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https://www.baeldung.com/cs/k-fold-cross-validation | 1,718,887,586,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861940.83/warc/CC-MAIN-20240620105805-20240620135805-00430.warc.gz | 581,439,337 | 56,983 | ## 1. Introduction
In this tutorial, we’ll explain the way how to validate neural networks or any other machine learning model. First, we’ll briefly introduce the term neural network. After that, we’ll describe what does validation means and different strategies for validation. Finally, we’ll explain a particular type of validation, called k-fold cross-validation, with some modifications.
In general, validation is a critical step in building a machine learning system since the validity of results directly depends on it.
## 2. Neural Networks
Neural networks are algorithms explicitly created as an inspiration for biological neural networks. The basis of neural networks is neurons interconnected according to the type of network. Initially, the idea was to create an artificial system that would function just like the human brain.
There are many types of neural networks, but they roughly fall into three main classes:
For the most part, the difference between them is the type of neurons that form them and how the information flows through the network. To test neural network predictions, we need to use appropriate methods that we’ll explain below.
## 3. Validation
After we train the neural network and generate results with a test set, we need to check how correct they are.
### 3.1. Machine Learning Metrics
Usually, in neural networks or machine learning methods, we measure the quality of the method using a metric that represents the error or correctness of the solution. Errors are used for problems such as regression, while correctness is more common for classification problems. Thus, the most commonly used metrics in classification problems are:
If a classification model, besides predicted class, outputs probability or confidence of the prediction, we can use measures:
• AUC
• Cross-entropy
Also, the most used metrics in regression problems are:
• Mean squared error (MSE)
• Root Mean Squared Error (RMSE)
• Mean absolute error (MAE)
Overall, these metrics are the most frequently used, but there are hundreds of different ones.
### 3.2. Underfitting and Overfitting
After choosing the metric, we’re going to set up the validation strategy, also known as cross-validation. One classic way of doing that is to split the whole data set into training and test set. Namely, it’s important to say that selecting the model with the highest accuracy on the training set doesn’t guarantee that it’ll perform similarly in the future with the new data.
Thus, the point of validation is to provide at least the approximate performance of the model for data that will appear in the future. In addition, we need to have in mind the importance of balancing between underfitting and overfitting.
Briefly, the underfitting means that the model doesn’t perform well on both training and test set. Most likely, the reason for underfitting is that model is not well-tuned on the training set or not trained enough. The consequence of that is high bias and low variance.
The overfitting implies that the model is too tuned to the training set. As a result, the model performs very well on the training set but poorly on the test set. The consequence of that is low bias and high variance:
## 4. K-Fold Cross-Validation
The most significant disadvantage of splitting the data into one training and test set is that the test set might not follow the same distribution of classes in general in the data. Also, some numerical features might not have the same distribution in the training and test set. The k-fold cross validation smartly solves this. Basically, it creates the process where every sample in the data will be included in the test set at some steps.
First, we need to define that represents a number of folds. Usually, it’s in the range of 3 to 10, but we can choose any positive integer. After that, we split the data into equal folds (parts). The algorithm has steps where at each step, we select different folds for the test set and the remaining folds we leave for the training set.
Using this method, we will train our model times independently and have scores measured by some of the selected metrics. Lastly, we can average all scores or even analyze their deviations. We presented the whole process in the image below:
Besides the classic k-fold cross-validation scheme, there are some modifications that we’ll mention below.
### 4.1. Leave-One-Out Cross-Validation
Leave-one-out cross-validation (LOOCV) is a special type of k-fold cross-validation. There will be only one sample in the test set. Basically, the only difference is that is equal to the number of samples in the data.
Instead of LOOCV, it is preferable to use the leave-p-out strategy, where defines several samples in the training set. Subsequently, the special case of leave-p-out for is LOOCV. The most significant advantage of this approach is that it uses almost all data in the training set but still requires building models that can be computationally expensive.
### 4.2. Stratified K-Fold Cross-Validation
This technique is a type of k-fold cross-validation, intended to solve the problem of imbalanced target classes. For instance, if the goal is to make a model that will predict if the e-mail is spam or not, likely, target classes in the data set won’t be balanced. This is because, in real life, most e-mails are non-spam.
Hence, stratified k-fold cross validation solves this problem by splitting the data set in folds, where each fold has approximately the same distribution of target classes. Similarly, in the case of regression, this approach creates folds that have approximately the same mean target value.
### 4.3. Repeated K-Fold Cross-Validation
Repeated k-fold cross-validation is a simple strategy that repeats the process of randomly splitting the data set into training and test set times. Unlike classic k-fold cross-validation, this method doesn’t divide data into folds but randomly splits the data times. It means that the proportion between training and test set doesn’t depend on the number of folds, but we can set it at any ratio.
Because of that, some samples might be selected multiple times for the test, while some samples might never be selected.
### 4.4. Nested K-Fold Cross-Validation
Nested k-fold cross-validation is an extension of classic k-fold cross-validation, and it’s mainly used for hyperparameter tuning. It solves two problems that we have in the normal cross-validation:
1. Possibility of information leakage.
2. The error estimation is made on the exact data for which we found the best hyperparameters, which might be biased.
It’s not best to use the same training and test sets for selecting hyperparameters and estimating error (score). Because of that, we’ll create two k-fold cross-validations, one inside another as nested loops. Through the inner loop, we search hyperparameters while the outer loop is for error estimation. The whole process is illustrated in the image below:
The algorithm for nested k-fold cross-validation is below:
## 5. Conclusion
In general, validation is an essential step in the machine learning pipeline. That is why we need to pay attention to validation since a small mistake can lead to biased and wrong models. This article explained some of the most common cross-validation techniques that we can use for training neural networks or any other machine learning models.
To conclude, If it’s not computationally too expensive, the suggestion is to use nested k-fold cross-validation. More complex models will most likely work well with classic k-fold cross-validation. | 1,545 | 7,595 | {"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} | 2.703125 | 3 | CC-MAIN-2024-26 | latest | en | 0.906768 |
https://courses.lumenlearning.com/mathforliberalartscorequisite/chapter/an-introduction-to-the-language-of-algebra/ | 1,660,165,596,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571210.98/warc/CC-MAIN-20220810191850-20220810221850-00508.warc.gz | 184,857,502 | 19,754 | ## Introduction to The Language of Algebra
### Learning Outcomes
• Use Variables and Algebraic Symbols
• Use variables to represent unknown quantities in algebraic expressions
• Identify the variables and constants in an algebraic expression
• Use words and symbols to represent algebraic operations on variables and constants
• Use inequality symbols to compare two quantities
• Translate between words and inequality notation
• Identifying Expressions and Equations
• Identify and write mathematical expressions using words and symbols
• Identify and write mathematical equations using words and symbols
• Use the order of operations to simplify mathematical expressions
• Simplifying Expressions Using the Order of Operations
• Use exponential notation
• Write an exponential expression in expanded form
## Use Variables and Algebraic Symbols
Greg and Alex have the same birthday, but they were born in different years. This year Greg is $20$ years old and Alex is $23$, so Alex is $3$ years older than Greg. When Greg was $12$, Alex was $15$. When Greg is $35$, Alex will be $38$. No matter what Greg’s age is, Alex’s age will always be $3$ years more, right?
In the language of algebra, we say that Greg’s age and Alex’s age are variable and the three is a constant. The ages change, or vary, so age is a variable. The $3$ years between them always stays the same, so the age difference is the constant.
In algebra, letters of the alphabet are used to represent variables. Suppose we call Greg’s age $g$. Then we could use $g+3$ to represent Alex’s age. See the table below.
Greg’s age Alex’s age
$12$ $15$
$20$ $23$
$35$ $38$
$g$ $g+3$
Letters are used to represent variables. Letters often used for variables are $x,y,a,b,\text{ and }c$.
### Variables and Constants
A variable is a letter that represents a number or quantity whose value may change.
A constant is a number whose value always stays the same.
To write algebraically, we need some symbols as well as numbers and variables. There are several types of symbols we will be using. In Whole Numbers, we introduced the symbols for the four basic arithmetic operations: addition, subtraction, multiplication, and division. We will summarize them here, along with words we use for the operations and the result.
Operation Notation Say: The result is…
Addition $a+b$ $a\text{ plus }b$ the sum of $a$ and $b$
Subtraction $a-b$ $a\text{ minus }b$ the difference of $a$ and $b$
Multiplication $a\cdot b,\left(a\right)\left(b\right),\left(a\right)b,a\left(b\right)$ $a\text{ times }b$ The product of $a$ and $b$
Division $a\div b,a/b,\frac{a}{b},b\overline{)a}$ $a$ divided by $b$ The quotient of $a$ and $b$
In algebra, the cross symbol, $\times$, is not used to show multiplication because that symbol may cause confusion. Does $3xy$ mean $3\times y$ (three times $y$ ) or $3\cdot x\cdot y$ (three times $x\text{ times }y$ )? To make it clear, use • or parentheses for multiplication.
We perform these operations on two numbers. When translating from symbolic form to words, or from words to symbolic form, pay attention to the words of or and to help you find the numbers.
• The sum of $5$ and $3$ means add $5$ plus $3$, which we write as $5+3$.
• The difference of $9$ and $2$ means subtract $9$ minus $2$, which we write as $9 - 2$.
• The product of $4$ and $8$ means multiply $4$ times $8$, which we can write as $4\cdot 8$.
• The quotient of $20$ and $5$ means divide $20$ by $5$, which we can write as $20\div 5$.
### Example
Translate from algebra to words:
1. $12+14$
2. $\left(30\right)\left(5\right)$
3. $64\div 8$
4. $x-y$
Solution:
1. $12+14$ $12$ plus $14$ the sum of twelve and fourteen
2. $\left(30\right)\left(5\right)$ $30$ times $5$ the product of thirty and five
3. $64\div 8$ $64$ divided by $8$ the quotient of sixty-four and eight
4. $x-y$ $x$ minus $y$ the difference of $x$ and $y$
### TRY IT
When two quantities have the same value, we say they are equal and connect them with an equal sign.
### Equality Symbol
$a=b$ is read $a$ is equal to $b$
The symbol $=$ is called the equal sign.
An inequality is used in algebra to compare two quantities that may have different values. The number line can help you understand inequalities. Remember that on the number line the numbers get larger as they go from left to right. So if we know that $b$ is greater than $a$, it means that $b$ is to the right of $a$ on the number line. We use the symbols $\text{<}$ and $\text{>}$ for inequalities.
$a<b$ is read $a$ is less than $b$
$a$ is to the left of $b$ on the number line
$a>b$ is read $a$ is greater than $b$
$a$ is to the right of $b$ on the number line
The expressions $a<b\text{ and }a>b$ can be read from left-to-right or right-to-left, though in English we usually read from left-to-right. In general,
$a<b\text{ is equivalent to }b>a$.
For example, $7<11\text{ is equivalent to }11>7$.
$a>b\text{ is equivalent to }b<a$.
.For example, $17>4\text{ is equivalent to }4<17$
When we write an inequality symbol with a line under it, such as $a\le b$, it means $a<b$ or $a=b$. We read this $a$ is less than or equal to $b$. Also, if we put a slash through an equal sign, $\ne$, it means not equal.
We summarize the symbols of equality and inequality in the table below.
Algebraic Notation Say
$a=b$ $a$ is equal to $b$
$a\ne b$ $a$ is not equal to $b$
$a<b$ $a$ is less than $b$
$a>b$ $a$ is greater than $b$
$a\le b$ $a$ is less than or equal to $b$
$a\ge b$ $a$ is greater than or equal to $b$
### Symbols $<$ and $>$
The symbols $<$ and $>$ each have a smaller side and a larger side.
smaller side $<$ larger side
larger side $>$ smaller side
The smaller side of the symbol faces the smaller number and the larger faces the larger number.
### Example
Translate from algebra to words:
1. $20\le 35$
2. $11\ne 15 - 3$
3. $9>10\div 2$
4. $x+2<10$
### TRY IT
In the following video we show more examples of how to write inequalities as words.
### Example
The information in the table below compares the fuel economy in miles-per-gallon (mpg) of several cars. Write the appropriate symbol $\text{=},\text{<},\text{ or }\text{>}$ in each expression to compare the fuel economy of the cars.
(credit: modification of work by Bernard Goldbach, Wikimedia Commons)
1. MPG of Prius_____ MPG of Mini Cooper
2. MPG of Versa_____ MPG of Fit
3. MPG of Mini Cooper_____ MPG of Fit
4. MPG of Corolla_____ MPG of Versa
5. MPG of Corolla_____ MPG of Prius
### TRY IT
Grouping symbols in algebra are much like the commas, colons, and other punctuation marks in written language. They indicate which expressions are to be kept together and separate from other expressions. The table below lists three of the most commonly used grouping symbols in algebra.
Common Grouping Symbols
parentheses ( )
brackets [ ]
braces { }
Here are some examples of expressions that include grouping symbols. We will simplify expressions like these later in this section.
$\begin{array}{cc}8\left(14 - 8\right)21 - 3\\\left[2+4\left(9 - 8\right)\right]\\24\div \left\{13 - 2\left[1\left(6 - 5\right)+4\right]\right\}\end{array}$
## Identify Expressions and Equations
What is the difference in English between a phrase and a sentence? A phrase expresses a single thought that is incomplete by itself, but a sentence makes a complete statement. “Running very fast” is a phrase, but “The football player was running very fast” is a sentence. A sentence has a subject and a verb.
In algebra, we have expressions and equations. An expression is like a phrase. Here are some examples of expressions and how they relate to word phrases:
Expression Words Phrase
$3+5$ $3\text{ plus }5$ the sum of three and five
$n - 1$ $n$ minus one the difference of $n$ and one
$6\cdot 7$ $6\text{ times }7$ the product of six and seven
$\frac{x}{y}$ $x$ divided by $y$ the quotient of $x$ and $y$
Notice that the phrases do not form a complete sentence because the phrase does not have a verb. An equation is two expressions linked with an equal sign. When you read the words the symbols represent in an equation, you have a complete sentence in English. The equal sign gives the verb. Here are some examples of equations:
Equation Sentence
$3+5=8$ The sum of three and five is equal to eight.
$n - 1=14$ $n$ minus one equals fourteen.
$6\cdot 7=42$ The product of six and seven is equal to forty-two.
$x=53$ $x$ is equal to fifty-three.
$y+9=2y - 3$ $y$ plus nine is equal to two $y$ minus three.
### Expressions and Equations
An expression is a number, a variable, or a combination of numbers and variables and operation symbols.
An equation is made up of two expressions connected by an equal sign.
### example
Determine if each is an expression or an equation:
1. $16 - 6=10$
2. $4\cdot 2+1$
3. $x\div 25$
4. $y+8=40$
Solution
1. $16 - 6=10$ This is an equation—two expressions are connected with an equal sign. 2. $4\cdot 2+1$ This is an expression—no equal sign. 3. $x\div 25$ This is an expression—no equal sign. 4. $y+8=40$ This is an equation—two expressions are connected with an equal sign.
### Learning Outcomes
• Use the order of operations to simplify mathematical expressions
• Simplify mathematical expressions involving addition, subtraction, multiplication, division, and exponents
## Simplify Expressions Containing Exponents
To simplify a numerical expression means to do all the math possible. For example, to simplify $4\cdot 2+1$ we’d first multiply $4\cdot 2$ to get $8$ and then add the $1$ to get $9$. A good habit to develop is to work down the page, writing each step of the process below the previous step. The example just described would look like this:
$4\cdot 2+1$
$8+1$
$9$
Suppose we have the expression $2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2$. We could write this more compactly using exponential notation. Exponential notation is used in algebra to represent a quantity multiplied by itself several times. We write $2\cdot 2\cdot 2$ as ${2}^{3}$ and $2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2$ as ${2}^{9}$. In expressions such as ${2}^{3}$, the $2$ is called the base and the $3$ is called the exponent. The exponent tells us how many factors of the base we have to multiply.
$\text{means multiply three factors of 2}$
We say ${2}^{3}$ is in exponential notation and $2\cdot 2\cdot 2$ is in expanded notation.
### Exponential Notation
For any expression ${a}^{n},a$ is a factor multiplied by itself $n$ times if $n$ is a positive integer.
${a}^{n}\text{ means multiply }n\text{ factors of }a$
The expression ${a}^{n}$ is read $a$ to the ${n}^{th}$ power.
For powers of $n=2$ and $n=3$, we have special names.
$a^2$ is read as “$a$ squared”
$a^3$ is read as “$a$ cubed”
The table below lists some examples of expressions written in exponential notation.
Exponential Notation In Words
${7}^{2}$ $7$ to the second power, or $7$ squared
${5}^{3}$ $5$ to the third power, or $5$ cubed
${9}^{4}$ $9$ to the fourth power
${12}^{5}$ $12$ to the fifth power
### example
Write each expression in exponential form:
1. $16\cdot 16\cdot 16\cdot 16\cdot 16\cdot 16\cdot 16$
2. $\text{9}\cdot \text{9}\cdot \text{9}\cdot \text{9}\cdot \text{9}$
3. $x\cdot x\cdot x\cdot x$
4. $a\cdot a\cdot a\cdot a\cdot a\cdot a\cdot a\cdot a$
### try it
In the video below we show more examples of how to write an expression of repeated multiplication in exponential form.
### example
Write each exponential expression in expanded form:
1. ${8}^{6}$
2. ${x}^{5}$
### try it
To simplify an exponential expression without using a calculator, we write it in expanded form and then multiply the factors.
### example
Simplify: ${3}^{4}$
## Simplify Expressions Using the Order of Operations
We’ve introduced most of the symbols and notation used in algebra, but now we need to clarify the order of operations. Otherwise, expressions may have different meanings, and they may result in different values.
For example, consider the expression:
$4+3\cdot 7$
$\begin{array}{cccc}\hfill \text{Some students say it simplifies to 49.}\hfill & & & \hfill \text{Some students say it simplifies to 25.}\hfill \\ \begin{array}{ccc}& & \hfill 4+3\cdot 7\hfill \\ \text{Since }4+3\text{ gives 7.}\hfill & & \hfill 7\cdot 7\hfill \\ \text{And }7\cdot 7\text{ is 49.}\hfill & & \hfill 49\hfill \end{array}& & & \begin{array}{ccc}& & \hfill 4+3\cdot 7\hfill \\ \text{Since }3\cdot 7\text{ is 21.}\hfill & & \hfill 4+21\hfill \\ \text{And }21+4\text{ makes 25.}\hfill & & \hfill 25\hfill \end{array}\hfill \end{array}$
Imagine the confusion that could result if every problem had several different correct answers. The same expression should give the same result. So mathematicians established some guidelines called the order of operations, which outlines the order in which parts of an expression must be simplified.
### Order of Operations
When simplifying mathematical expressions perform the operations in the following order:
1. Parentheses and other Grouping Symbols
• Simplify all expressions inside the parentheses or other grouping symbols, working on the innermost parentheses first.
2. Exponents
• Simplify all expressions with exponents.
3. Multiplication and Division
• Perform all multiplication and division in order from left to right. These operations have equal priority.
• Perform all addition and subtraction in order from left to right. These operations have equal priority.
Students often ask, “How will I remember the order?” Here is a way to help you remember: Take the first letter of each key word and substitute the silly phrase. Please Excuse My Dear Aunt Sally.
Order of Operations
Excuse Exponents
My Dear Multiplication and Division
It’s good that ‘My Dear’ goes together, as this reminds us that multiplication and division have equal priority. We do not always do multiplication before division or always do division before multiplication. We do them in order from left to right.
Similarly, ‘Aunt Sally’ goes together and so reminds us that addition and subtraction also have equal priority and we do them in order from left to right.
### example
Simplify the expressions:
1. $4+3\cdot 7$
2. $\left(4+3\right)\cdot 7$
Solution:
1. $4+3\cdot 7$ Are there any parentheses? No. Are there any exponents? No. Is there any multiplication or division? Yes. Multiply first. $4+\color{red}{3\cdot 7}$ Add. $4+21$ $25$
2. $(4+3)\cdot 7$ Are there any parentheses? Yes. $\color{red}{(4+3)}\cdot 7$ Simplify inside the parentheses. $(7)7$ Are there any exponents? No. Is there any multiplication or division? Yes. Multiply. $49$
### example
Simplify:
1. $\text{18}\div \text{9}\cdot \text{2}$
2. $\text{18}\cdot \text{9}\div \text{2}$
### example
Simplify: $18\div 6+4\left(5 - 2\right)$.
### try it
In the video below we show another example of how to use the order of operations to simplify a mathematical expression.
When there are multiple grouping symbols, we simplify the innermost parentheses first and work outward.
### example
$\text{Simplify: }5+{2}^{3}+3\left[6 - 3\left(4 - 2\right)\right]$.
### try it
In the video below we show another example of how to use the order of operations to simplify an expression that contains exponents and grouping symbols.
### example
Simplify: ${2}^{3}+{3}^{4}\div 3-{5}^{2}$. | 4,271 | 15,325 | {"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} | 4.78125 | 5 | CC-MAIN-2022-33 | latest | en | 0.893986 |
http://www.braingle.com/brainteasers/teaser.php?id=21101&comm=0 | 1,406,336,849,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997894931.59/warc/CC-MAIN-20140722025814-00211-ip-10-33-131-23.ec2.internal.warc.gz | 548,584,409 | 6,967 | Browse Teasers
Search Teasers
## A Giant Puzzle (Part 1)
Logic puzzles require you to think. You will have to be logical in your reasoning.
Puzzle ID: #21101 Fun: (2.93) Difficulty: (2.26) Category: Logic Submitted By: cdrock Corrected By: boodler
I climbed an unusual beanstalk to another world where I found a gigantic house. I went inside the house and I was picked up by the biggest thing I have ever seen. He threw me inside a messy room and told me that if I could clean it by the time he gets back, he would let me go, but if I didn't I would have to clean his house for the rest of eternity. He told me to put all the different things in separate boxes in a certain order and then the boxes in a certain order. He will leave me with a list of clues so I can get them in order.
First I will sort through the shoes and put them in order.
There are red, blue, green, yellow, brown, orange, and purple pairs of shoes, not necessarily in that order.
1. The brown shoes are 4 pairs of shoes away from the purple pair. (3 pairs in between)
2. The blue pair isn't next to the orange pair.
3. The distance from the green pair to the blue pair isn't the same distance as the distance from the yellow pair to the orange pair.
4. The red pair is next to the brown pair.
5. Either the blue or orange pair is between the green and yellow pairs.
6. The yellow pair is next to the purple pair.
7. The red pair is the first pair on the left.
8. The orange pair isn't next to one of the colors with a second letter of "r"
## What Next?
See another brain teaser just like this one...
Or, just get a random brain teaser
If you become a registered user you can vote on this brain teaser, keep track of
which ones you have seen, and even make your own. | 426 | 1,755 | {"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} | 3 | 3 | CC-MAIN-2014-23 | longest | en | 0.956477 |
https://www.jiskha.com/questions/196078/Find-the-student-who-looks-sad-Find-a-student-who-looks-sad-Which-one-is-correct | 1,548,148,898,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583831770.96/warc/CC-MAIN-20190122074945-20190122100945-00371.warc.gz | 846,282,046 | 5,393 | # English
Find the student who looks sad.
Find a student who looks sad.
(Which one is correct and common?)
How does Junsu look?
He looks sad.
Do you know why?
Yes, I do.
He got 100 points in English.
He got 60 points in math.
He got the best points in
1. π 0
2. π 0
3. π 28
asked by John
1. Find the student who looks sad.
Find a student who looks sad.
(Which one is correct and common?)
Both are correct, depending on the context. The first one implies that's there's only one student in the group who looks sad; the second one implies that several look sad, and that one of them should be identified.
How does Junsu look?
He looks sad.
Do you know why?
Yes, I do.
He got 100 points in English.
He got 60 points in math.
He got the best points in
These all look fine except that the last sentence isn't finished!
1. π 0
2. π 0
posted by Writeacher
## Similar Questions
1. ### Math
*Ms. Sue or Reed, please check my other question as well if you have time, thank you :) 1. Which sentence has proper subject-verb agreement? A) Even my cousin Steve feel sad today. B) Even my cousin Steve look sad today. C) Even
asked by Tim on April 21, 2016
2. ### Language
which sentence has proper subject-verb agreement? even my cousin steve feel sad today. even my cousin steve look sad today. even my cousin steve feels sad today.*** even my cousin steve looking sad today.
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3. ### English
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asked by rfvv on August 23, 2010
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5. ### english
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6. ### Student supervision
What is a good technique to follow when you'return supervising a reading group? A. Ignore the mistake a student makes. B. When a student comes to an unknown word, say what it is. ( my answer) C. Concentrate on correct words,
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7. ### stats
a student randomly guesses at five multiple-choice questions, each question has five possible choices (suppose the choice for each question is independent): a) Find the probability that the student gets exactly three correct. b)
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8. ### english
What is the adverb and then what is the adjective the adverb modifies in this sentence. In spite of her defeat, she was not sad at all? Is is not sad?
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9. ### Math Help
Suppose that each student in stat. has a 9% chance of missing class on any given day and that student attendance is independent. In a stat. class of 20 students, find the probability that: 1) no student is absent 2) one student is
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10. ### Check answers please Statistics
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asked by Chris on April 8, 2013
More Similar Questions | 949 | 3,585 | {"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-2019-04 | latest | en | 0.968761 |
http://www.onlinemath4all.com/trigonometric-ratio-table.html | 1,513,177,812,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948527279.33/warc/CC-MAIN-20171213143307-20171213163307-00770.warc.gz | 438,955,537 | 11,848 | TRIGONOMETRIC RATIO TABLE
Trigonometric ratio table will help us to find the values of trigonometric ratios for standard angles.
The standard angles are 0°, 30°, 45°, 60° and 90°.
The values of trigonometric ratios of standard angles are very important to solve many problems in trigonometry. Therefore, it is important to remember the values of the trigonometric ratios of these standard angles. The sin, cos, tan, csc, sec and cot of the standard angles are given the table below.
Now, let us see, how the above values of trigonometric ratios are determined for the standard angles.
First, let us see how the values are determined for the angles 30° and 60°
Trigonometric ratios of 30° and 60°
Let ABC be an equilateral triangle whose sides have length a (see the figure given below). Draw AD perpendicular to BC, then D bisects the side BC.
So, BD = DC = a/2 and < BAD = <DAC = 30°.
Now, in right triangle ADB, <BAD = 30° and BD = a/2.
In right triangle ADB, by Pythagorean theorem,
Hence, we can find the trigonometric ratios of angle 30° from the right triangle ADB.
In right triangle ADB, <ABD = 60°. So, we can determine the trigonometric ratios of angle 60°.
Now, let us see how the values are determined for the angle 45°.
Trigonometric ratio of 45°
If an acute angle of a right triangle is 45°, then the other acute angle is also 45°.
Thus the triangle is isosceles. Let us consider the triangle ABC with <B = 90°, <A = <C = 45°
Then AB = BC. Let AB = BC = a.
By Pythagorean theorem,
AC² = AB² + BC²
AC² = a² + a²
AC² = 2a²
AC = √2 x a
Hence, we can find the trigonometric ratios of angle 45° from the right triangle ABC.
Now, let us see how the values are determined for the angles 0° and 90°.
Trigonometric ratios of 0° and 90°
Consider the figure given below which shows a circle of radius 1 unit centered at the origin.
Let P be a point on the circle in the first quadrant with coordinates (x, y).
We drop a perpendicular PQ from P to the x-axis in order to form the right triangle OPQ.
Let <POQ = θ, then
sin θ = PQ / OP = y/1 = y (y coordinate of P)
cos θ = OQ / OP = x/1 = x (x coordinate of P)
tan θ = PQ / OQ = y/x
If OP coincides with OA, then angle θ = 0°.
Since, the coordinates of A are (1, 0), we have
If OP coincides with OB, then angle θ = 90°.
Since, the coordinates of B are (0, 1), we have
The six trigonometric ratios of angles 0°, 30°, 45°, 60° and 90° are provided in the following table.
After having gone through the stuff given above, we hope that the students would have understood "Trigonometric ratio table"
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Remainder when 2 power 256 is divided by 17
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Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
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Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 | 1,341 | 4,957 | {"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} | 4.78125 | 5 | CC-MAIN-2017-51 | latest | en | 0.696742 |
https://proofwiki.org/wiki/Axiom_of_Choice_Implies_Law_of_Excluded_Middle | 1,597,374,458,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739134.49/warc/CC-MAIN-20200814011517-20200814041517-00206.warc.gz | 450,487,812 | 10,928 | # Diaconescu-Goodman-Myhill Theorem
## Theorem
The axiom of choice implies the law of excluded middle.
## Proof
Let $\mathbb B = \set {0, 1}$.
Let $p$ be a proposition.
Let the following two sets be defined:
$A = \set {x \in \mathbb B: x = 0 \lor p}$
$B = \set {x \in \mathbb B: x = 1 \lor p}$
where $\lor$ denotes the disjunction operator.
We have that:
$0 \in A$
and:
$1 \in B$
so both $A$ and $B$ are non-empty
Then the set:
$X = \set {A, B}$
is a set of non-empty sets:
By the axiom of choice, there exists a choice function:
$f: X \to \mathbb B$
since $\displaystyle \bigcup X = \mathbb B$.
There are four cases:
$(1): \quad \map f A = \map f B = 0$
This means that $0 \in B$.
But for that to happen, $\paren {0 = 1} \vee p$ must be true.
So by Disjunctive Syllogism, $p$ is true.
$(2): \quad \map f A = \map f B = 1$
This means that $1 \in A$.
Arguing similarly to case $(1)$, it follows that $p$ is true in this case also.
$(3): \quad \map f A = 1 \ne \map f B = 0$
This means that $A \ne B$ (or otherwise $f$ would pick the same element).
But if $p$ is true, that means:
$A = B = \mathbb B$
Therefore in this case:
$\neg p$
$(4): \quad \map f A = 0 \ne \map f B = 1$
Using the same reasoning as in case $(3)$, it is seen that in this case:
$\neg p$
So by Proof by Cases:
$\paren {p \vee \neg p}$
That is the Law of Excluded Middle.
$\blacksquare$
## Source of Name
This entry was named for Radu DiaconescuNoah D. Goodman and John R. Myhill.
## Also known as
The Diaconescu-Goodman-Myhill Theorem is also known as Diaconescu's Theorem and the Goodman-Myhill Theorem.
