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https://www.geeksforgeeks.org/pentagonal-pyramidal-number/ | 1,576,436,328,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541309137.92/warc/CC-MAIN-20191215173718-20191215201718-00189.warc.gz | 714,609,502 | 29,420 | # Pentagonal Pyramidal Number
Given a number n, find the nth pentagonal pyramidal number.
A Pentagonal Pyramidal Number belongs to the figurate number class. It is the number of objects in a pyramid with a pentagonal base. The nth pentagonal pyramidal number is equal to sum of first n pentagonal numbers.
Examples:
```Input : n = 3
Output : 18
Input : n = 7
Output : 196
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Method 1: (Naive Approach) :
This approach is simple. It says to add all the pentagonal numbers up to n (by running loop) to get nth Pentagonal pyramidal number.
Below is the implementation of this approach:
## C++
`// CPP Program to get nth Pentagonal ` `// pyramidal number. ` `#include ` `using` `namespace` `std; ` ` ` `// function to get nth Pentagonal ` `// pyramidal number. ` `int` `pentagon_pyramidal(``int` `n) ` `{ ` ` ``int` `sum = 0; ` ` ` ` ``// Running loop from 1 to n ` ` ``for` `(``int` `i = 1; i <= n; i++) { ` ` ` ` ``// get nth pentagonal number ` ` ``int` `p = (3 * i * i - i) / 2; ` ` ` ` ``// add to sum ` ` ``sum = sum + p; ` ` ``} ` ` ``return` `sum; ` `} ` ` ` `// Driver Program ` `int` `main() ` `{ ` ` ``int` `n = 4; ` ` ``cout << pentagon_pyramidal(n) << endl; ` ` ``return` `0; ` `} `
## Java
`// Java Program to get nth ` `// Pentagonal pyramidal number. ` `import` `java.io.*; ` ` ` `class` `GFG ` `{ ` ` ` `// function to get nth ` `// Pentagonal pyramidal number. ` `static` `int` `pentagon_pyramidal(``int` `n) ` `{ ` ` ``int` `sum = ``0``; ` ` ` ` ``// Running loop from 1 to n ` ` ``for` `(``int` `i = ``1``; i <= n; i++) ` ` ``{ ` ` ` ` ``// get nth pentagonal number ` ` ``int` `p = (``3` `* i * i - i) / ``2``; ` ` ` ` ``// add to sum ` ` ``sum = sum + p; ` ` ``} ` ` ``return` `sum; ` `} ` ` ` `// Driver Code ` `public` `static` `void` `main (String[] args) ` `{ ` ` ``int` `n = ``4``; ` ` ``System.out.println(pentagon_pyramidal(n)); ` `} ` `} ` ` ` `// This code is contributed by anuj_67. `
## Python3
`# Python3 Program to get nth Pentagonal ` `# pyramidal number. ` ` ` `# function to get nth Pentagonal ` `# pyramidal number. ` `def` `pentagon_pyramidal(n): ` ` ``sum` `=` `0` ` ` ` ``# Running loop from 1 to n ` ` ``for` `i ``in` `range``(``1``, n ``+` `1``): ` ` ` ` ``# get nth pentagonal number ` ` ``p ``=` `( ``3` `*` `i ``*` `i ``-` `i ) ``/` `2` ` ` ` ``# add to sum ` ` ``sum` `=` `sum` `+` `p ` ` ` ` ``return` `sum` ` ` ` ` `# Driver Program ` `n ``=` `4` `print``(``int``(pentagon_pyramidal(n))) `
## C#
`// C# Program to get nth ` `// Pentagonal pyramidal number. ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// function to get nth ` `// Pentagonal pyramidal number. ` `static` `int` `pentagon_pyramidal(``int` `n) ` `{ ` ` ``int` `sum = 0; ` ` ` ` ``// Running loop from 1 to n ` ` ``for` `(``int` `i = 1; i <= n; i++) ` ` ``{ ` ` ` ` ``// get nth pentagonal number ` ` ``int` `p = (3 * i * ` ` ``i - i) / 2; ` ` ` ` ``// add to sum ` ` ``sum = sum + p; ` ` ``} ` ` ``return` `sum; ` `} ` ` ` `// Driver Code ` `static` `public` `void` `Main () ` `{ ` ` ``int` `n = 4; ` ` ``Console.WriteLine(pentagon_pyramidal(n)); ` `} ` `} ` ` ` `// This code is contributed by ajit. `
## PHP
` `
Output :
```40
```
Time Complexity : O(n)
Method 2: (Efficient Approach) :
In this approach, we use formula to get nth Pentagonal pyramidal number in O(1) time.
nth Pentagonal pyramidal number = n2 (n + 1) / 2
Below is the implementation of this approach:
## C++
`// CPP Program to get nth Pentagonal ` `// pyramidal number. ` `#include ` `using` `namespace` `std; ` ` ` `// function to get nth Pentagonal ` `// pyramidal number. ` `int` `pentagon_pyramidal(``int` `n) ` `{ ` ` ``return` `n * n * (n + 1) / 2; ` `} ` ` ` `// Driver Program ` `int` `main() ` `{ ` ` ``int` `n = 4; ` ` ``cout << pentagon_pyramidal(n) << endl; ` ` ``return` `0; ` `} `
## Java
`// Java Program to get nth ` `// Pentagonal pyramidal number. ` `import` `java.io.*; ` ` ` `class` `GFG ` `{ ` ` ` `// function to get nth ` `// Pentagonal pyramidal number. ` `static` `int` `pentagon_pyramidal(``int` `n) ` `{ ` ` ``return` `n * n * ` ` ``(n + ``1``) / ``2``; ` `} ` ` ` `// Driver Code ` `public` `static` `void` `main (String[] args) ` `{ ` ` ``int` `n = ``4``; ` ` ``System.out.println(pentagon_pyramidal(n)); ` `} ` `} ` ` ` `// This code is contributed by ajit `
## Python3
`# Python3 Program to get nth Pentagonal ` `# pyramidal number. ` ` ` `# function to get nth Pentagonal ` `# pyramidal number. ` `def` `pentagon_pyramidal(n): ` ` ``return` `n ``*` `n ``*` `(n ``+` `1``) ``/` `2` ` ` ` ` `# Driver Program ` `n ``=` `4` `print``(``int``(pentagon_pyramidal(n))) `
## C#
`// C# Program to get nth ` `// Pentagonal pyramidal number. ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// function to get nth ` `// Pentagonal pyramidal number. ` `static` `int` `pentagon_pyramidal(``int` `n) ` `{ ` ` ``return` `n * n * ` ` ``(n + 1) / 2; ` `} ` ` ` `// Driver Code ` `static` `public` `void` `Main () ` `{ ` ` ``int` `n = 4; ` ` ``Console.WriteLine( ` ` ``pentagon_pyramidal(n)); ` `} ` `} ` ` ` `// This code is contributed ` `// by ajit `
## PHP
` `
Output :
```40
```
Time Complexity : O(1)
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Improved By : vt_m, jit_t | 2,210 | 6,040 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2019-51 | latest | en | 0.518269 |
https://www.kidsacademy.mobi/printable-worksheets/online/age-8/math/multiplication-and-division-word-problems/ | 1,716,288,537,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058442.20/warc/CC-MAIN-20240521091208-20240521121208-00825.warc.gz | 734,998,323 | 61,653 | # Multiplication and Division Word Problems Worksheets for 8-Year-Olds
Unlock the world of numbers with our Multiplication and Division Word Problems worksheets, expertly crafted for eight-year-olds! These engaging Learning Online Printables are designed to strengthen your child's arithmetic skills through real-world scenarios. Each worksheet is filled with a variety of challenges that make learning exciting and help young learners master these essential math operations. Perfect for at-home practice or classroom use, our printables ensure a solid foundation in problem-solving and critical thinking. Empower your child's math journey today with our accessible and educational worksheets!
Favorites
Interactive
• 8
• Interactive
• Multiplication and Division Word Problems
## Subtracting Socks Worksheet
Before beginning this exercise with your children, warm them up with a counting game. If math is not their favorite subject, use this worksheet. Help them read the two word problems, then use their fingers to count and subtract. Ask them to select the correct answer and check the box.
Subtracting Socks Worksheet
Worksheet
## Visiting a Volcano Word Problems Worksheet
Read the word problems to your kids, note down the numbers and help them solve. Check the box under the correct answer. Word problems are sentences posed as math problems; just like regular number problems, first you must understand and interpret the sentence.
Visiting a Volcano Word Problems Worksheet
Worksheet
## Ancient Tomb Division and Multiplication Word Problems Worksheet
This worksheet teaches kids math, plus a bit about ancient Egypt. Read the text and point to the pictures to explain them. There are three word problems - help kids interpret and solve each one, then find and check the answer. 80 words
Ancient Tomb Division and Multiplication Word Problems Worksheet
Worksheet
## Magnet Multiplication: 1-Step Word Problems Worksheet
Refresh your students' knowledge about magnets with a simple math worksheet. Download the PDF with two word problems, along with pictures to help. Read the problems to the class, and help them identify the correct answer by checking the box. This activity will help them harness their science knowledge and practice their math skills.
Magnet Multiplication: 1-Step Word Problems Worksheet
Worksheet
## Going Bananas Worksheet
Marla needs to send 56 bananas in batches of 8. Help your child practice division and learn about fact families with this free worksheet. It'll help boost their multiplication skills and increase speed when solving problems.
Going Bananas Worksheet
Worksheet
## Pollinator Positions Worksheet
Kids will love learning about pollinators and how vital they are. Help their favorite friends get to the right flowers with this fun PDF. Children can solve the problems without even knowing they're doing division; just read the word problems and use the highlighted numbers. Finish by tracing the lines to the correct quotient!
Pollinator Positions Worksheet
Worksheet
## Rainforest Animal Division Worksheet
Children can save the rainforest and learn about division with this worksheet! It helps kids understand that division involves creating equal groups of specific numbers, and with its visual representation, they can grasp the concept more easily. They'll have fun learning about their place in the world and their role in protecting the environment.
Rainforest Animal Division Worksheet
Worksheet
## Sailing to the New World Division Worksheet
This worksheet combines history and math, helping kids make sense of problem-solving. Christopher Columbus sailing the ocean blue is combined with picture representations of division word problems in bold colors and highlighted numbers. Kids can learn while having fun figuring out the problems, regardless of whether it's math or history.
Sailing to the New World Division Worksheet
Worksheet
## Dividing with Landforms
Mixing subjects to help your child learn is always nice. This worksheet combines landforms, problem-solving and division. The PDF highlights numbers, uses bold colors and provides pictures to help your child understand the questions and answer choices, making them feel empowered, not intimidated.
Dividing with Landforms
Worksheet
## Water Division Word Problems Worksheet
Kids can sharpen their math skills with this fun PDF! They'll see pictures of friends at the beach, plus bold numbers and colorful pictures. All they need to do is read and understand the problem to find the answer - without even realizing they're doing division. A great way to make math enjoyable!
Water Division Word Problems Worksheet
Worksheet
## Visiting the USA Worksheet
Do your kids love travelling? Ask them why and find out what they look forward to the most. People travel for a variety of reasons, such as adventure, holiday, or business. Juan and Jenn in the picture travel for adventure and have a goal they want to achieve. Read the word problems to your kids and help them pick the right equation and match the total.
Visiting the USA Worksheet
Worksheet
## Archeology Word Problems Worksheet
Encourage your child to explore their career options! Show them an archeologist's job with this worksheet - featuring a picture of a dinosaur bone discovery. Read the accompanying text, then solve the word problems. Help your kids circle the correct answers to better understand this profession.
Archeology Word Problems Worksheet
Worksheet
## Multiplication Facts: Assessment 3 Worksheet
Test your kid's maths skills with this easy to use worksheet! Help them check the box that matches the equation in the first part, then read each word problem and underline the right answers to the second part. Assess your child's muliplication knowledge and find out where they need extra help.
Multiplication Facts: Assessment 3 Worksheet
Worksheet
## Growing Jamestown Worksheet
This exciting multiplication worksheet from Kids Academy uses American history facts as its theme. Kids learn about Native Americans helping early Americans at Jamestown, then read the word problems, match the equations and solve for the product. Finally, circle the correct answer!
Growing Jamestown Worksheet
Worksheet
## Presidential Duties Worksheet
Presidents have many duties. Utilize this worksheet to review the important ones while solving multiplication word problems. Read each passage, determine the equation, then find the product and select the correct answer.
Presidential Duties Worksheet
Worksheet
## Money Word Problems Printable
Boost your 3rd grader's math skills with money word problems. This worksheet takes them to a witch's shop for a fun way to practice multiplication and division.
Money Word Problems Printable
Worksheet
## Two Step Word Problems Worksheet
Boost your child's math skills with this fun and realistic two-step word problems worksheet! Kids will learn to break down complex multiplication word problems into easier steps to solve. Download this free PDF worksheet and help them reach the next level.
Two Step Word Problems Worksheet
Worksheet
Learning Skills
### Harnessing the Power of Learning Interactive Worksheets: The Importance of Multiplication and Division Word Problems for 8-Year-Olds
In the journey of learning mathematics, mastery of fundamental concepts like multiplication and division is crucial. For eight-year-old children, these concepts are gateways to more advanced mathematical skills. However, achieving proficiency in these areas involves not just understanding numbers but also applying them in real-world scenarios. This is where learning interactive worksheets on multiplication and division word problems prove to be immensely useful.
Engaging Young Minds: Eight-year-olds are at a stage where their cognitive abilities are rapidly expanding, and they are more open to learning through interactive and engaging content. Learning interactive worksheets are designed with this in mind, incorporating colorful graphics, intuitive layouts, and interactive elements that make learning not just educational but also entertaining. This engagement is crucial in maintaining a child’s interest and focus, especially in subjects that might otherwise seem daunting.
Building Practical Skills: Multiplication and division word problems help children connect mathematical concepts with everyday situations. These worksheets often present scenarios involving times tables, equal sharing, and distribution, which mirror real-life problems. By solving these, children learn not just to calculate but also to interpret and solve problems they may encounter in their daily lives. This practical application enhances their analytical skills and prepares them for more complex problem-solving scenarios.
Improving Language and Reading Comprehension: Learning interactive worksheets on word problems require children to read and comprehend the problem before attempting to solve it. This inadvertently boosts their reading skills and enhances their ability to understand and process information. The integration of math and language skills in these worksheets ensures a comprehensive cognitive development that transcends numerical ability.
Encouraging Independent Learning: The design of interactive worksheets often allows children to receive immediate feedback on their answers, which is crucial for self-assessment and improvement. This feature encourages learners to engage with the material independently, fostering a sense of responsibility and self-motivation. The immediate feedback mechanism helps them identify their areas of weakness and work on them promptly, promoting a healthy learning loop.
Catering to Diverse Learning Styles: Every child is unique, with different learning preferences and paces. Learning interactive worksheets are versatile and can be used in various ways to cater to individual needs. Whether it is group work, one-on-one tutoring, or self-paced learning at home, these worksheets are adaptable and can be integrated into any learning environment effectively. This flexibility makes them an invaluable tool for educators and parents alike | 1,836 | 10,146 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-22 | latest | en | 0.876636 |
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Please briefly explain why you feel this question should be reported.
# What is the value of 1 million?
What is the value of 1 million? Now, we know that 1 million = 1,000,000 in the international place value system. 1 million = 10,00,000 in the Indian place value system. Hence, 1 million is equivalent to 1000 thousands.
## What is the meaning of 1.9 million?
1.9 million = 19 lakh.
## How many millions is 10 lakhs?
10 Lakhs = 1 Million = 1 followed by 6 Zeros = 1,000,000. Similarly here, 1 Crore = 10 Million = 1 followed by 7 Zeros = 10,000,000.
## How can I make 1 billion?
1,000,000,000 (one billion, short scale; one thousand million or milliard, yard, long scale) is the natural number following 999,999,999 and preceding 1,000,000,001. One billion can also be written as b or bn. In standard form, it is written as 1 × 109.
## How is 1million written?
One million (1,000,000), or one thousand thousand, is the natural number following 999,999 and preceding 1,000,001. The word is derived from the early Italian millione (milione in modern Italian), from mille, « thousand », plus the augmentative suffix -one.
## What is 2.1 million as a number?
2.1 million in words can be written as two point one million. 2.1 million is also the same as two million one hundred thousand.
## How is 1.9 billion written in numbers?
We start by showing you how to write 1.9 billion, or one point nine billion, in words. one billion nine hundred million.
## How many millions is 2 lakhs?
One million is equal to ten lakhs.
Million to Lakhs Table.
Million (M) Lakhs (Lac)
0.1 1
0.2 2
0.3 3
0.4 4
## How many dollars is 5 lakhs?
Convert Indian Rupee to US Dollar
INR USD
1,000 INR 13.6756 USD
5,000 INR 68.3779 USD
10,000 INR 136.756 USD
50,000 INR 683.779 USD
## How much is a 100 million?
100 million or 0.1 billion is equal to 10 crores or 1000 lakhs.
## What is 1000000000000000000000000 called?
Some Very Big, and Very Small Numbers
Name The Number Symbol
septillion 1,000,000,000,000,000,000,000,000 Y
sextillion 1,000,000,000,000,000,000,000 Z
quintillion 1,000,000,000,000,000,000 E
## How much is a zillion?
Zillion may represent ANY very large power of a thousand, certainly larger than a trillion, and maybe even a vigintillion or centillion ! Just as a million had spawned the Chuquet illions, the « zillion » also had many follow ups.
## How do you write 1.5 billion?
1.5 billion in numbers is 1,500,000,000. Think of this as 1 billion plus 500 million.
## How do you write million in short?
This guide will, MM (or lowercase “mm”) denotes that the units of figures presented are in millions. The Latin numeral M denotes thousands. Thus, MM is the same as writing “M multiplied by M,” which is equal to “1,000 times 1,000”, which equals 1,000,000 (one million).
## How do you write 100k?
100000 in Words
1. 100000 in Words = One Hundred Thousand.
2. One Hundred Thousand in Numbers = 100000.
## What does 9 zeros mean?
Billion 1,000,000,000 (9 zeros) Trillion 1,000,000,000,000 (12 zeros) Quadrillion 1,000,000,000,000,000 (15 zeros) Quintillion 1,000,000,000,000,000,000 (18 zeros) Sextillion 1,000,000,000,000,000,000,000 (21 zeros)
## What is 2.4 million written out?
Standard notation is the normal way of writing the numbers. Therefore, we can write the 2.4 million as 2400000 as 1 million contains six zeros. Hence, the standard notation of 2.4 million is 2400000 .
## What is 1.1 million as a number?
1.1 million in words can be written as one point one million. 1.1 million is also the same as one million one hundred thousand.
## What number is 1million?
One million (1,000,000), or one thousand thousand, is the natural number following 999,999 and preceding 1,000,001. The word is derived from the early Italian millione (milione in modern Italian), from mille, « thousand », plus the augmentative suffix -one.
## What is this number 1000000000000000000000000?
Some Very Big, and Very Small Numbers
Name The Number Symbol
septillion 1,000,000,000,000,000,000,000,000 Y
sextillion 1,000,000,000,000,000,000,000 Z
quintillion 1,000,000,000,000,000,000 E
## How do you write 1.3 billion?
We start by showing you how to write 1.3 billion, or one point three billion, in words. one billion three hundred million.
## What is the number for 2 million?
Verify Number Formats: English
Numeric Format Compact-Short Compact-Long
1,000,000 1M 1 million
1,100,000 1.1M 1.1 million
1,500,000 1.5M 1.5 million
2,000,000 2M 2 million
## How many millions is 1 billion?
A billion is a number with two distinct definitions: 1,000,000,000, i.e. one thousand million, or 109 (ten to the ninth power), as defined on the short scale. This is now the meaning in all English dialects. 1,000,000,000,000, i.e. one million million, or 1012 (ten to the twelfth power), as defined on the long scale.
## How much does 2 million means?
1 Million = 10 Lakhs (10,000,00) 2 Million = 20 Lakhs (20,000,00) | 1,520 | 4,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-30 | latest | en | 0.911193 |
http://www.mcqslearn.com/business-statistics/mcq/data-classification-tabulation-presentation.php | 1,498,269,705,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320209.66/warc/CC-MAIN-20170624013626-20170624033626-00553.warc.gz | 594,056,775 | 8,377 | Learn Statistics Online Notes & Technology Articles
# Data Classification, Tabulation and Presentation Multiple Choice Questions Test 1 Tests pdf Download
Practice data classification, tabulation and presentation multiple choice questions (MCQs), business statistics test 1 online to learn. Practice data tables and types MCQs questions and answers on data tables and types, data classification with answers. Free data classification, tabulation and presentation study guide has answer key with choices as length diagram, width diagram, histogram and dimensional bar charts of multiple choice questions (MCQ) as if vertical lines are drawn at every point of straight line in frequency polygon then by this way frequency polygon is transformed into to test learning skills. Study to learn data tables and types quiz questions to practice MCQ based online exam preparation test.
## MCQ on Data Classification, Tabulation and Presentation Quiz pdf Download Test 1
MCQ. If vertical lines are drawn at every point of straight line in frequency polygon then by this way frequency polygon is transformed into
1. width diagram
2. length diagram
3. histogram
4. dimensional bar charts
C
MCQ. Diagrams such as cubes and cylinders are classified as
1. one dimension diagrams
2. two dimension diagram
3. three dimensional diagrams
4. dispersion diagrams
C
MCQ. Discrete variables and continuous variables are two types of
1. open end classification
2. time series classification
3. qualitative classification
4. quantitative classification
D
MCQ. In stem and leaf display diagrams used in exploratory analysis, stems are considered as
1. central digits
2. trailing digits | 330 | 1,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-26 | longest | en | 0.836837 |
https://origin.geeksforgeeks.org/javascript-program-for-swapping-nodes-in-a-linked-list-without-swapping-data/ | 1,680,190,153,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949331.26/warc/CC-MAIN-20230330132508-20230330162508-00606.warc.gz | 504,877,958 | 39,810 | Open in App
Not now
# Javascript Program For Swapping Nodes In A Linked List Without Swapping Data
• Last Updated : 30 Mar, 2022
Given a linked list and two keys in it, swap nodes for two given keys. Nodes should be swapped by changing links. Swapping data of nodes may be expensive in many situations when data contains many fields.
It may be assumed that all keys in the linked list are distinct.
Examples:
```Input : 10->15->12->13->20->14, x = 12, y = 20
Output: 10->15->20->13->12->14
Input : 10->15->12->13->20->14, x = 10, y = 20
Output: 20->15->12->13->10->14
Input : 10->15->12->13->20->14, x = 12, y = 13
Output: 10->15->13->12->20->14```
This may look a simple problem, but is an interesting question as it has the following cases to be handled.
1. x and y may or may not be adjacent.
2. Either x or y may be a head node.
3. Either x or y may be the last node.
4. x and/or y may not be present in the linked list.
How to write a clean working code that handles all the above possibilities.
The idea is to first search x and y in the given linked list. If any of them is not present, then return. While searching for x and y, keep track of current and previous pointers. First change next of previous pointers, then change next of current pointers.
Below is the implementation of the above approach.
## Javascript
``
Output:
```Linked list before calling swapNodes() 1 2 3 4 5 6 7
Linked list after calling swapNodes() 1 2 4 3 5 6 7 ```
Time Complexity: O(n)
Auxiliary Space: O(1)
Optimizations: The above code can be optimized to search x and y in single traversal. Two loops are used to keep program simple.
Simpler approach:
## Javascript
``
Output:
```Linked list before calling swapNodes() 1 2 3 4 5 6 7
Linked list after calling swapNodes() 6 2 3 4 5 1 7 ```
Time Complexity: O(n)
Auxiliary Space: O(1)
Please refer complete article on Swap nodes in a linked list without swapping data for more details!
My Personal Notes arrow_drop_up
Related Articles | 564 | 2,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-14 | latest | en | 0.789442 |
http://gamedev.stackexchange.com/questions/631/what-to-consider-when-deciding-on-2d-vs-3d-for-a-game/634 | 1,369,156,220,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700264179/warc/CC-MAIN-20130516103104-00025-ip-10-60-113-184.ec2.internal.warc.gz | 109,083,184 | 16,294 | # What to consider when deciding on 2D vs 3D for a game?
How much "harder" is 3D than 2D in terms of:
• Amount/complexity of the code
• Level of math skills required
• Time involved in making art assets
Original title: How hard is 3D game development versus 2D?
-
One dimension harder :) – Ólafur Waage Jul 16 '10 at 16:49
Waay subjective... – Cyclops Jul 16 '10 at 17:31
Hmm... having said that, I actually really like @munificent's answer. Now I'm not sure how I want to categorize this question. :) It's not quite right for meta, either... Dang these edge cases. :) – Cyclops Jul 16 '10 at 19:07
Code complexity, math level, and time seem like relatively objective measures of difficulty to me. – Brian Ortiz Jul 16 '10 at 19:16
@Brian, yeah, I wish we could undo close votes. :) Re-evaluating it, I think part of the problem is the title - "How hard is X versus Y", sounds subjective. A better title might be, "What are the differences in 3D development versus 2D?" (maybe change it?). Which is actually the Question that @munificent answered, although with a one-liner that addressed the title. :) – Cyclops Jul 16 '10 at 19:25
3D is an order of magnitude harder than 2D:
Programming:
• The math is significantly more complex for rendering, physics, collision, etc. Hope you like matrices and vectors!
• Because of the previous point, good performance is much more difficult to attain. With today's hardware, you can make a nice-looking 2D game without having to think about performance at all beyond not being actively stupid. With 3D, you will have to do some optimization.
• The data structures are much more complex. Because of the previous point, you'll need to think about culling, space partitioning, etc. all of which are more challenging then a simple "here's a list of everything in the level".
• Animation is much more complicated. Animation in 2D is just a filmstrip of frames with possibly different positions for each frame. With 3D, you'll need to deal with separate animation assets, bones, skinning, etc.
• The volume of data is much higher. You'll have to do intelligent resource management. Games ship with gigs of content, but consoles sure as hell don't have gigs of memory.
• The pipelines are more complex to develop and maintain. You'll need code to get assets into your engine's preferred format. That code doesn't write itself.
Art:
• The assets are, of course, much more complex. You'll need textures, models, rigs/skeletons, animation, etc. The tools are much more complex and expensive, and the skills to use them harder to find.
• The set of skills needed is wider. Good animators aren't often good texture artists. Good lighters may not be good riggers.
• The dependencies between the assets are more complex. With 2D, you can partition your assets across different artists cleanly: this guy does level one, this guy does enemies, etc. With 3D, the animation affects the rig which affects the skeleton which affects the model which affects the textures which affect the lighting... Your art team will have to coordinate carefully and constantly.
• The technical limitations are more complex to deal with. With 2D it's basically "here's your palette and your max sprite size". With 3D, your artists will have to balance texture size (for multiple textures: specular, color, normal, etc.), polygon count, keyframe count, bone count, etc. The particulars of the engine will place random weird requirements on them ("Engine X blows up if you have more than 23 bones!").
• Asset processing takes longer. Pipelines to convert 3D assets to game-ready format are complex, slow, and often buggy. This makes it take much longer for artists to see their changes in game, which slows them down.
Design:
• User input is bitch. You have to deal with camera tracking, converting user input into the character's space intuitively, projecting 2D selections into world space, etc.
• Levels are hard to author. Your level designers basically need the skills of a game designer and an architect. They have to take into account players getting lost, visibility, etc. when building levels.
• Level physics is tedious to author. You'll have to check and recheck and recheck again to make sure there aren't gaps and bugs in the level physics where players can get stuck or fall through the world.
• Tools are much harder. Most games need their own tools for authoring things like levels. Since the content is so much more complex, the tools are more work to create. That usually results in tools that are buggier, incomplete, and harder to use.
-
Excellent answer. I was thinking this would be more subjective, but you've laid it out very clearly. While there might not be an actual number (3.5x harder), you've made a good case for order of magnitude harder. – Cyclops Jul 16 '10 at 19:14
+1 Nailed it. This question is hardly subjective. We're talking about two completely different worlds. There is an inherent increase in difficulty. Don't start your thinking at the genius level... – David McGraw Jul 17 '10 at 0:25
Superb answer ! – zebrabox Jul 18 '10 at 22:05
I see my answer is beaten by several (to use the authors words) magnitudes. It is so superb that I leaned a lot of things myself that I'd never thought of previously. Nice work! – Toby Jul 19 '10 at 18:45
Great answer. One additional thought I did have - in 3D's favour - is that, when I was getting started, it was much, much easier to find resources relating to 3D development than 2D. But that well over 5 years ago, and the landscape has changed quite a bit since then. – Andrew Russell Jul 20 '10 at 16:02
show 1 more comment
Lots harder. If you're not comfortable making a 2D game, you will REALLY not like what it takes to make a 3D game.
The good news: 99% of the time, you don't really need it. Think of any 3D game you can. Take the camera, fix it on the ceiling looking down so that you're now looking at a 2D plane. Doom becomes Gauntlet. Civ IV becomes Civ I. Metal Gear Solid becomes the original Metal Gear. None of these games are "bad" just because they're 2D; they are perfectly playable and generally have much of the same gameplay.
-
This is a highly subjective question, since the answer depends on personal preference/experience/knowledge/intelligence.
I will try to answer neutrally, but since I am only a programmer and not an artist i can only hypothesize for the last point.
Code complexity should not be so very different, except in maths and maybe rendering/physics. Game Logic isn't so much different if you take a healthy level of abstraction (not too much - you're trying to make a game not an engine, at least I guess from your question.) Obviously its a lot easier to calculate movement in 2D because you have limited perspective. Physics is WAY more difficult when dealing in three axes. Also, loading a Sprite from a Bitmap is a lot easier than loading a 3D Model (and possibly texturing).
Maths is more complicated for 3D (a no-brainer really - quaternions, vectors, matrices. 'nuff said)
For art, I think it must be more difficult for 3D too, since you need to create art that looks good from every possible viewing angle (or at least a wide range), and you usually want to texture things too. Animating a mesh is no picnic to get realistic, and getting the texture to play along isn't either.
-
Loading a sprite is easier than loading a model? Models involve skeletons, skinning, shaders, UV coordinates, textures, normal maps, etc. A sprite is an image. Following along that vain, 3D art is much much harder, because of the things listed above. – Sean James Jul 18 '10 at 19:37 I'm sorry, that was an error. quite a big one too, thanks for pointing it out. It must have been so bland that it didn't register when I proofread. – Toby Jul 19 '10 at 18:40
The other thing to consider ... with a 3D game you probably want to consider using a pre-existing engine and concentrate on making a game not an engine. It can go someway to reducing the time and difficulty taken in a 3D game (as excellently identified by munificent).
It's much easier to build a 2D game ground up. But obviously you can (and should consider) using sprite, sound, portability libraries as well. No point in reinventing the wheel except for education purposes.
Seems obvious - but I thought it was worth saying.
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Transcript
```2.4 & 2.5
Absolute Value Inequalities and
Equations
Learning Goals
• Interpret complicated expressions by viewing one or
more of their parts as a single entity
• Create equations and inequalities in one variable and
use them to solve problems
absolute value: distance from zero on a number line
Ex 1
15 3x 6
Ex 2
2 x9 37
extraneous solution: a solution to a transformed
equation but not the original equation
Ex 3
5x 2 7x 14
Ex 4
3x 4 4x 1
Ex 5
x x 1
Absolute Value Inequalities
a b
b a b
x 5 distance within 5 units in both directions
a b
a b or a b
x 5 distance outside 5 units in both directions
Ex 6
3x 4 8
Ex 7
5x 10 15
Ex 8
2 x 1 5 3
Ex 9
x 5 2
Ex 10
x 5 2
Write each compound inequality as
an absolute value inequality.
Ex 11
12 m 20
Ex 12
x 1 or x 4
Absolute Value Statements
symbols
definition
distance from x to 0 is a
units
distance from x to 0 is
less than a units
distance from x to 0 is
greater than a units
graph
With a partner, graph each solution
x 5 and x 6
x 6 or x 5
x5 x
```
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https://scienceblogs.com/startswithabang/2016/01/13/the-science-of-powerball-synopsis | 1,726,463,358,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651676.3/warc/CC-MAIN-20240916044225-20240916074225-00423.warc.gz | 464,789,309 | 20,811 | # The Science of Powerball (Synopsis)
“I’ve done the calculation and your chances of winning the lottery are identical whether you play or not.” -Fran Lebowitz
Later today, the richest lottery drawing in history -- the \$1.5 billion Powerball jackpot -- will take place. While many outlets are encouraging people to purchase as many tickets as possible, it's important to run through the mathematics and find out what your expected value is for each ticket.
Image credit: E. Siegel, 2016.
While a naive analysis shows that a jackpot in excess of about \$245 million would lead to a break-even-or-better result, when you factor in taxes and split jackpots, you find that even for the \$1.5 billion jackpot, your \$2 ticket is only worth about \$0.85.
Image credit: E. Siegel, 2016.
Tags
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Didn't we already cover this the last time it got big? Obviously not this big though. This is quite an astounding historical event just because of the amount.
While there's no way to make it a good bet (i.e., above the breakeven point), you can make it a better bet by selecting numbers superstitious people are unlikely to select. That generally means numbers >30 since many people play birthdays or other dates, and numbers in sequence since laypeople tend to think sequences are less likely than scattered values. 46,47,48,49,50, 51 has the same crappy odds of winning as any other combo, but at least if you pick a ticket like that, you can console yourself with the idea that you likely won't be sharing the winnings with many people if it hits. And don't worry about sharing with other people who take my advice, the pool of "people who take eric's advice" is quite small. Probably one. Somedays, not even one. :)
Well, in any event good luck and best wishes to those who play.
I've always wondered about the appropriateness of the "TAX on the mathematically challenged" line. A lottery ticket purchase is a choice people make: they can buy tickets or not as they wish. That isn't the case with a tax.
You can choose with taxes too. Get a crappy low paid job and you won't have to pay tax. Or move to somewhere where there is no tax.
However, unlike the lottery, where you get nothing if you don't win, you get something with your taxes every time you have paid, and even when you haven't.
Paying taxes is no different from paying the butcher or baker for their produce, or paying the landlord the rent.
My comment wasn't a statement against taxes - it was a statement about the comparison of lottery cost with taxes.
Having a crappy low paying job doesn't keep you from paying taxes - you still pay taxes on the items you purchase.
Yeah, but the comparison is wrong for reasons other than the ones you cried off on.
You get taken for a ride with the lottery, getting nothing back.
You get something back for your taxes.
But in both cases you can decide to not pay.
You claimed you could decide to pay with the lottery but not taxes. This is not the case at all. You have two methods at least, and one of them allows you to get stuff FREE.
"you still pay taxes on the items you purchase."
So don't purchase anything. This is still a choice.
"So don’t purchase anything. This is still a choice."
No, purchasing food (for example) is not a choice. The fact that there are other ways to examine the difference between purchasing tickets for a lottery and paying taxes than the one I mentioned might be true, but it doesn't change my comment.
"No, purchasing food (for example) is not a choice"
(Ignoring that the response to this is "Yes it is.")
All you're REALLY saying here is that buying food is like paying taxes in your opinion: you don't get a choice of whether you do pay.
Don't hear "food" as being related to taxes anywhere.
"The fact that there are other ways to examine the difference between purchasing tickets for a lottery and paying taxes than the one I mentioned might be true,"
It is.
" but it doesn’t change my comment."
Nonsequitur.
Some other differences being true doesn't make your comment right either.
Nothing you've said explains why the choice of paying for a lottery ticket should be viewed in any way the same as paying taxes. The notion that purchases of food (or other necessities I didn't mention) are choices doesn't hold water, nor does your implication that there is some subclass of society that pays no taxes at all.
"Nothing you’ve said explains why the choice of paying for a lottery ticket should be viewed in any way the same as paying taxes."
Nothing about not being looked at in the same way makes the difference that you can choose to pay the lottery ticket but can't choose to pay taxes.
"The notion that purchases of food (or other necessities I didn’t mention) are choices doesn’t hold water"
Yes it does.
But even if it weren't valid, your problem is now that you think of taxes being like buying food.
"nor does your implication that there is some subclass of society that pays no taxes at all."
I've not said that there is. Though I can posit that many don't. However, become a subsistence farmer in Somalia. No taxes and you grow your own food.
There's nothing stopping you doing that except preferring to be a US citizen paying US taxes (and doing the job you are employed in that doesn't supply you with all your basic needs) and not having to live in Somalia as a subsistence farmer.
You CHOOSE to "have to" pay those taxes.
Just like you choose to pay the lottery.
“Nothing you’ve said explains why the choice of paying for a lottery ticket should be viewed in any way the same as paying taxes.”
And you're really not listening. I've not said they're the same. I've said they're DIFFERENT. The difference is that if you pay the lottery, you probably won't get anything at all for it. If you pay your tax, you DO *definitely* get something for it.
You don't get to "choose" to pay taxes because you can't choose not to use the roads, the protection of the building codes protecting your home from being dangerous, from the EPA making sure that the area you live in isn't polluted, the police and justice system seeing that there's some limit on criminality, the army from ensuring that no mexican invasion takes over your country, regulations that ensure the banks can't just take your money or your boss fails to pay you.
You can't choose NOT to have those, so why should you choose not to pay for them?
If you're susbsistence and in the middle of nowhere, therefore not using the roads, clean water, power or courts system, then you won't be paying taxes.
You can choose to live with other humans and not slave away to get the bare minimum of food and have to make everything you need yourself and fix everything that breaks, or you can choose not to pay tax. And you choose the former.
So don't go crying how you can't choose: you're lying.
If you're worried about choosing to pay taxes on necessities, then get the law changed so that sales taxes (or just those on necessities) disappear and that the various income taxes (or estate taxes) are increased to make the shortfall up.
The UK has a zero rate VAT on food and similar. Only luxury classed foods are taxed with VAT. And, yes, we do have slightly higher income taxes, but you choose what level of tax you want when you go for a particular wage. You don't have to be "forced" to pay "tax" because you'll starve to death otherwise.
My numbers have been 1, 2, 3, 4, 5, and 6 for every lottery on the planet, for over 20 years now (shrink sequence to fit the # of balls of varying lotteries). I don't buy a ticket, never have.
My approach is that I can ignore them - if my winning numbers ever hit, it will be front page news and riots are likely. Then I'll say "Dang it! If only I'd bought a ticket!". But until that time, I can multiply the # of weeks by the cost / week to play, and rightfully claim to be in the black by that amount.
In my home state lottery, I'm up by over \$3,000!
Oh, and dean - you are completely right. The lottery is not a tax.
wow must be having fun trolling you, since he often tells posters to Google for available information before bothering people here. Taxes are compulsory. Google it, wow.
I'm coincidentally reminded of the recent "calculation" that virological gain-of-function research carries a risk of 54 deaths per lab-year.* What this figure conceals (among other things) is a roughly 55,000 yr timescale.
* Not to get too far into the minutiae, but this doesn't seem to actually appear in the RBA, but rather to follow from the public comments of Lynn C. Foltz, who also included the odd statement that "the 0.4% value [of a single laboratory-acquired infection seeding a pandemic] is likely 1/0.02 = 50-times higher due to eliminating this intermediate local-outbreak step." So it's 200%?
The RBA is here, for anyone whose appetite for probabilities has been whetted by Poweball:
h[]tp://osp.od.nih.gov/sites/default/files/Risk%20and%20Benefit%20Analysis%20of%20Gain%20of%20Function%20Research%20-%20Draft%20Final%20Report.pdf
The public comments are here (Klotz and Lipsitch are the standouts):
^ Mostly chapter 6 of the RBA, and for mammalian-transmissible highly pathogenic avian influenza, BTW.
I’ve always wondered about the appropriateness of the “TAX on the mathematically challenged” line.
*shrug*
It's a metaphor (although I hear it more often referred to as a "stupidity tax"). The basic point is that in the real world, lotteries tend to be a regressive way of pulling money into state coffers. If you want something that's more directly comparable, look at sin taxes.
IMO, a lottery ticket is entertainment. For \$3, you can fantasize about some uber-wealthy future state. Similar to paying to see a movie, except this one is in your imagination.
From that perspective, it's not regressive. It's a cheap form of entertainment that appeals to poor people more than it does to well-off people. It isn't sinister, except in those cases where people suffer from a gambling addiction.
"It’s a metaphor"
I understand that - my comment was simply that I don't think it's a good one. Sheesh.
I just covered the probability and expected value calculations in my intro class on Monday. Yesterday one student gave me an article that talked about the "coverage" of the different possible plays: as Ethan says, there are 292,000,000 possible ticket plays. Last Saturday night, according to lottery officials, 75% of those possibilities were played. The forecast is that tonight approximately 80% will be played. That is a metric butt-load of tickets.
"Lottery expert" Dawn Nettles is hilariously clueless, BTW.
"Everyone should choose their own numbers.... If players had created their own Quick Picks on Saturday night, I think all combinations would have been sold and we would have had a winner."
I understand that – my comment was simply that I don’t think it’s a good one. Sheesh.
@Carl:
IMO, a lottery ticket is entertainment. For \$3, you can fantasize about some uber-wealthy future state. Similar to paying to see a movie, except this one is in your imagination.
Ideally, sure. I've bought Powerball tickets about twice in my life, and I can also find hours of entertainment in devising screwball numerological schemes for making the picks. This time around I also got to review my counting statistics and have some reading material about the relevant tax law* (e.g., here; "[k]eep in mind how unused to complexity most lottery winners are," which seems to be a bit haughty). But...
From that perspective, it’s not regressive. It’s a cheap form of entertainment that appeals to poor people more than it does to well-off people. It isn’t sinister, except in those cases where people suffer from a gambling addiction.
I don't think that either prong really holds, based on the anecdata of my having lived in a diverse urban neighborhood for a long time (oh, and having repeatedly failed to quit smoking). The people that I see buying lottery tickets are generally playing the lower stakes games such as pick 3 (with six bet types) and scratch-off cards.
There's no need to invoke "gambling addiction" to explain routinized behavior, and once it's routinized, the "them po' folks get a kick out of it" prong falls as well. Powerball is a multistate extravaganza, but I'm willing to venture – without actually looking at the books – that my state's lottery lives and breathes on the "penny ante" games.
This is where the social Darwinism unwraps itself from the metaphors, which is where I was trying to go in the first place. I'd never heard "tax on the mathematically challenged" before, but it seems all the more obnoxious for its cutesiness.
^ Back to Powerball's entertainment value, though, I left off that the few times I've played, there was a pretty consistent payout in terms of chatting with pleasantly interesting characters while waiting in line. YMMV.
“Everyone should choose their own numbers…. If players had created their own Quick Picks on Saturday night, I think all combinations would have been sold and we would have had a winner.”
I do wonder whether the quick pick machines are truly random or just pseudo random. I think if they were truly random, we would've heard some news story about someone suing because the machine gave them 1,2,3,4,5,6. I bet that just as weather.com makes their forecasts less accurate for public acceptance purposes, the quick pick machines are somewhat less than random for public acceptance purposes.
"Oh, and dean – you are completely right. The lottery is not a tax. "
So I'm right too. I say its not a tax:
The difference is that if you pay the lottery, you probably won’t get anything at all for it. If you pay your tax, you DO *definitely* get something for it.
"wow must be having fun trolling you"
By agreeing they're not the same???? Obviously you must be trolling dean...
Or an asshat.
"the few times I’ve played, there was a pretty consistent payout in terms of chatting with pleasantly interesting characters while waiting in line"
You can do that with any shopping visit. Or just walking in the street.
"Taxes are compulsory. Google it, wow."
Taxes are not compulsory. Google it,Carl. Or read the evidence I have given above.
The various "tax" metaphors are actually helpful, where they remind people that their lottery ticket money is probably wasted.
"If it only helps one person make a better fiscal decision, it's worth it." Isn't that how the metaphrase goes?
Wow, in this case you are definitely trolling. dean @2 wondered at the "tax" angle of the lottery metaphor, and you argued @3 that taxes were optional.
You spend the rest of your posts fighting to prove you're right, digging a deeper hole as you go. By @7 you are needlessly rude, @10 you start name calling (i.e. dean is a liar). None of that behavior was provoked.
The dictionary definition of "tax" is clear; they are compulsory. Here are the ones that I reference (search phrase "definition of tax"):
dictionary.reference.com/browse/tax
dictionary.reference.com/browse/taxes
www.merriam-webster.com/dictionary/tax
www.investopedia.com/terms/t/taxes.asp
Those are the top 5 search returns, and I've only stopped posting more because their is no point - they all agree.
My analysis - you like to argue and call names. You like to "win" arguments, and try to provoke them. You think you are smart enough to twist logic and language to "win" arguments, even when your position is wrong. Other than the name calling, you like to play "debate team." The name calling thing is just you.
You seem to be smart on many science topics that come up here. You've helped answer questions and educated posters, though you are often abrasive in the process (e.g. "Google it"). In this case you are just plain wrong.
I do wonder whether the quick pick machines are truly random or just pseudo random. I think if they were truly random, we would’ve heard some news story about someone suing because the machine gave them 1,2,3,4,5,6. I bet that just as weather.com makes their forecasts less accurate for public acceptance purposes, the quick pick machines are somewhat less than random for public acceptance purposes.
They are psuedo-random, since they are generated by a mechanical process. That said, the occurrences of the numbers pass tests for randomness, so they are as "random" as we can make them.
That said, the occurrences of the numbers pass tests for randomness, so they are as “random” as we can make them.
Well then, I'm surprised nobody's sued about getting a sequence. Maybe I'm cynical about numeracy, but I expect a lot of ticket buyers would interpret a 1,2,3,4,5,6 ticket as "machine broken, I want my money back."
Hey! That 1,2,3,4,5,6 ticket is mine! (see #12)
That line about playing the lottery indicating stupidity, comes out easily the stupidest stupid of them all. People don't buy a lottery ticket for the odds. They do it for inexpensive fun. Or hope. For home entertainment of the children. For facilitating god's miracle gift their way, just in case he wants to do that, only there's never any good openings.
The latter reason has really caught on in a big way since hundreds of worshippers heard God's booming words from the mouth of an evangelical speaking in tongues. Like totally for real and someone caught the audio on their Galaxy S5, sending it totally crazily viral with 40 million views on the same day. A spinster had got wind of the stupidest stupid clever turnip 8th grade probabiliificationing and took true umbrage in the direction of it and then took it out on God who had not miracle her a win eve, after playing with high fidelity since she was 37. She was screaming and crying out and banging her head against the vicars pocketed St James edition Holy New Testimant. How could He have forsaken her and now see her sodden in clever clog statistical peroratives She threw her arms up with her best profile side pumping a plum silhohette in afore mentioned Galaxy S5, which right at that point did God boom out "you never did buy a ticket you stingy old spinster"
You've been shaggy dogged baby
By Chris Mannering (not verified) on 16 Jan 2016 #permalink
"That line about playing the lottery indicating stupidity, comes out easily the stupidest stupid of them all"
No it doesn't.
Does spending \$3 to expect ~50c back sound like a sensible thing to do?
The best provable way not to loose at the lotteries is NOT TO BUY TICKEYS. And you can even win!
By Vincenzo Romano (not verified) on 20 Jan 2016 #permalink | 4,536 | 19,986 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-38 | latest | en | 0.926293 |
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## Delhi State Cancer Institute (DSCI) Staff Nurse Syllabus 2017
Download Pdf available for Delhi State Cancer Institute Syllabus 2017. Aspirants those participating in the New Delhi Cancer Institute can get Syllabus and Exam Pattern. Therefore, Download DSCI Sr/Jr Resident Exam Syllabus from this article. – www.dsci.nic.in
Applicants of New Delhi Government Jobs in Cancer Institute are looking for Exam Syllabus and Paper Pattern for Written test. In this article, you can get Staff Nurse, Jr/ Sr Resident, DSCI Computer Operator Syllabus and other syllabi. The Syllabus consists the topics that needed for Exam preparation. To make preparation efficient, we also given the Previous papers for this Written examination. You may also check the official website www.dsci.nic.in for more details on Cancer Institute Written test. Candidates must refer the Delhi State Cancer Institute Syllabus before they begin preparation. Practice the Model Papers of DSCI Exam from below enclosed link. So, Download these Sample papers as a reference to know the difficulty level of the examination.
### Delhi State Cancer Institute Sr/ Jr Resident Exam Pattern 2017
S.No Subject 1. Quantitative Aptitude 2. General knowledge 3. English 4. Reasoning 5. Discipline related questions
### DSCI Computer Operator Syllabus & Test Pattern
• Excepted New Delhi Cancer Institute exam will be of Objective type.
• For Assistant Engineer the questions will be asked on related discipline.
### DSCI Jr/Sr Resident Syllabus 2017 Pdf
The Syllabi consists the topics that may ask in the examination. Hence, aspirants need to check the required topics that needed to prepare for the Written test. So, Note down the topics in below sections.
### Delhi State Cancer Institute LDC Syllabus – Quantitative Aptitude
• Ratio and Time.
• Ratio and Proportion.
• Averages.
• Time and Distance.
• Mensuration.
• The relationship between Numbers.
• Use of Tables and Graphs.
• Computation of Whole Numbers.
• Time and Work.
• Decimals and Fractions.
• Number Systems.
• Profit and Loss.
• Discount.
• Fundamental arithmetical operations
• Percentages.
• Interest etc.
### DSCI Syllabus for Jr/ Sr Resident Exam 2017 – General Knowledge
• Economic Science
• Geography.
• Indian Constitution.
• History.
• Sports.
• Culture.
• Scientific Research.
• Current Events – International & National.
• General Polity.
Delhi State Cancer Institute Technologist Syllabus – English
• Grammar.
• Articles.
• Subject-Verb Agreement.
• Tenses.
• Error Correction.
• Word Formation.
• Conclusion.
• Theme detection.
• Vocabulary.
• Passage Completion.
• Synonyms.
• Idioms & Phrases.
• Unseen Passages.
• Antonyms.
• Sentence Completion.
• Sentence Rearrangement.
• Comprehension.
• Fill in the Blanks.
### Delhi Hospital DSCI Syllabus – Reasoning
• Space Visualization.
• Similarities and Differences.
• Non-Verbal Series.
• Analysis.
• Problem Solving.
• Decision Making.
• Judgment.
• Arithmetical Computation.
• Figure Classification.
• Visual Memory.
• Observation.
• Analytical Functions
• Relationship Concepts.
• Number Series etc.
### DSCI Professor Syllabus
• Surgical Oncology.
• Plastic & Reconstructive Surgery.
• Onco-Anaesthesia.
• Hemato-Oncology.
• Clinical Oncology.
• Intensive and Critical Care.
• Palliative Care and Pain Relief.
• Onco Imaging.
• Integrated Medicine including Yoga & Meditation.
• Onco-Pathology.
• Medical Physics.
• Lab Medicine.
• Internal Medicine.
• Chest & Respiratory Medicine.
• Nuclear Medicine.
• Cancer Research.
• Pediatric Oncology.
• Gastroenterology.
• Preventive Oncology & Onco-Epidemiology.
• Onco-Prosthodontics.
• Transfusion Medicine & Blood Bank.
• Psycho-Oncology.
### Delhi Cancer Nursing Institute Officer Syllabus 2017:
• Expected Syllabus for Staff Nurse Exam 2017
• Pharmaceutics.
• Pharmacology.
• Hospital & Clinical Pharmacy.
• Pharmaceutical Chemistry.
• Biochemistry.
• Health Education & Community Pharmacy.
• Clinical Pathology.
• Anatomy & Physiology
• Drug Store Management.
• First Aid.
• Pharmacognosy.
• Paediatric Nursing.
• Toxicology
• Human Anatomy & Physiology.
• Psychology.
• Midwifery & Gynaecological Nursing
• Microbiology.
• Environmental Hygiene.
• Sociology.
• Pharmaceutical Jurisprudence.
• Fundamentals of Nursing.
• Medical Surgical Nursing.
• Community Health Nursing.
• Psychiatric Nursing.
• Personal Hygiene.
• Computers in Nursing.
• Health Education & Communication Skills.
• Nursing Management.
• Mental Health.
• Nutrition.
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# Dajun
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# Science1
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centripetal force with ring chamber array
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Page 1 / 1
An introduction to the if then else control structure.
## Introduction to two way selection
We are going to introduce the control structure from the selection category that is available in every high level language. It is called the if then else structure. Asking a question that has a true or false answer controls the if then else structure. It looks like this:
```if the answer to the question is true then do thiselse because it's false do this```
In most languages the question (called a test expression ) is a Boolean expression . The Boolean data type has two values – true and false. Let's rewrite the structure to consider this:
```if expression is true then do thiselse because it's false do this```
Some languages use reserved words of: "if", "then" and "else". Many eliminate the "then". Additionally the "do this" can be tied to true and false. You might see it as:
```if expression is true action trueelse action false```
And most languages infer the "is true" you might see it as:
```if expression action trueelse action false```
The above four forms of the control structure are saying the same thing. The else word is often not used in our English speaking today. However, consider the following conversation between a mother and her child.
Child asks, "Mommy, may I go out side and play?"
Mother answers, "If your room is clean then you may go outside and play or else you may go sit on a chair for five minutes as punishment for asking me the question when you knew your room was dirty."
Let's note that all of the elements are present to determine the action (or flow) that the child will be doing. Because the question (your room is clean) has only two possible answers (true or false) the actions are mutually exclusive . Either the child 1) goes outside and plays or 2) sits on a chair for five minutes. One of the actions is executed; never both of the actions.
## One choice – implied two way selection
Often the programmer will want to do something only if the expression is true, that is with no false action. The lack of a false action is also referred to as a "null else" and would be written as:
```if expression action trueelse do nothing```
Because the "else do nothing" is implied, it is usually written in short form like:
```if expression action true```
## Two way selection within c++
The syntax for the if then else control structure within the C++ programming language is:
```if (expression) statement;else statement;```
Note: The test expression is within the parentheses, but this is not a function call. The parentheses are part of the control structure. Additionally, there is no semicolon after the parenthesis following the expression.
