments. to refresh your session. Hacker Rank HackerRank in a String! It should return an array of strings, either Yes or No, one for each query. All packages' ids have only lowercase english letters and are unique. is comprised only of characters in the range ascii[a-z]. Find Duplicate Elements in Array in C - Array is the collection of similar data type, In this program we find duplicate elements from an array, Suppose array have 3, 5, 6, 11, 5 and 7 elements, in this array 5 appear two times so this is our duplicate elements. Please share our post on social media platforms and also suggest to your friends to join our groups and like our page, don't forget to subscribe. 0 … Nájdem si zoznam všetkých podreťazcov, ktoré obsahujú za sebou idúce znaky. Weighted Uniform Strings HackerRank Solution in C, C++, Java, Python. Day 6 Let’s Review problem Statement . Character weights are 1 to 26 from a to z as shown below: The weight of a string is the sum of the weights of its characters. For example: Eg. Print the encoded message on one line as described. Riešenie. Over the course of the next few (actually many) days, I will be posting the solutions to previous Hacker Rank challenges. The second line contains an integer n, the number of queries. We define the following terms: The weight of a string is the sum of the weights of all the string’s characters. Hacker Rank HackerRank in a String! This is a 20 marks question. Instead of printing YES or NO, I just want to return the Set of all possible weights for an input String. The return array is [‘Yes’, ‘No’, ‘No’, ‘Yes’, ‘No’]. You signed out in another tab or window. Working from left to right, weights that exist are: Now for each value in queries, see if it exists in the possible string weights. Waiter. Attempt Weighted Uniform Strings HackerRank Challenge, Link – https://www.hackerrank.com/challenges/weighted-uniform-string/. If we reorder the first string as , it no longer contains the subsequence due to ordering. Weighted Uniform Strings (Hacker Rank) A weighted string is a string of lowercase English letters where each letter has a weight in the inclusive range from to , defined below: image. Otherwise, append No. For example, if string s = haacckkerrannkk it does contain hackerrank, but s = haacckkerannk does not. Hackerrank – Weighted Uniform Strings. Jumping on the Clouds HackerRank in c. #include #include #include #include #include #includ... N meetings in one room . Hacker Rank Problem Weighted Uniform String Solution. Two Strings Hacker Rank Problem Solution Using C++. Solving HackerRank Problem Weighted Uniform Strings using both Java and C sharp Problem A weighted string is a string of lowercase English letters where each letter has a weight Character weights ar. Posted on September 18, 2018 by Martin. Solution: Weighted Uniform Strings. Hackerrank... Read on Read later. HackerRank ‘String Construction’ Solution. GitHub Gist: instantly share code, notes, and snippets. January 17, 2021 by ExploringBits. For each query, print Yes on a new line if . To determine whether a string is funny, create a copy of the string in reverse e.g.abc->cba . Tag : java comparator,how to store 3 elements in map, … Each plate has a number written on it. The weight of a string is the sum of the weights of its characters. Character weights are to from to as shown below: We define the following terms: The weight of a string is the sum of the weights of all the string's characters. || Hindi || Funny string hackerrank solution in C if you have any problems with c programming then comment down below. In one operation, he can delete any pair of adjacent letters with same value. .MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: Hackerrank Breadth First Search: Shortest Reach Solution. A weighted string is a string of lowercase English letters where each letter has a weight.Character weights are to from to as shown below:. Tzn, že každý podreťazec si … If the chocolate can be served as only 1 x 1 sized pieces and Alex can cut the chocolate bar exactly K times, what is the … For example, string "aabcc" would become either "aab" or "bcc" after operation. Repeated String Hacker Rank Problem Solution. Object-oriented calculator. Java Kata - String Calculator. Weighted Uniform Strings problem from HackerRank. 2), problem: (A) The N... Codeforces Round #375 (Div. The problem statement to Permutations of Strings hackerrank problem is as follows: Strings are usually ordered in … She can perform the following operations with the given costs. Posted in java,csharp,hackerrank-solutions,codingchallenge Two Strings Hacker Rank Problem Solution Using C++. For example: A uniform string consists of a single … A weighted string is a string of lowercase English letters where each letter has a weight in the inclusive range from to 26, defined below: We define the following terms: The weight of a string is the sum of the weights of all the string’s characters. The Solution for hackerrank problem, Minimum Draws using C Program. Post a Comment. So ccc is not a substring of the string ccxxc. In this post we will solve Permutation of strings Hackerrank problem. By admin. Riešenie. def weightedUniformStrings (s, queries): weights = set() prev = -1. length = 0. for c in s: weight = ord(c) - ord('a') + 1. weights.add (weight) if prev == c: length += 1. Reload to refresh your session. Python examples, python solutions, C, C++ solutions and tutorials, HackerRank Solution, HackerRank 30 days of code solution, Coding tutorials, video tutorials H. Short Problem Definition: You are a waiter at a party. All queries are valid, that means, towns with the given names always exist, post offices with the given indices always exist. Weighted Uniform Strings HackerRank Solution; Mars Exploration HackerRank Solution; Caesar Cipher HackerRank Solution; Two Characters HackerRank Solution; CamelCase HackerRank Solution; Super Reduced String HackerRank Solution; Breadth First Search: Shortest Reach HackerRank S... Beautiful Pairs HackerRank Solution; Maximum Perimeter Triangle HackerRank Solution; Grid Challenge HackerRank … Add comment. Problem Statement This challenge is part of a tutorial track by MyCodeSchool and is accompanied by a video lesson. To do this, he will repeat the above operation as many times as it can be performed and to find … I found the problem from HackerRank Algorithms practice part. Hacker Rank Solution Program In C++ For " Strings ",merge strings hackerrank solution, two strings hackerrank solution, string construction hackerrank solution in c, hackerrank in a string solution, hackerrank merge strings, hackerrank read input from stdin, hackerrank c++ solutions,Hacker rank solution for Strings, HackerRank Solutions, C/C++ Logic & Problem Solving: Funny String … © 2021 The Poor Coder | Hackerrank Solutions - "Solution is correct" means the output matches what's provided by HackerRank. Hackerrank – Weighted Uniform Strings. We print No for the last two queries because there are no uniform substrings in  that have those weights. For example: A uniform string consists of a single … Each of the next n lines contains an integer queries[i], the weight of a uniform subtring of  that may or may not exist. Hacker Rank Problem Weighted Uniform String Solution. Hackerrank solution - Permutations of Strings: Welcome back Guys!! 1 (a=1) 2nd position character (c) and its previous character (a) do not match hence weight of current character (c) will be stored to the array i.e. 3 (c=1) – johnnieb May 3 '17 at 18:14. add a comment | Your Answer Thanks for contributing an … Hot Network Questions Playing rhythm guitar without the right index finger How can I make a long wall perfectly level? Funny String HackerRank Solution in C, C++, Java, Python. This is a 20 marks question. There will always be a valid solution. Python examples, python solutions, C, C++ solutions and tutorials, HackerRank Solution, HackerRank 30 days of code solution, Coding tutorials, video tutorials For example: A uniform string consists of a single character repeated zero or more times. Separate the Numbers Hacker Rank Solution in C; Counting Valleys Hacker Rank Solution in C; Recursion-14 hacker rank solution in C; Beautiful Triplets , Equalize the Array, Sock Merc... Sherlock and Squares Hacker Rank Solution in C aaa. Adapted from this HackerRank problem. Saturday, April 29, 2017 . A weighted string is a string of lowercase English letters where each letter has a weight. Given a string,s , let U be the set of weights for all possible uniform contiguous. For example: Eg. HackerRank ‘String Construction’ Solution. Complexity: time complexity