It is ever $\mathsf{Pr} \infty \mathsf{fWiki}$'s endeavour to attest a theorem to as many contributors as appropriate.
## Historical Note
The proof of the Diaconescu-Goodman-Myhill Theorem was first published in $1975$ by Radu Diaconescu.
It was later independently rediscovered by Noah D. Goodman and John R. Myhill and published in $1978$.
However, the first appearance of the result itself was in Errett Albert Bishop's $1967$ work Foundations of Constructive Analysis, where he set it as an exercise, without including a solution. | 686 | 2,158 | {"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} | 4.4375 | 4 | CC-MAIN-2020-34 | latest | en | 0.899045 |
https://es.pinterest.com/educacioplastic/animaciones/ | 1,495,890,382,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608953.88/warc/CC-MAIN-20170527113807-20170527133807-00322.warc.gz | 955,903,084 | 58,150 | Mathematica code: G[X_, Y_, Z_, S_, p_, q_, r_, a_, b_, c_, N_, e_, t_, PR_, IS_] := Graphics[ Table[ Disk[ {X*Cos[a (e*t + n)*2 Pi/N + p], Y*Cos[b (e*t + n)*2 Pi/N + q]}, Z*Cos[c (e*t + n)*2 Pi/N + r*t*2 Pi] + S], {n, 1, N}], PlotRange - PR, ImageSize - IS] Manipulate[ G[.5, 1, .02, .03, 0, Pi/4, -.05, 2, 1, 1, 40, 1, .1, t, {{-.786, .786}, {-1.1, 1.1}}, 500], {t, 1, 20, 1}] ?
Mathematica code: RotAxis = Table[Table[ Table[ R[o, {.01 + x, .01 + y, 0}, {0, 0, 0}], {o, 0, 2 Pi, 2 Pi/80}], {x, -10, 10, 1}], {y, -10, 10, 1}]Edge := {1, 2, 4, 3, 7, 8, 6, 5, 1, 3, 4, 8, 7, 5, 6, 2}CubeProjections[color_, pr_, b_, s_, h_, w_, m_, o_] :=Graphics[ Table[ Translate[ {AbsoluteThickness[h], If[color == 0, Black, White], Line[ Table[ Table[ RotAxis[[11 + y]][[11 + x]] [[1 + Mod[Round[ (Pi + ArcTan[.01 + x, .01 + y])/2Pi] + o, 80]]]…
Mathematica code:Rot80 = Table[ Table[ RotationTransform[a, {1, 1, 0}, {0, 0, 0}][Tuples[{-1, 1}, 3][[v]]], {v, 1, 8, 1}],{a, 0, 2 Pi, ...
Inspired by Visual illusions based on single-field contrast asynchronies and by beesandbombs. Mathematica code: v[a_] := {{Cos[a], 0}, {0,...
Mathematica code: P[A_, f_, w_, h_, M_, Y_, t_] := Plot[ Table[ A*Sin[f*x + t + n*2 Pi/w] + h*n, {n, 1, M, 1}], {x, 0, 4 Pi}, PlotSty...
Inspired by Intermodular Distortion by Volume2a. Mathematica code: ListAnimate[Table[ Graphics[ Table[ Table[ {Red, Disk[{(1 + 2 n) .5/N + Mod[((-1)^N) t/N, ((-1)^N) 1], If[N == 3, 0, Sum[.5/j, {j, 3, N - 1}]]}, .5/N, {0, Pi}]}, {n, -1, N, 1}], {N, 3, 35}], PlotRange - {{0, 1}, {0, 1.33}}, ImageSize - {500, 652}, Background - Blue],{t, 0, .95, .05}]
intothecontinuum.tumbler.com - a perspective on mathematics, the pattern, and the abstract with animated GIFs and code to reproduce.
### Tokyo type gif animation
Shamil Karim Kristina Udovichenko Tokyo type
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Pinterest | 840 | 1,927 | {"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-2017-22 | latest | en | 0.522082 |
https://git.gmu.edu/zrajabi/pymdptoolbox/-/commit/df7f9e557c38f06498d1c198269be624078ea37b | 1,656,224,063,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103037089.4/warc/CC-MAIN-20220626040948-20220626070948-00374.warc.gz | 330,664,221 | 28,157 | Commit df7f9e55 by Steven Cordwell
### fix some typos in docstrings
parent 118a2fa8
... ... @@ -70,46 +70,50 @@ class MDP(object): """A Markov Decision Problem. Let S = the number of states, and A = the number of acions. Let ``S`` = the number of states, and ``A`` = the number of acions. Parameters ---------- transitions : array Transition probability matrices. These can be defined in a variety of ways. The simplest is a numpy array that has the shape (A, S, S), ways. The simplest is a numpy array that has the shape ``(A, S, S)``, though there are other possibilities. It can be a tuple or list or numpy object array of length A, where each element contains a numpy array or matrix that has the shape (S, S). This "list of matrices" form is useful when the transition matrices are sparse as scipy.sparse.csr_matrix matrices can be used. In summary, each action's transition matrix must be indexable like ``P[a]`` where ``a`` ∈ {0, 1...A-1}. numpy object array of length ``A``, where each element contains a numpy array or matrix that has the shape ``(S, S)``. This "list of matrices" form is useful when the transition matrices are sparse as ``scipy.sparse.csr_matrix`` matrices can be used. In summary, each action's transition matrix must be indexable like ``transitions[a]`` where ``a`` ∈ {0, 1...A-1}, and ``transitions[a]`` returns an ``S`` × ``S`` array-like object. reward : array Reward matrices or vectors. Like the transition matrices, these can also be defined in a variety of ways. Again the simplest is a numpy array that has the shape (S, A), (S,) or (A, S, S). A list of lists can be used, where each inner list has length S. A list of numpy arrays is possible where each inner array can be of the shape (S,), (S, 1), (1, S) or (S, S). Also scipy.sparse.csr_matrix can be used instead of numpy arrays. In addition, the outer list can be replaced with a tuple or numpy object array can be used. array that has the shape ``(S, A)``, ``(S,)`` or ``(A, S, S)``. A list of lists can be used, where each inner list has length ``S`` and the outer list has length ``A``. A list of numpy arrays is possible where each inner array can be of the shape ``(S,)``, ``(S, 1)``, ``(1, S)`` or ``(S, S)``. Also ``scipy.sparse.csr_matrix`` can be used instead of numpy arrays. In addition, the outer list can be replaced by any object that can be indexed like ``reward[a]`` such as a tuple or numpy object array of length ``A``. discount : float Discount factor. The per time-step discount factor on future rewards. Valid values are greater than 0 upto and including 1. If the discount factor is 1, then convergence is cannot be assumed and a warning will be displayed. Subclasses of ``MDP`` may pass None in the case where the algorithm does not use a discount factor. be displayed. Subclasses of ``MDP`` may pass ``None`` in the case where the algorithm does not use a discount factor. epsilon : float Stopping criterion. The maximum change in the value function at each iteration is compared against ``epsilon``. Once the change falls below this value, then the value function is considered to have converged to the optimal value function. Subclasses of ``MDP`` may pass None in the case where the algorithm does not use a stopping criterion. the optimal value function. Subclasses of ``MDP`` may pass ``None`` in the case where the algorithm does not use an epsilon-optimal stopping criterion. max_iter : int Maximum number of iterations. The algorithm will be terminated once this many iterations have elapsed. This must be greater than 0 if specified. Subclasses of ``MDP`` may pass None in the case where the algorithm does not use a maximum number of iterations. specified. Subclasses of ``MDP`` may pass ``None`` in the case where the algorithm does not use a maximum number of iterations. Attributes ---------- ... ... @@ -130,12 +134,12 @@ class MDP(object): time : float The time used to converge to the optimal policy. verbose : boolean Whether verbose output should be displayed in not. Whether verbose output should be displayed or not. Methods ------- run Implemented in child classes as the main algorithm loop. Raises and Implemented in child classes as the main algorithm loop. Raises an exception if it has not been overridden. setSilent Turn the verbosity off ... ... @@ -314,11 +318,11 @@ class FiniteHorizon(MDP): --------------- V : array Optimal value function. Shape = (S, N+1). ``V[:, n]`` = optimal value function at stage ``n`` with stage in (0, 1...N-1). ``V[:, N]`` value function at stage ``n`` with stage in {0, 1...N-1}. ``V[:, N]`` value function for terminal stage. policy : array Optimal policy. ``policy[:, n]`` = optimal policy at stage ``n`` with stage in (0, 1...N). ``policy[:, N]`` = policy for stage ``N``. stage in {0, 1...N}. ``policy[:, N]`` = policy for stage ``N``. time : float used CPU time ... ...
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Snapdeal Interview Experience | Set 4 (On Campus)
• Difficulty Level : Hard
• Last Updated : 29 Aug, 2019
Hi, Recently Snapdeal visited my campus and I got an offer from Snapdeal,here is my interview experience:
Round 1:
First round was online written round.It consists of 25 Questions.
22 MCQ’s and 3 Coding questions
1) Check for balanced parentheses in an expression.
2) Find next greater number with same set of digits.
3) Given an array where each element is the money a person have and there is only Rs. 3 note. We need to check whether it is possible to divide the money equally among all the persons or not. If it is possible then find Minimum number of transactions needed.
Round 2:(45 min.)
F2F Technical Interview 1:
1) Largest contiguous subarray sum in an array containing both positive and negative numbers.
2) Merge two sorted arrays of integers.
3) Merge Sort for integer array.
4) Counting Sort.
5) Given 100 weights having weights 1,2,3,4—-100 and also given a number K.
You have to choose the minimum number of weights such That you can measure each quantity from 1 to K.
6) 3 couples crossing the river puzzle with many more constraints.
7) Dbms queries SELECT,CREATE TABLE,JOIN,VIEWS etc.
Round 2: (1 hr 45 min.)
F2F Technical Interview 2:
2) Difference between Process and Thread.
3) Find 3 elements in an array having sum K.(different approaches)
4) Implement stack using Queue. I gave him 2 queues solution.he said do it using 1 queue only.Lot of discussion
on it ,Different approaches
5) Given an array having 0,1,2 only sort it.Different approaches.
7) How many permutions are there for a string.
8) Mathematical proof for the above question.
9) Print all the permutations of string.Also dry run on given test case.
10) Binary tree to DLL.
Round 3: (20-25 min.)
F2F HR Interview:
Basic behavioral questions and Some situational questions.
I would like to thank GeeksForGeeks which helped me to improve my knowledge and understanding of Data structures and Algorithms 🙂
If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Write your Interview Experience or mail it to contribute@geeksforgeeks.org
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If a number between 0 and 1/2 is selected at random, which of the following will the number most likely be between?
A. 0 and 3/20
B. 3/20 and 1/5
C. 1/5 and 1/4
D. 1/4 and 3/10
E. 3/10 and 1/2
[Reveal] Spoiler: OA
Last edited by Bunuel on 14 Aug 2014, 01:03, edited 1 time in total.
Added the OA.
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Re: If a number between 0 and 1/2 is selected at random, which [#permalink]
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17 Oct 2008, 11:11
A. 0 and 3/20 - here difference is 0.15
B. 3/20 and 1/5 - here difference is 0.05
C. 1/5 and 1/4 - here difference is 0.05
D. 1/4 and 3/10 - here difference is 0.05
E. 3/10 and 1/2 - here difference is 0.2
Hence, E.
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Re: If a number between 0 and 1/2 is selected at random, which [#permalink]
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17 Oct 2008, 13:06
rishi2377 wrote:
GMAT TIGER wrote:
rishi2377 wrote:
If a number between 0 and 1/2 is selected at random, which of the following will the number most likely be between?
A. 0 and 3/20
B. 3/20 and 1/5
C. 1/5 and 1/4
D. 1/4 and 3/10
E. 3/10 and 1/2
E. 3/10 and 1/2
GMAT TIGER. I guess it is understood that I have OA. It wont hurt if you explain your answer.
Definitely not. I am always happy to explain a given question or issue if I can. However, I sometime donot put my explanation if I am answering first because I want to put others first for their explanation.
Now you have excellent explanantion by scthakur. In fact the question is asking the likelyness of or high chances/probability of having a given number. Hope you already got it.
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Re: If a number between 0 and 1/2 is selected at random, which [#permalink]
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11 Aug 2014, 07:56
scthakur wrote:
A. 0 and 3/20 - here difference is 0.15
B. 3/20 and 1/5 - here difference is 0.05
C. 1/5 and 1/4 - here difference is 0.05
D. 1/4 and 3/10 - here difference is 0.05
E. 3/10 and 1/2 - here difference is 0.2
Hence, E.
Can you please explain why did you eliminate first four choices?
as per my understanding 0 and 1/2 include these numbers also --> 0.15, 0.05, 0.05, 0.05
please correct me if i am missing something
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Re: If a number between 0 and 1/2 is selected at random, which [#permalink]
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11 Aug 2014, 20:31
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AkshayDavid wrote:
scthakur wrote:
A. 0 and 3/20 - here difference is 0.15
B. 3/20 and 1/5 - here difference is 0.05
C. 1/5 and 1/4 - here difference is 0.05
D. 1/4 and 3/10 - here difference is 0.05
E. 3/10 and 1/2 - here difference is 0.2
Hence, E.
Can you please explain why did you eliminate first four choices?
as per my understanding 0 and 1/2 include these numbers also --> 0.15, 0.05, 0.05, 0.05
please correct me if i am missing something
The question is one of probability. You select a number between 0 and 1/2 at random.
The number is between 0 and 10/20 (for ease of calculations)
Now, wider the desired range of the options, more is the probability that the number will lie in that range.
The probability that the number will lie between 0 and 10/20 is 1 i.e. the number is in this range. The probability that the number will lie between 0 and 9/20 is a little less than 1 since it is quite possible that the number is in this range but there is a small chance that it lies between 9/20 and 10/20 which this range doesn't cover. The probability that the number will lie between 0 and 1/20 is very small since this range covers a small part of the total range.
A. 0 and 3/20 - covers 0 to 3/20 but leaves out 3/20 to 10/20. So in essence, covers only 3 parts out of 10 parts
B. 3/20 and 1/5 - covers 3/20 to 4/20 (= 1/5). So in essence, covers only 1 part out of 10 parts
C. 1/5 and 1/4 - covers 4/20 to 5/20. So in essence, covers only 1 part out of 10 parts
D. 1/4 and 3/10 - covers 5/20 to 6/20. So in essence, covers only 1 part out of 10 parts
E. 3/10 and 1/2 - covers 6/20 to 10/20. So in essence, covers 4 parts out of 10 parts. Since this range is the widest, so the probability of the number lying in this range is the highest.
Answer (E)
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Re: If a number between 0 and 1/2 is selected at random, which [#permalink]
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14 Aug 2014, 01:05
rishi2377 wrote:
If a number between 0 and 1/2 is selected at random, which of the following will the number most likely be between?
A. 0 and 3/20
B. 3/20 and 1/5
C. 1/5 and 1/4
D. 1/4 and 3/10
E. 3/10 and 1/2
Similar questions to practice:
if-a-number-between-0-and-1-is-selected-at-random-which-of-130819.html
if-a-number-between-1-and-1-2-is-selected-at-random-which-of-101786.html
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If a number between 0 and 1 2 is selected at random, which of the foll [#permalink]
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13 Jul 2016, 19:23
If a number between 0 and 1/2 is selected at random, which of the following will the number most likely be between?
(A) 0 and 1/9
(B) 1/9 and 1/7
(C) 1/7 and 1/5
(D) 1/5 and 1/3
(E) 1/3 and 1/2
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Re: If a number between 0 and 1/2 is selected at random, which [#permalink]
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13 Jul 2016, 23:03
HARRY113 wrote:
If a number between 0 and 1/2 is selected at random, which of the following will the number most likely be between?
(A) 0 and 1/9
(B) 1/9 and 1/7
(C) 1/7 and 1/5
(D) 1/5 and 1/3
(E) 1/3 and 1/2
Topics merged. Please refer to the discussion above.
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Re: If a number between 0 and 1/2 is selected at random, which [#permalink] 13 Jul 2016, 23:03
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 3,094 | 9,395 | {"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-2016-50 | latest | en | 0.882542 |
http://www.edurite.com/kbase/disadvantages-line-graphs | 1,464,160,359,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049274191.57/warc/CC-MAIN-20160524002114-00147-ip-10-185-217-139.ec2.internal.warc.gz | 474,959,271 | 14,191 | #### • Class 11 Physics Demo
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#### • on line graph
Question:and where would you use them?
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Answers:advantages to graphing is that it is faster if the numbers are easy to work with, like whole small numbers. The disadvantage would be for graphing equations with decimals or really big numbers would be hard to do accurately. Substitution is easier to use when the numbers or equations are hard to graph or there is no graph paper handy
Question:
Answers:Advantages: It gives you a nice visual representation of a function or equation. You can get a good idea of how the function behaves by looking at the graph. Disadvantages: you can't necessarily use it to solve a function for a particular value, because eyeballing a picture isn't as accurate as doing the algebra. Also, not all functions can be graphed easily
Question:Explain how a line graph is similar to a bar graph.
Answers:There is information on it, it has a title, labels, and more. | 236 | 1,087 | {"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} | 2.78125 | 3 | CC-MAIN-2016-22 | longest | en | 0.936795 |
http://math.stackexchange.com/questions/638233/functors-and-groups | 1,467,134,166,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783396949.33/warc/CC-MAIN-20160624154956-00041-ip-10-164-35-72.ec2.internal.warc.gz | 189,389,987 | 19,293 | # Functors and Groups
Let $\alpha$ be a functor from the category of groups in the category of groups which assigns to every group $G$ a characteristic subgroup $\alpha (G)$ of $G$ and to every homomorphism $\theta : H \rightarrow K$ the restriction $\theta|_{\alpha(H)}:\alpha(H)\rightarrow \alpha(K)$ (in other words, $\alpha(H)^\theta \leq \alpha(K)$). So we have, in particular, that $\alpha(H)^\theta \leq \alpha(H^\theta)$. Is it true that also happen that $\alpha(H)^\theta$ is equal to $\alpha(H^\theta)$? I think it holds, but why?
References
D.J.S. Robinson, "Finiteness Conditions and Generalized Soluble Groups", Vol. 1 Pag. 18, (1.3, at the bottom of the page)
-
There are many characteristic subgroups (e.g. all subgroups of $G$ is $G$ is cyclic), so $\alpha$ might make different "choices" for different $G$. Do I understand the statement right that the only restriction to these choices is that $\alpha(H)^\theta \le \alpha(K)$ for each homomorphism $\theta\colon H\to K$? – Hagen von Eitzen Jan 14 '14 at 15:55
Yes, but I think that this isn't a real restriction since otherwise $\alpha$ could not be a functor assigning the restriction to each homomorphism. – W4cc0 Jan 14 '14 at 15:57
Yes but you need this assumption (which is stronger than being characteristic!) in order to get a functor. – Martin Brandenburg Jan 14 '14 at 16:33
ams.org/mathscinet-getitem?mr=173699 and ams.org/mathscinet-getitem?mr=188277 are the papers of Baer mentioned. Baer appears to have switched from functor in the first to Robinson's definition in the second. Presumably Baer also found the equivalence obvious. – Jack Schmidt Jan 14 '14 at 16:42
@JackSchmidt this is why I asked my question. It seems that both Robinson and Baer found the equivalence obvious. But to me is not so obvious :(. I "read" the paper of Baer but I cannot realize anything new. – W4cc0 Jan 14 '14 at 16:51
It is not true at least for the category of Abelian groups. Take $\alpha (G)$ the subgroup of all elements of finite order. Set $H=\mathbb{Z}$ and $K=\mathbb{Z}_n$.
-
But in this case: $\alpha(H)^i=[H,H]^i=\{1\}^i=\{1\}=[H^i,H^i]$ since $H$ and $H^i$ are both abelian (where $i$ is the identity function of $H$ in $K$). So this isn't a counterexample I think. – W4cc0 Jan 14 '14 at 16:08
Sorry, I read the question wrong. I changed the answer. – Boris Novikov Jan 14 '14 at 16:39
The subgroup generated by the elements of finite order might work in general. – Jack Schmidt Jan 14 '14 at 16:54
In your counterexample (following my notation) I think must be $K$ changed by $H$. Anyway it seems to work also in general. So the statment in Robinson's book is false? – W4cc0 Jan 14 '14 at 17:08
I don't know, I have not this book. – Boris Novikov Jan 14 '14 at 17:17
If you look at $\alpha(G)=G'$, the commutator subgroup, then it is true. In general verbal subgroup will do.
-
Yes but only the verbal subgroup will do this? This means that if $\alpha$ satisfies to my requirements then $\alpha$ is a "verbal" mapping? – W4cc0 Jan 14 '14 at 16:31
Well, that is not clear. With respect to any variety of groups in any case you will have a lot of characteristic subgroups that are eligible, see for example groupprops.subwiki.org/wiki/Verbal_subgroup – Nicky Hekster Jan 14 '14 at 16:34
Well, I know about verbal subgroup; my problem is the fact that Robinson in the text I cited says that: a group theoretical function (a function $\alpha$ assigning to each group $G$ a subgroup $\alpha(G)$ s.t. if $\theta : G \rightarrow K$ is an isomorphism of $G$ then $\alpha(G)^\theta=\alpha(G^\theta)$ is a functor of the category of groups if and only if for every homomorphism $\theta: G \rightarrow H$ it follows that $\alpha(G)^\theta=\alpha(G^\theta)$ - here the image of a homormorphism $G\rightarrow H$ under $\alpha$ is the restriction to $\alpha(G)$) – W4cc0 Jan 14 '14 at 16:41
I proved that if the condition holds then $\alpha$ is a functor but I cannot see why if $\alpha$ is a functor the condition hold. If the condition hold then it can be proved that $\alpha$ is a "verbal" mapping. @JackSchmidt Thanks, I removed it – W4cc0 Jan 14 '14 at 16:43 | 1,196 | 4,135 | {"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} | 2.640625 | 3 | CC-MAIN-2016-26 | latest | en | 0.890121 |
https://www.nag.com/numeric/nl/nagdoc_27/flhtml/f08/f08fcf.html | 1,632,402,278,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057421.82/warc/CC-MAIN-20210923104706-20210923134706-00046.warc.gz | 928,370,847 | 6,732 | # NAG FL Interfacef08fcf (dsyevd)
## 1Purpose
f08fcf computes all the eigenvalues and, optionally, all the eigenvectors of a real symmetric matrix. If the eigenvectors are requested, then it uses a divide-and-conquer algorithm to compute eigenvalues and eigenvectors. However, if only eigenvalues are required, then it uses the Pal–Walker–Kahan variant of the $QL$ or $QR$ algorithm.
## 2Specification
Fortran Interface
Subroutine f08fcf ( job, uplo, n, a, lda, w, work, info)
Integer, Intent (In) :: n, lda, lwork, liwork Integer, Intent (Out) :: iwork(max(1,liwork)), info Real (Kind=nag_wp), Intent (Inout) :: a(lda,*), w(*) Real (Kind=nag_wp), Intent (Out) :: work(max(1,lwork)) Character (1), Intent (In) :: job, uplo
#include <nag.h>
void f08fcf_ (const char *job, const char *uplo, const Integer *n, double a[], const Integer *lda, double w[], double work[], const Integer *lwork, Integer iwork[], const Integer *liwork, Integer *info, const Charlen length_job, const Charlen length_uplo)
The routine may be called by the names f08fcf, nagf_lapackeig_dsyevd or its LAPACK name dsyevd.
## 3Description
f08fcf computes all the eigenvalues and, optionally, all the eigenvectors of a real symmetric matrix $A$. In other words, it can compute the spectral factorization of $A$ as
$A=ZΛZT,$
where $\Lambda$ is a diagonal matrix whose diagonal elements are the eigenvalues ${\lambda }_{i}$, and $Z$ is the orthogonal matrix whose columns are the eigenvectors ${z}_{i}$. Thus
$Azi=λizi, i=1,2,…,n.$
## 4References
Anderson E, Bai Z, Bischof C, Blackford S, Demmel J, Dongarra J J, Du Croz J J, Greenbaum A, Hammarling S, McKenney A and Sorensen D (1999) LAPACK Users' Guide (3rd Edition) SIAM, Philadelphia https://www.netlib.org/lapack/lug
Golub G H and Van Loan C F (1996) Matrix Computations (3rd Edition) Johns Hopkins University Press, Baltimore
## 5Arguments
1: $\mathbf{job}$Character(1) Input
On entry: indicates whether eigenvectors are computed.
${\mathbf{job}}=\text{'N'}$
Only eigenvalues are computed.
${\mathbf{job}}=\text{'V'}$
Eigenvalues and eigenvectors are computed.
Constraint: ${\mathbf{job}}=\text{'N'}$ or $\text{'V'}$.
2: $\mathbf{uplo}$Character(1) Input
On entry: indicates whether the upper or lower triangular part of $A$ is stored.
${\mathbf{uplo}}=\text{'U'}$
The upper triangular part of $A$ is stored.
${\mathbf{uplo}}=\text{'L'}$
The lower triangular part of $A$ is stored.
Constraint: ${\mathbf{uplo}}=\text{'U'}$ or $\text{'L'}$.
3: $\mathbf{n}$Integer Input
On entry: $n$, the order of the matrix $A$.
Constraint: ${\mathbf{n}}\ge 0$.
4: $\mathbf{a}\left({\mathbf{lda}},*\right)$Real (Kind=nag_wp) array Input/Output
Note: the second dimension of the array a must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
On entry: the $n$ by $n$ symmetric matrix $A$.
• If ${\mathbf{uplo}}=\text{'U'}$, the upper triangular part of $A$ must be stored and the elements of the array below the diagonal are not referenced.
• If ${\mathbf{uplo}}=\text{'L'}$, the lower triangular part of $A$ must be stored and the elements of the array above the diagonal are not referenced.
On exit: if ${\mathbf{job}}=\text{'V'}$, a is overwritten by the orthogonal matrix $Z$ which contains the eigenvectors of $A$.
5: $\mathbf{lda}$Integer Input
On entry: the first dimension of the array a as declared in the (sub)program from which f08fcf is called.
Constraint: ${\mathbf{lda}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
6: $\mathbf{w}\left(*\right)$Real (Kind=nag_wp) array Output
Note: the dimension of the array w must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
On exit: the eigenvalues of the matrix $A$ in ascending order.
7: $\mathbf{work}\left(\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{lwork}}\right)\right)$Real (Kind=nag_wp) array Workspace
On exit: if ${\mathbf{info}}={\mathbf{0}}$, ${\mathbf{work}}\left(1\right)$ contains the required minimal size of lwork.
8: $\mathbf{lwork}$Integer Input
On entry: the dimension of the array work as declared in the (sub)program from which f08fcf is called.
If ${\mathbf{lwork}}=-1$, a workspace query is assumed; the routine only calculates the minimum dimension of the work array, returns this value as the first entry of the work array, and no error message related to lwork is issued.
Constraints:
• if ${\mathbf{n}}\le 1$, ${\mathbf{lwork}}\ge 1$ or ${\mathbf{lwork}}=-1$;
• if ${\mathbf{job}}=\text{'N'}$ and ${\mathbf{n}}>1$, ${\mathbf{lwork}}\ge 2×{\mathbf{n}}+1$ or ${\mathbf{lwork}}=-1$;
• if ${\mathbf{job}}=\text{'V'}$ and ${\mathbf{n}}>1$, ${\mathbf{lwork}}\ge 2×{{\mathbf{n}}}^{2}+6×{\mathbf{n}}+1$ or ${\mathbf{lwork}}=-1$.
9: $\mathbf{iwork}\left(\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{liwork}}\right)\right)$Integer array Workspace
On exit: if ${\mathbf{info}}={\mathbf{0}}$, ${\mathbf{iwork}}\left(1\right)$ contains the required minimal size of liwork.
10: $\mathbf{liwork}$Integer Input
On entry: the dimension of the array iwork as declared in the (sub)program from which f08fcf is called.
If ${\mathbf{liwork}}=-1$, a workspace query is assumed; the routine only calculates the minimum dimension of the iwork array, returns this value as the first entry of the iwork array, and no error message related to liwork is issued.
Constraints:
• if ${\mathbf{n}}\le 1$, ${\mathbf{liwork}}\ge 1$ or ${\mathbf{liwork}}=-1$;
• if ${\mathbf{job}}=\text{'N'}$ and ${\mathbf{n}}>1$, ${\mathbf{liwork}}\ge 1$ or ${\mathbf{liwork}}=-1$;
• if ${\mathbf{job}}=\text{'V'}$ and ${\mathbf{n}}>1$, ${\mathbf{liwork}}\ge 5×{\mathbf{n}}+3$ or ${\mathbf{liwork}}=-1$.
11: $\mathbf{info}$Integer Output
On exit: ${\mathbf{info}}=0$ unless the routine detects an error (see Section 6).
## 6Error Indicators and Warnings
${\mathbf{info}}<0$
If ${\mathbf{info}}=-i$, argument $i$ had an illegal value. An explanatory message is output, and execution of the program is terminated.
${\mathbf{info}}>0$
If ${\mathbf{info}}=〈\mathit{\text{value}}〉$ and ${\mathbf{job}}=\text{'N'}$, the algorithm failed to converge; $〈\mathit{\text{value}}〉$ elements of an intermediate tridiagonal form did not converge to zero; if ${\mathbf{info}}=〈\mathit{\text{value}}〉$ and ${\mathbf{job}}=\text{'V'}$, then the algorithm failed to compute an eigenvalue while working on the submatrix lying in rows and column $〈\mathit{\text{value}}〉/\left({\mathbf{n}}+1\right)$ through .
## 7Accuracy
The computed eigenvalues and eigenvectors are exact for a nearby matrix $\left(A+E\right)$, where
$E2 = Oε A2 ,$
and $\epsilon$ is the machine precision. See Section 4.7 of Anderson et al. (1999) for further details.
## 8Parallelism and Performance
f08fcf is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library.
f08fcf makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.
Please consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this routine. Please also consult the Users' Note for your implementation for any additional implementation-specific information.
The complex analogue of this routine is f08fqf.