## Definitions
if then else
A two way selection control structure.
mutually exclusive
Items that do not overlap. Example: true and false.
how can chip be made from sand
is this allso about nanoscale material
Almas
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where is the latest information on a no technology how can I find it
William
currently
William
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
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Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
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Nejxty Joljan
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Registered: 14.10.2002
From:
Posted: Saturday 30th of Dec 10:26 Yes ! I agree with you! The money back guarantee that comes with the purchase of Algebrator is one of the attractive options. In case you are dissatisfied with the help offered to you on any math topic, you can get a refund of the payment you made towards the purchase of Algebrator within the number of days specified on the label. Do take a look at http://www.mathfraction.com/equivalent-fractions-2.html before you place the order since that gives a lot of information about the areas on which you can expect to get assisted.
sxAoc
Registered: 16.01.2002
From: Australia
Posted: Sunday 31st of Dec 21:08 Algebrator is the program that I have used through several math classes - Pre Algebra, Pre Algebra and Intermediate algebra. It is a truly a great piece of math software. I remember of going through difficulties with distance of points, subtracting exponents and rational equations. I would simply type in a problem from the workbook , click on Solve – and step by step solution to my algebra homework. I highly recommend the program.
cucei
Registered: 08.03.2002
From: UK
Posted: Monday 01st of Jan 11:18 Thanks guys. There is no harm in trying it once. Please give me the link to the software.
Bet
Registered: 13.10.2001
From: kµlt øƒ Ø™
Posted: Tuesday 02nd of Jan 10:43 You can find the program here http://www.mathfraction.com/multiplying-and-dividing-fractions.html. | 832 | 3,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-09 | latest | en | 0.912626 |
https://community.filemaker.com/thread/121476 | 1,516,090,165,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886237.6/warc/CC-MAIN-20180116070444-20180116090444-00031.warc.gz | 654,767,939 | 31,952 | # Repost: Script help needed to determine shipping boxes needed
Question asked by JasonO'Berry on Jun 24, 2013
Latest reply on Jul 2, 2013 by philmodjunk
### Title
Repost: Script help needed to determine shipping boxes needed
### Post
Hi,
I am in need of some help crafting, what is to me, a complicated calculation. My company ships Art prints in different sized tubes. Depending on the size of the print and the quantity, I need a calc that can determine which box to use, and the final weight.
Each of our tubes can hold up to 3 prints. So an easy example, if a customer orders 1, 2, or 3 small prints, the calc should determine that 1 small tube can be used. But if they order 4 then 2 small tubes are required.
But if someone orders 1 extra large print and 2 small prints, it should determine that 1 extra large tube can be used.
It get more complicated if say a customer orders 1 extra large, 1 large, and 3 small prints. In that case I always want to group the larger prints into the largest tube. So the X-large, Large and 1 small print should use 1 extra large tube, and the remaining 2 small prints should use 1 small tube.
Is this an example of where some recursive function should be used or can this be achieved with standard operations.
I enter all products in on a line item table and that is where the size field is also. I've also created a boxes table to hold the tube sizes, dimensions, and max quantity.
UPDATE: I forgot to mention that I have a quantity field that needed to be figured into the script as well.
@Ninja & PhilModJunk
"I've been trying to get your previous examples to work for me but so far unsuccessfully. Mostly due to the quantity field and using a larger tube if it still has space available. I guess I'm not very experienced with looping and placing commands in the right place. I can get the number of the sizes of prints into variable, but I can't figure out how to correctly step through the variables to get the needed tubes with a 3 print maximum. I've put in the work, I just can't get the results I need. Your help is appreciated.maybe a more detailed walk through is what I need. Thanks."
"This is getting more complicated by the second. I thought I had a working loop but it would not take into account if the max had not been reached for a larger tube, when it started working on a smaller tube. It would just create another tube and start using the smallest size tube that would accomodate that print, leaving the previous tube with remaining space.
Please Help. I would post what I came up with, but i think it's was too complicated and probably not a good starting point for help. So back to the chalkboard." | 610 | 2,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-05 | longest | en | 0.941911 |
http://www.greenemath.com/Algebra%20II/43/FactoringTrinomialswithaleadingcoefficientthatisnot1factoringbygroupingmethodLesson.html | 1,493,151,631,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917120878.96/warc/CC-MAIN-20170423031200-00189-ip-10-145-167-34.ec2.internal.warc.gz | 552,272,341 | 5,121 | GreeneMath.com - Factoring Trinomials with a Leading Coefficient that is not 1 using the Factoring by Grouping Method Lesson
# In this Section:
In this lesson, we review how to factor a trinomial into the product of two binomials when the leading coefficient is not one. For this scenario, the process is much more tedious. We generally can use two different methods: factoring by grouping, or reverse FOIL. In order to factor a trinomial with grouping, we first re-write the trinomial as a four term polynomial. We do this by finding two integers whose product is a • c and whose sum is b. We use those two integers to expand the middle term; we can then use factoring by grouping to attain the product of two binomials. The alternative method uses reverse FOIL. When we use reverse FOIL, we must undo the FOIL process. In most cases, this process is more tedious than using the factoring by grouping method. We will also look at some special case scenarios that require us to factor out the GCF before we begin or factor when two variables are involved.
Sections:
# In this Section:
In this lesson, we review how to factor a trinomial into the product of two binomials when the leading coefficient is not one. For this scenario, the process is much more tedious. We generally can use two different methods: factoring by grouping, or reverse FOIL. In order to factor a trinomial with grouping, we first re-write the trinomial as a four term polynomial. We do this by finding two integers whose product is a • c and whose sum is b. We use those two integers to expand the middle term; we can then use factoring by grouping to attain the product of two binomials. The alternative method uses reverse FOIL. When we use reverse FOIL, we must undo the FOIL process. In most cases, this process is more tedious than using the factoring by grouping method. We will also look at some special case scenarios that require us to factor out the GCF before we begin or factor when two variables are involved. | 446 | 2,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2017-17 | latest | en | 0.92803 |
http://www.felienne.com/archives/358 | 1,511,259,286,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806338.36/warc/CC-MAIN-20171121094039-20171121114039-00752.warc.gz | 408,002,352 | 10,332 | The excel CHOOSE function
Today I learnt another cool Excel function I was previously unaware of, the CHOOSE function. As arguments it takes one value n, followed by a list of values that. CHOOSE will return the nth value of that list.
=CHOOSE(A1,”A”,”B”,”C”,”D”,”E”,”F”,”G”) returns “C” if A1 contains 3
You get the idea. I do not know for what scenario’s this function comes in handy.
4 Comments
1. Well this is basically an array access in Java, C#, C, you name it…
int A1 = 3;
char[] letters = new char[]{‘A’,’B’,’C’,’D’,’E’,’F’,’G’};
letters[A1] // returns “C”
I don’t know in which language this particular piece of code works, but it will work in at least one language I’m sure 🙂
P.S. I also don’t know for what Excel use case this will be useful but people can get *really* creative with Excel… (which you probably know better than me).
1. admin (Post author)
You are right, it is a bit like an array operation, but with funky syntax. And I am still not sure on how this is used. That calls for a paper! 🙂
2. admin (Post author)
I dug into this formula a bit more, and it turns out there is another, quite similar, way to do this:
=VLOOKUP(A1,{1,”A”;2,”B”;3,”C”;4,”D”;5,”E”;6,”F”;7,”G”},2,0)
What do you think, is this better or even worse?
Pro: The VLOOKUP is a common one, so it is more clear that something is to be found here.
Con: This version is longer
3. ska
I’m also quite intrigued with this formula. Been looking for a useful scenario, but find it hard to find one 🙂
Much as I dislike VLOOKUP (prefer INDEX(MATCH) any day 🙂 ) using CHOOSE in array format can enable one to do a “reverse VLOOKUP”, which is not possible with the normal version of VLOOKUP. (This is not a problem with INDEX(MATCH), by the way.)
Lifting a comment from here (1):
“CHOOSE can also be used to piece together an array of non-contiguous ranges. This can be very useful if you want to do a VLOOKUP to the left. (I.e. looking up the value in a column on the right side of a table and retrieving the value from a column on the left side of the table.) For example if your table is the range A1:C10 and you want to lookup a value in column C and return the corresponding value in column A you could use the following formula:
=VLOOKUP(lookup_value,CHOOSE({1,2},C1:C10,A1:A10),2,0)”
I also came across a really cool way of calculating the first Monday of the month here: http://www.globaliconnect.com/excel/index.php?option=com_content&view=article&id=106:choose-function-in-excel&catid=78&Itemid=474
Comments are closed. | 692 | 2,532 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-47 | longest | en | 0.891754 |
http://www.evi.com/q/.8_lbs_equals_how_many_ounces | 1,394,176,735,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999636575/warc/CC-MAIN-20140305060716-00039-ip-10-183-142-35.ec2.internal.warc.gz | 329,150,136 | 14,043 | You asked:
# .8 lbs equals how many ounces
• the mass 12.8 ounces
• tk10publ tk10ncanl
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• what's .8lbs in oz | 496 | 1,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2014-10 | longest | en | 0.926165 |
https://math.libretexts.org/Bookshelves/Precalculus/Book%3A_Precalculus_-_An_Investigation_of_Functions_(Lippman_and_Rasmussen)/2%3A_Linear_Functions/2.1%3A_Linear_Functions/2.1E%3A_Linear_Functions_(Exercises) | 1,571,736,517,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987813307.73/warc/CC-MAIN-20191022081307-20191022104807-00295.warc.gz | 609,979,711 | 21,289 | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 2.1E: Linear Functions (Exercises)
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
Section 2.1 exercise
1. A town's population has been growing linearly. In 2003, the population was 45,000, and the population has been growing by 1700 people each year. Write an equation,
$$P(t)$$, for the population $$t$$ years after 2003.
2. A town's population has been growing linearly. In 2005, the population was 69,000, and the population has been growing by 2500 people each year. Write an equation,
$$P(t)$$, for the population $$t$$ years after 2005.
3. Sonya is currently 10 miles from home, and is walking further away at 2 miles per hour. Write an equation for her distance from home $$t$$ hours from now.
4. A boat is 100 miles away from the marina, sailing directly towards it at 10 miles per hour. Write an equation for the distance of the boat from the marina after $$t$$ hours.
5. Timmy goes to the fair with $40. Each ride costs$2. How much money will he have left after riding n rides?
6. At noon, a barista notices she has $20 in her tip jar. If she makes an average of$0.50 from each customer, how much will she have in her tip jar if she serves $$n$$ more customers during her shift?
Determine if each function is increasing or decreasing
7. $$f(x) = 4x + 3$$
8. $$g(x) = 5x + 6$$
9. $$a(x) = -2x + 4$$
10. $$b(x) = 8 - 3x$$
11. $$h(x) = -2x + 4$$
12. $$h(x) = -4x + 1$$
13. $$j(x) = \dfrac{1}{2}x - 3$$
14. $$p(x) = \dfrac{1}{4} x - 5$$
15. $$n(x) = -\dfrac{1}{3} x - 2$$
16. $$m(x) = -\dfrac{3}{8} x + 3$$
Find the slope of the line that passes through the two given points
17. (2, 4) and (4, 10)
18. (1, 5) and (4, 11)
19. (-1, 4) and (5, 2)
20. (-2, 8) and (4, 6)
21. (6, 11) and (-4, 3)
22. (9, 10) and (-6, -12)
Find the slope of the lines graphed
23. 24.
25. Sonya is walking home from a friend’s house. After 2 minutes she is 1.4 miles from home. Twelve minutes after leaving, she is 0.9 miles from home. What is her rate?
26. A gym membership with two personal training sessions costs $125, while gym membership with 5 personal training sessions costs$260. What is the rate for personal training sessions?
27. A city's population in the year 1960 was 287,500. In 1989 the population was 275,900. Compute the slope of the population growth (or decline) and make a statement about the population rate of change in people per year.
28. A city's population in the year 1958 was 2,113,000. In 1991 the population was 2,099,800. Compute the slope of the population growth (or decline) and make a statement about the population rate of change in people per year.
29. A phone company charges for service according to the formula: $$C(n) = 24 0.+1n$$, where $$n$$ is the number of minutes talked, and $$C(n)$$ is the monthly charge, in dollars. Find and interpret the rate of change and initial value.
30. A phone company charges for service according to the formula: $$C(n) = 26 0.+04n$$, where $$n$$ is the number of minutes talked, and $$C(n)$$ is the monthly charge, in dollars. Find and interpret the rate of change and initial value.
31. Terry is skiing down a steep hill. Terry's elevation, $$E(t)$$, in feet after $$t$$ seconds is given by $$E(t) = 3000 - 70t$$.Write a complete sentence describing Terry’s starting elevation and how it is changing over time.
32. Maria is climbing a mountain. Maria's elevation, $$E(t)$$, in feet after $$t$$ minutes is given by $$E(t) =1200 + 40t$$. Write a complete sentence describing Maria’s starting elevation and how it is changing over time.
Given each set of information, find a linear equation satisfying the conditions, if possible
33. $$f(-5) = -4$$, and $$f(5) = 2$$
34. $$f(-1) = 4$$, and $$f(5) = 1$$
35. Passes through (2, 4) and (4, 10)
36. Passes through (1, 5) and (4, 11)
37. Passes through (-1, 4) and (5, 2)
38. Passes through (-2, 8) and (4, 6)
39. $$x$$ intercept at (-2, 0) and $$y$$ intercept at (0, -3)
40. $$x$$ intercept at (-5, 00 and $$y$$ intercept at (0, 4)
Find an equation for the function graphed
41. 42.
43. 44.
45. A clothing business finds there is a linear relationship between the number of shirts, $$n$$, it can sell and the price, $$p$$, it can charge per shirt. In particular, historical data shows that 1000 shirts can be sold at a price of $30, while 3000 shirts can be sold at a price of$22 . Find a linear equation in the form $$p = mn + b$$ that gives the price $$p$$ they can charge for $$n$$ shirts.
46. A farmer finds there is a linear relationship between the number of bean stalks, $$n$$, she plants and the yield, $$y$$, each plant produces. When she plants 30 stalks, each plant yields 30 oz of beans. When she plants 34 stalks, each plant produces 28 oz of beans. Find a linear relationships in the form $$y = mn + b$$ that gives the yield when $$n$$ stalks are planted.
47. Which of the following tables could represent a linear function? For each that could be linear, find a linear equation models the data.
48. Which of the following tables could represent a linear function? For each that could be linear, find a linear equation models the data.
49. While speaking on the phone to a friend in Oslo, Norway, you learned that the current temperature there was -23 Celsius (-23$${}^{o}$$C). After the phone conversation, you wanted to convert this temperature to Fahrenheit degrees, oF, but you could not find a reference with the correct formulas. You then remembered that the relationship between $${}^{o}$$F and $${}^{o}$$C is linear. [UW]
Using this and the knowledge that 32$${}^{o}$$F = 0 $${}^{o}$$C and 212 $${}^{o}$$F = 100 $${}^{o}$$C, find an equation that computes Celsius temperature in terms of Fahrenheit; i.e. an equation of the form C = “an expression involving only the variable F.”
Likewise, find an equation that computes Fahrenheit temperature in terms of Celsius temperature; i.e. an equation of the form F = “an expression involving only the variable C.”
How cold was it in Oslo in $${}^{o}$$F?
1. $$P(t) = 1700t + 45000$$
3. $$D(t) = 10 + 2t$$
5. $$M(n) = 40 - 2n$$
7. Increasing
9. Decreasing
11. Decreasing
13. Increasing
15. Decreasing
17. 3
19. $$-\dfrac{1}{3}$$
21. $$\dfrac{4}{3}$$
23. $$\dfrac{2}{3}$$
25. -0.05 mph (or 0.05 miles per hour toward her home)
27. Population is decreasing by 400 people per year
29. Monthly charge in dollars has an initial base charge of $24, and increases by$0.10 for each minute talked
31. Terry started at an elevation of 3,000 ft and is descending by 70ft per second.
33. $$y = \dfrac{3}{5} x - 1$$
35. $$y = 3x - 2$$
37. $$y = -\dfrac{1}{3}x + \dfrac{11}{3}$$
39. $$y = -1.5x - 3$$
41. $$y = \dfrac{2}{3} x + 1$$
43. $$y = -2x + 3$$
45. $$P(n) = -0.004n + 34$$
47. The $$1^{\text{st}}$$, $$3^{\text{rd}}$$ & $$4^{\text{th}}$$ tables are linear: respectively
1. $$g(x) = -3x + 5$$
3. $$f(x) = 5x - 5$$
4. $$k(x) = 3x - 2$$
49a. $$C = \dfrac{5}{9} F - \dfrac{160}{9}$$
b. $$F = \dfrac{9}{5} C + 32$$
c. $$-9.4^{\cric} F$$ | 2,427 | 7,505 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2019-43 | latest | en | 0.829237 |
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Home » Tip Bank » Visual Basic » Windows
Language: C++
Expertise: Beginner
Apr 24, 2000
### WEBINAR:On-Demand
Building the Right Environment to Support AI, Machine Learning and Deep Learning
# The unique() Algorithm
STL's unique() algorithm eliminates all but the first element from every consecutive group of equal elements in a sequence of elements. unique() takes two forward iterators, the first of which marks the sequence's beginning and the second marks its end. The algorithm scans the sequence and removes repeated identical values. The duplicate values are moved to the sequence's end. For example, applying unique to the following sequence of integers:
``````
1,0,0,9,2
``````
changes their order to:
``````
1,0,9,2,0
``````
unique() doesn't really delete the duplicate values; it moves them to the container's end. The third element in the original container is moved one position past the final 2. unique() returns an iterator pointing to the logical end of the container. In other words, it returns an iterator pointing to the element 2, not 0. You can delete all the duplicate elements that were moved past the logical end using the iterator returned from unique(). For example:
``````
int main()
{
vector <int> vi;
vi.push_back(1); // insert elements into vector
vi.push_back(0);
vi.push_back(0);
vi.push_back(9);
vi.push_back(2);
//move consecutive duplicates past the end; store new end
vector<int>::iterator new_end=
unique(vi.begin(), vi.end());
// delete all elements past new_end
vi.erase(new_end, vi.end());}
}
``````
unique() is useful when you want to make sure that only a single instance of a given value exists in a range, regardless of its distribution. For example, suppose you want to write an automated dictionary containing all the words Shakespeare's texts. First, you scan all the words in his texts and store them in a vector. Next, you sort the vector using the sort() algorithm. Finally, apply the unique() algorithm to the sorted vector to move duplicate words past its logical end and erase the duplicates as shown above.
Danny Kalev
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Thanks for your registration, follow us on our social networks to keep up-to-date | 658 | 2,877 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-39 | longest | en | 0.83129 |
http://de.metamath.org/ileuni/exists2.html | 1,529,397,874,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861981.50/warc/CC-MAIN-20180619080121-20180619100121-00468.warc.gz | 79,338,266 | 3,702 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > exists2 GIF version
Theorem exists2 1851
Description: A condition implying that at least two things exist. (Contributed by NM, 10-Apr-2004.) (Proof shortened by Andrew Salmon, 9-Jul-2011.)
Assertion
Ref Expression
exists2 ((xφ x ¬ φ) → ¬ ∃!x x = x)
Proof of Theorem exists2
Dummy variable y is distinct from all other variables.
StepHypRef Expression
1 hbeu1 1776 . . . . . 6 (∃!x x = xx∃!x x = x)
2 hba1 1364 . . . . . 6 (xφxxφ)
3 exists1 1850 . . . . . . 7 (∃!x x = xx x = y)
4 ax-16 1570 . . . . . . 7 (x x = y → (φxφ))
53, 4sylbi 112 . . . . . 6 (∃!x x = x → (φxφ))
61, 2, 5exlimd 1407 . . . . 5 (∃!x x = x → (xφxφ))
76com12 25 . . . 4 (xφ → (∃!x x = xxφ))
8 alex 1759 . . . 4 (xφ ↔ ¬ x ¬ φ)
97, 8syl6ib 148 . . 3 (xφ → (∃!x x = x → ¬ x ¬ φ))
109con2d 535 . 2 (xφ → (x ¬ φ → ¬ ∃!x x = x))
1110imp 113 1 ((xφ x ¬ φ) → ¬ ∃!x x = x)
Colors of variables: wff set class Syntax hints: ¬ wn 3 → wi 4 ∧ wa 95 ∀wal 1266 ∃wex 1313 = wceq 1324 ∃!weu 1766 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 97 ax-ia2 98 ax-ia3 99 ax-in1 526 ax-in2 527 ax-io 606 ax-3 719 ax-5 1267 ax-7 1268 ax-gen 1269 ax-ie1 1314 ax-ie2 1315 ax-8 1328 ax-10 1329 ax-11 1330 ax-i12 1331 ax-4 1333 ax-17 1349 ax-i9 1353 ax-ial 1358 ax-16 1570 This theorem depends on definitions: df-bi 108 df-tru 1201 df-fal 1202 df-eu 1770
Copyright terms: Public domain W3C validator | 728 | 1,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-26 | latest | en | 0.190791 |
https://bugs.dojotoolkit.org/ticket/8108 | 1,642,397,188,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300289.37/warc/CC-MAIN-20220117031001-20220117061001-00554.warc.gz | 223,287,554 | 7,055 | Opened 13 years ago
Closed 12 years ago
# straight line graphs with no axes don't show up
Reported by: Owned by: nazanin Eugene Lazutkin high 1.5 Charting 1.2.1 straight line graph [email protected]…
### Description
1. Create a graph with no axes with these values: (0,10), (1,10),(2,10),(3,10),..
Result: the graph doesn't show up since the max y value minus min y value becomes zero and scale which is span/(max - min) will become NaN.
Fixed it with this patch:
```--- a/public/javascripts/dojotoolkit/dojox/charting/scaler/primitive.js
+++ b/public/javascripts/dojotoolkit/dojox/charting/scaler/primitive.js
@@ -2,13 +2,17 @@ dojo.provide("dojox.charting.scaler.primitive");
dojox.charting.scaler.primitive = {
buildScaler: function(/*Number*/ min, /*Number*/ max, /*Number*/ span, /*Object*/ kwArgs){
+ if (min == max)
+ scale = min;
+ else
+ scale = span/(max - min);
return {
bounds: {
lower: min,
upper: max,
from: min,
to: max,
- scale: span / (max - min),
+ scale: scale,
span: span
},
scaler: dojox.charting.scaler.primitive
```
### comment:1 Changed 13 years ago by Eugene Lazutkin
Resolution: → invalid new → closed
Use Dojo 1.2.2 or the trunk. Set min/max parameters on the axis.
### comment:2 Changed 13 years ago by nazanin
Resolution: invalid closed → reopened
well that's the point. This happens when you graph with no axis''' in dojo/dojox/charting/plot2d/Base.js _calc function decides whether the graph has axis or not. So this problems doesn't occur when there are axis.
### comment:3 Changed 13 years ago by Eugene Lazutkin
Milestone: tbd → 1.3
Yep, missed the point. Thank you for reopening.
### comment:4 Changed 13 years ago by Eugene Lazutkin
Milestone: 1.3 → future
Moving open tickets to the future.
### comment:5 Changed 12 years ago by Eugene Lazutkin
Milestone: future → 1.4
### comment:6 Changed 12 years ago by Eugene Lazutkin
Milestone: 1.4 → 1.5
bumping tickets that didn't make the 1.4 cut, but most likely to go in the next point release. | 589 | 2,103 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-05 | latest | en | 0.728806 |
https://justaaa.com/statistics-and-probability/787863-weight-of-men-is-normal-distributed-with-mean-of | 1,721,748,206,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518058.23/warc/CC-MAIN-20240723133408-20240723163408-00184.warc.gz | 281,666,378 | 9,972 | Question
# weight of men is normal distributed with mean of 185 lb and standard deviation of 39...
weight of men is normal distributed with mean of 185 lb and standard deviation of 39 lb. A water taxi that holds 30 passengers has a maximum capacity of 6000 lb. If taxi is fully loaded with men, what is the probability that it goes over the maximum capacity? Round to 4 decimal places.
average capacity for a man =6000/30 =200
for normal distribution z score =(X-μ)/σ here mean= μ= 185 std deviation =σ= 39.0000 sample size =n= 30 std error=σx̅=σ/√n= 7.1204
probability that it goes over the maximum capacity:
probability = P(X>200) = P(Z>2.11)= 1-P(Z<2.11)= 1-0.9826= 0.0174
( please try 0.0176 if this comes wrong due to rounding)
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 233 | 815 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2024-30 | latest | en | 0.862108 |
https://www.coursehero.com/file/6237489/8/ | 1,495,714,940,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608058.57/warc/CC-MAIN-20170525102240-20170525122240-00135.warc.gz | 858,767,257 | 61,354 | # 8 - 18.05 Lecture 8 3.1 Random Variables and Distributions...
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18.05 Lecture 8 February 22, 2005 § 3.1 -Random Variables and Distributions Transforms the outcome of an experiment into a number. De±nitions: Probability Space: (S, A , P ) S -samp le space, A -events , P -probab i l ity Random variable is a function on S with values in real numbers, X:S R Examples: Toss a coin 10 times, Sample Space = { HTH. ..HT, .... } , all con±gurations of H T. Random Variable X = number of heads, X: S R X: S →{ 0 , 1 , ..., 10 } for this example. There are fewer outcomes than in S, you need to give the distribution of the random variable in order to get the entire picture. Probabilities are therefore given. De±nition: The distribution of a random variable X:S R , is de±ned by: A R , P ( A ) = P ( X A ) = P ( s S : X ( s ) A ) The random variable maps outcomes and probabilities to real numbers. This simpli±es the problem, as you only need to de±ne the mapped R , P , not the
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## This note was uploaded on 05/02/2011 for the course DYNAM 101 taught by Professor Matuka during the Spring '11 term at MIT.
### Page1 / 3
8 - 18.05 Lecture 8 3.1 Random Variables and Distributions...
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Ask a homework question - tutors are online | 426 | 1,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-22 | longest | en | 0.865728 |
https://www.omnicalculator.com/health/bac | 1,726,597,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00655.warc.gz | 858,503,834 | 43,028 | # BAC Calculator (Blood Alcohol Content)
Created by Mateusz Mucha, Jack Bowater, Maria Kluziak and Łucja Zaborowska, MD, PhD candidate
Reviewed by Dominik Czernia, PhD
Based on research by
Hustad JT, Carey KB. Using calculations to estimate blood alcohol concentrations for naturally occurring drinking episodes: a validity study.; Journal of Studies on Alcohol and Drugs; January 2005See 1 more source
Peck RC, Gebers MA, Voas RB, Romano E. The relationship between blood alcohol concentration (BAC), age, and crash risk; Journal of Safety Research; May 2008
Last updated: May 16, 2024
The BAC Calculator is a tool that can help you assess the concentration of alcohol in your blood. The calculation is based on what kind of alcohol you drank, how much of it you consumed, and how long ago.
As your blood alcohol concentration depends on a variety of aspects, this sobriety calculator takes into consideration your body weight and sex as well. Have you ever considered why one person gets drunk more easily than another? Or wondered how this changes depending on the weight, sex, or type of alcohol? If so, keep reading.
## How to use blood alcohol calculator?
To find out what your blood alcohol content actually is, you only need to fill the blanks with the number of beers, wine glasses, and/or vodka shots you drank. Do not ask friends; the calculator can only work with the exact amount that you drank. If you can't, you don't need to assess it; you are certainly drunk. When you have done that, plug in your weight, sex, and the time when you started drinking. The BAC Calculator will estimate the percentage of your blood comprised of alcohol.
Let’s take an example:
1. Large beers: 3
2. Wine glasses: 0
3. Vodka shots: 5
4. Body weight: 90
5. Sex: male
6. Started drinking: 8 hrs ago
The calculator shows your blood as 0.6‰ alcohol.
You can also use the advanced version of the calculator, which contains an additional “raw alcohol” input. You can either fill it with the amount of alcohol you drank, typing it in grams, or fill in the first three blank fields, and the calculator will display your raw blood alcohol content. For our 3 large beers and 5 vodka shots, which we have plugged in, we get 165 grams of raw alcohol.
## Why use the drunk calculator?
Sometimes, after drinking alcohol, you feel horrible. If this is the case, then you are still under the influence of alcohol. However, not everyone's hangover is a disaster. In such a situation, you may want to drive home in the morning (or afternoon) but are unsure if the breathalyzer test would show any alcohol content in your blood. That's why you can use the BAC Calculator to measure your blood alcohol concentration before you set off. However, it's worth confirming the result with another tool, like a store-bought breathalyzer test, which can be found in most pharmacies.
You can use the drunk calculator to appease your curiosity or to compare your result with those of your friends. If you are interested in other calculators related to the topic, you should take a look at our BMR calculator and wedding alcohol calculator.
## Sobriety calculator
This blood alcohol calculator can not only be used to check the concentration of alcohol in your blood right now but also to estimate when your blood alcohol content is going to equal 0. This is when you're fully sober again — it might be longer than you imagine! This number is presented below your current blood alcohol level. Thanks to this function, you can find out how much time you need to get a particular amount of alcohol out of your system and how it varies depending on the type of drinks you consume.
## How BAC affects your behavior
It is good to know how exactly a certain concentration of alcohol affects your organism, especially if you're trying to estimate when it will be okay to drive again.
BAC
Effects
0.02%
Relaxation, slight impairment of visual functions, slight impairment to divided attention.
0.05%
All above and loss of small-muscle control leading to difficulty focusing eyes, impaired judgment, lowered alertness, and reduced coordination.
0.08%
All above plus poor muscle coordination, impaired speech, impaired self-control, reason, and memory.
0.10%
All above and poor coordination, slurred speech, slowed thinking.
0.15%
All above and potential vomiting, major loss of balance.
## In conclusion, the BAC calculator can tell you:
• How drunk you are (regardless of how you feel).
• How long you need to get back to total sobriety.
• How the three most common alcohol types affect your blood alcohol content.
🙋 Are you worried about your alcohol consumption pattern? Check if you're an alcoholic with a test developed by WHO in our audit score calculator.
## FAQ
### What does BAC stand for?
BAC stands for blood alcohol concentration. It quantifies the result of a test to measure the percentage of alcohol in the blood. This level can be determined by a breath test or by chemical tests.
### How can I calculate BAC?
You can use the Widmark's formula to calculate BAC:
1. Determine the amount of alcohol consumed in grams, A.
2. State your body weight in kilograms, Wt.
3. Find the ratio of body water to total weight, bw.
4. Choose a time (t) in hours since alcohol consumption began.
5. Estimate the metabolism rate (mr): typically, it's 0.017% per hour.
6. Use the Widmark's formula:
BAC = A/(bw × Wt) × 100% − mr × t
### Which factor is the only way to lower BAC?
The only factor that reduces BAC is spending time without drinking. You need to give your body time to absorb and get rid of the alcohol. Popular beliefs say that coffee, a cold shower, or drinking a glass of water will help you sober up faster, but this is not true.
### What BAC means that I am drunk?
You are likely to start feeling the effects of being drunk above 0.08% BAC. At this point, you may experience mild impairment of speech, vision, coordination, and reaction times. This is why in the United States (except Utah) you cannot drive under the influence of alcohol at 0.08% BAC or higher.
### Which factors affect BAC levels?
Factors affecting BAC are:
• Weight — The heavier the person, the lower the BAC;
• Sex — Men generally metabolize alcohol at a faster rate than women;
• Medications — Some drugs can interact with alcohol and damage the liver;
• Speed of sipping — Drink slowly to keep your intoxication under control;
• Food — Drinking on an empty stomach results in faster alcohol absorption and intoxication; and
• Number of standard drinks you drank.
### What is my BAC after 2 beers?
0.06% or 0.6‰, assuming you are a US man weighing 180 lbs and just had two large beers (50 grams of raw alcohol). You will need 4 hours for your body to metabolize this amount of alcohol.
Mateusz Mucha, Jack Bowater, Maria Kluziak and Łucja Zaborowska, MD, PhD candidate
I am from
Body weight
lb
Sex
Male
Started drinking
hrs
ago
I drank
large beers
...and
wine glasses
BAC
3
Time until sober
20
hrs
1
min
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# Presenting teaching with variation
In this video we will introduce teaching with variation.
0
Teaching with Variation is being used in mathematics classrooms for a long time. Based on the previous experiences and his own experiment in Shanghai mathematics classrooms, Professor Gu who’s one of the most well known mathematics educators in China, systematically analyzed and synthesized the concept of Teaching with Variation. You will hear what he has to say later in this course about Teaching with Variation. And according to Professor Gu, there are two forms of variations, conceptual variation and procedural variation. Conceptual variation is about understanding concepts from multiple perspectives. As Professor Gu said, so it is to illustrate the essential features by demonstrating different forms of visual materials and instances, or highlight the essence of a concept by varying the non-essential features.
91.9
So here is one example to help students understand the concept of the height, or altitude of a triangle from the top vertices. So let’s look at the first triangle. Normally we will use this kind of position, so it is easy for student to understand the concept of the height. So this is the height. Then in this triangle, the height and one side of the triangle are the same. So that is height, so that is a bit more challenging. And the following three are variations to understand the concept. So the third one is this one. So you can see some students will make mistakes like this.
160.3
They draw a horizontal line, and then they draw a vertical line from the top vertex of the triangle to this horizontal line. Of course this is incorrect. The correct one is this. Okay, the next one is another variation. The situation for students to learn the concept. So you should extend this side beyond this point. Then you draw a height, that is a perpendicular line from the top vertex of this triangle to this horizontal line. So that might not be so difficult for students. So the most difficult one is this one. So many students will likely draw a horizontal line passing through this vertex of the triangle.
239.5
Then they will draw a line from the top vertex here, perpendicular to the horizontal line. So that is the height they might get, but this is incorrect. So the correct one should be this one. You extend this side of the triangle, go beyond this point. Then you draw a perpendicular line to this side, the extension of the side from the top vertex of the triangle, so that is height. The next is procedural variation. It means to progressively unfold mathematical and activities. The variations for constructing a particular experience system derived from three dimensions of problem solving.
308.6
First, varying a problem which means varying the original one as a scaffolding or extending the original problem by varying the conditions changing the results and generalization. The second is multiple methods of solving a problem by varying the different processes of solving a problem and associating different methods of solving a problem. So you vary the process of solving the problem. And the third one is multiple applications of a method by applying the same method to a group of similar problems. So in Chinese we say you use one method to solve three problems or even more. So here is an example, and you will think it’s a very simple example. So we ask students to calculate 2 plus 8 equals how much.
383.2
Of course student will know it’s 10. In terms of variation theory, we can vary the problem into different scenarios. So variation one is 2 plus what equals 10. And variation two is what plus 8 equals 10. Variation three is more difficult. So what number plus what number equals 10. So students are asked to find two numbers whose sum is 10. And variation four is 10 equals 2 plus what. So I’m sure you can find more ways to vary the problems. According to researchers in this area, that Teaching with Variation is a main feature of Chinese mathematics classroom.
456.9
And by adopting Teaching with Variation, even with a large classes like in Shanghai and in many other Asian classrooms, students still can actively involve themselves in the process of learning. And they can achieve excellent results.
This video explains what teaching with variation entails and will demonstrate the principle with some concrete examples.
In China, teaching with variation has been used in classrooms for a long time.
Professor Gu is one of the most well-known mathematics educators in China. Based on previous experience and his own experiments in Shanghai maths classrooms, he systematically analysed and synthesised the concepts of teaching with variation (1981). According to Gu, there are two forms of variation: ‘conceptual variation’ and ‘procedural variation’.
According to researchers (Gu, Huang, & Marton, 2004), teaching with variation characterises mathematics teaching in China and by adopting teaching with variation, even with large classes, students can actively involve themselves in the process of learning and achieve excellent results.
References
Gu, L. (1981). The visual effect and psychological implication of transformation of figures in geometry. Paper presented at annual conference of Shanghai Mathematics Association. Shanghai, China.
Gu, L., Huang, R., & Marton, F. (2004). Teaching with variation: A Chinese way of promoting effective mathematics learning. In Fan, L., Wong, N-Y., Cai, J., & Li, S. (Eds.) (2004). How Chinese learn mathematics: Perspectives from insiders. Singapore: World Scientific. | 1,121 | 5,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2021-04 | latest | en | 0.948483 |
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(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Consider the statements:
$\textbf{(1)}\ p\text{ }\wedge\sim q\wedge r\qquad\textbf{(2)}\ \sim p\text{ }\wedge\sim q\wedge r\qquad\textbf{(3)}\ p\text{ }\wedge\sim q\text{ }\wedge\sim r\qquad\textbf{(4)}\ \sim p\text{ }\wedge q\text{ }\wedge r$
where $p,q$, and $r$ are propositions. How many of these imply the truth of $(p\rightarrow q)\rightarrow r$?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$
## Solution
Statement $1$ states that $p$ is true and $q$ is false. Therefore, $p \rightarrow q$ is false, because a premise being true and a conclusion being false is, itself, false. This means that $(p \rightarrow q) \rightarrow X$, where $X$ is any logical statement (or series of logical statements) must be true - if your premise is false, then the implication is automatically true. So statement $1$ implies the truth of the given statement.
Statement $3$ similarly has $p$ as true and $q$ is false, so it also implies the truth of the given statement.
Statement $2$ states that $p$ and $q$ are both false. This in turn means that $p \rightarrow q$ is true. Since $r$ is also true from statement $2$, this means that $(p \rightarrow q) \rightarrow r$ is true, since $T \rightarrow T$ is $T$. Thus statement $2$ implies the truth of the given statement.
Statement $4$ states that $p$ is false and $q$ is true. In this case, $p \rightarrow q$ is true - your conclusion can be true even if your premise is false. And, since $r$ is also true from statement $4$, this means $(p \rightarrow q) \rightarrow r$ is true. Thus, statement $4$ implies the truth of the given statement.
All four statements imply the truth of the given statement, so the answer is (Error compiling LaTeX. ! Missing $inserted.)\boxed{\textbf{(E)}}$ | 573 | 1,917 | {"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": 33, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2020-50 | latest | en | 0.767729 |
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[index] Polynomial / PolynomialFactorization / Algebra::SplittingField / Algebra::Galois
# Polynomial
(Class of Polynomial Ring)
This class expresses the polynomial ring over arbitrary ring. For creating the actual class, use the class method ::create or Algebra.Polynomial, giving the coefficient ring.
## File Name:
• polynomial.rb
• Object
## Included Modules
• Enumerable
• Comparable
• Algebra::EuclidianRing
## Associated Functions:
`Algebra.Polynomial(ring [, obj0 , obj1 [, ...]])`
Same as ::create(ring, obj0[, obj1 [, ...]]).
## Class Methods:
`::create(ring, obj0[, obj1[, ...]])`
Creates a polynomial ring class over the coefficient ring expressed by the class: ring.
The objects designated in `obj0, obj1, ...` express the variables, and the polynomial ring on the polynomial ring is recursively created, if this is multiple.
The value of this method is the subclass of Polynomial class. This subclass has the class methods: ground, var, and vars, which return the coefficient ring ring, the primary variable object(the latest one) and all variables, respectively.
The objects `obj0, obj1, ...` are to utilize for the name (the value of to_s ) of the variable.
Example: Polynomial ring over Integer
```require "polynomial"
P = Algebra::Polynomial.create(Integer, "x")
x = P.var
p((x + 1)**100) #=> x^100 + 100x^99 + ... + 100x + 1
p P.ground #=> integer
```
Example: Multi variate Polynomial ring over Integer
```require "polynomial"
P = Algebra::Polynomial.create(Integer, "x", "y", "z")
x, y, z = P.vars
p((-x + y + z)*(x + y - z)*(x - y + z))
#=> -z^3 + (y + x)z^2 + (y^2 - 2xy + x^2)z - y^3 + xy^2 + x^2y - x^3
p P.var #=> z
```
This `P` is equal to
```Algebra::Polynomial.create(
Algebra::Polynomial.create(
Algebra::Polynomial.create(
Integer,
"x"),
"y"),
"z")
```
and the last variable z is the primary variable.
`::var`
Returns the (primary) variable of the polynomial ring.
`::vars`
Returns the array of the variables of the polynomial rings, collecting recursively.
`::mvar`
Same as ::vars.
`::to_ary`
Returns `[self, *vars]`.
Example: Define Polynomial ring and variables simulteniously
```P, x, y, z = Algebra::Polynomial.create(Integer, "x", "y", "z")
```
`::variable`
Returns the object which expresses the (primary) variable of the polynomial ring.
`::variables`
Returns the array of the objects which express the variables of the polynomial rings, collecting recursively.
`::indeterminate(obj)`
Returns the variable expressed by obj.
`::monomial([n])`
Returns the monomial of degree n.
Example:
```P = Polynomial(Integer, "x")
P.monomial(3) #=> x^3
```
`::const(c)`
Returns the constant value c.
Example:
```P = Polynomial(Integer, "x")
P.const(3) #=> 3
P.const(3).type #=> P
```
`::zero`
Returns the zero.
`::unity`
Returns the unity.
## Methods:
`var`
Same as ::var.
`variable`
Same as ::variable.
`each(&b)`
Iterates of coefficients in the ascendant power series.
Example:
```P = Polynomial(Integer, "x")
x = P.var
(x**3 + 2*x**2 + 4).each do |c|
p c #=> 4, 0, 2, 1
end
```
`reverse_each(&b)`
Iterates of coefficients in the descendent power series.
Example:
```P = Polynomial(Integer, "x")
x = P.var
(x**3 + 2*x**2 + 4).reverse_each do |c|
p c #=> 1, 2, 0, 4
end
```
`[n]`
Returns the coefficient of degree n.
`[n] = v`
Sets the coefficient of degree n into v.
`monomial`
Same as ::monomial.
`monomial?`
Returns true if self is a monomial.
`zero?`
Returns true if self is the zero.
`zero`
Returns the zero.
`unity`
Returns the unity.
`==(other)`
Returns true if self is equal to other.
`<=>(other)`
Returns positive if self is greater than other.
`+(other)`
Returns the sum of self and other.
`-(other)`
Returns the difference of self from other.
`*(other)`
Returns the product of self and other.
`**(n)`
Returns the n-th power of self.
`/(other)`
Returns the quotient of self by other. Same as div.
`divmod(other)`
Returns the array [quotient, remainder] by other.
`div(other)`
Returns the quotient of self by other. Same as `divmod(other).first`.
`%(other)`
Returns the remainder of self by other. Same as `divmod(other).last`.
`divide?(other)`
Returns true if self is divisible by other. Same as `divmod(other).last == zero?`.
`deg`
Returns the degree.
Example:
```P = Polynomial(Integer, "x")
x = P.var
(5*x**3 + 2*x + 1).deg #=> 3
```
`lc`
Returns the leading coefficient.
Example:
```(5*x**3 + 2*x + 1).lc #=> 5
```
`lm`
Returns the leading monomial.
Example:
```(5*x**3 + 2*x + 1).lm #=> x**3
```
`lt`
Returns the leading term). Same as `lc * lm`.
Example:
```(5*x**3 + 2*x + 1).lt #=> 5*x**3
```
`rt`
Returns the rest term, which has the same value as `self - lt`.
Example:
```(5*x**3 + 2*x + 1).rt #=> 2*x + 1
```
`monic`
Returns the polynomial, which corrected the maximum order coefficient in 1. Same as `self / lc` .
`cont`
Returns the content (i.e. L.C.M of coefficients).
`pp`
Returns the primitive part. Same as`self / cont`.
`to_s`
Returns the expression in strings. Use display_type in order to change the display format. The possible value of display_type is :norm(default) and :code.
Example:
```P = Polynomial(Integer, "x")
x = P.var
p 5*x**3 + 2*x + 1 #=>5x^3 + 2x + 1
P.display_type = :code
p 5*x**3 + 2*x + 1 #=> 5*x**3 + 2*x + 1
```
`derivate`
Return the derivative.
Example:
```(5*x**3 + 2*x + 1).derivate #=> 15*x**2 + 2
```
`sylvester_matrix(other)`
Return the Sylvester matrix of self and other.
`resultant(other)`
Return the resultant of self with other
`project(ring[, obj]){|c, n| ... }`
Returns the sum of the evaluations of ... for each monomial of coefficient c and degree n. If obj is omitted, it is assumed to be `ring.var`.
Example:
```require "polynomial"
require "rational"
P = Algebra::Polynomial(Integer, "x")
PQ = Algebra::Polynomial(Rational, "y")
x = P.var
f = 5*x**3 + 2*x + 1
p f.convert_to(PQ) #=> 5y^3 + 2y + 1
p f.project(PQ) {|c, n| Rational(c) / (n + 1)} #=> 5/4y^3 + y + 1
```
`evaluate(obj)`
Returns the value of self at obj. This is equivalent to ` project(ground, obj){|c, n| c} `.
Example:
```require "polynomial"
P = Algebra::Polynomial(Integer, "x")
x = P.var
f = x**3 - 3*x**2 + 1
p f.evaluate(-1) #=> -3 (in Integer)
p f.evaluate(x + 1) #=> x^3 - 3x - 1 (in P)
```
`call(obj)`
Same as evaluate.
`convert_to(ring)`
Returns the polynomial the one converted on ring. This is equivalent to ` project(ring){|c, n| c} `.
# PolynomialFactorization
(Module of Factorization)
The module of factorization of polynomials.
## File Name:
polynomial-factor.rb
## Methods:
`sqfree`
Returns the square free parts.
`sqfree?`
Returns true if square free.
`irreducible?`
Returns true if irreducible
`factorize`
Returns the factorization.
The following type can be factorized:
• Integer
• Rational
• prime field
• Algebraic Field
# Algebra::SplittingField
(Module of Splitting Field)
The module of the minimal splitting field of polynomials.
## File Name:
• splitting-field.rb
## Methods:
`decompose([fac0])`
Returns
```[field, modulus, facts, roots, addelems]
```
Here the elements are: field the mimimal splitting field of poly, def_polys the irreducible polynomial needed for the splitting, facts the linear factors of poly over field, roots the roots of poly and addelems the elements to extend the base field to field.
fac0 makes the factorization fast. (facts and fact0 are the instance of Algebra::Factors). Generally, field is the object of AlgebraicExtensionField. If self is splitted linearlly, that is the ground ring own.
Example:
```require "algebra"
PQ = Polynomial(Rational, "x")
x = PQ.var
f = x**4 + 2
field, def_polys, facts, roots, addelems = f.decompose
p def_polys #=> [a^4 + 2, b^2 + a^2]
p facts #=> (x - a)(x + a)(x - b)(x + b)
p roots #=> [a, b, a, -b]
p addelems #=> [a, b]
fp = Polynomial(field, "x")
x = fp.var
facts1 = Factors.new(facts.collect{|g, n| [g.call(x), n]})
p facts1.pi == f.convert_to(fp) #=> true
```
`splitting_field([fac0]))`
Returns the infomation of the splitting field of self. Each field corresponds to the return value of decompose:
poly, field, roots, def_polys, poly_exps
,except poly_exps. poly_exp is the array of roots which are representable by other roots.
Example:
```require "algebra"
PQ = Polynomial(Rational, "x")
x = PQ.var
f = x**4 + 2
sf = f.splitting_field
p sf.roots #=> [a, b, -a, -b]
p sf.def_polys #=> [a^4 + 2, b^2 + a^2]
p sf.poly_exps #=> [-a, -b]
```
# Algebra::Galois
(Module of Galois Group)
The module of Galois Group of polynomials
## File Name:
• galois-group.rb
(none)
## Associated Method
`GaloisGroup.galois_group(poly)`
Same as galois_group.
## Method:
`galois_group`
Retuns the galois group of self. Each elements of this is the object of FiniteGroup of which elements are in PermutationGroup.
Example:
```require "rational"
require "polynomial"
P = Algebra.Polynomial(Rational, "x")
x = P.var
p( (x**3 - 3*x + 1).galois_group.to_a )
#=>[[0, 1, 2], [1, 2, 0], [2, 0, 1]]
(x**3 - x + 1).galois_group.each do |g|
p g
end
#=> [0, 1, 2]
# [1, 0, 2]
# [2, 0, 1]
# [0, 2, 1]
# [1, 2, 0]
# [2, 1, 0]
``` | 2,825 | 9,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2022-49 | latest | en | 0.703562 |
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Fractions are mathematical notations that show the amount of something in the whole.
## Solution: 0.3 as a fraction could write as 3/10
Explanation:
First, we write the number we want to write in its most basic form: of p/q. In this form, the terms p and Q can be positive numbers.
To write it write down the number as 0.3/1 Then, multiply it and divide by 10.
Since there is a number immediately after the decimal Therefore, we multiplied and divided by 10.
We get, 0.3/1 x 10/10 = 3/10.
### Thus, 0.3 when written as a fraction is equivalent to 3/10.
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Top | 2,022 | 10,255 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2020-34 | latest | en | 0.919721 |
https://www.physicsforums.com/threads/forces-in-equilibrium.750833/ | 1,508,229,317,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820930.11/warc/CC-MAIN-20171017072323-20171017092323-00144.warc.gz | 999,079,773 | 17,157 | # Forces in Equilibrium.
1. Apr 26, 2014
### ----md
Hi All- Hope you can help me :|
1. The problem statement, all variables and given/known data
The person in the drawing is standing on crutches. Assume that the force exerted on each crutch by the ground is directed along the crutch, as the force vectors in the drawing indicate. If the coefficient of static friction between a crutch and the ground is 0.90, determine the largest angle θMAX that the crutch can have just before it begins to slip on the floor.
2. Relevant equations
Fup= Fdown
Fright= Fleft
Fnet=0
3. The attempt at a solution
I figured the for each crutch
Fy = N= 1/2mg + FcosΘ
Fx= μsN=FsinΘ
But I end up with mg and no way to eliminate it?
I know I am missing some important detail- but as with many force problems, I can't visualize it.
Thanks
2. Apr 26, 2014
### goraemon
It appears from the picture that the person is standing on one of his legs in addition to the two crutches. So wouldn't there be 3 upward forces: normal force on his leg, and 2F cos $\theta$?
3. Apr 26, 2014
### ----md
Does that mean that because he is standing- his weight isn't a factor in the upwards forces- because it gets canceled out? and the only net force is the vertical component of the crutch?
4. Apr 26, 2014
### goraemon
Upon thinking some more, assuming that he's supporting himself on one leg plus 2 crutches doesn't seem to help. For the moment, let's assume that he's supporting himself entirely on the two crutches only. With this assumption, let's take a look at the vertical net force:
Fnet(y) = 0 = 2F cos $\theta$ - mg
Now, I'm unsure how you derived the following: Fy = N= 1/2mg + FcosΘ. Can you explain your thought process?
5. Apr 26, 2014
### ----md
I realized my mistake. SO I was assuming that the F of the crutches are like an external force. But its just a component of the mg. So
I set it up as μmgsinθ=mgcosθ -> .9 tan-1= θ
Thanks.
6. Apr 26, 2014
### goraemon
Just a sec, $.9 tan^{-1}= θ$ isn't a valid equation. Do you mean $tan^{-1}(.9)=\theta$? | 588 | 2,051 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2017-43 | longest | en | 0.954676 |
https://studylib.net/doc/10785041/--- | 1,680,163,115,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949107.48/warc/CC-MAIN-20230330070451-20230330100451-00129.warc.gz | 574,982,353 | 12,430 | # ( )
```Stat 511 Lecture 5 "Handout"
The following are equivalent regarding c ∈ ℜk
1) ∃ a ∈ℜn such that a′Xβ = c′β ∀β
2) c ∈ C ( X′ ) ( c′ is a linear combination of the rows of X )
3) Xβ1 = Xβ 2 ⇒ c′β1 = c′β 2
Condition 3) is a condition about the lack of ambiguity of the value of c′β . Condition 1) is
c′β = a′Xβ = a′EY = Ea′Y
so that a′Y can be used to estimate c′β in an unbiased fashion.
Proof: First suppose that 2) holds. c ∈ C ( X′ ) ⇒ ∃ a such that c′ = aX . So for this a ,
c′β = a′Xβ ∀β and 1) holds.
Next suppose that 1) holds, i.e. that ∃ a ∈ ℜn such that a′Xβ = c′β ∀β . Suppose Xβ1 = Xβ 2 .
For this a ,
c′β1 = a′Xβ1 = a′Xβ 2 = c′β 2
and 3) holds.
Finally, suppose that 3) holds. It is "obviously" equivalent to write
X ( β1 − β 2 ) = 0 ⇒ c′ ( β1 − β 2 ) = 0 ∀ β1 , β 2
That is, it is equivalent to 3) to write
Xd = 0 ⇒ c′d = 0 ∀ d
Then, the claim that 3) ⇒ 2) is the claim that
{c | [ Xd = 0 ⇒ c′d = 0 ∀ d] holds} ⊂ C ( X′)
Suppose that c is such that [ Xd = 0 ⇒ c′d = 0 ∀ d ] holds. Write
c = PX′c + ( I − PX′ ) c
and let d* = ( I − PX′ ) c . We can then argue that d* = 0 as follows. Clearly, d* ∈ C ( X′ ) so it
⊥
must be that c′d* = 0 from the condition
[ ].
But
c′d* = c′ ( I − Px′ ) c = c′c − c′Px′c
so that c′c = c′Px′c . But
c′c = c′ ( PX′ + ( I − PX′ ) ) c
= c′PX′c + c′ ( I − PX′ ) c
= c′PX′c + c′ ( I − PX′ )′ ( I − PX′ ) c
Then since c′ ( I − PX′ )′ ( I − PX′ ) c = d*′d* , we have d*′d* = 0 so that d* = 0 . Thus c = PX′c so that
c ∈ C ( X′ ) . ,
``` | 673 | 1,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-14 | latest | en | 0.892608 |
https://discuss.leetcode.com/topic/24571/possible-c-solution | 1,516,111,045,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886436.25/warc/CC-MAIN-20180116125134-20180116145134-00084.warc.gz | 682,371,807 | 8,282 | # Possible C++ Solution
• ``````class Solution {
public:
bool isNumber(string s) {
int i = 0;
for (; i < s.size() && s[i] == ' '; i++);
if (i < s.size() && s[i] == '-' || s[i] == '+') i++;
bool digit = false;
for (; i < s.size() && isdigit(s[i]); i++) digit = true;
if (i < s.size() && s[i] == '.') i++;
for (; i < s.size() && isdigit(s[i]); i++) digit = true;
if (i < s.size() && (s[i] == 'e' || s[i] == 'E') && digit) {
i++;
digit = false;
if (i < s.size() && (s[i] == '-' || s[i] == '+')) i++;
}
for (; i < s.size() && isdigit(s[i]); i++) digit = true;
for (; i < s.size() && s[i] == ' '; i++);
return i == s.size() && digit;
}
};``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 254 | 746 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-05 | latest | en | 0.17659 |
http://www.jiskha.com/display.cgi?id=1296687398 | 1,495,968,878,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609613.73/warc/CC-MAIN-20170528101305-20170528121305-00371.warc.gz | 698,669,517 | 3,607 | # algebra
posted by on .