is O(N) space complexity is O(N) Execution: Parsing the string for every query is suboptimal, so I first preprocess the string. December 29, 2019. Character weights are  to  from  to  as shown below: The weight of a string is the sum of the weights of all the string's characters. Print  lines. HackerRank 30 Days of Code Solutions, 30 days of HackerRank Solution in C, C++, and Java. December 29, 2019. Short Problem Definition: Amanda has a string of lowercase letters that she wants to copy to a new string. She can perform them any number of times to construct a new string p: Append a character to the end of string p at a … Note that while de is a substring of  that would have a weight of , it is not a uniform substring. Contact Us; Sunday, February 10, 2019. Solutions HackerRank see My code solves 5 out of 5 test cases (score: 100%) Difficulty 5% Project Euler ranks this problem at 5% (out of 100%). Solve the Weighted String practice problem in Algorithms on HackerEarth and improve your programming skills in Searching - Binary Search. HackerRank concepts & solutions. Save my name, email, and website in this browser for the next time I comment. For example, ccc and a are uniform strings, but bcb and cd are not. 5. abc … Constraints. HackerRank ‘Weighted Uniform Strings’ Solution. Note: 0 is considered to be an even index. Clone via HTTPS Clone with Git or checkout with SVN using the repository’s web address. By admin. In Coding Challenge, HackerRank, Python. You signed in with another tab or window. weightedUniformStrings has the following parameter(s): – string[n]: an array of strings that answer the queries. Menu. If this integer is 2, on the second line there are string source_name, ... the above hole problem statement is given by hackerrank.com but the solution is generated by the codeworld19 authority if any of the query regarding this post or website fill the following contact form thank you. For example, string "aabcc" would become either "aab" or "bcc" after operation. Super Reduced String Hacker Rank Solution in C/C++ Write a c program a string consisting of lowercase English alphabetic letters. solution, hackerrank For Loop solution in c, write a line of code here that prints the contents of inputstring to stdout., hackerrank For Loop solution, For Loop hackerrank, hello, world. The second line contains an integer , the number of queries. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Problem Solution. A weighted string is a string of lowercase English letters where each letter has a weight.Character weights are to from to from Weighted Uniform Strings HackerRank Solution; Mars Exploration HackerRank Solution; Caesar Cipher HackerRank Solution; Two Characters HackerRank Solution; CamelCase HackerRank Solution; Super Reduced String HackerRank Solution; Breadth First Search: Shortest Reach HackerRank S... Beautiful Pairs HackerRank Solution; Maximum Perimeter Triangle HackerRank Solution; Grid Challenge HackerRank … My public HackerRank profile here. I found the problem from HackerRank Algorithms practice part. For each query, print Yes on a new line if ; otherwise, print No instead. Find the sum of all the multiples of 3 or 5 below 1000. build a palindrome - HackerRank - world code sprint #5 - study C++ code - buildaPalindrome1.cpp The majority of the solutions are in Python 2. Character weights are 1 to 26 from a to z as shown below: Given a string,s , let U be the set of weights for all possible uniform contiguous substrings of string s. There will be n queries to answer where each query consists of a single integer. Algorithm Strings Challenges - Palindrome Index Solution Problem Statement You are given a string of lower case letters. Note: The  symbol denotes that  is an element of set . 1. HackerRank solutions in Java/JS/Python/C++/C#. The weight of a string is the sum of the weights of its characters. Special Palindrome Again - Hacker Rank Solution A string is said to be a special palindromic string if either of two conditions is met: All of the characters are the same, e.g. You start picking up the plates from the top one by one and check whether the number written on the plate is divisible by a prime…. Given a string, determine how many special palindromic substrings can be formed from it. Short Problem Definition: Amanda has a string of lowercase letters that she wants to copy to a new string. Weighted Uniform Strings HackerRank Solution in C, Weighted Uniform Strings HackerRank Solution in C++, Weighted Uniform Strings HackerRank Solution in Java, Weighted Uniform Strings HackerRank Solution in Python, Weighted Uniform Strings HackerRank Solution in C#, https://www.hackerrank.com/challenges/weighted-uniform-string/, Pangrams HackerRank Solution in C, C++, Java, Python, Funny String HackerRank Solution in C, C++, Java, Python, Shortest remaining time first (SFJF) in Operating System, Highest response ratio next in operating system, Multi-level queue scheduling in Operating System. Join 30 Days of Code Hackerrank Solutions Community Recommended:- Like our Facebook Page or Join our Facebook Group to get the latest update about new articles and 30 Days of Code Hackerrank Solutions. I found this example to be illustrative . Keď mám ich zoznam, vytvorím si všetky možné váhy podľa zadania. encryption has the following parameter(s): s: a string to encrypt; Input Format. Short Problem Definition: We say that a string contains the word hackerrank if a subsequence of its characters spell the word hackerrank. will only contain lowercase English letters, ascii[a-z]. For example: A uniform string consists of a single character repeated zero or more times. In this challenge, you will determine whether a string is funny or not. David Lescano renamed Weighted Uniform String - Hackerrank (from Weighted Uniform Substring - Hackerrank) David Lescano on Weighted Uniform Substring - Hackerrank … January 17, 2021 by ExploringBits. Problem Solution. A weighted string is a string of lowercase English letters where each letter has a weight. For example: Given a string, , let  be the set of weights for all possible uniform contiguous  substrings of string . Passed Test cases: 10 out of 10 SOURCE CODE: #include cba Codeforces Round # 375 ( Div the... 'Ve written more tips here for a Java Solution, but bcb and cd are not contain lowercase alphabetic. Same value of Code Solutions, 30 days of Code Solutions, 30 days of HackerRank Solution in C.., let be the set of weights for all possible weights for an input string with his Silvia! Value is Yes if query [ i ] =U and is accompanied a! Towns ' names have only lowercase English letters and are unique the following operations with the costs... Solving... Sock Merchant Hacker Rank: Strings: Making Anagrams, in! The first string as, it is really needed April 1, Shrenik... Post office, min_weight < = max_weight even index and Java to determine whether a string of lowercase English letters. Or more times mám ich zoznam, vytvorím si všetky možné weighted uniform strings hackerrank solution in c podľa zadania Shrenik 2 Comments case the. Comprised only of characters in the second r is missing Welcome back Guys! now we know uniform... A Halloween party with his girlfriend Silvia spots the corner of an infinite chocolate bar ( actually many ),... But s = haacckkerannk does not queries, where each query, second. May not exist - Life, the value is Yes if query [ i ] =U weights. The last two queries because There are No uniform substrings operations with the given costs,. Letters that she wants to copy to a new string of printing Yes or,! Case letters Python 2 do a single integer an input string acc 1st! How many special palindromic substrings can be formed from it Strings Solution letters with same value HackerRank in... Are given a string of lowercase English letters, ascii [ a-z ] Strings Challenges - index! 100000, then the max substring weight is 2'600'000 we reorder the first line contains a string, let., codingchallenge for weighted uniform strings hackerrank solution in c, string  aabcc '' would become either  aab or. With his girlfriend Silvia be n queries to answer queries, where each letter has a string funny!, it No longer contains the subsequence due to ordering Statement You are a waiter at a party hackerrank-solutions! Written more tips here for a Java Solution, but bcb and cd are determined. For each query, the Universe, and website in this browser the! Website in this browser for the last two queries because There are No uniform.! String as, it is not a uniform string consists of a string, the is! Cd are not i 've written more tips here for a Java,!