## 10Example
This example computes all the eigenvalues and eigenvectors of the symmetric matrix $A$, where
$A = 1.0 2.0 3.0 4.0 2.0 2.0 3.0 4.0 3.0 3.0 3.0 4.0 4.0 4.0 4.0 4.0 .$
### 10.1Program Text
Program Text (f08fcfe.f90)
### 10.2Program Data
Program Data (f08fcfe.d)
### 10.3Program Results
Program Results (f08fcfe.r) | 2,468 | 7,685 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 93, "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-2021-39 | latest | en | 0.59678 |
https://slopeinterceptform.net/rags-to-riches-slope-intercept-form/ | 1,653,168,259,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00208.warc.gz | 593,859,682 | 13,739 | # Rags To Riches Slope Intercept Form
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
Rags To Riches Slope Intercept Form – One of the many forms employed to depict a linear equation, one of the most frequently found is the slope intercept form. You may use the formula of the slope-intercept to identify a line equation when that you have the straight line’s slope , and the y-intercept. This is the point’s y-coordinate where the y-axis is intersected by the line. Learn more about this particular line equation form below.
## What Is The Slope Intercept Form?
There are three main forms of linear equations, namely the standard, slope-intercept, and point-slope. While they all provide similar results when used, you can extract the information line produced faster with an equation that uses the slope-intercept form. It is a form that, as the name suggests, this form makes use of a sloped line in which its “steepness” of the line is a reflection of its worth.
The formula can be used to find a straight line’s slope, the y-intercept (also known as the x-intercept), which can be calculated using a variety of available formulas. The line equation of this particular formula is y = mx + b. The straight line’s slope is indicated with “m”, while its y-intercept is signified with “b”. Each point of the straight line is represented as an (x, y). Note that in the y = mx + b equation formula the “x” and the “y” have to remain as variables.
## An Example of Applied Slope Intercept Form in Problems
The real-world in the real world, the slope intercept form is frequently used to illustrate how an item or issue changes over its course. The value given by the vertical axis is a representation of how the equation tackles the degree of change over the amount of time indicated through the horizontal axis (typically the time).
An easy example of the application of this formula is to discover the rate at which population increases in a specific area in the course of time. In the event that the area’s population increases yearly by a fixed amount, the point amount of the horizontal line will rise one point at a moment each year and the values of the vertical axis will grow to represent the growing population by the amount fixed.
You can also note the starting point of a challenge. The starting value occurs at the y-value of the y-intercept. The Y-intercept represents the point at which x equals zero. In the case of the problem mentioned above the beginning value will be at the time the population reading begins or when time tracking begins , along with the related changes.
Thus, the y-intercept represents the location that the population begins to be recorded by the researcher. Let’s assume that the researcher begins to do the calculation or measurement in the year 1995. This year will represent considered to be the “base” year, and the x 0 points will be observed in 1995. Thus, you could say that the population of 1995 is the y-intercept.
Linear equation problems that utilize straight-line formulas can be solved in this manner. The initial value is represented by the yintercept and the rate of change is expressed by the slope. The primary complication of the slope-intercept form generally lies in the interpretation of horizontal variables, particularly if the variable is accorded to one particular year (or any other kind in any kind of measurement). The trick to overcoming them is to ensure that you know the meaning of the variables. | 734 | 3,507 | {"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} | 4.59375 | 5 | CC-MAIN-2022-21 | latest | en | 0.946048 |
https://lists.freedesktop.org/archives/pixman/2016-January/004229.html | 1,695,331,003,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506045.12/warc/CC-MAIN-20230921210007-20230922000007-00621.warc.gz | 426,255,544 | 2,465 | # [Pixman] [PATCH 06/13] pixman-filter: Correct Simpsons integration
Oded Gabbay oded.gabbay at gmail.com
Mon Jan 4 01:32:52 PST 2016
```On Mon, Jan 4, 2016 at 5:12 AM, <spitzak at gmail.com> wrote:
> From: Bill Spitzak <spitzak at gmail.com>
>
> Simpsons uses cubic curve fitting, with 3 samples defining each cubic. This
> makes the weights of the samples be in a pattern of 1,4,2,4,2...4,1, and then
> dividing the result by 3.
>
> The previous code was using weights of 1,2,6,6...6,2,1. Since it divided by
> 3 this produced about 2x the desired value (the normalization fixed this).
> Also this is effectively a linear interpolation, not Simpsons integration.
>
> With this fix the integration is accurate enough that the number of samples
> could be reduced a lot. Likely even 16 samples is too many.
You forgot to sign-off
> ---
> pixman/pixman-filter.c | 17 +++++++++++------
> 1 file changed, 11 insertions(+), 6 deletions(-)
>
> diff --git a/pixman/pixman-filter.c b/pixman/pixman-filter.c
> index 15f9069..5677431 100644
> --- a/pixman/pixman-filter.c
> +++ b/pixman/pixman-filter.c
> @@ -189,8 +189,10 @@ integral (pixman_kernel_t reconstruct, double x1,
> }
> else
> {
> - /* Integration via Simpson's rule */
> -#define N_SEGMENTS 128
> + /* Integration via Simpson's rule
> + * See http://www.intmath.com/integration/6-simpsons-rule.php
> + */
> +#define N_SEGMENTS 16
> #define SAMPLE(a1, a2) \
> (filters[reconstruct].func ((a1)) * filters[sample].func ((a2) / scale))
>
> @@ -204,11 +206,14 @@ integral (pixman_kernel_t reconstruct, double x1,
> {
> double a1 = x1 + h * i;
> double a2 = x2 + h * i;
> + s += 4 * SAMPLE(a1, a2);
> + }
>
> - s += 2 * SAMPLE (a1, a2);
> -
> - if (i >= 2 && i < N_SEGMENTS - 1)
> - s += 4 * SAMPLE (a1, a2);
> + for (i = 2; i < N_SEGMENTS; i += 2)
> + {
> + double a1 = x1 + h * i;
> + double a2 = x2 + h * i;
> + s += 2 * SAMPLE(a1, a2);
> }
>
> s += SAMPLE (x1 + width, x2 + width);
> --
> 1.9.1
>
> _______________________________________________
> Pixman mailing list
> Pixman at lists.freedesktop.org
> http://lists.freedesktop.org/mailman/listinfo/pixman
Patch is:
Reviewed-by: Oded Gabbay <oded.gabbay at gmail.com>
``` | 766 | 2,422 | {"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-2023-40 | latest | en | 0.676582 |
https://octave.1599824.n4.nabble.com/Vectorization-of-matrix-td4693139.html | 1,571,530,895,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986700560.62/warc/CC-MAIN-20191020001515-20191020025015-00254.warc.gz | 630,681,307 | 12,861 | # Vectorization of matrix
7 messages
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## Vectorization of matrix
X_norm = (X - (t * mu)) ./ (t * sigma); %vectorized In above Octave code, if X is 97 x 3 matrix then will it implicitly get converted to column vector? -- Sent from: http://octave.1599824.n4.nabble.com/Octave-General-f1599825.html
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## Re: Vectorization of matrix
> X_norm = (X - (t * mu)) ./ (t * sigma); %vectorized > > In above Octave code, if X is 97 x 3 matrix then will it implicitly get > converted to column vector? I do not see any operation there that would reduce the size of the X matrix. so i dare say no, that wont happen. If this is a norm, you probably need some matrix product against a transpose of X or something on that line.
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## Re: Vectorization of matrix
In reply to this post by mayank9889 On Mon, Jun 17, 2019, 8:50 AM mayank9889 <[hidden email]> wrote:X_norm = (X - (t * mu)) ./ (t * sigma); %vectorized In above Octave code, if X is 97 x 3 matrix then will it implicitly get converted to column vector? What are the shapes of t mu and sigma
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## Re: Vectorization of matrix
%X is 97 x 3 matrix X_norm = X; mu = zeros(1, size(X, 2)); sigma = zeros(1, size(X, 2)); mu = mean(X); sigma = std(X); t = ones(length(X), 1); X_norm = (X - (t * mu)) ./ (t * sigma); % Vectorized -- Sent from: http://octave.1599824.n4.nabble.com/Octave-General-f1599825.html
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## Re: Vectorization of matrix
In reply to this post by Juan Pablo Carbajal-2 This is the complete code %X is 97 x 3 matrix X_norm = X; mu = zeros(1, size(X, 2)); sigma = zeros(1, size(X, 2)); mu = mean(X); sigma = std(X); t = ones(length(X), 1); X_norm = (X - (t * mu)) ./ (t * sigma); % Vectorized -- Sent from: http://octave.1599824.n4.nabble.com/Octave-General-f1599825.html | 582 | 1,910 | {"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} | 3.40625 | 3 | CC-MAIN-2019-43 | latest | en | 0.816871 |
https://forums.ankiweb.net/t/insertion-order-insert-just-before-the-new-card-queue/43892 | 1,716,678,523,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058858.81/warc/CC-MAIN-20240525222734-20240526012734-00250.warc.gz | 214,918,545 | 8,985 | # Insertion Order: Insert just before the new card queue
Background: I’m trying to edit a deck for a standardised exam. The deck is based on some textbook which I read along with the deck. Now the deck author have not included every piece of information from the books which I’m trying to include now.
Issue: When I encounter that some potential flashcard hasn’t been included in the deck I want to include it in the deck, but I want to maintain the order of the Textbooks. This would mean assigning a due number to the card that is just below the lowest due number of a new card. (I’m explaining this in next paragraph)
Solution: A new Insertion Order that would do the afore-mentioned. Ideally it would assign the new cards the lowest due number that a new card has and reposition the rest of the cards. So for instance, if the cards are with due number 1, 2, 3, 4, … etc., and we’ve already gone thorough 99 of them—and the next new card has a due number of 100— then the newly created card will get a due number of 100 and the rest of the cards will get repositioned.
Repositioning the existing cards would modify every new card in your collection, leading to longer syncs.
2 Likes
Not many people will be using this. For those who will, this is still better than nothing. If there is a simple way of solving my issue I’d be happy to hear.
The only other way that comes to my mind is reposition cards once but leave at least 10 unused due numbers between the used due numbers and then fill it up later whenever cards are added. It would get the job done for at least my case.
There’s no reason to move any of the other cards though. It doesn’t hurt anything for multiple cards to have the same New queue number. And the number stops mattering as soon as the card is introduced.
If you’re trying to see a card soon, just Reposition it to any arbitrarily low number. If you know where you are in your New queue, and you want to see it soon-but-not-immediately, you can pick something appropriate among those numbers – but otherwise, `0` works just fine.
When I encounter new vocabulary in the wild, and add an note or unsuspend it in the pre-made part of my deck – it’s easy to decide right then whether I want it near the front of my queue.
@Danika_Dakika @dae I think I was not clear enough. If I was doing a deck for myself this wouldn’t have been necessary. I am instead trying to edit a deck for everyone prepping for the NEET UG. Anubis Nekhet has made some really nice decks many are using. But he cracked the test last year and probabaly won’t be making any updates as he’s really busy. I thought I would share mine with everyone but having the cards sorted properly becomes important. It also means others can easily find missing information in decks and add them on their own. Recent changes in syllabus also mean some content have gotten irrelevant and some have been added a new. The syllabus has been changed multiple times recently and I expect more turbulence as States across India brace for the New Education Policy.
If I add cards normally what would end up happening is all newly added cards would be at the end of the deck instead of the order it should be according to the Textbooks. Textbook order is also a good-to-have feature for large community maintained decks as others can easily see what’s there and what’s not. Hopefully, my idea or some others’ is able to achieve that.
That was definitely not clear from your posts.
But the answer is pretty much the same – and what I would suggest to anyone sharing or publishing a deck. Before you release the file, import your apkg into a fresh profile – put the cards in the right order, Reposition as necessary, and then make your final export apkg for sharing.
To deal with the complexities of the changing syllabus, make sure you’re using tags (or at least subdecks), to keep the material organized. Then you’ll be able to use those to get the cards in the right queue order too.
Or a variant of this: keep the cards as new in the source profile, and import them into a separate profile for studying (either by you, or others using the deck)
Well repositioning later, manually, is what I dread.
For illustration if the deck author misses some information about blue-green algae, I first need to find the cards that cover Kingdom Monera → Eubacteria and the flashcard that should come before it so that I can put it in right place. Maybe I am being too pedantic about precision here but it is just that I was thrown off by wrong order a couple times by pre-made decks. But even if I compromise on precision still it stands that repositioning this way will take a lot of time and effort even after categorising things as much as I can.
What you’re proposing would have to be done manually too though – whenever you want to add a card at the “front” of the queue, you’ll have to change the deck options to that.
The easiest thing would be to just reposition the cards to the generally correct area at the time you create them, and then fine-tune that when you’re done. You can section the current cards out into 100s or 1000s, and keep a list for yourself – if that’s easier. Otherwise, just add your note > search `monera OR Eubacteria` [or the tags that you’re using, as already suggested] > Reposition the new ones to match something already in that section. And then move on.
If we get that feature, I’ll set my Insertion Order to that, and I am never changing it. I mean I don’t need to. I’ll only add cards when I’m studying the deck and then don’t see something that I should.
@dae Can the reposition feature get changed so that only the deck I’m adding cards to is affected and others aren’t.That would partly solve worries of longer sync times.
Maybe in the future.
1 Like
Hey @dae, If I’m not wrong your issue with this was that repositioning cards would make syncs longer, right? As Danika mentioned, two cards can have the same queue number. If that’s correct then repositioning the cards become completely unnecessary. The new cards just needs to be in the top of the queue. Please let me know if you think this is still not workable.
It’s not clear to me why you’d want two cards to have the same due number, instead of the new card getting a smaller number.
Well then I can reposition the cards later and it will have the correct ordering and I wouldn’t have to go through the rigamarole of Card info of present card → Note down due number → Add card → Card Browser → Reposition the newly added card
Ideally, I would want the new cards to get a lower due number (than the already studied card; so that it is on the start of the new card queue) but I understood that such a feature repositioning cards each and every time is something you do not like. I think I’ve explained my situation in the previous posts. TL;DR: I’m adding missing cards to a deck. I want them to have the same ordering as in the textbook. If studying the deck and I discover a particular factoid was skipped by the deck author I’d want it to be added but I’d also want it to be added to the correct position. | 1,585 | 7,094 | {"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} | 3.40625 | 3 | CC-MAIN-2024-22 | latest | en | 0.966523 |
https://www.studypool.com/services/90619/the-phenomenon-of-newton-s-rings | 1,544,472,681,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823442.17/warc/CC-MAIN-20181210191406-20181210212906-00022.warc.gz | 1,066,800,708 | 22,103 | # The phenomenon of Newton's rings
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The phenomenon of Newton's rings, named after Isaac Newton who first studied them in 1717, is an interference pattern caused by the reflection of light between two surfaces - a spherical surface and an adjacent flat surface. When viewed with monochromatic light it appears as a series of concentric
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Measuring the Moment of Inertia of a Bicycle WheelAlexandre Dub, (# 110234667)Paris Hubbard Davis, (# 110222984)Philippe Roy, (# 110235920)Guillaume Rivest, (# 110227286)McGill University, February 3, 2002ABSTRACTFIGURE 1:Measuring the Moment of Inertia of a Bicycle Wheel. Illustration of the basic setup for the experiment. A more detailed scheme is presented in the third section, Apparatus and Procedure.Moment of inertia is a quantity which varies as the axis rotation varies, or, more clearly, as the distance from the axis varies. Our goals were two: 1) experimentally measure the moment of inertia and then check this result with a theoretical measure, and 2) play with the bicycle wheel. The former was conducted with weights exerting torques about the rim for a revolution of the wheel. The resulting period was measured, and via conservation of energy, moment of inertia was derived. The latter consisted of a summation: the weight and distance from the centre of the various parts of the wheel: hub, spokes, nipples, and rim. The conservation of energy method and a good approximation of the theoretical value were consistent within error.I. INTRODUCTIONII. THEORYThe experimental derivation of moment of inertia was conducted by equating energy at two different times. Our first approximation is that g is constant. We set the zero of potential energy to be at the floor, and energy is given by:(1)1The period of the wheels half revolution ( radi
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https://aprove.informatik.rwth-aachen.de/eval/JAR06/JAR_TERM/TRS/AG01/%233.6a.trs.Thm17:POLO_7172_DP:NO.html.lzma | 1,719,244,574,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865401.1/warc/CC-MAIN-20240624151022-20240624181022-00732.warc.gz | 78,506,898 | 2,378 | Term Rewriting System R:
[y, x]
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))
Termination of R to be shown.
R
Dependency Pair Analysis
R contains the following Dependency Pairs:
LE(s(x), s(y)) -> LE(x, y)
MINUS(s(x), s(y)) -> MINUS(x, y)
GCD(s(x), s(y)) -> IFGCD(le(y, x), s(x), s(y))
GCD(s(x), s(y)) -> LE(y, x)
IFGCD(true, s(x), s(y)) -> GCD(minus(x, y), s(y))
IFGCD(true, s(x), s(y)) -> MINUS(x, y)
IFGCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))
IFGCD(false, s(x), s(y)) -> MINUS(y, x)
Furthermore, R contains three SCCs.
R
DPs
→DP Problem 1
Polynomial Ordering
→DP Problem 2
Polo
→DP Problem 3
Polo
Dependency Pair:
LE(s(x), s(y)) -> LE(x, y)
Rules:
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))
The following dependency pair can be strictly oriented:
LE(s(x), s(y)) -> LE(x, y)
There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(LE(x1, x2)) = x1 POL(s(x1)) = 1 + x1
resulting in one new DP problem.
R
DPs
→DP Problem 1
Polo
→DP Problem 4
Dependency Graph
→DP Problem 2
Polo
→DP Problem 3
Polo
Dependency Pair:
Rules:
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))
Using the Dependency Graph resulted in no new DP problems.
R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polynomial Ordering
→DP Problem 3
Polo
Dependency Pair:
MINUS(s(x), s(y)) -> MINUS(x, y)
Rules:
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))
The following dependency pair can be strictly oriented:
MINUS(s(x), s(y)) -> MINUS(x, y)
There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(MINUS(x1, x2)) = x1 POL(s(x1)) = 1 + x1
resulting in one new DP problem.
R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 5
Dependency Graph
→DP Problem 3
Polo
Dependency Pair:
Rules:
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))
Using the Dependency Graph resulted in no new DP problems.
R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 3
Polynomial Ordering
Dependency Pairs:
IFGCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))
IFGCD(true, s(x), s(y)) -> GCD(minus(x, y), s(y))
GCD(s(x), s(y)) -> IFGCD(le(y, x), s(x), s(y))
Rules:
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))
The following dependency pairs can be strictly oriented:
IFGCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))
IFGCD(true, s(x), s(y)) -> GCD(minus(x, y), s(y))
Additionally, the following usable rules using the Ce-refinement can be oriented:
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(0) = 0 POL(GCD(x1, x2)) = x1 + x2 POL(false) = 0 POL(minus(x1, x2)) = x1 POL(true) = 0 POL(IF_GCD(x1, x2, x3)) = x2 + x3 POL(s(x1)) = 1 + x1 POL(le(x1, x2)) = 0
resulting in one new DP problem.
R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 3
Polo
→DP Problem 6
Dependency Graph
Dependency Pair:
GCD(s(x), s(y)) -> IFGCD(le(y, x), s(x), s(y))
Rules:
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))
Using the Dependency Graph resulted in no new DP problems.
Termination of R successfully shown.
Duration:
0:00 minutes | 1,973 | 5,025 | {"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-2024-26 | latest | en | 0.628443 |
https://techytok.com/lesson-variables-and-types/ | 1,582,608,087,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146033.50/warc/CC-MAIN-20200225045438-20200225075438-00184.warc.gz | 555,879,389 | 8,903 | # Variables and Types
In this lesson you will learn what are variables and how to perform simple mathematical operations. Furthermore we will deal with the concept of “types” and their role in Julia.
# Variables
If we want to really simplify what a program is and what it does, we may say that a program is a series of instructions to be performed on various kind of data. Such data may be the name of a place, the height of a person or a series of measurements made by a scientist. In programming, when we want to “label” a piece of information we give it a name and call it a variable, so we may say that a variable is a box which contains some kind of data.
In Julia, when we want to assign some data to a variable, we do it by typing:
my_name = "Aurelio"
my_favourite_number = 42
my_favourite_pie = 3.1415
Here my_name contains a string, which is a piece of text, while my_favourite_number contains an integer and my_favourite_pie a floating point number. Contrary to other programming languages like C++ and similarly to Python, Julia can infer the type of the object on the right side of the equal sign, so there is no need to specify the type of the variable as you would do in C++.
If we want to print the value of a variable, we use print:
>>>print(my_name)
Aurelio
It is possible to perform mathematical operations on numbers and variables and assign the results to new variables:
a = 2
b = 3
sum = a + b
difference = a - b
product = a * b
quotient = b / a
power = a^3
modulus = b % a
It is also possible to use rational numbers, if it is needed for exact numerical computations, and they take the form of numerator//denominator, e.g. 1//3.
For a complete list of the mathematical operations see the official guide.
# Types
In Julia every element has a type. The type system is a hierarchical structure: at the top of the tree there is the type Any, which means that every element belongs to it, then there are many other sub-types, for example Number which includes Real and Complex, and Real contains for example Int (integer) numbers and Float64 numbers.
We will deal with types again later, for now what we need to remember is that since every variable has a type (which can be inferred by the compiler or specified by the programmer), Julia can build machine code optimised for the specific types which are being used. This is one of the main reasons why Julia is a really fast language! Furthermore, we can expect that if a function can work on a Type, for example a Number, it will also work on all of its Subtypes, like Float64 or Int64.
To determine the type of a variable, we can use the typeof function:
>>>typeof(0.1)
Float64
>>>typeof(42)
Int64
>>>typeof("TechyTok")
String
Although it is possible to change the value of a variable inside a program (it is a variable, after all) it is good programming practice and is also critical for performance that inside a program a variable is “type stable”. This means that if we have assigned a = 42 it is better not to assign a new value to a which cannot be converted into an Int without losing information, like a Float64 a = 0.42 (if we convert a Float64 to an Int, the decimal part gets truncated).
If we know that a variable (such as a) will have to contain values of type Float64 it is better to initialise it with a value that is already of that type.
a = 2 # if we need to operate with ints
b = 2.0 # if we need to operate with floats
It is possible, if needed, to convert a value from a type to another using the function convert, for example:
>>>convert(Float64, 2)
2.00
a = 2
b = convert(Float64, a)
>>>a
2
>>>b
2.00
# Conclusion
We have learned what variables are, how to perform basic operations and we have dealt about types.
In the next lesson we will try to understand what functions are and how they can be used to write versatile and reusable code!
If you liked this lesson and you would like to receive further updates on what is being published on this website, I encourage you to subscribe to the newsletter! If you have any question or suggestion, please post them in the discussion below!
Thank you for reading this lesson and see you soon on TechyTok!
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http://life.familyeducation.com/nutrition-and-diet/weight/35881.html?page=1 | 1,464,429,927,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049277592.42/warc/CC-MAIN-20160524002117-00062-ip-10-185-217-139.ec2.internal.warc.gz | 175,718,020 | 13,516 | Home > Mom's Life > Mom's Health and Fitness > Weight Management > Calorie Guidelines for Women
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# Calorie Guidelines for Women
Calorie counting can be a source of frustration and focus for many girls and women. Although formulas have been standardized, calories might need to be adjusted for each individual based on genetics, amount of muscle or fat, and other activity in addition to athletic activities. The amount of calories required to maintain weight for one woman can be more or less than for another; and, just like sleep, training, and other aspects of life, the amounts can never be exact, as all bodies are different. Still, estimates of calorie needs can provide helpful guidelines.
Nutritionists have devised various complicated formulas for estimating calorie needs. The standards are all in kilograms (kg) of body weight, which can be converted as 1 kg= 2.2 pounds. There are three formulas commonly used by nutritionists, the Harris Benedict, the Total Energy Expenditure Method, and Activity Factors. The first two formulas do not account for additional calories burned in exercise, and they should only be used as a guideline for basic calorie needs.
Harris Benedict Equation
Daily calories (kcal) = [447.6 + 9.2 (Weight in kg) + 3.1 (Height in cm) -4.3 (age)] x 1.7
Total Energy Expenditure Method
Daily calories (kcal) = Weight in kg x 40 kcal/kg
To make it easier and more specific to your athletic activity level, measure your weight in pounds by the following numbers based on activity level. Light, for which you multiply your body weight by 13, implies 20 to 30 minutes total aerobic activity a day (heart rate at 65 to 80 percent). Moderate, a factor of 16, implies 45 to 60 minutes of aerobic activity a day. Heavy, a factor of 19, implies 75 to 90 minutes. Exceptional, a factor of 22, implies more than 2 hours a day.
Activity Factor
Daily Calories = Weight in lbs. x Activity Factor (AF)
Calories Required at Various Body Weights, Based on Activity Level
Weight Light Moderate Heavy Exceptional Pounds AF* = 13, Light AF* = 16, Moderate AF* = 19, Heavy AF* = 22, Exceptional 115 1,495 1,840 2,185 2,530 120 1,560 1,920 2,280 2,640 125 1,625 2,000 2,375 2,750 130 1,690 2,080 2,470 2,860 135 1,755 2,160 2,565 2,970 140 1,820 2,240 2,660 3,080 145 1,885 2,320 2,755 3,190 150 1,950 2,400 2,850 3,300 155 2,015 2,480 2,945 3,410 160 2,080 2,560 3,040 3,520 165 2,145 2,640 3,135 3,630 170 2,210 2,720 3,230 3,740 175 2,275 2,800 3,325 3,850 180 2,340 2,880 3,420 3,960 185 2,405 2,960 3,515 4,070
Modern technology also makes it possible to have your baseline calorie burning rate measured. This test is now being done in some health clubs. It is done by monitoring breathing to measure oxygen used during a short time. This gives an estimate of your basal metabolic rate, or how many calories your body burns in a day of rest. Unless this testing is done in a specific physiology lab, the results can vary, and, just like the formulas on previous pages, should be used only as a guideline. The healthiest assessment of body composition, weight, and appropriate caloric intake and activity levels specific for you are made with the combined efforts of your physician and nutritionist. It is highly recommended to consult with both of these health professionals if you are seeking help with weight changes.
A Balanced Diet
The ideal diet for an active woman includes 50 to 60 percent carbohydrates, 20 to 30 percent protein, and 20 to 30 percent fats. To fulfill your vitamin and mineral requirements, a multivitamin is always recommended, and if you do not eat at least four servings of dairy or calcium-fortified foods a day, add a calcium supplement. The best brands are those made by reputable drug companies.
Essential Components of a Healthy Diet
• 50 to 60 percent carbohydrates
• 20 to 30 percent protein
• 20 to 30 percent fat
• Enough calories
• A daily multivitamin
• Four servings of calcium-rich foods or supplements
To make it easy, the following chart breaks down the recommended calories and grams (in parentheses) of each nutrient required for a diet that is 55 percent carbohydrate, 25 percent protein, and 20 percent fat. These numbers are only guidelines, not strict dietary rules, and should be adjusted if you require extra carbohydrates for training (eat slightly less fats and proteins).
Guidelines for a Healthy Athletic Diet
Total cal/d Protein Fat Carbo cal(g)—25% cal(g)—20% cal(g)/day—55% 1,200 300 (75) 240 (27) 660 (165) 1,500 375 (94) 300 (33) 825 (206) 1,800 450 (113) 360 (40) 990 (248) 2,000 500 (125) 400 (44) 1,100 (275) 2,200 550 (138) 440 (49) 1,210 (303) 2,400 600 (150) 480 (53) 1,320 (330) 2,600 650 (163) 520 (58) 1,430 (358) 2,800 700 (175) 560 (62) 1,540 (385) 3,000 750 (188) 600 (67) 1,650 (413) 3,200 800 (200) 640 (71) 1,760 (440) 3,400 850 (213) 680 (76) 1,870 (468) 3,600 900 (225) 720 (80) 1,980 (495) 3,800 950 (238) 760 (84) 2,090 (523)
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From The Active Woman's Health and Fitness Handbook by Nadya Swedan. Copyright © 2003 by Nadya Swedan. Used by arrangement with Perigee, a member of Penguin Group (USA) Inc.
To order this book visit Amazon.
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Family meditation and mindfulness can help reduce anxiety and promote health and happiness. Learn some fun and easy mindfulness activities for kids, and set them on the path to inner peace! | 1,625 | 5,799 | {"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} | 2.859375 | 3 | CC-MAIN-2016-22 | longest | en | 0.920465 |
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MAP2302 Final (2 pts) 1. Give the solution to y 0 + (1 /x ) y = 1 , y (2) = 1. (2 pts) 2. If the equation y 00 + y 0 - 6 y = tan t is to be solved using the Method of Variation of Parameters, write the resulting system of equations used in the technique. (Do not solve the system!) (2 pts) 3. Find a particular solution to y 00 + 2 y 0 + y = te 2 t . (2 pts) 4. Find the solution to the following IVP, expressing the answer in terms of a convolution: y 00 + 2 y 0 + 2 y = g ( t ) , y (0) = 0 , y 0 (0) = 0 .
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(2 pts) 5. Solve: (2 x + y ) dx + ( x + y - 2) dy = 0. (2 pts) 6. Classify (separable, etc.) the following equation: y 0 = 2 x + 3 y 5 x - 2 y . (2 pts) 7. If y 1 and y 2 are solutions to the equation ay 00 + by 0 + cy = 0, then what is the name of the the function y 1 y 0 2 - y 0 1 y 2 ? (2 pts) 8. Determine the inverse transform of F ( s ) = e - 2 s ( s 2 - 5 s +13) s 3 - 6 s 2 +13 s .
(2 pts) 9. Determine the Laplace Transform of f ( t ) = te t u ( t - 2) . (2 pts) 10. Solve: x 2 y 00 - xy 0 + 4 y = 0 , x > 0 . (2 pts) 11. Given y 00 + e x y 0 + y = 0 , y (0) = - 2 , y 0 (0) = 2 , e x = X n =0 x n n ! , find the value of the coefficient a 3 in a series expansion about x 0 = 0 for y . (2 pts) 12. Given the equation ( x 2 - 9) 2 y 00 +( x - 3) y 0 +5 y = 0 and using the nomenclature from Section 8.6, categorize the point x 0 = 3 .