Anna Can Paint The Garage In 12 Hours. Jenna Can Paint The Same Garage In 7 Hours. How Long Will It Take If They Both Work Together?
• algebra - ,
4.42 hrs.
1/7 +1/12 = 19/84
84/19= 4.42... | 71 | 220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-22 | latest | en | 0.702646 |
https://oeis.org/A005564 | 1,695,447,687,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506479.32/warc/CC-MAIN-20230923030601-20230923060601-00662.warc.gz | 492,955,325 | 6,030 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A005564 Number of n-step walks on square lattice in the first quadrant which finish at distance n-3 from the x-axis. (Formerly M4134) 11
6, 20, 45, 84, 140, 216, 315, 440, 594, 780, 1001, 1260, 1560, 1904, 2295, 2736, 3230, 3780, 4389, 5060, 5796, 6600, 7475, 8424, 9450, 10556, 11745, 13020, 14384, 15840, 17391, 19040, 20790, 22644, 24605, 26676, 28860, 31160, 33579, 36120, 38786, 41580, 44505 (list; graph; refs; listen; history; text; internal format)
OFFSET 3,1 COMMENTS The steps are N, S, E or W. For n>=4, a(n-1)/2 is the coefficient c(n-2) of the m^(n-2) term of P(m,n) = (c(m-1)* m^(n-1) + c(m-2)* m^(n-2) +...+ c(0)* m^0)/((a!)* (a-1)!), the polynomial for the number of partitions of m with exactly n parts. - Gregory L. Simay, Jun 28 2016 2a(n) is the denominator of formula 207 in Jolleys' "Summation of Series." 2/(1*3*4)+3/(2*4*5)+...n terms. Sum_{k = 1..n} (k+1)/(k*(k+2)*(k+3)). This summation has a closed form of 17/36-(6*n^2+21*n+17)/(6*(n+1)*(n+2)*(n+3)). - Gary Detlefs, Mar 15 2018 a(n) is the number of degrees of freedom in a tetrahedral cell for a Nédélec first kind finite element space of order n-2. - Matthew Scroggs, Jan 02 2021 REFERENCES L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 38. N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS Vincenzo Librandi, Table of n, a(n) for n = 3..1000 DefElement, Nédélec first kind R. K. Guy, Letter to N. J. A. Sloane, May 1990 R. K. Guy, Catwalks, Sandsteps and Pascal Pyramids, J. Integer Seqs., Vol. 3 (2000), #00.1.6. See figure 4, sum of terms in (n-2)-nd row. Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009. Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992 Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1). FORMULA G.f.: x^3 * ( 6 - 4*x + x^2 ) / ( 1 - x )^4. [Simon Plouffe in his 1992 dissertation] a(n) = (n-2)*n*(n+1)/2 = (n-2)*A000217(n). a(n) = Sum_{j = 0..n} ((n+j-1)^2-(n-j+1)^2)/4. - Zerinvary Lajos, Sep 13 2006 a(n) = Sum_{k = 2..n-1} k*n. - Zerinvary Lajos, Jan 29 2008 a(n) = 4*binomial(n+1,2)*binomial(n+1,4)/binomial(n+1,3) = (n-2)*binomial(n+1,2). - Gary Detlefs, Dec 08 2011 a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 18 2012 E.g.f.: x - x*(2 - 2*x - x^2)*exp(x)/2. - Ilya Gutkovskiy, Jun 29 2016 a(n) = 6*Sum_{i = 1..n-1} A000217(i) - n*A000217(n). - Bruno Berselli, Jul 03 2018 Sum_{n>=3} 1/a(n) = 5/18. - Amiram Eldar, Oct 07 2020 EXAMPLE The n=4 diagram in Fig. 4 of Guy's paper is: 1 0 4 9 0 6 0 16 0 4 10 0 9 0 1 Adding 16+4 we get a(4)=20. The a(3) = 6 walks are EEN, ENE, ENW, NEW, NSN, NNS. - Michael Somos, Jun 09 2014 G.f. = 6*x^3 + 20*x^4 + 45*x^5 + 84*x^6 + 140*x^7 + 216*x^8 + 315*x^9 + ... From Gregory L. Simay Jun 28 2016: (Start) P(m,4) = (m^3 + 3*m^2 + ...)/(3!*4!) with 3 = a(3)/2 = 6/2. P(m,5) = (m^4 + 10*m^3 + ...)/(4!*5!) with 10 = a(4)/2 = 20/2. P(m,6) = (m^5 + (45/2)*m^4 +...)/(5!*6!) with 45/2 = a(5)/2. P(m,7) = (m^6 + 42*m^5 +...)/(6!*7!) with 42 = a(6)/2 = 84/2. (End) MAPLE A005564 := proc(n) (n-2)*(n)*(n+1)/2 ; end proc: seq(A005564(n), n=0..10) ; # R. J. Mathar, Dec 09 2011 MATHEMATICA Table[(n-2)*Binomial[n+1, 2], {n, 3, 40}] LinearRecurrence[{4, -6, 4, -1}, {6, 20, 45, 84}, 50] (* Vincenzo Librandi, Jun 18 2012 *) PROG (PARI) a(n)=(n-2)*(n)*(n+1)/2 \\ Charles R Greathouse IV, Dec 12 2011 (Magma) I:=[6, 20, 45, 84]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..45]]; // Vincenzo Librandi, Jun 18 2012 (GAP) a:=List([0..45], n->(n+1)*Binomial(n+4, 2)); # Muniru A Asiru, Feb 15 2018 CROSSREFS Cf. A000217. First differences of A001701. Fourth column of A093768. Sequence in context: A225269 A048969 A353692 * A011928 A055455 A203552 Adjacent sequences: A005561 A005562 A005563 * A005565 A005566 A005567 KEYWORD nonn,walk,easy AUTHOR N. J. A. Sloane EXTENSIONS Entry revised by N. J. A. Sloane, Jul 06 2012 STATUS approved
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Last modified September 23 01:19 EDT 2023. Contains 365532 sequences. (Running on oeis4.) | 1,948 | 4,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2023-40 | latest | en | 0.638055 |
http://www.chegg.com/homework-help/questions-and-answers/bowler-releases-bowling-ball-spin-sending-slidingstraight-alley-pins-ball-continues-toslid-q393447 | 1,394,311,503,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999662994/warc/CC-MAIN-20140305060742-00051-ip-10-183-142-35.ec2.internal.warc.gz | 284,849,056 | 7,176 | # i dont understand this problem
0 pts ended
A bowler releases a bowling ball with no spin, sending it slidingstraight down the alley toward the pins. The ball continues toslide for some distance before its motion becomes rolling withoutslipping; what is the magnitude of this distance? Assume the ballmaintains an essentially constant speed of 4.60 m/s and that the coefficient of kinetic frictionfor the polished alley is 0.105.
m | 93 | 433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2014-10 | latest | en | 0.905566 |
https://editthis.info/biolk483/09/13/06 | 1,701,491,950,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100327.70/warc/CC-MAIN-20231202042052-20231202072052-00541.warc.gz | 272,047,071 | 8,336 | # 09/13/06
• note: much of the notes on pKa here is omitted as it is very difficult to put into web-text
## pKa
• there are about 8 pKas that we'll need to know
• pKa of H on N of amino acid is 9.1
• pKa of H on OH of Carboxy group of amino acid is 2.1
• pKa of H on OH of Glutamic acid's r-group is 4.0
• we limit our study of pKa to labeling charges as +1, +1/2, and 0 for N and 0, -1/2, and -1 for O
• so:
• if the pH is above 2.1, then the O (with the H removed because the pH is high so there are lots of H+ floating around) will have a charge of -1.
• if the pH is exactly 2.1, then the O with the H removed will have a charge of -1/2 (because approximately half of the Os have their H and half do not)
• if the pH is below 2.1, then the O with the H removed will have a charge of 0.
• these rules are true also for the OH group of the R-group of Glutamic acid (using a 4.0 pH)
• and:
• if the pH is above 9.1 the N will have a charge of 0 because the H+ has been removed from the N
• if the pH is exactly 9.1, then we say N has a charge of +1/2 because half have lost their H and half have not.
• if the pH is below 9.1, then we say the N has a charge of +1 because it has all its H
• so, we add up all the charges on all the dissociable groups to get the entire chain's charge
• all this is to show that pKa never changes, but the charge of the chain can be changed by changing the pH in which it resides
• Here are all the pKas we need to know:
GROUP Amino Acid pKA alpha-COOH all 2.1 r-group COOH Asp, Glu 4.0 imidazole His 6.1 alpha-amine all 9.1 -SH group Cys 8.3 phenalic group Tyr 10.1 r-group Epslion Lys 10.8 adeno group Arg 12.5
##### Isoelectric Point
• this is the pH at which the net charge on the polypeptide is 0
• this is the minimum water solubility a polypeptide can have because there is zero charge (and charge helps make things water soluble)
• at the Isoelectric point, a polypeptide will not move in an electric field, therefore, this phenom can be used to separate out one's desired polypeptide from others because each has a very specific isoelectric point.
##### Relative Abundance of Amino Acid Forms
• we use glycine as a generalization of all amino acids:
• 99.58% of glycine in the world is in the zwitterion form: it has formal charges but no net charge.
• 0.41% is anionic: having the H from the Carboxyl group removed
• 0.00019% is cationic: having an extra H on the amine group
• 0.0000037% is neutral
• how do we know?
• we compare the dipole moment of water and glycine
• water's dipole moment is 1.8 and glycine's is 11.2, so we know that LOTS of the glycines in the bottle must have formal charges
## The 20 Common Amino Acids
• remember that common is not the same as abundant: there are tons of other amino acids, often in greater abundance than the twenty our genetic code includes
• amino acids in proteins that are non-common are the result of post-translation modification of common amino acids
• essential amino acids are those that must be consumed via diet because our bodies cannot make them
• there are 10 of these
• Argenine we actually learn how to make as we become adults.
• we have 19 amino acids and 1 imino acid: proline
• the ring structure of proline makes it very restrained
• we can identify proline easily because it shines a brilliant yellow while all the others shine bluish-reddish.
### Categories of Amino Acids
• amino acids are categorized based on r-group
• we can divide the amino acids into four groups based on whether they prefer a water or lipid environment.
##### Non-polar (Hydrophobic)
• Inert. Can't form H-bonds.
• about 50% of all amino acids fall in this group
• have hydrophobic side chains, don't have dissociable groups, often have CH2s and Aromatic groups
• mostly found on inside of proteins
• these guys are responsible for folding the protein
##### Neutral, Polar
• neutral side chains that are polar, they like water
• mostly found on outside of protein
###### Aliphatic
• Ala, Leu, Val, (Pro), Ile
• no catalysis but they fold the chain because of the hydrophobic effect
###### Aromatic
• Phe, Trp
• Terrible H20 solubility
###### Contain Sulphur
• Met
• disulphide bonds
• hold protein in position
##### Anionic
• mostly found on outside of protein
##### Cationic
• mostly found on outside of protein
##### Aromatic
• there are only three, and we could classify them in the other groups
### Protein Shape
1. Chain: all amino acids are involved in the chain
2. Folding: some amino acids are involved in folding because of hydrophobic effect, some because of disulfide bonds (see Non-polar amino acids)
3. Function: some amino acids cause function because theya re dynamic.
• Proline's involvment in shape
• not flexible (because of ring)
• no enzymatic activity on alpha-helix
• if proline is present, alpha helix won't work so we call it an alpha helix terminator
• this is an exmaple of an unique function
• Phenylalanine
• a great binding site for the helper prophin which also has Pi electrons with a metal in the middle toggline between 2+ and 3+
• so these two (phenylalanine and porphin) are planar molecules
• prophin must be protected on one side of porphin's plane (this keeps H20 from oxidizing the metal and is useful in biochemical reactions). | 1,439 | 5,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-50 | latest | en | 0.938826 |
https://za.pinterest.com/brittanybryce0701/place-value-of-decimals/ | 1,603,931,623,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107902038.86/warc/CC-MAIN-20201028221148-20201029011148-00497.warc.gz | 981,373,722 | 47,094 | # Place Value of Decimals
## Collection by Mrs. Bryce
7
Pins
This board will help students understand the relationship between a number and how it compares to the number in the place to the left and to the right.
Anchors Away!!! Need help teaching your kiddos about place value and the powers of ten? Me too!! So I created this anchor chart for standard 5.NBT.1 to post in my room for students to reference easily and daily! I hope it is just what you are looking for for your classroom as well!! How to print...
The cut-out is a review of 5.NBT.1. It allows students to practice 1/10 of and 10 times as much. I have used this activity for a station. This is an excellent review of place value as well. ...
I use this quiz with my 5th grades while learning place value. The quiz focuses upon recognizing that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1/10 of what it represents in the place to its left.
Free math worksheets for almost every subject. Create your own daily (spiral) reviews, test, worksheets and even flash cards. All for free! No signup or app to download.
Browse the Khan Academy math skills by Common Core standard. With over 50,000 unique questions, we provide complete coverage.
In this lesson you will learn that the digit in one place is 1/10 the value of the digit to the left by using base ten blocks.
This document includes 12 questions appropriate for 5.NBT.1 - understanding place values in multi-digit numbers. An answer key is included! Common Core Aligned! | 361 | 1,576 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-45 | latest | en | 0.943188 |
https://nrich.maths.org/public/leg.php?code=-56&cl=4&cldcmpid=2686 | 1,508,798,516,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826840.85/warc/CC-MAIN-20171023221059-20171024001059-00864.warc.gz | 783,104,051 | 6,506 | # Search by Topic
#### Resources tagged with Golden ratio similar to Golden Eggs:
Filter by: Content type:
Stage:
Challenge level:
### There are 20 results
Broad Topics > Fractions, Decimals, Percentages, Ratio and Proportion > Golden ratio
### Golden Eggs
##### Stage: 5 Challenge Level:
Find a connection between the shape of a special ellipse and an infinite string of nested square roots.
### Pythagorean Golden Means
##### Stage: 5 Challenge Level:
Show that the arithmetic mean, geometric mean and harmonic mean of a and b can be the lengths of the sides of a right-angles triangle if and only if a = bx^3, where x is the Golden Ratio.
### The Golden Ratio, Fibonacci Numbers and Continued Fractions.
##### Stage: 4
An iterative method for finding the value of the Golden Ratio with explanations of how this involves the ratios of Fibonacci numbers and continued fractions.
### Golden Mathematics
##### Stage: 5
A voyage of discovery through a sequence of challenges exploring properties of the Golden Ratio and Fibonacci numbers.
### Pent
##### Stage: 4 and 5 Challenge Level:
The diagram shows a regular pentagon with sides of unit length. Find all the angles in the diagram. Prove that the quadrilateral shown in red is a rhombus.
### Golden Fractions
##### Stage: 5 Challenge Level:
Find the link between a sequence of continued fractions and the ratio of succesive Fibonacci numbers.
### Gold Yet Again
##### Stage: 5 Challenge Level:
Nick Lord says "This problem encapsulates for me the best features of the NRICH collection."
### Golden Construction
##### Stage: 5 Challenge Level:
Draw a square and an arc of a circle and construct the Golden rectangle. Find the value of the Golden Ratio.
### Golden Fibs
##### Stage: 5 Challenge Level:
When is a Fibonacci sequence also a geometric sequence? When the ratio of successive terms is the golden ratio!
### Golden Powers
##### Stage: 5 Challenge Level:
You add 1 to the golden ratio to get its square. How do you find higher powers?
##### Stage: 5
What is the relationship between the arithmetic, geometric and harmonic means of two numbers, the sides of a right angled triangle and the Golden Ratio?
### Whirling Fibonacci Squares
##### Stage: 3 and 4
Draw whirling squares and see how Fibonacci sequences and golden rectangles are connected.
### Golden Thoughts
##### Stage: 4 Challenge Level:
Rectangle PQRS has X and Y on the edges. Triangles PQY, YRX and XSP have equal areas. Prove X and Y divide the sides of PQRS in the golden ratio.
### Pentakite
##### Stage: 4 and 5 Challenge Level:
ABCDE is a regular pentagon of side length one unit. BC produced meets ED produced at F. Show that triangle CDF is congruent to triangle EDB. Find the length of BE.
### Darts and Kites
##### Stage: 4 Challenge Level:
Explore the geometry of these dart and kite shapes!
### Golden Ratio
##### Stage: 5 Challenge Level:
Solve an equation involving the Golden Ratio phi where the unknown occurs as a power of phi.
### Leonardo of Pisa and the Golden Rectangle
##### Stage: 2, 3 and 4
Leonardo who?! Well, Leonardo is better known as Fibonacci and this article will tell you some of fascinating things about his famous sequence.
### Pentabuild
##### Stage: 5 Challenge Level:
Explain how to construct a regular pentagon accurately using a straight edge and compass.
### Golden Triangle
##### Stage: 5 Challenge Level:
Three triangles ABC, CBD and ABD (where D is a point on AC) are all isosceles. Find all the angles. Prove that the ratio of AB to BC is equal to the golden ratio.
### Gold Again
##### Stage: 5 Challenge Level:
Without using a calculator, computer or tables find the exact values of cos36cos72 and also cos36 - cos72. | 834 | 3,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2017-43 | longest | en | 0.814549 |
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4 we discuss the calibration. In Section 4 we present the results and perform some sensitivity analysis in Section 5. In Section 6 we discuss exchange rate pass-through and we conclude in Section 7. 2 The Model The world economy consists of two countries, denominated home and foreign. Each country is populated by a continuum of identical households, firms, and a monetary authority. Households consume two types of final goods, a tradable good T and a nontradable good N. The production of nontradable goods requires capital and labor and the production of tradable consumption goods requires the use of home and foreign tradable inputs as well as nontradable goods. Therefore, consumer markets of tradable consumption goods are segmented and consumers are unable to arbitrage price differentials for these goods across countries. Households own the capital stock and rent labor and capital services to firms. Households also hold domestic currency and trade a riskless bond denominated in home currency with foreign households. Each firm is a monopolistic supplier of a differentiated variety of a good and sets the price for the good it produces in a staggered fashion. In what follows, we describe the home country economy. The foreign country economy is analogous. Asterisks denote foreign country variables. 2.1 Households The representative consumer in the home country maximizes the expected value of lifetime utility, given by U 0 = E 0 t=0 ( β t u c t, 1 h t, M ) t+1, (1) P t where c t denotes consumption of a composite good to be defined below, h t denotes hours worked, M t+1 /P t denotes real money balances held from period t to period t + 1, and u represents the momentary utility function. The composite good c t is an aggregate of consumption of a tradable good c T,t and a 4
7 where ρ denotes the elasticity of substitution between X T,t (i) and X N,t (i) and ω is a weight. We interpret this sector as a retail sector. Thus, X N,t (i) can be interpreted as retail services used by firm i. 7 For simplicity, we assume that the local nontradable good used for retail services X N,t is given by the same Dixit-Stiglitz aggregator (2) as the nontradable consumption good c N. Thus, P N,t is the price of one unit of X N,t. The composite of home and foreign intermediate tradable inputs X T,t is given by X T,t = [ ω 1 ξ ξ 1 X X ξ h,t ] ξ + (1 ω X ) 1 ξ 1 ξ X ξ f,t ξ 1, (5) where X h,t and X f,t denote home and foreign intermediate traded goods, respectively. These goods X h and X f are each a Dixit-Stiglitz aggregate, as in (2), of all the varieties of each good produced in the home and foreign intermediate traded goods sector, X h (j) and X f (j), j [0, 1]. Let the unit price (in home-currency units) of X h,t and X f,t be denoted by P h,t and P f,t, respectively. Then, the price of one unit of the composite tradable good X T,t is given by P X,t = [ ω X P 1 ξ h,t ] 1 + (1 ω X )P 1 ξ 1 ξ f,t. (6) Given these prices, the real marginal cost of production, common to all firms in this sector, is ψ T, ψ T,t = [ ω ( PXN,t P t Firms in this sector set prices for J T ) 1 ρ + (1 ω) ( PXT,t P t ) 1 ρ ] 1 1 ρ. (7) periods in a staggered way. That is, each period, a fraction 1/J T of these firms optimally chooses prices that are set for J T periods. The 7 Note that we assume that the retail sector is a monopolistic competitive sector where each firm produces a differentiated good, by combining retail services X T with a tradable composite X N. With this assumption, the market structure of this sector mirrors that of the other sectors in our model. This assumption differs from that in other models that incorporate distribution/retail services, such as Burstein, Neves, and Rebelo (2003), Corsetti and Dedola (2005), and Corsetti, Dedola, and Leduc (2004a), which assume a perfectly competitive distribution sector where distribution costs are applied to each traded good separately. 7
8 problem of a firm i adjusting its price in period t is given by J T 1 [ max E t ϑt+i t (P T,t (0) P t+i ψ T,t+i ) y T,t+i (i) ], P T,t (0) i=0 where y T,t+i (i) = c T,t+i (i) + i t+i (i) represents the demand (for consumption and investment purposes) faced by this firm in period t+i. The term ϑ t+i t denotes the pricing kernel, used to value profits at date t + i which are random as of t. In equilibrium ϑ t+i t is given by the consumer s intertemporal marginal rate of substitution in consumption, β i (u c,t+i /u c,t )P t /P t+i Intermediate Traded Goods Sector There is a continuum of firms in the intermediate traded goods sector, each producing a differentiated variety of the intermediate traded input, X h (i), i [0, 1], to be used by local and foreign firms in the retail sector. The production of each intermediate tradable input requires the use of capital and labor. The production function is y h,t (i) = z h,t k h,t (i) α l h,t (i) 1 α. The term z h,t represents a productivity shock specific to this sector and k h,t and l h,t denote the use of capital and labor services by firm i. Each firm chooses one price, denominated in units of domestic currency, for the home and foreign markets. Thus, the law of one price holds for intermediate traded inputs. 8 Like retailers, intermediate goods firms set prices in a staggered fashion. The problem of an intermediate goods firm in the traded sector setting its price in period t is described by J h 1 [ max E t ϑt+i t (P h,t (0) P t+i ψ h,t+i ) (X h,t+i (i) + Xh,t+i(i)) ], (8) P h,t (0) i=0 where X h,t+i (i) + Xh,t+i (i) denotes total demand (from home and foreign markets) faced by this firm in period t + i. The term ψ h denotes the real marginal cost of production (common 8 Thus, in our benchmark model, the pass-through of exchange rate changes to import prices at the wholesale level is one. This pricing assumption makes our model consistent with the finding that the exchange rate pass-through is higher at the wholesale than at the retail level. Empirical evidence, however, suggests that exchange rate pass-through is lower than one even at the wholesale level (for instance, Goldberg and Knetter, 1997). Below we investigate the implications of alternative pricing assumptions for intermediate goods producers. 8
9 to all firm in this sector) and is given by Nontradable Goods Sector ψ h,t = 1 ( rt ) ( ) α 1 α wt. (9) z h,t α 1 α This sector has a structure analogous to the intermediate traded sector. Each firm operates the production function y N,t (i) = z N,t k N,t (i) α l N,t (i) 1 α, where all the variables have analogous interpretations. The price-setting problem for a firm in this sector is J N 1 [ max E t ϑt+i t (P N,t (0) P t+i ψ N,t+i ) y N,t+i (i) ], P N,t (0) i=0 where y N,t+i (i) = X N,t+i (i)+c N,t+i (i) denotes demand (from the retail sector and consumers) faced by this firm in period t + i. The real marginal cost of production in this sector is given by ψ N,t = ψ h,t z h,t /z N,t. 2.3 The Monetary Authority The monetary authority issues domestic currency. Additions to the money stock are distributed to consumers through lump-sum transfers T t = Mt s Mt 1. s The monetary authority is assumed to follow an interest rate rule similar to those studied in the literature. In particular, the interest rate is given by R t = ρ R R t 1 + (1 ρ R ) [ R + ρ R,π (E t π t+1 π) + ρ R,y ln (y t /y) ], (10) where π t denotes CPI-inflation, y t denotes real GDP, and barred variables represent their target value. 9
10 2.4 Market Clearing Conditions and Model Solution We close the model by imposing market clearing conditions for labor, capital, and bonds, h t = k t = J h 1 i=0 i=0 l h,t (i) + J N 1 i=0 i=0 l N,t (i), J h 1 J N 1 k h,t (i) + k N,t (i), 0 = B t + B t. We focus on the symmetric and stationary equilibrium of the model. We solve the model by linearizing the equations characterizing equilibrium around the steady-state and solving numerically the resulting system of linear difference equations. We now define some variables of interest. The real exchange rate q, defined as the relative price of the reference basket of goods across countries, is given by q = ep /P. The terms of trade τ represent the relative price of imports in terms of exports in the home country and are given by τ = P f /(eph ). Nominal GDP in the home country is given by Y = P c+p T i+nx, where NX = eph X h P fx f represents nominal net exports. We obtain real GDP by constructing a chain-weighted index as in the National Income and Product Accounts. 3 Calibration In this section we report the parameter values used in solving the model. Our benchmark calibration assumes that the world economy is symmetric so that the two countries share the same structure and parameter values. The model is calibrated largely using US data as well as productivity data from the OECD Stan data base. We assume that a period in our model corresponds to one quarter. Our benchmark calibration is summarized in Table 1. 10
11 3.1 Preferences and Production We assume a momentary utility function of the form U ( c, l, M P ) { ( = 1 ( ) η ) 1 σ } M ac η η + (1 a) exp { v(h)(1 σ)} 1. (11) 1 σ P The discount factor β is set to 0.99, implying a 4% annual real rate in the stationary economy. We set the curvature parameter σ equal to two. The parameters a and η are obtained from estimating the money demand equation implied by the first-order condition for bond and money holdings. Using the utility function defined above, this equation can be written as log M t = 1 P t η 1 log a 1 a + log c t + 1 η 1 log R t 1. (12) R t We use data on M 1, the three-month interest rate on T-bills, consumption of non-durables and services, and the price index is the deflator on personal consumption expenditures. The sample period is 1959:1-2004:3. The parameter estimation is carried out in two steps. Because real M 1 is non-stationary and not co-integrated with consumption, equation (12) is first differenced. The coefficient estimate on consumption is and is not statistically different from one, so the assumption of a unitary consumption elasticity implied by the utility function is consistent with the data. The coefficient on the interest rate term is 0.021, and we calibrate η to be 32, which implies an interest elasticity of Next, we form a residual u t = log(m t /P t ) log c t 1 log R t 1 η 1 R t. This residual is a random walk with drift and we use a Kalman filter to estimate the drift term, which is the constant in equation (12). The resulting estimate of a is very close to one and we set a equal to Therefore, our calibration is close to imposing separability between consumption and real money balances. 9 The estimation procedure neglects sampling error, because in the second stage we are treating η as a parameter rather than as an estimate. 11
13 for all goods j = T, N, h. As usual, this elasticity is related to the markup chosen when firms adjust their prices, which is γ j / (γ j 1). Our choice for γ j implies a markup of 1.11, which is consistent with the empirical work of Basu and Fernald (1997). In our benchmark calibration, we assume that all firms set prices for 4 quarters (J j = 4). Regarding production, we take the standard value of α = 1/3, implying that one-third of payments to factors of production goes to capital services. 3.2 Monetary Policy Rule The parameters of the nominal interest rate rule are taken from the estimates in Clarida, Galí, and Gertler (1998) for the US. We set ρ R = 0.9, α p,r = 1.8, and α y,r = The target values for R, π, and y are their steady-state values, and we have assume a steady-state inflation rate of 2 percent per year. 3.3 Capital Adjustment and Bond Holding Costs We model capital adjustment costs as an increasing convex function of the investment to capital stock ratio. Specifically, Φ k (i/k) = φ 0 + φ 1 (i/k) φ 2. We parameterize this function so that Φ k (δ) = δ, Φ k (δ) = 1, and the volatility of HP-filtered consumption relative to that of HP-filtered private GDP is approximately The bond holdings cost function is Φ b (B t /P t ) = θ b /2 (B t /P t ) 2. The parameter θ b is set to 0.001, the lowest value that guarantees that the solution of the model is stationary, without affecting the short-run properties of the model. 3.4 Productivity Shocks The technology shocks are assumed to follow independent AR(1) processes zi,t k = Azi,t 1 k + ε k i,t, where i = {U.S., ROW } and k = {mf, sv}; ROW stands for rest of world, mf for manufacturing and sv for services. ε k i, represents the innovation to zi k and has standard deviation σi k. The data are taken from the OECD STAN data set on total factor productivity (TFP) for manufacturing and for wholesale and retail services. The data is annual and runs from making for a very short sample in which to infer the time series characteristics 13
14 of these measures. We cannot reject a unit root for any of the series, which is consistent with other data series on productivity in manufacturing, namely that constructed by the BLS or Basu, Fernald, and Kimball (2004). The shortness of the time series on TFP prevents us from estimating any richer characterization of TFP with any precision. 11 In looking at the univariate autoregressive estimates we found coefficients ranging from 0.9 for U.S. manufacturing to 1.05 for rest of world services. Therefore, we use as a benchmark a stationary, but highly persistent processes for each of the technology shocks. Based on these simple regressions, we set A = 0.98 and we set the standard deviations of the TFP on manufacturing and services to and respectively. 4 Findings In this section we assess the role of nontradable goods in our model. We report HP-filtered population moments for our model under the benchmark and alternative parameterizations in Table We find that the presence of either nontradable consumption goods or retail services raises the volatility of real and nominal exchange rates relative to GDP by a factor of about 1.5. In addition the presence of nontradable goods also lowers the correlation between exchange rates and other macro variables. Therefore, nontradable goods bring a standard two-country open economy model closer to the data. Finally, in the presence of nontradable goods, the asset structure of the model (and whether agents have access to a complete set of state-contingent assets or not) matters for the adjustment to country-specific shocks. This result is in sharp contrast with many other two-country models, where agents are able to optimally share risk across states and dates with one discount bond only. 14
15 Table 1: Calibration Preferences Coefficient of risk aversion (σ) 2 Elasticity of labor supply 2 Time spent working 0.25 Interest elasticity of money demand (1/(ν 1)) Weight on consumption (a) 0.99 Aggregates Elast. of substitution C N and C T (γ) 0.74 Elast. of substitution X and Ω (ρ) Elast. of substitution X h and X f (ξ) 0.85 Elast. of substitution individual varieties 10 Share of imports in GDP 0.13 Share of retail services in GDP 0.19 Share of C N in GDP 0.44 Production and Adjustment Functions Capital share (α) 1/3 Price stickiness (J) 4 Depreciation rate (δ) Relative volatility of investment 2.5 Bond Holdings (b) Monetary Policy Coeff. on lagged interest rate (ρ R ) 0.9 Coeff. on expected inflation (ρ p,r ) 1.8 Coeff. on output (ρ y,r ) 0.07 Productivity Shocks Autocorrelation coeff. (A) 0.98 Std. dev. of innovations to z T &z N & The Benchmark Economy The benchmark model implies that nominal and real exchange rates are about 1.6 times as volatile as real GDP. In the data, dollar nominal and real exchange rates are about 4.5 times as volatile as real GDP. 13 The volatility of nominal and real exchange rates in our model is accounted mostly by productivity shocks to the nontraded goods sector. Shocks to 11 We estimated a VAR to investigate the relationship across the four TFP series. It was hard to make sense of the results. In this regard our results are similar to those of Baxter and Farr (2001) who analyze the relationship between total factor productivity in manufacturing between the U.S. and Canada. 12 We thank Robert G. King for providing the algorithms that compute population moments. 13 We report data values from Chari, Kehoe, and McGrattan (2002). 15
16 Table 2: Model results Benchmark No No No Complete Statistic Economy Retail C NT NT Markets Stand. Dev. Relative to GDP Consumption Investment Employment Nominal E.R Real E.R Terms of trade Net Exports Autocorrelations GDP Nominal E.R Real E.R Terms of trade Net Exports Cross-correlations Between nominal and real E.R Between real exchange rates and GDP Terms of trade Relative consumptions Between foreign and domestic GDP Consumption Investment Employment productivity in the traded goods sector imply minimal responses of exchange rates in the benchmark model. As in the data, exchange rates in our model are much more volatile than the price ratio P /P (about 7 times) and are highly correlated with each other (0.99). In general, movements in the real exchange rate can be decomposed into deviations from the law of one price for traded goods and movements in the relative prices of nontraded to traded goods across countries. 14 Let q T denote the real exchange rate for traded goods, defined as q T = ept /P T. Then, the real exchange rate can be written as q = q T p, where p is a 14 See, for example, Engel (1999). 16 | 4,330 | 17,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-47 | longest | en | 0.913545 |
http://library.thinkquest.org/C0126598/eng/pq_eng.html | 1,386,946,211,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164949664/warc/CC-MAIN-20131204134909-00018-ip-10-33-133-15.ec2.internal.warc.gz | 110,034,558 | 8,153 | Science Matemathics Physics Physical quantities Kinematics Dynamics Work and energy Collisions Revolving Balance Gravitation Elasticity The static of the fluids Dynamics of fluids The surface suspense Technology ThinkQuest team About us
# Physical quantities
Perception of the physical quantities
The experiment is the basis of every objective perception of nature. The aim of the experiment is to determine parameters on which the given natural concept depends. Those parameters we call physical quantities which can be understood as abstractions, we use in order to describe the concepts. The first step to understand physical concepts is determining parameters on which those concepts depend on. After that we can perform the measuring, comparing with given values of quantities which we call units.
The further step in the initiation with the physical concepts is to connect the derived physical quantities into mathematic formulas, which show the physical laws. So, for example, to describe the drop we use the equation v=g*t so the different physical quantities are connected into one physical law. It is understandable that the number of physical quantities can be very big. Because of that physical quantities are divided into basic physical quantities and derived physical quantities.
The basic physical quantities
Today's development of physics shows that the whole range of concepts researched by modern physics can be shown through the following basic physical quantities:
3 mechanical quantities
1 electric quantity
1 thermic quantity
1 photometric quantity
1 atomistic quantity
In mechanics we have the following physical quantities: length, time, mass and sometimes force.
In electric measurements there is a fourth physical quantity introduced in 1901. by Giorgi and is called the strength of electricity. All other quantities are derived and can be expressed by means of three mechanical quantities and strength of electricity.
For the field of thermodynamics it is necessary to introduce another physical quantity: temperature.
For the photometric measurements 3 basic quantities are necessary: two kinetic, the length and time and the forth the light flow. The other possibility is to define the analogue strength in electrodynamics. That quantity will be called light strength.
When performing measurements in the nuclear physics we except for the mechanical quantities also need another specific quantity, a number of equal for the given problem equivalent units, for example atom, molecule, radicals, electrons etc.
It is necessary to know the number of individual particles. As that number can be determined only in the exceptional cases, we instead of that take the quantity of material , proportional to the number of particles. As the consequence of discontinuity of structure of the material, the factor of the proportionality among the quantity of material and number of particles is the universal constant (Avangard's number AO = NO/m). So, as the basic quantity that is based on the number of the particles is introduced the quantity of material, which will be defined as 1 mol.
The 11th General conference for the weights and measures, held in October 1960. accepted the following basic quantities as quantities which will form the International system of measures (Systeme International d' Unites, symbol SI):
- length (mechanics)
- time (mechanics)
- mass (mechanics)
- electricity strength (electrodynamics)
- temperature (thermodynamics)
- light strength (photometry)
The 14th General international conference for weights and measures added in 1971. the seventh basic physical quantity:
- the quantity of material (nuclear physic)
The measuring systems; units
The measuring systems
The set of units of basic physical quantities we call the measuring system. The measuring system generally comprises the units of basic physical quantities determined by the given field of physics. So, we are talking about the measuring system in mechanics or in thermodynamics; each of those systems includes certain number of units.
The history of metric measuring systems
The necessity to measure and compare the quantities is above all the practical problems that appeared already in the beginning of civilization. It is natural that the fist quantities for length, weight and time were connected to objects or time intervals for example parts of the body. With the broader development of international exchange there appeared the necessity to standardize measuring units. The social revolution was necessary in order to throw away the traditional units and accept that what is today called the international unit system.
On the 23rd of August 1790., the French constituent assembly, which aroused during the French revolution, gives a task to the French Academy of Science to prepare the unified system of measures for length (as well as surface and volume). On the 9th of March, next year the Academy proposes that the system is decimal and that it's basis be the tenmilionth part of the squate of the Earth's meridian. Delambre and Mchain have measured that the square of the Parisian meridian between Dunquerque and Barcelona is approx. 9°30′, it was by Law accepted on the 7th of April 1975, by the French government that the measure for length is meter, defined as tenmilionth part of the square of the Earth's meridian. At the same time the units for surface and volume were defined, as well as for mass (kilogram). The samples of the ancient kilogram and ancient meter were kept in the French national archive in 1799. The metric system has spread into most European countries in the 19th century. In the meantime the unit for measuring time still stay in the old system. At the same time in other branches of physics (electricity and magnetism) the international systems of units are accepted.
The international system of measuring units (SI)
The international system of measuring units consists of the following units:
- for length - meter (m)
- for mass - kilogram (kg)
- for time - second (s)
- for the electricity strength - ampere (A)
- for thermodynamic temperature - Calvin (K)
- for the light strength - Candel (cd)
- for the quantity of material - mol (mol)
The definition of the SI units
1 meter is the length equal to 1 650763,73 wavelength in the vacuum radiation which is correspondent to the transforming of the atom nuklid 86 Kr from the state 5 d5 in the state 2p10.
1 kilogram is the mass of the international ancient kilogram which is kept in Sevres near Paris.
1 second lasts 9 192 631 770 periods of radiation which corresponds to transforming between two superfine level of basic atom nuklid 133Cs.
1 ampere is the strength of the stabile electricity which passes through two straight line unmeasured long conductors of very small intersection, distanced for 1 meter in vacuum, cause among them electro- dynamic force 2*10-7 N/m.
1 Calvin is 273,16th part of thermo-dynamic temperature of the inertial point of the chemically pure water in the natural isotopic mixture.
1 Candel is the light strength in the direction vertical to the leak of the surface of the 1/600 000 m2 if the black body on the temperature of melting point of platinum under the pressure of 101 325 m-1-kgs-2 (later that unit was abandoned as being imprecise).
1 mol is the quantity of material in the system which withholds as many equal individual particles as there are atoms in the 0,012 kg of isotope of carbon 12C.
Other measuring systems in physics and technical sciences
CGS system was once generally accepted and is still often used today. This system is based on three metric units: centimeter for length, gram for mass and second for time.
Technical or M Kp S system. In the original definition of the metric system kilogram was defined as the unit for mass. The concept difference , important in physics, that exists among mass and weight was not always noticed in practice. That's why kilogram is often used as unit for weight - force. At that case we call the unit for force kilopond (kp) - 1 kilopond is the weight of the mass 1 kg. On the place of normal Earth acceleration - 1 kp = 9,806 66 N = 1000 ponds.
Some physical units out of measuring systems
Although the unit of SI system are accepted in most countries it is often that in general use are some units out of this system. The reason for this is tradition, but also the fact that some of SI units being rather unpractical.
Pressure 1 physical (normal) atmosphere (atm) is defined as the pressure 0,76 m high tower of liquid of 13 595,1 kg m-3 density (approx. mercury at 0°C) on the place of the normal Earth acceleration - 1 atm= 101 325 Pa
Work and energy 1 kilowatt hour is the work done on the constant power of 1 kW in one hour time. - 1kWh = 3,6*10 on 6J.
The quantity of heat 1 thermo-chemical calorie is defined as 4,1840 J. 1 international calorie was originally defined as 1/880 international Wh - 1 caliT= 4,1868 J. 1 water calorie is defined as the quantity of heat needed to heat 1 kg of pure water from 14,5° to 15.5 °C under the normal pressure of p0 = 1 atm. - 1 cal= 4,1855 J.
cal = 4,1855 J.
ThinkQuest ThinkQuest Internet Challenge 2001 Team C0126598 - Interconnecting science with technology Thanks to: Zagrebacki Racunalni Savez I. Tehnicka skola III. Gimnazija X. Gimnazija prof. Andreja Stancl prof. Hrvoje Negovec Our parents: Mario, Ljerka, Drazen, Tanja, Jasminka. . . | 2,051 | 9,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2013-48 | latest | en | 0.918826 |
https://www.esaral.com/q/the-number-of-elements-in-94849 | 1,721,369,775,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00233.warc.gz | 666,280,946 | 11,366 | # The number of elements in
Question:
The number of elements in the set $\{x \in \mathbb{R}:(|x|-3)|x+4|=6\}$ is equal to
1. 3
2. 2
3. 4
4. 1
Correct Option: , 2
Solution:
$x \neq-480$
$(|x|-3)(|x+4|)=6$
$\Rightarrow \quad|x|-3=\frac{6}{|x+4|}$
No. of solutions $=2$ | 122 | 278 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-30 | latest | en | 0.500064 |
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# amilby30 view on 100% lotto system by kola
Topic closed. 99 replies. Last post 11 years ago by tntea.
Page 5 of 7
Blundering Time Traveler
United States
Member #28945
December 25, 2005
1532 Posts
Offline
Posted: May 25, 2006, 2:42 pm - IP Logged
John
Do you have a chart with these boxed numbers for each letter? I think that could help visually, not to mention allow some back testing. If I got 4.72 box hits per week, I could make a nice side income for sure. Especially the way I play with the progressive betting strategy.
Thanks for this revelation!
CPS,
I guess I am using the original chart posted to come up with that elimination formula.
Kola's original post gives 64 columns with 10 numbers each for a total of 640 numbers. What I described was eliminating the columns that have a starting letter of S, U or M. That eliminates 48 columns right off the bat. Then, looking at the 2nd Letter of the remaining columns, you eliminate the letter that is out the longest. This will eliminate an additional 4 columns. Finally, you look at the 3rd Letter of the remaining columns and eliminate the letter that is out the longest. This eliminates an additional 3 columns leaving 9 remaining columns with 90 total numbers.
I hope that makes sense.
Good Luck,
John
Its great you adapted it to your own usage. Very Good Strategy. Tenaj had a good point with it being user driver. It is. But not only.
I've come to realize that the DSUM GRID is like an "X,Y" Grid. According to how I arranged the grid in my posts(you can arrange anyway you like)
1. X = string-row
2. Y = the column
The winning number is where they converge. I'm getting nice results, still have to check to see its not a fluke. One thing is becoming clear, I've been getting consistent result finding "X" which is the row that contains the winning number. problem is that pointer or number I'm using may be found in multiple rows. So I would have to play at least 1 to 2 row or even three rows. But that's okay. Because I can then cross out the duplicates, and be left with not so many numbers. Stay tuned though...I'll be forthcoming - no smoke and mirrors.
The Carolinas - Charlotte
United States
Member #21627
September 12, 2005
4138 Posts
Offline
Posted: May 25, 2006, 2:52 pm - IP Logged
Good to hear Kola...keep up the good work!
The North Carolina Education Lottery - so much a joke that here are their mascots:
East of Atlanta
United States
Member #6191
August 11, 2004
1389 Posts
Offline
Posted: May 25, 2006, 3:08 pm - IP Logged
After a "very general" overview of Kola's "system", I see the format is actually relatively simple. So much so, it only took 20 minutes to set it up in my spreadsheet. The reason is, it's 3 columns wide by 4 rows high. The trick to "simplifying" it was change the position of everything.
D D D S S S U U U M M M
Once the above was established, it meshed in nicely with my existing spreadsheets. The end results is this. One item you may notice is that the S & D was Switched. This helps to simplify my table setup and keeps it automated so it changes each time I update my spreadsheet without intervention from me. With Last nite's and Today's Cash 3 hits, here are the results.
Eve Mid S 0 5 9 8 1 3 D 9 4 8 7 0 2 U 1 6 0 9 2 4 M 5 0 4 3 6 8
Evening Results
059 069 959 969 159 169 559 569 058 068 958 968 158 168 558 568 050 060 950 960 150 160 550 560 054 064 954 964 154 164 554 564 049 009 949 909 149 109 549 509 048 008 948 908 148 108 548 508 040 000 940 900 140 100 540 500 044 004 944 904 144 104 544 504
Midday Results
813 823 713 723 913 923 313 323 812 822 712 722 912 922 312 322 814 824 714 724 914 924 314 324 818 828 718 728 918 928 318 328 803 863 703 763 903 963 303 363 802 862 702 762 902 962 302 362 804 864 704 764 904 964 304 364 808 868 708 768 908 968 308 368
Now, this combined with my other tables should help me narrow down my choices (then I will sprout my wings and fly home...sarcasm intended).
Sir Metro
Blundering Time Traveler
United States
Member #28945
December 25, 2005
1532 Posts
Offline
Posted: May 25, 2006, 3:41 pm - IP Logged
After a "very general" overview of Kola's "system", I see the format is actually relatively simple. So much so, it only took 20 minutes to set it up in my spreadsheet. The reason is, it's 3 columns wide by 4 rows high. The trick to "simplifying" it was change the position of everything.
D D D S S S U U U M M M
Once the above was established, it meshed in nicely with my existing spreadsheets. The end results is this. One item you may notice is that the S & D was Switched. This helps to simplify my table setup and keeps it automated so it changes each time I update my spreadsheet without intervention from me. With Last nite's and Today's Cash 3 hits, here are the results.
Eve Mid S 0 5 9 8 1 3 D 9 4 8 7 0 2 U 1 6 0 9 2 4 M 5 0 4 3 6 8
Evening Results
059 069 959 969 159 169 559 569 058 068 958 968 158 168 558 568 050 060 950 960 150 160 550 560 054 064 954 964 154 164 554 564 049 009 949 909 149 109 549 509 048 008 948 908 148 108 548 508 040 000 940 900 140 100 540 500 044 004 944 904 144 104 544 504
Midday Results
813 823 713 723 913 923 313 323 812 822 712 722 912 922 312 322 814 824 714 724 914 924 314 324 818 828 718 728 918 928 318 328 803 863 703 763 903 963 303 363 802 862 702 762 902 962 302 362 804 864 704 764 904 964 304 364 808 868 708 768 908 968 308 368
Now, this combined with my other tables should help me narrow down my choices (then I will sprout my wings and fly home...sarcasm intended).
Sir Metro
Its interesting. When I first worked the system out, it was more natural for met o put SDUM instead of DSUM. I though it would be more simple as far as the workout was concerned. More fluid. But I liked the HARD "D" in the beginning for a stronger sound. So I sold out. I'm glad you've adapted the DSUM code to suit you. Ciao.
Michigan
United States
Member #22395
September 24, 2005
1583 Posts
Offline
Posted: May 26, 2006, 12:42 am - IP Logged
I hope I am not too anxious?
Friday has been here over ½ hour!
United States
Member #14
November 9, 2001
31540 Posts
Offline
Posted: May 26, 2006, 12:44 am - IP Logged
I hope I am not too anxious?
Friday has been here over ½ hour!
love to nibble those micey feet.
Cobb
United States
Member #26586
November 19, 2005
9765 Posts
Offline
Posted: May 26, 2006, 6:39 am - IP Logged
after work today ill get with you as promised, i just stepped in for a second to see if i had any hit in ga ,and wow at some of the bad post about me and this system when people dont get things the way they want , like lil mad children
when they dont get what they want for christmas
later and good luck
I LOVE THIS SPORT !! BY30PREDICTIONS
Sunny Georgia
United States
Member #38894
May 7, 2006
41 Posts
Offline
Posted: May 26, 2006, 7:06 am - IP Logged
FRIDAY IS HERE!!!
HE HE HE ! YES! HOOHOO !
HOORAY!
Seven hours into it to be exact. Let's play CASH 3 in Georgia.
happy friday and good luck to everyone!
Laurlye
Oklahoma
United States
Member #33770
February 24, 2006
3146 Posts
Offline
Posted: May 26, 2006, 7:46 am - IP Logged
I'm all eyes and ready to get three hits a week! YaHoo!!!
Steve
The Carolinas - Charlotte
United States
Member #21627
September 12, 2005
4138 Posts
Offline
Posted: May 26, 2006, 8:16 am - IP Logged
Me too!! WOO HOO!! :)
The North Carolina Education Lottery - so much a joke that here are their mascots:
United States
Member #173
April 8, 2002
6114 Posts
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Posted: May 26, 2006, 10:58 am - IP Logged
I'am still waiting.
WTG all Winner\$.
Poway CA (San Diego County)
United States
Member #3489
January 25, 2004
14120 Posts
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Posted: May 26, 2006, 11:03 am - IP Logged
She went to work and will reveal the winning method when she gets back. But, didn't she say she made a living at the lottery? Oh, maybe she works in a 7-11 selling lotto tickets.
Anyway, I gave up a fishing trip with my buddies to stay home today to get this method and start betting it right away. It looks like it will shut down the online betting services within hours, so there is no time to waste once she reveals the method.
I will be right here all day waiting.
Tx
United States
Member #4570
May 4, 2004
5180 Posts
Offline
Posted: May 26, 2006, 11:28 am - IP Logged
CD
You should had gone fishing, at least you would have something by now instead of nothing.
The pick 3 lotteries are not about to close down no time soon, system or no system.
20 more systems might come by and they still won't shut-down.
Maybe you can still go fishing, it is not too late yet.
Minnesota
United States
Member #13028
March 28, 2005
870 Posts
Offline
Posted: May 26, 2006, 11:34 am - IP Logged
She went to work and will reveal the winning method when she gets back. But, didn't she say she made a living at the lottery? Oh, maybe she works in a 7-11 selling lotto tickets.
Anyway, I gave up a fishing trip with my buddies to stay home today to get this method and start betting it right away. It looks like it will shut down the online betting services within hours, so there is no time to waste once she reveals the method.
I will be right here all day waiting.
CD,
I admire your vigilance in getting the scoop on this system. Anyone who gives up a Memorial weekend fishing trip must be extremely excited for this system.
Personally, I am already skeptical. I hate feeling this way, but I do. We're running up against the midday drawing deadlines and she still hasn't performed the "GRAND UNVAILING" of this system. I don't know ...... but how long does it take to post something? I don't mean to be harsh or to criticize, but man, come out with it already.
I lay odds that Friday will pass without the "GRAND REVELATION".
Anyway, Happy Memorial Day! My hat goes off to all Vets. To them, I say thank you and God Bless You!
Godd Luck,
John
BOW WOW WOW ......
...... YIPPY YOH YIPPY YAY!!!
Poway CA (San Diego County)
United States
Member #3489
January 25, 2004
14120 Posts
Offline
Posted: May 26, 2006, 11:42 am - IP Logged
She went to work and will reveal the winning method when she gets back. But, didn't she say she made a living at the lottery? Oh, maybe she works in a 7-11 selling lotto tickets.
Anyway, I gave up a fishing trip with my buddies to stay home today to get this method and start betting it right away. It looks like it will shut down the online betting services within hours, so there is no time to waste once she reveals the method.
I will be right here all day waiting.
CD,
I admire your vigilance in getting the scoop on this system. Anyone who gives up a Memorial weekend fishing trip must be extremely excited for this system.
Personally, I am already skeptical. I hate feeling this way, but I do. We're running up against the midday drawing deadlines and she still hasn't performed the "GRAND UNVAILING" of this system. I don't know ...... but how long does it take to post something? I don't mean to be harsh or to criticize, but man, come out with it already.
I lay odds that Friday will pass without the "GRAND REVELATION".
Anyway, Happy Memorial Day! My hat goes off to all Vets. To them, I say thank you and God Bless You!
Godd Luck,
John
As a Viet Nam veteran, I thank you.
As everyone knows I have been critical of those that had "systems" that didn't work. I often was told to back off. But, you will have to admit that I was always right in the end. I have tried to bite my tongue on this one because I could see that this might work and I wouldn't want to be wrong for the first time!! We have had people that had a fool-proof (pun intended) system that after they were challenged by me and others, they said that they would not reveal the secret and they took their ball and went home. I did not want that to happen this time. If someone can show me how to play 64 combos and hit a box every time, then I don't want to run them off. We have plenty of time to be critical late Friday or starting on Saturday. Remember ... follow the rules ... criticize the method, not the person.
Page 5 of 7 | 3,612 | 12,266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-04 | latest | en | 0.9229 |
http://excel.bigresource.com/Maximum-text-length-with-VLOOKUP-5EXOK74E.html | 1,516,495,113,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889798.67/warc/CC-MAIN-20180121001412-20180121021412-00539.warc.gz | 111,593,243 | 13,423 | # Maximum Text Length With VLOOKUP
Dec 7, 2009
I use VLOOKUP with text (to return comments made by people that I have copied in another sheet). The problem is that sometimes, it doesnt copy the whole comment.
Apparently there is a limit for the amount of text VLOOKUP can copy: after a LEN() test I have found that I cant copy texts longer than 255 characters.
Is there a simple way to make the VLOOKUP work even if the text is more than 255 characters long ?
## Using Data Validation To Control Text Length W/Vlookup
Mar 26, 2009
Can anyone provide a formula to be used in data validation that will control text length (6 digits) and restrict duplicate entries. The best formula will prevent anything other that 6 digits, but question the user regarding a duplicate entry.
For example: if the user enters 123456 no problem, but if 12345 is entered, Excel validation would not allow. If the user enters 123456 again, Excel's validation window would allow but the window will pop-up and ask to confirm.