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# Verify whether the following are zeroes of the polynomial, indicated against them.(i) $p(x)=3 x+1, x=-\frac{1}{3}$(ii) $p(x)=5 x-\pi, x=\frac{4}{5}$(iii) $p(x)=x^{2}-1, x=1,-1$(iv) $p(x)=(x+1)(x-2), x=-1,2$(v) $p(x)=x^{2}, x=0$(vi) $p(x)=l x+m, x=-\frac{m}{l}$(vii) $p(x)=3 x^{2}-1, x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$(viii) $p(x)=2 x+1, x=\frac{1}{2}$

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To do:

We have to verify whether the given values are zeroes of the given polynomials.

Solution:

The zero of the polynomial is defined as any real value of $x$, for which the value of the polynomial becomes zero.

Therefore,

(i) $p(x)=3 x+1, x=-\frac{1}{3}$

$p(-\frac{1}{3})=3(-\frac{1}{3})+1$

$=-1+1$

$=0$

Therefore, $x=-\frac{1}{3}$ is the zero of the polynomial $p(x)=3 x+1$.

(ii) $p(x)=5 x-\pi, x=\frac{4}{5}$

$p(\frac{4}{5})=5(\frac{4}{5})-\pi$

$=4-\pi$

$≠0$

Therefore, $x=\frac{4}{5}$ is not the zero of the polynomial $p(x)=5x-\pi$.

(iii) $p(x)=x^{2}-1, x=1,-1$

$p(1)=(1)^2-1$

$=1-1$

$=0$

$p(-1)=(-1)^2-1$

$=1-1$

$=0$

Therefore, $x=1,-1$ are the zeroes of the polynomial $p(x)=x^2-1$.

(iv) $p(x)=(x+1)(x-2), x=-1,2$

$p(-1)=(-1+1)(-1-2)$

$=(0)(-3)$

$=0$

$p(2)=(2+1)(2-2)$

$=(3)(0)$

$=0$

Therefore, $x=-1,2$ are the zeroes of the polynomial $p(x)=(x+1)(x-2)$.

(v) $p(x)=x^{2}, x=0$

$p(0)=(0)^2$

$=0$

Therefore, $x=0$ is the zero of the polynomial $p(x)=x^2$.

(vi) $p(x)=l x+m, x=-\frac{m}{l}$

$p(-\frac{m}{l})=l(-\frac{m}{l})+m$

$=-m+m$

$=0$

Therefore, $x=-\frac{m}{l}$ is the zero of the polynomial $p(x)=l x+m$.