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(2 pts) 13. Give the general solution to y 00 + 3 y 0 - y = 0. (2 pts) 14. Given 3 x 2 y 00 + xy = 0, use the Method of Frobenius to find a recurrence relation for a series solution about x 0 = 0 for y . (2 pts) 15. If a particular solution to the equation
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## This note was uploaded on 12/15/2011 for the course MAP 2302 taught by Professor Tuncer during the Spring '08 term at University of Florida.
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ODE4 - MAP2302 Final (2 pts) 1. Give the solution to y +...
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# NCERT Exemplar: Conic Sections (Exercise) | Mathematics (Maths) Class 11 - Commerce PDF Download
Table of contents Long Answer Type Questions Objective Answer Type Questions Fill in the Blanks Multiple Choice Types Questions
### Short Answer Type Questions
Q.1: Find the equation of the circle which touches the both axes in first quadrant and whose radius is a.
Ans:
Clearly centre of the circle = (a, a)
and radius = a
Equation of circle with radius r and centre (h, k) is
(x – h)2 + (y – k)2 = r2
So, the equation of the required circle
⇒ (x – a)2 + (y – a)2 = a2
⇒ x2 – 2ax + a2 + y2 – 2ay + a2 = a2
⇒ x2 + y2 – 2ax – 2ay + a2 = 0
Hence, the required equation is
x2 + y2 – 2ax – 2ay + a2 = 0
Q.2.:Show that the point (x, y) given by x = lies on a circle for all real values of t such that –1 ≤ t ≤ 1 where a is any given real numbers.
Ans:
Given
= a2
∴ x2 + y2 = a2 which is the equation of a circle.
Hence, the given points lie on a circle.
Q.3: If a circle passes through the point (0, 0) (a, 0), (0, b) then find the coordinates of its centre.
Ans:
Given points are (0, 0), (a, 0) and (0, b)
General equation of the circle is x2 + y2 + 2gx + 2fy + c = 0
where the centre is (– g, – f) and radius =
If it passes through (0, 0)
∴ c = 0
If it passes through (a, 0) and (0, b) then
a2 + 2ga + c = 0 ⇒ a2 + 2ga = 0 [∵ c = 0]
and 0 + b2 + 0 + 2fb + c = 0 ⇒ b2 + 2fb = 0 [∵ c = 0]
Hence, the coordinates of centre of the circle are (– g, – f)
Q.4: Find the equation of the circle which touches x-axis and whose centre is (1, 2).
Ans: Since the circle whose centre is (1, 2) touch x-axis
∴ r = 2
So, the equation of the circle is
(x – h)2 + (y – k)2 = r2
(x – 1)2 + (y – 2)2 = (2)2
⇒ x2 – 2x + 1 + y2 – 4y + 4 = 4
x2 + y2 – 2x – 4y + 1 = 0
Hence, the required equation is
x2 + y2 – 2x – 4y + 1 = 0.
Q.5: If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then find the radius of the circle.
[Hint: Distance between given parallel lines gives the diameter of the circle.]
Ans: Given equation are 3x – 4y + 4 = 0
and 6x – 8y – 7 = 0 ⇒
Sincethen the lines are parallel.
So, the distance between the parallel lines
Diameter = 3/2
∴ Radius = 3/4
Hence, the required radius = 3/4.
Q.6. Find the equation of a circle which touches both the axes and the line 3x – 4y + 8 = 0 and lies in the third quadrant.
[Hint: Let a be the radius of the circle, then (– a, – a) will be centre and perpendicular distance from the centre to the given line gives the radius of the circle.]
Ans. Let a be the radius of the circle.
Centre of the circle = (– a, – a)
Distance of the line 3x – 4y + 8 = 0
From the centre = Radius of the circle
⇒ a = 5a – 8
⇒ 5a – a = 8
⇒ 4a = 8 ⇒ a = 2
and
∴ The equation of the circle is
(x + 2)2 + (y + 2)2 = (2)2
x2 + 4x + 4 + y2 + 4y + 4 = 4
x2 + y2 + 4x + 4y + 4 = 0
Hence, the required equation of the circle
x2 + y2 + 4x + 4y + 4 = 0.
Q.7. If one end of a diameter of the circle x2 + y2 – 4x – 6y + 11 = 0 is (3, 4), then find the coordinate of the other end of the diameter.
Ans. Let the other end of the diameter is (x1, y1).
Equation of given circle is
x2 + y2 – 4x – 6y + 11 = 0
Centre = (– g, – f) = (2, 3)
Hence, the required coordinates are (1, 2).
Q.8. Find the equation of the circle having (1, –2) as its centre and passing through 3x + y = 14, 2x + 5y = 18
Ans. Given equations are
3x + y = 14 ...(i)
and 2x + 5y = 18 ...(ii)
From eq. (i) we get y = 14 – 3x ...(iii)
Putting the value of y in eq. (ii) we get
⇒ 2x + 5(14 – 3x) = 18
⇒ 2x + 70 – 15x = 18
⇒ – 13x = – 70 + 18
⇒ – 13x = – 52
∴ x = 4
From eq. (iii) we get,
y = 14 – 3 x 4 = 2
∴ Point of intersection is (4, 2)
Now, radius r =
So, the equation of circle is (x – h)2 + (y – k)2 = r2
(x – 1)2 + ( + 2)2 = (5)2
⇒ x2 – 2x + 1 + y2 + 4y + 4 = 25
⇒ x2 + y2 – 2x + 4y – 20 = 0
Hence, the required equation is x2 + y2 – 2x + 4y – 20 = 0
Q.9. If the line y = √3x + k touches the circle x2 + y2 = 16, then find the value of k.
[Hint: Equate perpendicular distance from the centre of the circle to its radius].
Ans. Given circle is x2 + y2 = 16
Centre = (0, 0)
radius r = 4
Perpendicular from the origin to the given line y = √3x + k is equal to the radius.
Hence, the required values of k are ± 8.
Q.10. Find the equation of a circle concentric with the circle x2 + y2 – 6x + 12y + 15 = 0 and has double of its area.
[Hint: concentric circles have the same centre.]
Ans. Given equation of the circle is
x2 + y2 – 6x + 12y + 15 = 0 ...(i)
Centre = (– g, – f) = (3, – 6)
Since the circle is concentric with the given circle
∴ Centre = (3, – 6)
Now let the radius of the circle is r
Area of the given circle (i) = πr2 = 30π sq unit
Area of the required circle = 2 x 30π = 60π sq. unit
If r1 be the radius of the required circle 2
πr12 = 60π ⇒ r21 = 60
So, the required equations of the circle is
(x - 3)2 + (y + 6)2 = 60
x2 + 9 - 6x + y2 + 36 + 12y - 60 = 0
⇒ x2 + y2 – 6x + 12y – 15 = 0
Hence, the required equation is x2 + y2 – 6x + 12y – 15 = 0.
Q.11. If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity.
Ans. Let the equation of an ellipse is
Length of major axis = 2a
Length of minor axis = 2b.
and the length of latus rectum =
According to the question, we have
Now b2 = a2(1 – e2),
where e is the eccentricity
⇒ b2 = 4b2(1 – e2)
⇒ 1 = 4(1 – e2)
Hence, the required value of eccentricity is
Q.12. Given the ellipse with equation 9x2 + 25y2 = 225, find the eccentricity and foci.
Ans. Given equation of ellipse is
9x2 + 25y2 = 225
Here a = 5 and b = 3
b2 = a2(1 – e2)
⇒ 9 = 25(1 – e2)
Now foci = (± ae, 0) =
Hence, eccentricity =
Q.13. If the eccentricity of an ellipse is 5/8 and the distance between its foci is 10, then find latus rectum of the ellipse.
Ans. Equation of an ellipse is
Eccentricity,
Distance between its foci = ae + ae = 2ae
∴ 2ae = 10
Now b2 = a2(1 – e2)
So, the length of the latus rectum =
Hence, the length of the latus rectum = 39/4.
Q.14. Find the equation of ellipse whose eccentricity is 2/3, latus rectum is 5 and the centre is (0, 0).
Ans. Equations of ellipse is ....(i)
Given that,e = 2/3
and latus rectum
...(ii)
We know that b2 = a2(1 – e2)
and
Hence, the required equation of ellipse is
Q.15. Find the distance between the directrices of the ellipse
Ans. Given equation of ellipse is
Here a2 = 36 ⇒ a = 6
b2 = 20 ⇒ b = 2 √5
We know that b2 = a2(1 – e2)
⇒ 20 = 36(1 – e2)
Now distance between the directrices is
Hence, the required distance = 18.
Q.16. Find the coordinates of a point on the parabola y2 = 8x whose focal distance is 4.
Ans. Given parabola is y2 = 8x ...(i)
Comparing with the equation of parabola y2 = 4ax
4a = 8 ⇒ a = 2
Now focal distance =
⇒ (x + a) = ± 4
⇒ x + 2 = ± 4
⇒ x = 4 – 2 = 2 and x = – 6
But x ≠ - 6 ∴ x = 2
Put x = 2 in equation (i) we get y2 = 8 x 2 = 16
∴ y = ± 4
So, the coordinates of the point are (2, 4), (2, – 4).
Hence, the required coordinates are (2, 4) and (2, – 4).
Q.17. Find the length of the line-segment joining the vertex of the parabola y2 = 4ax and a point on the parabola where the line-segment makes an angle θ to the xaxis.
Ans. Equation of parabola is y2 = 4ax
Let P(at2, 2at) be any point on the parabola.
In ΔPOA, we have
⇒ t = 2 cot θ ...(i)
[∴ t = 2 cot θ]
Hence, the required length =
Q.18. If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.
Ans. Given that: Vertex = (0, 4) and Focus = (0, 2)
Let P(x, y) be any point on the parabola. PB is perpendicular to the directrix.
According to the definition of parabola, we have
PF = PB
[Equation of directrix is y = 6]
Squaring both sides, we have
x2 + (y – 2)2 = (y – 6)2
⇒ x2 + y2 + 4 – 4y = y2 + 36 – 12y
⇒ x2 – 4y + 12y – 32 = 0
⇒ x2 + 8y – 32 = 0
Hence, the required equation is x2 + 8y = 32.
Q.19. If the line y = mx + 1 is tangent to the parabola y2 = 4x then find the value of m.
[Hint: Solving the equation of line and parabola, we obtain a quadratic equation and then apply the tangency condition giving the value of m]
Ans. Given that y2 = 4x ...(i)
and y = mx + 1 ...(ii)
From eq. (i) and (ii) we get
(mx + 1)2 = 4x
⇒ m2x2 + 1 + 2mx – 4x = 0
m2x2 + (2m – 4)x + 1 = 0
Applying condition of tangency, we have
(2m – 4)2 – 4m2 x 1 = 0
⇒ 4m2 + 16 - 16m - 4m2 = 0
⇒ – 16m = – 16
⇒ m = 1
Hence, the required value of m is 1.
Q.20. If the distance between the foci of a hyperbola is 16 and its eccentricity is √2 , then obtain the equation of the hyperbola.
Ans. Equation of hyperbola is
Distance between the foci = 2ae
2ae = 16 ⇒ ae = 8
⇒ a x √2 = 8
Now, b2 = a2(e2 – 1) [for hyperbola]
⇒ b2 = (4 √2)2 (2 - 1)
⇒ b2 = 32
a = 4√2 ⇒ a= 32
Hence, the required equation is
⇒ x2 – y2 = 32.
Q.21. Find the eccentricity of the hyperbola 9y2 – 4x2 = 36.
Ans. Given equation is 9y2 – 4x2 = 36
Clearly it is a vertical hyperbola .
Where a = 3 and b = 2
We know that b2 = a2(e2 – 1)
⇒ 4 = 9(e2 – 1)
Hence, the required value of e is
Q.22. Find the equation of the hyperbola with eccentricity 3/2 and foci at (± 2, 0).
Ans. Given that e = 3/2 and foci = (± 2, 0)
We know that foci = (± ae, 0)
∴ ae = 2
We know that b2 = a2(e2 – 1)
So, the equation of the hyperbola is
Hence, the required equation is
## Long Answer Type Questions
Q.23. If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.
Ans. We know that the intersection point of the diameter gives the centre of the circle.
Given equations of diameters are
2x – 3y = 5 ...(i)
3x – 4y = 7 ...(ii)
From eq. (i) we have ...(iii)
Putting the value of x in eq. (ii) we have
⇒ 15 + 9y – 8y = 14
⇒ y = 14 – 15 ⇒ y = – 1
Now from eq. (iii) we have
So, the centre of the circle = (1, – 1)
Given that area of the circle = 154
⇒ πr2 = 154
⇒ r2 = 7 x 7
⇒ r = 7
So, the equation of the circle is
(x – 1)2 + (y + 1)2 = (7)2
⇒ x2 + 1 – 2x + y2 + 1 + 2y = 49
⇒ x2 + y2 – 2x + 2y = 47
Hence, the required equation of the circle is
x2 + y2 – 2x + 2y = 47
Q.24. Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y – 4x + 3 = 0.
Ans. Let the equation of the circle be
(x – h)2 + (y – k)2 = r2 ...(i)
If the circle passes through (2, 3) and (4, 5) then
(2 – h)2 + (3 – k)2 = r2 ...(ii)
and (4 – h)2 + (5 – k)2 = r2 ...(iii)
Subtracting eq. (iii) from eq. (ii) we have
(2 – h)2 – (4 – h)2 + (3 – k)2 – (5 – k)2 = 0
⇒ 4 + h2 - 4h - 16 - h2 + 8h + 9 + k2 - 6k - 25 - k2 + 10k = 0
⇒ 4h + 4k – 28 = 0
⇒ h + k = 7 ....(iv)
Since, the centre (h, k) lies on the line y – 4x + 3 = 0
then k – 4h + 3 = 0
⇒ k = 4h – 3
Putting the value of k in eq. (iv) we get
h + 4h – 3 = 7
⇒ 5h = 10 ⇒ h = 2
From (iv) we get k = 5
Putting the value of h and k in eq. (ii) we have
(2 – 2)2 + (3 – 5)2 = r2
⇒ r2 = 4
So, the equation of the circle is
(x – 2)2 + (y – 5)2 = 4
x2 + 4 – 4x + y2 + 25 – 10y = 4
⇒ x2 + y2 – 4x – 10y + 25 = 0
Hence, the required equation is
x2 + y2 – 4x – 10y + 25 = 0.
Q.25. Find the equation of a circle whose centre is (3, –1) and which cuts off a chord of length 6 units on the line 2x – 5y + 18 = 0.
[Hint: To determine the radius of the circle, find the perpendicular distance from the centre to the given line.]
Ans. Given that:
Centre of the circle = (3, – 1)
Length of chord AB = 6 units
Now AB = 6 units.
In ΔCPA, AC2 = CP2 + AP2
= (√29)2 + (3)2 = 29 + 9 = 38
∴ AC = √38
So, the radius of the circle, r = √38
∴ Equation of the circle is
(x – 3)2 + (y + 1)2 = (√38)2
⇒ (x – 3)2 + (y + 1)2 = 38
⇒ x2 + 9 – 6x + y2 + 1 + 2y = 38
⇒ x2 + y2 – 6x + 2y = 28
Hence, the required equation is x2 + y2 – 6x + 2y = 28.
Q.26. Find the equation of a circle of radius 5 which is touching another circle x2 + y2 – 2x – 4y – 20 = 0 at (5, 5).
Ans. Given circle is
x2 + y2 – 2x – 4y – 20 = 0
2g = – 2 ⇒ g = – 1
2f = – 4 ⇒ f = – 2
∴ Centre C1 = (1, 2)
and
Let the centre of the required circle be (h, k).
Clearly, P is the mid-point of C1C2
Radius of the required circle = 5
∴ Eq. of the circle is (x – 9)2 + (y – 8)2 = (5)2
⇒ x2 + 81 – 18x + y2 + 64 – 16y = 25
⇒ x2 + y2 – 18x – 16y + 145 – 25 = 0
⇒ x2 + y2 – 18x – 16y + 120 = 0
Hence, the required equation is x2 + y2 – 18x – 16y + 120 = 0.
Q.27. Find the equation of a circle passing through the point (7, 3) having radius 3 units and whose centre lies on the line y = x – 1.
Ans. Let the equation of the circle be
(x – h)2 + (y – k)2 = r2
If it passes through (7, 3) then
(7 – h)2 + (3 – k)2 = (3)2 [∵ r = 3]
⇒ 49 + h2 – 14h + 9 + k2 – 6k = 9
⇒ h2 + k2 – 14h – 6k + 49 = 0 ...(i)
If centre (h, k) lies on the line y = x – 1 then
k = h – 1 ...(ii)
Putting the value of k in eq. (i) we get
h2 + (h – 1)2 – 14h – 6(h – 1) + 49 = 0
⇒ h2 + h2 + 1 – 2h – 14h – 6h + 6 + 49 = 0
⇒ 2h2 – 22h + 56 = 0
⇒ h2 – 11h + 28 = 0
⇒ h2 – 7h – 4h + 28 = 0
⇒ h(h – 7) – 4(h – 7) = 0
⇒ (h – 4)(h – 7) = 0
∴ h = 4, h = 7
From eq. (ii) we get k = 4 – 1 = 3 and k = 7 – 1 = 6.
So, the centres are (4, 3) and (7, 6).
∴ Equation of the circle is
Taking centre (4, 3),
(x – 4)2 + (y – 3)2 = 9
x2 + 16 – 8x + y2 + 9 – 6y = 9
⇒ x2 + y2 – 8x – 6y + 16 = 0
Taking centre (7, 6)
(x – 7)2 + (y – 6)2 = 9
⇒ x2 + 49 – 14x + y2 + 36 – 12y = 9
⇒ x2 + y2 – 14x – 12y + 76 = 0
Hence, the required equations are
x2 + y2 – 8x – 6y + 16 = 0
and x2 + y2 – 14x – 12y + 76 = 0.
Q.28. Find the equation of each of the following parabolas
(a) Directrix x = 0, focus at (6, 0)
(b) Vertex at (0, 4), focus at (0, 2)
(c) Focus at (–1, –2), directrix x – 2y + 3 = 0
Ans.
(a) Given that directrix = 0 and focus (6, 0)
∴ The equation of the parabola is
(x – 6)2 + y2 = x2
x2 + 36 – 12x + y2 = x2
⇒ y2 – 12x + 36 = 0
Hence, the required equations is y2 – 12x + 36 = 0
(b) Given that vertex at (0, 4) and focus at (0, 2).
So, the equation of directrix is y – 6 = 0
According to the definition of the parabola
PF = PM.
Squaring both the sides, we get
x2 + y2 + 4 – 4y = y2 + 36 – 12y
x2 + 4 – 4y = 36 – 12y
⇒ x2 + 8y – 32 = 0
⇒ x2 = 32 – 8y
Hence, the required equation is x2 = 32 – 8y.
(c) Given that focus at (– 1, – 2) and directrix x – 2y + 3 = 0
Let (x, y) be any point on the parabola.
According to the definition of the parabola, we have
PF = PM
Squaring both sides, we get
x2 + 1 + 2x + y2 + 4 + 4y =
⇒ 5x2 + 5 + 10x + 5y2 + 20 + 20y
= x2 + 4y2 + 9 – 4xy – 12y + 6x
4x2 + y2 + 4xy + 4x + 32y + 16 = 0
Hence, the required equation is
4x2 + 4xy + y2 + 4x + 32y + 16 = 0
Q.29. Find the equation of the set of all points the sum of whose distances from the points (3, 0) and (9, 0) is 12.
Ans. Let (x, y) be any point.
Given points are (3, 0) and (9, 0)
According to the question, we have
Putting x2 + 9 – 6x + y2 = k
Squaring both sides, we have
⇒ 72 – 12x + k = 144 + k - 24√k
⇒ 24 √k = 144 – 72 + 12x
⇒ 24 √k = 72 + 12x
⇒ 2 √k = 6 + x
Again squaring both sides, we get
4k = 36 + x2 + 12x
Putting the value of k, we have
4(x2 + 9 – 6x + y2) = 36 + x2 + 12x
⇒ 4x2 + 36 – 24x + 4y2 = 36 + x2 + 12x
⇒ 3x2 + 4y2 – 36x = 0
Hence, the required equation is 3x2 + 4y2 – 36x = 0
Q.30. Find the equation of the set of all points whose distance from (0, 4) are 2/3 of their distance from the line y = 9.
Ans.
Let P(x, y) be a point.
According to question, we have
Squaring both sides, we have
⇒ 9x2 + 9(y – 4)2 = 4y2 + 324 – 72y
⇒ 9x2 + 9y2 + 144 – 72y = 4y2 + 324 – 72y
⇒ 9x2 + 5y2 + 144 – 324 = 0
⇒ 9x2 + 5y2 – 180 = 0
Hence, the required equation is 9x2 + 5y2 – 180 = 0.
Q.31. Show that the set of all points such that the difference of their distances from (4, 0) and (– 4, 0) is always equal to 2 represent a hyperbola.
Ans. Let P(x, y) be any point.
According to the question, we have
Putting the x2 + y+ 16 = z ...(i)
Squaring both sides, we get
Again squaring both sides, we have
z2 + 4 – 4z = z2 – 64x2
⇒ 4 – 4z + 64x2 = 0
Putting the value of z, we have
⇒ 4 – 4(x2 + y2 + 16) + 64x2 = 0
⇒ 4 – 4x2 – 4y2 – 64 + 64x2 = 0
⇒ 60x2 – 4y2 – 60 = 0
⇒ 60x2 – 4y2 = 60
Which represent a hyperbola. Hence proved.
Q.32. Find the equation of the hyperbola with
(a) Vertices (± 5, 0), foci (± 7, 0)
(b) Vertices (0, ± 7), e = 4/3
(c) Foci (0, ± √10), passing through (2, 3)
Ans. (a) Given that vertices (± 5, 0), foci (± 7, 0)
Vertex of hyperbola = (± a, 0) and foci (± ae, 0)
∴ a = 5 and ae = 7 ⇒ 5 x e = 7 ⇒ e = 7/5
Now b2 = a2 (e– 1)
The equation of the hyperbola is
(b) Given that vertices (0, ± 7), e = 4/3
Clearly, the hyperbola is vertical.
Vertices = (± 0, a)
∴ a = 7 and e = 4/3
We know that b2 = a2(e2 – 1)
Hence, the equation of the hyperbola is
⇒ 9x2 – 7y2 + 343 = 0
(c) Given that: foci = (0 , ± √10)
∴ ae = √10 ⇒ a2e2 = 10
We know that b2 = a2(e2 – 1)
⇒ b2 = a2e2 – a2
⇒ b2 = 10 – a2
Equation of hyperbola is
If it passes through the point (2, 3) then
⇒ 90 – 13a2 = a2(10 – a2)
⇒ 90 – 13a2 = 10a2 – a4
a4 – 23a2 + 90 = 0
⇒ a4 – 18a2 – 5a2 + 90 = 0
⇒ a2(a2 – 18) – 5(a2 – 18) = 0
⇒ (a2 – 18)(a2 – 5) = 0
⇒ a2 = 18, a2 = 5
∴ b2 = 10 –18 = – 8 and b2 = 10 – 5 = 5
b ≠ – 8 ∴ b2 = 5
Here, the required equation is
## Objective Answer Type Questions
Q.33. The line x + 3y = 0 is a diameter of the circle x2 + y2 + 6x + 2y = 0.
Ans. Given equation of the circle is
x2 + y2 + 6x + 2y = 0
Centre is (– 3, – 1)
If x + 3y = 0 is the equation of diameter, then the centre (– 3, – 1) will lie on x + 3y = 0
– 3 + 3(– 1) = 0
⇒ – 6 ≠ 0
So, x + 3y = 0 is not the diameter of the circle.
Hence, the given statement is False.
Q.34. The shortest distance from the point (2, –7) to the circle x+ y2 – 14x – 10y – 151 = 0 is equal to 5.
[Hint: The shortest distance is equal to the difference of the radius and the distance between the centre and the given point.]
Ans. Given equation of circle is x2 + y2 – 14x – 10y – 151 = 0
Shortest distance = distance between the point (2, – 7)
and the centre – radius of the circle
Centre of the given circle is
2g = – 14 ⇒ g = – 7
2f = – 10 ⇒ f = – 5
∴ Centre = (– g, – f) = (7, 5)
and
∴ Shortest distance
Hence, the given statement is False.
Q.35. If the line lx + my = 1 is a tangent to the circle x2 + y= a2, then the point (l, m) lies on a circle.
[Hint: Use that distance from the centre of the circle to the given line is equal to radius of the circle.]
Ans. Given equation of circle is x2 + y= a2
and the tangent is lx + my = 1
Here centre is (0, 0) and radius = a
If (l, m) lies on the circle
⇒ l2 + m2 = a2 (which is a circle)
So, the point (l, m) lies on the circle.
Hence, the given statement is True.
Q.36. The point (1, 2) lies inside the circle x2 + y2 – 2x + 6y + 1 = 0.
Ans. Given equation of circle is x2 + y2 – 2x + 6y + 1 = 0
Here 2g = – 2 ⇒ g = – 1
2f = 6 ⇒ f = 3
∴ Centre = (– g, – f) = (1, – 3)
and
∴ Distance between the point (1, 2) and the centre (1, – 3)
Here 5 > 3, so the point lies out side the circle.
Hence, the given statement is False.
Q.37. The line lx + my + n = 0 will touch the parabola y2 = 4ax if ln = am2.
Ans. Given equation of parabola is y2 = 4ax ...(i)
and the equation of line is lx + my + n = 0 ...(ii)
From eq. (ii), we have
Putting the value of y in eq. (i) we get
⇒ l2x2 + n2 + 2lnx – 4am2x = 0
⇒ l2x2 + (2ln – 4am2)x + n2 = 0
If the line is the tangent to the circle, then
b2 – 4ac = 0
(2ln – 4am2)2 – 4l2n2 = 0
⇒ 4l2n2 + 16a2m4 – 16lnm2a – 4l2n2 = 0
⇒ 16a2m4 – 16lnm2a = 0
⇒ 16am2(am2 – ln) = 0
⇒ am2(am2 – ln) = 0
⇒ am2 ≠ 0 ∴ am2 - ln = 0
∴ ln = am2
Hence, the given statement is True.
Q.38. If P is a point on the ellipsewhose foci are S and S′, then PS + PS′ = 8.
Ans. Let P(x1, y1) be a point on the ellipse. foci = (± ae, 0)
Here a2 = 25 ⇒ a = 5
b2 = 16 ⇒ b = 4
b2 = a2(1 – e2)
16 = 25(1 – e2)
So, the foci are S(3, 0) and S’(- 3, 0).
Since PS + PS’ = 2a = 2 x 5 = 10.
Hence, the given statement is False.
Q.39. The line 2x + 3y = 12 touches the ellipseat the point (3, 2).
Ans. If line 2x + 3y = 12 touches the ellipsethen the point (3, 2) satisfies both line and ellipse.
∴ For line 2x + 3y = 12
2(3) + 3(2) = 12
6 + 6 = 12
12 = 12 True
For ellipse
1 + 1 = 2
2 = 2 True
Hence, the given statement is True.
Q.40. The locus of the point of intersection of lines √3 x − y − 4 √3k = 0 and
√3kx + ky – 4√3 = 0 for different value of k is a hyperbola whose eccentricity is 2.
[Hint:Eliminate k between the given equations]
Ans. The given equations are
√3x - y - 4 √3k = 0 ...(i)
and √3kx + ky - 4 √3 = 0 ...(ii)
From eq. (i) we get
4√3k = √3x - y
Putting the value of k in eq. (ii), we get
⇒ 3x2 - √3xy + √3xy - y2 - 48 = 0
⇒ 3x2 – y2 = 48
which is a hyperbola.
Here a2 = 16, b2 = 48
We know that b2 = a2(e2 – 1)
⇒ 48 = 16(e2 – 1)
⇒ 3 = e2 – 1
⇒ e2 = 4 ⇒ e = 2
Hence, the given statement is True.
## Fill in the Blanks
Q.41. The equation of the circle having centre at (3, – 4) and touching the line 5x + 12y – 12 = 0 is _______.
[Hint: To determine radius find the perpendicular distance from the centre of the circle to the line.]
Ans. Given equation of the line is 5x + 12y – 12 = 0 and the centre is (3, – 4)
CP = radius of the circle
So, the equation of the circle is
Hence, the value of the filler is (x – 3)2 + (y + 4)2 =
Q.42. The equation of the circle circumscribing the triangle whose sides are the lines y = x + 2, 3y = 4x, 2y = 3x is _______ .
Ans. Let AB represents 2y = 3x ...(i)
BC represents 3y = 4x ...(ii)
and AC represents y = x + 2 ...(iii)
From eq. (i) and (ii)
Putting the value of y in eq. (ii) we get
⇒ 9x = 8x
⇒ x = 0 and y = 0
∴ Coordinates of B = (0, 0)
From eq. (i) and (iii) we get
y = x + 2
Putting y = x + 2 in eq. (i) we get
2(x + 2) = 3x
⇒ 2x + 4 = 3x
⇒ x = 4 and y = 6
∴ Coordinates of A = (4, 6)
Solving eq. (ii) and (iii) we get
y = x + 2
Putting the value of y in eq. (ii) we get
3(x + 2) = 4x ⇒ 3x + 6 = 4x ⇒ x = 6 and y = 8
∴ Coordinates of C = (6, 8)
It implies that the circle is passing through (0, 0), (4, 6) and (6, 8).
We know that the general equation of the circle is
x2 + y2 + 2gx + 2fy + c = 0 ...(i)
Since the points (0, 0), (4, 6) and (6, 8) lie on the circle then
0 + 0 + 0 + 0 + c = 0 ⇒ c = 0
⇒ 16 + 36 + 8g + 12f + c = 0
⇒ 8g + 12f + 0 = – 52
⇒ 2g + 3f = – 13 ...(ii)
and 36 + 64 + 12g + 16f + c = 0
⇒ 12g + 16f + 0 = – 100
⇒ 3g + 4f = – 25 ...(iii)
Solving eq. (ii) and (iii) we get
2g + 3f = – 13
3g + 4f = – 25
Putting the value of f in eq. (ii) we get
2g + 3 x 11 = - 13
⇒ 2g + 33 = – 13
⇒ 2g = – 46 ⇒ g = – 23
Putting the values of g, f and c in eq. (i) we get
x2 + y2 + 2(– 23)x + 2(11)y + 0 = 0
⇒ x2 + y– 46x + 22y = 0
Hence, the value of the filler is x2 + y2 – 46x + 22y = 0.