## Excceds The Maximum Length
Mar 5, 2007
Below is a formula that I use which now excceds the maximum length. =Consumables!S56/('Price list'!S56+'Price list'!S57+'Price list'!S58+'Price list'!S59+'Price list'!S60+'Price list'!S61+'Price list'!S62+'Price list'!S63+'Price list'!S64+'Price list'!S65 "all the way up to" +'Price list'!S118) This formula is also repeated is cells =Consumables!T57/('Price list'!S56+'Price list'!T57...... etc all the way =Consumables!AE56/('Price list'!AE56+'Price list'!AE57.............
Is there any way I can condense this formula? I have tried:
=Consumables!S56/('Price list'!S56:'Price list'!S118)
=Consumables!S56/('Price list'!S56,'Price list'!S118)
## Maximum Length For A Macro
Aug 29, 2007
Anyone know the maximum lenght/size limitations for a macro in excel?
hit this limitation a couple times, and it serves to be quite a annoynace... end up having to link them together with a Application.Run command.
Would be good information to have if anyone knows...
my guess is 2000 lines, but havnt tested, and am not sure if it is limited to lines necessarily... could be size too,
## Maximum Length Of Variant Or String
Aug 10, 2007
I have looked into the maximum length of a variant/string in vba and it appears to be 250 characters. I am running a macro which first lists all excel files in a folder, returning them to a sheet, then using a loop statement opens each one in turn extracting the information to a second summary sheet before closing it. The file path to the folder is ridiculously long and the macro stumbles. I used the =LEN(A1) formula to check if the file names were too long for the string, but the maximum file name length was 226 characters. I've tried both String and Variant to collect the file names but both have the same effect.
## Vlookup Maximum
Jan 26, 2010
I am using the vlookup function for a reference that has two values. I want to choose the max. See attached file. In essence, I would like the vlookup for "a" to output "2" and "e" to output "7".
## Multiple VLOOKUP Conditions And MAXIMUM Value
Aug 16, 2009
I have written a snooker scorebaord spreadsheet which keeps a history of highest scores and highest breaks, but am unable to find a way of showing these key values in a table.
Attached is a cut down version to demonstrate the problem. If you can provide a solution it can be either in formula or code.
## Pull Varying Length Text From Cell Text
Jul 2, 2007
I need to find text within middle of a string.
Character before required text is say AAA
Character after required text is say BBB
Text required can vary in length.
Extract text and place in another column.
All text in a single column, required text not in every line. but
does repeat.
## Validate Cell For Text Length & Characters In Text
Mar 13, 2008
I have a cell (B2) I would like to apply multiple data validations to.
I know I need to use the custom formula option but don't know how to write the formula.
I don't even know if it is possible, but here is what I'm after
I need to make sure the cell is 4 digits long
I need to make sure the cell starts with a zero (Because the cell starts with a zero I have it as a text cell)
I need to make sure the 2nd number is not 0 if A2 begins with 5 (A2 is also a text cell).
## Shorten String Length If Exceeds 31 Characters (max Sheet Name Length)
Mar 27, 2014
I've set up a filing system which saves sheets/ workbooks based on the value of a cell - Range("B1") Everything works great apart from when ThisFile String length exceeds 31 characters which you may know is the max useable character length for a sheet name - I had no idea! 8-0
Is there a way i can check if string length exceeds 31 characters then, if it does, shorten it to 31 characters?
[Code] .....
## Get Text To The Right - Length Will Vary
Feb 8, 2011
If I have a cell with this info: [6126]BOB SMITH
What formula can I use to get BOB SMITH. The length of the name will vary. The number will change, but will always be 4 digits and will have the brackets. [XXXX]
## Text Box Length Limitation
May 8, 2008
i have tried to create a userform thru which data is to be entered. so that the data will be printed on a worksheet. here there are few text boxes, in which the number of digits should be equal to 14. after filling the userform when i click the print button if it is less than or more than 14 a pop up msgbox should be displayed with OK button and the cursor should go back to that particular text box. i have written the following code, but it has a problem. even if the total no of digits are 14 the msgbox is displayed.
say the text box name is Roll no
If txtRollNo.MaxLength 14 Then
MsgBox "Roll No should be of 14 digits", vbInformation + vbOKOnly
txtRollNo.SetFocus
Exit Sub
End If
## Formula Code To Find Arc Length From Chord Length
Sep 4, 2012
How to create a code formula to calculate the arc length from a given chord length?
If you know the radius of the major circle.
Say the chord is 50mm and major circle dia is 72mm (radius 36mm)
arc from chord.jpg
## Give Cell A Value Based On Length Of Row (variable Row Length)
Jan 9, 2010
My worksheet contains data with the reaction times on a psychological test. Each respondent in the test has 280 rows in my excel sheet.
The 'perfect' length of the row, is from A to M. When an error is made in the test, the length of the row will increase. So the error length can be A to AA.
For me it is important to analyse the error. So I would like to give a perfect row length, the value 1, and an error row length a value 2.
So, in conclusion:
If:
Cell length = A1 - M1? --> Copy A1 B1 C1 (A B Cof that row) to Sheet3, and give D1 in sheet 3 the value 1
Cell length >= A1 - M1? --> Copy A1 B1 C1 (A B C of that row) to Sheet3, and give D1 in sheet 3 the value 2
## Text Parsing With Variable Length?
Jun 10, 2014
Column M:
##/##/#### | Variable Length Text-####
Example:
01/06/2014 | Daniel Trimble-4048
I need to parse out the different parts of Column M.
In Column R -- "Close Date", I'm successfully using:
=LEFT(M2,FIND(" | ",M2)-1)
...to extract the close date of the donation.
In Column S, I want to list the donor name--which is all of the text after " | ", and before the "-".
I don't need anything after the hyphen, and fortunately in this data, no one's name has a hyphen in it.
The Close Date is working fine for the LEFT and FIND functions, but for the life of me, I can't seem to get MID to work for the variable-length text. The text will always start in the same position -- 14, as the date and delimiter are standardized. And the last 5 characters of the text are not variable in length, so they can be cut out completely.
How do I use MID to extract everything starting at position 14, and stopping 5 characters short of the end of the text?
## Format To Display Set Text Length
Sep 17, 2013
I'm trying to avoid using merged cells or text wrapping with altered row height to display some text in multiple rows - similar to using centre across selection to have a header across multiple columns without merging. The guys that use the sheet type a comment that is relevant to five rows and the easy solution here would be to just type the first portion of the comment in the first row (about 30 characters will display in the column width available), then put the next 30 characters in the next row and so on but the guys keep getting lazy and merging the cells so they can type the comments more easily. I can lock the sheet or force validation but I think there's a better solution.
I can effectively "wrap" the text across the five rows the header is relevant to by using a formula to pick up everything except the first 30 characters of each cell. Ie if they type whatever they want in cell B16 then I can use this formula to break it in to 30 character lengths to "wrap" it in to the next four rows:
=IF(LEN(B17)>30,RIGHT(B17,(LEN(B17)-30)),""),
And I'm sure it would be easy enough to use search with the formula to break it where there is a space in the text so partial words don't flow over.
BUT because the text ends up slightly different widths I want to use formats to force only 30 characters to display (whilst keeping the remainder of the text string intact). I can't figure out the syntax to format only 30 text characters to display but you can easily do it with numbers and dates and so on.
## Cut Text Data To Length Down Column
Dec 25, 2006
I have a range of text data in a column and need to get the text lengths to no longer than 60 characters.
The remaining data then need to be copied in a cell inserted below the original.
I have been playing with the following code…
## Formula To Select Text Length
Jan 23, 2007
I am having difficulty with creating an IF formula that will only show 9 digit numbers. If the cell the formula is looking at has less or more than a 9 digit number in it, or the cell ha no value, the result will be blank.
## Cut/Reduce Cell Text Length
May 2, 2008
I have a cell with 200+ character, I only want the 40 first character is there a function that will give me only those 40 first character or do I have to use a "=len" and manually remove the extra characters?)
## Split Text Into Fixed Length Columns
Jul 20, 2012
I have 5 columns with data in each
I want to create a 6th column that looks to the columns on the left with data in ti and concatenates all data in the 5 columns and puts it into one cell in the 6th column however put a space between each break of data so that it can be distinguished which bit of data was in what column previously.
The challenge is the new 6th column can only contain 30 characters - When it exceeds 30 characters then create a 7th column and put the rest of data in the 7th column, again the 7th column can only have 30 characters so if exceeds this then put the remaining characters in a 8th column
There will never be more than a total of 90 characters in the original 5 columns so there will only need to be scope for a maximum of 3 additional columns
So for example
Column A had two words in it that totaled 20 characters (the space between the two words is also counted as a character)
Column B had two words in it that totaled 20 characters (the space between the two words is also counted as a character)
Column C had a word that contained 10 characters
Column D had a word that contained 5 characters
Column E had a word that contained 10 characters
Then the result would be
Column F would only have the data originally held in Column A (because it can't include Column B's data as this would exceed the 30 characters)
Column G would have data that was originally held in column B and column C - with a space between B and C data
Column H would have data that was originally held C, D and E - with a space between C, D and E data
Another point to consider is if in one of the orginal 5 columns had say 3 words in it and lets say the 3rd word is the word that exceeds the 30 character limit, then the whole of the third word is to be carried oved to the next new column, I can't have words cut in hlaf with one half in Column 'F' and the other half in Column 'H' for example.
## Importing Fixed Length Text Into Excel?
Apr 15, 2014
I am using the code below to import a fixed-length text file into Excel. As the macro is written, it imports starting at the first line of the text file. How do i tell it to start importing at line 1000 and above?
## VBA Macro - If Statement Based On Length Of Text
Feb 11, 2014
I'm having a bit of trouble with a macro designed to read the length text in a cell and if it = a certain length then perform an action (in this case Text to column)
Here is a small sample of the data I'm working with:
Tue 02/11/2014
LastBootUpTime
20140211082244.222441+000 <<
Tue 02/11/2014
LastBootUpTime
20140211082244.222441+000 <<
Tue 02/11/2014
LastBootUpTime
20140211082244.222441+000 <<
Tue 02/11/2014
LastBootUpTime
20140211082244.222441+000 <<
Tue 02/11/2014
LastBootUpTime
Tue 02/11/2014
LastBootUpTime
20140211082244.222441+000 <<
*End Sample*
The text length I want it to perform the action on is highlighted with "<<" if the length of text does not meet the required number then I want the statement to skip and move onto the next one.
I have the text to column code already done with relative references however the long text string I want the statement activated on is not always present which means that the pattern (0,3) is not always consistent.
## Measure The Length Of Text In 10 Boxes Consecutivly.
Feb 13, 2009
I am trying to use a For loop to measure the lenght of text in 10 text boxes on a user form so I can run a check but can't think of a way to do it. This is the best I have so far...
## Set Font Size Depending On Length Of Text
Sep 3, 2007
When a number is entered in cell S3, it triggers formulas throughout the worksheet to populate the it with information from another sheet. If the length of the text in B6 is greater than 80, the font size for B6 only should be 8; if the length of the text in B6 is less than or equal to 80, the font size for B6 should be 10. Regardless of the length of the text in B6, the font size for the rest of the sheet should not be changed.
I tried the following
Option Explicit
Private Sub Worksheet_Change(ByVal Target As Range)
If Target.Address "\$B\$6" And Target.Count > 1 Then Exit Sub
Dim cnt As Long
cnt = Len(Cells(6, 2).Value)
If cnt > 80 Then Cells(6, 2).Font.Size = 8
ElseIf cnt < 80 Then
Cells(6, 2).Font.Size = 10
End If
End Sub
## Extracting Date From Text Strings Of Different Length
May 9, 2008
Am trying to devise a formula that will allow me to extract a date (not stored in date format or recognizable by excel as such: "DEC1/09") and then manipulate it so that excel can recognize it and change it to a "1-Jan-01" form. Problem for me is that functions such as LEFT MID and RIGHT are very specific and sensitive obviously to any additional characters. some of my dates are preceded by "CAN BND 4.25/09"; "CANB BND 4.25/09"; "BC BND 4.25/09" and so on, you get the idea im sure that they are of differing lengths. The dates are equally strewn around as some (these are bond maturities) are 1st of the month while others may hold dates in the middle-end...15th, 30th, 31st etc.
is there a formula that will recognize the dates in the text strings regardless of string length and then a subsequent formula to manipulate the date to proper format?
i.e. "CAN BND 4.25/09 DEC1/09" and "CANB BND 4.25/07 JUN15/07"
Converted into: "01-Dec-09" and "15-Jun-07"
Not that the other parts of the string don't matter, already have macro that can recognize and rip bond coupons.
## Extracting Text From A String With Variable Length
Jan 13, 2009
Hey I got a long String like this "[...] increase of x.xx% [...]".
I am trying to extract only the percentage number which can be of variable length, so maybe 900.99% or 9.99%.
I tried this formula:
=MID(G14,SEARCH("%",G14)-5,5)
but this one doesnt bring the right results as the percentage figure is often not exactly 5 characters long.
## Limit Cell Entries Text Length
Aug 23, 2007
I think it's an easy one but for the life of me I can't get around it. ccasionally while re-typing data in fields with the intention of saving-as a new file, an error message appears saying "no more than 20 characters" This happens even when the number of characters is LESS than 20! I have tried clearing cell contents, copying & pasting, re-setting conditions/parameters for columns/rows etc., and nothing seems to work with any consistency. Instead of looking for a work-around solution I'd like to find the actual corrective action.
## Sort Cells Based On Text Length
Dec 13, 2007
I need to write a Excel VBA (2003) code that can arrange the cells of one column based on the character length. An example is this:
Column Data before Running VBA
(Column A)
cat
oranges
apple
Column Data after Running VBA
(Column A)
cat
apple
oranges
I been doing this manually by using the LEN() command in an adjacent column (Column B) and sorting Column A. based on Column B. However, I wish to do this all in a VBA code that does not rely on how many rows are in Column A. I have dozens of excel sheets with various number of rows for Column A. Therefore, the VBA code has to also figure out when the last non-empty row is in Column A. Can anyone offer their suggestions to my problem?
## Determine Text Length Of Lookup Result
Mar 24, 2008
i need a formula that looks up a range on sheet 1 coloum a and returns the value in column b, unless the value in column c is less than 5 letters long/ or not equal to a time format if this is the case it should return the value in column c
[code]
=IF(LEN(VLOOKUP(A6,'look up'!A6:C18,3,FALSE)>6),VLOOKUP(A6,'look up'!A6:C18,3,FALSE),VLOOKUP(A6,'look up'!A6:C18,2,FALSE))
[code]
this is my effort but it returns value in the the middle every time
## Excel 2010 :: Match - Text Length Limitation
May 8, 2014
Excel 2007-2010. I'm using match(string, range,0) but there must be a limitation on the length of the string since I know the string is in the range but it returns #value as if it is not found. Is there a VBA solution to get around this without having to loop/cycle through the entire range? | 4,685 | 18,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-05 | latest | en | 0.824546 |
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### Homework_4_Solutions
Course: MATH 1110 1110, Fall 2011
School: Cornell
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Word Count: 700
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1110 Name: Homework Math 4 Due 9/22 or 9/23 in class Please print out these pages. Write your answers to the text exercises on separate paper and staple it to these pages. You should include computational details. These problems will be assessed for completeness. Always write neatly and legibly. Please answer the presentation problems in the spaces provided. Include full explanations and write your answers in...
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1110 Name: Homework Math 4 Due 9/22 or 9/23 in class Please print out these pages. Write your answers to the text exercises on separate paper and staple it to these pages. You should include computational details. These problems will be assessed for completeness. Always write neatly and legibly. Please answer the presentation problems in the spaces provided. Include full explanations and write your answers in complete, mathematically and grammatically correct sentences. Your answers will be assessed for style and accuracy, and you will be given written feedback on these problems. GRADES Text exercises / 20 Pres probs / 20 Staple Text exercises. Please do the following problems from the book. 2.5 # 4, 6, 7, 18, 20, 32, 33, 40, 44, 56, 60 2.6 #2, 14, 29, 38, 47, 50, 54, 66, 70, 73, 84, 100 /1 Math 1110 (Fall 2011) HW4 Presentation Problems 2 Question 1. (10 points) On a recent trip to the Department of Motor Vehicles (DMV), Professor Holm noticed a sign describing the ne for writing the DMV a bad check: one that is subsequently returned for insufcient funds in the bank account. The sign indicated that if the bad check was for \$200.00 or less, there would be a \$35.00 ne; and if the bad check was in excess of \$200.00, the ne would be 15% of the total amount of the check. Let f(x) be the function whose value at x is the ne (in dollars) imposed on a bad check of x dollars. (a) Complete the equation of this piece-wise dened function: 35 0 < x < 200; f(x) = x = 200; and 35 .15x x > 200. Note that the units of f(x) are dollars. (b) Is this function continuous at x = 200? Why or why not? We need to check that limx 200 f(x) = f(200). This is the same as checking that limx 200+ f(x) = limx 200 f(x) = f(200). Note that limx 200+ f(x) = limx 200+ .15x = 30, but f(200) = 35. We conclude that f is not continuous at x = 200. (c) Is this function non-decreasing for 0 < x < 300? Why or why not? We need to determine if for 0 < < x1 x2 < 300 it is always the case that f(x1) f(x2). Note that f(100) = 35 > 30.15 = f(201), but 100 < 201. So the function is not non-decreasing for 0 < x < 300. I nstructor Math 1 110 Name: Sample T rue/False To d emonstrate i n c lass Presentation p roblem 1 . D etermine w hether t he f ollowing s tatements a re ( always) t rue o r ( at l east sometimes) f alse, a nd c ircle y our r esponse. P leaseg ive a b rief e xplanation ( in c omplete s entences!) (b) I f f Math 1110 (Fall 2011) HW4 Presentation Problems( x) a nd g (x) b oth o ne-to-one f un 3 - a r eason w hy i t's t rue, o r a n e xample w here i t f ails. Tnun I (x - 4) Z (a) I f f (x) i s a (10 venf unction,t hen swhether :the 'following+statements are true or false, F alss Question 2. n e points) Determine o i s 9 (x) 2 f and circle your response. Please give a brief explanation (in a complete sentence!) a reason why its true, or an example where it fails. (a) Suppose f(x) is a function dened for all real numbers, and that x2 f(x) x2 for every x. Then f(0) = 0 and f(x) is continuous at x = 0. Evaluating the expression x2 f(x) x2 at x = 0, we get 02 f(0) 02. In particular, 0 f(0) 0, if f(0) is dened. Since we are given that f is dened for all real numbers, we may conclude that f(0) = 0. Further, note that lim x2 = 0 = lim x2. We may conclude by the sandwich theorem that x 0 x 0 lim f(x) = 0. In particular, lim f(x) = 0 = f(0), so f(x) is continuous at x = 0. o t ( x) (b) I f f x 0a nd g (x) b oth o ne-to-one 0unctions d efined o n a ll o f l R., hen f x f g i s a lso o ne-to-one. Tnus I F elss (b) If lim+ f(x) = lim f(x), then f(x) is continuous at x = a. x a x a Consider the example: f(x) = 0 x < 0; 1 x = 0; and 0 x > 0. We have lim f(x) = 0 = lim f(x), and so lim f(x) = 0. The function is not continuous at + x 0 x 0 x = 0, however, because f(0) = 1. x 0 Homework 4 Book Problem Answers Section 2..5: Section 2.6:
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1.Today's nanotechnology-produced computer transistors are roughly equivalent in size to:A) a human hair.B) a virus.C) the width of a fingernail.D) an atom.2.Larry is the IT manager at his retail firm. He has seen IT purchases go up dramatically ov | 6,190 | 22,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2013-20 | latest | en | 0.890671 |
http://aimsciences.org/journals/displayArticles.jsp?paperID=628 | 1,544,644,201,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824115.18/warc/CC-MAIN-20181212181507-20181212203007-00578.warc.gz | 10,353,584 | 10,544 | # American Institute of Mathematical Sciences
2004, 4(4): 961-982. doi: 10.3934/dcdsb.2004.4.961
## On the $L^2$-moment closure of transport equations: The Cattaneo approximation
1 Department of Mathematical and Statistical Sciences, University of Alberta, Edmonton, AB, T6G 2G1, Canada
Received June 2003 Revised April 2004 Published August 2004
We consider the moment-closure approach to transport equations which arise in Mathematical Biology. We show that the negative $L^2$-norm is an entropy in the sense of thermodynamics, and it satisfies an $H$-theorem. With an $L^2$-norm minimization procedure we formally close the moment hierarchy for the first two moments. The closure leads to semilinear Cattaneo systems, which are closely related to damped wave equations. In the linear case we derive estimates for the accuracy of this moment approximation. The method is used to study reaction-transport models and transport models for chemosensitive movement. With this method also order one perturbations of the turning kernel can be treated - in extension of an earlier theory on the parabolic limit of transport equations (Hillen and Othmer 2000). Moreover, this closure procedure allows us to derive appropriate boundary conditions for the Cattaneo approximation. Finally, we illustrate that the Cattaneo system is the gradient flow of a weighted Dirichlet integral and we show simulations.
The moment closure for higher order moments and for general transport models will be studied in a second paper.
Citation: T. Hillen. On the $L^2$-moment closure of transport equations: The Cattaneo approximation. Discrete & Continuous Dynamical Systems - B, 2004, 4 (4) : 961-982. doi: 10.3934/dcdsb.2004.4.961
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# If the circus were to sell all of its 220 tickets for this
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23 Jan 2012, 13:33
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If the circus were to sell all of its 220 tickets for this month's performance at its usual price, the revenue from sales would be 10% greater than that collected last month. If the circus raised the ticket price by 5% and sold only 200 tickets as a result, what percent less would last month's revenue be compared to this month's revenue?
A. 2
B. 5
C. 100/21
D. 110/20
E. 9/4
[Reveal] Spoiler: OA
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23 Jan 2012, 16:38
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manalq8 wrote:
If the circus were to sell all of its 220 tickets for this month's performance at its usual price, the revenue from sales would be 10% greater than that collected last month. If the circus raised the ticket price by 5% and sold only 200 tickets as a result, what percent less would last month's revenue be compared to this month's revenue?
A. 2
B. 5
C. 100/21
D. 110/20
E. 9/4
For a percentage questions like this one, it's almost always better to plug some smart numbers.
Let the usual price of a ticket be $20 (I chose$20 because $20+5%=$21=integer, which will make calculations easier). This month's revenue for this price would be 220*$20=$4,400 and we are told that it's 10% greater than the revenue collected last month, hence the last month's revenue was $4,400/1.1=$4,000;
Circus raised the ticket price by 5%, so the new price was $21 and the actual revenue from 200 tickets was 200*$21=$4,200; What percent less would last month's revenue be compared to this month's revenue: $$percent=\frac{4,200-4,000}{4,200}*100=\frac{100}{21}%$$. General formula for percent increase or decrease, (percent change): $$percent=\frac{Change}{Original}*100$$, so as we are comparing the difference to this month's revenue we should put this value ($4,200) in the denominator.
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07 Feb 2012, 22:56
I'm not sure what the question is asking:
"what percent less would last month's revenue be compared to this month's revenue?"
The way I interpreted the question was "last month's revenue is less than this month's revenue by how many percent?"
I got 5%,
(this month)-(last month)] / (last month)
but the OA is C. Please explain. Thanks.
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08 Feb 2012, 00:18
I think the answer is B.
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08 Feb 2012, 01:04
Samwong wrote:
I'm not sure what the question is asking:
"what percent less would last month's revenue be compared to this month's revenue?"
The way I interpreted the question was "last month's revenue is less than this month's revenue by how many percent?"
I got 5%,
(this month)-(last month)] / (last month)
but the OA is C. Please explain. Thanks.
I merged similar topics. Please refer to the solutions above. Hope it helps. Please let me know if you have any questions.
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08 Feb 2012, 01:08
It's B for sure, human error in inputing OA.
1.1 Y = 220*X
Y = 200X
This month ticket price went up by 5% total number of people is 200. You need to find out how much LESS last month's revenue is compare to this month.
(last month-this month)/this month.
(200X-(200*1.05*X))/(200*1.05*X)
(-200*.05)/(200*1.5*X) ==> (-10/210X) ==> 1/21 (ignored negative since they only want the percent).
1/21 is a real number n not a percent, so in order to get a percent you need to times it by 100 (.03 = 3%, 1= 100%) ==. 100/21.
The wording is pretty crappy... Took me a while to get what it meant
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Re: If the circus were to sell all of its 220 tickets for this [#permalink]
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08 Feb 2012, 09:55
Bunuel wrote:
manalq8 wrote:
If the circus were to sell all of its 220 tickets for this month's performance at its usual price, the revenue from sales would be 10% greater than that collected last month. If the circus raised the ticket price by 5% and sold only 200 tickets as a result, what percent less would last month's revenue be compared to this month's revenue?
A. 2
B. 5
C. 100/21
D. 110/20
E. 9/4
For a percentage questions like this one, it's almost always better to plug some smart numbers.
Let the usual price of a ticket be $20 (I chose$20 because $20+5%=$21=integer, which will make calculations easier). This month's revenue for this price would be 220*$20=$4,400 and we are told that it's 10% greater than the revenue collected last month, hence the last month's revenue was $4,400/1.1=$4,000;
Circus raised the ticket price by 5%, so the new price was $21 and the actual revenue from 200 tickets was 200*$21=$4,200; What percent less would last month's revenue be compared to this month's revenue: $$percent=\frac{4,200-4,000}{4,200}*100=\frac{100}{21}%$$. General formula for percent increase or decrease, (percent change): $$percent=\frac{Change}{Original}*100$$, so as we are comparing the difference to this month's revenue we should put this value ($4,200) in the denominator.
Let's suppose that if question is "what percent increase would this month's revenue be compare to last month's revenue?"
Would the answer then be 5%?
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Re: If the circus were to sell all of its 220 tickets for this [#permalink]
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08 Feb 2012, 11:13
Yup. The denominator will be 20 instead of 21. 1/20 = 5%.
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Re: If the circus were to sell all of its 220 tickets for this [#permalink]
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09 Feb 2012, 08:12
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manalq8 wrote:
If the circus were to sell all of its 220 tickets for this month's performance at its usual price, the revenue from sales would be 10% greater than that collected last month. If the circus raised the ticket price by 5% and sold only 200 tickets as a result, what percent less would last month's revenue be compared to this month's revenue?
A. 2
B. 5
C. 100/21
D. 110/20
E. 9/4
This is what comes to mind when I read this question:
"If the circus were to sell all of its 220 tickets for this month's performance at its usual price, the revenue from sales would be 10% greater than that collected last month."
220 tickets at usual price make 10% extra revenue. In case last month the tickets were sold at usual price, they would have sold 200 tickets (10% extra revenue would be due to 10% extra tickets sold if the price is the same)
"If the circus raised the ticket price by 5% and sold only 200 tickets as a result"
If the circus sold the same number of tickets as sold last month but increased the price by 5%, the revenue would also increase by 5%. (Revenue increases by either increasing the number of tickets or increasing the price or both)
"what percent less would last month's revenue be compared to this month's revenue"
The question isn't the % by which this month's revenue is greater (which we know is 5%). The question is by what percent is last month's revenue less?
If last month's revenue = 100, this month's revenue = 105
Last revenue is less by 5/105 * 100 = 100/21%
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What percent less would last month's revenue be compared to this month's revenue: $$percent=\frac{4,200-4,000}{4,200}*100=\frac{100}{21}%$$. General formula for percent increase or decrease, (percent change): $$percent=\frac{Change}{Original}*100$$, so as we are comparing the difference to this month's revenue we should put this value ($4,200) in the denominator. Answer: C. Thank you for your comment. I don't quite agree with the % change formula. Common formula for % change is Change % = (Current month less Last Month)/Last Month *100% (this approach is consistent with 13th OE, please refer to question 177 problem solving). But even if you would like to use current month as a reference value then the formula should be Change % = (Last month less Current Month)/Current Month *100 You always deduct and divide by the same amount (i.e. reference value) Common approach is to use prior month as a reference value. Thus the correct answer should be either negative (if current month is used as a reference value) or recalculated using last month as a reference value Math Expert Joined: 02 Sep 2009 Posts: 37102 Followers: 7251 Kudos [?]: 96457 [0], given: 10751 Re: If the circus were to sell all of its 220 tickets for this [#permalink] ### Show Tags 29 Jun 2013, 04:17 NikRu wrote: Bunuel wrote: manalq8 wrote: If the circus were to sell all of its 220 tickets for this month's performance at its usual price, the revenue from sales would be 10% greater than that collected last month. If the circus raised the ticket price by 5% and sold only 200 tickets as a result, what percent less would last month's revenue be compared to this month's revenue? A. 2 B. 5 C. 100/21 D. 110/20 E. 9/4 For a percentage questions like this one, it's almost always better to plug some smart numbers. Let the usual price of a ticket be$20 (I chose $20 because$20+5%=$21=integer, which will make calculations easier). This month's revenue for this price would be 220*$20=$4,400 and we are told that it's 10% greater than the revenue collected last month, hence the last month's revenue was$4,400/1.1=$4,000; Circus raised the ticket price by 5%, so the new price was$21 and the actual revenue from 200 tickets was 200*$21=$4,200;
What percent less would last month's revenue be compared to this month's revenue: $$percent=\frac{4,200-4,000}{4,200}*100=\frac{100}{21}%$$. General formula for percent increase or decrease, (percent change): $$percent=\frac{Change}{Original}*100$$, so as we are comparing the difference to this month's revenue we should put this value ($4,200) in the denominator. Answer: C. Thank you for your comment. I don't quite agree with the % change formula. Common formula for % change is Change % = (Current month less Last Month)/Last Month *100% (this approach is consistent with 13th OE, please refer to question 177 problem solving). But even if you would like to use current month as a reference value then the formula should be Change % = (Last month less Current Month)/Current Month *100 You always deduct and divide by the same amount (i.e. reference value) Common approach is to use prior month as a reference value. Thus the correct answer should be either negative (if current month is used as a reference value) or recalculated using last month as a reference value Not quite understand what you mean here. Anyway, check here: if-the-circus-were-to-sell-all-of-its-220-tickets-for-this-126466.html#p1041966 Might help. Revenue this month =$4,200.
Revenue last month = $4,000. 4,000 is ~4.8% (100/21%) less than 4,200. _________________ GMAT Club Legend Joined: 09 Sep 2013 Posts: 13940 Followers: 590 Kudos [?]: 167 [0], given: 0 Re: If the circus were to sell all of its 220 tickets for this [#permalink] ### Show Tags 03 Sep 2014, 08:33 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ GMAT Club Legend Joined: 09 Sep 2013 Posts: 13940 Followers: 590 Kudos [?]: 167 [0], given: 0 Re: If the circus were to sell all of its 220 tickets for this [#permalink] ### Show Tags 07 Sep 2015, 05:54 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Senior Manager Joined: 20 Feb 2015 Posts: 389 Concentration: Strategy, General Management Followers: 2 Kudos [?]: 66 [0], given: 10 Re: If the circus were to sell all of its 220 tickets for this [#permalink] ### Show Tags 07 Sep 2015, 08:18 say last month the revenue was x this month it increased by 10 % =1.1x now as per the question 1.1x=220 x=200, thus the revenue last month was 200 As per the question :1.05x=200 therefore x=210 and hence increase from last month =(210-200/210)*100 which gives 100/21 Director Joined: 23 Jan 2013 Posts: 582 Schools: Cambridge'16 Followers: 1 Kudos [?]: 21 [0], given: 40 Re: If the circus were to sell all of its 220 tickets for this [#permalink] ### Show Tags 05 Oct 2015, 21:41 y - last month price x - last month revenue Price x Number = Revenue --y---------220--------1.1x --1.05y----200---------? 220y=1.1x 210y=?x 220y/210y=1.1x/?x => 210y=1.05x (105-100)/105=(5/105)*100=500/105=100/21 C Director Joined: 05 Mar 2015 Posts: 713 Followers: 8 Kudos [?]: 116 [0], given: 37 Re: If the circus were to sell all of its 220 tickets for this [#permalink] ### Show Tags 05 Jun 2016, 19:59 manalq8 wrote: If the circus were to sell all of its 220 tickets for this month's performance at its usual price, the revenue from sales would be 10% greater than that collected last month. If the circus raised the ticket price by 5% and sold only 200 tickets as a result, what percent less would last month's revenue be compared to this month's revenue? A. 2 B. 5 C. 100/21 D. 110/20 E. 9/4 Let price be 10/ticket then 220 tickets for this month's performance at its usual price=2200 which is 10%greater than last month(let X)---->x(1+10/100)=2200----->x=2000 now new revenue this month=200*10.5=2100 as per question asked 2000=2100(1-A/100)----->A=100/21% (Note:- if question asked for what % greater is this month revenue of last month,then =2000((1+A/100)=2100,will make ans choice B i.e 5% as correct answer) Ans C Senior Manager Joined: 18 Jan 2010 Posts: 257 Followers: 5 Kudos [?]: 90 [0], given: 9 If the circus were to sell all of its 220 tickets for this [#permalink] ### Show Tags 05 Jun 2016, 20:36 manalq8 wrote: If the circus were to sell all of its 220 tickets for this month's performance at its usual price, the revenue from sales would be 10% greater than that collected last month. If the circus raised the ticket price by 5% and sold only 200 tickets as a result, what percent less would last month's revenue be compared to this month's revenue? A. 2 B. 5 C. 100/21 D. 110/20 E. 9/4 Price remains unchanged and still revenue is increasing. This means that ticket sales were increased. This increase would be 10 % of the tickets sold last month Now increased ticket sales were 220. So last month if tickets sold were x, then 1.1 x = 220; x = 200. So last month 200 tickets were sold. Now suppose the usual ticket price was p. So last month revenue = 200p This month ticket price has been increased by 5 %. New price = 1.05p Revenue of this month = 1.05p*200 = 210p How much % is last month revenue less than this month's revenue. Reduction: 210p - 200p = 10 p Key point: Reduction is with respect to this month so reduction % = $$\frac{10p}{210p}$$ * 100 = $$\frac{100}{21}$$ Answer: C Senior Manager Joined: 08 Dec 2015 Posts: 275 GMAT 1: 600 Q44 V27 Followers: 1 Kudos [?]: 15 [0], given: 35 Re: If the circus were to sell all of its 220 tickets for this [#permalink] ### Show Tags 06 Feb 2017, 10:04 my equations: 220P=11/10R R=revenue last month 105/100P * 200 = x/100R 200P=R 105P/100*200=x/100*200P x=105 so: answer B 5% change. I know something is wrong, but this thing makes sense. help please? Re: If the circus were to sell all of its 220 tickets for this [#permalink] 06 Feb 2017, 10:04 Go to page 1 2 Next [ 21 posts ] Similar topics Replies Last post Similar Topics: 1 Tickets for all but 100 seats in a 10,000-seat stadium were sold. 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6 If the circus were to sell all of its 220 tickets for this 3 29 Jul 2010, 12:41
22 If the circus were to sell all of its 220 tickets for this 13 07 Jan 2008, 08:20
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IABLE3-1 Critical Values of thex" Distribution 0.995 0.975 0.05 0.025 0.005 O0o 0o0 0.016 0.455 2.706 3.841 5,024 6.635 7.879 0.010 0.051 0.211 1.386 4.605 5.991 7.378 9.210 10.597 0.072 0.216 0.584 2.366 6.251 7.815 9.348 11.345 12.838 0.207 0.484 1.064 3.357 7,.779 9.488 11,143 13,277 14.860 ...
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The concentration for Na2S2O4 2. Determining the oxidizing capacity of a household cleanser or liquid bleach (a) Record the following data for the titration of a household cleanser (or bleach). Brand name of cleanser or bleach Data Trial 1 Trial 2 Mass of sample Trial 3 O.15 g 015 Volume of NA2S2O3 ...
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Homework: Section 7.3-7.4 HW Score: An"9 commete HW Score: 58.3396 , 10.5 of 18 pts kma Culeui 7.4.21-T Queston Heip Crcilntor Achin Ikhlaenvotonin antennlaln probabilty eual Iliant ure on tINt at last diants ar (untr Dar ahb are bcawccn 87 In0 703 udusit onune Supooxe 119 Jighe mndcmb accnd Le...
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(a) find the critical numbers of $f$ (if any), (b) find the open interval(s) on which the function is increasing or decreasing, (c) apply the First Derivative Test to identify all relative extrema, and (d) use a graphing utility to confirm your results. $f(x)=(x-1) e^{x}$...
##### What is the value of each of these sums of terms of a geometric progression?a) $sum_{j=0}^{8} 3 cdot 2^{j}$b) $sum_{j=1}^{8} 2^{j}$c) $sum_{j=2}^{8}(-3)^{j}$d) $sum_{j=0}^{8} 2 cdot(-3)^{j}$
What is the value of each of these sums of terms of a geometric progression? a) $sum_{j=0}^{8} 3 cdot 2^{j}$ b) $sum_{j=1}^{8} 2^{j}$ c) $sum_{j=2}^{8}(-3)^{j}$ d) $sum_{j=0}^{8} 2 cdot(-3)^{j}$... | 3,250 | 10,247 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.740476 |
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# Is the average (arithmetic mean) of a certain series of
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Is the average (arithmetic mean) of a certain series of [#permalink]
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18 Apr 2010, 13:45
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Is the average (arithmetic mean) of a certain series of consecutive integers an integer?
(1) The number of terms in the series is even.
(2) The median of the terms in the series is not an integer.
[Reveal] Spoiler:
D
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18 Apr 2010, 13:58
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IMO D
Statement 1. if number of terms is even , AM cannot be integer, hence sufficient.
take eg of 1,2,3 and 1,2,3,4
sum = n(A+L)/2 here d=1
AM = (A+L)/2 , now if A is odd , and its having even terms then L is even and vice versa .
Hence AM is not an integer.
Statement 2. if median is not an integer then , it cannot have odd number of terms, it must have even number of terms.. hence sufficient.
Thus D
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### Show Tags
18 Apr 2010, 19:03
sudai wrote:
Is the average (arithmetic mean) of a certain series of consecutive integers an integer?
(1) The number of terms in the series is even.
(2) The median of the terms in the series is not an integer.
[Reveal] Spoiler:
D
statement 1:sum of series is given by: S = n/2[2a+(n-1).d]
hence arithmetic mean A.M.= S/n => 1/2[2a+(n-1)] as d=1( consecutive integer)
=> A.M.= a+(n-1)/2
now, since n is even so n-1 is odd hence (n-1)/2 will not be integer => hence A.M. will not be an integer ....suff.
2nd statement: as gurpreetsingh has mentioned ...suff.
hence both statements are sufficient to answer.
Re: Math [#permalink] 18 Apr 2010, 19:03
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# Interactive video lesson plan for: Incredible Formula - Numberphile
#### Activity overview:
Dr James Grime discusses a couple of clever formulas which are pandigital - using all the numbers from 1-9.
More links & stuff in full description below ↓↓↓
More on pandigital numbers: https://youtu.be/gaVMrqzb91w
More on e: https://youtu.be/AuA2EAgAegE
More James Grime videos: http://bit.ly/grimevideos
Book James for a talk: http://jamesgrime.com
The contest which gave us these formulas: http://www2.stetson.edu/~efriedma/mathmagic/0804.html
Pi Playlist: http://bit.ly/PiPlaylist
Support us on Patreon: http://www.patreon.com/numberphile
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https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Slew_rate.html | 1,717,105,473,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971684053.99/warc/CC-MAIN-20240530210614-20240531000614-00817.warc.gz | 267,635,817 | 5,452 | # Slew rate
For other uses, see Slew (disambiguation).
slew rate effect on a square wave: red=desired output, green=actual output
In electronics, slew rate is defined as the change of voltage per unit of time. Expressed in SI units, the unit of measurement is volts/second, (but is usually measured in V/μs).
Electronic circuits may specify minimum or maximum limits on the slew rates for their inputs or outputs, with these limits only valid under some set of given conditions (e.g. output loading). When given for the output of a circuit, such as an amplifier, the slew rate specification guarantees that the speed of the output signal transition will be at least the given minimum, or at most the given maximum. When applied to the input of a circuit, it instead indicates that the external driving circuitry needs to meet those limits in order to guarantee the correct operation of the receiving device. If these limits are violated, some error might occur and correct operation is no longer guaranteed. For example, when the input to a digital circuit is driven too slowly, the digital input value registered by the circuit may oscillate between 0 and 1 during the signal transition.[1] In other cases, a maximum slew rate is specified[2] in order to limit the high frequency content present in the signal, thereby preventing such undesirable effects as ringing or radiated EMI.[3]
In amplifiers, limitations in slew rate capability can give rise to non-linear effects. For a sinusoidal waveform not to be subject to slew rate limitation, the slew rate capability (in volts per second) at all points in an amplifier must satisfy the following condition:
where f is the operating frequency, and is the peak amplitude of the waveform.
In mechanics the slew rate is given in dimensions 1/T and is associated with the change in position over time of an object which orbits around the observer. Slew rate can also be measured in degrees per second.
## Definition
The slew rate of an electronic circuit is defined as the rate of change of the voltage per unit time. Slew rate is usually expressed in units of V/µs.
where is the output produced by the amplifier as a function of time t.
## Measurement
The slew rate can be measured using a function generator (usually square wave) and an oscilloscope. The slew rate is the same, regardless of whether feedback is considered.
## Slew rate limiting in amplifiers
There are slight differences between different amplifier designs in how the slewing phenomenon occurs. However, the general principles are the same as in this illustration.
The input stage of modern amplifiers is usually a differential amplifier with a transconductance characteristic. This means the input stage takes a differential input voltage and produces an output current into the second stage.
The transconductance is typically very high — this is where the large open loop gain of the amplifier is generated. This also means that a fairly small input voltage can cause the input stage to saturate. In saturation, the stage produces a nearly constant output current.
The second stage of modern power amplifiers is, among other things, where frequency compensation is accomplished. The low pass characteristic of this stage approximates an integrator. A constant current input will therefore produce a linearly increasing output. If the second stage has an effective input capacitance and voltage gain , then slew rate in this example can be expressed as:
where is the output current of the first stage in saturation.
Slew rate helps us identify the maximum input frequency and amplitude applicable to the amplifier such that the output is not significantly distorted. Thus it becomes imperative to check the datasheet for the device's slew rate before using it for high-frequency applications.
## Musical applications
In electronic musical instruments, slew circuitry or software functions are used deliberately to provide a portamento (also called glide or lag) feature, where an initial digital value or analog control voltage is slowly transitioned to a new value over a period of time. | 811 | 4,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-22 | latest | en | 0.916013 |
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# Interactive video lesson plan for: Constructing, solving two-step inequality example | Linear inequalities | Algebra I | Khan Academy
#### Activity overview:
Let's tackle this word problem together. We'll interpret the information and then construct a linear inequality to solve it.
Practice this lesson yourself on KhanAcademy.org right now:
https://www.khanacademy.org/math/algebra/linear_inequalities/inequalities/e/interpretting-solving-linear-inequalities?utm_source=YT&utm_medium=Desc&utm_campaign=AlgebraI
Watch the next lesson: https://www.khanacademy.org/math/algebra/linear_inequalities/compound_absolute_value_inequali/v/compound-inequalities?utm_source=YT&utm_medium=Desc&utm_campaign=AlgebraI
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Algebra I on Khan Academy: Algebra is the language through which we describe patterns. Think of it as a shorthand, of sorts. As opposed to having to do something over and over again, algebra gives you a simple way to express that repetitive process. It's also seen as a "gatekeeper" subject. Once you achieve an understanding of algebra, the higher-level math subjects become accessible to you. Without it, it's impossible to move forward. It's used by people with lots of different jobs, like carpentry, engineering, and fashion design. In these tutorials, we'll cover a lot of ground. Some of the topics include linear equations, linear inequalities, linear functions, systems of equations, factoring expressions, quadratic expressions, exponents, functions, and ratios.
About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content.
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Maya API / 14 Jun 2016
Maya API: Shortest path between vertices
Have you ever wondered how to find the shortest path between vertices on mesh? I have. Here’s the code.
I have recently decided to update few of my custom Maya tools from MEL to API. The most useful one, Planarize Edge Tool, was frankensteined from ZenTools, which contains all sort of useful functions for working with vertices. Now that I moved to C++, I had to come up with my own solution for finding the shortest path between multiple vertices. Before I could do that, I needed to cover a simple two-point scenario.
### Shortest path between two vertices
Lets say we have even square mesh and vertices A and B. Since all edges have the same length (weight), the shortest path between vertices A and B will be the one with least iterations. In the following example it is four. In every iteration the path grafts to all connected points, increasing the search radius until the end vertex is reached.
As you probably noticed, this method itself does’t produce a single path, rather a whole tree. To narrow it down we have to apply some rule. Following illustration shows two paths, both are results of the previous example. By measuring deviation of each path from a straight line between the two end-vertices, we can figure out which one is the most “direct” path. As you can see total deviation (a+b+c) is smaller with the second path. The deciding factor could be actually anything – corner angle, edge length, vertex index or color.
### Implementation
We begin by initializing the storage variables and by extracting the base mesh and the end vertices from active selection.
Now we have to convert the vertex data in correct format. For certain operations we’ll use vertices in form of a component object, in other cases it is easier to work with vertex indices in numeric format.
### ‘Inception’
The extendPath function does all the magic thanks to MItMeshVertex::getConnectedVertices() method, which lists point on the other side of each edge connected to given vertex. In the first loop we call this function using the start point, in every next round it uses vertices collected in the previous iteration. This repeats until we reach the target vertex. It is a good idea to limit the number of iterations to number of all mesh vertices, to prevent infinite loop. Notice how we use the path deviation to choose between multiple paths pointing to the same vertex.
This function generates path in a string format containing vertex indices separated by semilocon e.g. “1;2;3;4;5;6”. For a bit more API friendly result use the following function to converted it into a numeric array.
### Conclusion
This fairly simple code is a great entry point for more complex solutions. By comparing paths generated by this method you can sort multiple vertices along an edge and find the shortest path between multiple vertices. Watch out for my future blog posts, it will definitely appear there.
1. #### Ravshan Gaziev
24 Apr 2022 - 1:17 pm
You can use just “polySelect” func. It will be much more short and faster
• #### Stepan Jirka
25 Apr 2022 - 8:17 am
Hi Ravshan,
thank you for your suggestion – you might be right. I was looking for a solution that would work directly with C++ API without the need to call MEL or Python command. | 751 | 3,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-50 | latest | en | 0.911635 |
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160b10_formulas - 1 2 Gamma: G ( k, ) , k, &amp;gt; f X...
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Pstat160b Spring 2010 Main formulas Discrete distributions name p X ( k ) = expected value variance Hypergeometric ( N 1 ,N 2 ,n ) N i of type i select n ( N 1 k )( N 2 n - k ) ( N 1 + N 2 n ) 0 k N 1 n - N 2 k n N 1 n N 1 + N 2 Binomial ( n, p ) ( n k ) p k (1 - p ) n - k 1 k n np np (1 - p ) Poisson ( λ ) e - λ λ k k ! k 0 λ λ Geometric ( p ) (1 - p ) k - 1 p k 1 1 p 1 - p p 2 Negative Binomial ( r,p ) ( k - 1 r - 1 ) p r (1 - p ) k - r , k = r,r + 1 , ··· r p r (1 - p ) p 2 Continuous distributions name f X ( x ) = expected value variance Uniform( a,b ) 1 b - a a x b a + b 2 ( b - a ) 2 12 Exponential( λ ) λe - λx x 0 1
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Unformatted text preview: 1 2 Gamma: G ( k, ) , k, > f X ( x ) = k ( k ) x k-1 e-x x k k 2 Normal ( , 2 ) 1 2 e-( x- ) 2 2 2 x R 2 Beta: B ( , ) , , > 1 B ( , ) x -1 (1-x ) -1 x 1 + ( + ) 2 ( + +1) 1...
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This note was uploaded on 01/10/2011 for the course STAT PStat 160b taught by Professor Bonnet during the Spring '10 term at UCSB.
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# 38 - Special Relativity II - Physics 212 Fall 2009...
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Physics 212 Fall 2009 12/02/2009 Special Relativity II Relativity in Practice Special Relativity - Clock on satellite is slower that stationary ground clock by 7 μ s/day General Relativity - Clock on satellite is faster that stationary ground clock by 45 μ s/day 38 μ s/day difference (if uncorrected) translates to a large positional difference!
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Special Relativity Observation of events in coordinate systems DeFne observer, event Synchronization of clocks Simultaneity Time Dilation Length Contraction Velocity Addition Paradoxes Implications Newton’s Laws, Maxwell’s Equations General Relativity
Summary 1. The laws of physics are the same for observers in all inertial frames. No one frame is preferred. 2. The speed of light in a vacuum has the same value c in all directions and in all inertial reference frames. Einstein’s Postulates of Special Relativity Light is Special Oscillations (of E & B) in time and space produce the propagation of an E&M wave (via Maxwell’s Equations) If we could transform to a system (via relative motion) where the wave was at rest - it would cease to exist -- no propagation - no oscillations -- The time interval between two events depends on how far apart they occur in both space and time; that is their spatial and temporal separations are entangled.
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## This note was uploaded on 10/03/2011 for the course PHYS 221 taught by Professor Tedeschi during the Fall '09 term at South Carolina.
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38 - Special Relativity II - Physics 212 Fall 2009...