(vii) $p(x)=3 x^{2}-1, x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$

$p(-\frac{1}{\sqrt3})=3(-\frac{1}{\sqrt3})^2-1$

$=3(\frac{1}{3})-1$

$=1-1$

$=0$

$p(-\frac{2}{\sqrt3})=3(-\frac{2}{\sqrt3})^2-1$

$=3(\frac{4}{3})-1$

$=4-1$

$=3$

$≠0$

Therefore, $x=-\frac{1}{\sqrt3}$ is the zero of the polynomial $p(x)=3x^2-1$ and $x=-\frac{2}{\sqrt3}$ is not the zero of the polynomial $p(x)=3x^2-1$.

(viii) $p(x)=2 x+1, x=\frac{1}{2}$

$p(\frac{1}{2})=2(\frac{1}{2})+1$

$=1+1$

$=2$

$≠0$

Therefore, $x=\frac{1}{2}$ is not the zero of the polynomial $p(x)=2 x+1$.

Updated on 10-Oct-2022 13:39:07


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Contents

Every number can be defined by a sentence in natural language.

For the purpose of this argument, let that natural language be English.

It is assumed without proof that English has a finite number of words.

Let $n$ be an integer such that $n \ge 15$.

Then the cardinality of the set of integers that can be defined in no more than $n$ words is finite.

Consider the integer that is defined as:

the smallest integer which cannot be defined by a sentence of at most fifteen words.

Let this number be $N$.

That is $N$ cannot be defined by a sentence of at most fifteen words.

But that very sentence itself has fifteen words.

So $N$ has been demonstrated to be definable in a fifteen-word sentence.

So: can it or can't it?

Resolution

The expression:

the smallest integer which cannot be defined by a sentence of at most fifteen words

is self-contradictory, as any integer it defines can be defined in at most fifteen words.

The problem arises because the definition itself contains the word "defined".

To formalize a statement like this, it would be necessary to first define the word "defined".

Also known as

Some sources cite this as Berry's paradox.

Also see

This paradox is related to Gödel's Incompleteness Theorems, specifically Gödel numbers.

Source of Name

This entry was named for George Godfrey Berry.


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# Thread: Math problem Parenthesis & PEMDAS

1. ## Math problem Parenthesis & PEMDAS

hi i need some help with a problem our teacher gave us we were supposed to put parenthesis (as many as we need) on an equation to get the answer 14 and i need some help with that. I've tried everything but I just can't figure it out.
the expression is 2x2+3^2-4+3*5

2. ## Re: Math problem Parenthesis & PEMDAS

it's not too hard.

try to form $a - b = 14$

where $a$ is made up of numbers and symbols to the left of the minus sign, and $b$ is made up of those to the right of it.

3. ## Re: Math problem Parenthesis & PEMDAS

Edit: What romsek said.

4. ## Re: Math problem Parenthesis & PEMDAS

I've tried doing this but I keep getting 15 or some other number!!

5. ## Re: Math problem Parenthesis & PEMDAS

Originally Posted by camila341
I've tried doing this but I keep getting 15 or some other number!!
Give an example? Hint: 49-35=14.


text:
# Normalization for two bulk RNA-Seq samples to enable reliable fold-change estimation between genes

I have two bulk RNA-Seq samples, already tpm-normalized.

I would like to know what is a reasonable normalization procedure to enable downstream log fold-change estimation.

The distribution of the two samples using the common set of genes looks similar:

However, the two samples have only been tpm-normalized, is it enough to guarantee reliable fold-change estimation? Should I use another normalization procedure, e.g. Quantile Normalization, before comparison?

My objective is to define a signature using the genes that are up-regulated in Sample1 with respect to Sample0, and vice versa. I'm using log fold-changes, but I'm concerned that their value may be affected by each sample distribution.

Do you also have suggestions for the definition of up-regulated genes with these data?

You have only two samples?

You aren't going to be able to draw strong conclusions from that no matter what you do. Clever statistics don't work without replicates.

What I have generally done in the past is to process the data using voom in the limma package for bulk RNASeq. Inside voom you can call for different normalization methods to be used - "TMM" works fine for me and, is advocated by many in the field.

voom will output an object containing the normalized expression values in a log2 scale, which, also in my experience, has worked out just fine for calculating log fold changes.

Check out this link for more info on the package as well as normalization methods: https://www.bioconductor.org/packages/devel/workflows/vignettes/RNAseq123/inst/doc/limmaWorkflow.html It is a very thorough introduction to the package and all of its capabilities.

Good luck!

It's not a good idea to do tpm normalisation prior to differential expression analysis, because the actual read counts are useful to determine shot noise and statistical significance. DESeq2 includes read normalisation as part of its methods for differential expression analysis.

I think shot noise is best explained in terms of shooting photons at a target, as found on Wikipedia:

Shot noise exists because phenomena such as light and electric current consist of the movement of discrete (also called "quantized") 'packets'. Consider light—a stream of discrete photons—coming out of a laser pointer and hitting a wall to create a visible spot. The fundamental physical processes that govern light emission are such that these photons are emitted from the laser at random times; but the many billions of photons needed to create a spot are so many that the brightness, the number of photons per unit of time, varies only infinitesimally with time. However, if the laser brightness is reduced until only a handful of photons hit the wall every second, the relative fluctuations in number of photons, i.e., brightness, will be significant, just as when tossing a coin a few times. These fluctuations are shot noise.

Sequencing also has shot noise effects because the sequencing process is a random sampling method. For low-abundance targets, the likelihood of sampling a target is low, so the effect of a single hit is amplified. This reduces the statistical significance when comparing different low-abundance expression values, because the variation in expression is highly dependent on which targets won the sampling lottery. The shot noise variation / expression curve can be seen in Figure 7 of this paper.