Q.43. An ellipse is described by using an endless string which is passed over two pins. If the axes are 6 cm and 4 cm, the length of the string and distance between the pins are _______.
Ans. Let equation of ellipse is
Here 2a = 6 ⇒ a = 3
and 2b = 4 ⇒ b = 2
We know that c2 = a2 – b2
= (3)2 – (2)2 = 9 – 4 = 5
Length of string = 2a + 2ae = 2a(1 + e)
Distance between the pins = CC’ = 2ae =
Hence the value of the filler are 6 + 2√5 cm and 2√5 cm.
Q.44. The equation of the ellipse having foci (0, 1), (0, –1) and minor axis of length 1 is_______.
Ans. We know that the foci of the ellipse are (0, ± ae) and given foci are (0, ± 1), so ae = 1
Length of minor axis = 2b = 1 ⇒
We know that b2 = a2(1 – e2)
∴ Equation of ellipse is
Hence, the value of the filler is
Q.45. The equation of the parabola having focus at (–1, –2) and the directrix x – 2y + 3 = 0 is ________.
Ans. Let (x1, y1) be any point on the parabola.
According to the definition of the parabola
Squaring both sides, we get
⇒ 5x12 + 5y12 + 10x1 + 20 y1 + 25
= x21 + 4 y12 - 4x1 y1 - 12 y1 + 6x1 + 9
4 x12 + y12 + 4 x1 + 32 y1 + 4 x1 y1 + 16 = 0
Hence, the value of the filler is 4x2 + 4xy + y2 + 4x + 32y + 16 = 0.
Q.46. The equation of the hyperbola with vertices at (0, ± 6) and eccentricity 5/3 is________and its foci are _______ .
Ans. Let equation of the hyperbola is
Vertices are (0, ± b) ∴ b = 6 and e = 5/3
We know that
⇒ a2 = 64
So the equation of the hyperbola is
and foci = (0, ± be) =
Hence, the value of the filler is
## Multiple Choice Types Questions
Choose the correct answer out of the given four options (M.C.Q.)
Q.47. The area of the circle centred at (1, 2) and passing through (4, 6) is
(a) 5 π
(b) 10π
(c) 25π
(d) None of these
Ans. (c)
Solution.
Given that the centre of the circle is (1, 2)
Radius of the circle =
So, the area of the circle = πr2
= π x (5)2 = 25π
Hence, the correct option is (c).
Q.48. Equation of a circle which passes through (3, 6) and touches the axes is
(a) x2 + y2 + 6x + 6y + 3 = 0
(b) x2 + y2 – 6x – 6y – 9 = 0
(c) x2 + y2 – 6x – 6y + 9 = 0
(d) None of these
Ans. (c)
Solution.
Let the required circle touch the axes at (a, 0) and (0, a)
∴ Centre is (a, a) and r = a
So the equation of the circle is (x – a)2 + (y – a)2 = a2
If it passes through a point P(3, 6) then
(3 – a)2 + (6 – a)2 = a2
9 + a2 – 6a + 36 + a2 – 12a = a2
⇒ a2 – 18a + 45 = 0
⇒ a2 – 15a – 3a + 45 = 0
⇒ a(a – 15) – 3(a – 15) = 0
⇒ (a – 3) (a – 15) = 0
a = 3 and a = 15 which is not possible
∴ a = 3
So, the required equation of the circle is
(x – 3)2 + (y – 3)2 = 9
⇒ x2 + 9 – 6x + y2 + 9 – 6y = 9
⇒ x2 + y2 – 6x – 6y + 9 = 0
Hence, the correct option is (c).
Q.49. Equation of the circle with centre on the y-axis and passing through the origin and the point (2, 3) is
(a) x2 + y2 + 13y = 0
(b) 3x2 + 3y2 + 13x + 3 = 0
(c) 6x2 + 6y2 – 13x = 0
(d) x2 + y2 + 13x + 3 = 0
Ans. (a)
Solution.
Let the equation of the circle be
(x – h)2 + (y – k)2 = r2
Let the centre be (0, a)
∴ Radius r = a
So, the equation of the circle is
(x – 0)2 + (y – a)2 = a2
⇒ x2 + (y – a)2 = a2
⇒ x2 + y2 + a2 – 2ay = a2
⇒ x2 + y2 – 2ay = 0 ...(i)
Now CP = r
⇒ 13 + a2 – 6a = a2
⇒ 13 – 6a = 0
∴ a = 13/6
Putting the value of a in eq. (i) we get
⇒ 3x2 + 3y2 – 13y = 0
(Note: (a) option is correct and it should be 3x2 + 3y2 – 13y = 0)
Hence, the correct option is (a).
Q.50.
The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is
(a) x2 + y2 = 9a2
(b) x2 + y2 = 16a2
(c) x2 + y2 = 4a2
(d) x2 + y2 = a2
[Hint: Centroid of the triangle coincides with the centre of the circle and the
radius of the circle is 2/3 of the length of the median]
Ans. (c)
Solution.
Let ABC be an equilateral triangle in which median AD = 3a.
Centre of the circle is same as the centroid of the triangle i.e., (0, 0)
AG : GD = 2 : 1
So,
∴ The equation of the circle is
(x – 0)2 + (y – 0)2 = (2a)
⇒ x2 + y= 4a2
Hence, the correct option is (c).
Q.51. If the focus of a parabola is (0, –3) and its directrix is y = 3, then its equation is
(a) x2 = –12y
(b) x2 = 12y
(c) y2 = –12x
(d) y2 = 12x
Ans. (a)
Solution.
According to the definition of parabola
Squaring both sides, we have
x2 + y2 + 9 + 6y = y2 + 9 – 6y
x2 + 9 + 6y = 9 – 6y
⇒ x2 = – 12y
Hence, the correct option is (a).
Q.52. If the parabola y2 = 4ax passes through the point (3, 2), then the length of its latus rectum is
(a) 2/3
(b) 4/3
(c) 1/3
(d) 4
Ans. (b)
Solution.
Given parabola is y2 = 4ax
If the parabola is passing through (3, 2) then
(2)2 = 4a x 3
⇒ 4 = 12a ⇒ a = 1/3
Now length of the latus rectum = 4a =
Hence, the correct option is (b).
Q.53. If the vertex of the parabola is the point (– 3, 0) and the directrix is the line x + 5 = 0, then the equation is
(a) y2 = 8(x + 3)
(b) x2 = 8(y + 3)
(c) y2 = – 8(x + 3)
(d) y2 = 8(x + 5)
Ans. (a)
Solution.
Given that vertex = (– 3, 0)
∴ a = – 3
and directrix is x + 5 = 0
According to the definition of the parabola, we get
AF = AD i.e., A is the mid-point of DF
and
∴ Focus F = (– 1, 0)
Squaring both sides, we get
(x + 1)2 + y2 = (x + 5)2
x2 + 1 + 2x + y2 = x2 + 25 + 10x
⇒ y2 = 10x – 2x + 24 ⇒ y2 = 8x + 24
⇒ y2 = 8(x + 3)
Hence, the correct option is (a).
Q.54. The equation of the ellipse, whose focus is (1, – 1), the directrix the line x – y – 3 = 0 and eccentricity 1/2, is
(a) 7x+ 2xy + 7y2 – 10x + 10y + 7 = 0
(b) 7x2 + 2xy + 7y2 + 7 = 0
(c) 7x2 + 2xy + 7y2 + 10x – 10y – 7 = 0
(d) None of the above
Ans. (a)
Solution.
Given that focus of the ellipse is (1, – 1) and the equation of the directrix is x – y – 3 = 0 and e = 1/2.
Let P(x, y) by any point on the parabola
Squaring both sides, we have
⇒ 8x2 + 8y2 – 16x + 16y + 16 = x+ y2 – 2xy + 6y – 6x + 9
⇒ 7x2 + 7y2 + 2xy – 10x + 10y + 7 = 0
Hence, the correct option is (a).
Q.55. The length of the latus rectum of the ellipse 3x2 + y2 = 12 is
(a) 4
(b) 3
(c) 8
(d) 4√3
Ans. (d)
Solution.
Equation of the ellipse is
3x2 + y= 12
Here a2 = 4 ⇒ a = 2
b2 = 12 ⇒ b = 2√ 3
Length of the latus rectum =
Hence, the correct option is (d).
Q.56. If e is the eccentricity of the ellipse then
(a) b2 = a2(1 – e2
(b) a2 = b2(1 – e2
(c) a2 = b2(e2 – 1)
(d) b2 = a2(e2 – 1)
Ans. (b)
Solution.
Given equation is
∴Eccentricity e
Hence, the correct option is (b).
Q.57. The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half of the distance between the foci is
(a) 4/3
(b) 4/√3
(c) 2/√3
(d) None of these
Ans. (c)
Solution.
Length of the latus rectum of the hyperbola
Distance between the foci = 2ae
Transverse axis = 2a
and Conjugate axis = 2b
)
[from eq. (i)]
Now b= a2(e2 – 1)
⇒ 4a = a2(e2 – 1)
Hence, the correct option is (c).
Q.58. The distance between the foci of a hyperbola is 16 and its eccentricity is 2 . Its equation is
(a) x2 – y2 = 32
(c) 2x2 – 3y2 = 7
(d) None of these
Ans. (a)
Solution.
We know that the distance between the foci = 2ae
∴ 2ae = 16 ⇒ ae = 8
Given that e = √2
∴ 2a = 8 ⇒ a = 4 √2
Now b2 = a2(e2 – 1)
⇒ b2 = 32(2 – 1)
⇒ b2 = 32
So, the equation of the hyperbola is
Hence, the correct option is (a).
Q.59. Equation of the hyperbola with eccentricty 3/2 and foci at (± 2, 0) is
(d) None of these
Ans. (a)
Solution.
Given that e = 3/2
and foci = (± ae, 0) = (± 2, 0)
∴ ae = 2
Now we know that b2 = a2(e2 – 1)
So, the equation of the hyperbola is
Hence, the correct option is (a).
The document NCERT Exemplar: Conic Sections (Exercise) | Mathematics (Maths) Class 11 - Commerce is a part of the Commerce Course Mathematics (Maths) Class 11.
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## Mathematics (Maths) Class 11
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## FAQs on NCERT Exemplar: Conic Sections (Exercise) - Mathematics (Maths) Class 11 - Commerce
1. What are the different types of conic sections?
Ans. The different types of conic sections are: circles, ellipses, parabolas, and hyperbolas. These are formed by intersecting a plane with a cone at different angles.
2. How are conic sections used in real-life applications?
Ans. Conic sections are used in various real-life applications such as designing satellite orbits, constructing bridges, designing optical instruments like telescopes and cameras, and in the field of architecture for designing arches and domes.
3. What is the general equation of a conic section?
Ans. The general equation of a conic section is given by Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, where A, B, C, D, E, and F are constants and x and y are variables.
4. How are conic sections related to the focus and directrix of a curve?
Ans. For each type of conic section, there is a specific relationship between the focus, directrix, and eccentricity of the curve. For example, in the case of a parabola, the focus lies on the axis of symmetry and the directrix is a line perpendicular to the axis of symmetry.
5. How can conic sections be used to solve geometric problems?
Ans. Conic sections can be used to solve geometric problems such as finding the intersection points of two curves, determining the center and radius of a circle, and finding the foci and eccentricity of an ellipse or hyperbola.
## Mathematics (Maths) Class 11
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; | 13,624 | 32,145 | {"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} | 4.65625 | 5 | CC-MAIN-2024-26 | latest | en | 0.812098 |
https://everything2.com/user/thax/writeups/Fundamental+theorem+of+calculus | 1,527,302,873,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867277.64/warc/CC-MAIN-20180526014543-20180526034543-00096.warc.gz | 565,181,722 | 6,882 | I'd like to try and provide a somewhat intuitive explanation of this theorem, since, if you're anything like me, when you first learn it, there seems to be nothing at all intuitive about it.
So, say we have some integrable function f defined on the real numbers. Then, if we assume f is positive, recall that the definite integral of f from a to b, written
``` /\b
|
| f(t) dt
|
\/ a
```
can be thought of as the area under f between a and b. OK, then consider the function F defined as follows:
``` /\x
|
F(x):= | f(t) dt
|
\/ a
```
where x > a. Then F(x) can be thought of as giving us the "area so far" under f from a to x. We can see this graphically (where the "curve" is the graph of the function f):
```|
| .-.
| ..--' \ __.--- f
| /* '--'
|__.' * *
| * F(x) *
| * *
0-----|----------|--------->
a x
```
Now the Fundamental Theorem of Calculus tells us that the derivative of F at x (let's call it F'(x) ), is equal to f(x). This does not seem obvious at first, but let's see if we can get an idea of why this is so. Remember that the derivative of F at x is defined to be the limit as h goes to zero of ((F(x+h)-F(x))/h); that is:
``` f(x+h) -f(x)
lim ------------
h->0 h
```
Now, consider the quantity ((F(x+h)-F(x))/h) for some small value of h. Now, F(x+h)-F(x) gives us the area under f between x and x+h:
``` __.f(x+h)
f(x)___..--' *
* *
* F(x+h) - *
* F(x) *
* *
-----|-----------|-------->
x x+h
<---- h ---->
```
But since h is small, we can approximate this area by the area of a rectangle with a height of f(x) and width h. So, we have the approximate equality, h*f(x) ~= F(x+h)-F(x). Dividing both sides by h, we have f(x) ~= (F(x+h)-F(x))/h. Now, if you make h smaller and smaller, then it seems like it would make sense that this approximation would get better and better. So, we haven't proven it, but I hope it's at least believable that when we take the limit as h goes to zero, the approximate equality above becomes actual equality, and we know (by definition) that the right hand side of the equality becomes F'(x), so f(x)=F'(x).
I feel like this makes it a little bit easier to understand what exactly is going on with the Fundamental Theorem (for proof, go to proof of the Fundamental Theorem of Calculus), but I'm open to any suggestions on how to make it more clear. /msg me! | 686 | 2,455 | {"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} | 4.25 | 4 | CC-MAIN-2018-22 | longest | en | 0.912269 |
http://functions.wolfram.com/HypergeometricFunctions/ChebyshevTGeneral/13/01/01/02/0002/ | 1,386,873,429,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164676172/warc/CC-MAIN-20131204134436-00074-ip-10-33-133-15.ec2.internal.warc.gz | 78,041,850 | 8,054 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
ChebyshevT
http://functions.wolfram.com/07.04.13.0002.01
Input Form
Wronskian[ChebyshevT[\[Nu], z], Sin[\[Nu] ArcCos[z]], z] == -(\[Nu]/Sqrt[1 - z^2])
Standard Form
Cell[BoxData[RowBox[List[RowBox[List["Wronskian", "[", RowBox[List[RowBox[List["ChebyshevT", "[", RowBox[List["\[Nu]", ",", "z"]], "]"]], ",", RowBox[List["Sin", "[", RowBox[List["\[Nu]", " ", RowBox[List["ArcCos", "[", "z", "]"]]]], "]"]], ",", "z"]], "]"]], "\[Equal]", RowBox[List["-", FractionBox["\[Nu]", SqrtBox[RowBox[List["1", "-", SuperscriptBox["z", "2"]]]]]]]]]]]
MathML Form
W z ( T ν ( z ) , sin ( ν cos - 1 ( z ) ) ) - ν 1 - z 2 Subscript W z ChebyshevT ν z ν z -1 ν 1 -1 z 2 1 2 -1 [/itex]
Rule Form
Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List["Wronskian", "[", RowBox[List[RowBox[List["ChebyshevT", "[", RowBox[List["\[Nu]_", ",", "z_"]], "]"]], ",", RowBox[List["Sin", "[", RowBox[List["\[Nu]_", " ", RowBox[List["ArcCos", "[", "z_", "]"]]]], "]"]], ",", "z_"]], "]"]], "]"]], "\[RuleDelayed]", RowBox[List["-", FractionBox["\[Nu]", SqrtBox[RowBox[List["1", "-", SuperscriptBox["z", "2"]]]]]]]]]]]
Date Added to functions.wolfram.com (modification date)
2001-10-29 | 588 | 1,640 | {"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-2013-48 | longest | en | 0.242354 |
http://physicshelpforum.com/periodic-circular-motion/10549-ice-skater-solid-cylinder-questions.html | 1,553,388,347,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203123.91/warc/CC-MAIN-20190324002035-20190324024035-00203.warc.gz | 162,488,004 | 9,787 | Physics Help Forum Ice skater and solid cylinder questions
Periodic and Circular Motion Periodic and Circular Motion Physics Help Forum
Oct 14th 2014, 06:32 PM #1 Junior Member Join Date: Aug 2014 Posts: 9 Ice skater and solid cylinder questions (a) An ice skater with an initial moment of inertia of 1.34 kgm2 with arms outstretched horizontally is initially spinning at 9.10 revs/s. She draws her arms vertically in by her side reducing her moment of inertia to 1.25 kgm2. What is her new angular velocity? (b) A solid cylinder with a diameter of 36.8 cm and mass 13.2 kg is rolling on a horizontal surface with a linear velocity of 1.76 m/s. Calculate the angular momentum of the cylinder. I got 61.3 rad/s for the final angular velocity for the skater and for b I got Moment of Inertia I= 0.22 and angular momentum L=2.1 Can someone please check if I did these correctly ? It would be greatly appreciated !
Oct 15th 2014, 07:58 AM #2 Physics Team Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,331 Very good - I get the same results. Just don't forget to include units.
Oct 15th 2014, 01:57 PM #3 Junior Member Join Date: Aug 2014 Posts: 9 Thanks so much !
Oct 15th 2014, 08:12 PM #4 Junior Member Join Date: Aug 2014 Posts: 9 angle of wheel turn OK this one has me stumped as not sure which way to go. A wheel was rotating at 11.5 rad/s when a torque was applied for 12.7 secs. The angular velocity increased to 47.0 rad/s. What angle did the wheel turn through in that time? I tried to use one of the kinematics formulas after working out he acceleration to get the angle but I don't think it is working.... help ? I got 370.84 ??
Tags cylinder, ice, questions, skater, solid | 467 | 1,711 | {"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} | 3.265625 | 3 | CC-MAIN-2019-13 | latest | en | 0.920256 |
https://wiki.colby.edu/pages/diffpages.action?pageId=371622860&originalId=371622859 | 1,555,649,472,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578527135.18/warc/CC-MAIN-20190419041415-20190419063415-00353.warc.gz | 596,980,185 | 12,513 | # Pages … Home Zena Abulhab CS151 - Computational Thinking: Visual Media Zena's CS151 Project 11 Page History
## Key
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Task 1 was to make at least four 3D Turtle shapes and add them to our shapes library. In my case, I made a cube, a pyramid, a triangular prism, and a "house" shape.
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Making shapes in 3D is just like making 2D shapes, except now when the turtle turns, it is not only limited to "left" and "right". Rotation around the z axis (turning left or right like before) is called "yaw", "roll" is rotation around the x axis (leaning left or right), and "pitch" is rotation around the y axis (like turning to face up or down). We needed to add cases to the turtle interpreter, so "&" was given pitch up, "^" was given pitch down, "\" was given roll up, and "/" was given roll down. Keeping this in mind, I drew the shapes shown below. The characters used to represent a pyramid, for example, were [F+[(45)(45)&F]F[(45)(45)&F]F[(45)(45)&F]F(45)+(45)&F]. The turtle moves forward, turns left (90 degrees), remembers that position because it will continue from there to make the square base, then turns left 45 degrees, up 45 degrees (pitches up), and goes forward. Then it goes back to the remembered position, turns left 90 degrees, and continues on in the same manner until the pyramid is made. Note that if string needs to be repeated and you don't want to write it all out again, you can use the string multiplication property; writing 'F' * 4, for instance, makes the turtle go forward forward and right left times, drawing a square. The picture below also shows the jitter3 and jitter styles from last week's project.
For Task 3, we needed to improve our code in some way, including doing one of the extensions, so I did extension 7, which was to make another "shape" that was actually a dynamic shape, drawing whatever the file it reads says, and, optionally, what the user inputs in the terminal. I didn't need to change the parent Shape class at all, I just needed to give this dynamic shape class instructions to read the file that was passed in in its init function, or to write the argv into another file (or possibly the same, depending on what is given) if it is given and read that instead. Then, once the parent init function is called, the only difference is that the string passed in is the string read from the file. This code is shown below: Image Added
This "argv" is passed in through whatever line of code is calling the shape to be made; for example, the next pictures show what code I input in a file and into the command line to draw the picture below it. I wrote into the command line the two shapes that I wanted to be drawn. The place the yellow shape expects to get its text from is the first index of the command line, and the blue one expects it from index two. If the command line is not to be used, the argv should be given the value of None instead, so the file will just be read instead.
For my first extension, extension 5, I made a way for the user to type instructions for an L system into the command line and have that be drawn. This involved passing two arguments through the shapes class init, L system class init, readString init (which I made fresh), and the Turtle Interpreter's drawString function: base string (bstring) and rule string (rstring). First off, these two were necessary to differentiate between the base and rule(s) of the command line turtle instructions. The way I organized them was to have index 1 of argv be where the base is, then the ones after just follow the pattern of "base to replace" and "rule to replace it with", so I took advantage of that by using the following code when calling an L system to be made through the tree.py file. Image Added
As you can see, it is organized into two categories: the base and all the rest, from which we will separate the bases to replace and the rules using the following code: | 945 | 4,110 | {"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} | 3.59375 | 4 | CC-MAIN-2019-18 | latest | en | 0.948177 |
https://enakai00.hatenablog.com/archive/2022/09 | 1,716,369,975,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058534.8/warc/CC-MAIN-20240522070747-20240522100747-00627.warc.gz | 205,371,985 | 10,453 | # Quantum Information and Quantum Optics with Superconducting Circuits - Exercise Solutions (Chapter 4)
4.1 For the thermal state : For the pure state : 4.2 By mathematical induction, you can prove: Hence: 4.4 Current conservation: Branch currents: Hence: Without the register: With the register: 4.5 4.6 This is very rough apporximation. I'm …
# Derivation of the first and second Josephson relation
In the following discussion, we assume that the quantum states are quasi-static. We solve the time-independent Schrödinger equation supposing that external potentials are independent of time . When we change them slowly enough, the corresp…
# Derivation of the gauge-independent relation between the phase and the electric field
The effective wavefunction and the charge current are given as: --- (3.4) ---(3.13)The wavefunction follows the Schrödinger equation: --- (3.5)Without losing the generality, we can take the Coulomb gauge: --- (1)Now, we assume that the cha…
# Study notes on "Quantum Information and Quantum Optics with Superconducting Circuits"
Quantum Information and Quantum Optics with Superconducting Circuits作者:García Ripoll, Juan JoséCambridge University PressAmazon Derivation of the Lindblad master equation (2.26) I found the following paper is useful to understand the gen…
# Gauge Invariance of superconducting circuits wavefunction model
Effective wavefunction describing the flow of superconducting elections (Cooper pairs): Schrödinger equation for : --- (3.5)I will show that this model is invariant under the gauge transformation: [Proof](3.5) is equivalent to --- (1)A sim…
# (Informal) Errata of "A short introduction to the Lindblad Master Equation"
arxiv.orgp.8 Equation (24) p.10 Equation (30) p.11 Equation (31) p.13 Derivation of (45) from (44)First, in (44), we change the integral variable from (without any approximation) to get: Then, we assume that the kernel in the integration i… | 468 | 1,934 | {"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-2024-22 | latest | en | 0.837005 |
https://forum.mibuso.com/discussion/72165/create-sql-query-based-flow-filter | 1,723,503,770,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641052535.77/warc/CC-MAIN-20240812221559-20240813011559-00773.warc.gz | 199,248,465 | 74,330 | #### Howdy, Stranger!
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# create sql query based flow filter
Member Posts: 158
edited 2018-09-18
HI everyone,
I'm trying to create the sql query code based on a flow filter calc formula.
The formula is "Sum("Detailed Vendor Ledg. Entry".Amount WHERE (Vendor Ledger Entry No.=FIELD(Entry No.),Posting Date=FIELD(Date Filter),Excluded from calculation=CONST(No)))" and I "convert" this to
Select Sum([E18375\$Detailed Vendor Ledg_ Entry].[Amount]) as 'Valor pendente' FROM [E18375\$Detailed Vendor Ledg_ Entry] INNER JOIN [E18375\$Vendor Ledger Entry] ON [E18375\$Detailed Vendor Ledg_ Entry].[Vendor Ledger Entry No_] = [E18375\$Vendor Ledger Entry].[Entry No_] WHERE [E18375\$Detailed Vendor Ledg_ Entry].[Excluded from calculation] = 0 AND [E18375\$Detailed Vendor Ledg_ Entry].[Document No_]='1707NC006'
AND YEAR([E18375\$Detailed Vendor Ledg_ Entry].[Posting Date])>=2017
The problem is I get the value "Valor pendente" equals to "1838" instead of "10" like the image below.
I follow this example i found online to create my sql query..
Sum("Detailed Cust. Ledg. Entry"."Amount (LCY)" WHERE (Cust. Ledger Entry No.=FIELD(Entry No.),Entry Type=FILTER(Initial Entry),Posting Date=FIELD(Date Filter)))
Select Sum([Amount (LCY)] FROM [Detailed Cust. Ledg. Entry]
INNER JOIN
[Cust. Ledg. Entry]
ON
[Detailed Cust. Ledg. Entry].[Entry No.] = [Cust. Ledg. Entry].[Entry No.]
WHERE
[Detailed Cust. Ledg. Entry].[Entry Type] = "Initial Entry"
• Member Posts: 1,690
edited 2018-09-14
As a starting point the predicate
```..ON [E18375\$Detailed Vendor Ledg_ Entry].[Entry No_] = [E18375\$Vendor Ledger Entry].[Entry No_] ..
```
should rather go like this
```ON [E18375\$Detailed Vendor Ledg_ Entry].[Vendor Ledger Entry No_] = [E18375\$Vendor Ledger Entry].[Entry No_]...
```
Slawek Guzek
Dynamics NAV, MS SQL Server, Wherescape RED;
GDPR Certified Data Protection Officer - PECB License DPCDPO1025070-2018-03
• Member Posts: 158
edited 2018-09-17
Even with this query I don't get the correct result
```select 'E18375' as Empresa, [E18375\$Detailed Vendor Ledg_ Entry].[Posting Date], [E18375\$Detailed Vendor Ledg_ Entry].[Document No_], [E18375\$Detailed Vendor Ledg_ Entry].[Vendor No_], [E18375\$Detailed Vendor Ledg_ Entry].Amount, (Select Sum([Amount (LCY)]) FROM [E18375\$Detailed Vendor Ledg_ Entry] INNER JOIN [E18375\$Vendor Ledger Entry] ON [E18375\$Detailed Vendor Ledg_ Entry].[Entry No_] = [E18375\$Vendor Ledger Entry].[Entry No_] WHERE [E18375\$Detailed Vendor Ledg_ Entry].[Excluded from calculation]=0 AND [E18375\$Detailed Vendor Ledg_ Entry].[Amount]>0) as 'Valor pendente' from [E18375\$Vendor Ledger Entry] inner join [E18375\$Detailed Vendor Ledg_ Entry] on [E18375\$Detailed Vendor Ledg_ Entry].[Vendor Ledger Entry No_]=[E18375\$Vendor Ledger Entry] .[Entry No_] where [E18375\$Vendor Ledger Entry].[Open]=1 and [E18375\$Detailed Vendor Ledg_ Entry].[Document No_]='1707NC006'
group by [E18375\$Detailed Vendor Ledg_ Entry].[Posting Date], [E18375\$Detailed Vendor Ledg_ Entry].[Document No_], [E18375\$Detailed Vendor Ledg_ Entry].[Vendor No_], [E18375\$Detailed Vendor Ledg_ Entry].[Vendor Ledger Entry No_], [E18375\$Detailed Vendor Ledg_ Entry].[Amount]
having sum([E18375\$Detailed Vendor Ledg_ Entry].[Amount])>0
```
It returns a amount of "3510098.20000000000000000000" for "Valor Pendente" instead of "10". Maybe the group by clause is wrong? or the having.. i don't get it
maybe the problem is the flow filter conversion...
• Member, Moderator Posts: 9,100
[Topic moved from 'NAV/Navision Classic Client' forum to 'SQL General' forum]
Regards,Alain Krikilion
No PM,please use the forum. || May the <SOLVED>-attribute be in your title!
• Member, Moderator Posts: 9,100
PS: AND [E18375\$Detailed Vendor Ledg_ Entry].[Document No_]='1707NC006'
Shouldn't that filter in the document no. in the vendor ledger entry? Because in the detailed it can have multiple values for 1 vendor ledger entry.
And I would turn around the query:
SELECT dvle.SUM(Amount)
FROM "Vendor Ledger Entry" vle
INNER JOIN "Detailed Vendor Legd_ Entry" dvle
ON (dvle."Vendor Ledger Entry No_" = vle."Entry No_")
AND (dvle."Excluded From Calculation" = 0)
AND (dvle.[Posting Date] >= '2017-01-01')
AND (dvle.[Posting Date] <= '2017-12-31');
WHERE vle.[Document No_]='1707NC006'
# AVOID AT ALL COST : YEAR("Some Date") = 2017. This forces SQL to scan the whole table or index!.
PS2: the INNER JOIN with the "Excluded from calculation" could have as a result that vendor ledger entries are skipped if No detailed ledger entries are selected for a vendor ledger entry
Regards,Alain Krikilion
No PM,please use the forum. || May the <SOLVED>-attribute be in your title!
• Member Posts: 158
Hi @kriki, I try your query and it returns 1838 (same value as Amount) and not the 10 (as the image below)
SELECT SUM([E18375\$Detailed Vendor Ledg_ Entry].[Amount]) FROM [E18375\$Vendor Ledger Entry] WITH(READUNCOMMITTED) INNER JOIN [E18375\$Detailed Vendor Ledg_ Entry] WITH(READUNCOMMITTED) ON ([E18375\$Detailed Vendor Ledg_ Entry].[Vendor Ledger Entry No_] = [E18375\$Vendor Ledger Entry].[Entry No_]) AND ([E18375\$Detailed Vendor Ledg_ Entry].[Excluded From Calculation] = 0) AND ([E18375\$Detailed Vendor Ledg_ Entry].[Posting Date] >= '2017-01-01') AND ([E18375\$Detailed Vendor Ledg_ Entry].[Posting Date] <= '2017-12-31') WHERE [E18375\$Vendor Ledger Entry].[Document No_]='1707NC006'
With green should be the correct value...