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1 Chapter 5 Chemical Quantities and Reactions 1
2 Avogadro's Number Small particles such as atoms, molecules, and ions are counted using the mole. 1 mole = 6.02 x items Avogadro s number = 6.02 x mole of an element = 6.02 x atoms of that element 1 mole of carbon = 6.02 x atoms of carbon 1 mole of sodium = 6.02 x atoms of sodium 2
3 Number of Particles in One-Mole Samples 3
4 Avogadro's Number Avogadro s number, 6.02 x 10 23, can be written as an equality and as two conversion factors. Equality: 1 mole = 6.02 x particles Conversion Factors: 6.02 x particles and 1 mole 1 mole 6.02 x particles 4
5 Guide to Calculating Atoms or Molecules 5
6 Converting Moles to Particles Avogadro s number is used to convert moles of a substance to particles. How many CO 2 molecules are in 0.50 mole of CO 2? Step 1 State the needed and given quantities: Given: 0.50 mole of CO 2 Needed: molecules of CO 2 6
7 Converting Moles to Particles Step 2 Write a plan to convert moles to atoms or molecules: Avogadro s number moles of CO 2 molecules of CO 2 Step 3 Use Avogadro s number to write conversion factors. 1 mole of CO 2 = 6.02 x molecules of CO x CO 2 molecules and 1 mole CO 2 1 mole CO x CO 2 molecules 7
8 Converting Moles to Particles Step 4 Set up the problem to calculate the number of particles mole CO 2 x 6.02 x molecules CO 2 1 mole CO 2 = 3.01 x molecules of CO 2 8
9 Learning Check The number of atoms in 2.0 mole of Al atoms is: A. 2.0 Al atoms B. 3.0 x Al atoms C. 1.2 x Al atoms 9
10 Solution The number of atoms in 2.0 moles of Al atoms is: Step 1 State the needed and given quantities: Given: 2.0 mole Al Needed: atoms of Al Step 2 Write a plan to convert moles to atoms or molecules: Avogadro s number Moles of Al Atoms of Al Step 3 Use Avogadro s number to write conversion factors. 1 mole of Al = 6.02 x atoms of Al 6.02 x atoms Al and 1 mole A 1 mole Al 6.02 x atoms Al 10
11 Solution The number of atoms in 2.0 moles of Al atoms is: Step 4 Set up the problem to calculate the number of particles. C. 2.0 moles Al x 6.02 x Al atoms 1 mole Al = 1.2 x Al atoms 11
12 Molar Mass from Periodic Table is the mass of 1 mole of an element is the atomic mass expressed in grams Molar mass is the atomic mass expressed in grams. 1 mole of Ag 1 mole of C 1 mole of S = g of Ag = 12.0 g of C = 32.1 g of S 12
13 Guide to Calculating Molar Mass 13
14 Molar Mass of the Compound C 2 H 6 O To calculate the molar mass of C 2 H 6 O: Step 1 Obtain the molar mass of each element. 1 mole of C = 12.0 g of C 1 mole of H = 1.01 g of H 1 mole of O = 16.0 g of O Step 2 Multiply each molar mass by the number of moles (subscripts) in the formula. 2 moles C x 12.0 g C = 24.0 g of C 1 mole C 6 moles H x 1.01 g H = 6.06 g of H 1 mole H 1 mole O x 16.0 g O = 16.0 g of O 1 mole O Step 3 Calculate the molar mass by adding the masses of the elements. 2 moles of C = 24.0 g of C 6 moles of H = 6.06 g of H 1 mole of O = 16.0 g of O Molar mass of C 2 H 6 O = 46.1 g of C 2 H 6 O 14
15 One-Mole Quantities 15
16 Learning Check What is the molar mass of each compound? A. K 2 O B. Al(OH) 3 16
17 Solution What is the molar mass of each compound? Step 1 Obtain the molar mass of each element. A. K 2 O K = 39.1 g/mole O = 16.0 g/mole B. Al(OH) 3 Al = 27.0 g/mole O = 16.0 g/mole H = 1.01 g/mole 17
18 Solution What is the molar mass of each compound? Step 2 Multiply each molar mass by the number of moles (subscripts) in the formula. A. K 2 O 2 moles K x 39.1 g K = 78.2 g of K 1 mole K 1 mole O x 16.0 g O = 16.0 g of O 1 mole O 18
19 Solution What is the molar mass of each compound? Step 2 Multiply each molar mass by the number of moles (subscripts) in the formula. B. Al(OH) 3 1 mole Al x 27.0 g Al = 27.0 g of Al 1 mole Al 3 moles O x 16.0 g O = 48.0 g of O 1 mole O 3 moles H x 1.01 g H = 3.03 g of H 1 mole H 19
20 Solution What is the molar mass of each compound? Step 3 Calculate the molar mass by adding the masses of the elements. A. K 2 O 2 moles of K = 78.2 g of C 1 mole of O = 16.0 g of O Molar mass of K 2 O = 94.2 g of K 2 O B. Al(OH) 3 1 mole of Al = 27.0 g of Al 3 moles of O = 48.0 g of O 3 moles of H = 3.03 g of H Molar mass of Al(OH) 3 = 78.0 g of Al(OH) 3 20
21 Calculations Using Molar Mass Molar mass conversion factors are fractions (ratios) written from the molar mass. relate grams and moles of an element or compound. for methane, CH 4, used in gas stoves and gas heaters, is 1 mole of CH 4 = 16.0 g of CH 4 (molar mass equality) Conversion factors: 16.0 g CH 4 and 1 mole CH 4 1 mole CH g CH 4 21
22 Guide to Calculating Moles from Mass 22
23 Converting Mass to Moles of Compound NaCl A box of table salt, NaCl, contains 737 g of NaCl. How many moles of NaCl are in the box? 23
24 Converting Mass to Moles of Compound NaCl A box of table salt, NaCl, contains 737 g of NaCl. How many moles of NaCl are in the box? Step 1 State the given and needed quantities. Given: 737 g of NaCl Need: moles of NaCl Step 2 Write a plan to convert grams to moles. molar mass Grams of NaCl Moles of NaCl 24
25 Converting Mass to Moles of Compound NaCl Step 3 Determine the molar mass and write conversion factors. Molar Mass 1 mole Na x 23.0 g Na = 23.0 g of Na 1 mole Na 1 mole Cl x 35.5 g Cl = 35.5 g of Cl 1 mole Cl 58.5 g of NaCl Conversion Factors 1 mole of NaCl = 58.5 g of NaCl 58.5 g NaCl and 1 mole NaCl 1 mole NaCl 58.5 g NaCl Step 4 Set up the problem to convert grams to moles. 737 g NaCl x 1 mole NaCl = 12.6 moles of NaCl 58.5 g NaCl 25
26 Map: Mass Moles Particles 26
27 Visible Evidence of a Chemical Reaction 27
28 Identifying a Balanced Equation In a balanced chemical equation, no atoms are lost or gained the number of reacting atoms is equal to the number of product atoms 28
29 Balancing a Chemical Equation 29
30 Balancing a Chemical Equation: Formation of Al 2 S 3 Step 1 Write an equation using the correct formulas of the reactants and products. Al(s) + S(s) Al 2 S 3 (s) Step 2 Count the atoms of each element in the reactants and products. Reactants Products 1 atom Al 2 atoms Al Not balanced 1 atoms S 3 atoms S Not balanced 30
31 Balancing a Chemical Equation: Formation of Al 2 S 3 Step 3 Use coefficients to balance each element. Starting with the most complex formula, change coefficients to balance equation. 2Al(s) + 3S(s) Al 2 S 3 (s) Step 4 Check the final equation to confirm it is balanced. Make sure coefficients are the lowest ratio. Reactants Products 2 atoms Al 2 atoms Al Balanced 3 atoms S 3 atoms S Balanced 31
32 Learning Check State the number of atoms of each element on the reactant side and the product side for each of the following balanced equations. A. P 4 (s) + 6Br 2 (l) 4PBr 3 (g) B. 2Al(s) + Fe 2 O 3 (s) 2Fe(s) + Al 2 O 3 (s) 32
33 Solution State the number of atoms of each element on the reactant side and the product side for each of the following balanced equations. A. P 4 (s) + 6Br 2 (l) 4PBr 3 (g) Reactants Products P atoms 4 4 Br atoms B. 2Al(s) + Fe 2 O 3 (s) 2Fe(s) + Al 2 O 3 (s) Reactants Products Al atoms 2 2 Fe atoms 2 2 O atoms
34 Balancing Equations with Polyatomic Ions When balancing equations with polyatomic ions, balance each polyatomic ion as a unit. 2Na 3 PO 4 (aq) + 3MgCl 2 (aq) Mg 3 (PO 4 ) 2 (s) + 6NaCl(aq) Reactants Products PO 3 4 ions 2 2 Na + ions 6 6 Mg 2+ ions 3 3 Cl ions
35 Learning Check Balance and list the coefficients from reactants to products. A. Fe 2 O 3 (s) + C(s) Fe(s) + CO 2 (g) 1) 2, 3, 2, 3 2) 2, 3, 4, 3 3) 1, 1, 2, 3 B. Al(s) + FeO(s) Fe(s) + Al 2 O 3 (s) 1) 2, 3, 3, 1 2) 2, 1, 1, 1 3) 3, 3, 3, 1 C. Al(s) + H 2 SO 4 (aq) Al 2 (SO 4 ) 3 (aq) + H 2 (g) 1) 3, 2, 1, 2 2) 2, 3, 1, 3 3) 2, 3, 2, 3 35
36 Solution A. 2) 2, 3, 4, 3 2Fe 2 O 3 (s) + 3C(s) B. 1) 2, 3, 3, 1 2Al(s) + 3FeO(s) 4Fe(s) + 3CO 2 (g) 3Fe(s) + 1Al 2 O 3 (s) C. 2) 2, 3, 1, 3 2Al(s) + 3H 2 SO 4 (aq) 1Al 2 (SO 4 ) 3 (aq) + 3H 2 (g) 36
37 Types of Reactions Chemical reactions can be classified as combination reactions decomposition reactions single replacement reactions double replacement reactions combustion reactions 37
38 Combination Reactions In a combination reaction, two or more elements form one product or simple compounds combine to form one product 2Mg(s) + O 2 (g) 2Na(s) + Cl 2 (g) SO 3 (g) + H 2 O(l) 2MgO(s) 2NaCl(s) H 2 SO 4 (aq) 38
39 Decomposition Reaction In a decomposition reaction, one substance splits into two or more simpler substances. 2HgO(s) 2KClO 3 (s) 2Hg(l) + O 2 (g) 2KCl(s) + 3O 2 (g) 39
40 Single Replacement Reaction In a single replacement reaction, one element takes the place of a different element in another reacting a compound. Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) Fe(s) + CuSO 4 (aq) FeSO 4 (aq) + Cu(s) 40
41 Double Replacement Reaction In a double replacement, two elements in the reactants exchange places. AgNO 3 (aq) + NaCl(aq) AgCl(s) + NaNO 3 (aq) ZnS(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 S(g) Boy 1/Girl 1 + Boy 2/Girl 2 Boy 1/Girl 2 + Boy 2/Girl 1 41
42 Combustion Reaction In a combustion reaction, a carbon-containing compound burns in oxygen gas to form carbon dioxide (CO 2 ) and water (H 2 O) energy is released as a product in the form of heat CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) + energy 42
43 Summary of Reaction Types 43
44 Learning Check Identify each reaction as combination, decomposition, combustion, single replacement, or double replacement. A. 3Ba(s) + N 2 (g) Ba 3 N 2 (s) B. 2Ag(s) + H 2 S(aq) Ag 2 S(s) + H 2 (g) C. 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(g) D. PbCl 2 (aq) + K 2 SO 4 (aq) 2KCl(aq) + PbSO 4 (s) E. K 2 CO 3 (s) K 2 O(aq) + CO 2 (g) 44
45 Solution Identify each reaction as combination, decomposition, combustion, single replacement, or double replacement. A. 3Ba(s) + N 2 (g) Ba 3 N 2 (s) Combination B. 2Ag(s) + H 2 S(aq) Ag 2 S(s) + H 2 (g) C. C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(g) Single Replacement Combustion 45
46 Solution Identify each reaction as combination, decomposition, combustion, single replacement, or double replacement. D. PbCl 2 (aq) + K 2 SO 4 (aq) 2KCl(aq) + PbSO 4 (s) Double Replacement E. K 2 CO 3 (s) K 2 O(aq) + CO 2 (g) Decomposition 46
47 Oxidation Reduction Reactions An oxidation reduction reaction provides us with energy from food provides electrical energy in batteries occurs when iron rusts: 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) 47
48 Oxidation Reduction In an oxidation reduction reaction, electrons are transferred from one substance to another. 48
49 Oxidation and Reduction 2Cu(s) + O 2 (g) 2CuO LEO says GER Loss of Electrons is Oxidation. 2Cu(s) 2Cu 2+ (s) + 4e Gain of Electrons is Reduction. O 2 (g) + 4e 2O 2 (s) The green patina on copper is due to oxidation 49
50 Oxidation Reduction in Biological Systems The oxidation of a typical biochemical molecule can involve the transfer of two hydrogen atoms (or 2H + and 2e ) to a proton acceptor such as the coenzyme FAD (flavin adenine dinucleotide). 50
51 Characteristics of Oxidation and Reduction 51
52 Characteristics of Oxidation and Reduction Methyl alcohol hydrogen gas) formaldehyde Oxidation (loss of H as CH 3 OH H 2 CO + 2H 2 Formaldehyde formic acid 2H 2 CO + O 2 2H 2 CO 2 Oxidation (addition of O) 52
53 Law of Conservation of Mass The law of conservation of mass indicates that in an ordinary chemical reaction, matter cannot be created or destroyed no change in total mass occurs the mass of products is equal to mass of reactants 53
54 Conservation of Mass 54
55 Information from a Balanced Equation 55
56 Reading Equations in Moles Consider the following equation: 2Fe(s) + 3S(s) Fe 2 S 3 (s) This equation can be read in moles by placing the word moles of between each coefficient and formula. 2 moles of Fe + 3 moles of S 1 mole of Fe 2 S 3 56
57 Learning Check Consider the following equation: 3H 2 (g) + N 2 (g) 2NH 3 (g) A. A mole mole factor for H 2 and N 2 is: 1) 3 moles N 2 2) 1 mole N 2 3) 1 mole N 2 1 mole H 2 3 moles H 2 2 moles H 2 B. A mole mole factor for NH 3 and H 2 is: 1) 1 mole H 2 2) 2 moles NH 3 3) 3 moles N 2 2 moles NH 3 3 moles H 2 2 moles NH 3 57
58 Solution 3H 2 (g) + N 2 (g) 2NH 3 (g) A. A mole mole factor for H 2 and N 2 is: 2) 1 mole N 2 3 moles H 2 B. A mole mole factor for NH 3 and H 2 is: 2) 2 moles NH 3 3 moles H 2 58
59 Calculations with Mole Factors How many moles of Fe 2 O 3 can form from 6.0 moles of O 2? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Relationship: 3 moles of O 2 = 2 moles of Fe 2 O 3 Use a mole mole factor to determine the moles of Fe 2 O moles O 2 x 2 moles Fe 2 O 3 = 4.0 moles of Fe 2 O 3 3 moles O 2 59
60 Guide to Using Mole Mole Factors 60
61 Learning Check How many moles of Fe are needed for the reaction of 12.0 moles of O 2? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) A moles of Fe B moles of Fe C moles of Fe 61
62 Solution In a problem, identify the compounds given and needed. How many moles of Fe are needed for the reaction of 12.0 moles of O 2? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Step 1 State the given and needed quantities. Given: 12.0 moles of O 2 The possible mole factors for the solution are: 4 moles Fe and 3 moles O 2 3 moles O 2 4 moles Fe Needed:? moles of Fe 62
63 Solution Step 2 Write a plan to convert the given to the needed moles. Mole mole factor Moles of O 2 Moles of Fe Step 3 Use coefficients to write relationships and mole mole factors. 4 moles of Fe = 3 moles of O 2 4 moles Fe and 3 moles O 2 3 moles O 2 4 moles Fe 63
64 Solution Step 4 Set up the problem using the mole mole factor that cancels given moles moles O 2 x 4 moles Fe = 16.0 moles of Fe 3 moles O 2 The answer is C, 16.0 moles of Fe. 64
65 Mass Calculations in Equations 65
66 Moles to Grams Suppose we want to determine the mass (g) of NH 3 that can be produced from 32 grams of N 2. N 2 (g) + 3H 2 (g) 2NH 3 (g) Step 1 Use molar mass to convert grams of given to moles. 1 mole of N 2 = 28.0 g of N 2 1 mole N 2 and 28.0 g N g N 2 1 mole N 2 32 g N 2 x 1 mole N 2 = 1.1 mole of N g N 2 66
67 Moles to Grams Step 2 Write a mole mole factor from the coefficients in the equation. 1 mole of N 2 = 2 moles of NH 3 1 mole N 2 and 2 moles NH 3 2 moles NH 3 1 mole N 2 Step 3 Convert moles of given to moles of needed using the mole mole factor. 1.1 mole N 2 x 2 moles NH 3 = 2.2 moles of NH 3 1 mole N 2 67
68 Moles to Grams Step 3 Convert moles of given to moles of needed using the mole mole factor. 1.1 mole N 2 x 2 moles NH 3 = 2.2 moles of NH 3 1 mole N 2 68
69 Moles to Grams Step 4 Convert moles of needed substance to grams using molar mass. 1 mole of NH 3 = 17.0 g of NH 3 1 mole NH 3 and 17.0 g NH g NH 3 1 mole NH moles NH 3 x 17.0 g NH 3 = 37 g of NH 3 1 mole NH 3 69
70 Learning Check How many grams of O 2 are needed to produce 45.8 grams of Fe 2 O 3 in the following reaction? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) A g of O 2 B g of O 2 C g of O 2 70
71 Solution How many grams of O 2 are needed to produce 45.8 grams of Fe 2 O 3 in the following reaction? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Step 1 Use molar mass to convert grams of given to moles. 1 mole of Fe 2 O 3 = g of Fe 2 O 3 1 mole Fe 2 O 3 and g Fe 2 O g Fe 2 O 3 1 mole Fe 2 O g Fe 2 O 3 x 1 mole Fe 2 O 3 = mole of Fe 2 O g Fe 2 O 3 71
72 Solution Step 2 Write a mole mole factor from the coefficients in the equation. 3 moles of O 2 = 2 moles of Fe 2 O 3 3 moles O 2 and 2 moles Fe 2 O 3 2 moles Fe 2 O 3 3 moles O 2 Step 3 Convert moles of given to moles of needed using the mole mole factor mole Fe 2 O 3 x 3 moles O 2 = mole of O 2 2 moles Fe 2 O 3 72
73 Solution Step 4 Convert moles of needed substance to grams using molar mass. 1 mole of O 2 = 32.0 g of O 2 1 mole O 2 and 32.0 g O go 2 1 mole O mole O 2 x 32.0 g O 2 = 13.8 g of O 2 1 mole O 2 The answer is B, 13.8 g of O 2. 73
74 Molecules Must Collide for Reaction Three conditions for a reaction to occur are: 1. collision: The reactants must collide. 2. orientation: The reactants must align properly to break and form bonds. 3. energy: The collision must provide the energy of activation. 74
75 Exothermic Reaction In an exothermic reaction, heat is released the energy of the products is less than the energy of the reactants heat is a product C(s) + 2H 2 (g) CH 4 (g) + 18 kcal 75
76 Endothermic Reactions In an endothermic reaction, heat is absorbed the energy of the products is greater than the energy of the reactants heat is a reactant (added) N 2 (g) + O 2 (g) kcal 2NO(g) 76
77 Reaction Rate The reaction rate is the speed at which reactant is used up is the speed at which product forms increases when temperature rises because reacting molecules move faster, providing more colliding molecules with energy of activation increases with increase in concentration of reactants 77
78 Catalyst A catalyst increases the rate of a reaction lowers the energy of activation is not used up during the reaction
79 Factors That Increase Reaction Rate 79
80 Learning Check State the effect of each on the rate of reaction as increases, decreases, or has no effect: A. increasing the temperature B. removing some of the reactants C. adding a catalyst D. placing the reaction flask in ice E. increasing the concentration of one of the reactants 80
81 Solution State the effect of each on the rate of reaction as increases, decreases, or has no effect: A. increasing the temperature increases B. removing some of the reactants decreases C. adding a catalyst increases D. placing the reaction flask in ice decreases E. increasing the concentration of increases one of the reactant 81
82 Learning Check Indicate the effect of each factor listed on the rate of the following reaction as increases, decreases, or no change. 2CO(g) + O 2 (g) 2CO 2 (g) A. raising the temperature B. adding O 2 C. adding a catalyst D. lowering the temperature 82
83 Solution Indicate the effect of each factor listed on the rate of the following reaction as increases, decreases, or no change. 2CO(g) + O 2 (g) 2CO 2 (g) A. raising the temperature increases B. adding O 2 increases C. adding a catalyst increases D. lowering the temperature decreases 83
84 Chemical Quantities and Reactions 84
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https://www.bartleby.com/questions-and-answers/feii-can-be-precipitated-from-a-slightly-basic-aqueous-solution-by-bubbling-oxygen-through-the-solut/89d7d83e-d07b-4717-a386-d02e84f3771b | 1,580,141,919,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251700988.64/warc/CC-MAIN-20200127143516-20200127173516-00503.warc.gz | 762,752,591 | 22,860 | # Fe(II) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts Fe(II) to insoluble Fe(III):\$\$4Fe(OH)+(aq)+4OH−(aq)+O2(g)+2H2O(l)4Fe(OH)3(s) How many grams of O2 are consumed to precipitate all of the iron in 75.0 mL of 0.0550 M Fe(II)?
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Fe(II) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts Fe(II) to insoluble Fe(III):
\$\$4Fe(OH)+(aq)+4OH−(aq)+O2(g)+2H2O(l)4Fe(OH)3(s)
How many grams of O2 are consumed to precipitate all of the iron in 75.0 mL of 0.0550 M Fe(II)?
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Tagged in | 299 | 956 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-05 | latest | en | 0.839753 |
https://www.wired.com/2008/02/how-much-does-a | 1,503,046,742,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104631.25/warc/CC-MAIN-20170818082911-20170818102911-00664.warc.gz | 1,000,087,948 | 29,921 | How Much Does a Kilogram Weigh?
# How Much Does a Kilogram Weigh?
Locked away in an underground Parisian vault, there is a very important and valuable piece of platinum-iridium alloy known as "Le Grand K." It's not a ring, or a necklace, and didn't belong to anyone particularly famous. Instead, Le Grand K is the piece of metal, created in the 1880s, that is the current definition of how much a kilogram weighs. Many copies of the object have been made around the world, but Le Grand K is Le Real McCoy.
The problem, Sandia Lab scientists are the latest to say, is that pegging the entire system of measurement to a physical object is inherently risky. What if the bar is stolen or destroyed? Or, more likely, what if the metal itself loses some mass? Scientists have said the official has lost up to 50 micrograms of mass. As one measurer told the BBC late last year:
Relative to the average of all the sister copies made over the last 100 years you could say it is losing [mass], but by definition it can't," explained Dr Richard Steiner of the National
Institute of Standards and technology (NIST) in the US. "So the others are really gaining mass."
The conundrum means that Le Grand K's days are numbered, but without an object, how are scientists going to measure the amount of mass known as a kilogram? Find out about the competing proposals after the jump.
The ideal would be to match the kilogram to known physical constants. For example, the meter is officially defined as the length of the path that light travels in a vacuum in during 1/299,792,458 of a second. That's not exactly as neat as a hunk of metal in a Parisian vault, but it is elegant in its immutability.
Right now, there are several proposals under consideration by weights and measures specialists:
• The longest running proposal is to use a watt-balance machine, which measures the exact force required to balance a 1 kilogram mass against the pull of Earth's gravity. The hitch here is getting a suitably accurate measurement of Planck's constant, and that's proving tougher than anticipated.
• As profiled in WIRED 15.09, Australian scientists are building an ultrapure silicon ball that should contain 215 x 1023 atoms, and would henceforth be the new definition of one kg.
• A Georgia Tech professor proposed that a kilogram be defined as 18 x 14074481^3 carbon-12 atoms, but carbon is apparently a bit out of fashion in the physics community.
The kilogram controversy, and Le Grand K's long career, could both end in 2011, when the body governing the metric system will look at redefining the kilo. | 587 | 2,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-34 | latest | en | 0.968606 |
https://www.indiabix.com/digital-electronics/logic-gates/discussion-133 | 1,591,255,778,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347439213.69/warc/CC-MAIN-20200604063532-20200604093532-00554.warc.gz | 726,851,175 | 5,614 | # Digital Electronics - Logic Gates - Discussion
### Discussion :: Logic Gates - Filling the Blanks (Q.No.2)
2.
The gates in this figure are implemented using TTL logic. If the output of the inverter is open, and you apply logic pulses to point B, the output of the AND gate will be ________.
[A]. a steady LOW [B]. a steady HIGH [C]. an undefined level [D]. pulses
Explanation:
No answer description available for this question.
Shobha Pathare said: (Jan 9, 2014) If the o/p of inverter is open that means 0 and for B I/p is pulse means 1 so the o/p become 0.
Shan said: (Nov 7, 2014) How to get pulses because one of the input is open it means "0" so output should be 0?
Anand said: (Nov 29, 2015) Doesn't 'open' mean no current in that wire i.e. HighZ, makes first input undriven. Answer should be "an Undefined level".
Anuj said: (Nov 28, 2019) In TTL an open terminal is always "high" 1. So B will be at the output. B= pulses. | 268 | 946 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-24 | latest | en | 0.856428 |
https://www.wonkeedonkeetools.co.uk/sockets-and-socket-sets/what-is-a-hex-socket | 1,723,449,675,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036271.72/warc/CC-MAIN-20240812061749-20240812091749-00764.warc.gz | 831,954,427 | 28,112 | # What is a Hex Socket?
The term ‘hex socket’ refers to a socket that fits around the outside of a hexagonal fastener head. However, its use has broadened and it is now sometimes used to refer to any shape of socket that fits around the outside of a fastener, instead of locating into the fastener head.
## 6 Point Hex Sockets
A ‘hex socket’ is a socket designed to fit the head of a hexagonal fastener such as a nut or bolt.
Hex sockets are also referred to as ‘6 point’ sockets as six points are formed where the internal walls from the hexagonal recess meet.
Better quality hex sockets have internal walls that are slightly curved. This helps apply the torque from the socket to the flats of the fastener instead of the corners, which reduces the chance of rounding the head of the fastener.
This also means that more torque can be safely applied to the fastener by the socket, allowing them to be tightened more or screwed into harder materials such as metal or hard woods.
### What is meant by regular 6 point?
When a socket is referred to as being a “regular 6 point” it means it is a 6 point hex socket. This term is sometimes used to avoid confusion with E Torx sockets that also have 6 points.
## 12 point bi-hex sockets
A ‘bi-hex’ or ‘bi-hexagon’ socket is also called a 12 point socket. This is because twelve points are formed where the internal walls of the recesses intersect.
The name ‘bi-hex’ comes from the internal shape of the socket recess being that of two hexagons laid over one another.
Just like ‘hex sockets’, ‘bi-hex’ sockets can also be used to tighten or loosen hexagonal shaped fasteners, but they have a couple of advantages:
They are able to fit 6 point fastener heads in twice as many positions as a regular ‘hex socket’. This means it is easier to locate the socket onto the fastener in tight spaces.
They can also be used to turn square (4 point) and bi-hex (12 point) fastener heads as well, giving them greater versatility.
The bi-hex shape allows the socket to fit over the head of squares fasteners, unlike a regular hex socket.
However, a bi-hex socket will apply the torque closer to the corners of a 6 point hex head fastener, which means the chances of the fastener head becoming rounded are greater.
## Hex Socket Sizes
Socket heads can be measured in both metric and imperial units. Metric socket sizes can range from 3mm to over 80mm, while imperial socket sizes range from 5/32″ to over 1-½”.
### Drive Socket
At the other end of the socket is the drive socket, which is used to attach the socket to a ratchet, torque wrench or other type of turning tool.
Unlike the head of hex and bi-hex sockets, which is available in both metric and imperial sizes, the drive socket is always sized in imperial measurements. The most common sizes are ¼”, ⅜”, ½”, ¾” and 1″.
## Other Types of Hex Socket
With the increasing variety of sockets, the term “hex socket” is often used to describe any socket that fits over the head of a fastener instead of fitting into the fastener head, as with a socket bit. This is despite many of these new forms not having a hexagon shape to them.
Other sockets exist too, see What are the different types of socket? to find out more!
### 8 point and square sockets
Whilst it still may be possible to get an 8 point or square socket, they are now considered a specialist item that is often only made to order in the size you specify, as 12 point sockets will usually be able to do the same job.
### E Torx Sockets
E Torx, or ‘external Torx sockets’ to give their full name, are sockets designed to fit and turn fasteners with an external Torx head.
E Torx sockets are sized from E1 (smallest) to E100 (largest), although most E Torx socket sets will range in size from E4 – E24. Like standard sockets, they have a square drive for attaching a turning tool that can range from ¼” – 1″ in size.
E Torx sockets and fasteners can apply more torque than regular hex or bi-hex sockets without damaging the fastener head.
This is because of the greater contact area between the socket and the fastener head, along with the fact that more of the torque is applied away from the corners of the fastener head, so there is less chance of it rounding off.
E Torx are often used in areas where high torque and clamping force is required.
One of the most common places to find E Torx fasteners is on the cylinder head bolts of car engines. | 1,022 | 4,424 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-33 | latest | en | 0.965165 |
https://math.stackexchange.com/questions/3029398/orthogonal-line-segment-between-line-segments-with-length | 1,585,887,062,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370510287.30/warc/CC-MAIN-20200403030659-20200403060659-00123.warc.gz | 568,342,308 | 34,766 | # Orthogonal line segment between line segments with length
I am given two line segments $$\overline{AB}$$ and $$\overline{CD}$$. I am looking for the line segment $$\overline{EF}$$ between $$\overline{AB}$$ and $$\overline{CD}$$ which has a desired length $$l$$ while also being orthogonal to $$\overline{AB}$$, whereas $$E$$ should lie on $$\overline{AB}$$ and $$F$$ on $$\overline{CD}$$. All of this is happening in $$\mathbb{R^3}$$. When multiple solutions exist, we want to know all of them. We also want to know when none exist.
To sum up: $$(\overrightarrow{b}-\overrightarrow{a})\bullet(\overrightarrow{f}-\overrightarrow{e})=0$$ $$|\overrightarrow{f}-\overrightarrow{e}| = l \quad,l\ge0$$ $$\overrightarrow{e} = \overrightarrow{a}+m(\overrightarrow{b}-\overrightarrow{a}) \quad,0\le m\le1$$ $$\overrightarrow{f} = \overrightarrow{c}+n(\overrightarrow{d}-\overrightarrow{c}) \quad,0\le n\le1$$
We can insert the third and fourth requirement into the first to get:
$$(\overrightarrow{b}-\overrightarrow{a})\bullet(\overrightarrow{c}+n(\overrightarrow{d}-\overrightarrow{c})-\overrightarrow{a}-m(\overrightarrow{b}-\overrightarrow{a}))=0$$
From here on, however, I struggle to incorporate the second requirement. How do I get the length constraint in there as well? Here is a different approach which also leaves me stranded:
The wanted line segment $$\overline{EF}$$ lies somewhere on the plane with normal vector $$\frac{\overrightarrow{b}-\overrightarrow{a}}{|\overrightarrow{b}-\overrightarrow{a}|}$$ and origin somewhere on $$\overline{AB}$$. This yields the plane equation $$\frac{\overrightarrow{b}-\overrightarrow{a}}{|\overrightarrow{b}-\overrightarrow{a}|}\bullet\overrightarrow{p}+d=0$$ where $$\overrightarrow{p}$$ is a point on the plane and $$d$$ is the (signed) distance of the plane from the origin. For $$\overrightarrow{e}$$, this would be: $$\frac{\overrightarrow{b}-\overrightarrow{a}}{|\overrightarrow{b}-\overrightarrow{a}|}\bullet(\overrightarrow{a}+m(\overrightarrow{b}-\overrightarrow{a}))+d=0$$ Since $$\overrightarrow{e}$$ lies on $$\overline{AB}$$, which is parallel to our plane normal, we can express it using $$d$$ instead of $$m$$, effectively getting rid of one unknown variable (mind the sign): $$\frac{\overrightarrow{b}-\overrightarrow{a}}{|\overrightarrow{b}-\overrightarrow{a}|}\bullet(-d\frac{\overrightarrow{b}-\overrightarrow{a}}{|\overrightarrow{b}-\overrightarrow{a}|})+d=0$$ For $$\overrightarrow{f}$$ the plane equation is: $$\frac{\overrightarrow{b}-\overrightarrow{a}}{|\overrightarrow{b}-\overrightarrow{a}|}\bullet(\overrightarrow{c}+n(\overrightarrow{d}-\overrightarrow{c}))+d=0$$ Here, we want to solve for $$n$$ to get rid of that unknown as well. This will allow us to define $$\overrightarrow{e}$$ and $$\overrightarrow{f}$$ using a single variable in such a way that $$\overrightarrow{f}-\overrightarrow{e}$$ will always express a vector which is orthogonal to the plane normal. To do this, we need to explode the equation: $$\frac{(b_x-a_x)(c_x+n(d_x-c_x))+(b_y-a_y)(c_y+n(d_y-c_y))+(b_z-a_z)(c_z+n(d_z-c_z))}{\sqrt{(b_x-a_x)^2+(b_y-a_y)^2+(b_z-a_z)^2}}+d=0$$ Here, I struggle to solve for $$n$$.
If I didn't, though, I could then solve the following for $$d$$ $$l=|\overrightarrow{f}-(-d\frac{\overrightarrow{b}-\overrightarrow{a}}{|\overrightarrow{b}-\overrightarrow{a}|})|$$ where $$\overrightarrow{f}$$ would be replaced by an equation which expresses $$\overrightarrow{f}$$ through $$d$$ instead of $$n$$. I could then solve $$\overrightarrow{e}$$: $$\overrightarrow{e}=-d\frac{\overrightarrow{b}-\overrightarrow{a}}{|\overrightarrow{b}-\overrightarrow{a}|}$$ and – if I had figured it out – $$\overrightarrow{f}$$.
• The orthogonality constraint gives you a one-parameter family of parallel planes on which $\overline{EF}$ must lie. Intersect these planes with $\overline{CD}$. – amd Dec 7 '18 at 3:20
• I tried a different approach upon your suggestion, but ended with an obscure equation which I can't seem to solve for $n$. I edited the question. – Zyl Dec 7 '18 at 16:54
What you’re essentially trying to find are the intersections of $$\overline{CD}$$ with a piece of a cylinder of radius $$l$$ and axis $$AB$$. There can be zero, one, two or an infinite number of intersection points. I would suggest breaking this down into two stages. First, find all of the points on the lines $$AB$$ and $$CD$$ that meet the criteria—that is, the intersection of the line $$CD$$ with the entire cylinder—and then cull those solutions that don’t lie on the corresponding line segments.
There are several ways to approach the unconstrained problem. A straightforward way is to parameterize $$CD$$ and then use a point-line distance formula. So, let $$P(\lambda)=(1-\lambda)C+\lambda D$$. From the distance formula we then get the equation $$\|(B-A)\times(P(\lambda)-A)\|^2 = l^2\|B-A\|^2. \tag 1$$ This will either be a quadratic equation in $$\lambda$$, or, if $$CD$$ is parallel to $$AB$$, an equality of two constants. Solutions in the interval $$[0,1]$$ lie on the segment $$\overline{CD}$$. The corresponding point on $$AB$$ is the intersection of this line with its perpendicular plane through $$P(\lambda)$$, which has equation $$(B-A)\cdot (X - P(\lambda))=0.\tag 2$$ This intersection point can be computed directly in various ways, such as the Plücker matrix of $$AB$$, but since you need to do a range check, anyway, parameterize $$\overline{AB}$$ as $$Q(\mu)=(1-\mu)A+\mu B$$ and substitute into (2) to get $$(B-A)\cdot Q(\mu)=(B-A)\cdot P(\lambda). \tag 3$$ This is a linear equation in $$\mu$$ that you can solve and again check that the solution lies in $$[0,1]$$. Alternatively, compare the signs of the values that you get by plugging $$A$$ and $$B$$ into the left-hand side of (2): if either is zero or they have opposite signs, then the intersection lies on the segment $$\overline{AB}$$ and you can go ahead and compute it using any convenient method.
You could instead have started with equation (2), which expresses the constraint that the two points both lie on a plane perpendicular to $$AB$$. This is a linear equation in $$\lambda$$ and $$\mu$$. The distance constraint can be expressed as the equation $$\|P(\lambda)-Q(\mu)\|^2=l^2, \tag 4$$ so the problem is reduced to the two-dimensional problem of finding the intersections of a conic and line. There are algorithms for doing this that are suitable for automation if you don’t have access to an equation solver. Once again, after you’ve found the solutions to this system of equations, the ones you want are those for which $$\lambda,\mu\in[0,1]$$. As a third approach, you could derive the equation of the cylinder and set up equations for its intersection with $$CD$$, but that’s eventually going to lead to equations similar to those above with a lot more work.
As amd suggested, a good starting point is the equation of an axis-aligned cylinder: $$r^2=x^2+y^2$$ I ended up solving the problem by transforming $$\overline{AB}$$ and $$\overline{CD}$$ so that $$A$$ is at $$(0,0,0)$$ and $$\overrightarrow{b}-\overrightarrow{a}$$ points into positive z. I could then insert the $$x$$ and $$y$$ components from the definitions of $$\overrightarrow{f}$$ from my question into above formula and solve for $$n$$, determine transformed $$\overrightarrow{f}$$ and then transform the result back. I also defined: $$\overrightarrow{v}=\overrightarrow{d}-\overrightarrow{c}$$ We then start with this: $$l^2=(c_x+nv_x)^2+(c_y+nv_y)^2$$ Here, $$r^2$$ became $$l^2$$ from my question. Solving for $$n$$ using solver of choice yields: $$n=\frac{\pm\sqrt{-c_x^2v_y^2+2c_xc_yv_xv_y-c_y^2v_x^2+l^2(v_x^2+v_y^2)}-c_xv_x-c_yv_y}{v_x^2+v_y^2}$$ A computational difficulty occurs when $$\overline{AB}$$ and $$\overline{CD}$$ are parallel, as it causes the divisor to approach zero. In or near these cases a distance test with any point on $$\overline{CD}$$ can determine whether it is at the wanted distance. Then we can determine the transformed $$\overrightarrow{f}$$, and transform it back.
Determining $$\overrightarrow{e}$$ is then as simple as projecting $$\overrightarrow{f}$$ onto $$\overline{AB}$$.
Finally, one needs to test that $$\overrightarrow{e}$$ lies on $$\overline{AB}$$ to validate the result. | 2,368 | 8,242 | {"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": 105, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2020-16 | latest | en | 0.624775 |
https://girls-beauty-tips.com/qa/question-will-110v-kill-you.html | 1,601,487,372,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402127397.84/warc/CC-MAIN-20200930172714-20200930202714-00508.warc.gz | 399,077,527 | 10,235 | # Question: Will 110v Kill You?
## What voltage kills the most?
120 volts120 volts is indeed the voltage that “kills” the most people, in that it’s the one that most people come into contact with and get killed.
However, as 480sparky correctly pointed out, it’s the current that kills..
## How many amps is a 120 volt?
20 AmpsVolts = Watts / Amps 2400 Watts / 20 Amps = 120 Volts.
## Can Home Electricity kill you?
Any electrical device used on a house wiring circuit can, under certain conditions, transmit a fatal current. While any amount of current over 10 milliamps (0.01 amp) is capable of producing painful to severe shock, currents between 100 and 200 mA (0.1 to 0.2 amp) are lethal.
## Why are electrical outlets installed upside down?
Electricians may position the outlet in an upside-down position so that you can quickly identify the switch-controlled receptacle. Since it stands out visually to most people right away – it provides convenience to the occupants to easily remember which outlet is switch controlled.
## How long does electricity stay in the body after a shock?
Your Care Instructions The shock can cause a burn where the current enters and leaves your body. The electricity may have injured blood vessels, nerves, and muscles. The electricity also could have affected your heart and lungs. You might not see all the damage the shock caused for up to 10 days after the shock.
## What does it feel like to get electrocuted?
The heart, in particular, will probably go into ventricular fibrillation, and it takes little actual current through it to cause this. You could die. ! AC shock is rather like a buzzing sensation, DC shock is extremely intense. muscular contraction, feels like you might snap something in either case.
## Can a 120 volts kill you?
Ordinary, household, 120 volts AC electricity is dangerous and it can kill. We can use a simple formula to calculate the current: Current in Amps = Voltage in Volts divided by Resistance in Ohms. … Using electrical tools or equipment in wet areas can be a hazard.
## Will I die if I stick a fork in an outlet?
Electrical outlets The Fear: If you stick a fork or a bobby pin in one of the sockets, you’ll be electrocuted. … The left slot is connected to the neutral wire, the right is connected to the hot one, and electricity flows from hot to neutral. Sticking something into either slot will disrupt the flow and send it into you.
## How much voltage is safe for humans?
In industry, 30 volts is generally considered to be a conservative threshold value for dangerous voltage. The cautious person should regard any voltage above 30 volts as threatening, not relying on normal body resistance for protection against shock.
## Is it 110 or 120?
The important difference is the amount of potential energy, or voltage, the socket is able to supply to the device plugged into it. The most common electrical outlet in any home is a 110 volt. Sometimes you may hear 110 volt plugs referred to as 120 volt. Do not be confused by this; think of them as one and the same.
## How many volts is lethal?
fifty voltsThe human body has an inherent high resistance to electric current, which means without sufficient voltage a dangerous amount of current cannot flow through the body and cause injury or death. As a rough rule of thumb, more than fifty volts is sufficient to drive a potentially lethal current through the body.
## How many volts is a lightning strike?
Lightning can have 100 million to 1 billion volts, and contains billions of watts.
## Can you die from being shocked by an outlet?
Electric shocks can cause injuries that are not always visible. Depending on how high the voltage was, the injury may be fatal. However, if a person survives the initial electrocution, a person should seek medical attention to ensure that no injuries have occurred.
## Is 110v DC dangerous?
110 DC is more dangerous as there is no frequency in dc and 110 V dc remains always at peak i.e, 110V DC.. While in Ac voltage there is concept of frequency and in every cycle, three times the value of ac voltages is zero and hence there is a chance of escaping from the electrocution….
## Is 110v more dangerous than 220v?
Meaning, higher current can be more dangerous than higher voltage; however, since voltage and amperage are directly proportional, 110v wiring is usually considered safer to work with because it uses fewer volts and as such can only carry half as much current as 220v wiring.
## How many volts are in a police taser?
50,000 voltsA police taser typically peaks at 50,000 volts, and by the time it reaches your skin it’s only around 1,200 volts… Not even close to a million volts.
## Can an electrical outlet kill a child?
If a child sticks wet fingers into an outlet, or even worse if a child sticks a piece of metal into an outlet, electocution is a definite possibility. … Electrocution frequently results in death. About 100 kids die every year by electrocution.
## Can 100000 volts kill you?
However, a person may not die from high-voltage electric shock if the electricity did not pass through the heart. … Therefore, for electricity simply to move current through 10 cm of air, the voltage required is 100,000 volts, and this is between the cloud generating the electricity and the earth below our feet.
## Can 120 volts stop your heart?
120 volts isn’t “strong” enough to push much current through your body which is why most 120 volt shocks are survivable. However, it’s still enough current to interfere with your nerves’ communication so if your heart happens to be part of this “current highway” it may start beating erratically, which can cause death.
## What does 120 volt shock feel like?
120-volt AC Wall Current. Suppose you put two wires into a wall outlet. A very nasty tingling sensation that will usually leave your hand and arm numb. … Those overhead lines you see running through your neighborhood are anywhere from 4000 to 50,000 volts.
## What does a minor electric shock feel like?
When you touch a light switch to turn on a light, you may receive a minor electrical shock. You may feel tingling in your hand or arm. Usually, this tingling goes away in a few minutes. If you do not have damage to the skin or other symptoms, there is no reason to worry. | 1,375 | 6,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-40 | latest | en | 0.920872 |
https://www.lidolearning.com/questions/m-bb-ncert7-ch2-ex2p5-q1/q1-which-is-greater-i-05-or-00/ | 1,600,798,021,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400206329.28/warc/CC-MAIN-20200922161302-20200922191302-00470.warc.gz | 932,527,794 | 10,294 | Ncert solutions
# Question 1
Q1) Which is greater?
i) 0.5 or 0.05
ii) 0.7 or 0.5
iii) 7 or 0.7
iv) 1.37 or 1.49
v) 2.03 or 2.30
vi) 0.8 or 0.88
Solution 1
i) Comparing the tenths place, we get 5>0
Hence, 0.5 > 0.05
ii) Comparing the tenths place, we get 7>5
Hence, 0.7>0.5
iii) Comparing the ones place, we get 7>0
Hence 7>0.7
iv) Comparing the tenths place, we get 0<3
Hence, 1.37<1.49
v) Comparing thr tenths place, we get 0<3
Hence, 2.03<2.30
vi) Comparing the tenths place, we get 0<8
Hence, 0.80<088
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https://gist.github.com/arik-so/2d228c3046c65ce2f73ee9c9ac819ce0 | 1,723,602,592,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641086966.85/warc/CC-MAIN-20240813235205-20240814025205-00230.warc.gz | 210,946,408 | 23,303 | Instantly share code, notes, and snippets.
# arik-so/rusty_share_trick.md Secret
Last active September 5, 2023 19:09
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# Rusty's Revocation Secret Threshold Trick
Rusty describes how given x shares, one can distribute different subsets of those shares to n participants such that any quorum of at least t participants would have the complete set, but no quorum of t-1 or fewer participants would.
In this exercise, we try to standardize how to determine the necessary number of shares x for any t/n threshold scenario, and how to distribute subsets of those shares in a manner that will ensure these constraints.
For the specific use case of threshold channel revocation, the assumption is that for a set of shares {A, B, C}, the actual secret that participants meeting the threshold requirement would need to be able to calculate is simply the XOR of all components, i. e. S = A xor B xor C. However, the share distribution mechanism is agnostic to the exact manner in which knowledge of the full set of shares may translate to specific applications.
Something I would like to explicitly point out:
• A participant may have multiple shares, though all users will have the same number of shares ("shares per user")
• Multiple participants may have the same share, though the number of participants with a given share will be equivalent across all shares ("users per share")
## Trivial Scenarios
Let's think through some trivial scenarios that are either 1/n or n/n, where the share distribution is rather obvious.
### 1/2
One of two users needs to come together to calculate the secret.
Participant Shares
Alice A
Bob A
Secret: A
Total shares 1
Shares per user 1
Users per share 2
### 2/2
Two of two users need to come together to calculate the secret.
Participant Shares
Alice A
Bob B
Secret: A xor B
Total shares 2
Shares per user 1
Users per share 1
### 1/3
One of three users needs to come together to calculate the secret.
Participant Shares
Alice A
Bob A
Charlie A
Secret: A
Total shares 1
Shares per user 1
Users per share 3
### 3/3
Three of three users need to come together to calculate the secret.
Participant Shares
Alice A
Bob B
Charlie C
Secret: A xor B xor C
Total shares 3
Shares per user 1
Users per share 1
### Summary
We can immediately see that for any 1/n, there is only one share, and all participants have it, meaning any individual participant comprises the full quorum.
On the other hand, for any n/n, we observe that each user simply has their own unique share, so the set can only be complete with a quorum of all users.
## Nontrivial Scenarios
### 2/3
Two of three users need to come together to calculate the secret.
Participant Shares
Alice A B
Bob B C
Charlie A C
Secret: A xor B xor C
Total shares 3
Shares per user 2
Users per share 2
### 2/4
Two of four users need to come together to calculate the secret.
Participant Shares
Alice A B C
Bob A B D
Charlie A C D
Dylan B C D
Secret: A xor B xor C xor D
Share Users
A Alice, Bob, Charlie
B Alice, Bob, Dylan
C Alice, Charlie, Dylan
D Bob, Charlie, Dylan
Total shares 4
Shares per user 3
Users per share 3
### 3/4
Three of four users need to come together to calculate the secret.
Participant Shares
Alice A B C
Bob A D E
Charlie B D F
Dylan C E F
Secret: A xor B xor C xor D xor E xor F
Share Users
A Alice, Bob
B Alice, Charlie
C Alice, Dylan
D Bob, Charlie
E Bob, Dylan
F Charlie, Dylan
Total shares 6
Shares per user 3
Users per share 2
## Tabulation
Scenario Total shares Shares per user Users per share
1/2 1 1 2
2/2 2 1 1
1/3 1 1 3
2/3 3 2 2
3/3 3 1 1
1/4 1 1 4
2/4 4 3 3
3/4 6 3 2
4/4 4 1 1
Observations:
Users per share $n - t + 1$
Total shares ${n \choose \text{{<}Users per share{>}}} = {n \choose n - t + 1}$
Shares per user ${n - 1 \choose t - 1}$
## Experiment
### 3/5
$$\text{Total shares} = {n \choose n - t + 1} = {5 \choose 5 - 3 + 1} = {5 \choose 3} = 10$$ $$\text{Shares per user} = {n - 1 \choose t - 1} = {5 - 1 \choose 3 - 1} = {4 \choose 2} = 6$$ $$\text{Users per share} = n - t + 1 = 5 - 3 + 1 = 3$$
So to summarize the result of our calculation:
Total shares 10
Shares per user 6
Users per share 3
A critical thing to point out is that there are exactly 10 ways of combining 3 users per share out of a total of 5 participants, which is the number of shares that we have. In fact, the reason the number of total shares is ${n \choose n - t + 1}$ is because it's actually ${n \choose \text{{<}users per share{>}}}$, and $\text{{<}users per share{>}} = n - t + 1$.
On the other hand, the number of combinations of 6 shares per user out of 10 is 210, which is drastically more than the number of participants. For that reason, finding correct subset distributions is easier based on the users per share than vice versa, so we'll go through each of the 10 shares and iterate over the 10 possible combinations of the 5 users in sets of 3.
Share Users
A Alice, Bob, Charlie
B Alice, Bob, Dylan
C Alice, Bob, Emily
D Alice, Charlie, Dylan
E Alice, Charlie, Emily
F Alice, Dylan, Emily
G Bob, Charlie, Dylan
H Bob, Charlie, Emily
I Bob, Dylan, Emily
J Charlie, Dylan, Emily
Let's construct a dim(<shares> X <participants>)-matrix to map the share-participant-pairings:
Share Alice Bob Charlie Dylan Emily
A X X X
B X X X
C X X X
D X X X
E X X X
F X X X
G X X X
H X X X
I X X X
J X X X
Based on this output, we now transpose from the shares to the users.
Participant Shares
Alice A B C D E F
Bob A B C G H I
Charlie A D E G H J
Dylan B D F G I J
Emily C E F H I J
The secret is S = A xor B xor C xor D xor E xor F xor G xor H xor I xor J. Any set of 3 participants should have sufficient information to calculate it, whereas no combination of 2 or fewer participants should.
Another trick for easy iteration would be thinking of the user allocation as a 5-digit binary number where exactly 3 bits must be 1, and going through all combinations between 00111 and 11100 where that is true. The bit index may indicate the participant, and that they would have that particular share if the bit is set.
## Script
Ugly script to generate the share distribution: https://gist.github.com/arik-so/f23c710c42da0fe5d06a528c49fb1e58
to join this conversation on GitHub. Already have an account? Sign in to comment | 1,822 | 6,396 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-33 | latest | en | 0.86944 |
https://www.physicsforums.com/threads/coulombs-law-net-electrostatic-force.303313/ | 1,566,463,985,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317037.24/warc/CC-MAIN-20190822084513-20190822110513-00548.warc.gz | 908,191,228 | 16,437 | # Coulomb's Law, net electrostatic force
#### nn3568
1. Homework Statement
A particle with charge −9 μC is located on the x-axis at the point 8 cm, and a second particle with charge 5 μC is placed on the x-axis at 6 cm. The Coulomb constant is 8.9875 × 109 N · m2/C2. What is the magnitude of the total electrostatic force on a third particle with charge 2 μC placed on the x-axis at −2 cm? Answer in units of N.
2. Homework Equations
Coulomb's Law
fe = (kq1q2)/(d2)
3. The Attempt at a Solution
(8.9875e9 * 2e-6 * 5e-6) / (0.08^2) = 14.04296875
(8.9875e9 * 5e-6 * -9e-6) / (0.02^2) = -983.0078125
14.04296875 + -983.0078125 = -997.0507813
magnitude 997.0507813
Why is this wrong? What can I do to make it right?
Related Introductory Physics Homework Help News on Phys.org
#### Doc Al
Mentor
(8.9875e9 * 5e-6 * -9e-6) / (0.02^2) = -983.0078125
You want the force on the 2μC charge.
#### nn3568
Thank you so much!
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• Solo and co-op problem solving | 410 | 1,245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2019-35 | longest | en | 0.791082 |
http://www.audifans.com/archives/1996/09/msg01585.html | 1,516,391,244,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888113.39/warc/CC-MAIN-20180119184632-20180119204632-00714.warc.gz | 400,614,288 | 3,060 | # Re: Wheelspin
```STEADIRIC@aol.com wrote:
>
> >No hard feelings, just hard facts. I don't disagree that you get better
> >acceleration in a roadcar by spinning the tires a little off the line -
> >I'd be an idiot to argue with this fact. But to say that traction is
> >increased with "wheelspin" is not right. "Slip", I will agree,
> >increases traction. "Slip" is not equal to "wheelspin".
>
> Go back and take a look at the Delta V number.... It's less than 10% for
> a Street compound So for max acceleration from a stand still the tire has
> to be spinning 10% faster than Road speed.... Race tires which operate a
> smaller slip angles (on the order of 2-4/deg or % depending on
> acceleration direction) spin even less for max traction. I think that
> the problem that your having with this is your equating optimum Wheelspin
> with tire smoking wheelspin. The other proof of this is that max braking
> occurs when the tire is working at optimum slip which means that the tire
> is rotating 10% slower than road speed. Try to kick that delta V up to
> 11% you get lock up. And BTW Dragsters spin their wheels off the line....
>
> Later!
>
> Eric Fletcher
> '87 5KCSTQIA2RSR2B
> St. Louis, MO
>
What your trying to say is right, eric. You're just using the wrong
terminology. In case you don't believe me (which is obviously the case), I'll
quote directly from the Milliken and Milliken book, "Race Car Vehicle Dynamics".
"Tractive force FT and braking force FB are a function of slip ratio. As the
slip ratio increases (numerically)from zero, the forces rise rapidly to a
maximum which usually occurs in the range of 0.10 to 0.15 slip ratio, after
which the forces fall off."
This is what you're saying. And quite honestly, I'm impressed with your
accuracy on the 10% slip ratio. But in the next paragraph:
"For the traction case, note that the force falls off rapidly after the onset of
spinning.", and "Once a tire exceeds the slip ratio for peak force in either
traction or braking, it becomes unstable and the wheel tends to either spin-up
(traction) or lock (braking)."
Thus my point - "slip" IS NOT THE SAME AS "wheelspin". It should be clear from
these quotes (from an 890 page SAE publication no less) that maximum tractive
force occurs at ~0.10 to 0.15 "slip ratio", but falls off dramatically with the
onset of "wheelspin."
---------------------------------------------------
Jeremy R. King Clemson University, S Carolina, USA
/////////// '86 VW Quantum GL5 \\\\\\\\\\\\
Rear Dynomax Glasspak, Hollow cat, Drilled Airbox
Boge / KYB struts, Round 100W Driving Lights
\\\\\\\\\\\ Still Slower'n Christmas ////////////
---------------------------------------------------
``` | 709 | 2,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-05 | latest | en | 0.944108 |
https://math-semester-final.wonderhowto.com/how-to/solve-equations-with-fractions-0147070/ | 1,723,136,486,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640736186.44/warc/CC-MAIN-20240808155812-20240808185812-00546.warc.gz | 315,722,539 | 26,211 | # How To: Solve Equations with Fractions
## Solving Equations with Fractions Using Addition
In our first example, our equation is b - 11/10 = -13/5
We need to find out what the variable, 'b', is. A variable is a symbol that stands for an unknown number. To do so, we have to isolate the variable, meaning that we have to have the variable on one side of the equal sign by itself. First, we add -11/10 + 11/10, which equals 0. Now remember, whatever you do on one side of the equal sign has to also be done on the other side of the equal sign. So we add -13/5 + 11/10, which is -15/10. The current equation is now b = -15/10. We have finally found out what 'b' is equal to! But, we have to simplify -15/10, which is -1 1/2. Now the correct answer is b = -1 1/2
## Solving Equations with Fractions Using Subtraction
In our second example, our equation is 5/6 = x + 2/3
Like we did in the first example, we need to isolate the variable. Subtract 2/3 - 2/3, which results in 0. Then, also subtract 5/6 - 2/3, equalling 1/6. The equation is now 1/6 = x. Since 1/6 can't be simplified, 1/6 = x is the final answer.
## Solving Equations with Fractions Using Division
In our final example, our equation is 3 1/4j = 3/2
Before we do anything at all, to make it easier on ourselves, we must convert all mixed numbers into improper fractions. The only number we need to convert is 3 1/4 and as a improper fraction, it is 13/4. The current equation is now 13/4j = 3/2. Again, we need to isolate the variable so we divided 13/4 by 13/4. On the other side of the equal sign, we do the same thing by doing 13/4 divided by 3/2, which is 6/13. We finally found out that j = 6/13 and that this is the final answer since 6/13 can't be simplified.
## Practice Time
Solve the equations below
1. a - 3/5 = -7/5
1. v + 3 4/9 = -46/9
1. 15/2 = 3/5x
1. a = -4/5
1. v = -77/9 = -8 5/9
1. x = 25/2 = 12 1/2
Just updated your iPhone? You'll find new features for TV, Messages, News, and Shortcuts, as well as important bug fixes and security patches. Find out what's new and changed on your iPhone with the iOS 17.6 update.
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• Latest | 671 | 2,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2024-33 | latest | en | 0.942864 |
http://www.bankersadda.com/2017/11/reasoning-questions-for-ibps-clerk-29.html | 1,527,003,356,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864798.12/warc/CC-MAIN-20180522151159-20180522171159-00026.warc.gz | 332,318,764 | 47,190 | Reasoning Questions for IBPS Clerk Prelims
The reasoning is a game of wits and presence of mind! Yes, it is true and it might seem as the greatest of the challenge after English Section’s surprises but yet this one can easily be dealt with. You just need correct practice and hardwire your brain to quickly make decisions about what to attempt and what to leave. And for same we are providing you questions of Reasoning Question and Answers. Solve these to Practice latest pattern reasoning question for bank exams.