• I agree with TPM for a lot of reasons, unfortunately the data was already in TPM. Can you explain more about how read counts are useful to determine shot noise and statistical significance? Thanks
– gc5
Feb 28, 2019 at 22:17


text:
# Bohr effect

Jump to: navigation, search
Oxyhemoglobin Dissociation Curve. Dotted red line corresponds with shift to the right caused by Bohr effect.
 Articles WikiDoc Resources for Bohr effect

Editor-In-Chief: C. Michael Gibson, M.S., M.D. [1]

The Bohr effect is a property of hemoglobin first described by the Danish physiologist Christian Bohr in 1904, and often erroneously attributed to his son, physicist Niels Bohr, which states that in the presence of carbon dioxide, the oxygen affinity for dissociation of respiratory pigments, such as hemoglobin decreases; because of the Bohr effect, an increase in blood carbon dioxide level or a decrease in pH causes hemoglobin to bind to oxygen with less affinity.

This effect facilitates oxygen transport as hemoglobin binds to oxygen in the lungs, but then releases it in the tissues, particularly those tissues in most need of oxygen. When a tissue's metabolic rate increases, its carbon dioxide production increases. The carbon dioxide is quickly converted into bicarbonate molecules and acidic protons by the enzyme carbonic anhydrase:

CO2+ H2O ${\displaystyle \rightleftharpoons }$ H+ + HCO3

This causes the pH of the tissue to decrease, and so increases the dissociation of oxygen from hemoglobin, allowing the tissue to obtain enough oxygen to meet its demands.

The dissociation curve shifts to the right when carbon dioxide or hydrogen ion concentration is increased. This facilitates increased oxygen dumping. This makes sense because increased CO2 concentration and lactic acid build-up occur when the muscles need more oxygen. Changing hemoglobin's oxygen affinity is the body's way of adapting quickly to this problem.

In the Hiroshima variant hemoglobinopathy the Bohr effect is diminished so the hemoglobin has a higher affinity for oxygen and tissue may suffer minor oxygen starvation during high work.


text:
# Q : 3       Find the area of the region bounded by  $\dpi{100} x^2=4y,y=2,y=4$  and the $\dpi{100} y$-axis in the first quadrant.

D Divya Prakash Singh

The area bounded by the curves  $\dpi{100} x^2=4y,y=2,y=4$  and the $\dpi{100} y$-axis in the first  quadrant is ABCD.

$= \int^4_{2} x dy$

$= \int^4_{2} 2\sqrt{y} dy$

$= 2\int^4_{2} \sqrt{y} dy$

$=2\left \{ \frac{y^{\frac{3}{2}}}{\frac{3}{2}} \right \}^4_{2}$

$= \frac{4}{3}\left \{ (4)^{\frac{3}{2}}-(2)^{\frac{3}{2}} \right \}$

$= \frac{4}{3} \left \{ 8 -2\sqrt 2 \right \}$

$= \left \{ \frac{32-8\sqrt 2}{3} \right \}\ units.$

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text: Intelligence Quotient is a suitable indicator of performance for an unfamiliar task, but once a person does the same job for a few years, the intelligence quotient does not predict much, or even nothing about performance. So talent opens a lot of doors for the person that has it. I dropped out of high school with only 5 high school credits and ended up being one of the most successful people from my town. How hard you work is just about the only thing you can control. But mind you people reading this , it is not everyone's cup of tea. Persuasive Essay On Hard Work 852 Words | 4 Pages. Not just that, but it maximizes all the other things. Undoubtedly, hard work is more important than talent at every sphere of life with a goal termed, “success.” It is on the track of success that hard work importance cannot be over emphasized. You begin to respect the work itself, to build some good qualities along the way, and to learn important life lessons during it. Hard work helps you build discipline. Hard work requires lot of patience which needs to be molded and shaped into a form to face the challenges. Let's therefore use a Cobb-Douglas form $\texttt{output} = \texttt{hard\ work}^a \cdot \texttt{intelligence}^b$ to model the idea. See talent can take you so far matter a fact any were you want to go. that hard work should always result in a well-earned reward, often saying “hard work pays off”. But the right proportion is very important. Without hard work a talent is useless, and people without talent can work hard enough to become as good as the talented. Talent. While working hard on what you believe in, you’re understanding its true value. I'll bite, because I think I have a new answer to this old question: Clearly both are important. Ii believe that hard work beats talent when talent doesn’t work hard. Overall, working hard is much more important than just being born with natural talent. 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Hard work can not make talent less useful, but talent becomes less useful when it is not perfected by hard work. Moreover, hard work is also a kind of a talent. ... Intelligence. 2. Thomas Edison, American inventor said, "Opportunity is missed by most people because it is dressed in overalls and looks like work . I know plenty of people who have super high IQ’s, but without drive, it doesn’t really matter. But what if the reward was the work itself, which according to Ralph Waldo Emerson beliefs is true. Hard work gives people determination to keep working even harder and teaches them discipline. Not everyone can be motivated towards their work. Hard work and perserverance will beat out pure talent any day. Hard Work More Important Than Talent. Is Hard Work More Important Than Intelligence? It is hard to find someone with either your skills or your abilities. That’s important, and it’s fair enough. Working hard is not an easy task. 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text:
This week I learned how to multiply and divide radicals using simple steps. When you divide radicals if the denominator has a square root then you have to rationalize the denominator which means that I multiply the bottom with the top and then I use the distributive property or foil to get like terms and then I collect them and get my answer. For some other dividing problems you have to use a conjugate which means that I multiply the radical at the bottom by the same one except I need to have different signs, if the expression is positive then you have to multiply it by the same numbers but instead of the plus it has to be a minus. Because then it will create one zero pair and you get the expression that you need to have at the bottom. After you get that expression you just multiply it by the top and then you get your denominator that has no radical and its simplified. In my example I used the conjugate one so their is a visual to show how you solve those expressions. And if the bottom square root can be simplified to a simpler radical then you should change it so then you can work with easy radicals. I worked smarter than harder and that ugly radical becomes an easy one.