Here's the nav table info | 1,586 | 5,529 | {"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} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.713759 |
https://www.phys.ksu.edu/reu2013/brendanh/research-project/research-project.htm | 1,545,008,898,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828018.77/warc/CC-MAIN-20181216234902-20181217020902-00511.warc.gz | 1,006,958,311 | 4,530 | Research Project
Calibrations for Light Backscatter from Fractal Aggregates
Research Project
My project is to design and calibrate an apparatus capable of investigating how light scatters from soot (a fractal aggregate). Soot is a black powdery substance formed by many different types of combustion, and as such is a concern to climatologists who study how particles in the atmosphere affect earth's global climate. Soot is an aggregation of many carbon particles, which bind together randomly. This randomness gives rise to the fractal aspect; the idea that the structure of the soot looks the same no matter what scale you view it from (within reason).
Goal
The goal of this project is to set up a device to be able to better understand how soot in the atmosphere effects global climate change. As of right now, aerosols like soot are the least understood factor in climate modeling and so impose limits on the certainty of predictions. By studying how soot scatters light backwards, we can get a better picture of what the future holds in store for humanity.
Basic Theory
Calibrations rely on being able to reproduce and measure known phenomenon with enough accuracy to have confidence in quantifying the unexplored. To do this, I scattered from spherical particles, because spheres are well understood. In fact, how light scatters from spheres of arbitrary size and refractive index has a definitive description, known as the Mie solution. Using Mie curves calculated by a computer is an important aspect of my work, as an unfavorable comparison of experimental data with these curves usually means something is amiss.
An important parameter in light scattering is q, sometimes called the momentum transfer, which is equal to 4*pi*sin(theta/2) all over wavelength. Dr. Sorensen has found patterns many have missed by plotting intensity vs. qR rather than theta.
Particles that have a diameter bigger than the wavelength of light exhibit what is known as "enhanced forward scattering" where they scatter light the most strongly in the forward direction. This is called the frontal lobe. This frontal lobe contains information on the size of a particle, as discussed below.
Calibrations
I started by forward scattering from 9.8 micron spheres and looking for ways to experimentally verify their size. Unfortunately, initial efforts to find so called Mie ripples resulted in failure, perhaps due to too large of a size distribution in the particles. However, I was able to obtain data for the frontal scattering lobe drop-off, an area in the Guinier regime that can be used to quickly determine the size of the particle using the 1/q~r relationship.
Backscatter
After that I was able to move to the backward direction and now scattered from a water vapor aerosol made from a household humidifier. I collected consistent data over 35 degrees, but they did not match both the intensity and ripples predicted by the Mie calculator.
A contributor to this disparity is almost certainly polydispersity, however a more interesting and informative explanation may lie in what's called the Opposition Effect, Opposition Surge, or retroflectance. Notice that the biggest difference between experiment and theory is the large intensity at near 180 degrees. This is consistent with the opposition effect, although I'm currently in the process of verifying this as the primary candidate.
Forward Scattering
Before finding out about opposition effects from the literature on backscattering, I was at a loss to explain the discrepancy, and so turned back to forward scattering to hopefully gain a better idea of the size of the water droplet aerosol. The data suggested a size that was much too big to also match the ripples in the backward direction, leading me to believe the aerosol was polydisperse in a big way. More evidence for polydispersity can be found in the fact that a droplet of diameter 6 microns matches the Guinier regime but has ripples that don't show in the data. If there are many sized particles, these ripples would cancel each other out leaving a smooth line like that seen in the data.
At this point, it's important to note that larger particles tend to dominate scattering in the forward direction. In fact, scattered intensity near zero degrees is dependant on the radius to the fourth power. This means that if a sample is polydisperse, a size measurement in the forward direction will usually only attain a size for the biggest particles. With this in mind, it makes sense that a forward scattering measurement would suggest a size much bigger than found in the back direction and again speaks volumes about the dispersity of the aerosol.
Dynamic Light Scattering
With two different measurements of the size we decided to try to use dynamic light scattering to perhaps settle the matter and further probe the size(s) of the droplets from the aerosol generator. When laser light is incident a on sample, the sample creates laser speckle. By measuring how these speckles change with time we can find a time constant, the correlation time, which relates to how the particles are diffusing through a medium. This diffusion depends on the size of the particle, and so by measuring the correlation time and knowing other aspects of the system, such as the viscosity of the medium, we can find the radius.
This plot show the correlation function vs. time. By finding the inverse of the slope of the natural logarithm of the correlation function we get the correlation time discussed above. For the best of these runs, we generally found a size of about 2.5 to 4 microns. This falls between the two sizes found from the other types of scattering.
The Future
I will continue to work through the next year on making these types of measurements more accurately and with new aerosol sources while studying the literature on opposition effects and backscattering in general. | 1,189 | 5,901 | {"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-2018-51 | latest | en | 0.952744 |
https://vocal.media/gamers/what-is-chess | 1,680,327,743,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949701.0/warc/CC-MAIN-20230401032604-20230401062604-00350.warc.gz | 672,494,725 | 31,526 | # What is Chess?
## Know everything about chess
By King of EarthPublished 2 months ago 5 min read
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I. Introduction
A. Brief history of chess:
Chess is a board game that has been played for centuries, with the earliest versions of the game dating back to the 6th century in India. It then spread to the Islamic world and eventually to Europe, where it evolved into the game we know today. The modern game of chess has been standardized for centuries, with the current standard rules being set in the 19th century.
B. Basic rules and concepts:
Chess is played on a board with 64 squares arranged in an 8x8 grid. Each player starts with 16 pieces: one king, one queen, two rooks, two knights, two bishops, and eight pawns. The goal of the game is to checkmate the opponent's king, which means to put the king in a position where it is under attack and cannot escape capture. Each piece has its own unique way of moving, and players must strategize and plan their moves in order to outmaneuver their opponent.
II. Gameplay
A. Setting up the board:
At the beginning of the game, the board is set up with the rooks in the corners, the knights next to them, the bishops next to them, the queen on her own color (white queen on a white square, black queen on a black square), and the king on the remaining square of the queen's color. The pawns are placed in front of the other pieces.
B. Object of the game:
The objective of the game is to checkmate the opponent's king. This occurs when the king is under attack and cannot escape capture.
C. How pieces move:
Each piece has its own unique way of moving. The king can move one square in any direction, the queen can move any number of squares in any direction, the rook can only move horizontally or vertically, the bishop can only move diagonally, the knight moves in an L-shape (two squares in one direction and then one square in a perpendicular direction), and the pawns have the most complex movement rules. Pawns can move forward one square, but capture diagonally.
D. Special moves (e.g. castling, en passant):
There are several special moves in chess that can be used to gain an advantage over the opponent. Castling is a move that allows the king to be moved two squares towards a rook on the player's first rank, then that rook moves to the square the king crossed. En passant is a special pawn capture that can only occur immediately after a pawn makes a move of two squares from its starting position, and an opponent's pawn could have captured it had it moved only one square forward.
E. Check and checkmate:
A check is a move that puts the opponent's king in a position where it could be captured in the next move. Checkmate is when a king is in check and cannot escape capture.
III. Strategies and Tactics
A. Opening moves:
The opening is the initial phase of the game where players develop their pieces and control the center of the board. The goal is to create a strong pawn structure, control the center, and develop the pieces towards the center. There are many different openings in chess, each with their own strengths and weaknesses. Some popular openings include the Sicilian Defense, the Ruy Lopez, and the Queen's Gambit.
B. Middle game:
The middle game is the phase of the game where players begin to attack and defend. Players use their pieces to gain control of key squares and launch attacks against the opponent's pieces and pawns. In this phase, players must use their knowledge of chess strategy and tactics to outmaneuver their opponent and gain an advantage.
C. Endgame:
The endgame is the final phase of the game, where there are fewer pieces left on the board. The goal is to convert an advantage into a win by checkmating the opponent's king or forcing the opponent to resign due to a lost position. In the endgame, players must be able to calculate and plan ahead accurately to win.
D. Common traps and pitfalls:
In chess, it is important to be aware of common traps and pitfalls that can lead to a loss. These can include falling for a bait piece, failing to protect a weak pawn, or moving the king into a dangerous position. It is important to be aware of these traps and to be able to recognize and avoid them in order to succeed in the game.
E. Importance of planning and foresight:
In order to be successful in chess, it is important to have a plan for the game and to think several moves ahead. This means considering not only the current move, but also potential future moves and how they will affect the position. It is also important to be aware of potential threats and to have a plan to deal with them.
IV. Variants of Chess
A. Different forms of chess (e.g. speed chess, blindfold chess):
There are many variations of chess, including speed chess where players have limited time to make their moves, and blindfold chess where players play the game without physically seeing the board. These variations add a new level of challenge and require different skills and strategies to succeed.
B. Chess960:
Chess960 is a variant of chess that starts with a random placement of the pieces on the back row, with the only restriction being that the bishops must be on opposite-colored squares. This variation adds an element of surprise and requires players to be adaptable and think on their feet.
C. Other chess variants (e.g. Shatranj, Makruk):
There are many other chess variants that have been developed throughout history, including Shatranj, which was the precursor to modern chess and Makruk, which is a popular variant in Thailand. These variants have their own unique rules and strategies, and can provide a new and challenging experience for chess players.
V. Competitive Chess
A. Tournaments and competitions:
Chess is a popular competitive sport, with tournaments and competitions taking place all over the world. These can range from local events to international tournaments, and can offer players the opportunity to test their skills against other players and to earn prizes.
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### King of Earth
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Be the first to respond and start the conversation. | 1,319 | 6,172 | {"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-2023-14 | latest | en | 0.961304 |
https://stats.arabpsychology.com/st/generate-random-numbers-in-r-with-examples/ | 1,701,637,151,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.53/warc/CC-MAIN-20231203193127-20231203223127-00241.warc.gz | 622,410,763 | 23,092 | # Generate Random Numbers in R (With Examples)
You can use the following methods to generate random numbers in R:
Method 1: Generate One Random Number in Range
```#generate one random number between 1 and 20
runif(n=1, min=1, max=20)
```
Method 2: Generate Multiple Random Numbers in Range
```#generate five random numbers between 1 and 20
runif(n=5, min=1, max=20)```
Method 3: Generate One Random Integer in Range
```#generate one random integer between 1 and 20
sample(1:20, 1)
```
Method 4: Generate Multiple Random Integers in Range
```#generate five random integers between 1 and 20 (sample with replacement)
sample(1:20, 5, replace=TRUE)
#generate five random integers between 1 and 20 (sample without replacement)
sample(1:20, 5, replace=FALSE)
```
The following examples show how to use each of these methods in practice.
## Method 1: Generate One Random Number in Range
The following code shows how to generate one random number between 1 and 20:
```#generate one random number between 1 and 20
runif(n=1, min=1, max=20)
[1] 8.651919
```
This function generates 8.651919 as the random number between 1 and 20.
## Method 2: Generate Multiple Random Numbers in Range
The following code shows how to generate five random numbers between 1 and 20:
```#generate five random numbers between 1 and 20
runif(n=5, min=1, max=20)
[1] 12.507360 6.719675 1.836038 17.685829 16.874723
```
## Method 3: Generate One Random Integer in Range
The following code shows how to generate one random integer between 1 and 20:
```#generate one random integer between 1 and 20
sample(1:20, 1)
[1] 7
```
This function generates 7 as the random integer between 1 and 20.
## Method 4: Generate Multiple Random Integers in Range
The following code shows how to generate five random integers between 1 and 20:
```#generate five random integers between 1 and 20 (sample with replacement)
sample(1:20, 5, replace=TRUE)
[1] 20 13 15 20 5
#generate five random integers between 1 and 20 (sample without replacement)
sample(0:20, 5, replace=FALSE)
[1] 6 15 5 16 19
```
Note that if we use replace=TRUE then we allow the same integer to be generated more than once.
However, if we use replace=FALSE then we do not allow the same integer to be generated more than once. | 641 | 2,279 | {"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} | 3.546875 | 4 | CC-MAIN-2023-50 | latest | en | 0.818354 |
http://ocw.mit.edu/courses/aeronautics-and-astronautics/16-901-computational-methods-in-aerospace-engineering-spring-2005/assignments/convect2d.m | 1,472,329,807,000,000,000 | text/plain | crawl-data/CC-MAIN-2016-36/segments/1471982925602.40/warc/CC-MAIN-20160823200845-00157-ip-10-153-172-175.ec2.internal.warc.gz | 184,134,257 | 1,851 | % This Matlab script solves the two-dimensional convection % equation using a finite volume algorithm on a mesh of (constant) % rectangular cells. The problem is assumed to be periodic and % have a constant velocity. close all; clear all; % Specify x range and number of points x0 = -2; x1 = 2; Nx = 40; % Specify y range and number of points y0 = -2; y1 = 2; Ny = 40; % Construct mesh x = linspace(x0,x1,Nx+1); y = linspace(y0,y1,Ny+1); [xg,yg] = ndgrid(x,y); % Construct mesh needed for plotting xp = zeros(4,Nx*Ny); yp = zeros(4,Nx*Ny); n = 0; for j = 1:Ny, for i = 1:Nx, n = n + 1; xp(1,n) = x(i); yp(1,n) = y(j); xp(2,n) = x(i+1); yp(2,n) = y(j); xp(3,n) = x(i+1); yp(3,n) = y(j+1); xp(4,n) = x(i); yp(4,n) = y(j+1); end end % Calculate midpoint values in each control volume xmid = 0.5*(x(1:Nx) + x(2:Nx+1)); ymid = 0.5*(y(1:Ny) + y(2:Ny+1)); [xmidg,ymidg] = ndgrid(xmid,ymid); % Calculate cell size in control volumes (assumed equal) dx = x(2) - x(1); dy = y(2) - y(1); A = dx*dy; % Set velocity u = 1; v = 1; % Set final time tfinal = 10; % Set timestep CFL = 1.0; dt = CFL/(abs(u)/dx + abs(v)/dy); % Set initial condition to U0 = exp(-x^2 - 20*y^2) % Note: technically, we should average the initial % distribution in each cell but I chose to just set % the value of U in each control volume to the midpoint % value of U0. U = exp(-xmidg.^2 - 20*ymidg.^2); t = 0; % Loop until t > tfinal while (t < tfinal), % The following implement the bc's by creating a larger array % for U and putting the appropriate values in the first and last % columns or rows to set the correct bc's Ubc(2:Nx+1,2:Ny+1) = U; % Copy U into Ubc Ubc( 1,2:Ny+1) = U(Nx, :); % Periodic bc Ubc(Nx+2,2:Ny+1) = U( 1, :); % Periodic bc Ubc(2:Nx+1, 1) = U( :,Ny); % Periodic bc Ubc(2:Nx+1,Ny+2) = U( :, 1); % Periodic bc % Calculate the flux at each interface % First the i interfaces F = 0.5* u *( Ubc(2:Nx+2,2:Ny+1) + Ubc(1:Nx+1,2:Ny+1)) ... - 0.5*abs(u)*( Ubc(2:Nx+2,2:Ny+1) - Ubc(1:Nx+1,2:Ny+1)); % Now the j interfaces G = 0.5* v *( Ubc(2:Nx+1,2:Ny+2) + Ubc(2:Nx+1,1:Ny+1)) ... - 0.5*abs(v)*( Ubc(2:Nx+1,2:Ny+2) - Ubc(2:Nx+1,1:Ny+1)); % Add contributions to residuals from fluxes R = (F(2:Nx+1,:) - F(1:Nx,:))*dy + (G(:,2:Ny+1) - G(:,1:Ny))*dx; % Forward Euler step U = U - (dt/A)*R; % Increment time t = t + dt; % Plot current solution Up = reshape(U,1,Nx*Ny); clf; [Hp] = patch(xp,yp,Up); set(Hp,'EdgeAlpha',0); axis('equal'); caxis([0,1]); colorbar; drawnow; end | 988 | 2,445 | {"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} | 3.421875 | 3 | CC-MAIN-2016-36 | latest | en | 0.661388 |
http://oeis.org/A305059 | 1,606,859,854,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141681524.75/warc/CC-MAIN-20201201200611-20201201230611-00197.warc.gz | 65,725,382 | 5,334 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A305059 Triangle read by rows: T(n,k) is the number of connected unicyclic graphs on n unlabeled nodes with cycle length k and all nodes having degree at most 4. 9
1, 1, 1, 3, 1, 1, 6, 4, 1, 1, 15, 8, 4, 1, 1, 33, 24, 9, 5, 1, 1, 83, 55, 28, 12, 5, 1, 1, 196, 147, 71, 40, 13, 6, 1, 1, 491, 365, 198, 106, 47, 16, 6, 1, 1, 1214, 954, 521, 317, 136, 63, 18, 7, 1, 1, 3068, 2431, 1418, 868, 428, 190, 73, 21, 7, 1, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 3,4 COMMENTS Equivalently, the number of monocyclic skeletons with n carbon atoms and a ring size of k. LINKS Andrew Howroyd, Table of n, a(n) for n = 3..1178 (rows n = 3..50) Camden A. Parks and James B. Hendrickson, Enumeration of monocyclic and bicyclic carbon skeletons, J. Chem. Inf. Comput. Sci., vol. 31, 334-339 (1991). Eric Weisstein's World of Mathematics, Unicyclic Graph EXAMPLE Triangle begins: 1; 1, 1; 3, 1, 1; 6, 4, 1, 1; 15, 8, 4, 1, 1; 33, 24, 9, 5, 1, 1; 83, 55, 28, 12, 5, 1, 1; 196, 147, 71, 40, 13, 6, 1, 1; 491, 365, 198, 106, 47, 16, 6, 1, 1; 1214, 954, 521, 317, 136, 63, 18, 7, 1, 1; ... MATHEMATICA G[n_] := Module[{g}, Do[g[x_] = 1 + x*(g[x]^3/6 + g[x^2]*g[x]/2 + g[x^3]/3) + O[x]^n // Normal, {n}]; g[x]]; T[n_, k_] := Module[{t = G[n], g}, t = x*((t^2 + (t /. x -> x^2))/2); g[e_] = (Normal[t + O[x]^Quotient[n, e]] /. x -> x^e) + O[x]^n // Normal; Coefficient[(Sum[EulerPhi[d]*g[d]^(k/d), {d, Divisors[k]}]/k + If[OddQ[ k], g[1]*g[2]^Quotient[k, 2], (g[1]^2 + g[2])*g[2]^(k/2-1)/2])/2, x, n]]; Table[T[n, k], {n, 3, 13}, {k, 3, n}] // Flatten (* Jean-François Alcover, Jul 03 2018, after Andrew Howroyd *) PROG (PARI) \\ here G is A000598 as series G(n)={my(g=O(x)); for(n=1, n, g = 1 + x*(g^3/6 + subst(g, x, x^2)*g/2 + subst(g, x, x^3)/3) + O(x^n)); g} T(n, k)={my(t=G(n)); t=x*(t^2+subst(t, x, x^2))/2; my(g(e)=subst(t + O(x*x^(n\e)), x, x^e) + O(x*x^n)); polcoeff((sumdiv(k, d, eulerphi(d)*g(d)^(k/d))/k + if(k%2, g(1)*g(2)^(k\2), (g(1)^2+g(2))*g(2)^(k/2-1)/2))/2, n)} CROSSREFS Columns 3..10 are A063832, A116719, A120333, A120779, A120790, A120795, A121156, A121157. Row sums are A036671. Cf. A000598. Sequence in context: A125230 A208334 A162430 * A128101 A211351 A124802 Adjacent sequences: A305056 A305057 A305058 * A305060 A305061 A305062 KEYWORD nonn,tabl AUTHOR Andrew Howroyd, May 24 2018 STATUS approved
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Last modified December 1 16:55 EST 2020. Contains 338846 sequences. (Running on oeis4.) | 1,351 | 3,174 | {"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-2020-50 | latest | en | 0.537168 |
https://mathskills4kids.com/practice-matching-shapes-with-pictures-game-online | 1,722,815,354,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640417235.15/warc/CC-MAIN-20240804230158-20240805020158-00025.warc.gz | 308,267,417 | 9,989 | # Identify the missing shape - Matching shapes with pictures game online
## Identifying shapes game
Identify the missing shape is an outstanding tool for kids that possesses a good observational skill.
Considering the fact that shapes are all round us, this concept will give kids the opportunity to differentiate between the different characters each shape has. As we all know, anything interesting usually attracts the interest of kids.
Therefore, our matching shapes with pictures game online are well designed and attractive for kids to easily identify the missing shapes.
although this identifying shapes game is somehow challenging, it is so fun, interactive and a basic educational math skill.
• ## IN THE SAME CATEGORY
### Identifying shapes game for critical reasoning and shapes recognition skills
Common shapes that kids will master in the identifying shapes game include the triangle, square, circle, and the heart. Considering the fact that number recognition is an early math skill, a perfect understanding of these shapes will help kids to better recognize numbers and how they look like.
As such, it suffices just for the kids to pay attention, then recognize the characteristics of all the different shapes. For instance, a triangle has three sides; a square has four equal sides and so on.
We've got very amazing shape monkey games that will drill kids to easily identify the missing shape, given the missing part of each monkey's face.
### Example – find the missing shape
This game has been made very easy for kids to find the missing shape. Just consider the following guides
• Firstly, two identical monkeys are presented, one with its complete face and the other with a missing part of the face.
• Secondly, if we observe keenly, we’ll notice that the missing part is covered with a particular shape, having three sides.
This particular shape is called a triangle.
As a matter of fact, I'm sure we can only see the monkey’s left side eye. Where then are the other parts, i.e. the right side eye, nose and mouth?
Below are multiple shapes for you to clear your doubts by choosing that which is triangular as the one above, and also possesses the missing parts.
Hey, I have observed that a triangle has three sides. So let me spot it out from the multiples below.
Yeeeaah! I got it. I have seen the triangular shape. It’s the second shape with three sides.
Moreover, this triangle is exactly the missing part of the monkey’s face, because it has the monkey’s right side eye, its nose, and its mouth.
Huraayyyy! I spotted it so easily. I want more exercises.
“Are you really sure you want more exercises? Ok. Just click on the triangle, then click on Next to continue with the fun game”
• | 552 | 2,730 | {"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} | 4.375 | 4 | CC-MAIN-2024-33 | latest | en | 0.941591 |
https://quantumtetrahedron.wordpress.com/tag/lqg/ | 1,516,318,872,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887660.30/warc/CC-MAIN-20180118230513-20180119010513-00386.warc.gz | 798,950,370 | 58,520 | # Calculations on Quantum Cuboids and the EPRL-FK path integral for quantum gravity
This week I have been studying a really great paper looking at Quantum Cuboids and path-integral calculations for the EPRL vertex in LQG and also beginning to write some calculational software tools for performing these calculations using Sagemath.
In this work the authors investigate the 4d path integral for Euclidean quantum gravity on a hypercubic lattice, as given by the EPRL-FK model. To tackle the problem, they restrict the path to a set of quantum geometries that reflects the lattice symmetries. In particular, the sum over intertwiners is restricted to quantum cuboids, that is, coherent intertwiners which describe a cuboidal
geometry in the large-j limit.
Using asymptotic expressions for the vertex amplitude, several interesting properties of the state sum are found.
• The value of coupling constants in the amplitude functions determines whether geometric or non-geometric configurations dominate the path integral.
• There is a critical value of the coupling constant α, which separates two phases. In one phase the main contribution
comes from very irregular and crumpled states. In the other phase, the dominant contribution comes from a highly regular configuration, which can be interpreted as flat Euclidean space, with small non-geometric perturbations around it.
• States which describe boundary geometry with high
torsion have exponentially suppressed physical norm.
The symmetry-restricted state sum
Will work on a regular hypercubic lattice in 4d. On this lattice consider only states which conform to the lattice symmetry. This is a condition on the intertwiners, which corresponds to cuboids.
A cuboid is completely determined by its three edge lengths, or equivalently by its three areas.
All internal angles are π/2 , and the condition of regular cuboids on all dual edges of the lattice result in a high degree of symmetries on the labels: The area and hence the spin on each two parallel squares of the lattice which are translations perpendicular to the squares, have to be equal.
The high degree of symmetry will make all quantum geometries flat. The analysis carried out here is therefore not suited for describing local curvature.
Introduction
The plan of the paper is as follows:
• Review of the EPRL-FK spin foam model
• Semiclassical regime of the path integral
• Construction of the quantum cuboid intertwiner
• Full vertex amplitude, in particular describe its asymptotic expression for large spins
• Numerical investigation of the quantum path integral
The spin foam state sum employed is the Euclidean EPRL-FK model with Barbero-Immirzi parameter γ < 1. The EPRL-FK model is defined on an arbitrary 2-complexes. A 2-complex is determined by its vertices v, its edges e connecting two vertices, and faces f which are bounded by the edges.
The path integral is formulated as a sum over states. A state in this context is given by a collection of spins – irreducible representations
jf ∈ 1/2 N of SU(2) to the faces, as well as a collection of intertwiners ιe on edges.
The actual sum is given by
where Af , Ae and Av are the face-, edge- and vertex- amplitude functions, depending on the state. The sum has to be carried out over all spins, and over an orthonormal orthonormal basis in the intertwiner space at each edge.
The allowed spins jf in the EPRL-FK model are such
that are both also half-integer spins.
The face amplitudes are either
The edge amplitudes Ae are usually taken to be equal to 1.
In Sagemath code this looks like:
Coherent intertwiners
In this paper, the space-time manifold used is M∼ T³×[0, 1] is the product of the 3-torus T3 and a closed interval. The space is compactified toroidally. M is covered by 4d hypercubes, which
form a regular hypercubic lattice H.There is a vertex for each hypercube, and two vertices are connected by an edge whenever two hypercubes intersect ina 3d cube. The faces of are dual to squares in H, on which four hypercubes meet.The geometry will be encoded in the state, by specification of spins jf
and intertwiners ιe.
Intertwiners ιe can be given a geometric interpretation in terms of polyhedra in R³. Given a collection of spins j1, . . . jn and vectors n1, . . . nn which close . Can define the coherent polyhedron
The geometric interpretation is that of a polyhedron, with face areas jf and face normals ni. The closure condition ensures that such a polyhedron exists.
We are interested in the large j-regime of the quantum cuboids. In this limit, these become classical cuboids which are completely specified by their three areas. Therefore, a
semiclassical configuration is given by an assignment of
areas a = lp² to the squares of the hypercubic lattice.
Denote the four directions in the lattice by x, y, z, t. The areas satisfy
The two constraints which reduce the twisted geometric
configurations to geometric configurations are given by:
For a non-geometric configuration, define the 4-volume of a hypercube as:
Define the four diameters to be:
then we have, V4 = dxdydzdt
We also define the non- geometricity as:
as a measure of the deviation from the constraints.
In sagemath code this looks like:
Quantum Cuboids
We let’s look at the quantum theory. In the 2-complex, every edge has six faces attached to it, corresponding to the six faces of the cubes. So any intertwiner in the state-sum will be six-valent, and therefore can be described by a coherent polyhedron with six faces. In our setup, we restrict the state-sum to coherent cuboids, or quantum cuboids. A cuboid is characterized by areas on opposite sides of the cuboid being equal, and the respective normals being negatives of one another
The state ιj1,j2,j3 is given by:
The vertex amplitude for a Barbero-Immirzi parameter γ < 1 factorizes as Av = A+vAv with
with the complex action
where, a is the source node of the link l, while b is its target node.
Large j asymptotics
The amplitudes A±v possess an asymptotic expression for large jl. There are two distinct stationary and critical points, satisfying the equations.
for all links ab . Using the convention shown below
having fixed g0 = 1, the two solutions Σ1 and Σ2 are
The amplitudes A±satisfy, in the large j limit,
In the large j-limit, the norm squared of the quantum cuboid states is given by:
For the state sum, in the large-j limit on a regular hypercubic lattice:
In sagemath code this looks like:
Related articles
# Group field theories generating polyhedral complexes by Thürigen
This week I have been studying recent developments in Group Field Theories. Group field theories are a generalization of matrix models which provide both a second quantized reformulation of loop quantum gravity as well as generating functions for spin foam models. Other posts looking at this include:
While states in canonical loop quantum gravity are based on graphs with vertices of arbitrary valence, group field theories have been defined so far in a simplicial setting such that states have support only on graphs of fixed valency. This has led to the question whether group field theory can indeed cover the whole state space of loop quantum gravity.
The paper discusses the combinatorial structure of the complexes generated by the group field theory partition function. These new group field theories strengthen the links between the various quantum gravity approaches and might also prove useful in the investigation of renormalizability.
The combinatorial structure of group field theory
The common notion of GFT is that of a quantum field theory on group manifolds with a particular kind of non-local interaction vertices. A group field is a function of a Lie group G and the GFT is defined by a partition function
the action is of the form:
The evaluation of expectation values of quantum observables O[φ], leads to a series of Gaussian integrals evaluated
through Wick contraction which are catalogued by Feynman diagrams Γ,
where sym(Γ) are the combinatorial factors related to the automorphism group of the Feynman diagram Γ:
The specific non-locality of each vertex is captured by a boundary graph. In the interaction term in each group field term can be represented by a graph consisting of a k-valent vertex connected to k univalent vertices. One may further understand the graph as
the boundary of a two-dimensional complex a with a single internal vertex v. Such a one-vertex two-complex a is called a spin foam atom.
The GFT Feynman diagrams in the perturbative sum have the structure of two complexes because Wick contractions effect bondings of such atoms along patches. The combinatorial
structure of a term in the perturbative sum is then a collection of spin foam atoms, one for each vertex kernel, quotiented by a set of bonding maps, one for each Wick contraction. Because of this construction such a two-complex will be called a spin foam molecule.
The crucial idea to create arbitrary boundary graphs in a more efficient way is to distinguish between virtual and real edges and obtain arbitrary graphs from regular ones by contraction of the
virtual edges.
In terms of these contractions, any spin foam molecule can be obtained from a molecule constructed from labelled regular graphs.