Directions (1-5): In each question below are given two/three statements followed by two conclusions numbered I and II. You have to take the given statements to be true even if they seem to be at variance with commonly known facts. Read all the conclusions and then decide which of the given conclusions logically follows from the given statements, disregarding commonly known facts. Give answer
(a) if only conclusion I follows.
(b) if only conclusion II follows.
(c) if either conclusion I or II follows.
(d) if neither conclusion I nor II follows.
(e) if both conclusions I and II follow.
Q1. Statements:
All plates are glasses.
Some cups are glasses.
Conclusions:
I. At least some cups are plates.
II. Some glasses are cups.
Q2. Statements:
All trolleys are lamps.
No lamp is a chair.
Conclusions:
I. At least some trolleys are chairs.
II. Some chairs are definitely not trolleys.
Q3. Statements:
Some clothes are shirts.
All shirts are paints.
Conclusions:
I. All paints being clothes is a possibility.
II. Some shirts are clothes.
Q4. Statements:
No sand is a stone
No sand is a tree.
Conclusions:
I. No stone is sand.
II. No tree is a stone.
Q5. Statements:
Some teachers are doctors.
No doctor is a lawyer.
Conclusions:
I. Some teachers are not lawyers.
II. Some lawyers are doctors.
Directions (6-10): Study the following information carefully and answer the questions given below:
A, B, C, D, E, F, G and H are sitting around a circular table facing the centre. H is third to the right of C and second to the left of E. B is not an immediate neighbour of H or C. F is second to the right of D and is an immediate neighbour of C. G is not the neighbour of E.
Q6. Who among the following is second to the right of C?
(a) H
(b) G
(c) F
(d) E
(e) None of these
Q7. Who among the following is an immediate neighbour of H and E(both)?
(a) A
(b) B
(c) C
(d) G
(e) None of these
Q8. In which of the following pairs the second person is sitting on the immediate right of the first person?
(a) A, H
(b) C, D
(c) G, H
(d) E, H
(e) F, C
Q9. Who among the following is second to the left of B?
(a) C
(b) H
(c) F
(d) A
(e) None of these
Q10. Who among the following is opposite D?
(a) A
(b) G
(c) H
(d) E
(e) None of these
Q11. If the letters of the word AMERICA are arranged in the English alphabetical order from left to right, the position of how many letters will remain unchanged?
(a) None
(b) One
(c) Two
(d) Three
(e) None of these
Q12. In a certain code language TREAT is written as UBFSU and HABIT is written as UJCBI. How is AGREE written in that code language?
(a) FSHBF
(b) FSHFB
(c) FFSHB
(d) FFQBH
(e) None of these
Q13. Four of the following five are alike in a certain way and hence form a group. Which is the one that does not belong to that group?
(a) EV
(b) KP
(c) IR
(d) OL
(e) CW
Q14. How many such pairs of letters are there in the word STARVATION each of which has as many letters between them in the word as in the English alphabet? (In both forward and backward directions)
(a) None
(b) One
(c) Two
(d) Three
(e) More than three.
Q15. What should come in place of question mark (?) in the following series?
JK, MN, QR, VW, ?
(a) BC
(b) XY
(c) YZ
(d) AB
(e) None of these
You May also like to Read: | 996 | 3,753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-22 | latest | en | 0.942731 |
https://www.atheistsforhumanrights.org/how-do-i-type-an-r-symbol/ | 1,679,362,913,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943589.10/warc/CC-MAIN-20230321002050-20230321032050-00068.warc.gz | 759,459,221 | 16,930 | # How do I type an R symbol?
## How do I type an R symbol?
Press and hold Alt while typing 0174 for the registered trademark (®) symbol. This code gives you the registered trademark symbol, which is the “R” in a circle.
What is the R in a circle symbol?
The ® on a product means that it’s a registered trademark, meaning the brand name or logo is protected by (officially registered in) the US Patent and Trademark Office, while plain old ™ trademarks have no legal backing.
What does the R symbol indicate?
The R (®) symbol on a product indicates that it is a registered trademark. This means that the logo enjoys legal protection as per the Trademarks Act, 1999.
### How do you make the R symbol in latex?
Set of real number is represented by the ℝ symbol. For this, you need to pass the argument R in \mathbb command in latex….How to write a real number(ℝ) symbol in LaTeX?
Symbol Real numbers
Command \mathbb{R}
Example \mathbb{R} → ℝ
How do you get the R with a circle around it?
Press and hold ALT then press 0, followed by 1, 7, 4. You will get ®. For MacBooks, press R while holding the Option key (ALT).
What is R and TM?
You do not have to have registered a trademark to use it and many companies will opt to use the TM symbol for new goods or services in advance of and during the application process. The R symbol indicates that this word, phrase, or logo is a registered trademark for the product or service.
## How do I insert mathematical symbols in LaTeX?
LaTeX allows two writing modes for mathematical expressions: the inline math mode and display math mode: inline math mode is used to write formulas that are part of a paragraph….Display math mode
1. $…$
2. \begin{displaymath}…\end{displaymath}
3. $$…$$
Can I use a registered trademark logo?
You may use the registration symbol anywhere around the trademark, although most trademark owners use the symbol in a superscript or subscript manner to the right of the trademark. You may only use the registration symbol with the trademark for the goods or services listed in the federal trademark registration.
What is the patent R?
What does the circle R symbol (®) mean? The circle R symbol (®) means that a mark is federally registered with the US Patent and Trademark Office, either on the Principal Register or Supplemental Register. | 531 | 2,324 | {"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": 1, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-14 | latest | en | 0.857666 |
https://www.exceldemy.com/author/rafi/page/6/ | 1,695,936,199,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510454.60/warc/CC-MAIN-20230928194838-20230928224838-00541.warc.gz | 825,681,635 | 66,295 | User Posts: Rafiul Hasan
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You may have created a document in Excel, scaled the sheet size and printed it. Usually, the page size in an Excel sheet for printing is selected to Letter by ...
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## How to Change Column Name from ABC to 1 2 3 in Excel
Usually, the column names in Excel remain in alphabetical form (A, B, C) by default, but you can change it to numerical order (1,2,3) too. And sometimes you ...
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## What Is Accounting Number Format in Excel?
In an Excel worksheet, there are some available number formats. You can transform your numerical values to any of the formats of your choice. The most common ...
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## How to Create Sparklines in Excel (2 Easy Ways)
You may have created a chart in Excel on the basis of some gathered data. But sometimes you may need to create Sparklines in your Excel worksheet to make your ...
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## How to Add Data to an Existing Chart in Excel (5 Easy Ways)
You may have created a Chart in Excel on the basis of some gathered data. But sometimes you may need to update your chart by adding data to the existing Chart ...
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## How to Edit Legend of a Pie Chart in Excel (3 Easy Methods)
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## How to Create Chart with Dynamic Date Range in Excel (2 Easy Ways)
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## How to Make a Stacked Bar Chart in Excel (2 Quick Methods)
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## How to Change Axis Labels in Excel (3 Easy Methods)
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## How to Add Axis Titles in Excel (2 Quick Methods)
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## How to Capitalize All Letters Without Formula in Excel (4 Methods)
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## How to Print Barcode Labels in Excel (with 4 Easy Steps)
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Browsing All Comments By: Rafiul Hasan
1. Hi BAGUS,
In the context of Exponential Smoothing in Excel’s Data Analysis Toolpak, the Damping Factor refers to a parameter used to control the impact of older observations on the forecasted values.
The damping factor has a value between 0 and 1. It determines the weight assigned to the most recent observation when calculating the forecast. A higher value (closer to 1) gives more weight to the most recent data point, making the forecast more responsive to recent changes in the data. On the other hand, a lower value (closer to 0) gives less weight to the most recent data point, making the forecast more stable and less responsive to short-term fluctuations.
The value of Damping Factor being 0.91 suggests relatively high importance to the most recent data while still considering some historical data. The choice of 0.91 is somewhat arbitrary and depends on the specific characteristics of the data and the desired balance between responsiveness and stability in the forecast. Different values of the damping factor may be chosen based on the analyst’s judgment and the nature of the data being analyzed.
To determine the most appropriate value for the Damping Factor, it’s often a good practice to experiment with different values and evaluate the forecast accuracy using techniques like Mean Absolute Error (MAE) or Mean Squared Error (MSE). The choice of Damping Factor should ideally be data-driven and selected based on how well it performs in forecasting historical data or predicting future values.
Regards
Rafiul Hasan
Team ExcelDemy
2. Hi ANTHONEY,
Thanks for your comment. You requirement needs the Excel file. Please share your Excel file and repost your problem on our official ExcelDemy Forum. Our experts will try to reach you out.
Regards
Rafiul Hasan
Team ExcelDemy
3. Hi GREG,
I appreciate your suggestion. In VBA, there isn’t a built-in function to directly evaluate a string as code like in some other programming languages. You cannot directly evaluate a string as executable code like you would with the eval() function in JavaScript. Here, alternative approach has been provided to achieve similar results. You can achieve the concept of “evaluating strings as code” through code manipulation and dynamic formula evaluation. If you are looking for a method to directly execute arbitrary VBA code contained within a string, Excel VBA doesn’t provide a direct “eval” function for this purpose. The technique shared here provides structured way to achieve dynamic behavior in VBA.
4. Hello KRISTY,
If you want to create a weight loss graph with values from 230 to 150, you’ll need to adjust the scale of the vertical (y-axis) to accommodate the new range. If you want to set your target weight as 150, you can create the weight loss graph in Excel like below:
Regards
Rafiul Hasan
Team ExcelDemy
5. Hello EDI,
the INDEX-MATCH formula can return a #VALUE! error under certain circumstances.
1. The MATCH function within the INDEX-MATCH formula couldn’t find a match for the lookup value in the specified array or range.
2. The INDEX function is expecting a numeric value as a result, but the matched value is non-numeric.
3. Incorrect arguments or ranges.
If you elaborate the problem you’re facing, then it will be easy to provide solution. Let us know if you face any difficulties.
Regards
Rafiul Hasan
Team ExcelDemy
6. Hi MASHAL,
You can use use VBA macro to automatically convert any entry you enter in specific cells to upper case by just only entering the text. Just follow these steps:
1. Press ALT + F11 to open the Visual Basic for Applications (VBA) editor.
2. In the Project Explorer window, locate and double-click on the “ThisWorkbook” module for the desired workbook.
3. In the appeared code window, paste the following code:
``````Private Sub Workbook_SheetChange(ByVal Sh As Object, ByVal Target As Range)
Dim rng As Range
Set rng = Sh.Range("A1:Z100") ' Replace with the desired range
If Not Intersect(Target, rng) Is Nothing Then
Application.EnableEvents = False
On Error Resume Next
Target.Value = UCase(Target.Value)
Application.EnableEvents = True
End If
End Sub``````
Replace “A1:Z100” with the range of cells you want to convert to uppercase.
4. Close the VBA editor.
Now, whenever you enter any text within the specified range, it will be automatically converted to uppercase without the need to manually run any macro. Adjust the range in the code as needed to match your desired cells.
Regards
Rafiul Hasan
Team ExcelDemy
7. Hi MARTIN PEREZ,
Thank you for your comment and for sharing your additional solution! I appreciate your input and the effort you put into finding a workaround for area chart markers. It sounds like you found an interesting method using a supporter series to manipulate the legend key font size. Your explanation will undoubtedly help other readers who come across your comment. Thank you again for your contribution!
Regards
Rafiul Hasan
Team ExcelDemy
8. Hi MARK,
Thanks for your comment. To learn different ways to print multiple sheets, you can follow this article of ExcelDemy: Print multiple sheets in Excel
If you want to print multiple sheets using VBA only, then follow this one: Print multiple Excel sheets to single PDF with VBA
Hope you will find this helpful.
Regards
Rafiul Hasan
Team ExcelDemy
9. Hi HANI,
Thanks for your query. The choice of epsilon depends on the desired accuracy of the eigenvalue estimate. It is typically a small positive value, and the algorithm terminates when the absolute difference between consecutive eigenvalue estimates falls below epsilon.
The specific value of epsilon will vary depending on the application and the desired level of accuracy. Using epsilon = 0.0005 indicates that the power method iterations will continue until the absolute difference between consecutive eigenvalue estimates falls below 0.0005. This level of tolerance can be suitable for many applications, especially when the eigenvalues of the matrix are relatively close in magnitude.
However, it’s important to note that the appropriate value of epsilon may vary depending on the specific problem and matrix being analyzed. If you find that the algorithm converges too quickly or does not reach the desired accuracy with epsilon = 0.0005, you may need to adjust the value accordingly.
Consider the scale and conditioning of the problem as well. If the matrix is ill-conditioned or has eigenvalues with large differences in magnitude, a smaller epsilon may be necessary to obtain accurate results.
So, it’s always important to assess the specific requirements of your problem and adjust the value of epsilon accordingly to achieve the desired accuracy.
Regards
Rafiul Hasan
Team ExcelDemy
10. Hi KEN,
This may encounter from several reasons.
1. If you have protected the worksheet or specific cells, it can prevent changes to the font settings. Try unprotecting the cells or the entire worksheet to see if that resolves the issue.
2. If there are any conflicting formatting applied to the cells, it might cause the barcode font to revert. Select the cells and clear any formatting such as bold, italic, or color changes. Then, set the font to “Code128” again.
3. Ensure that the page setup and print preview are correctly configured. Check the scaling options to make sure that the labels fit properly on the A4 sheet. Adjust the margins and other settings as needed to accommodate 51 labels per sheet.
4. If the above steps don’t solve the problem, try creating a new Excel file and follow the barcode printing steps from scratch. This can help identify if the issue is specific to your existing file.
Regards,
Rafiul Hasan
ExcelDemy Team
11. Hi ALLISON,
If you want to use wildcards with the second formula, you need to modify the dataset a little bit. Let’s say, we have short form of country name in the “Country” column. Our aim is to find the name of the person aged “38” and whose country is “India“. You can use the following formula:
=INDEX(\$B\$5:\$B\$17, SMALL(IF(COUNTIF(\$F\$6, \$C\$5:\$C\$17)*COUNTIF(\$G\$6,\$D\$5:\$D\$17&”*”),ROW(\$B\$5:\$D\$17)-MIN(ROW(\$B\$5:\$D\$17))+1), ROW(A1)), COLUMN(A1))
We have used the wildcards character ampersand (*) in the second COUNTIF portion: COUNTIF(\$G\$6,\$D\$5:\$D\$17&”*”)
As we want to find multiple output so it will result from an array. So, the \$D\$5:\$D\$17 acts as criteria as we can use wildcards character in criteria. This formula will match the short form mentioned in the country column and match it with criteria and extract the Name.
Regards
Rafiul Hasan
Team ExcelDemy
12. Hi SINGGIH WAHYU N,
You can use VBA code for serving your requirement.
1. Select the merged cell(s) that contain data you want to copy.
2. Open the Visual Basic Editor by pressing Alt + F11.
3. Insert a new module by selecting “Insert” -> “Module” from the menu bar.
4. In the new module, enter the following code:
``````Sub CopyMergedCells()
Dim mergedRange As Range, cell As Range
Dim unmergedRange As Range, destCell As Range
Set mergedRange = Selection 'Assuming the merged cells are currently selected
Set unmergedRange = mergedRange.Cells(1).Resize(mergedRange.Rows.Count * mergedRange.Columns.Count, 1)
Set destCell = Application.InputBox(prompt:="Select the top-left cell of the destination range", Type:=8)
If destCell Is Nothing Then Exit Sub 'User cancelled the selection
For Each cell In unmergedRange.Cells
If cell.Value <> "" Then
destCell.Value = cell.Value
Set destCell = destCell.Offset(1, 0)
End If
Next cell
End Sub``````
5. Run the code and a prompt will ask you to select the first cell of destination range. Select a cell where you want to paste values and you will get output.
Hope this help you. If you don’t want to use VBA, you have to first paste values and then use the “Go To Special” (pressing CTRL+G) dialog box and then select “Blanks”> click “OK” to remove blank cells.
Regards.
Rafiul Hasan
Team ExcelDemy
13. Hi DARRELL,
Congratulations on your new project. Be confident, you can make it. Based on your selections, you may use any of the following to get resulting list:
1. IF function: You can use the “IF” function to create a formula that checks if certain criteria are met and return the appropriate value. For example, if you have a pick list for materials and another for color, you can use an “IF” function to generate a list of materials that match the selected color.
2. VLOOKUP function: The “VLOOKUP” function can be used to search for a value in a table and return a corresponding value. You can set up a table with all the possible combinations of selections and use “VLOOKUP” to generate the resulting list based on the selections made.
3. FILTER function: The “FILTER” function can be used to filter a list based on certain criteria. You can set up a table with all the materials and their attributes (such as color, size, etc.) and use “FILTER” to generate a list of materials that match the selected attributes.
4. PivotTable: Pivot tables can be used to analyze and summarize data in a table. you can set up a table with all the selections made and your corresponding materials and use a pivot table to generate a list of materials based on the selections made.
These are just a few options you can consider. You will need to choose the one that works best for your specific situation. Don’t hesitate to contact with us if you face any problem. Best of luck.
Regards
Rafiul Hasan
Team ExcelDemy
14. Hi NOUR,
Thanks for the appreciation.
As per your requirement, it is possible to allow multiple users to access and edit a workbook on a network computer. Here are some steps you can follow in this regard:
2. Restrict editing access for other users to ensure only authorized users can edit the workbook.
3. Create a login form that allows users to enter their credentials to access the userform. You can store user credentials in a separate worksheet within the workbook or use an external database.
4. Use VBA code to pull the necessary data from the shared workbook and display it to the userform.
5. When a user makes changes to the data in the userform, use VBA code to write the changes back to the shared workbook.
6. Implement a save feature that automatically saves changes made to the shared workbook when the userform is closed.
7. Create a backup plan to ensure data is not lost in case of system failure or network issues. This can include regular backups or saving a copy of the workbook in a secure location.
8. Test the userform with different user accounts to ensure it works as intended and all users can access and edit the shared workbook.
Hope this guide helps you to approach and solve the issue you’re facing. Actually it is not possible to give an exact solution without checking the Excel file. Most probably you need physical assistance in this regard. Let us know if you have any additional questions or concerns. You can also send your Excel file to our official mail address: [email protected]
Have a great day!
Regards
Rafiul Hasan | ExcelDemy Team
15. Hi DIEP TRAN,
The “Yes to all” button is a feature in Excel that allows you to apply a selected action to all the occurrences of the same conflict during a process. This button is usually available in the Name Manager dialog box, which pops up when you create, edit or delete names in your workbook.
The availability of the “Yes to all” button may vary depending on the version of Excel you are using. However, in most versions of Excel, including the latest version, which is Excel 2021, the “Yes to all” button should be available in the Name Manager dialog box.
If you are not seeing the “Yes to all” button in the Name Manager dialog box, it could be that the dialog box is not showing all the options. You can try expanding the dialog box to see if the “Yes to all” button is hidden or check your Excel settings to ensure that the option is enabled.
Overall, it’s best to consult the specific version of Excel you are using or refer to the documentation to learn more about the availability and functionality of the “Yes to all” button in Excel.
Regards
Rafiul | ExcelDemy Team
16. Hi JEFF,
Thanks for your concern. Actually, all the methods here explained are very easy to understand. Considering the situation, you have to choose the best-suited method for serving your purpose. If you don’t want to use any formula, then you can use Method 1.
Regards,
Rafiul | ExcelDemy Team
17. Hello AARON MWALE,
Thanks for your comment. Sorry to hear that you’re facing issues with Excel.
When Excel hangs and starts an Undo action, that means Excel is trying to process a large amount of data or executing a complex operation. This may cause the program to be unresponsive for a period of time while it completes the task.
In order to stop the undo action, you can try pressing the “Esc” key on your keyboard. If this does not work, you can try pressing “Ctrl” & “Break” on your keyboard. If don’t find these helpful, you may need to wait for Excel to finish the operation before you can regain control of the program.
To prevent Excel from hanging and starting an Undo action in the future, you can try the following:
1. Limit the amount of data you are working with at one time. If you are working with a large amount of data, try breaking it up into smaller chunks that are easier for Excel to handle.
2. Close any unnecessary programs or applications running in the background. This can free up system resources and improve Excel’s performance.
3. Disable any add-ins or macros that may be causing Excel to slow down or become unresponsive.
4. Check for and install any available updates for Excel. Updates often include bug fixes and performance improvements that can help prevent Excel from hanging in the future.
5. Consider upgrading your computer hardware if it is outdated or underpowered. Excel can be resource-intensive, and having a fast and capable computer can make a big difference in its performance.
Regards,
Rafi
ExcelDemy team
18. Dear BILLY,
Regards
ExcelDemy Team
19. Hi TOM,
Thank you so much for letting us know. The formula has been edited. Thanks for your concern. Stay connected!
Regards,
Rafi
Author: ExcelDemy Team
20. Hello STEPH,
Did you enter any newer data into your pivot table field? With newer or re-entry of data, you need to go to the PivotTable Fields window (at the right corner of the worksheet)-> unmark the corresponding field of your data (i.e. Ship Date for this article) -> right-click on the PivotTable and click Refresh-> again mark the corresponding field that you unmarked a little bit ago.
Thanks and Regards
21. Hello SG,
You can customize the error message. In the Data Validation message box=> Error Alert icon, you can choose “Warning” style from the dropdown list. Now, if you try to input invalid data, the error message will show 3 options asking you whether you want to continue: Yes/No/Cancel/Help. Clicking “Yes” will allow you to proceed with the current value and will not show the error again.
22. Hello ANN HALL,
This may happen for different reasons.
1. Putting two statements in a single line. Check whether this is encountered and send the second statement in a new line.
2. Absence of parentheses.
3. Lack of whitespace in front of the ampersand(&) interpret as a long variable. So, put space between the variable and operator(&).
23. Hi KAREN LING,
Glad that you liked the template. This template has got fixed interest rate. You can add variable extra payments manually to your Excel file if you need it.
Regards,
Rafi (ExcelDemy Team)
24. Hi JON,
What I can say is that paying an extra amount from the minimum will reduce the next payments. You can proceed with your data. You can also share your Excel file with us and we will look into it.
Regards.
Rafi (ExcelDemy Team)
25. Hi ABDUS SALAM,
An equation is not like a formula as it isn’t supposed to perform any calculation. You can’t enter an equation in a particular cell in Excel just like you do for a formula. What you can do here is that, you can adjust height and width of the “Equation Editor” box or the cell as if it looks like the equation stays in the cell. This can serve your purpose to some extent.
Regards.
26. Hi KRISHNENDU,
Thanks for your comment. An equation is not like a formula as it isn’t supposed to perform any calculation. You can’t enter an equation in a particular cell just like you do for a formula. What you can do here is that, you can adjust height and width of your “Equation Editor” box or the cell as if it looks like the equation stays in the cell. I think this can serve your purpose to some extent.
Regards.
Advanced Excel Exercises with Solutions PDF | 4,913 | 22,111 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-40 | latest | en | 0.899162 |
http://www.osti.gov/eprints/topicpages/documents/record/244/1432800.html | 1,441,340,033,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645335509.77/warc/CC-MAIN-20150827031535-00238-ip-10-171-96-226.ec2.internal.warc.gz | 613,839,786 | 3,168 | ISyE 2027 Exam # 2 Please be neat and show all your work so that I can give you partial credit. Summary: ISyE 2027 Exam # 2 Fall 1999 Name Please be neat and show all your work so that I can give you partial credit. GOOD LUCK. Question 1 Question 2 Question 3 Question 4 Total 1 (25) 1. Suppose we roll two dice and let X and Y be the two numbers that appear on the dice. (a) Find the probability mass function of |X - Y |. (b) Find the expected value of |X - Y |. 2 (25) 2. Suppose that the probability of a defect in a foot of magnetic tape is 0.002. Use the Poisson approximation to binomial to compute the prob- ability that a 1500 foot long tape has no defects. (b) If X has a uniform distribution on (1, 5) compute E(X3).? 3 Collections: Engineering | 218 | 755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2015-35 | longest | en | 0.873129 |
http://compcogscisydney.org/lsr/solutions-2.4.html | 1,542,511,906,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743963.32/warc/CC-MAIN-20181118031826-20181118053126-00047.warc.gz | 63,489,754 | 351,510 | # Exercises for 2.4: Linear models
## Preliminaries
• load the `driving.Rdata` file.
• type `head( driving )` to look at the first few observations
• load the following packages: `lsr`, `car`
``````load( "~/Work/Research/Rbook/workshop_dsto/datasets/driving.Rdata")
``````## id gender age distractor peak.hour errors_time1 errors_time2 rt_time1
## 1 s.1 male 19 radio yes 7 7 346
## 2 s.2 female 42 toddler no 15 16 424
## 3 s.3 male 27 none no 10 7 415
## 4 s.4 female 22 radio yes 5 1 266
## 5 s.5 female 33 none no 4 9 302
## 6 s.6 female 35 toddler yes 15 12 423
## rt_time2
## 1 636
## 2 787
## 3 580
## 4 459
## 5 513
## 6 767``````
``````library(lsr)
library(car)``````
## Exercise 2.4.1: Multiple regression
• use `lm()` to fit a regression model with the RT at time 1 as the outcome variable, including age and errors at time 1 as predictors. Save the results to a variable called `mod.1`
• use `summary()` to run the hypothesis tests etc associated with `mod.1`
• use `standardCoefs()` to extract standardised regression coefficients
## Solution 2.4.1:
``````mod.1 <- lm(
formula = rt_time1 ~ errors_time1 + age,
data = driving
)
summary( mod.1 )``````
``````##
## Call:
## lm(formula = rt_time1 ~ errors_time1 + age, data = driving)
##
## Residuals:
## Min 1Q Median 3Q Max
## -81.501 -24.829 -4.703 26.731 117.324
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 209.033 53.092 3.937 0.00149 **
## errors_time1 8.196 2.661 3.080 0.00814 **
## age 2.782 1.917 1.451 0.16878
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 49.78 on 14 degrees of freedom
## Multiple R-squared: 0.566, Adjusted R-squared: 0.504
## F-statistic: 9.127 on 2 and 14 DF, p-value: 0.002902``````
``standardCoefs( mod.1 )``
``````## b beta
## errors_time1 8.196376 0.5937550
## age 2.782137 0.2797177``````
## Exercise 2.4.2: Regression diagnostic plots
• use `plot()` to draw the standard regression diagnostic plots associated with `mod.1`
## Solution 2.4.2:
``plot( mod.1, which = 1 )`` | 830 | 2,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-47 | latest | en | 0.624864 |
https://ask.sagemath.org/question/41426/extracting-info-from-diff_map-definition/ | 1,726,824,071,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652246.93/warc/CC-MAIN-20240920090502-20240920120502-00508.warc.gz | 95,193,653 | 13,390 | Ask Your Question
# extracting info from diff_map definition
I've defined the following diff_map
R2 = Manifold(2, 'R2', start_index=1, latex_name=r'\mathbb{R}^2')
cartesian2d.<x, y> = R2.chart()
R1 = Manifold(1, 'R1', start_index=1, latex_name=r'\mathbb{R}^1')
cartesian1d.<t> = R1.chart()
h = R1.diff_map(R2, [1-t, sqrt(2*t-t^2)])
h.display()
I would like to extract the elements of the diff_map, namely 1-t and sqrt(2*t-t^2) for propose of plotting the resulting curve or for other manipulations. How can I do that?
Thanks,
Daniel
edit retag close merge delete
## 1 Answer
Sort by ยป oldest newest most voted
Given the Sage object h you can see which methods apply to it, with the tab completion:
sage: h.<TAB>
You will see that there is an expression method:
sage: h.expression()
(-t + 1, sqrt(-t^2 + 2*t))
This is a tuple, and you can get its entrie as follows:
sage: h.expression()[0]
-t + 1
sage: h.expression()[1]
sqrt(-t^2 + 2*t)
more
## Comments
By the same method I also found h.coord_functions()[0]
Thanks!
( 2018-03-08 11:03:33 +0200 )edit
## Your Answer
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## Stats
Asked: 2018-03-07 21:15:24 +0200
Seen: 199 times
Last updated: Mar 07 '18 | 401 | 1,296 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-38 | latest | en | 0.778079 |
https://cl-su-ai.cddddr.org/msg04951.html | 1,603,791,323,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107893845.76/warc/CC-MAIN-20201027082056-20201027112056-00190.warc.gz | 260,444,837 | 3,043 | # Re: constant folding/smashing
• To: NGALL@G.BBN.COM
• Subject: Re: constant folding/smashing
• From: David A. Moon <Moon@STONY-BROOK.SCRC.Symbolics.COM>
• Date: Fri, 10 Jun 88 14:28 EDT
• Cc: common-lisp@SAIL.STANFORD.EDU
``` Date: 10 Jun 1988 11:03-EDT
From: NGALL@G.BBN.COM
....
And this brings up the second example of the confusion in constant
folding that no one has yet addressed: Why are lists and strings
(among others) constant folded, but not general vectors, since all of
them are composite objects?
What makes you think that is true? (I assume you mean "coalesced", that
is, two distinct pieces of source text resulting in a single object in the
compiled program, rather than "constant folded", which means a pure function
with constant arguments replaced by its constant result.) I don't of
anything that says that Common Lisp must treat quoted general vectors
differently from quoted strings. If some implementations do so, that's
merely an implementation.
[3] (EQ (QUOTE #1=(1 2 3)) (QUOTE #1#)) => T or NIL, wow!!!!!
I see no evidence that any interpreter or compiler is permitted to
return NIL for the above form. What you're probably thinking of
is (EQ (QUOTE #(1 2 3)) (QUOTE #(1 2 3))), which might be true or false.
Once you (or anyone else in this somewhat rambling and inconclusive
discussion) think you understand the issues above, figure out what
you think about these three forms:
(EQ (FUNCTION (LAMBDA () 1)) (FUNCTION (LAMBDA () 1)))
(EQ (FUNCTION (LAMBDA (N) (DO ((I 1 (1+ I)) (A 0 (+ A I)))
((>= I N) A))))
(FUNCTION (LAMBDA (N) (DO ((I 0 (1+ I)) (A 0 (+ A I)))
((>= I N) A)))))
(EQ (FUNCTION (LAMBDA (N) (DO ((I 1 (1+ I)) (A 0 (+ A I)))
((>= I N) A))))
(FUNCTION (LAMBDA (N) (* N (1- N) 1/2))))
Kudos to the Scheme community for elucidating this undecidable
issue some years ago.
I couldn't get Symbolics CL
(which I thought WOULD do it), Vax CL, or KCL to put a "constant" into
read-only space. What implementations DO put constants in RO space?
Try (set :foo 105) in Symbolics. It will signal a write in read only error.
That's one kind of constant.
The Symbolics compiler doesn't currently put quoted constants in
compiled functions into read-only space, because it puts the constants
and the code on the same virtual memory page and there is currently a
stupid reason why compiled functions can't normally be put into
read-only space. I believe it used to make constants read-only a few
years ago. At some point the stupid reason will be fixed and then all
constants and compiled code will be read-only (and a certain
programmer/technical manager, who I won't name but who will recognize
himself, will have to stop writing self-modifying code!).
I think you'll find that in some systems compiled code and its constants
become read-only during a system building procedure, but not when you | 791 | 2,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-45 | longest | en | 0.886984 |
https://discuss.leetcode.com/topic/15329/shortest-c-solution-in-0ms | 1,513,437,159,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948588251.76/warc/CC-MAIN-20171216143011-20171216165011-00503.warc.gz | 532,562,224 | 10,062 | # Shortest C++ solution in 0ms
• Idea is to use vectors to keep track of invalid positions , so validity can be checked in O(1) and put a queen in each column
``````#include<vector>
using namespace std;
class Solution {
public:
int find(int n, int left, int i, int r, vector<int>&rows,vector<int>&d1,vector<int>&d2){
if (left == 0)
return 1;
int j,sum=0;
for (j=r; j<n; j++){
if (rows[j] || d1[i+j] || d2[n-1+i-j])
continue;
rows[j]=d1[i+j]=d2[n-1+i-j]=1;
sum += find(n, left-1, i+1, 0,rows,d1,d2 );
rows[j]=d1[i+j]=d2[n-1+i-j]=0;
}
return sum;
}
int totalNQueens(int n) {
vector<int> rows(n),d1(2*n-1),d2(2*n-1);
return find(n,n,0,0,rows,d1,d2);
}
};``````
• I share you with the same idea.
Here is my 0ms c++ solution:
``````class Solution {
public:
int totalNQueens(int n) {
std::vector<int> flag(5 * n - 2, 1);
int res = 0;
totalNQueens(flag, 0, n, res);
return res;
}
private:
void totalNQueens(std::vector<int> &flag, int row, int &n, int &res) {
if (row == n) {
++res;
return;
}
for (int col = 0; col != n; ++col)
if (flag[col] && flag[col + row + n] && flag[col - row + 4 * n - 2]) {
flag[col] = flag[col + row + n] = flag[col - row + 4 * n - 2] = 0;
totalNQueens(flag, row + 1, n, res);
flag[col] = flag[col + row + n] = flag[col - row + 4 * n - 2] = 1;
}
}
};
``````
Or:
``````class Solution {
public:
int totalNQueens(int n) {
std::vector<int> flag(5 * n - 2, 1);
}
private:
int totalNQueens(std::vector<int> &flag, int row, int &n) {
if (row == n)
return 1;
int res = 0;
for (int col = 0; col != n; ++col)
if (flag[col] && flag[col + row + n] && flag[col - row + 4 * n - 2]) {
flag[col] = flag[col + row + n] = flag[col - row + 4 * n - 2] = 0;
res += totalNQueens(flag, row + 1, n);
flag[col] = flag[col + row + n] = flag[col - row + 4 * n - 2] = 1;
}
return res;
}
};
``````
• The fourth argument r in find is unnecessary and you can remove it.
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 728 | 1,963 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2017-51 | latest | en | 0.374393 |
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# William
Last seen: Today Active since 2012
Semi-retired physicist with interests in numerical modeling and mathematics.
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https://www.physicsforums.com/threads/sailing-downwind-faster-than-the-wind-resolved.272437/ | 1,518,987,749,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812259.18/warc/CC-MAIN-20180218192636-20180218212636-00664.warc.gz | 942,166,517 | 24,978 | # Sailing downwind faster than the wind: resolved?
1. Nov 16, 2008
### mender
I've been following this on another forum and am planning to build a non-propeller design to sidestep the sailing references. I accept that the vehicles in the videos linked in the other two threads presented on this forum are doing what it looks like they are doing without any trickery. I also feel that the treadmill test is a valid way to test and refine a design. However, it appeared that there was some disagreement about that before the previous thread here was locked.
My understanding is that general tone of the disagreement is what caused the thread to be locked, not the topic or actual disagreement, so I'm hoping that this thread won't be automatically shut down. I've read and agreed to the conditions of this forum and will abide by the guidelines.
I feel I have a pretty good understanding of what is happening but would like to make sure that I haven't overlooked something that could affect the outcome of my experiments with a non-propeller design. I want to observe and measure the various forces and interactions of this under controlled conditions to optimize a small device before attempting to scale this up.
Is the treadmill test a valid substitute for an outdoor test? If not, why not?
2. Nov 16, 2008
### rcgldr
One way to explain why these carts work is to note that the power input is equal to the force at the driving wheels times the forwards speed of the car relative to the ground. After losses, the power output is equal to the thrust times the relative air flow through the prop, which is much slower than the ground speed. Via gearing, prop diameter, and prop pitch, the torque at the wheels is multiplied so that the prop generates more thrust than the force from the wheels, but at a much lower speed, so that power output remains well below power input. As long as the difference between wind speed and ground speed is large enough, (and perhaps not too large), the cart can go downwind faster than the wind, depending on the ratio of power output versus power input (efficiency factor), and the ratio of air flow speed through the prop versus ground speed.
3. Nov 16, 2008
### Staff: Mentor
Yes, I think so. One thing to note: on the treadmill, you start with the wheels moving. With the videos of these devices, they need a push to get them to work, which is the same thing.
4. Nov 16, 2008
### mender
I don't think it's required to get the cart started but is more dependent on the wind speed and the surface area of the prop. The Jack Goodman cart did start to roll on its own after the brake was released, but once the propeller is spinning it does seem to help noticeably.
I did notice that the smallest cart that was tested on the street in the wind caught enough of a gust that the wheels and prop were turning backwards briefly but that might have more to do with the traction of the drive wheels and the strength of the gust.
By the way, a nice long extended version of the the treadmill is available at some airports, enough to go from a stop relative to the moving surface without running out of room too quickly. I don't know if the speed of those walkways is sufficient though.
5. Nov 17, 2008
### yoavraz
Mender:
Yes, this has been resolved. In order to get there you have to tack downwind. If the best "velocity made good" (VMG) is greater than wind speed, at a given wind, then you "get there before the wind." Otherwise, no. Also, even having faster VMG you cannot make it if you go straight downwind (wind direction 180 degrees). You find best VMG by finding the point where the tangent to the sailboat-speed Vs. real-wind-angle graph is parallel to the horizontal axis of the graph (prove this). A positive example is with the 18ft Skiff class that has best VMG of ~12kn at 10kn wind. Data are from such polar diagram for skiff in the book "The symmetry of sailing" 1996. I believe that by now other fast classes can make it too. It is done easily by ice and land sailboats.
Last edited: Nov 17, 2008
6. Nov 17, 2008
### ThinAirDesign
Our smallest device works in winds down to 2.7mph. I believe most moving walkways are above that, but not by much.
I'd love to find a moving walkway that had a surface suitable to our small light wheels. Most of those walkways are a myriad of slots.
JB
7. Nov 19, 2008
### yoavraz
With regular sails or with other means the result is the same (my previous comment):
At 180 degrees wind (wind from behind), when reaching wind speed, the relative wind at the vehicle is 0. This is a fact that cannot be changed. With wind 0 no lift can be generated on the vehicle: either by sail, or wing, or wind turbine. If passing wind speed by some external means, e.g., push, immediately the relative wind is from front, and will apply a stopping force that will reduce speed, and so forth.
The only way to pass wind speed in the direction of the wind (180) is to tack (zigzag) downwind. If the vehicle has high lift and low drag, it is possible that the best VMG (velocity component in 180) is larger than wind speed (my previous comment), and the vehicle "gets to target before the wind." In this case, when zigzagging, always a side wind component, even small, must exist, i.e., wind direction is <180.
This applies to all videos with wind turbines that I have seen, with, or without treadmills. Neither magic nor unfamiliar physics.
8. Nov 19, 2008
### schroder
I agree with this 100%. Going Directly Downwind, Faster than the wind, (180 Deg) is not possible. This has been my stand all along. The best data I have seen was for an iceboat in a 35 mph wind doing 34.9 mph DDW. That is very impressive, but only proves my point that it is not possible. The treadmills are very confusing, in that merely advancing against the tread has convinced some people that this is equivalent to outrunning the wind. I was confused about this also. In retrospect, advancing against the tread by employing another medium, such as air, is no more amazing than a right-angled drill; just another way of redirecting force! I hope this issue is finally resolved.
9. Nov 19, 2008
### ThinAirDesign
Schroder, of course a traditionally equipped ice-boat can only do 34.9mph when going DDW in a 35mph wind. There's no lift involved when going DDW - it's always going to be Wind Speed minus Overall Drag with a traditional rig.
Now, let the ice-boat zig zag and their VMG (velocity made good, or the downwind component of their path) can exceed the speed of the wind dramatically
In land yachting, VMGs of 3x to 4x of true wind speed are common.
http://sports.groups.yahoo.com/group/2nalsa/message/161
Unfortunately, your point that "it is not possible", doesn't hold up to much scrutiny.
Perhaps they are, but here on a physics forum, where the principles of 'the equivalency of inertial frames of reference' should rule, I'm surprised to find you stlll clinging to the notion that a treadmill running at 10mph relative to the air is somehow different than a street with 10mph of of relative air passing over it.
And apparently, you still are.
The device in the videos is powered *only* by the relative motion between air and a rolling surface.
If the device advances on the treadmill running in a still air room, it is now running DDWFTTW (directly downwind faster than the wind).
If the device is hovering on the treadmill (that is no forward or rearward motion relative to the belt), it is running DDWTSSATW (directly downwind the same speed as the wind).
I'm not concerned about how "amazing" you do or don't find it. I'm am concerned when you say that it's not what it claims, which is DDWFTTW.
Again, a treadmill running at 10mph relative to the air is exactly the same as a street with 10mph of of relative air passing over it. I'm sticking with Galileo on this one.
I wish it was, but alas DDWFTTW is such a maddening brainteaser to some that it wouldn't be over if they were run over by a vehicle doing it.
Russ ... could you help our gentlemen friend Schroder here to a lesson on inertial frames of reference? He's not going to listen to me.
Thanks
JB
Last edited by a moderator: Sep 25, 2014
10. Nov 19, 2008
### Phrak
Actually, you could sale downwind, 180 degrees, faster than the wind--just not for very long. I have a computer model with variable L/Ds and L_sale/L_keel options. I was very suprised to find that it could sail downwind, and when setup, would perpetually sale in circles with a net downwind drift each cycle.
11. Nov 19, 2008
### schroder
12. Nov 19, 2008
### ThinAirDesign
Excellent. Glad to hear that. It's amazing how many folks will *not* concede that point even after 90+ years of it being physically demonstrated on ice and a couple decades on water.
For the record, I will state my two claims:
A: I have built and can demonstrate on demand a vehicle which will go directly downwind, faster than the wind, powered only by the wind, steady state.
B: Based on the principle of the equivalency of intertial frames of reference (IFOR), a treadmill in a still air room is a technically perfect environment to prove or disprove claim "A".
One can extract energy from the difference in speed between the air and the ground. One can use Galiliean relativity to prove that physics of the cart at fixed position on the treadmill are identical to the physics of the cart moving at wind speed on level ground. From that it follows that steady advancement against the rotation of the rolling surface constitutes a perfect and valid demonstration of claim "A".
And yet I have a device that regularly accomplishes it sitting right here on my desk and have demonstrated it countless times now in front of all sorts of folk.
That *is* where the issue is, and fortunately for me I have time tested principles of IFOR behind me. 10mph of relative airflow is just that "relative". I can create that relative movement anywhere I wish -- in the back of a truck, on the back of truck, on the highway, in a gymnasium, inside a space station, on the equator -- ad nauseam. The cart won't care and there isn't a single scientific test available to tell those apart.
Excellent, you have now devised a mechanism which will successfully climb a treadmill using relative motion between to solid surfaces. Congrats. Next time someone asks me if that can be done I will tell them "yes", Schroder devised such a system quite some time ago. Unfortunately, to go DDWFTTW, we need a device which exploits the relative motion between a solid and a gas, not two solids.
I didn't "extrapolate", I did it.
I did do it in the wind. 10mph of it to be exact.
JB
Last edited by a moderator: Nov 19, 2008
13. Nov 19, 2008
### M Grandin
I also didn't understand why earlier thread on this issue was locked - the discussion climate was rather polite, I thought.
Regarding this downwind device, I also was a little duped and doubting at first - but after thinking a while I realized it must work. But I think it is not necessary compare with sailing, tacking and so on. Although it could be accomplished using vectorized aerodynamics.
The theory behind is rather simple: Assume vehicle velocity = V1, wind velocity = V2 and "propeller" backwards projected pushing velocity V3 at force F. Received power Pr =
F x V1 , picked up by rotating wheels. Consumed power Pc = F x (V3 - V2) at driving "propeller". Net power received is Pr - Pc = F x (V1 +V2-V3) . All velocities related to +Z direction. Net power > 0 as long as V3 < V1 + V2. Or relative velocity
V3-V2 < V1.
Some error may have occured - but the core is that consumed power is lower than
received power because the propeller acts att lower relative velocity toward wind than the wheels against ground - at the same but opposite directed force F.
14. Nov 19, 2008
### mender
It appears that the treadmill test is in question again. That was the original question that I asked to start this thread.
Is the treadmill test a valid substitute for an outdoor test? If not, why not? Please state your view and explain your reasoning. I am not addressing DDWFTTW.
For the record, I feel that the treadmill test is valid because of the replication of a moving ground/air interface as experienced in a wind. The difference in speed and direction between the two is identical. The energy available is a result of the relative motion between the ground and the air.
A test of this would be to interchange the observer's perspective. Increasing the scale of the treadmill test in a still room would allow the observer to ride on the treadmill surface and measure the velocity of the devices mentioned as well as the air flow relative to the observer.
What would the riding observer get for measurements? If the treadmill were running at 10 mph and the air in the room was still, the observer would measure a wind of 10 mph and a ground speed of zero when the observer is sitting on the treadmill surface. If the cart were to move at the same speed as the walls of the room, it would appear to be moving at the same speed as the wind since it would be stationary relative to the air. The cart's speed would be measured as 10 mph forward relative to the viewer on the treadmill surface.
Let's freeze this for a moment and add a second observer outside the room watching through the window. What would the second observer see when we unfreeze the scene? They'd see the first observer moving backwards at 10 mph and the cart holding station in front of them. If the cart starts to move relative to the second observer, that movement is either added or subtracted from the speed that the first observer would be measuring the cart's progress at.
Have I presented this correctly?
Last edited: Nov 19, 2008
15. Nov 19, 2008
### ThinAirDesign
Thanks for keeping the thread on track mender.
Like you I am interested to hear peoples arguments for and against the application of IFOR in this case.
JB
PS: M_Grandin, I'm also happy to discuss DDWFTTW if you wish to PM me. My above comments were not meant to insult your thoughtful post.
16. Nov 19, 2008
### ThinAirDesign
mender, I certainly don't wish to patronize anyone but at the same time I don't wish to leave a stone unturned.
If you are not sure you understand the basics of IFOR -- the issue at the heart of the treadmill matter, I am happy to give you some good examples and explanations.
JB
17. Nov 19, 2008
### M Grandin
To me it is obviously the same thing - if you understand the "theory" behind the machine.
Imagine for instance the thread-mill moving at extremely high velocity. The slightest force
from propeller transferred to apparatus would generate a corresponding enormous power generated from apparatus wheeels. Power = Force x Velocity. While that slight force
would claim very small power extracted from wheel generator to rotate the propeller giving that slight force.
If vehicle is hold still the wheels are rotating att speed of thread-mill. Already a fraction of max power obtained that way may be sufficient running the propeller holding the vehicle still or accelerating passing the velocity of ambient wind = faster than downwind .
Last edited: Nov 19, 2008
18. Nov 19, 2008
### mender
I agree. The force that is available is from the relative movement between the surfaces. A method of harnessing and redirecting that force is all that is needed to provide movement of a device. It's a matter of gearing and total drag vs force as to how fast the device moves and in which direction.
Shroder provided an explanation of a similar interface. By using the relative motion between the treadmill and the stationary frame of the treadmill or the ground beside the treadmill, energy can be extracted and through the appropriate gearing cause the device to advance on the treadmill. The forces involved can be measured and the experiment is repeatable by anyone with the same equipment under the same conditions.
I accept this solid to solid or ground/ground interface as a valid test. If the device moved forward on the treadmill, it would be moving forward faster than both the treadmill surface and the treadmill frame or the supporting ground around it. It is an exchange of force in one direction for movement in another.
An intermediary step between ground/ground and ground/air would be ground/water. If a trough of water were to be placed around the treadmill and paddlewheels substituted for the wheels in shroder's device, the device would have a less direct link but would still move forward as long as the total drag from the device was less than the force generated.
To me, having a less obvious and more tenuous medium to work with does not negate the principles involved. Nor does interchanging which surface is moving relative to the observer, since the energy is extracted from the relative movement of the media. The ground/air interface works just like the ground/ground and ground/water but with less obvious interaction visible.
Last edited: Nov 19, 2008
19. Nov 19, 2008
### ThinAirDesign
You have presented it perfectly.
Well done.
Again perfectly put. Thanks.
JB
20. Nov 19, 2008
### yoavraz
Amazing reactions.
All experiments I have seen on videos do not prove the 180 possibility, to my opinion.
It is impossible in such conditions to keep the wind at 180 all the time. Even small fluctuations generate deviation from 180, which makes it equivalent to tacking at <180.
Even a fully controlled wind tunnel experiment with a treadmill, with laminar stable flow in the exact direction of treadmill and vehicle wheels, will be hard to convince me that motion above wind speed is generated without wind direction (possibly only minor) fluctuations around treadmill and vehicle direction. | 4,031 | 17,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-09 | longest | en | 0.969354 |
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# Activity 23B Teacher Notes: Voltage in a Parallel Circuit PS-2826
## Teacher Notes – Activity 23B: Voltage in a Parallel Circuit
Time Estimates Preparation: 20 min Activity: 30 min
Objectives
Students will be able to…
use a Voltage Probe to measure the voltage across small light bulbs and the voltage source in a
parallel circuit
compare the voltages across the light bulbs in a parallel circuit to the voltage of the voltage source
describe the relationship between the brightness of light bulbs in a parallel circuit and the number of
bulbs in the circuit
describe what happens to a parallel circuit when one of the light bulbs in the circuit is removed from
the circuit
Notes
It is possible to combine this activity with 23A Voltage in a Series Circuit. Both activities
use the same equipment.
Theoretically, the voltage of a simple circuit consisting of light bulbs and ‘D’ cells remains
constant. For light bulbs in parallel, the voltage across individual bulbs is the same as the
source voltage.
You do not need to calibrate the Voltage Probe for this activity.
Every simple circuit starts out as a series circuit, so the setup of one light bulb connected to
the voltage source for the first part of this activity is the same as the circuit setup in the
first part of the previous activity.
Background
An array of resistors will have different measured resistances depending on how they are
connected. If they are connected in series (end-to-end), their total resistance equals the sum of all
of their individual resistances. If light bulbs are connected in series to a voltage source, the
brightness of the individual bulbs diminishes as more and more bulbs are added to the “chain”.
The current decreases as the overall resistance increases. In addition, if one bulb is removed from
the “chain” the other bulbs go out.
If resistors are connected in parallel (side-by-side), their total
resistance is less than the sum of their individual resistances. In fact, 1 1 1 1
= + + + ...
the total resistance is related to the individual resistances as shown in R p R1 R2 R3
the equation where Rp is the total resistance:
If light bulbs are connected in parallel to a voltage source, the brightness of the individual bulbs
remains more-or-less constant as more and more bulbs are added to the “ladder”. The current
increases as more bulbs are added to the circuit and the overall resistance decreases. In addition,
if one bulb is removed from the “ladder” the other bulbs do not go out. Each bulb is
independently linked to the voltage source.
## Introductory Physics with the Xplorer GLX © 2006 PASCO p. 87
Activity 23B Teacher Notes: Voltage in a Parallel Circuit PS-2826
Sample Data
The first screenshot shows a sample of voltage across a light bulb. The second shows voltage
across the D cells.
## Introductory Physics with the Xplorer GLX © 2006 PASCO p. 88
Activity 23B Teacher Notes: Voltage in a Parallel Circuit PS-2826
## Lab Report - Activity 23B: Voltage in a Parallel Circuit
Predict
1. How would the voltages across each light bulb in a parallel circuit change as more bulbs
Since each light bulb in a parallel circuit is connected to the voltage source, the voltage for
each bulb should remain the same as more bulbs are added to the circuit.
2. How would the brightness of the light bulbs in a parallel circuit change as more bulbs are
The brightness of the light bulbs in a parallel circuit will not change as more bulbs are added to
the circuit.
3. If one bulb in a parallel circuit is removed, what happens to the rest of the bulbs?
In one bulb in a parallel circuit is removed, nothing happens to the rest of the bulbs.
Data
1. How bright were the two light bulbs in parallel compared to the first light bulb?
The two bulbs in parallel were each as bright as the first bulb.
2. How bright were the three light bulbs in parallel compared to the first light bulb?
All three bulbs in parallel were each as bright as the first bulb.
3. What happened in the three-light bulb parallel circuit when you removed the light bulb
from the middle lamp socket?
When the light bulb was removed from the middle lamp socket of the parallel circuit, the other
bulbs remained lit.
Data Table
One Light Bulb
Voltage Across Light Bulb: 2.81
Voltage Across Batteries: 2.82
## Two Light Bulbs in Parallel
Voltage Across Light Bulb 1: 2.62
Voltage Across Light Bulb 2: 2.59
Voltage Across Batteries: 2.68
## Three Light Bulbs in Parallel
Voltage Across Light Bulb 1: 2.30
Voltage Across Light Bulb 2: 2.29
Voltage Across Light Bulb 3: 2.28
Voltage Across Batteries: 2.51
## Introductory Physics with the Xplorer GLX © 2006 PASCO p. 89
Activity 23B Teacher Notes: Voltage in a Parallel Circuit PS-2826
Questions
1. How did the voltage across the two D cells compare to the voltage across the first light
bulb?
The voltage across the voltage source and the voltage across the first light bulb were almost the
same.
2. How did the voltage across each of the two light bulbs in parallel compare to the voltage
across the D cells?
The voltage across each of the two light bulbs in parallel was almost the same as the voltage
across the voltage source.
3. What did you notice about the voltage across each light bulb and the voltage across the D
cells when three bulbs are in parallel?
The voltage across each bulb in a parallel circuit is the same as the voltage source.
## 4. What can you say about the voltage in a parallel circuit?
The voltage across any branch in a parallel circuit is the same as the voltage source.
5. What happened to the light bulbs when you removed the middle bulb from the socket?
Why?
Nothing happens to the other light bulbs in a parallel circuit when the middle bulb is removed
because the other light bulbs still have a connection to the voltage source.
6. If all the lights in a house are connected together in parallel and they are all turned on,
what would happen to the lights when you turn one of them off (or it burns out)?