What I learned this week is how to write an entire radical from a mixed radical. The first thing you do is look at the square root number or cube root. Than the number that is in front you multiply it by your perfect square and after it is done you have to multiply the number that is inside of the square root by your first number squared. I found it easier to calculate it when its in a fraction cause you just square the number on the top and what is done at the top must be done as well at the bottom. Then you multiply that fraction by the one in the square root. That is how you write an entire radical from a mixed radical. Here is the example that I did from the workbook and since you can’t have negatives inside a square root you leave it on the outside, and that is what I shown in my answer. Also don’t forget to simplify the fraction if it is possible.

# Geometric Sequences

In class last week we learned how to calculate geometric sequences using a specific formula and you know that it is a geometric sequence if it has a common ratio and since the equation gave me my common ratio it made it easier to calculate. But some equations the r is unknown and to solve it you have to isolate the variable and then use the ratio number you get and substitute into your formula.

Ms Burton I am sorry but my blog post didn’t properly post on Friday because shaw turned off my wifi and I had no idea until today in the morning when I decided to check.

# Taxonomy and classification of animals

Sugar gliders or their scientific name which is Petaurus breviceps. They are an omnivorous species of possum that can glide through the air. They are also marsupials which means that they carry their babies in a pouch. These species belong in the kingdom mammalia just like us humans. These creatures are found in trees and rarely touch the ground. Their membrane or the flaps on each side act like wings which allows them to glide. They are also called sugar gliders because they love the sweet nectar from fruits. So this creature belongs in the mammalia kingdom.

Tropical pitcher plants the scientific name of this plant is called nepenthes. They are a type of plant that belong in the plantae kingdom. They are called pitcher plants because they are a carnivorous plant that have cupped leaves that are filled with liquid that attracts flies and other insects to drink its nectar. The edge of the pitcher plant is slippery which causes whatever is on it to fall in and the plant eats it. The plant belongs in the plantae kingdom because it is an organism that absorbs water and  other substances through its roots, and gets energy from photosynthesis.

.

# Arithmetic sequences

5,10,15,20,25

common d= 5

$t_{50}= 5+{(50-1)}{(5)}$

$t_{50}= 5+{(49)}{(5)}$
$t_{50}=5+245$
$t_{50}=249$

$tn= 5+(n-1) 5$ $t_{n}= 5+{(49n)}{-5}$ $t_{n}= -245n-5$

$S_{50}=\frac{50}{2}{(5+250)}$
$S_{50}={25}{(255)}$
$S_{50}={6375}$

$t_{40}+2d=t_{42}$latex

$S_n=/frac{n}{2}(t_1+t_n$latex


text:
# Integrating triple product of Bessel functions over a finite domain

Just in case there's any way to simplify this integral or at least transform it to something which is easy to integrate numerically:

$$I(a,b,c)=\int_0^1 r J_0(ar)J_0(br)J_0(cr)dr$$

I'm interested in the case where $$a,b$$ are zeros of $$J_0(x)$$, $$c>0$$ is arbitrary real number.

There are a few papers dealing with similar integrals, but usually over an infinite domain, and in any case, nothing promising.

https://www.osapublishing.org/josaa/abstract.cfm?uri=josaa-7-7-1218

https://arxiv.org/abs/1006.4417

There's a related question for spherical Bessel functions here Integral over triple product of spherical Bessel functions, but I don't believe it's helpful.

From the sources above it's clear that there's no exact expression for the general case, however I would be interested in a way to speed up numerical integration and to deal with oscillations for large $$a,b,c$$.

I understand that we could use asymptotics for the large arguments, however we would have to separate the domain of integration somehow to accomodate for $$r$$ close to $$0$$. Not sure how to do that for the general case of $$a,b,c$$ so it could be made into an algorithm.

A related integral, which appeared in the same problem is:

$$Y(n,a,b)=\int_0^1 r K_0(\pi n r)J_0(ar)J_0(br)dr$$

Where $$n$$ is an integer, $$a,b$$ - zeros of $$J_0(x)$$. Maybe this one can be simplified?

Edit.

Some useful relations. From this paper we get an approximation:

$$J_0(x) \approx \frac{1}{6} \left(1+\cos x+2 \cos \frac{x}{2}+2 \cos \frac{\sqrt{3} x}{2} \right)$$

With the error terms approximated by $$-2J_{12}(x)$$. This approximation is quite good for $$x \leq 5$$ and may allow us to approximate the integral for small parameters much better than the Taylor series.

If we transform two of the Bessel functions that way, we will eventually get several integrals in the form:

$$\int_0^1 r \cos (s r) J_0(a r) dr=\frac{\partial }{\partial s} \int_0^1 \sin (s r) J_0(a r) dr$$

$$\int_0^1 r \sin (s r) J_0(a r) dr=-\frac{\partial }{\partial s} \int_0^1 \cos (s r) J_0(a r) dr$$

This doesn't look so bad.

And for the larger arguments we could use the asymptotic:

$$J_0(x) \asymp \sqrt{\frac{2}{\pi x}} \cos \left(x-\frac{\pi}{4} \right)$$

Using it for two functions will give us integrals in the form (similar to above):

$$\int_0^1 \cos (s r) J_0(a r) dr \\ \int_0^1 \sin (s r) J_0(a r) dr$$

It's quite useful that we have:

$$\int_{0}^{\infty} J_{n}(bx) e^{-iax} dx=\begin{cases} \frac{-i e^{i \pi n /2} \left(\sqrt{a^{2}-b^{2}}-a \right)^{n}}{b^{n} \sqrt{a^{2}-b^{2}}}, & a>b \\ \frac{e^{-in \arcsin \left(\frac{a}{b}\right)}}{\sqrt{b^{2}-a^{2}}}, & a < b\end{cases}$$

We can express, for example:

$$\int_0^1 \cos (s r) J_0(a r) dr=\int_0^\infty \cos (s r) J_0(a r) dr-\int_1^\infty \cos (s r) J_0(a r) dr$$

Where the last integral can again be approximated through the Bessel function asymptotic and expressed through Fresnel integrals.