Conclusions
This paper has aimed to generalization of GFT to be compatible with LQG. It has clarified the combinatorial structure underlying the amplitudes of perturbative GFT using the notion of spin foam atoms and molecules and discussed their possible spacetime interpretation.
Related articles
# Group field theory as the 2nd quantization of Loop Quantum Gravity by Daniele Oriti
This week I have been reviewing Daniele Oriti’s work, reading his Frontiers of Fundamental Physics 14 conference paper – Group field theory: A quantum field theory for the atoms of space and making notes on an earlier paper, Group field theory as the 2nd quantization of Loop Quantum Gravity. I’m quite interested in Oriti’s work as can be seen in the posts:
Introduction
We know that there exist a one-to-one correspondence between spin foam models and group field theories, in the sense that for any assignment of a spin foam amplitude for a given cellular complex,
there exist a group field theory, specified by a choice of field and action, that reproduces the same amplitude for the GFT Feynman diagram dual to the given cellular complex. Conversely, any given group field theory is also a definition of a spin foam model in that it specifies uniquely the Feynman amplitudes associated to the cellular complexes appearing in its perturbative expansion. Thus group field theories encode the same information and thus
define the same dynamics of quantum geometry as spin foam models.
That group field theories are a second quantized version of loop quantum gravity is shown to be the result of a straightforward second quantization of spin networks kinematics and dynamics, which allows to map any definition of a canonical
dynamics of spin networks, thus of loop quantum gravity, to a specific group field theory encoding the same content in field-theoretic language. This map is very general and exact, on top of being rather simple. It puts in one-to-one correspondence the Hilbert space of the canonical theory and its associated algebra of quantum observables, including any operator defining the quantum dynamics, with a GFT Fock space of states and algebra of operators and its dynamics, defined in terms of a classical action and quantum equations for its n-point functions.
GFT is often presented as the 2nd quantized version of LQG. This is true in a precise sense: reformulation of LQG as GFT very general correspondence both kinematical and dynamical. Do not need to pass through Spin Foams . The LQG Spinfoam correspondence is obtained via GFT. This reformulation provides powerful new tools to address open issues in LQG, including GFT renormalization and Effective quantum cosmology from GFT condensates.
Group field theory from the Loop Quantum Gravity perspective:a QFT of spin networks
Lets look at the second quantization of spin networks states and the correspondence between loop quantum gravity and group field theory. LQG states or spin network states can be understood as many-particle states analogously to those found in particle physics and condensed matter theory.
As an example consider the tetrahedral graph formed by four vertices and six links joining them pairwise
The group elements Gij are assigned to each link of the graph, with Gij=Gij-1. Assume gauge invariance at each vertex i of the graph. The basic point is that any loop quantum gravity state can be seen as a linear combination of states describing disconnected open spin network vertices, of arbitrary number, with additional conditions enforcing gluing conditions and encoding the connectivity of the graph.
Spin networks in 2nd quantization
A Fock vacuum is the no-space” (“emptiest”) state |0〉 , this is the LQG vacuum – the natural background independent, diffeo-invariant vacuum state.
The 2nd quantization of LQG kinematics leads to a definition of quantum fields that is very close to the standard non-relativistic one used in condensed matter theory, and that is fully compatible
with the kinematical scalar product of the canonical theory. In turn, this can be seen as coming directly from the definition of the Hilbert space of a single tetrahedron or more generally a quantum polyhedron.
The single field quantum is the spin network vertex or tetrahedron – the so called building block of space.
A generic quantum state is anarbitrary collection of spin network vertices including glued ones or tetrahedra including glued ones.
The natural quanta of space in the 2nd quantized language are open spin network vertices. We know from the canonical theory that they carry area and volume information, and know their pre geometric properties from results in quantum simplicial geometry.
Related articles
# Semiclassical states in quantum gravity: Curvature associated to a Voronoi graph by Daz-Polo and Garay
This week I’ returning to a much more fundamental level and reviewing a paper on Voronoi graphs. These are a method of dividing up a space into triangles and my very early work on this blog was looking at random triangulations. This paper outlines an attempt to compute the curvature of a surface that didn’t work – that’s science, but we can build on that to find a method that does work.
The building blocks of a quantum theory of general relativity are
expected to be discrete structures. Loop quantum gravity is formulated using a basis of spin networkswave functions over oriented graphs with coloured edges. Semiclassical states should,
however, reproduce the classical smooth geometry in the appropriate limits. The question of how to recover a continuous geometry from these discrete structures is, therefore, relevant in this context. The authors explore this problem from a rather general mathematical perspective using properties of Voronoi graphs to search for their compatible continuous geometries. They test the previously proposed methods for computing the curvature associated to such graphs and analyse the framework in detail in the light of the
results obtained.
Introduction
General relativity describes the gravitational interaction as a consequence of the curvature of space-time, a 4-dimensional Lorentzian manifold. Given this geometric nature of gravity, it is expected that, when quantizing, a prescription for quantum
geometry would arise based on more fundamental discrete structures, rather than on smooth differential manifolds.
Loop quantum gravity (LQG) is a candidate theory for such a quantization of general relativity. The fundamental objects , thebasis of the kinematical Hilbert space are the so-called spin network states, which are defined as wave functions constructed over oriented coloured graphs. The building blocks of the theory are, therefore, discrete combinatorial structures – graphs. The theory provides quantum operators with a direct geometric interpretation, areas and volumes, which happen to have discrete eigenvalues, reinforcing the idea of a discrete geometry.
This perspective of considering abstract combinatorial structures as the fundamental objects of the theory is also adopted in other approaches to quantum gravity, such as spin-foam models, causal dynamical triangulations and causal sets . Also, the algebraic quantum gravity approach follows the same spirit of constructing a quantum theory of gravity from an abstract combinatorial structure.
Despite the variety of successful results obtained in LQG, the search for a semiclassical sector of the theory that would connect with the classical description given by general relativity in terms of a smooth manifold is still under research. An interesting question for the description of a semiclassical sector, given the combinatorial nature of the building blocks of the theory, would be whether there
is any correspondence between certain types of graphs and the continuous classical geometries. Tentatively, this would allow for the construction of gravitational coherent states corresponding to solutions of Einstein eld equations. While it is certainly true
that spin networks are a particular basis, and coherent states constructed from them might resemble nothing like a graph, some works seem to indicate that these graphs do actually represent the structure of space-time at the fundamental level.
This raises a very interesting question. How does the transition between a fundamental discrete geometry, encoded in a graph structure, and the continuous geometry we experience in every-day life happen? In particular, how does a one- dimensional structure give rise to 3-dimensional smooth space? A step towards answering these questions could be to think of these graphs as embedded in the
corresponding continuous geometries they represent. However, the situation is rather the opposite, being the smooth continuous structure an effective structure, emerging from the more fundamental discrete one, and not the other way around. Therefore,
a very relevant question to ask would be: Is there any information, contained in the abstract structure of a graph, that determine the compatible continuous geometries? Can we determine what types of manifolds a certain graph can be embedded in? One could even go further and ask whether any additional geometric information, like curvature, can be extracted from the very abstract structure of the graph itself. The goal is, therefore, to construct a unique correspondence between the discrete structures given by graphs, which in general do not carry geometric information, and smooth manifolds.
This problem was studied in the context of quantum gravity by Bombelli, Corichi and Winkler, who proposed a statistical method to compute the curvature of the manifold that would be associated to a certain class of graphs, based on Voronoi diagrams, giving a new step towards the semiclassical limit of LQG. Indeed, due to their properties, Voronoi diagrams appear naturally when addressing this kind of problems.They also play an important role in the discrete approach to general relativity provided by Regge calculus.
Voronoi diagrams are generated from a metric manifold and, by construction, contain geometric information from it. What was proposed, however, is to throw away all additional geometric information and to keep only the abstract structure of
the one-dimensional graph that forms the skeleton of the Voronoi diagram. Then, the task is to study if there are any imprints of the original geometry which remain in this abstract graph structure. Although the work is somewhat preliminary and, for the
most part, restricted to 2-dimensional surfacesz, it tackles very interesting questions and explores a novel path towards a semiclassical regime in LQG. The results obtained could also provide a useful tool for the causal dynamical triangulations approach.
Curvature associated to a graph
A Voronoi diagram is constructed in the following way. For a set of points -seeds, on a metric space, each highest-dimensional cell of the Voronoi diagram contains only one seed, and comprises the region of space closer to that one seed than to any of the others. Then, co-dimension n cells are made by sets of points equidistant to n + 1 seeds, e.g., in 2 dimensions, the edges (1-dimensional cells) of the Voronoi diagram are the lines separating two of these regions, and are therefore equidistant to two seeds. In the same way, vertices (0-dimensional cells) are equidistant to three seeds.
Therefore, except in degenerate situations which are avoided by randomly sprinkling the seed, the valence of all vertices in a D-dimensional Voronoi diagram is D + 1. Another interesting property of Voronoi diagrams is that their dual graph is the so- called Delaunay triangulation, whose vertices are the Voronoi seeds. By construction, for a given set of seeds on a metric space the corresponding Voronoi diagram is uniquely defined.
The starting point is to consider a given surface on which we randomly sprinkle a set of points, that will be the seeds
for the Voronoi construction. A Voronoi cell-complex is constructed, containing zero, one, and two-dimensional cells (vertices, edges and faces). We keep, then, the abstract structure of the one-dimensional graph encoded, for instance, in an adjacency matrix. We are, thus, left with an abstract graph.
All vertices of the Voronoi graph are tri-valent. This gives rise
to the following relation between the total number V of vertices in the graph and the total number of edges E:
since every vertex is shared by three edges and every edge contains two vertices. We will also use the definition of the Euler-Poincare characteristic χ in the two-dimensional case
where F is the total number of faces in the graph (that equals the number of seeds). Finally, one can define the number p of sides of a face (its perimeter in the graph). Taking into account that every edge is shared by two faces, the average p over a set of faces satisfies
The following expression for the Euler-Poincare characteristic χ
can be obtained
in terms of the total number of faces F and the average number of sides of the faces p.
On the other hand, if there is a manifold M associated to the Voronoi diagram, this manifold should have the same topology as the diagram. The Gauss-Bonnet theorem can be used then to relate χ with the integral of the curvature over the manifold. If M is a manifold without boundary (like a sphere), the theorem takes the form
where dA is the area measure and R is the Ricci scalar.
Assume that the region of the graph one is looking at is small enough
so that the curvature can be considered constant. In that case
where As is the total surface area of the sphere.
this formula can also be applied to the sphere patch by defining a density of faces ρ= F=As = Fp =Ap , where Fp and Ap are respectively the number of faces and area of the patch.
Implementation and results
Conclusions and outlook
The problem of reconstructing a continuous geometry starting from a discrete, more fundamental combinatorial structure, like a graph, is interesting for a wide range of research fields. In the case of LQG theory whose Hilbert space is constructed using wave functions defined over graphsthe solution to this problem could provide interesting hints on the construction of semiclassical states, moving toward a connection with classical solutions of the Einstein equations.
In this article the authors discussed and implemented the method proposed to compute the curvature of a manifold from an abstract Voronoi graph associated to it. By making use of some topological arguments involving the Gauss-Bonnet theorem, a method to statistically compute the curvature in terms of the average number of sides of the faces in the graph is suggested. They tested this
method for the simplest geometries: the unit sphere – constant positive curvature and the plane – zero curvature. They
found highly unsatisfactory results for the value of the curvature in both cases.
# Properties of the Volume Operator in Loop Quantum Gravity by Brunnemann and Rideout
This week I’ve returned more strongly to my work on the numerical spectra of quantum geometrical operators. This post looks at an analysis of the Ashtekar and Lewandowski version of the volume operator.
The authors analyze the spectral properties of the volume operator of Ashtekar and Lewandowski in Loop Quantum Gravity, which is the quantum analogue of the classical volume expression for regions in three dimensional Riemannian space. The analysis also considers generic graph vertices of valence greater than four. The authors find that the geometry of the underlying vertex characterizes the spectral properties of the volume operator, in particular the presence of a volume gap – a smallest non-zero eigenvalue in the spectrum is found to depend on the vertex embedding.
The authors compute the set of all non-spatially diffeomorphic non-coplanar vertex embeddings for vertices of valence 5–7 and argue that these sets can be used to label spatial diffeomorphism invariant states. it is seen that gauge invariance connects vertex geometry and the representation properties of the underlying gauge group in a natural way. Analytical results on the spectrum of 4-valent vertices show the presence of a volume gap.
Loop Quantum Gravity (LQG) is a candidate for a quantum theory of gravity. It is an attempt to canonically quantize General Relativity. The resulting quantum theory is formulated as an SU(2) gauge theory.
General Relativity is put into a Hamiltonian formulation by introducing a foliation of four dimensional spacetime into spatial three dimensional hypersurfaces ∑, with the orthogonal timelike
direction parametrized by t. first class constraints which have to be imposed on the theory so that it obeys the dynamics of Einstein’s equations and is independent of the particular choice of foliation. These constraints are the three spatial diffeomorphism or vector
constraints which generate diffeomorphisms and the so called Hamilton constraint which generates deformations of the hypersurfaces in the t- foliation direction. In addition there are three Gauss constraints due to the introduction of additional SU(2) gauge degrees of freedom.
The theory is then quantized on the kinematical level in terms of holonomies h and electric fluxes E. Kinematical states are defined over collection of edges of embedded graphs. The physical states have to be constructed by imposing the operator version of the constraints on the thus defined kinematical theory.
There are well defined operators in the kinematical quantum theory which correspond to differential geometric objects, such as the length of curves, the area of surfaces, and the volume of regions in the spatial foliation hypersurfaces. All these quantities have discrete spectra, which can be traced back to the compactness of the gauge group SU(2). A coherent state framework address questions on the correct semiclassical limit of the theory.
A central role in investigations is played by the area and the volume operators.The volume operator is a crucial object not only in order to analyze matter coupling to LQG, but also for evaluating the action of the Hamilton constraint operator in order to construct the physical sector of LQG. By construction the spectral properties of the constraint operators are driven by the spectrum of the volume operator.
Loop Quantum Gravity
Hamiltonian Formulation
In order to cast General Relativity into the Hamiltonian formalism, one has to perform a foliation M ≅ Rx∑ of the four dimensional spacetime manifold M into three dimensional spacelike hypersurfaces with transverse time direction labelled by a foliation parameter t ∈ R. The four dimensional metric g can then be decomposed into the three metric q(x) on the spatial slices ∑ and its extrinsic curvature K, which serve as canonical variables. Introducing new variables due to Ashtekar may take
as canonical variables electric fields E and connections A as used in the canonical formulation of Yang Mills theories. Here a, b = 1 . . . 3 denote spatial (tensor) indices, i, j = 1 . . . 3 denote su(2)-
indices.
The occurrence of SU(2) as a gauge group comes from the fact that it is necessary for the coupling of spinorial matter and it is the universal covering group of SO(3) which arises naturally when one
rewrites the three metric q in terms of cotriads as
The pair (A,E) is related to (q,K) as
with Γ being the spin connection.
The pair (A,E) obeys the Poisson bracket:
Constraints
The theory is subject to constraints which arise due to the background independence of General Relativity. It can be treated as suggested by Dirac . There are:
• three vector (spatial diffeomorphism) constraints
• a scalar – Hamilton constraint
• three Gauss constraints
The Volume Operator
Definition of the Volume Operator
The operator corresponding to the volume V (R) of a spatial region R ⊂ ∑
Acting on gauge invariant spin network states, is defined as
The sum in has to be taken over all vertices v of the underlying graph γ. At each vertex v one has to sum over all possible ordered triples (ei , ej , ek).
Matrix Formulation
Consider,
where Q is a totally antisymmetric matrix with purely real elements. Its eigenvalues λ are purely imaginary and come in pairs λQ = ±iλ. Choosing Z = 1 the volume operator V has the same eigenstates as Q and its eigenvalues λ = |λQ|½
We are left with the task of calculating the spectra of totally antisymmetric real matrices of the form:
Analytical Results on the Gauge Invariant 4-Vertex
The spectrum of the volume operator at a given gauge invariant 4-vertex is simple, that is all its eigenvalues, except zero, come in pairs, and there are no further degeneracies or accumulation points in the spectrum. An explicit expression for the eigenstates of the volume operator in terms of polynomials of its matrix elements and its eigenvalues can be given as:
where,
The specific matrix realization of the volume operator plays a crucial role here since it can be written as an antisymmetric purely imaginary matrix having non-zero entries only on the first off-diagonal. This makes it possible to apply techniques from orthogonal polynomials and Jacobi-matrices.
# Quantum cosmology of loop quantum gravity condensates: An example by Gielen
This week I have mainly been studying the work done during the Google Summer of Code workshops, in particular that on sagemath knot theory at:
This work looks great and I’ll be using the results in some of my calculations later in the summer.
Another topic I’ve been reviewing is the idea of spacetime as a Bose -Einstein condensate. This together with emergent, entropic and thermodynamic gravitation seem to be an area into which the quantum tetrahedron approach could naturally fit via statistical mechanics.
In the paper, Quantum cosmology of loop quantum gravity condensates, the author reviews the idea that spatially homogeneous universes can be described in loop quantum gravity as condensates of elementary excitations of space. Their treatment by second-quantised group field theory formalism allows the adaptation of techniques from the description of Bose–Einstein condensates in condensed matter physics. Dynamical equations for the states can be derived directly from the underlying quantum gravity dynamics. The analogue of the Gross–Pitaevskii equation defines an anisotropic quantum cosmology model, in which the condensate wavefunction becomes a quantum cosmology wavefunction on minisuperspace.
Introduction
The spacetimes relevant for cosmology are to a very good approximation spatially homogeneous. One can use this fact and perform a symmetry reduction of the classical theory – general relativity coupled to a scalar field or other matter – assuming spatial
homogeneity, followed by a quantisation of the reduced system. Inhomogeneities are usually added perturbatively. This leads to models of quantum cosmology which can be studied without the need for a full theory of quantum gravity.
Loop quantum gravity (LQG) has some of the structures one would expect in a full theory of quantum gravity: kinematical states corresponding to functionals of the Ashtekar–Barbero connection can be rigorously defined, and geometric observables such
as areas and volumes exist as well-defined operators, typically with discrete spectrum. The use of the LQG formalism in quantising symmetry-reduced gravity leads to loop quantum cosmology (LQC).
Because of the well-defined structures of LQG, LQC allows a rigorous analysis of issues that could not be addressed within the Wheeler– DeWitt quantisation used in conventional quantum cosmology, such as a definition of the physical inner product. More recently, LQC has made closer contact with CMB observations, and the usual inflationary scenario is now discussed within LQC.
A new approach towards addressing the issue of how to describe cosmologically relevant universes in loop quantum gravity uses the group field theory (GFT) formalism, itself a second quantisation formulation of the kinematics and dynamics of LQG: one has a Fock space of LQG spin network vertices or tetrahedra, as building blocks of a simplicial complex, annihilated and created by the field operator ϕ and its Hermitian conjugate ϕ†, respectively. The advantage of using this reformulation is that field-theoretic techniques are available, as a GFT is a standard quantum field theory on a curved group manifold. In particular, one can define coherent or squeezed states for the GFT field, analogous to states used in the physics of Bose– Einstein condensates or in quantum optics; these represent quantum gravity condensates. They describe a large number of degrees of freedom of quantum geometry in the same microscopic quantum state, which is the analogue of homogeneity for a differentiable metric geometry. After embedding a condensate of tetrahedra into a smooth manifold representing a spatial hypersurface, one shows that the spatial metric in a fixed frame reconstructed from the quantum state is compatible with spatial homogeneity. As the number of tetrahedra is taken to infinity, a continuum homogeneous metric can be approximated to a better and better degree.
At this stage, the condensate states defined in this way are kinematical. They are gauge-invariant by construction, and represent geometric data invariant under spatial diffeomorphisms. The strategy followed for extracting information about the dynamics of these states is the use of Schwinger–Dyson equations of a given GFT model. These give constraints on the n point functions of the theory evaluated in a given condensate state – approximating a non-perturbative vacuum, which can be translated into differential equations for the condensate wavefunction used in the definition of the state. This is analogous to condensate states in many-body quantum physics, where such an expectation value gives, in the simplest case, the Gross–Pitaevskii equation for the condensate
wavefunction. The truncation of the infinite tower of such equations to the simplest ones is part of the approximations made. The effective dynamical equations obtained can be viewed as defining a quantum cosmology model, with the condensate wavefunction interpreted as a quantum cosmology wavefunction. This provides a general procedure for deriving an effective cosmological dynamics directly from the underlying theory of quantum gravity. It canbe shown that a particular quantum cosmology equation of this type, in a semiclassical WKB limit and for isotropic universes, reduces to the classical Friedmann equation of homogeneous,
isotropic universes in general relativity.
See posts:
Let’s analyse more carefully the quantum cosmological models derived from quantum gravity condensate states in GFT. In particular, the formalism identifies the gauge-invariant configuration space of a tetrahedron with the minisuperspace of homogeneous generally anisotropic geometries.
Using a convenient set of variables the gauge-invariant geometric data, can be mapped to the variables of a general anisotropic Bianchi model it is possible to find simple solutions to the full quantum equation, corresponding to isotropic universes.
They can only satisfy the condition of rapid oscillation of the WKB approximation for large positive values of the coupling μ in the GFT model. For μ < 0, states are sharply peaked on small values for the curvature, describing a condensate of near-flat building blocks, but these do not oscillate. This supports the view that rather than requiring semiclassical behaviour at the Planck scale, semiclassicality should be imposed only on large-scale observables.
From quantum gravity condensates to quantum cosmology
Review the relevant steps in the construction of effective quantum cosmology equations for quantum gravity condensates. Use group field theory (GFT) formalism, which is a second quantisation formulation of loop quantum gravity spin networks of fixed valency, or their dual interpretation as simplicial geometries.
The basic structures of the GFT formalism in four dimensions are a complex-valued field ϕ : G⁴ → C, satisfying a gauge invariance property
and the basic non-relativistic commutation relations imposed in the quantum theory
These relations are analogous to those of non-relativistic scalar field theory, where the mode expansion of the field operator defines annihilation operators.
In GFT, the domain of the field is four copies of a Lie group G, interpreted as the local gauge group of gravity, which can be taken to be G = Spin(4) for Riemannian and G = SL(2,C) for Lorentzian models. In loop quantum gravity, the gauge group is the one given by the classical Ashtekar–Barbero formulation, G = SU(2). This property encodes invariance under gauge transformations acting on spin network vertices.
The Fock vacuum |Ø〉 is analogous to the diffeomorphism-invariant Ashtekar–Lewandowski vacuum of LQG, with zero expectation value for all area or volume operators. The conjugate ϕ acting on the Fock vacuum |Ø〉 creates a GFT particle, interpreted as a 4-valent spin network vertex or a dual tetrahedron:
The geometric data attached to this tetrahedron, four group elements gI ∈ G, is interpreted as parallel transports of a gravitational connection along links dual to the four faces. The LQG interpretation of this is that of a state that fixes the parallel transports of the Ashtekar–Barbero connection to be gI along the four links given by the spin network, while they are undetermined everywhere else.
In the canonical formalism of Ashtekar and Barbero, the canonically conjugate variable to the connection is a densitised inverse triad, with dimensions of area, that encodes the spatial metric. The GFT formalism can be translated into this momentum space formulation by use of a non-commutative Fourier transform
The geometric interpretation of the variables B ∈ g is as geometric bivectors associated to a spatial triad e, defined by the integral over a face △ of the tetrahedron. Hence, the one-particle state
Defines a tetrahedron with minimal uncertainty in the fluxes, that is the oriented area elements given by B . In the LQG interpretation this state completely determines the metric variables for one tetrahedron, while being independent of all other degrees of freedom of geometry in a spatial hypersurface.
The idea of quantum gravity condensates is to use many excitations over the Fock space vacuum all in the same microscopic configuration, to better and better approximate a smooth homogeneous metric or connection, as a many-particle state can contain information about the connection and the metric at many different points in space. Choosing this information such that it is compatible with a spatially homogeneous metric while leaving the particle number N free, the limit N → ∞ corresponds to a continuum limit in which a homogeneous metric geometry is recovered.
In the simplest case, the definition for GFT condensate states is
where N(σ) is a normalisation factor. The exponential creates a coherent configuration of many building blocks of geometry. At fixed particle number N, a state of the form σⁿ|Ø〉 would be interpreted as defining a metric (or connection) that looks spatially homogeneous when measured at the N positions of the tetrahedra, given an embedding into space usually there is a sum over all possible particle numbers. The condensate picture does not use a fixed graph or discretisation of space.
The GFT condensate is defined in terms of a wavefunction on G⁴
invariant under separate left and right actions of G on G⁴ . The strategy is then to demand that the condensate solves the GFT quantum dynamics, expressed in terms of the Schwinger–Dyson equations which relate different n-point functions for the condensate. An important approximation is to only consider the simplest Schwinger– Dyson equations, which will give equations of the form
This is analogous to the case of the Bose–Einstein condensate where the simplest equation of this typegives the Gross–Pitaevskii equation.
In the case of a real condensate, the condensate wavefunction Ψ (x), corresponding to a nonzero expectation value of the field operator, has a direct physical interpretation: expressing it in terms of amplitude and phase, one can rewrite the
Gross–Pitaevskii equation to discover that ρ(x) and v(x) = ∇θ(x) satisfy hydrodynamic equations in which they correspond to the density and the velocity of the quantum fluid defined by the condensate. Microscopic quantum variables and macroscopic classical variables are directly related.
The wavefunction σ or ξ of the GFT condensate should play a similar role. It is not just a function of the geometric data for a single tetrahedron, but equivalently a function on a minisuperspace of spatially homogeneous universes. The effective dynamics for it, extracted from the fundamental quantum gravity dynamics given by a GFT model, can then be interpreted as a quantum cosmology model.
Minisuperspace – gauge-invariant configuration space of a tetrahedron
Condensate states are determined by a wavefunction σ, which is
a complex-valued function on the space of four group elements for given gauge group G which is invariant under
is a function on G\G⁴/G. This quotient space is a smooth manifold
with boundary, without a group structure. It is the gauge-invariant configuration space of the geometric data associated to a tetrahedron. When the effective quantum dynamics of GFT condensate states is reinterpreted as quantum cosmology equations, G\G⁴/G becomes a minisuperspace of spatially homogeneous geometries.
Conclusion
Condensate states in group field theory can be used to derive effective quantum cosmology models directly from the dynamics of a quantum theory of discrete geometries. This can be illustrated by the interpretation of the configuration space of gauge-invariant geometric data of a tetrahedron, the domain of the condensate
wavefunction, as a minisuperspace of spatially homogeneous 3-metrics.
I’ll also looking at the calculations behind this paper in more detail in a later post.
# Linking covariant and canonical LQG: new solutions to the Euclidean Scalar Constraint by Alesci, Thiemann, and Zipfel
This week I have been continuing my work on the Hamiltonian constraint in Loop Quantum Gravity, The main paper I’ve been studying this week is ‘Linking covariant and canonical LQG: new solutions to the Euclidean Scalar Constraint’. Fortunately enough linking covariant and canonical LQG was also the topic of a recent seminar by Zipfel in the ilqgs spring program.
The authors of this paper emphasize that spin-foam models could realize a projection on the physical Hilbert space of canonical Loop Quantum Gravity (LQG). As a test the authors analyze the one-vertex expansion of a simple Euclidean spin-foam. They find that for fixed Barbero-Immirzi parameter γ= 1 the one vertex-amplitude in the KKL prescription annihilates the Euclidean Hamiltonian constraint of LQG. Since for γ = 1 the Lorentzian part of the Hamiltonian constraint does not contribute this gives rise to new solutions of the Euclidean theory. Furthermore, they fi nd that the new states only depend on the diagonal matrix elements of the volume. This seems to be a generic property when applying the spin-foam projector.
To circumvent the problems of the canonical theory, a
covariant formulation of Quantum Gravity, the so-called spin-foam model was introduced. This model is mainly based on the observation that the Holst action for GR de fines a constrained BF-theory. The strategy is first to quantize discrete BF-theory and then to implement the so called simplicity constraints. The main building block of the model is a linear two-complex κ embedded into 4-dimensional space-time M whose boundary is given by an initial and final gauge invariant spin-network, Ψi respectively Ψf , living on the initial respectively final spatial hyper surface of a
foliation of M. The physical information is encoded in the spin-foam amplitude.
where Af , Ae and Av are the amplitudes associated to the internal faces, edges and vertices of κ and B contains the boundary amplitudes.
Each spin-foam can be thought of as generalized
Feynman diagram contributing to the transition amplitude from an ingoing spin-network to an outgoing spin-network. By summing over all possible two-complexes one obtains the complete transition amplitude between ψi and ψf .
The main idea in this paper is that if f spin-foams provide a rigging map the physical inner product would be given by
and the rigging map would correspond to
Since all constraints are satis ed in Hphys the physical scalar product must obey
for all ψout, ψn ∈ Hkin.
As a test the authors consider an easy spin-foam amplitude and show that
where κ is a two-complex with only one internal vertex such that Φ is a spin-network induced on the boundary of κ and Hn is the Hamiltonian constraint acting on the node n.
Hamiltonian constraint
The classical Hamiltonian constraint is
where,
The constraint can be split into its Euclidean part H = Tr[F∧e] and Lorentzian part HL = C- H.
Using,
where V is the volume of an arbitrary region ∑ containing the point x. Smearing the constraints with lapse function N(x) gives
This expression requires a regularization in order to obtain a well-de fined operator on Hkin. Using a triangulation T of the manifold into elementary tetrahedra with analytic links adapted to the graph Γ of an arbitrary spin-network.
Three non-planar links de fine a tetrahedron . Now decompose H[N] into a sum of one term per each tetrahedron of the triangulation,
To define the classical regularized Hamiltonian constraint as,
The connection A and the curvature are regularized by the holonomy h in SU (2), where in the fundamental representation m = ½. This gives,
which converges to the Hamiltonian constraint if the triangulation is sufficiently fine.
As seen in the post
This can be generalized with a trace in an arbitrary irreducible representation m leading to
this converges to H[N] as well.
Properties
The important properties of the Euclidean Hamiltonian constraint are;
when acting on a spin-network state, the operator reduces to a sum over terms each acting on individual nodes. Acting on nodes of valence n the operator gives
The Hamiltonian constraint on di ffeomorphism invariant states is independent from the refi nement of the triangulation.