If all the lights in a house are connected in parallel and they are all turned on, they remain
turned on if one light is turned off (or it burns out). | 1,419 | 6,167 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2019-35 | latest | en | 0.914984 |
http://www.solutioninn.com/in-an-article-in-the-journal-of-advertising-weinberger-and-spotts | 1,503,344,912,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109525.95/warc/CC-MAIN-20170821191703-20170821211703-00132.warc.gz | 675,307,231 | 7,935 | # Question: In an article in the Journal of Advertising Weinberger and
In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and in the United Kingdom. Suppose that independent random samples of television ads are taken in the two countries. Arandom sample of 400 television ads in the United Kingdom reveals that 142 use humor, while a random sample of 500 television ads in the United States reveals that 122 use humor.
a. Set up the null and alternative hypotheses needed to determine whether the proportion of ads using humor in the United Kingdom differs from the proportion of ads using humor in the United States.
b. Test the hypotheses you set up in part a by using critical values and by setting a equal to .10, .05, .01, and .001. How much evidence is there that the proportions of U. K. and U. S. ads using humor are different?
c. Set up the hypotheses needed to attempt to establish that the difference between the proportions of U.K. and U.S. ads using humor is more than .05 ( five percentage points). Test these hypotheses by using a p- value and by setting α equal to .10, .05, .01, and .001. How much evidence is there that the difference between the proportions exceeds .05?
d. Calculate a 95 percent confidence interval for the difference between the proportion of U. K. ads using humor and the proportion of U.S. ads using humor. Interpret this interval. Can we be 95 percent confident that the proportion of U.K. ads using humor is greater than the proportion of U. S. ads using humor?
View Solution:
Sales1
Views223 | 354 | 1,617 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-34 | latest | en | 0.911215 |
https://www.physicsforums.com/threads/best-mathematica-tensor-general-relativity-package.872492/ | 1,709,195,498,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474795.48/warc/CC-MAIN-20240229071243-20240229101243-00348.warc.gz | 916,038,768 | 16,249 | # Best Mathematica Tensor/General Relativity Package?
• Mathematica
• xdrgnh
In summary, Mathematica Tensor/General Relativity Package is a software package that allows users to perform calculations and simulations related to tensor analysis and general relativity using the Mathematica programming language. Its main features include the ability to define and manipulate tensors, perform tensor calculus, solve tensorial equations, and simulate physical systems in curved space-time. While some basic knowledge is required to use it effectively, it offers a user-friendly interface and comprehensive documentation for beginners. It is also widely used for research purposes and there are alternatives available, but Mathematica Tensor/General Relativity Package remains a popular and well-established option.
#### xdrgnh
Hello can anyone recommend me a good mathematics package for solving the Einstein Field equations. You know one that can easily compute covariant derivatives and calculate the Reinman curvature tensor and can also minimize the Einstien Hilbert Action?
## 1. What is Mathematica Tensor/General Relativity Package?
Mathematica Tensor/General Relativity Package is a software package that allows users to perform calculations and simulations related to tensor analysis and general relativity using the Mathematica programming language. It provides a comprehensive set of tools for working with tensors and solving equations in curved space-time.
## 2. What are the main features of Mathematica Tensor/General Relativity Package?
The main features of Mathematica Tensor/General Relativity Package include the ability to define and manipulate tensors, perform tensor calculus, solve tensorial equations, and simulate physical systems in curved space-time using Einstein's field equations. It also offers visualization tools for displaying tensor fields and geometric objects.
## 3. Is Mathematica Tensor/General Relativity Package suitable for beginners?
While some basic knowledge of tensor analysis and general relativity is necessary to use Mathematica Tensor/General Relativity Package effectively, it provides a user-friendly interface and comprehensive documentation that can help beginners get started with the software. It also offers tutorials and examples to demonstrate its capabilities.
## 4. Can Mathematica Tensor/General Relativity Package be used for research purposes?
Yes, Mathematica Tensor/General Relativity Package is a powerful and versatile tool that is widely used by researchers and scientists for various applications in the fields of physics, astronomy, and mathematics. It provides a reliable and efficient platform for performing complex calculations and simulations related to tensor analysis and general relativity.
## 5. Are there any alternatives to Mathematica Tensor/General Relativity Package?
Yes, there are other software packages available for performing tensor analysis and simulating general relativity, such as SageMath, Cadabra, and xAct. However, Mathematica Tensor/General Relativity Package is a popular and well-established option, offering a comprehensive set of features and a user-friendly interface. The choice of software ultimately depends on the specific needs and preferences of the user. | 585 | 3,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-10 | latest | en | 0.858542 |
http://math.stackexchange.com/questions/21957/dy-dx-when-x-and-y-are-functions | 1,469,571,384,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825124.55/warc/CC-MAIN-20160723071025-00300-ip-10-185-27-174.ec2.internal.warc.gz | 160,800,233 | 20,184 | # dy/dx when x and y are functions
Let $x$ and $y$ be continuously differentiable real-valued functions defined on $\mathbf{R}$ and suppose that $x'(t) = f\,(x(t),y(t))$ and $y'(t) = g(x(t),y(t))$ where $f$ and $g$ are continuously differentiable on $\mathbf{R}^2$. According to the ODE book I am reading, if $f$ is never 0, then $\frac{dy}{dx} = \frac{g(x,y)}{f(x,y)}$ $y(x_0) = y_0$ where $x_0,y_0$ are constants, is an initial value problem.
Now what does $dy/dx$ mean here? Is it just the function $y'(x)$? Or is the author being sloppy and $x$ is actually a dummy variable? If the former is true, how is this an initial value problem? Normally, an initial value problem is given as $y'(t) = h(y(t),t)$, $y(t_0) = y_0$ where $t$ is variable.
PS. Why is it that ODE books are sloppy with notation and unshamefully nonrigorous?
-
Possible duplicate: math.stackexchange.com/questions/8040/… – Hans Lundmark Feb 14 '11 at 6:29
If $x$ and $y$ are functions of $t$, $\frac{\mathrm dy}{\mathrm dx}$ means, in all likelihood, the following: where possible by the inverse function theorem, write $t$ in terms of $x$; then $y=y(t)=y(t(x))$ is a function of $x$: differentiate that.
If you think about it a bit, you'll see that $\frac{\mathrm dy}{\mathrm dx}$ is in fact the ratio of change of $y$ with respect to changes in $x$, as it should be!
-
I did not think about the inverse function theorem. In any case, when you write $y(t(x))$, are you treating $x$ as a dummy variable? Your last statement gave me an idea: Perhaps $dy/dx$ is the function whose value at $t$ is just $y'(t)/x'(t)$? – echoone Feb 14 '11 at 15:29
@echoone, when I write $y(t(x))$ I mean the function resulting from composing $y$ with the inverse function of $t$. As for the last sentence of your comment: it is true that the function $dy/dx$ is equal to $y'(t)/x'(t)$, but not by definition: you actually have to prove that. – Mariano Suárez-Alvarez Feb 14 '11 at 16:15
The condition that f is never 0 is important .by the continuity of f,f must be always positive or negative ,which means x(t) is monotonous,so it has inverse function t(x),which is also smooth.so y(t)=y(t(x)).
by
-
Just wanted to comment ( but I don't have enough points), that , most likely, when you see a reference to "f never being 0" in problems of this sort; other than the obvious not wanting to divide by 0, this is a reference to the implicit/inverse function theorem, which guarantees a(n) (at least) local differentiable inverse. It then often ends up being a case of playing around with the chain rule until you get the right expressions.
-
I asked my ODE professor and he said that $dy/dx$ is the function whose value at $t$ gives the slope of the graph of $(x,y)$ at the point $(x(t), y(t))$. This means that $dy/dx$ is the function defined by $\frac{dy}{dx}(t) = \lim_{s \to t} \frac{y(t+s) - y(t)}{x(t+s) - x(t)}$ and this is why $(dy/dx)(t) = y'(t)/x'(t) = g(x(t),y(t))/f(x(t),y(t))$.
- | 899 | 2,955 | {"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} | 3.828125 | 4 | CC-MAIN-2016-30 | latest | en | 0.907805 |
https://justaaa.com/statistics-and-probability/443543-a-professor-wishes-to-estimate-the-variance-of | 1,721,448,968,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514981.25/warc/CC-MAIN-20240720021925-20240720051925-00543.warc.gz | 312,910,600 | 10,214 | Question
# A professor wishes to estimate the variance of student test scores. A random sample of 18...
A professor wishes to estimate the variance of student test scores. A random sample of 18 scores had a sample standard deviation of 10.4. Find a 98% confidence interval for the population variance and interpret it. The lower confidence limit is _________ and the upper confidence limit is ________ . What assumption was necessary to calculate this interval estimate? Answer to 2 decimal places (PLEASE MAKE THE ANSWER ACCURATE)
Given that, sample size (n) = 18 and
sample standard deviation (s) = 10.4
confidence level = 0.98
=> significance level = 1 - 98 = 0.02
Degrees of freedom = 18 - 1 = 17
Lower critical value is,
Upper critical value is,
The 98% confidence interval for population variance is,
Therefore, The lower confidence limit is 55.04 and the upper confidence limit is 286.94. | 214 | 906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-30 | latest | en | 0.825256 |
https://www.coursehero.com/file/6663034/410ahw11/ | 1,493,035,049,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917119356.19/warc/CC-MAIN-20170423031159-00317-ip-10-145-167-34.ec2.internal.warc.gz | 891,513,071 | 22,172 | 410ahw11
# 410ahw11 - HW#11 —Phys410—Fall 2011 Prof. Ted Jacobson...
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Unformatted text preview: HW#11 —Phys410—Fall 2011 Prof. Ted Jacobson Room 4115, (301)405-6020 www.physics.umd.edu/grt/taj/410a/ [email protected] S11.1 Harmonic oscillator using complex phase space coordiantes The Hamiltonian for a simple harmonic oscillator is H = p 2 / 2 + ω 2 x 2 / 2 in units where the mass m = 1. Let a be the complex phase space coordinate a = p ω/ 2( x + ip/ω ), and let a * be its complex conjugate. (a) Express H in terms of a and a * . (b) Evaluate the Poisson bracket { a,a * } , and use that to evaluate { a,H } and { a * ,H } . (c) Write and solve the equations of motion for a and a * using the Poisson bracket form of Hamilton’s equations. S11.2 Particle in a box with moving wall A particle of mass m moves in one dimension x between rigid walls at x = 0 and x = ‘ . (a) Using elementary mechanics, show that the average (outward) force on one of the walls is 2 E/‘ , where E is the (kinetic) energy of the particle. (b) Suppose now that the wall at x = ‘ is moved adiabatically. The energy of the particle then changes as a result of its collisions with the moving wall. Find the relation between δE and δ‘ , and use this to show that E‘ 2 is an adiabatic invariant. (c) Derive the same result instead using adiabatic invariance if H pdx . (Note that you could also find the invariant by dimensional analysis:....
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## This note was uploaded on 12/29/2011 for the course PHYSICS 410 taught by Professor Jacobson during the Fall '11 term at Maryland.
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Ask a homework question - tutors are online | 562 | 2,033 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-17 | longest | en | 0.862911 |
https://edurev.in/course/quiz/attempt/5737_Physics-Test-10-Heat-And-Thermodynamics/4f154f5b-2de7-4fe2-aaa5-c3b4fa210e61 | 1,656,840,415,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104215805.66/warc/CC-MAIN-20220703073750-20220703103750-00286.warc.gz | 277,507,302 | 40,113 | # Physics Test 10 - Heat And Thermodynamics
## 25 Questions MCQ Test Mock Test Series for JEE Main & Advanced 2022 | Physics Test 10 - Heat And Thermodynamics
Description
Attempt Physics Test 10 - Heat And Thermodynamics | 25 questions in 60 minutes | Mock test for JEE preparation | Free important questions MCQ to study Mock Test Series for JEE Main & Advanced 2022 for JEE Exam | Download free PDF with solutions
QUESTION: 1
### For Boyle’s law to hold good, the gas should be
Solution: For Boyle's law gases must obey ideal gas equation
PV=nRT
Boyles law states that for a fixed amount of gas at constant temperature pressure is inversely proportional to its volume.i.e.PV=constant
QUESTION: 2
### A bulb contains 1 mole of hydrogen and 1 mole of oxygen at temperature T. The ratio of rms values of velocity of hydrogen molecules to that of oxygen molecule is
Solution: RMS speed sqroot of 3RT/M M=molar mass T=temp R=cont used it
QUESTION: 3
### An ideal gas is heated from 270 to 6270 C at constant pressure. If initial volume was 4m3,then the final volume of gas will be
Solution:
QUESTION: 4
Critical temperature can be difined as the temperature
Solution:
QUESTION: 5
A molecule of mass m of an ideal gas collides with the wall of the vessel with the same velocity. The change in the linear momentum of the molecule will be
Solution:
Change in linear momentum = Final
Momentum-Initial momentum.
Momentum = mass x velocity
Initial velocity = v
Since molecule returns back with same velocity, final velocity = -v
Initial momentum = mv
Final momentum = -mv
Change in linear momentum is = -mv-(mv)=-2mv
QUESTION: 6
Rate of diffusion is
Solution:
QUESTION: 7
Which one of the following is not a thermodynamical co-ordinate ?
Solution:
QUESTION: 8
The mean kinetic energy of a molecule of a gas at 300 K is 6.21 x 10-21 The root mean square velocity of a molecule of H2 gas at this temperature will be(R=8.3 J/mole K,N= 6.02 x 1023 /mole, mass of the
hydrogen moleucle = 2 x 1.67 x 10-27 Kg)
Solution:
QUESTION: 9
At what temperautre will the rms velocity of oxygen moleucle be sufficient, so as to escape from the earth ? Escape velocity from the earth is 11.0 km/s and the mass of 1 moleucle of oxygen 5.34 x 10-26 (Boltzman constant k =1.38 x 10-23 J/K)
Solution:
QUESTION: 10
The temperature of inside and outside of a refrigerator are 273 K and 303 K respectively. Assuming that the regrigerator cycle is reverisble,for every joule of work done, the heat delivered to the surrounding will be nearly
Solution:
QUESTION: 11
A sample of gas expands from volume V1 to V2. The amount of work done by the gas is greatest, when the expansion is
Solution: While expanding the PV diagram shows drop in pressure for isothermal and even more in adiabatic process, but in isobaric process pressure remains constant ie parallel to V axis. the work done in P V diagram is area under PV curve between the limits of volumes. hence the area is maximum for isobaric then comes isothermal and adiabatic has the least work done for a given range of volumes.
QUESTION: 12
During the adiabatic expansion of 2 moles of a gas the internal energy of a gas is found to decrease by 2 J. The work done during the process on gas will be equal to
Solution: In any adiabatic process heat is not given or taken out of system , so there will be adiabatic expansion it will be on expense of its own internal energy so we see decrease in internal energy by first law of thermodynamics that when heat is given to a system it either is used in doing work or increasing internal energy, when no heat is given and system is expanding it has to spend it's own internal energy appearing as work done.
QUESTION: 13
Magnitude of slope of an adiabatic curve is always
Solution:
QUESTION: 14
Air is filled in a bottle at atmoshperic pressure and it is cordked at 350C. If the cork can come out at 3 atmospheric pressure then upto what temperature should the bottle be heated in order to remove the cork?
Solution:
P1 V1 / T1 = P2 V2 / T2
Volume is constant (bottle volume). So,
P1 / T1 = P2 / T2
1 / (273+35) = 3 / T2
So temperature to be heated is,
651 C.
At which bottle would have vaporized.
QUESTION: 15
A perfect gas goes from state A to another state B by absorbing 8 x 105 J of heat and doing 6.5 x 105 J of expernal work. It is now transfered between the same two states in another process in which it absorbs 105 J of heat. Then, in the second process
Solution: Now ΔQ=ΔW +ΔE
For first case, 8 x 10'5=6.5 x 10'5+ΔE
Therefore, ΔE=1.5 x 10'5 J
In the second case, ΔQ=10'5 J
Therefore ΔW=ΔQ-ΔE
=-0.5 x 10'5 so answer option A correct
QUESTION: 16
If 150 J of heat is added to a system and the work done by the system is 110 J. Then, change in internal energy will be
Solution:
QUESTION: 17
An ideal refirgator has a freezer at a temperature of -130C. The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) will be
Solution:
QUESTION: 18
A Carnot reversible engine converts 1/6 of heat input work. When the temperature of the sink is reduced by 62 K, the efficiency of Carnot’s cycle becomes 1/3. The temperature of the source and sink will be
Solution:
QUESTION: 19
When a solid is converted into a gas,directly by heating then this process is known as
Solution:
QUESTION: 20
The latent heat of vapourization of water is 2240 J.If the work done in the process of vapourization of 1 g is 168 J, then increase in internal energy will be
Solution:
*Answer can only contain numeric values
QUESTION: 21
A vessel contains a mixture of 7 g nitrogen and 11 g of carbondioxide at temperature T = 290K. If the pressure of the mixture is 1 atm (1.01 × 105 N/m2). Calculate its density in kg/m3:-
Solution:
*Answer can only contain numeric values
QUESTION: 22
To form a composite 16 μF, 1000V capacitor from a supply of identical capacitors marked 8 μF, 250 V, we require a minimum number of capacitors.
Solution:
*Answer can only contain numeric values
QUESTION: 23
If the height of transmitting and receiving antenna are 32 m and 50 m respectively, then for line of sight (LOS) propagation maximum distance between both antenna is (If radius of earth = 6400 km) (in km) :-
Solution:
*Answer can only contain numeric values
QUESTION: 24
Figure, shows a circuit in which three identical diodes are used. Each diode has forward resistance of 20Ω and infinite backward resistance. Resistors R1 = R2 = R3 = 50 Ω. Battery voltage is 6V. Find the value of current in mA through R3?
Solution:
Since D3 is reverse biased, there is not current through that branch
*Answer can only contain numeric values
QUESTION: 25
A horizontal pipe has a cross section of 10 cm2 in one region and of 5 cm2 in another. The water velocity at the first is 5 ms–1 and the pressure in the second is 2 × 105 Nm–2. Find the pressure of water in 1st region (in N/m2).
Solution:
Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code | 1,862 | 6,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2022-27 | latest | en | 0.807921 |
https://www.patriotichackers.com/2024/01/55-tools-of-geometry-answer-key.html | 1,708,739,986,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474482.98/warc/CC-MAIN-20240224012912-20240224042912-00081.warc.gz | 946,463,385 | 13,380 | # 55 Tools Of Geometry Answer Key
## Introduction
Geometry is a branch of mathematics that deals with the study of shapes, sizes, and properties of figures and spaces. It plays a significant role in various fields such as engineering, architecture, and design. To master geometry, it is crucial to have a thorough understanding of the concepts and principles involved. One useful resource that can aid in this process is the "Tools of Geometry" answer key, which provides step-by-step solutions to problems and exercises.
### The Importance of the "Tools of Geometry" Answer Key
The "Tools of Geometry" answer key serves as a valuable tool for students and educators alike. It offers comprehensive solutions to problems found in geometry textbooks, enabling students to check their work and learn from their mistakes. This resource not only helps students gain a deeper understanding of geometric concepts but also enhances their problem-solving skills. Educators can also utilize the answer key to create lesson plans and assessments, ensuring that students grasp the fundamental principles of geometry.
## Using the "Tools of Geometry" Answer Key
### 1. Finding Answers to Practice Problems
One of the primary uses of the "Tools of Geometry" answer key is to find answers to practice problems. Students can refer to the answer key after attempting a problem to check if their solution is correct. This helps in identifying any errors and allows for immediate feedback, promoting a deeper understanding of the concepts being tested.
### 2. Understanding the Steps and Techniques
The answer key not only provides the final answers but also includes the step-by-step process of arriving at the solution. This enables students to understand the methods and techniques used to solve geometric problems. By studying the solutions in the answer key, students can grasp the underlying principles and apply them to similar problems in the future.
### 3. Identifying Common Mistakes
Geometry can be a challenging subject, and students often make similar mistakes when solving problems. The answer key allows students to compare their approach with the correct solution, helping them identify common mistakes and misconceptions. By recognizing these errors, students can work on improving their problem-solving skills and avoid repeating the same mistakes in the future.
### 4. Enhancing Problem-Solving Skills
Geometry requires logical reasoning and critical thinking skills. The answer key provides students with additional practice opportunities, allowing them to apply the concepts they have learned. By solving a variety of problems and comparing their solutions with the answer key, students can develop their problem-solving skills and become more proficient in geometry.
## Maximizing the Benefits of the "Tools of Geometry" Answer Key
### 1. Actively Engage with the Material
Instead of simply relying on the answer key to check the answers, it is crucial for students to actively engage with the material. They should attempt the problems on their own first, and then refer to the answer key to verify their solutions. By actively participating in the problem-solving process, students can enhance their understanding and retention of geometric concepts.
### 2. Analyze the Solutions
After checking their answers, students should carefully analyze the solutions provided in the answer key. They should pay attention to the steps, techniques, and strategies used to arrive at the correct answers. By analyzing the solutions, students can gain insights into alternative approaches and develop a deeper understanding of the subject matter.
### 3. Seek Help if Needed
If students encounter difficulties or have questions while using the answer key, it is essential for them to seek help from their teachers or peers. Geometry can be complex, and it is normal to have doubts or uncertainties. By seeking assistance, students can clarify their doubts and gain a clearer understanding of the concepts.
### 4. Practice Regularly
Consistent practice is key to mastering geometry. Students should make use of the answer key to practice regularly and reinforce their understanding of the subject. By solving a wide range of problems and checking their answers using the answer key, students can build confidence in their abilities and improve their overall performance in geometry.
## Conclusion
The "Tools of Geometry" answer key is a valuable resource that can significantly aid students in their journey to master geometry. By utilizing this tool effectively, students can enhance their problem-solving skills, deepen their understanding of geometric concepts, and achieve success in their geometry studies. It is important to remember that the answer key should be used as a tool for learning and improvement, rather than a shortcut to obtaining answers. With consistent practice, active engagement, and a willingness to seek help when needed, students can maximize the benefits of the "Tools of Geometry" answer key and excel in their geometry studies. | 922 | 5,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-10 | longest | en | 0.941179 |
denilukman.blogspot.com | 1,503,242,331,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106779.68/warc/CC-MAIN-20170820150632-20170820170632-00509.warc.gz | 102,684,222 | 11,303 | ## Sunday, 31 May 2015
### Commutative, Associative, and Distributive Law in Modulo
I still remember the old days when we learned about the Distributive, Associative, and Commutative Law in Addition, Subtraction, Multiplication, and Division in the Elementary School. If you already forget what the law stated, here I describe again:
Commutative:
Commutative Law means we can change the position of the numbers and still get the same result.
Working on:
Addition: 5 + 6 = 6 + 5
Multiplication: 5 x 6 = 6 x 5
Not Working On:
Subtraction: 5-6 != 6-5
Division: 5/6 != 6/5
Associative:
Associative Law means we can change the group of the numbers and still get the same result.
Working on:
Addition: (5 + 6) + 7 = 5 + (6 + 7)
Multiplication: (5 x 6) x 7 = 5 x (6 x 7)
Division: (5 / 6) / 7 != 5 / (6 / 7)
Not Working On:
Subtraction: (5 - 6) - 7 != 5 - (6 - 7)
Distributive:
Distributed Law means we can distribute the number into group of the numbers and still get the same result.
Example:
5 x (6 + 5) = 5 x 6 + 5 x 5
How about the Modulo? Is the same law can working in the Modulo?
Commutative:
Not working: 7 % 5 % 3 != 7 % 3 % 5
Associative:
Not working: (7 % 5) % 3 != 7 % (5 % 3)
Distributive:
Not working if the group come second: 7 % (5 + 3) != (7 % 5) + (7 % 3)
But working if the group come first : (7 + 5) % 3 = (7 % 3) + (5 % 3) | 463 | 1,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2017-34 | longest | en | 0.824438 |
https://in.mathworks.com/matlabcentral/cody/problems/45392-convert-a-temperature-reading-from-celsius-to-an-unknown-scale/solutions/2179248 | 1,591,217,466,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347435987.85/warc/CC-MAIN-20200603175139-20200603205139-00537.warc.gz | 386,897,247 | 20,221 | Cody
# Problem 45392. Convert a temperature reading from Celsius to an unknown scale
Solution 2179248
Submitted on 28 Mar 2020 by Mehmed Saad
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
assert(isequal(celsius_to_franklin(605.86,942.86,701.08,873.83,981.92),670.23))
2 Pass
assert(isequal(celsius_to_franklin(-283.48,-820.99,34.93,-540.53,578.22),-61.99))
3 Pass
assert(isequal(celsius_to_franklin(-642.38,-545.91,-236.27,259.69,641.57),2001.06))
4 Pass
assert(isequal(celsius_to_franklin(-388.76,740.12,-355.52,996.42,156.00),4940.54))
5 Pass
assert(isequal(celsius_to_franklin(-424.57,-136.40,-544.47,-598.13,-454.91),-253.24))
6 Pass
assert(isequal(celsius_to_franklin(-943.67,428.22,-381.09,-96.63,823.88),-1220.79))
7 Pass
assert(isequal(celsius_to_franklin(205.93,437.77,-539.18,6.82,-447.39),59.91))
8 Pass
assert(isequal(celsius_to_franklin(863.18,284.69,-263.58,368.62,926.41),279.98))
9 Pass
assert(isequal(celsius_to_franklin(-147.74,127.01,-672.12,-960.23,-492.42),-587.64))
10 Pass
assert(isequal(celsius_to_franklin(-470.00,-330.01,245.92,32.44,368.50),94.50))
11 Pass
assert(isequal(celsius_to_franklin(-953.62,-685.32,-111.79,461.55,-660.24),-285.63))
12 Pass
assert(isequal(celsius_to_franklin(657.17,897.22,-335.17,803.76,-866.72),753.70))
13 Pass
assert(isequal(celsius_to_franklin(584.48,-166.12,259.41,-70.18,555.04),-157.43))
14 Pass
assert(isequal(celsius_to_franklin(-409.79,-416.75,-96.33,609.90,-841.00),-1829.06))
15 Pass
assert(isequal(celsius_to_franklin(-307.20,-48.77,366.97,569.72,-590.24),-308.43))
16 Pass
assert(isequal(celsius_to_franklin(-640.68,-365.85,-741.44,-757.17,-230.91),1225.57))
17 Pass
assert(isequal(celsius_to_franklin(-132.47,214.18,-277.77,782.82,-612.67),2093.47))
18 Pass
assert(isequal(celsius_to_franklin(-690.34,-308.03,216.70,-736.01,-355.91),-465.83))
19 Pass
assert(isequal(celsius_to_franklin(927.61,379.39,698.87,962.06,-538.43),4113.84))
20 Pass
assert(isequal(celsius_to_franklin(-886.01,-463.51,756.77,803.12,87.47),287.07))
21 Pass
assert(isequal(celsius_to_franklin(-502.42,-588.56,-206.72,-98.65,321.02),775.70))
22 Pass
assert(isequal(celsius_to_franklin(-153.74,7.78,-682.05,-719.25,120.16),384.71))
23 Pass
assert(isequal(celsius_to_franklin(-144.83,-134.94,-189.12,-37.86,-515.74),678.06))
24 Pass
assert(isequal(celsius_to_franklin(-995.20,151.44,-741.17,-470.43,-406.85),-1288.85))
25 Pass
assert(isequal(celsius_to_franklin(871.26,-14.30,236.99,-926.20,-443.03),-1903.88))
26 Pass
assert(isequal(celsius_to_franklin(715.15,782.47,47.57,-466.79,44.72),-472.12))
27 Pass
assert(isequal(celsius_to_franklin(899.12,-837.45,-191.19,-256.33,-293.28),-201.92))
28 Pass
assert(isequal(celsius_to_franklin(-202.59,-537.15,-192.74,407.01,299.90),47628.43))
29 Pass
assert(isequal(celsius_to_franklin(913.66,334.21,33.59,-112.18,-55.21),-157.22))
30 Pass
assert(isequal(celsius_to_franklin(955.44,756.25,-738.91,-848.13,114.84),-39.71))
31 Pass
assert(isequal(celsius_to_franklin(-666.83,-718.81,55.93,-298.83,-586.09),-671.89))
32 Pass
assert(isequal(celsius_to_franklin(147.36,-107.49,37.96,373.14,543.46),-1847.69))
33 Pass
assert(isequal(celsius_to_franklin(-187.90,-485.31,-936.87,953.10,-349.60),-174.76))
34 Pass
assert(isequal(celsius_to_franklin(-341.01,93.29,-190.04,507.03,-51.48),886.76))
35 Pass
assert(isequal(celsius_to_franklin(584.80,435.13,-16.48,-899.54,29.33),-797.85))
36 Pass
assert(isequal(celsius_to_franklin(-340.40,903.99,-371.65,-204.97,884.36),44366.71))
37 Pass
assert(isequal(celsius_to_franklin(-781.65,-583.60,127.07,910.74,-822.91),-651.45))
38 Pass
assert(isequal(celsius_to_franklin(-386.75,-935.79,531.29,-280.34,-408.39),-951.24))
39 Pass
assert(isequal(celsius_to_franklin(161.90,440.48,-210.43,-49.91,269.72),582.49))
40 Pass
assert(isequal(celsius_to_franklin(-748.00,558.83,611.62,-70.60,-33.59),228.10))
41 Pass
assert(isequal(celsius_to_franklin(657.96,-975.96,777.84,21.10,-407.55),-9837.97))
42 Pass
assert(isequal(celsius_to_franklin(-230.46,-919.89,-284.33,499.32,-234.82),-805.03))
43 Pass
assert(isequal(celsius_to_franklin(-301.12,-825.93,814.58,-552.94,507.82),-628.00))
44 Pass
assert(isequal(celsius_to_franklin(697.75,701.18,10.89,34.27,653.33),658.05))
45 Pass
assert(isequal(celsius_to_franklin(280.00,888.78,-786.06,403.46,708.18),1083.71))
46 Pass
assert(isequal(celsius_to_franklin(-10.67,543.28,264.36,637.92,894.65),854.81))
47 Pass
assert(isequal(celsius_to_franklin(-206.85,153.56,-128.64,-453.56,89.79),-2149.16))
48 Pass
assert(isequal(celsius_to_franklin(-76.51,-747.71,305.05,982.05,-576.62),-3014.90))
49 Pass
assert(isequal(celsius_to_franklin(292.10,-573.74,-958.20,-149.66,113.10),-513.03))
50 Pass
assert(isequal(celsius_to_franklin(792.56,-79.19,-775.41,-838.95,-698.15),-801.51))
51 Pass
assert(isequal(celsius_to_franklin(-396.81,922.69,629.52,-216.29,678.00),-270.09))
52 Pass
assert(isequal(celsius_to_franklin(-517.35,852.83,-16.57,-944.12,849.97),-4053.53))
53 Pass
assert(isequal(celsius_to_franklin(-434.10,504.36,-908.25,-132.70,317.96),1514.82))
54 Pass
assert(isequal(celsius_to_franklin(829.18,913.01,168.51,348.27,731.39),829.42))
55 Pass
assert(isequal(celsius_to_franklin(-333.40,-166.89,-456.37,639.93,-427.43),450.05))
56 Pass
assert(isequal(celsius_to_franklin(-294.90,-60.17,550.47,304.60,671.10),356.65))
57 Pass
assert(isequal(celsius_to_franklin(485.15,789.20,766.49,210.31,465.73),829.16))
58 Pass
assert(isequal(celsius_to_franklin(203.09,-17.32,914.47,-533.31,-199.58),274.75))
59 Pass
assert(isequal(celsius_to_franklin(966.57,445.31,-794.28,-130.59,-831.11),-142.64))
60 Pass
assert(isequal(celsius_to_franklin(-738.77,-731.47,-714.39,984.04,-269.54),32286.12))
61 Pass
assert(isequal(celsius_to_franklin(930.51,64.10,-449.17,775.23,87.89),498.41))
62 Pass
assert(isequal(celsius_to_franklin(-868.15,640.06,347.72,-454.46,17.03),-156.77))
63 Pass
assert(isequal(celsius_to_franklin(-937.82,569.83,-404.13,866.50,-171.47),995.83))
64 Pass
assert(isequal(celsius_to_franklin(-169.08,901.62,-131.93,-661.45,-597.44),18924.68))
65 Pass
assert(isequal(celsius_to_franklin(-189.62,-713.35,-270.02,-220.67,637.73),-5783.24))
66 Pass
assert(isequal(celsius_to_franklin(492.42,-319.45,141.79,288.15,113.21),337.68))
67 Pass
assert(isequal(celsius_to_franklin(283.90,-538.88,-437.08,918.34,-115.93),269.24))
68 Pass
assert(isequal(celsius_to_franklin(-306.06,561.45,469.95,418.77,357.52),439.44))
69 Pass
assert(isequal(celsius_to_franklin(-750.15,755.17,-347.75,-855.09,549.57),-4445.84))
70 Pass
assert(isequal(celsius_to_franklin(-522.01,440.91,261.38,-459.71,195.94),-384.48))
71 Pass
assert(isequal(celsius_to_franklin(741.61,107.09,454.92,-904.42,-603.83),-4639.94))
72 Pass
assert(isequal(celsius_to_franklin(91.21,547.62,235.88,78.98,176.30),271.98))
73 Pass
assert(isequal(celsius_to_franklin(970.32,-331.81,24.95,989.19,396.94),469.39))
74 Pass
assert(isequal(celsius_to_franklin(573.18,-145.55,-501.14,406.38,-809.38),564.74))
75 Pass
assert(isequal(celsius_to_franklin(674.12,182.10,-769.93,-438.99,216.83),-14.58))
76 Pass
assert(isequal(celsius_to_franklin(-501.54,364.36,122.84,736.62,105.33),726.18))
77 Pass
assert(isequal(celsius_to_franklin(423.69,-98.78,-153.62,-130.92,663.06),-85.45))
78 Pass
assert(isequal(celsius_to_franklin(796.67,-66.87,908.68,-989.81,987.42),-1638.61))
79 Pass
assert(isequal(celsius_to_franklin(652.08,797.79,-377.24,-59.91,-383.42),-65.06))
80 Pass
assert(isequal(celsius_to_franklin(-965.37,-955.60,-397.18,361.85,-445.66),249.44))
81 Pass
assert(isequal(celsius_to_franklin(47.99,766.77,932.43,-754.90,521.98),-48.72))
82 Pass
assert(isequal(celsius_to_franklin(511.24,231.31,-85.92,-985.28,-611.87),-2056.79))
83 Pass
assert(isequal(celsius_to_franklin(768.55,-217.35,-262.56,-220.12,515.30),-218.03))
84 Pass
assert(isequal(celsius_to_franklin(-413.15,-952.69,133.75,-922.77,-505.69),-957.75))
85 Pass
assert(isequal(celsius_to_franklin(188.47,610.83,837.90,81.43,134.78),654.60))
86 Pass
assert(isequal(celsius_to_franklin(-574.69,934.83,-668.57,-702.77,408.89),18091.95))
87 Pass
assert(isequal(celsius_to_franklin(-485.25,-858.09,-316.38,869.88,-25.16),3849.80))
88 Pass
assert(isequal(celsius_to_franklin(-723.68,-261.45,-214.48,-33.00,-356.68),-96.80))
89 Pass
assert(isequal(celsius_to_franklin(264.40,131.68,-304.40,-659.34,772.14),837.78))
90 Pass
assert(isequal(celsius_to_franklin(-927.25,687.47,846.53,-842.40,758.80),-766.73))
91 Pass
assert(isequal(celsius_to_franklin(-874.43,-518.50,179.50,46.43,232.70),74.95))
92 Pass
assert(isequal(celsius_to_franklin(-541.26,-857.08,-142.04,-777.46,25.32),-744.08))
93 Pass
assert(isequal(celsius_to_franklin(363.20,879.66,545.24,-99.49,-34.24),3017.40))
94 Pass
assert(isequal(celsius_to_franklin(166.07,415.49,693.31,-912.26,-204.72),1349.25))
95 Pass
assert(isequal(celsius_to_franklin(587.20,644.09,-450.02,764.92,143.15),695.82))
96 Pass
assert(isequal(celsius_to_franklin(-881.13,477.87,733.74,533.48,346.53),520.15))
97 Pass
assert(isequal(celsius_to_franklin(21.10,833.97,33.12,94.93,-522.20),34238.33))
98 Pass
assert(isequal(celsius_to_franklin(129.18,-721.79,-176.17,715.01,589.38),-2887.22))
99 Pass
assert(isequal(celsius_to_franklin(453.99,259.96,-596.74,279.21,-811.28),283.14))
100 Pass
assert(isequal(celsius_to_franklin(369.37,958.33,-425.57,-338.45,769.98),1611.84)) | 4,055 | 9,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-24 | latest | en | 0.547283 |
http://www.careerride.com/mcq-daily/dynamics-of-machinery-test-questions-set-1-328.aspx | 1,487,890,528,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171251.27/warc/CC-MAIN-20170219104611-00399-ip-10-171-10-108.ec2.internal.warc.gz | 353,730,386 | 11,304 | Dynamics of Machinery Test Questions - Set - 1
1) Which of the following conditions should be satisfied for static balancing?
1. Dynamic forces acting on the system should be zero
2. Couple acting on the system due to dynamic force should be zero
3. Centrifugal forces acting on the system should be zero
4. Couple acting on the system due to centrifugal forces should be zero
a. Condition 1 and condition 2
b. Condition 1 and condition 3
c. Condition 3 and condition 4
d. All of the above
Answer Explanation Related Ques ANSWER: Condition 1 and condition 3 Explanation: No explanation is available for this question!
2) Which of the following factors are not responsible for unbalancing in rotating systems?
a. Errors
b. Tolerances
c. Shape of the rotor
d. None of the above
Answer Explanation Related Ques ANSWER: None of the above Explanation: No explanation is available for this question!
3) Determine magnitude of balancing mass required if 250 mm is the radius of rotation. Masses of A, B and C are 300 kg, 250 kg and 100 kg which have radii of rotation as 50 mm, 80 mm and 100 mm respectively . The angles between the consecutive masses are 110o and 270o respectively.
a. 45.36 kg
b. 47.98 kg
c. 40.50 kg
d. None of the above
Answer Explanation Related Ques ANSWER: 47.98 kg Explanation: No explanation is available for this question!
4) The unbalanced force caused due to reciprocating mass is given by the equation
a. mrω2 sin θ + mrω2 (sin 2θ/n)
b. mrω2 sin θ + mrω2 (cos 2θ/n)
c. mrω2 cos θ + mrω2 (cos 2θ/n)
d. mrω2 (sin θ + sin 2θ/n)
Answer Explanation Related Ques ANSWER: mrω2 cos θ + mrω2 (cos 2θ/n) Explanation: No explanation is available for this question!
5) Which among the following is the fundamental equation of S.H.M.?
a. x + (k / m) x =0
b. x + ω2x =0
c. x + (k/ m)2 x =0
d. x2 + ωx2 =0
Answer Explanation Related Ques ANSWER: x + ω2x =0 Explanation: No explanation is available for this question!
6) What are discrete parameter systems?
a. Systems which have infinite number of degree of freedom
b. Systems which have finite number of degree of freedom
c. Systems which have no degree of freedom
d. None of the above
Answer Explanation Related Ques ANSWER: Systems which have finite number of degree of freedom Explanation: No explanation is available for this question!
7) In which type of vibrations, amplitude of vibration goes on decreasing every cycle?
a. Damped vibrations
b. Undamped vibrations
c. Both a. and b.
d. None of the above
Answer Explanation Related Ques ANSWER: Damped vibrations Explanation: No explanation is available for this question!
8) Which of the following vibrations are classified according to magnitude of actuating force?
a. Torsional vibrations
b. Deterministic vibrations
c. Transverse vibrations
d. All of the above
Answer Explanation Related Ques ANSWER: Deterministic vibrations Explanation: No explanation is available for this question!
9) What are deterministic vibrations?
a. Vibrations caused due to known exciting force
b. Vibrations caused due to unknown exciting force
c. Vibrations which are aperiodic in nature
d. None of the above
Answer Explanation Related Ques ANSWER: Vibrations caused due to known exciting force Explanation: No explanation is available for this question!
10) According to which method, maximum kinetic energy at mean position is equal to maximum potential energy at extreme position?
a. Energy method
b. Rayleigh's method
c. Equilibrium method
d. All of the above
Answer Explanation Related Ques ANSWER: Rayleigh's method Explanation: No explanation is available for this question!
11) Which among the following is the value of static deflection (δ) for a fixed beam with central point load?
a. (Wl3) /(192 EI)
b. (Wl2) /(192 EI)
c. (Wl3) /(384 EI)
d. None of the above
Answer Explanation Related Ques ANSWER: (Wl3) /(192 EI) Explanation: No explanation is available for this question!
12) δ = (W a2b2) / (3 EIl) is the value of deflection for ______
a. simply supported beam which has central point load
b. simply supported beam which has eccentric point load
c. simply supported beam which has U.D.L. point load per unit length
d. fixed beam which has central point load
Answer Explanation Related Ques ANSWER: simply supported beam which has eccentric point load Explanation: No explanation is available for this question!
13) Determine natural frequency of a system, which has equivalent spring stiffness of 30000 N/m and mass of 20 kg?
a. 12.32 Hz
b. 4.10 Hz
c. 6.16 Hz
d. None of the above
Answer Explanation Related Ques ANSWER: 6.16 Hz Explanation: No explanation is available for this question!
14) Which formula is used to calculate mass moment of inertia (IG) of a circular rim about the axis through centre of gravity?
a. mr2/2
b. mr2/12
c. mr2/4
d. mr2
Answer Explanation Related Ques ANSWER: mr2 Explanation: No explanation is available for this question!
15) The equation m(d2x/ dt2) + c (dx/dt) + Kx = F0 sin ωt is a second order differential equation. The solution of this linear equation is given as
a. complementary function
b. particular function
c. sum of complementary and particular function
d. difference of complementary and particular function
Answer Explanation Related Ques ANSWER: sum of complementary and particular function Explanation: No explanation is available for this question!
16) What is meant by phase difference or phase angle in forced vibrations?
a. Difference between displacement vector (xp) and velocity vector Vp
b. Angle in which displacement vector leads force vector by (F0 sinωt)
c. Angle in which displacement vector (xp) lags force vector (F0 sinωt)
d. None of the above
Answer Explanation Related Ques ANSWER: Angle in which displacement vector (xp) lags force vector (F0 sinωt) Explanation: No explanation is available for this question!
17) Consider the steady-state absolute amplitude equation shown below, if ω / ωn = √2 then amplitude ratio (X/Y) =?
(X/Y) = √{1 + [ 2ξ (ω/ωn)]2} / √{[1 – (ω/ωn)2]2 + {2ξ (ω/ωn)2}
a. 0
b. 1
c. less than 1
d. greater than 1
Answer Explanation Related Ques ANSWER: 1 Explanation: No explanation is available for this question!
18) Magnification factor is the ratio of ______
a. zero frequency deflection and amplitude of steady state vibrations
b. amplitude of steady state vibrations and zero frequency deflection
c. amplitude of unsteady state vibrations and zero frequency distribution
d. none of the above
Answer Explanation Related Ques ANSWER: amplitude of steady state vibrations and zero frequency deflection Explanation: No explanation is available for this question!
19) Which of the following statements is/are true?
1. Magnification factor is minimum at resonance
2. The maximum value of amplification factor increases as damping factor decreases
3. The maximum value of amplification factor increases as damping factor increases
4. Magnification factor is maximum at resonance
a. Statement 1 and statement 2
b. Statements 1,2 and 3
c. Statement 2 and statement 4
d. All the above statements are true
Answer Explanation Related Ques ANSWER: Statement 2 and statement 4 Explanation: No explanation is available for this question!
20) A vertical circular disc is supported by a horizontal stepped shaft as shown below. Determine equivalent length of shaft when equivalent diameter is 20 mm.
a. 1.559 m
b. 0.559 m
c. 0.633 m
d. None of the above
Answer Explanation Related Ques ANSWER: 0.559 m Explanation: No explanation is available for this question!
21) What is meant by geometric modeling?
a. Representation of an object with graphical information
b. Representation of an object with non-graphical information
c. Both a. and b.
d. None of the above
Answer Explanation Related Ques ANSWER: Both a. and b. Explanation: No explanation is available for this question!
22) Simulation is a process which _________
a. involves formation of a prototype
b. explores behavior of a model by varying input variables
c. develops geometry of an object
d. all of the above
Answer Explanation Related Ques ANSWER: explores behavior of a model by varying input variables Explanation: No explanation is available for this question!
23) In the diagram shown below, if rotor X and rotor Z rotate in same direction and rotor Y rotates in opposite direction, then specify the type of node vibration.
a. Three node vibration
b. Two node vibration
c. Single node vibration
d. None of the above
Answer Explanation Related Ques ANSWER: Two node vibration Explanation: No explanation is available for this question!
24) Which of the following statements is/are true?
a. Torsional vibrations do not occur in a three rotor system, if rotors rotate in same direction
b. Shaft vibrates with maximum frequency when rotors rotate in same direction
c. Zero node behavior is observed in rotors rotating in opposite direction
d. All of the above
Answer Explanation Related Ques ANSWER: Torsional vibrations do not occur in a three rotor system, if rotors rotate in same direction Explanation: No explanation is available for this question!
25) Seismometer can be used to measure acceleration of any instrument only if _____
a. it's natural frequency is high
b. it generates output signal which is proportional to relative acceleration of the vibrating object
c. both a. and b.
d. none of the above
Answer Explanation Related Ques ANSWER: both a. and b. Explanation: No explanation is available for this question!
26) Which type of monitoring system uses stroboscope to measure speed of the machine?
a. Portable condition monitoring system
b. Basic condition monitoring system
c. Computer based condition monitoring system
d. None of the above
Answer Explanation Related Ques ANSWER: Basic condition monitoring system Explanation: No explanation is available for this question!
27) In graph of (Transmissibility vs Frequency Ratio) shown below, the shaded region is called as____
a. Spring controlled region
b. Damping controlled region
c. Mass controlled region
d. None of the above
Answer Explanation Related Ques ANSWER: Mass controlled region Explanation: No explanation is available for this question!
28) Which of the following vibrometers have frequency ratio (ω/ωn) << 1?
a. Accelerometers
b. Velometers
c. Both a. and b.
d. None of the above
Answer Explanation Related Ques ANSWER: Accelerometers Explanation: No explanation is available for this question!
29) Which of the following conditions is/are to be satisfied by the seismometer for it to be used as velometer?
a. It's natural frequency should be large
b. It's natural frequency should be small
c. It's output signal should be proportional to relative acceleration of the vibrating body
d. None of the above
Answer Explanation Related Ques ANSWER: It's natural frequency should be small Explanation: No explanation is available for this question!
30) Which of the following methods can be used to reduce excitation level at the source?
a. Lubrication of joints
b. Balancing inertia forces
c. Both a. and b.
d. None of the above
Answer Explanation Related Ques ANSWER: Both a. and b. Explanation: No explanation is available for this question!
31) Which of the following is a type of untuned vibration absorber?
a. Houdaille damper
b. Torsional vibration absorber
c. Centrifugal pendulum absorber
d. All of the above
Answer Explanation Related Ques ANSWER: Houdaille damper Explanation: No explanation is available for this question!
32) In the graph shown below, the region in which frequency ratio (ω/ωn) > √2 is known as____
a. Amplification region
b. Isolation region
c. Spring controlled region
d. None of the above
Answer Explanation Related Ques ANSWER: Isolation region Explanation: No explanation is available for this question!
33) Which basic document describes general requirements for measurement and evaluation of machine vibrations using shaft measurements?
a. ISO – 10816-1
b. ISO – 7919-1
c. Both a. and b.
d. None of the above
Answer Explanation Related Ques ANSWER: ISO – 7919-1 Explanation: No explanation is available for this question!
34) Which among the following is not considered when reference standards are used in the field of mechanical vibration and shock, monitoring and analysis of machines?
a. Terminology
b. Methods of testing
c. Methods of measurement
d. None of the above
Answer Explanation Related Ques ANSWER: None of the above Explanation: No explanation is available for this question! | 3,098 | 12,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-09 | longest | en | 0.727161 |
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# Correlation Objectives i Calculate correlations i Calculate correlations for subgroups using split file i Create scatterplots with lines of best fit for subgroups and multiple correlations Correlati
Correlation Objectives i Calculate correlations i Calculate correlations for subgroups using split file i Create scatterplots with lines of best fit for subgroups and multiple correlations Correlation The first infer
## Correlation Objectives i Calculate correlations i Calculate correlations for subgroups using split file i Create scatterplots with lines of best fit for subgroups and multiple correlations Correlati
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## Presentation on theme: "Correlation Objectives i Calculate correlations i Calculate correlations for subgroups using split file i Create scatterplots with lines of best fit for subgroups and multiple correlations Correlati"— Presentation transcript:
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5. Correlation Objectives i Calculate correlations i Calculate correlations for subgroups using split file i Create scatterplots with lines of best fit for subgroups and multiple correlations Correlation The first inferential statistic we will focu s on is correlation. As noted in the text, correlation is used to test the degree of association between variables. All of the LQIHUHQWLDOVWDWLVWLFVFRPPDQGVLQ6366DUHDFFHVVHGIURPWKH\$QDO\]HPHQX/HWVRSHQ SPSS and
replicate the correlation betw een height and we ight presented in the text. D Open HeightWeight .sav . Take a moment to review the data file. D Under Analyze , select Correlate/Bivariate . Bivariate means we are examining the simple association between 2 variables. D In the dialog box, select height and weight f or Variables . Select Pearson for Correlation Coefficients since the data are continuous. The default for Tests of Significance is Two tailed . You could change it to One tailed if you have a directional hypothesis. Selecting Flag significant correlation s means that the significant
correlations will be noted in the output by asterisks. This is a nice feature. Then click Options
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For example I may run correlations between height, weight, and blood pressure. One subject may be missing blood pressure data. If I check Exclude cases listwise , SPSS will QRWLQFOXGHWKDWSHUVRQV data in the correlation between height and weight, even though those data are not missing. If I check Exclude cases pairwise, SPSS will include that
SHUVRQVGDWDWRFDOFXODWHDQ\FRUUHODWLRQVWKDWGRQRWLQYROYHGEORRGSUHVVXUH,QWKLV case, the perso QVGDWDZRXOGVWLOOEHUHIOHFWHGLQWKHFRUUHODWLRQEHWZHHQKHLJKWDQG weight. You have to decide whether or not you want to exclude cases that are missing any data from all analyses. (Normally it is much safer to go with listwise deletion, even though i t will reduce your sample size.) In this
case, it GRHVQWPDWWHUEHFDXVHWKHUHDUH no missing data. Click Continue . When you return to the previous dialog box, click Ok . The output follow. Correlations Descriptive Statistics 68.72 3.66 92 145.15 23.74 92 HEIGHT WEIGHT Mean Std. Dev iation Correlations 1.000 .785 ** .000 92 92 .785 ** 1.000 .000 92 92 Pearson Correlation Sig. (2-tailed) Pearson Correlation Sig. (2-tailed) HEIGHT WEIGHT HEIGHT WEIGHT Correlation is signif icant at the 0.01 lev el (2-tailed). **. Notice, the correlation coefficient is .785 and is statistically significant, just as
reported in the text. In the text , Howell made the po int that heterogeneous samples a ffect FRUUHODWLRQFRHIILFLHQWV,QWKLVH[DPSOHZHLQFOXGHGERWKPDOHVDQGIHPDOHV/HWV examine the correlation separately for males and females as was done in the text. D Now you can see how descriptive statistics are built into other menus. Select Means and standard deviations under Statist ics . Missing Values are important. In large data sets, pieces of data are often missing for
some variables.
Page 3
Subgroup Correlations We need to get SPSS to calculate the correlation between height and weight separately for males and females. The easiest way to do this is to split our data file by VH[/HWVWU\WKL s together. D In the Data Editor window, select Data/Split file D Notice that the order of the data file has been changed. It is now sorted by Gender , with males at the top of the file. D Now, select Analyze/Correlation/Bivariate . The same variables and options you selected last time are still in the dialog box.
Take a moment to check to see for yourself. Then, click Ok . The output follow broken down by males and females. Correlations SEX = Male Descriptive Statistics 70.75 2.58 57 158.26 18.64 57 HEIGHT WEIGHT Mean Std. Dev iation SEX = Male a. D Select Organize output by groups and Groups Based on Gender . This means that any analyses you specify will be run separately for males and fema les. Then, click Ok
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SEX = Female Descriptive Statistics 65.40 2.56 35 123.80 13.37 35 HEIGHT WEIGHT Mean Std. Dev iation SEX = Female a. As before, our results rep licate those in the text . The
correlation between height DQGZHLJKWLVVWURQJHUIRUPDOHVWKDQIHPDOHV1RZOHWVVHHLIZHFDQFUHDWHDPRUH complicated scatterplot that illustrates the pattern of correlation for males and females on one graph. First, we need to turn off split file. D Select Data/Split file from the Data Editor window. Then select Analyze all cases, do not compare groups and click Ok . Now, we can proceed. Scatterplots of Data by Subgroups D Select Graphs/ Legacy/ Scatter .
Then, select Simple and click Define
Page 5
When your graph appears, you will see that the only way males and females are GLVWLQFWIURPRQHDQRWKHULVE\FRORU7KLVGLVWLQFWLRQPD\QRWVKRZXSZHOOVROHWVHGLW the graph. D Double click the graph to activate the Ch art Editor. Then double click on one of the female dots on the plot. SPSS will highlight them. (I often have trouble with this. If it selects all the points, click again on a
female one. That should do it.) The click the Marker menu . D Click on Cha rt/Options D To be consistent with the graph in the text book, select weight as the Y Axis and height as the X Axis . Then, select sex for Set Markers by . This means SPSS will distinguish the males dots from the female dots on the graph. Then, click Ok D Select the circle under Marker Type and chose a Fill color. Then click Apply . Then click on the male dots, and select the open circle in Marker Type and click Apply . Then, close the dialog box. The resulting graph should look just like the o ne in the textbook. I
would like to alter our graph to include the line of best fit for both groups. D Under Elements, select Fit Line at Subgroups . Then select Linear and click Continue . (I had to select something else and then go back to Lin ear to highlight the Apply button.) The resulting graph follows. I think it looks pretty good.