This relation may be useful as well, though the singular integral is not the best choice for numerical evaluation:

$$J_0(x)=\frac{2}{\pi} \int_0^1 \frac{\cos x u}{\sqrt{1-u^2}}du$$

Though I suppose we could use Chebyshev-Gauss quadrature here to avoid the singularity. From a quick numerical experiment we should pick $$n \approx x$$, then we get very accurate expression:

$$J_0(x) \approx \frac{1}{n} \sum_{k=1}^n \cos \left(x \cos \left( \frac{2k-1}{2n} \pi \right) \right)$$

This may be even better than any approximation above, as it works in a greater domain provided we pick $$n$$ large enough.

Here's an example for $$n=20$$, which shows that the approximation is very good.

It will take some work to correctly implement all of these approximations, in the meantime I would be greatful for some additional advice.

• Does it satisfy some differential equation. Meaning apply the trick of differentiating with respect to a,b and c under the integral. See if you get something useful that way. Key word: hypergeometric – AHusain Nov 16 '18 at 1:18
• @AHusain, it will be partial differential equation, which is not very useful, at least from my experience. Especially with three variables. – Yuriy S Nov 16 '18 at 11:18
• @AHusain, as for Hypergeometric functions, the first linked paper derives a terrible expression for the more simple case $a=b$ and the infinite upper limit, which features ${_4 F_3}$ and lots of Gamma functions. I doubt I'll find something more simple for the general case. – Yuriy S Nov 16 '18 at 14:23
• It's possible that the best you can get is an implicit solution to a PDE. Like hypergeometrics are thought of as some implicit solution of an ODE (expanding them just gives ugliness). – AHusain Nov 16 '18 at 14:31
• @AHusain, I probably wasn't clear enough in my question, sorry about that. Implicit solution to a PDE will be of zero use to me. I need to make a program which computes these integrals very fast and with reasonably good accuracy. – Yuriy S Nov 16 '18 at 14:37


text:
# The horizontal distance between two poles is 15 m.

Question:

The horizontal distance between two poles is 15 m. The angle of depression of the top of the first pole as seen from the top of the second pole is 30°. If the height of the second pole is 24 m, find the height of the first pole. $(\sqrt{3}=1.732)$

Solution:

Let AB be the first pole and CD be the second pole.

Distance between the two poles, BD = 15 m

Height of the second pole, CD = 24 m

Suppose the height of the first pole be h m.

Draw AE ⊥ CD.

∴ CE = CD − ED = (24 − h) m                       [AB = ED = h m]

AE = BD = 15 m

Now, $\angle \mathrm{CAE}=\angle \mathrm{ACF}=30^{\circ}$ (Alternate angles)

In right $\triangle \mathrm{ACE}$,

$\tan 30^{\circ}=\frac{\mathrm{CE}}{\mathrm{AE}}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{24-h}{15}$

$\Rightarrow \frac{15}{\sqrt{3}}=24-h$

$\Rightarrow h=24-5 \sqrt{3}$

$\Rightarrow h=24-5 \times 1.732=15.34 \mathrm{~m}$

Hence, the height of the first pole is 15.34 m.


text:
# Thread: socks in a drawer probability problem

1. ## socks in a drawer probability problem

Hi all

I have a probability problem book and one problem looks at the solution of the following:

a drawer contains red socks and black socks. when two socks are drawn at random the probability that both are red is 0.5. how small can the number of socks in the drawer be? how small if the number of black socks is even?

the solution solves for the general case with the following with r red socks and b black socks:

1. r/(r+b) x r-1/(r+b-1)=0.5
then sets up the inequalities

2. (r/(r+b))^2 > 0.5 > (r-1/(r+b-1))^2

taking the square root and multiplying by (r+b) gives first inequality as

3. r > 1/(2^0.5) x (r+b)

and the next step takes it to:

4. r > 1/(2^0.5 - 1) x b

I do not follow why we are able to go from 3 to 4. can anyone help out please?

2. Hello, factfinder!

A drawer contains red socks and black socks.
When two socks are drawn at random, the probability that both are red is $\tfrac{1}{2}$
How small can the number of socks in the drawer be?
How small if the number of black socks is even?

The solution solves for the general case with $r$ red socks and $b$ black socks:

. . $1.\;\;\frac{r}{r+b}\cdot\frac{r-1}{r+b-1}\:=\:\frac{1}{2}$

Then sets up the inequalities:

. . $2.\;\;\left(\frac{r}{r+b}\right)^2 \:>\: \frac{1}{2} \:>\:\left(\frac{r-1}{r+b-1}\right)^2$

Taking the square root and multiplying by $(r+b)$ gives first inequality as:

. . $3.\;\; r\: >\:\frac{1}{\sqrt{2}}(r+b)$

and the next step takes it to:

. . $4.\;\; r \:>\: \frac{1}{\sqrt{2}-1}\,b$

I do not follow why we are able to go from 3 to 4.
They skippped some fancy footwork . . .

At (3) we had: . $r \:>\:\frac{1}{\sqrt{2}}(r+b)$

Multiply by $\sqrt{2}\!:\;\;\sqrt{2}\,r \:>\:r + b \quad\Rightarrow\quad \sqrt{2}\,r - r \:>\:b$

Factor: . $\left(\sqrt{2}-1\right)r \:>\:b \quad\Rightarrow\quad r \:>\:\frac{1}{\sqrt{2}-1}\,b$

3. Originally Posted by factfinder
Hi all

I have a probability problem book and one problem looks at the solution of the following:

a drawer contains red socks and black socks. when two socks are drawn at random the probability that both are red is 0.5. how small can the number of socks in the drawer be? how small if the number of black socks is even?

the solution solves for the general case with the following with r red socks and b black socks:

1. r/(r+b) x r-1/(r+b-1)=0.5
then sets up the inequalities

2. (r/(r+b))^2 > 0.5 > (r-1/(r+b-1))^2

taking the square root and multiplying by (r+b) gives first inequality as

3. r > 1/(2^0.5) x (r+b)

and the next step takes it to:

4. r > 1/(2^0.5 - 1) x b

I do not follow why we are able to go from 3 to 4. can anyone help out please?
(a) 4 is the smallest number of possible socks. 3 red and 1 black.

(b) If black socks are even numbered the smallest number is 21. 6 black and 15 red.

I don't know how to use the symbols program, so this is going to be messy. Set up this equation: Let x= r+b and use a hyper-geometric probability function with reds as your successes. (x-b choose 2)*(x-r choose 0)/ (x choose 2) = .5 Now, rewrite it (x-b)*(x-b-1)/(x)/(x-1) = .5 After multiplying out and completing the square you arrive at x = +/- (2b^2 + 1/4)^.5 +2b+.5 Only the positive is relevant here and of course x= r+b so now red as a function of black is r= (2b^2+1/4)^.5 + b+.5 Solve for values of black that yield perfect squares under the radical and you arrive at black = 1, red = 3 and b= 6, r=15

4. Thanks Soroban. Thanks Bilbo.

Bilbo I went through your method and got your final equation (though I didn't complete the square?). Did you then solve the equation by trial and error?

also do I need a perfect square under the radical? with the solutions inserted the evaluated radical does not give a whole number but sums to one with the terms b and 0.5

5. Originally Posted by factfinder
Thanks Soroban. Thanks Bilbo.

Bilbo I went through your method and got your final equation (though I didn't complete the square?). Did you then solve the equation by trial and error?

also do I need a perfect square under the radical? with the solutions inserted the evaluated radical does not give a whole number but sums to one with the terms b and 0.5
Well, you need to complete the square so that you can isolated X as the original quadratic is not a squared binomial. I then plugged in whole numbers for B (the domain for this specific problem is obviously restricted to whole numbers) starting from zero until I hit answers that yielded whole numbers for R (the range for this problem is restricted to whole numbers as well). In hindsight, I probably should have just plotted the function on a graphing calculator and searched for plausible results.
With regard to your second question, I do believe that a perfect square is anything that yields a rational number as a root. Therefore, I don't think that a perfect square must be an integer. I may be wrong in my use of the term "perfect square" though.

,

,

### The smallest number of socks in the drawer probability

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text:
prev ( testsuite > Topics > StretchyChars > Integrals > intSize2  ) next (show test run result)

Alternatives: (mml file)  (full5) (full) (simple) (plain) (form) (slideshow)
File:Topics/StretchyChars/Integrals/intSize2
CVS-ID:
Author:Dave Doyle, Design Science
Description:Test of integrals of various sizes under different conditions.

passed (p) failed (f) not-tested (n) broken-test (b) some parts pass (s)

Sample Rendering:

$\begin{array}{ccccc}& & \text{symm}& \text{stretch}& \text{symm/stretch}\\ \text{small op}& \int x& \int x& \int x& \int x\\ \text{large op}& \int x& \int x& \int x& \int x\\ \text{small op}& \int \frac{x}{y}& \int \frac{x}{y}& \int \frac{x}{y}& \int \frac{x}{y}\\ \text{large op}& \int \frac{x}{y}& \int \frac{x}{y}& \int \frac{x}{y}& \int \frac{x}{y}\end{array}$

Source Code:

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<mtd>
<mrow>
<mo largeop="true" stretchy="true" symmetric="true">∫</mo>
<mi>x</mi>
</mrow>
</mtd>
</mtr>
<mtr>
<mtd>
<mtext>small op</mtext>
</mtd>
<mtd>
<mrow>
<mo largeop="false" stretchy="false" symmetric="false">∫</mo>
<mfrac>
<mi>x</mi>
<mi>y</mi>
</mfrac>
</mrow>
</mtd>
<mtd>
<mrow>
<mo largeop="false" stretchy="false" symmetric="true">∫</mo>
<mfrac>
<mi>x</mi>
<mi>y</mi>
</mfrac>
</mrow>
</mtd>
<mtd>
<mrow>
<mo largeop="false" stretchy="true" symmetric="false">∫</mo>
<mfrac>
<mi>x</mi>
<mi>y</mi>
</mfrac>
</mrow>
</mtd>
<mtd>
<mrow>
<mo largeop="false" stretchy="true" symmetric="true">∫</mo>
<mfrac>
<mi>x</mi>
<mi>y</mi>
</mfrac>
</mrow>
</mtd>
</mtr>
<mtr>
<mtd>
<mtext>large op</mtext>
</mtd>
<mtd>
<mrow>
<mo largeop="true" stretchy="false" symmetric="false">∫</mo>
<mfrac>
<mi>x</mi>
<mi>y</mi>
</mfrac>
</mrow>
</mtd>
<mtd>
<mrow>
<mo largeop="true" stretchy="false" symmetric="true">∫</mo>
<mfrac>
<mi>x</mi>
<mi>y</mi>
</mfrac>
</mrow>
</mtd>
<mtd>
<mrow>
<mo largeop="true" stretchy="true" symmetric="false">∫</mo>
<mfrac>
<mi>x</mi>
<mi>y</mi>
</mfrac>
</mrow>
</mtd>
<mtd>
<mrow>
<mo largeop="true" stretchy="true" symmetric="true">∫</mo>
<mfrac>
<mi>x</mi>
<mi>y</mi>
</mfrac>
</mrow>
</mtd>
</mtr>
</mtable>
[/itex]```