Page 6
This more complex scatterplot nicely illustrates the difference in the correlation
EHWZHHQKHLJKWDQGZHLJKWIRUPDOHVDQGIHPDOHV/HWVPRYHRQWRDPRUHFRPSOLFDWHG example. Overlay Scatterplots Another kind of scatterpl ot that might be useful is one that displays the association between different independent variables with the same dependant variable. Above, we compared the same correlation for different groups. This time, we want to compare different correlations. Le WVXVHWKHFRXUVHHYDO uation example from the
text It looks like expected grade is more strongly related to ratings of fairness of the exam than ratings of instructor knowledge is related to the exam ,GOLNHWRSORWERWKFRUUHODWLRQV, can reasona bly plot them on the same graph since all of the questions were rated on the same scale. D Open courseevaluation.sav . You do not need to save HeightW eight.sav since you did not change it. So click No D Edit the graph to suit your style as you learned in Chapter 3 (e.g., add a title, change the axes titles
and legend).
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D )LUVWOHWVPDNHVXUHWKHFRUUHODWLRQVUHSRUWHGLQ the text are accurate. Click Analyze/Correlation/Bivariate and select all of the variables. Click Ok . The output follow. Do they agree with the text Correlations 1.000 .804 ** .596 ** .682 ** .301 -.240 .000 .000 .000 .034 .094 50 50 50 50 50 50 .804 ** 1.000 .720 ** .526 ** .469 ** -.451 ** .000 .000 .000 .001 .001 50 50 50 50 50 50 .596 ** .720 ** 1.000 .451 ** .610 ** -.558 ** .000 .000 .001 .000 .000 50 50 50 50 50 50 .682 ** .526 ** .451 **
1.000 .224 -.128 .000 .000 .001 .118 .376 50 50 50 50 50 50 .301 .469 ** .610 ** .224 1.000 -.337 .034 .001 .000 .118 .017 50 50 50 50 50 50 -.240 -.451 ** -.558 ** -.128 -.337 1.000 .094 .001 .000 .376 .017 50 50 50 50 50 50 Pearson Correlation Sig. (2-tailed) Pearson Correlation Sig. (2-tailed) Pearson Correlation Sig. (2-tailed) Pearson Correlation Sig. (2-tailed) Pearson Correlation Sig. (2-tailed) Pearson Correlation Sig. (2-tailed) OVERALL TEACH EXAM KNOWLEDG GRADE ENROLL OVERALL TEACH EXAM KNOWLEDG GRADE ENROLL Correlation is signif icant at the 0.01 lev el (2-tailed). **. Correlation
is signif icant at the 0.05 lev el (2-tailed). *. 1RZOHWVPDNHRXUVFDWWHUSORW D Select Graphs /Legacy /Scatter . Then select Overlay and click Define D Click on exam and grade and shift the m into X Pairs . Then click on exam and knowledge and click them into X pairs . Since exam is the commonality between ERWKSDLUV,GOLNHLWWREHRQWKH<D[LV If it is not listed as Y, highlight the pair and click on the two headed arrow. It will reverse the ordering. Exam
should then appear first for both. Then, click Ok
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D As in the previous example, the dots are distinguished by color. Double click the graph and use the Marker icon to make them more distinct as you learned above. Also use the Elements menu to Fit line at total. It will draw a line f or each set of data. As you can see, the association between expected grade and fairness of the exam LVVWURQJHUWKDQWKHFRUUHODWLRQEHWZHHQLQVWUXFWRUVNQRZOHGJH and the fairness of the exam . Now, you should have the tools necessary to
calculate Person Correlations and to create various scatterplots that compliment those correlations. Complete the following exercises to help you internalize these steps. Exercises Exercises 1 through 3 are based on appendixd.sav 1. Calculate the correlat ions between Add symptoms, IQ, GPA, and English grade twice, once using a one tailed test and once using a two tailed test. Does this make a difference? Typically, when would this make a difference. Note that the axes are not labeled. You could label the Y Axis Grade. But you could not label the X axis because it represents two different
variables exam and knowledge. That is why the legend is necessary. (If you figure out how to label that axis, please let me know. It should be so easy.)
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2. Calculate the same correlations separately for those wh o did and did not drop out, using a two tailed test. Are they similar or different? 3. Create a scatterplot illustrating the correlation between IQ score and GPA for those who did and did not drop out. Be sure to include the line of best fit for each grou p. 4. Open courseevaluation.sav . Create a scatterplot for fairness of exams and teacher skills and exam and
instructor knowledge on one graph. Be sure to include the lines of best fit. Describe your graph. | 3,556 | 11,779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-40 | latest | en | 0.782821 |
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Math Central Quandaries & Queries
Question from Zaheer: solve simultaneous equations and give answer in fractional form 3x - 2 = 4y +5/3 y + 7 = 2x + 4 would really appreciate some help on this please
Hi,
I would start by eliminating the fraction in the first equation. To accomplish this multiply both sides of the first equation by 3.
Can you solve the problem now?
Write back if you need more assistance,
Penny
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. | 125 | 543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-13 | longest | en | 0.936562 |
https://stats.stackexchange.com/questions/639877/rho-a-b-0-9999-rho-x-y-0-9999-sigma-a-sigma-x-sigma | 1,723,401,451,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008125.69/warc/CC-MAIN-20240811172916-20240811202916-00521.warc.gz | 422,228,917 | 39,548 | # $\rho( A, B ) = 0.9999$, $\rho( X, Y ) = 0.9999$, $\sigma_A = \sigma_X$, $\sigma_B = \sigma_Y$, but $\rho( A - B, X - Y) = 0.88$ [closed]
I have four time series $$A,B,X,Y$$. I found that somehow, $$\rho( A, B ) = 0.9999$$, $$\rho( X, Y ) = 0.9999$$, I have also validated that $$\sigma_A = \sigma_X$$, $$\sigma_B = \sigma_Y$$
In other words $$A$$ and $$X$$, $$B$$ and $$Y$$ are really almost just two pairs of near-identical time series.
However I observe that $$\rho( A - B, X - Y) = 0.85$$
[EDITED after Alex's answer] I defined a "modified correlation coefficients" $$\phi(X, Y) = \frac{X \cdot Y}{ \sqrt{X \cdot X} \sqrt{Y \cdot Y } }$$ here $$\cdot$$ means a dot product.
I still observe that $$\phi( A, B ) = 0.9999$$, $$\phi( X, Y ) = 0.9999$$. so I think I can really conlude that they are near-identical time series ... But I still observe that $$\rho( A - B, X - Y) = 0.85$$. $$\phi( A - B, X - Y) = 0.87$$ slightly higher but not much.
What could be the cause that there is such big drop in correlation (or modified correlation) after I subtract them?
• Hi @Mattt Frank, what do you mean by $\rho( A - B ) = \rho( X - Y) = 0.85$? Do you mean $\rho( A - B , X - Y) = 0.85$ Commented Feb 22 at 1:55
• Given any possible correlation coefficient $-1\lt r\lt 1,$ you can construct an example like this. Simply find two variables $\epsilon$ and $\delta$ with correlation $r,$ start with $A$ and $X,$ and define $B = u^\prime A+u\epsilon$ and $Y=v^\prime X+v\delta$ for suitably small values of $u$ and $v.$ (The coefficients $u^\prime$ and $v^\prime$ will both be close to $1$ and are included to guarantee equality of variances.) You can check that as $u$ and $v$ shrink to $0,$ the correlation of $A-B$ with $X-Y$ approaches $r$ in the limit.
– whuber
Commented Feb 22 at 20:20
It's worth noticing what $$A-B$$ and $$X-Y$$ refer to. Each is essentially the random noise that differs between the two members of each pair. It's frankly quite surprising to me that the random noise of the first pair is so highly correlated with the random noise of the second pair. That suggests that any perturbations in the sync within pairs happens across pairs.
For example, let's say $$A$$ and $$B$$ are the sales of chocolate and vanilla ice cream at store 1, and $$X$$ and $$Y$$ are the sales of chocolate and vanilla ice cream at store 2, and let's say that chocolate and vanilla sales are extremely highly correlated. Let's also say that chocolate sales are slightly influenced by the presence of a commercial playing one day for chocolate ice cream. On days the commercial airs, chocolate sells a tiny bit better at both stores.
This would yield a scenario very similar to what you observe: $$A$$ (chocolate ice cream sales at store 1) and $$B$$ (vanilla ice cream sales at store 1) are highly correlated, and $$X$$ (chocolate ice cream sales at store 2) and $$Y$$ (vanilla ice cream sales at store 2) are highly correlated, but on the few days $$A$$ is higher than $$B$$, $$X$$ is also higher than $$Y$$, which leads to a high correlation between $$A-B$$ and $$X-Y$$.
In the absence of a commercial or other shock affecting both pairs of time series, you might expect $$A-B$$ and $$X-Y$$ to be completely uncorrelated. That is, sometimes $$A$$ is higher than $$B$$ and sometimes $$X$$ is higher than $$Y$$, but there doesn't have to be anything linking those two events. Such a high correlation suggests there is a linking of those two events.
Your second sentence $$A$$ and $$X$$, $$B$$ and $$Y$$ are really almost just two pairs of near-identical time series does not necessarily follow from the first statement.
Consider this example in R. I define the four time series that satisfy the conditions on $$\rho(A,B)$$, $$\rho(X,Y)$$, $$\sigma_A=\sigma_X$$, $$\sigma_B=\sigma_Y$$.
But if you look at what the four time series actually are, they're obviously not identical. $$A$$ and $$B$$ are increasing over time, but $$X$$ and $$Y$$ are decreasing. Remember that correlation quantifies how one variable changes with another, not absolutely what the variables are doing.
A <- seq(0, 100, by = 1)
B <- seq(0, 5, length.out = 101)
X <- seq(0, -100, by = -1)
Y <- seq(0, -5, length.out = 101)
cor(X, Y) # ~ 1 as required
cor(A, B) # ~ 1 as required
sd(A); sd(X) # same as required
sd(B); sd(Y) # same as required
cor(A-B, X-Y) # the sign has reversed
• Thanks Alex. I edited my question to address your counter-example (which is very informative) Commented Feb 22 at 2:24
• Do you have a link somewhere about your "modified correlation coefficient"? It's not something I am familiar with. Commented Feb 22 at 2:43
• I kind of made that up to measure the "magnitude" problem you identified ... Commented Feb 22 at 2:48
• What's it supposed to be doing? Just asking because it's not immediately clear to me Commented Feb 22 at 2:54
• Re the magnitude thing - have you tried plotting the time series? That to me would be the quickest way to see if they are actually the same or not, and would probably provide some insight on any difference Commented Feb 22 at 2:55 | 1,452 | 5,080 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 50, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-33 | latest | en | 0.901322 |
https://lists.racket-lang.org/users/archive/2010-November/043192.html | 1,624,238,673,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488259200.84/warc/CC-MAIN-20210620235118-20210621025118-00496.warc.gz | 321,649,290 | 3,547 | # [racket] what initial velocity to use when calculating velocity from acceleration (peripherally moby-related...)
From: Will M. Farr (wmfarr at gmail.com) Date: Tue Nov 30 16:02:01 EST 2010
```On Nov 30, 2010, at 2:52 PM, Joe Marshall wrote:
> An integrator is likely to `drift' over time. In theory, when you
> stop moving you
> should have to apply exactly the same acceleration in reverse from when you
> started. In practice, however, the two accelerations won't quite
> cancel (there will
> be errors in measurement and error accumulation in the integrator).
> You'll have to
> re-zero the integration every now and then to keep it at a sane value.
>
> Alternatively, you could use GPS or some other position data to supply an upper
> bound on the velocity and acceleration and then try to correct the integration
> that way.
>
> In any case, this is an officially difficult problem which has no
> known `solution'.
> You're going to learn a lot!
In particular, combining noisy measurements in the context of an ODE that describes the evolution of a system (in this case, you measure a = dv/dt = d^x/dt^2, and want to "integrate" to find x(t)) is often done using a Kalman filter:
http://en.wikipedia.org/wiki/Kalman_filter
This is also almost certainly the approach you would take if you want to combine data from a GPS unit with the accelerometer data. Kalman filters are often used in commercial inertial navigation systems (i.e. in planes) to track position as well. The literature on the subject is *very* extensive, if you enjoy that sort of reading. Alternately, from the basic description it can be fun to work out a lot of the simple results yourself (depending, of course, on how much you enjoy math and what your level of experience with statistics and differential equations are). In practice (from someone who is not in the field of inertial navigation, but has heard talks about it) it seems like the "tuning" of the filter is as much art as science, so I wouldn't necessarily assign too much weight to the prior literature in your case.
Have fun!
Will
```
Posted on the users mailing list. | 507 | 2,127 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-25 | latest | en | 0.955035 |
https://www.stem.org.uk/news-and-views/opinions/how-find-happy-number | 1,618,803,364,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038863420.65/warc/CC-MAIN-20210419015157-20210419045157-00175.warc.gz | 1,129,654,802 | 11,109 | # How to find a happy number
Published: Mar 15, 2018 3 min read
#### STEM learning
How to find a happy number
The World Happiness Report 2017 states that Norway is the happiest nation in the world, with the United Kingdom ranking 19th out of the 51 countries surveyed.
20 March 2018 marks the United Nations International Day of Happiness, recognising the relevance of happiness and well-being as universal goals and aspirations in the lives of human beings around the world.
### Happiness and mathematics: the perfect formula
Mathematics is widely recognised as a key component in achieving many of these goals, both at a personal and national level. Happy numbers offer a great way to bring International Day of Happiness into the classroom.
To find out if a number is happy, replace the number by the sum of the squares of its digits. Repeat the process until either the number equals 1 or loops endlessly in a cycle that does not include 1. When the process ends with 1 then the number is described as a happy number, otherwise, it is an unhappy number.
Using 23 as an example:
22 = 4
32 = 9
4 + 9 = 13
Hence 23 is replaced by 13. The process is now repeated:
12 = 1
32 = 9
1 + 9 = 10
Hence 13 is replaced by 10. The process is now repeated:
12 = 1
02 = 0
1 + 0 = 1
The process ends at this point since 12 = 1. From this we conclude that 23 (and 13 and 10) are happy numbers.
Unhappy numbers migrate towards a cyclical loop that includes 4, 16, 37, 58, 89, 145, 42, and 20. When considering numbers between 1 and 100, some of the paths for unhappy numbers can be quite lengthy, for example, 60 takes nine intermediate steps before migrating to the cyclical loop. The image below illustrates the connections between unhappy numbers.
There are a number of ways this investigation can be used in the classroom. There are twenty happy numbers between 1 and 100, finding two or three of them is a task that is accessible to a wide range of students during a lesson introducing square numbers.
There are ways to introduce more problem-solving skills into the work. A strategy for reducing work is to realise that multiplying a number by 10 gives the same result. So, the fact that 6 is an unhappy number means that 60 is also unhappy. Reversing digits also yields the same result, so the fact that 24 is not a happy number means that neither is 42.
These strategies can be combined and used to work backwards from 1 to find happy numbers. It follows that 10 and 100 are also happy. Focussing for now on 100, the task becomes one of ‘finding squares that sum to 100’.
62 + 82 = 100
Hence the next numbers in the diagram are 68 and 86. There are no squares that sum to 86, so that branch comes to an end. However,
82 + 22 = 68
hence the next numbers in the diagram are 82 and 28. There are no squares that sum to 28, so that branch comes to an end. However,
92 + 12 = 82
Hence the next numbers in the diagram are 91 and 19. There are no squares that sum to either 91 or 19, and hence both branches come to an end. The thinking behind constructing the branch from 10 works in a similar way.
### Bursary-supported CPD to boost your confidence
If you are looking for inspiration for your mathematics teaching, join us at the National STEM Learning Centre and get involved with our bursary-supported CPD. | 811 | 3,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2021-17 | longest | en | 0.937217 |
https://lengthconverter.intemodino.com/conversion/fur-to-dm-and-dm-to-fur.html | 1,713,220,574,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00257.warc.gz | 337,239,941 | 5,286 | # Online Converter to Convert Furlongs to Decimeters and Decimeters to Furlongs
Length Conversion Settings
Number of decimal places
Round fraction to the nearest
## Fur to dm Conversion Formula and Conversion Factor
The fur to dm conversion factor is 2011.68
1 fur = 2011.68 dm
To convert X furlongs to decimeters you should use the following conversion formula:
X furlongs × 2011.68 = Result in decimeters
## Furlongs to Decimeters Conversion Chart
1 fur = 2011.68 dm
2 fur = 4023.36 dm
3 fur = 6035.04 dm
4 fur = 8046.72 dm
5 fur = 10058.4 dm
6 fur = 12070.08 dm
7 fur = 14081.76 dm
8 fur = 16093.44 dm
9 fur = 18105.12 dm
10 fur = 20116.8 dm
11 fur = 22128.48 dm
12 fur = 24140.16 dm
13 fur = 26151.84 dm
14 fur = 28163.52 dm
15 fur = 30175.2 dm
16 fur = 32186.88 dm
17 fur = 34198.56 dm
18 fur = 36210.24 dm
19 fur = 38221.92 dm
20 fur = 40233.6 dm
21 fur = 42245.28 dm
22 fur = 44256.96 dm
23 fur = 46268.64 dm
24 fur = 48280.32 dm
25 fur = 50292 dm
26 fur = 52303.68 dm
27 fur = 54315.36 dm
28 fur = 56327.04 dm
29 fur = 58338.72 dm
30 fur = 60350.4 dm
31 fur = 62362.08 dm
32 fur = 64373.76 dm
33 fur = 66385.44 dm
34 fur = 68397.12 dm
35 fur = 70408.8 dm
36 fur = 72420.48 dm
37 fur = 74432.16 dm
38 fur = 76443.84 dm
39 fur = 78455.52 dm
40 fur = 80467.2 dm
41 fur = 82478.88 dm
42 fur = 84490.56 dm
43 fur = 86502.24 dm
44 fur = 88513.92 dm
45 fur = 90525.6 dm
46 fur = 92537.28 dm
47 fur = 94548.96 dm
48 fur = 96560.64 dm
49 fur = 98572.32 dm
50 fur = 100584 dm
51 fur = 102595.68 dm
52 fur = 104607.36 dm
53 fur = 106619.04 dm
54 fur = 108630.72 dm
55 fur = 110642.4 dm
56 fur = 112654.08 dm
57 fur = 114665.76 dm
58 fur = 116677.44 dm
59 fur = 118689.12 dm
60 fur = 120700.8 dm
61 fur = 122712.48 dm
62 fur = 124724.16 dm
63 fur = 126735.84 dm
64 fur = 128747.52 dm
65 fur = 130759.2 dm
66 fur = 132770.88 dm
67 fur = 134782.56 dm
68 fur = 136794.24 dm
69 fur = 138805.92 dm
70 fur = 140817.6 dm
71 fur = 142829.28 dm
72 fur = 144840.96 dm
73 fur = 146852.64 dm
74 fur = 148864.32 dm
75 fur = 150876 dm
76 fur = 152887.68 dm
77 fur = 154899.36 dm
78 fur = 156911.04 dm
79 fur = 158922.72 dm
80 fur = 160934.4 dm
81 fur = 162946.08 dm
82 fur = 164957.76 dm
83 fur = 166969.44 dm
84 fur = 168981.12 dm
85 fur = 170992.8 dm
86 fur = 173004.48 dm
87 fur = 175016.16 dm
88 fur = 177027.84 dm
89 fur = 179039.52 dm
90 fur = 181051.2 dm
91 fur = 183062.88 dm
92 fur = 185074.56 dm
93 fur = 187086.24 dm
94 fur = 189097.92 dm
95 fur = 191109.6 dm
96 fur = 193121.28 dm
97 fur = 195132.96 dm
98 fur = 197144.64 dm
99 fur = 199156.32 dm
100 fur = 201168 dm
## Dm to fur Conversion Factor and Conversion Formula
The dm to fur conversion factor is 0.0004971
1 dm = 0.0004971 fur
To convert X decimeters to furlongs you should use the following conversion formula:
X decimeters × 0.0004971 = Result in furlongs
## Decimeters to Furlongs Conversion Table
1 dm = 0.0004971 fur
2 dm = 0.00099419 fur
3 dm = 0.00149129 fur
4 dm = 0.00198839 fur
5 dm = 0.00248548 fur
6 dm = 0.00298258 fur
7 dm = 0.00347968 fur
8 dm = 0.00397678 fur
9 dm = 0.00447387 fur
10 dm = 0.00497097 fur
11 dm = 0.00546807 fur
12 dm = 0.00596516 fur
13 dm = 0.00646226 fur
14 dm = 0.00695936 fur
15 dm = 0.00745645 fur
16 dm = 0.00795355 fur
17 dm = 0.00845065 fur
18 dm = 0.00894775 fur
19 dm = 0.00944484 fur
20 dm = 0.00994194 fur
21 dm = 0.01043904 fur
22 dm = 0.01093613 fur
23 dm = 0.01143323 fur
24 dm = 0.01193033 fur
25 dm = 0.01242742 fur
26 dm = 0.01292452 fur
27 dm = 0.01342162 fur
28 dm = 0.01391871 fur
29 dm = 0.01441581 fur
30 dm = 0.01491291 fur
31 dm = 0.01541001 fur
32 dm = 0.0159071 fur
33 dm = 0.0164042 fur
34 dm = 0.0169013 fur
35 dm = 0.01739839 fur
36 dm = 0.01789549 fur
37 dm = 0.01839259 fur
38 dm = 0.01888968 fur
39 dm = 0.01938678 fur
40 dm = 0.01988388 fur
41 dm = 0.02038098 fur
42 dm = 0.02087807 fur
43 dm = 0.02137517 fur
44 dm = 0.02187227 fur
45 dm = 0.02236936 fur
46 dm = 0.02286646 fur
47 dm = 0.02336356 fur
48 dm = 0.02386065 fur
49 dm = 0.02435775 fur
50 dm = 0.02485485 fur
51 dm = 0.02535194 fur
52 dm = 0.02584904 fur
53 dm = 0.02634614 fur
54 dm = 0.02684324 fur
55 dm = 0.02734033 fur
56 dm = 0.02783743 fur
57 dm = 0.02833453 fur
58 dm = 0.02883162 fur
59 dm = 0.02932872 fur
60 dm = 0.02982582 fur
61 dm = 0.03032291 fur
62 dm = 0.03082001 fur
63 dm = 0.03131711 fur
64 dm = 0.03181421 fur
65 dm = 0.0323113 fur
66 dm = 0.0328084 fur
67 dm = 0.0333055 fur
68 dm = 0.03380259 fur
69 dm = 0.03429969 fur
70 dm = 0.03479679 fur
71 dm = 0.03529388 fur
72 dm = 0.03579098 fur
73 dm = 0.03628808 fur
74 dm = 0.03678517 fur
75 dm = 0.03728227 fur
76 dm = 0.03777937 fur
77 dm = 0.03827647 fur
78 dm = 0.03877356 fur
79 dm = 0.03927066 fur
80 dm = 0.03976776 fur
81 dm = 0.04026485 fur
82 dm = 0.04076195 fur
83 dm = 0.04125905 fur
84 dm = 0.04175614 fur
85 dm = 0.04225324 fur
86 dm = 0.04275034 fur
87 dm = 0.04324743 fur
88 dm = 0.04374453 fur
89 dm = 0.04424163 fur
90 dm = 0.04473873 fur
91 dm = 0.04523582 fur
92 dm = 0.04573292 fur
93 dm = 0.04623002 fur
94 dm = 0.04672711 fur
95 dm = 0.04722421 fur
96 dm = 0.04772131 fur
97 dm = 0.0482184 fur
98 dm = 0.0487155 fur
99 dm = 0.0492126 fur
100 dm = 0.0497097 fur
Wednesday, December 16, 2020 | 2,292 | 5,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-18 | latest | en | 0.486273 |
https://quizizz.com/en-in/ordering-three-digit-numbers-worksheets-class-4 | 1,721,427,733,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514928.31/warc/CC-MAIN-20240719200730-20240719230730-00404.warc.gz | 415,165,928 | 27,493 | Class IV Tier 1 Revision
15 Q
4th
Place Value, Comparing, Ordering, & Rounding
10 Q
3rd - 4th
Chapter 1
20 Q
4th
4-Digit Numbers
15 Q
3rd - 4th
LARGE NUMBERS
20 Q
3rd - 4th
Maths Rapid Fair
14 Q
4th
Ordering Numbers
10 Q
3rd - 4th
Ordering Numbers
20 Q
3rd - 4th
Place Value, Comparing, and Ordering Practice
10 Q
4th
Divisibility tests
17 Q
4th
Three Digit Subtraction
10 Q
2nd - 4th
large numbers - grade 4
15 Q
4th
Numeration system
12 Q
4th - 6th
Multiplying Three-Digit Numbers by One-Digit: lesson 2
11 Q
4th
Grade 5 Numbers
12 Q
4th - 5th
Ordering Negative Numbers
20 Q
4th
Comparing & Ordering Numbers
10 Q
3rd - 4th
Ordering Whole Numbers
10 Q
4th
4 digit numbers
12 Q
4th
Multiplying three and four digit numbers by one digit number
12 Q
4th
Addition Class-4
10 Q
4th
4 digit numbers
10 Q
3rd - 4th
Large Numbers
10 Q
4th - 5th
Multiplying Multi-digit Numbers by 1-digit Number
10 Q
4th
Explore printable Ordering Three-Digit Numbers worksheets for 4th Class
Ordering Three-Digit Numbers worksheets for Class 4 are an essential tool for teachers to help their students develop a strong foundation in Math and Number Sense. These worksheets focus on teaching students how to compare and order three-digit numbers, which is a crucial skill for understanding place value and number relationships. By incorporating these worksheets into their lesson plans, teachers can provide their students with engaging and targeted practice that will help them master this important concept. Additionally, these worksheets can be easily adapted to suit the individual needs of each student, making them a versatile and valuable resource for any Class 4 Math classroom.
Quizizz is an excellent platform for teachers to access a wide range of resources, including Ordering Three-Digit Numbers worksheets for Class 4, as well as other Math and Number Sense materials. This platform offers a variety of interactive quizzes and games that can be used alongside the worksheets to create a comprehensive and engaging learning experience for students. Teachers can also utilize Quizizz's analytics and reporting features to track their students' progress and identify areas where they may need additional support. By incorporating Quizizz into their teaching strategies, educators can ensure that their Class 4 students develop a strong understanding of three-digit numbers and the skills necessary to succeed in Math. | 603 | 2,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-30 | latest | en | 0.935726 |
https://homeworkdave.com/product/lab-5-function-and-recursion/ | 1,660,116,808,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571150.88/warc/CC-MAIN-20220810070501-20220810100501-00272.warc.gz | 322,013,545 | 32,529 | Sale!
# Lab 5 function and recursion
\$30.00
Category:
CSC230 Lab 5
Goal: This lab includes function and recursion.
Please try your best to finish the lab in class, and submit it to CANVAS.
In this lab, please write a Lab5.cpp file. There is main function in this file. Please define
one more function in the lab.
In the lecture, we know that a recursive definition has two parts:
• Base case
• Reduction step
A good example of recursive definition is factorial:
0! = 1
�! = � ∗ � − 1 ! , �� � > 0
The corresponding recursive program can be:
int factorial(int number) {
int temp;
if(number <= 1) return 1;
temp = number * factorial(number – 1);
return temp;
}
The iteration (loop) solution can be:
int factorial(int n) {
int temp = 1;
for (int counter = 1; counter <= n; counter++)
temp = temp * counter;
return fact;
}
As you can see, the recursion and iteration perform the same task. Both of them solve a
large task one piece at a time, and combine the results. However, the emphasis of
iteration is to keep repeating until the task is “done”. While the emphasis of recursion is
to break up the large problem into smaller pieces until you can solve it, then combine the
results.
There is no clear answer to which one is better. But usually mathematicians often prefer
recursive definition; programmers often prefer iterative solutions. As a programmer,
when you define recursive functions, please keep in mind that whenever a function is
called, it will incur some overhead (prepare for the stack, switch control from one
function to another function, clean up the stack frame, etc.). If a recursive function causes
the Fibonacci sequence as an example.
An important notation related to recursion is backtracking. “Backtracking is a general
algorithm for finding all (or some) solutions to some computational problem, notably
constraint satisfaction problems, that incrementally builds candidates to the solutions, and
abandons each partial candidate c (backtracking) as soon as it determines that c cannot be
completed to a valid solution.” —– wikipedia
A classic example of using backtracking is the eight queen puzzle, that asks for all
arrangements of eight chess queens on a standard chessboard so that no queen attacks any
other. Google Developer has very good explanation of a generalized eight queen puzzle –
N-queen Problem at https://developers.google.com/optimization/puzzles/queens . A C++
solution can be found at
https://en.wikibooks.org/wiki/Algorithm_Implementation/Miscellaneous/N-Queens.
Please take some time to read these two webpages, understand the idea of the algorithm
and the code. It is a popular interview question.
In this lab, define a global array called A with size 100. The data type of the array is int.
Design a recursive function called com(int n, int t) inside Lab5.cpp file. The return type
is void. The com(int n, int t) function prints out all strings with size t. Each char in the
string is letter between 0 and 9. That means each letter has 10 choices. The generated
string should be stored in array A. Of course, at any given time, array A can store at most
one complete string.
The first parameter of the function com() indicates which letter the function is working
on. The second parameter of the function com() indicates the length of a desired string.
The following texts are several execution examples,
jli\$ ./a.out 1
0
1
2
3
4
5
6
7
8
9
jli\$ ./a.out 2
00
10
20
30
40
50
60
70
80
90
01
11
21
31
41
51
In above examples, the argument of the command is the string length. The first example
prints out ten strings, the second one prints out one hundred strings (partial results
displayed).
Hints:
• If n < 1, print out the string stored in array A.
• If n >=1, uses a for loop to give different values to A[n-1], recursively call com(n1, t).
Wrap up
When you’re done, jar three files to lab5.jar
jar –cvf lab5.jar *
Submit lab5.jar to Canvas.
Make sure you logout before you leave!
If you cannot finish lab in class, please save all your files. Next time you login the
computer in the lab, you can continue work on your files. Please save them before you
logout. If you work in a Linux lab, please save the file to your machine. However, if you
are working in the Mac lab, please save the file to a CLOUD. The Mac machine will
erase everything you saved once you logout.
## Reviews
There are no reviews yet. | 1,057 | 4,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-33 | latest | en | 0.893692 |
https://discuss.leetcode.com/topic/90061/small-simple-c-java-python | 1,513,456,736,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948589177.70/warc/CC-MAIN-20171216201436-20171216223436-00460.warc.gz | 537,358,227 | 18,875 | # Small simple C++/Java/Python
• Keep the overall result in `A / B`, read the next fraction into `a / b`. Their sum is `(Ab + aB) / Bb` (but cancel their greatest common divisor).
C++:
``````string fractionAddition(string expression) {
istringstream in(expression);
int A = 0, B = 1, a, b;
char _;
while (in >> a >> _ >> b) {
A = A * b + a * B;
B *= b;
int g = abs(__gcd(A, B));
A /= g;
B /= g;
}
}
``````
Java:
``````public String fractionAddition(String expression) {
Scanner sc = new Scanner(expression).useDelimiter("/|(?=[-+])");
int A = 0, B = 1;
while (sc.hasNext()) {
int a = sc.nextInt(), b = sc.nextInt();
A = A * b + a * B;
B *= b;
int g = gcd(A, B);
A /= g;
B /= g;
}
return A + "/" + B;
}
private int gcd(int a, int b) {
return a != 0 ? gcd(b % a, a) : Math.abs(b);
}
``````
Python 3:
Added this after @lee215 reminded me about Python 3's `math.gcd` with his solution in the comments.
``````def fractionAddition(self, expression):
ints = map(int, re.findall('[+-]?\d+', expression))
A, B = 0, 1
for a in ints:
b = next(ints)
A = A * b + a * B
B *= b
g = math.gcd(A, B)
A //= g
B //= g
return '%d/%d' % (A, B)``````
• @StefanPochmann I like the way you use `Scanner`. I spent quite a while to write my own parser during contest :(
• Perhaps we should declare the variable to long to avoid integer overflow.
• @Jim With the given limits, I'm certain that's not necessary. But I'm not sure exactly how large the numbers can get.
• Hi, Could you please elaborate/explain the Pattern: ( "/|(?=[-+])" ).
Thanks
• Super smart way to use Scanner and regex!
Since the test cases are small, I used long to avoid overflow and calculating GCD every iteration. This is my code after stealing your idea of using Scanner:
``````public String fractionAddition1(String expression) {
Scanner s = new Scanner(expression).useDelimiter("/|(?=[-+])");
long num = 0, den = 1;
while (s.hasNext()) {
long a = s.nextLong(), b = s.nextLong();
num = num * b + a * den;
den *= b;
}
long gcd = gcd(num, den);
return (num / gcd) + "/" + (den / gcd);
}
private long gcd(long a, long b) {
if (b == 0)
return a < 0 ? -a : a; // always return positive gcd
return gcd(b, a % b);
}``````
• Hello，I wonder what happened when `a<0` in `gcd` function in Java? Will it causes more calculation?
• Python3 version:
``````def fractionAddition(self, exp):
i, j, a, b, n = 0, 0, 0, 1, len(exp)
while j < len(exp):
i,j = j,exp.find('/', j)
a2 = int(exp[i:j])
i, j = j + 1, min(exp.find('+', j) % (n + 1), exp.find('-', j) % (n + 1))
b2 = int(exp[i:j])
a, b = a * b2 + b * a2, b * b2
a, b = a // math.gcd(a, b), b // math.gcd(a, b)
return str(a) + '/' + str(b)``````
• @lee215 Nice, I forgot they added `math.gcd`. Another Python one, I prefer the parsing to be cleanly separated:
``````def fractionAddition(self, expression):
ints = map(int, re.findall('[+-]?\d+', expression))
A, B = 0, 1
for a in ints:
b = next(ints)
A = A * b + a * B
B *= b
g = math.gcd(A, B)
A //= g
B //= g
return '%d/%d' % (A, B)
``````
Though really my first Python solution (actually my first solution overall) was this:
``````import fractions
class Solution(object):
x = eval(re.sub('(\d+)', r'fractions.Fraction(\1)', expression))
return '%d/%d' % (x.numerator, x.denominator)
``````
• @StefanPochmann
I did the same here using fractions:
``````from fractions import Fraction
class Solution(object):
res = sum(map(Fraction, exp.replace('+', ' +').replace('-', ' -').split()))
return str(res.numerator) + '/' + str(res.denominator)``````
• Could anybody please put in words on how the regex: ( "/|(?=[-+])" ) , works in this scenario.
Thanks
• _gcd
how can you use the __gcd function in C++ example.
It doesn't work in my computer.
• @guettawah Use a compiler that has it :-P (The compiler used by LeetCode does have it). Or with C++17, try this: http://en.cppreference.com/w/cpp/numeric/gcd
• @aayushgarg The (?=) part is a zero-width positive lookahead. Since [-,+] means - or +, the regex (?=[-,+]) means the next element is either - or +.
Since | is logical or operator, "/|(?=[-+])" means the element is "/", or the next element is either - or +. For example, when expression = "-1/2+1/2-1/3",
``````Scanner sc = new Scanner(expression).useDelimiter("/|(?=[-+])")
``````
generates [-1, 2, +1, 2, -1, 3 ]. Note that the signs - and + are preserved.
• This post is deleted!
• @StefanPochmann Can you please explain why we need to cancel the gcd ?
• @gshirish Because we have read the problem description. (Really I don't know what else to say)
• @StefanPochmann I got it, I didn't read the irreducible fraction. Thanks
• java code not robust:
test this one: "-1/2-1/2+---+1/3".
the above test case would throw an exception, which should work, even if it is not included in the test cases. But the judge would give a correct answer to this test case | 1,508 | 4,836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-51 | latest | en | 0.700415 |
http://www.lconvert.com/area/square_meter/acre/ | 1,555,949,306,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578558125.45/warc/CC-MAIN-20190422155337-20190422181337-00263.warc.gz | 260,042,304 | 6,477 | # Square Meter to Acre Converter
1 Square Meter = 0.0002471055 Acre
## How to convert from Square Meter to Acre?
Every 1 Square Meter equals 0.0002471055 Acre. For example, 100 Square Meters equal 100 * 0.0002471055 = 0.02471055 Acres and so on..
## Square Meter to Acre Convesions Table
1 Square Meter = 0.0002471055 Acre
2 Square Meter = 0.000494211 Acre
4 Square Meter = 0.000988422 Acre
5 Square Meter = 0.0012355275 Acre
10 Square Meter = 0.002471055 Acre
20 Square Meter = 0.00494211 Acre
25 Square Meter = 0.0061776375 Acre
50 Square Meter = 0.012355275 Acre
100 Square Meter = 0.02471055 Acre
200 Square Meter = 0.0494211 Acre
250 Square Meter = 0.061776375 Acre
500 Square Meter = 0.12355275 Acre
1000 Square Meter = 0.2471055 Acre
2000 Square Meter = 0.494211 Acre
2500 Square Meter = 0.61776375 Acre
5000 Square Meter = 1.2355275 Acre
10000 Square Meter = 2.471055 Acre
20000 Square Meter = 4.94211 Acre
25000 Square Meter = 6.1776375 Acre
50000 Square Meter = 12.355275 Acre
100000 Square Meter = 24.71055 Acre
200000 Square Meter = 49.4211 Acre
500000 Square Meter = 123.55275 Acre
1000000 Square Meter = 247.1055 Acre
1 Acre = 4046.8544811831 Square Meter
2 Acre = 8093.7089623663 Square Meter
4 Acre = 16187.417924733 Square Meter
5 Acre = 20234.272405916 Square Meter
10 Acre = 40468.544811831 Square Meter
20 Acre = 80937.089623663 Square Meter
25 Acre = 101171.36202958 Square Meter
50 Acre = 202342.72405916 Square Meter
100 Acre = 404685.44811831 Square Meter
200 Acre = 809370.89623663 Square Meter
250 Acre = 1011713.6202958 Square Meter
500 Acre = 2023427.2405916 Square Meter
1000 Acre = 4046854.4811831 Square Meter
2000 Acre = 8093708.9623663 Square Meter
2500 Acre = 10117136.202958 Square Meter
5000 Acre = 20234272.405916 Square Meter
10000 Acre = 40468544.811831 Square Meter
20000 Acre = 80937089.623663 Square Meter
25000 Acre = 101171362.02958 Square Meter
50000 Acre = 202342724.05916 Square Meter
100000 Acre = 404685448.11831 Square Meter
200000 Acre = 809370896.23663 Square Meter
500000 Acre = 2023427240.5916 Square Meter
1000000 Acre = 4046854481.1831 Square Meter
## Acre
What does an acre mean?
Acre is a unit of area which is used commonly for measuring tracts of land.
The expression of acre was first introduced during the Middle Ages, it was defined as the amount of land plowed in one day by one man and an ox. This definition was not clear, because the rate of work differs from an ox to another. Nowadays acre has a specific definition and it equals 43,560 square feet.
Acre can measure any area- rectangles, circles or even hexagons- regardless of it’s shape.
To make it clear for you, a soccer pitch has approximately an area of 1.7 acre. 300,000 acres is the total area of New York City.
United Sates is using another acre to measure roads and alleyways, it is called the commercial acre. Commercial acre equals 36,000 square feet, this means that it is smaller the residential acre. It is approximately 82.6 percent of an international acre.
Now we will show you what an acre equals comparing with some of other other area units:
• 1 acre = 43,560 square feet
• 1 acre = 4046.86 square meters
• 1 acre = 4840 square yards
• 1 acre = 0.404686 hectares
• 1 acre = 1/640th of a square mile | 1,054 | 3,253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-18 | latest | en | 0.506371 |
https://www.askiitians.com/forums/7-grade-maths/the-heights-of-10-girls-were-measured-in-cm-and-th_270176.htm | 1,701,402,131,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100264.9/warc/CC-MAIN-20231201021234-20231201051234-00396.warc.gz | 739,886,344 | 42,366 | # The heights of 10 girls were measured in cm and the results are as follows: 135, 150, 139, 128, 151, 132, 146, 149, 143, 141.How many girls have heights more than the mean height.
Harshit Singh
3 years ago
Dear Student
First we have to arrange the given data in an ascending order,
= 128, 132, 135, 139, 141, 143, 146, 149, 150, 151
5 girls have heights more than the mean height
Thanks | 125 | 391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2023-50 | latest | en | 0.934772 |
http://www.physicsforums.com/showthread.php?t=572441 | 1,408,821,713,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500826343.66/warc/CC-MAIN-20140820021346-00211-ip-10-180-136-8.ec2.internal.warc.gz | 421,937,253 | 8,178 | # Reference frame for harmonic motion.
by rollcast
Tags: frame, harmonic, motion, reference
P: 418 If you a mass being accelerated by a force which is acting upon a spring attached to the mass it will exhibit harmonic motion. However unlike a fixed harmonic oscillator there is no explicit solution to the equation which describes the motion of the mass in a reference frame outside of the system. However what if you took the reference frame and imagined a reference frame where the mass or spring was stationary would this allow you to calculate an explicit solution for the equation? Thanks AL
P: 5,462 Perhaps a diagram to make your intent more clear? Why would the force cause SHM unless it is variable (oscillatory) or impulsive itself? What other forces are acting?
P: 418 Here's a diagram which might make it a bit more clear as to the set up which I'm referring to. I forgot to say that the force is stopped after a certain amount of time and the system s allowed to deccelerate freely. That shows the velocity against time for the system. The red curve shows the harmonic motion - there are better graphs but I haven't got excel on this laptop. Attached Thumbnails
P: 5,462 Reference frame for harmonic motion. I still can't see it. Assume your system is in free space so there are no other forces acting. Assume the force is applied at t=o as shown. What causes the spring to extend? ie what opposes F?
Engineering Sci Advisor HW Helper Thanks P: 7,121 I don't understand why you are saying "there is no explicit solution to the equations of motion". Have you studied ordinary differential equations yet? If you have, you should know how to solve this. And if you know about Laplace Transforms, you can almost write down the solution without doing any "plug and chug" algebra.
P: 418
Quote by Studiot I still can't see it. Assume your system is in free space so there are no other forces acting. Assume the force is applied at t=o as shown. What causes the spring to extend? ie what opposes F?
The inertia of the mass on one end of the spring opposes the force being applied to the other end of the spring.
Quote by AlephZero I don't understand why you are saying "there is no explicit solution to the equations of motion". Have you studied ordinary differential equations yet? If you have, you should know how to solve this. And if you know about Laplace Transforms, you can almost write down the solution without doing any "plug and chug" algebra.
I understand 1st and 2nd ODEs but I'm not familiar with the formula that is being used to describe the motion of the mass. A poster on another forum has put it into an excel doc. which calculates the solution to the ODE using a Runge Kutta method - which from what I can figure out about it means the formula has no explicit solution?
P: 5,462 Are you quite sure that your description of the system is correct? With the system as currently described what happens depends upon how long F is applied for and if it can move its point of application, since it is the only external force acting in the system. Runge Kutta methods are simply numerical methods which may be more convenient for the calculation of a differential equation. They were developed long before we could put complicated formulae into computers and have the answer in microseconds for many mesh points. In particular their use does not mean the equation has no analytic solution. However without some more boundary conditions the equation has no solution of any description. What is this differential equation to which you refer? It is usual to offer a system similar to yours, but with the spring connected to a fixed point at one end, the mass at the other, to demonstrate simple harmonic motion. Is this what you really mean?
Related Discussions Advanced Physics Homework 0 Advanced Physics Homework 7 Advanced Physics Homework 23 Cosmology 16 Introductory Physics Homework 4 | 828 | 3,918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2014-35 | latest | en | 0.956748 |
http://cfd.mace.manchester.ac.uk/ercoftac/doku.php?id=cases:case007 | 1,618,694,102,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038464045.54/warc/CC-MAIN-20210417192821-20210417222821-00565.warc.gz | 21,873,496 | 11,654 | cases:case007
Circular to Rectangular Transition Duct
Turbulent flow in a circular to rectangular transition duct. The flow remains wholly attached along the transition. A 3D flow with constant temperature.
Geometry of the Computational Domain
The inlet duct is of length $0.5D$ and has a circular cross-section of diameter $D = 2R = 204.3$ mm. The outlet duct is of length $2D$ with a rectangular cross-section of aspect ratio 3:1. In the transition region, the wall coordinates are given by the superellipse:
$\left(y/a \right)^{\eta} + \left( z/b\right)^{\eta} = 1$
$a$, $b$ and $\eta$ are functions of the axial distance along the duct, $x$. The values for $a$, $b$ and $\eta$ are given in the file geom.dat.
The geometry, coordinates, and measurement locations are shown in figure 1.
Fig. 1: Duct geometry
Flow Characteristics
A secondary flow vortex pair develops in the transition duct which distorts the primary mean velocity and Reynolds stress fields. Analysis of their results by the experimenters shows that in this region conventional wall functions, while applicable, must be used with caution. In the downstream straight duct, the longitudinal vorticity diffuses very rapidly.
Flow Parameters
• Air at atmospheric pressure and temperature $T = 298.3$ K
• At the inlet:
• axial bulk velocity $U_b=29.95$ m/s
• kinematic viscosity $\nu =1.57 \times 10^{-5}$ m2s
• Reynolds number $Re=U_b D/\nu =3.9 \times 10^5$.
Inflow Conditions
Partially developed turbulent pipe flow at station 1 ($x/D=-0.5$).
• Boundary layer features:
• boundary-layer thickness $\delta/R=0.2855$
• axisymmetric displacement thickness $\delta_1/R=0.0383$
• axisymmetric momentum thickness $\delta_2/R=0.0281$
• axisymmetric energy thickness $\delta_3/R=0.0497$
• free stream turbulence level of 0.3%
• Friction velocity $U^*/U_b=0.04063$
Measured profiles are available for:
• Axial velocity $U/U_b$
• Second moments $\overline{u^2}/U_b^2$, $\overline{v^2}/U_b^2$, $\overline{w^2}/U_b^2$, $\overline{uv}/U_b^2$, $k/U_b^2$
Measured quantities:
Wall static pressures have been measured through 0.508 mm tappings. Wall shear stresses have been measured using Preston tubes with different diameters ranging from 1.067 mm to 3.073 mm. Velocities have been measured close to the wall using a single-hot-wire probe rotation technique.
Measurement Errors:
• $\delta(\text{probe positions}) \pm 0.025$ m.
• $\delta(\text{angles}) \pm 0.5^o$.
Variable Error Variable Error
$\delta(C_p)$ $\pm 0.002$ $\delta(C_f)$ $\pm 0.0005$
$\delta(U/U_b)$ $\pm 0.01$ $\delta(V/U_b)$ $\pm 0.002$
$\delta(\overline{u^2}/U_b^2)$ $\pm 0.0001$ $\delta(\overline{uv}/U_b^2)$ $\pm 0.00015$
$\delta(\overline{v^2}/U_b^2)$ $\pm 0.0002$ $\delta(\overline{uw}/U_b^2)$ $\pm 0.00015$
$\delta(\overline{w^2}/U_b^2)$ $\pm 0.0002$ $\delta(\overline{vw}/U_b^2)$ $\pm 0.0001$
Measurements were taken at 6 streamwise locations, at $x/D$ positions given in the table below.
Station 1 2 3 4 5 6
$x/D$ -0.5 0.5 1.1 1.4 2.0 4.0
• Profiles along semi-major and semi-minor axes of:
• Mean velocity $U$ and pressure at $x/D=-0.5$, $1.1$, $1.4$, $2.0$, $4.0$
• Second moments at $x/D=-0.5$, $2.0$, $4.0$
• Profiles of circumferential pressure coefficient $C_p$ at $x/D=1.1$, $1.4$, $2.0$, $4.0$
• Profiles of circumferential skin friction coefficient $C_f$ at $x/D=2.0$, $4.0$
• 2-D maps of
• Mean velocities $U$, $V$, $W$ and pressure at $x/D=1.1$, $1.4$, $2.0$, $4.0$
• Second moments at $x/D=2.0$, $4.0$
Sample plots of selected quantities are available.
The data can be downloaded as compressed archive files from the links below, or as individual files by selecting those required from the tables.
Profile Data
Profile Data at Station 1, $x/D=-0.5$
$y_1$ $y_2$ $y_3$ $y_4$
$U$ trd_pr_x1_mu_y1.dat trd_pr_x1_mu_y2.dat trd_pr_x1_mu_y3.dat trd_pr_x1_mu_y4.dat
Total Pressure trd_pr_x1_pt_y1.dat trd_pr_x1_pt_y2.dat trd_pr_x1_pt_y3.dat trd_pr_x1_pt_y4.dat
$\overline{u^2}$ trd_pr_x1_uu_y3.dat
$\overline{v^2}$ trd_pr_x1_vv_y3.dat
$\overline{w^2}$ trd_pr_x1_ww_y3.dat
$k$ trd_pr_x1_ke_y3.dat
$\overline{uv}$ trd_pr_x1_uv_y3.dat
Profile Data at Station 3, $x/D=1.1$
$y_2$ (Semi-major axis) $y_3$ (Semi-minor axis)
$U$ trd_pr_x3_mu_y2.dat trd_pr_x3_mu_y3.dat
Total Pressure trd_pr_x3_pt_y2.dat trd_pr_x3_pt_y3.dat
$C_p$ around circumference: trd_pr_x3_cp.dat
Profile Data at Station 4, $x/D=1.4$
$y_2$ (Semi-major axis) $y_3$ (Semi-minor axis)
$U$ trd_pr_x4_mu_y2.dat trd_pr_x4_mu_y3.dat
Total Pressure trd_pr_x4_pt_y2.dat trd_pr_x4_pt_y3.dat
$C_p$ around circumference: trd_pr_x4_cp.dat
Profile Data at Station 5, $x/D=2.0$
$y_2$ (Semi-major axis) $y_3$ (Semi-minor axis)
$U$ trd_pr_x5_mu_y2.dat trd_pr_x5_mu_y3.dat
Total Pressure trd_pr_x5_pt_y2.dat trd_pr_x5_pt_y3.dat
$\overline{u^2}$ trd_pr_x5_uu_y2.dat trd_pr_x5_uu_y3.dat
$\overline{v^2}$ trd_pr_x5_vv_y2.dat trd_pr_x5_vv_y3.dat
$\overline{w^2}$ trd_pr_x5_ww_y2.dat trd_pr_x5_ww_y3.dat
$k$ trd_pr_x5_ke_y2.dat trd_pr_x5_ke_y3.dat
$\overline{uv}$ trd_pr_x5_uv_y2.dat
$\overline{uw}$ trd_pr_x5_uw_y3.dat
$C_p$ around circumference: trd_pr_x5_cp.dat
$Cf$ around circumference: trd_pr_x5_cf.dat
Profile Data at Station 6, $x/D=4.0$
$y_2$ (Semi-major axis) $y_3$ (Semi-minor axis)
$U$ trd_pr_x6_mu_y2.dat trd_pr_x6_mu_y3.dat
Total Pressure trd_pr_x6_pt_y2.dat trd_pr_x6_pt_y3.dat
$\overline{u^2}$ trd_pr_x6_uu_y2.dat trd_pr_x6_uu_y3.dat
$\overline{v^2}$ trd_pr_x6_vv_y2.dat trd_pr_x6_vv_y3.dat
$\overline{w^2}$ trd_pr_x6_ww_y2.dat trd_pr_x6_ww_y3.dat
$k$ trd_pr_x6_ke_y2.dat trd_pr_x6_ke_y3.dat
$\overline{uv}$ trd_pr_x6_uv_y2.dat
$\overline{uw}$ trd_pr_x6_uw_y3.dat
$C_p$ around circumference: trd_pr_x6_cp.dat
$C_f$ around circumference: trd_pr_x6_cf.dat
$C_p$ Along Duct Centerline
trd_pr_cl_cp.dat
2-D Map Data
The results of both the calculations of Sotiropoulos and Patel (1993), using a two-layer $k$-$\varepsilon$ model, and Demuen, using a full Reynolds stress closure with wall functions, predict a weaker secondary motion than the measured one.
On the other hand, the results of both the calculations of Lien and Leschziner (1993), using a low-Reynolds $k$-$\varepsilon$ model, and Sotiropoulos and Patel (1993), using the near-wall full-Reynolds stress closure of Launder and Shima, give a good representation of the flow inside the transition and a little bit too rapid decay of the Reynolds stress downstream.
Description of Experiments
1. Davis, D.O. (1992). Experimental investigation of turbulent flow through a circular-to-rectangular transition duct. NASA Technical Memorandum 105210.
2. Davis, D.O., Gessner, F.B. (1992). Experimental investigation of turbulent flow through a circular-to-rectangular transition duct. AIAA J., Vol. 30, p. 367.
3. Davis, D.O., Gessner, F.B. (1990). Experimental investigation of turbulent flow through a circular-to-rectangular transition duct. AIAA Paper 90-1505.
Previous Numerical Calculations
1. Lien, F.S., Leschziner, M.A. (1993). Computational modelling of 3D turbulent flow in S. diffuser and transition ducts. 2nd Int. Symp. on Engineering Turbulence Modelling and Measurements, May 31 - June 2.
2. Sotiropoulos, F., Patel, V.C. (1993). Numerical calculation of turbulent flow through a circular-to-rectangular transition duct using advanced turbulent closures. AIAA paper 93-3030.
3. Lien, F.S., Leschziner, M.A. (1993). Modelling the flow in a transition duct with a non orthogonal FV procedure and low-Re turbulence-transport models. UMIST report TFD/93/10.
4. Sotiropoulos, F., Patel, V.C. (1994). Prediction of turbulent flow through a transition duct using a second-moment closure. AIAA J., Vol. 32, p. 2194.
Indexed data:
case007 (dbcase, confined, flow)
case007
titleCircular to Rectangular Transition Duct
authorDavis, Gessner
year1990
typeEXP
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