Since the Ashtekar-Lewandowsk volume operator annihilates coplanar nodes and gauge invariant nodes of valence three H does not act on the new nodes – the so called extraordinary nodes.
Action on a trivalent node
To ompute the action of the operator on a trivalent node where all links are outgoing, denote a trivalent node by |n(ji,jj , jk)> ≡ |n3>, whereas ji, jj ,jk are the spins of the adjacent links ei, ej , ek:
To quantize [N] the holonomies and the volume are replaced by their corresponding operators and the Poisson bracket is replaced by a commutator. Since the volume operator vanishes on a gauge invariant trivalent node only need to compute;
so, h(m) creates a free index in the m-representation located at the node , making it non-gauge invariant and a new node on the link ek:
so we get,
where the range of the sums over a, b is determined by the Clebsch-Gordan conditions and
The complete action of the operator on a trivalent state |n(ji, jj , jk)> can be obtained by contracting the trace part with εijk. So, H projects on a linear combination of three spin networks which differ by exactly one new link labelled by m between each couple of the oldlinks at the node.
Action on a 4-valent node
The computation for a 4-valent node |n4> is
where i labels the intertwiner – inner link.
The holonomy h(m) changes the valency of the node and the Volume therefore acts on the 5-valent non-gauge invariant node. Graphically this corresponds to
and finally we get:
This can be simpli ed to;
SPIN-FOAM
Using the de finition of an Euclidean spin-foam models as suggested by Kaminski, Kisielowski and Lewandowski (KKL) and since we,re only interested in the evaluation of a spin-foam amplitude we choose a combinatorial de finition of the model:
Consider an oriented two-complex de fined as the union of the set of faces (2-cells) F, edges (1-cells) E and vertices (0-cells) V such that every edge e is a 1-face of at least one face f (e ∈ ∂f) and every vertex V is a 0-face of at least one edge e (v ∈∂e).
We call edges which are contained in more than one face f internal and denote the set of all internal edges by Eint. All vertices adjacent to more than one internal edge are also called internal and denote the set of these vertices by Vint. The boundary ∂κ is the union of all external vertices (-nodes) n ∉ Vint and external edges (links) l ∉Eint.
A spin-foam is a triple (κ, ρ; I) consisting of a proper foam whose faces are labelled by irreducible representations of a Lie-group G, in this case here SO(4) and whose internal edges are labeled by intertwiners I. This induces a spin-network structure ∂(κ, ρ; I) on the boundary of κ.
The BF partition function can be rewritten as
Av de fines an SO (4) invariant function on the graph Γ induced on the boundary of the vertex v
In the EPRL model the simplicity constraint is imposed weakly so,
It follows immediately that
defi nes the EPRL vertex amplitude with
Expanding the delta function in terms of spin-network function and integrating over the group elements gives
In order to evaluate the fusion coefficients by graphical calculus it is convenient to work with 3j-symbols instead of Clebsch-Gordan coefficients. When replacing the Clebsch-Gordan coefficients we have to multiply by an overall factor .
Spin-foam projector
Given any couple of ingoing and outgoing kinematical states ψout, ψin, the Physical scalar product can be
formally de fined by
where η is a projector – Rigging map onto the Kernel of the Hamiltonian constraint.
Suppose that the transition amplitude Z
can be expressed in terms of a sum of spin-foams, then
this can be interpreted as a function on the boundary graph ∂κ;
in the EPRL sector.
Restricting the boundary elements h ∈ SU (2) ⊂ SO (4) then;
where |S>N is a normalized spin-network function on SU(2). This fi nally implies;
NEW SOLUTIONS TO THE EUCLIDEAN HAMILTONIAN CONSTRAINT
Compute new solutions to the Euclidean Hamiltonian constraint by employing spin-foam methods. Show that
in the Euclidean sector with γ= 1 and s = 1, where κ is a 2-complex with only one internal vertex.
Trivalent nodes
Consider the simplest possible case given by an initial and final state |Θ>, characterized by two trivalent nodes joined by three links:
the only states produced by the Hamiltonian acting on a node, are given by a linear combination of spin-networks that diff er from the original one by the presence of an extraordinary (new) link. In particular the term will be non vanishing only if |s> is of the kind:
The simplest two-complex κ(Θ,s) with only one internal vertex de fining a cobordism between |Θ> and |s> is a tube Θ x[0,1] with an additional face between the internal vertex and the new link m,
Since the space of three-valent intertwiners is one-dimensional and all labelings jf are fixed by the states |Θ> , |s> the fi rst sum is trivial.
Γv= s and therefore we have,
where the sign factor is due to the orientation of s, The fusion coefficients contribute four 9j symbols since,
The full amplitude is
and
This yields,
The last two terms are equivalent to the first term when exchanging jk↔ jj. The EPRL spin-foam reduces just to the SU(2) BF amplitude that is just the single 6j in the first line.
Now using the defi nition of a 9j in terms of three 6j’s,
The 9j’s involved in this expression can be reordered using the permutation symmetries giving,
the statesare solutions of the Euclidean Hamiltonian constraint
The spin-foam amplitude selects only those terms which depend on the diagonal elements on the volume. This simplifies the calculation since we do not have to evaluate the volume explicitly.
Four valent nodes
The case with ψ in = ψout = |n4>where
The matrix element <s| |n4> is non-vanishing if |s>is of the form
Choose again a complex of the form
The vertex trace can be evaluated by graphical calculus
The fusion coefficients give two 9j symbols for the two trivalent edges and two 15j- symbols for the two four-valent edges.The fusion coefficients reduce to 1 when γ=1.Taking the scalar product we obtain,
Taking the scalar product with the Hamiltonian gives,
Summing over a and using the orthogonality relation gives,
The three 6j’s in the two terms defi ne a 9j ,summing over the indexes and b respectively gives:
the final result is,
As for the trivalent vertex the spin-foam amplitude just takes those elements into account which depend on the diagonal Volume elements.
CONCLUSIONS
LQG is grounded on two parallel constructions; the canonical and the covariant ones. In this paper the authors construct a simple spin-foam amplitude which annihilates the Hamiltonian constraint .They found that in the euclidean sector with signature s = 1 and Barbero-Immirzi parameter γ = 1 the Euclidean Hamiltonian constraint is annihilated by a spin-foam amplitude Z for a simple two-complex with only one internal vertex. The one vertex amplitudes of BF theory are explicit analytic solutions of the Hamiltonian theory.
Also the 6j symbol associated to every face is annihilated by the Euclidean scalar constraint. This is a generalization of the work by Bonzom-Freidel in the context of 3d gravity where they found that the 6jis annihilated by a suitable quantization of the 3d scalar constraint F = 0 . The spin-foam amplitude diagonalizes the Volume.
Related articles | 12,495 | 55,251 | {"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} | 2.734375 | 3 | CC-MAIN-2018-05 | latest | en | 0.910674 |
http://www.floridatoday.com/article/CD/20130714/POLITICSPOLICY08/307140034/New-old-standards-Common-Core-vs-Sunshine-State | 1,398,056,900,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609539493.17/warc/CC-MAIN-20140416005219-00544-ip-10-147-4-33.ec2.internal.warc.gz | 413,065,338 | 29,289 | # New and old standards: Common Core vs. Sunshine State
###### 12:25 AM, Jul. 14, 2013 | Comments
The Common Core State Standards contain fewer individual subject-matter requirements for students. They emphasize applying students' knowledge. Here's a sampling of the new Common Core standards and the outgoing Sunshine State standards.
Common Core (new): With guidance and support from adults, use technology to produce and publish writing (using keyboard skills) as well as to interact and collaborate with others.
Sunshine State (old): The student will write a final product for the intended audience.
Common Core: Distinguish comparisons of absolute value from statements about order. For example, recognize that an account balance less than -\$30 represents a debt greater than \$30.
Sunshine State: Use and justify the rules for adding, subtracting, multiplying, dividing and finding the absolute value of integers. (Note: Absolute value is not mentioned until seventh grade.)
Common Core: Trace and evaluate the argument and specific claims in a text, assessing whether the reasoning is sound and the evidence is relevant and sufficient to support the claims.
Sunshine State: The student uses a variety of strategies to comprehend grade-level text.
### High School Math:
Common Core: Recognize and explain the concepts of conditional probability and independence in everyday language and everyday situations. For example, compare the chance of having lung cancer if you are a smoker with the chance of being a smoker if you have lung cancer.
Sunshine State: Determine probabilities of independent events. Understand and use the concept of conditional probability, including: understanding how conditioning affects the probability of events and finding conditional probabilities from a two-way frequency table.
Source: Florida Department of Education
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Join the conversation | 411 | 2,122 | {"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} | 2.859375 | 3 | CC-MAIN-2014-15 | longest | en | 0.886238 |
http://www.mathwarehouse.com/algebra/complex-number/how-to-simplify-negative-radicals.php | 1,512,978,209,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512584.10/warc/CC-MAIN-20171211071340-20171211091340-00563.warc.gz | 420,762,276 | 13,085 | How to simplify negative radicals. Video, examples, interactive calculator and practice problems
# How to Simplify Negative Radicals $$\sqrt{-36} \text{= ?}$$
II. Simplifying negative radicands
$$\sqrt{-25} =?$$
### Video Tutorial
#### A trip down memory lane
The rules for simplifying square roots is critical prerequisite knowledge for this topic. So, let's see if you remember these rules
(take some good mental notes if you don't)
TRUE!, you can do this and you'll need to do it today!
FALSE this rule does not apply to negative radicands!
$$\red{ \sqrt{a} \sqrt{b} = \sqrt{a \cdot b} }$$ only works if a > 0 and b > 0.
In other words, the product of two radicals does not equal the radical of their products when you are dealing with imaginary numbers.
##### Example 1
$$\sqrt{-7} = \sqrt{-1 \cdot7} = \sqrt{-1} \cdot\sqrt{7} = i\sqrt{7}$$
In example 1, the only type of work that you can do is to remove the $$-1$$ from the radicand
##### Example 2
$$\sqrt{-9} = \sqrt{-1 \cdot 9} = \sqrt{-1} \cdot\sqrt{9} = i \cdot 3 = 3i$$
In this one, you can actually reduce the $$\sqrt{ 9 }$$
##### Example 3
$$\sqrt{-12} = \sqrt{-1 \cdot 4 \cdot 3} = \sqrt{-1} \cdot \sqrt{4} \cdot \sqrt{3} = i\sqrt{4} \cdot \sqrt{3} = 2 i \sqrt{3}$$
Similar to example 2, you can actually reduce the $$\sqrt{ 12 }$$
### General Formula
$$\sqrt{-a} = \sqrt{-1 \cdot a} = \sqrt{-1} \cdot \sqrt{a} = i\sqrt{a}$$
### Negative Radical Reducer
${}$
### Practice Problems
Step 1
"extract" $$i$$
$$\sqrt{-5} \\= \red{\sqrt{-1}} \cdot \sqrt{5} \\ = \red{i} \sqrt{5}$$
Step 2
Simplify $$\red{ \text{radicand}}$$
$$i \sqrt { \red{ 5 }}$$
In this case, the radicand is already in simplest form so there's really only 1 step to perform.
Step 1
"extract" $$i$$
$$\sqrt{-36} \\= \red{\sqrt{-1}} \cdot \sqrt{36} \\ = \red{i} \sqrt{36}$$
Step 2
Simplify $$\red{ \text{radicand}}$$
$$i \sqrt { \red{ 36 }} \\ i \cdot 6 \\ 6i$$
As you can see from step 2, most of the work for these kinds of problems involves simplifying the square root!
Step 1
"extract" $$i$$
$$\sqrt{-48} \\= \red{\sqrt{-1}} \cdot \sqrt{48} \\ = \red{i} \sqrt{48}$$
Step 2
Simplify $$\red{ \text{radicand}}$$
$$i \sqrt { \red{ 48 }} \\ i \sqrt { \red{ 16 \cdot 3 }} \\ i \sqrt { \red{ 16}} \cdot \sqrt { \red{ 3 }} \\ i \cdot 4 \cdot \sqrt { \red{3 }} \\ 4 i \sqrt { 3}$$
As you can see from step 2, most of the work for these kinds of problems involves simplifying the square root!
Step 1
"extract" $$i$$
$$\sqrt{-50} \\= \red{\sqrt{-1}} \cdot \sqrt{50} \\ = \red{i} \sqrt{50}$$
Step 2
Simplify $$\red{ \text{radicand}}$$
$$i \sqrt { \red{ 50 }} \\ i \sqrt { \red{ 25 \cdot 2 }} \\ i \sqrt { \red{ 25}} \cdot \sqrt { \red{2 }} \\ i 5 \cdot \sqrt { \red{2 }} \\ 5 i \sqrt { 2}$$
### Ultimate Math Solver (Free)
Free Algebra Solver ... type anything in there! | 1,010 | 2,845 | {"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} | 4.65625 | 5 | CC-MAIN-2017-51 | longest | en | 0.690803 |
https://www.quizzes.cc/calculator/length/meters/330 | 1,669,596,394,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710462.59/warc/CC-MAIN-20221128002256-20221128032256-00564.warc.gz | 1,032,325,173 | 2,671 | ### How much is 330 meters?
Convert 330 meters. How far is 330 meters? What is 330 meters in other units? Convert to cm, km, in, ft, meters, mm, yards, and miles. To calculate, enter your desired inputs, then click calculate. Some units are rounded since conversions between metric and imperial can be messy.
### Summary
Convert 330 meters to cm, km, in, ft, meters, mm, yards, and miles.
#### More Conversions below
330 meters to meters 330 meters to miles 330 meters to millimeters 330 meters to yards
#### 330 meters to Other Units
330 meters equals 33000 centimeters 330 meters equals 1082.677165 feet 330 meters equals 12992.12598 inches 330 meters equals 0.33 kilometers
330 meters equals 330 meters 330 meters equals 0.2050524934 miles 330 meters equals 330000 millimeters 330 meters equals 360.8923885 yards | 216 | 825 | {"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-2022-49 | latest | en | 0.881694 |
https://openhome.cc/eGossip/OpenSCAD/lib3x-curve.html | 1,653,254,038,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662546071.13/warc/CC-MAIN-20220522190453-20220522220453-00298.warc.gz | 491,358,645 | 3,394 | # curve
Draws a curved line from control points. The curve is drawn only from the 2nd control point to the second-last control point. It's an implementation of Centripetal Catmull-Rom spline.
Since: 2.5
## Parameters
• `t_step` : 0 ~ 1. Control the distance between two points of the generated curve.
• `points` : A list of `[x, y]` or `[x, y, z]` control points.
• `tightness` : You can view it as the curve tigntness if you provide a value between 0.0 and 1.0. The default value is 0.0. The value 1.0 connects all the points with straight lines. The value greater than 1.0 or less than 0.0 is also acceptable because it defines how to generate a bezier curve every four control points.
## Examples
``````use <curve.scad>;
pts = [
[28, 2, 1],
[15, 8, -10],
[2, 14, 5],
[28, 14, 2],
[15, 21, 9],
[2, 28, 0]
];
t_step = 0.05;
tightness = 0;
points = curve(t_step, pts, tightness);
polyline_join(points)
sphere(.5);
#for(pt = pts) {
translate(pt)
sphere(1);
}
#polyline_join(pts)
sphere(.05);
`````` | 325 | 1,007 | {"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-2022-21 | latest | en | 0.8345 |
http://www.docstoc.com/docs/125887921/Reducing-Congestion-at-Intersection-of-Alameda-and-Atherton-Avenue | 1,386,583,169,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163955638/warc/CC-MAIN-20131204133235-00025-ip-10-33-133-15.ec2.internal.warc.gz | 307,888,553 | 18,093 | # Reducing Congestion at Intersection of Alameda & Atherton Avenue
Document Sample
Improving the Intersection of
Alameda & Atherton Avenue
Tuesday, 5/8/12
Atherton Transportation Committee Meeting
by Gary Lauder
Introduction
Who I am
• Atherton resident with 36-year fascination with traffic patterns
• No formal training in field...that doesn’t mean I’m wrong
• International travel shown what’s possible w/roundabouts
• The CA Department of Transportation is enthusiastic
- Most other states’ DOTs are too
Why me?
• Who else?
Why this is worth doing
Problem affects thousands of people each day
Avoid time wastage (time = \$, so hidden tax)
Better for the environment
Better fuel efficiency
• Another tax
• Geopolitical implications
Safer
Create a model to be imitated elsewhere
This is 8:30AM last week
This is 8:30 AM last week
This is 5:40 PM last week
This is 5:40 PM last week
Main Problem: Rush hour congestion
I measured 4.5 minutes to get through northbound at 5:30pm
Traffic was backed up to Camino Al Lago = 1800 feet
Compared to speed limit, that means delay of 3.8 minutes/car
There were ≈ 56 cars in the jam
≈ 750 cars/hour get through at rate of 12.5 cars/minute
Q: What is the cost of that lost time?
Cost of lost time DURING JAM
3.8 minutes/car
x 750 cars/hour
= 48 car hours per hour of jam (in that direction)
x 1.2 occupants per car
= 57 person hours per hour of jam (in that direction)
x \$20/hour (average wage in USA)
= \$1,145/hour opportunity cost of time lost x 260 delay hrs/yr
= \$297,818/year per direction (assuming jam is only 1 hour)
x 2 directions = \$595,636 per year
Lost time from stopping at sign
when no queue
7,000 cars/day – 1,500 cars already counted during jam = 5,500
x 10 Seconds/car = 15.3 Hours/day = 5,580 Hours/year
x \$20/hour (average wage in USA) x 1.2 occupants/car
= \$133,925 opportunity cost of time lost / year just from stopping
Add to that the time lost waiting in the jams of \$595,636
= \$729,561 of people’s time wasted
Each year wastes the equvalent of 8.5 person-years
Steve Jobs story from bio (saving lives by saving time)
Cost of Gas
> 7,000 cars/day on Alameda
≈2 ounces of gas ≈ 5¢ to accelerate
X 7,000 cars/day = \$350/day = \$127,838 /year
But that is only the gasoline cost
Other costs include pollution and wear on car
Added to the cost of lost time, total = \$857,399 per year!!!
PV of that annuity discounted @ 5% = \$17,147,970 !!!
Why have people not complained?
Why would people if they did not imagine a solution?
Not enough roundabouts here for people to realize it’s a solution
No roundabouts in San Mateo County nor SF, only one in Santa
Clara.
mentioned in this book
I bought in 1989
Safety:
Traffic Lights & Stop Signs vs.
A study of 24 intersections converted to roundabouts found:
• Crashes dropped 40%
• Injury crashes dropped 76%
• Fatal crashes dropped 90%
Main reasons:
• Slower speeds
• Lower angles
• Fewer conflict points
Capacity in cars/hour is ≈ 1,200 per direction minus cars that
turn left in opposite direction
That’s improvement of >50%, so would eliminate congestion
Cost
• Mini-roundabout can be as little as \$50K
• Standard ≈ \$100K (very approximate)
- Might require some adjacent corner land outside of fence
Even if only 10% of the people inconvenienced are
Athertonians, still worth it.
Other issues
Major source of complaints: shortcuts to go around congestion
This should alleviate the underlying congestion
Shift jam elsewhere?
• A: Neither 100% nor 0%, but have to start somewhere
Recommendations
Traffic study to quantify extent of problem
Retain expert on roundabouts to opine
Thank you
Gary Lauder
Gary@LauderPartners.com
Thoroughfare Plan Map Map of Carmel, Indiana, roundabout capital of America
E 146th St
W 146th St
Hazel Dell Pkwy
Carey Rd
Gray Rd
Oak Ridge Rd
W 141st St
E 136th St
W 136th St
Range Line Rd
Ditch Rd
Main St E 131st St
St
W 131st St
n
dia
ri
Me
River Rd
Old
West Rd
E 126th St
W 126th St
Illinois St
122nd St
Carmel Dr
S Guilford Rd
Pennsylvania St
Clay Center Rd
W 116th St E 116th St
Carmel, IN map
Meridian St
111th St
showing the many
US 4
21
W 106th St
Spring Mill Rd
ll Pkwy
College Ave
installed. Only one
Gray Rd
Keystone Parkway
Shelborne Rd
Westfield Blvd
Hazel De
Ditch Rd
Towne Rd
US 31
I-465
Mi c h
traffic light left.
i g an
Rd
W 96th St
MAP LEGEND THOROUGHFAREPLAN MAP
te te
In rsta e n a rte l
Sco d ryA ria e e tia a a
Rsid n l P rkw y ff-S e o mte ra
O tre t Cm u r T il
.S ta ig w y
U ./S teH h a e n a rte l ro o d
Sco d ryA ria (P p se ) e e tia a a ro o d
Rsid n l P rkw y(P p se ) ra e e a te ro g
G d -Sp ra dC ssin
1/4 1/2 3/4 1
rimry rte l
P a A ria e na a a
Sco d ryPrkw y o cto tre t
Clle r S e In rch n eL ca n
te a g o tio MILE MILE MILE MILE
rimry a a
P a Prkw y e n a a a ro o d
Sco d ryPrkw y(P p se ) o cto tre t ro o d
Clle r S e (P p se ) v rp ss o tio
Oe a L ca n
rb n rte l
U a A ria rb n o cto
U a Clle r o cto tre t e O g e e t
Clle r S e p r N AXA re mn o n a o t te ctio x g
Ru d b u In rse n(Eistin )
rb n rte l ro o d
U a A ria (P p se ) rb n o cto ro o d
U a Clle r (P p se ) e e tia tre t
Rsid n l S e o n a o t te cto ro o d
Ru d b u In rse n(P p se )
e e tia tre t ro o d
Rsid n l S e (P p se ) iv r
Re
Traffic keeps flowing
Less braking = less accelerating = less gas = less pollution
Less time wasted (time = \$)
Partly accounts for Europe’s better MPG than in USA
Better than traffic lights
Better than 4-way stop signs
Expensive to install
More expensive not to
Not applicable in all situations
Solving the problem can be worth
Cut down the shrubbery
Improve the corner sight distance
Then sell it & still come out ahead
(compared w/putting in a traffic
control to deal w/limited visibility)
Full and Complete Stop?
Few people do
The law should not compel what is bad for:
• the environment
• the driver
• economy
• and serves no one
Police know this and take advantage when they need to
raise money quickly
Cutting Room Floor: Quotations
Useless laws weaken necessary laws.
– Charles de Secondat, Baron de Montesquieu
Discovery is seeing what everybody else has seen, and thinking
what nobody else has thought.
– Albert Szent-Gyorgyi
The world is only changed by people who are naïve enough to
think that they can change the world.
— Said by Eric Lander at the Aspen Ideas Festival 7/2/09.
The reasonable man adapts himself to the world; the
unreasonable one persists in trying to adapt the world to
himself. Therefore all progress depends on the unreasonable
man. – George Bernard Shaw, Man and Superman (1903)
"Maxims for Revolutionists” (Irish dramatist & socialist (1856 –
1950))
DOCUMENT INFO
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views: 5 posted: 8/6/2012 language: pages: 25
How are you planning on using Docstoc? | 2,095 | 8,988 | {"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} | 2.734375 | 3 | CC-MAIN-2013-48 | longest | en | 0.911504 |
https://www.convert-measurement-units.com/convert+Decisteradian+to+Square+degree.php | 1,624,256,091,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488268274.66/warc/CC-MAIN-20210621055537-20210621085537-00120.warc.gz | 630,984,375 | 17,023 | Convert dsr to deg² (Decisteradian to Square degree)
numbers in scientific notation
## How many Square degree make 1 Decisteradian?
1 Decisteradian [dsr] = 328.280 635 001 17 Square degree [deg²] - Measurement calculator that can be used to convert Decisteradian to Square degree, among others.
# Convert Decisteradian to Square degree (dsr to deg²):
1. Choose the right category from the selection list, in this case 'Solid angle'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Decisteradian [dsr]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Square degree [deg²]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '796 Decisteradian'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Decisteradian' or 'dsr'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Solid angle'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '73 dsr to deg2' or '67 dsr into deg2' or '94 Decisteradian -> Square degree' or '82 dsr = deg2' or '21 Decisteradian to deg2' or '15 dsr to Square degree' or '20 Decisteradian into Square degree'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(3 * 88) dsr'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '796 Decisteradian + 2388 Square degree' or '72mm x 70cm x 5dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 8.466 044 490 860 2×1031. For this form of presentation, the number will be segmented into an exponent, here 31, and the actual number, here 8.466 044 490 860 2. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 8.466 044 490 860 2E+31. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 84 660 444 908 602 000 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 880 | 3,773 | {"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} | 3.46875 | 3 | CC-MAIN-2021-25 | longest | en | 0.848401 |
https://www.geeksforgeeks.org/fuzzy-c-means-clustering-in-matlab/?ref=lbp | 1,702,262,030,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103464.86/warc/CC-MAIN-20231211013452-20231211043452-00229.warc.gz | 885,641,869 | 54,325 | # Fuzzy C-means Clustering in MATLAB
Fuzzy C-means (FCM) is a method of clustering that allows points to be more than one cluster. The (FCM) is a kind of data clustering technique in which the data set is grouped into N numbers of clusters with every data point corresponding to each cluster on the basis. which is to differentiate the distance between the cluster center and the data point.
The FCM function performs Fuzzy C-means clustering, and fuzzy C-means clustering is soft skills clustering technique in Machine Learning.
### Clustering
Clustering is a process of grouping the objects(data) having higher measure similarities than the object(data) in any other cluster which full fills the requirement of our criteria. it has been done by the Clustering Algorithms, various similarity measures can be used, including Euclidean, probabilistic, cosine distance, and correlation.
### Dataset
The Dataset is a kind of group or storage location where we can hold(contain) the require the structured collection of information data, which is having some numeric values. And a Database has the functionality to store multiple datasets.
```Note: Here we use the Iris dataset
for Fuzzy C-means Clustering in MATLAB.```
### Syntax
[centers, U] = fcm(data Nc)
% perform the fcm function clustering on data and return Nc cluster center.
[centers, U] = fcm(data Nc,options)
% option means using specific additional clustering options.
[centers, U, objFunc] = fcm( __ )
% Return objective function value and optimise iteration for above syntax.
## Algorithm of FCM Clustering
So first Let the set of data points X= { x1, x2, x3, —–, xn } and the set of clusters centers V= { v1, v2, v3,——, vc } .
• So first we will select the random c cluster center from the set of clusters center.
• And then find the fuzzy membership using the given formula –
Here ‘m’ is the fuzziness index, which lies between [1, ∞ ], and ‘n’ is a number of data points.
• Computing the fuzzy centers Vj using the formula.
• Optimize the Vj and fuzzy membership until the minimum j value is achieved.
In the above iteration –
```K is iteration step.
β is termination value lies between [0,1]
J is the objective function.
U=(µij)n*c is fuzzy membership matrix.```
### Steps for Fuzzy C-means Clustering
Step 1: First we create the MATLAB file in an editor and take the simple data sets and select the data set to N number cluster. So you have to make sure that whatever data set we will use here that data set should be inside MATLAB.
Example:
## Matlab
`% Define the number of clusters and the fuzziness parameter` `numClusters = 3;` `fuzziness = 2;` `% Generate some data for clustering` `X = rand(100, 2);` `% Perform Fuzzy C-means Clustering` `[cluster_idx, cluster_center] = fcm(X, numClusters, [2, 100, 1e-5, 0]);` `% Plot the results` `figure;` `[c, h] = contourf(cluster_idx);` `colorbar;` `hold on;` `plot(X(:,1), X(:,2), ``'ko'``);` `title(``'Fuzzy C-Means Clustering Results'``);` `xlabel(``'x'``);` `ylabel(``'y'``);` `hold off;`
• n_clusters is the length of the cluster like how many numbers of clusters we taking.
• colorbar – it’s used to displays a vertical colorbar to the right of the current axes in figure. and it mapping the data values.
• [center,U,obj_fcn]=fcm(__) it’s one of the syntaxes of the FCM function which returns the center, U matrix, and objective function value.
• Save (filename) saves all the variables from the current workshop.
• Plot() is used to create a graphical representation of some data.
Output:
FCM Result
#### Explanation-
The fcm function performs Fuzzy C-means Clustering on the input data X, with numClusters clusters and fuzziness parameter. The output, cluster_idx, is a matrix of the same size as X, where each element represents the cluster index of the corresponding element in X. The cluster_center is a matrix where each row contains the cluster center of each cluster.
The contourf function is used to plot the clustering results by showing the cluster index of each data point. The plot function is used to plot the data points on top of the contour plot.
It is important to note that the input data X must be a matrix where each row represents a data point and each column represents a feature, and the number of clusters should be defined according to the data set and the problem to solve.
The Fuzzy C-means Clustering Algorithm gives the best result for manipulated, overlapped data set, and FCM is better than K-means clustering.
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https://serc.carleton.edu/sp/library/models/EqStBOT.html | 1,527,470,698,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870604.65/warc/CC-MAIN-20180528004814-20180528024814-00324.warc.gz | 625,930,346 | 11,169 | Pedagogy in Action > Library > Models > Why are Models Useful > Systems Thinking > Equilibrium, Stability and Behavior Over Time
# Equilibrium, Stability, and Behavior over time
#### Equilibrium, Stability, Behavior Over Time, Linear Process, and Time Delays
Equilibrium and Stability
To understand feedback processes it is important to understand the concepts of equilibrium and stability. Negative feedback loops are linked to stable equilibrium states and positive feedback loops are linked to unstable equilibrium. The stability of a system is intimately connected to its equilibrium state. If a system in equilibrium is disturbed slightly, then if it is stable it tends to return to or oscillate about its original equilibrium state. An unstable system tends to continue to move away from its original equilibrium state when perturbed from it. The figure at left shows a red ball in the bottom of a bowl, on top of an inverted bowl, or on a flat surface corresponding to stable, unstable, and neutral equilibrium states respectively.
Behavior over time (temporal behavior) There are many situations where the static balance (equilibrium) of a system is of central interest: for example in studies of geologic structures, bridges, or buildings. Modeling such systems and thinking in terms of systems can be extremely helpful in our understanding of how different components of such systems are interconnected. However, for our discussion regarding system dynamics and systems thinking our primary focus will be to understand the temporal behavior of dynamic systems. Understanding a system's expected behavior over time is a primary objective of system modeling and system dynamics. The graph below is an example of S-shaped growth obtained for a system that obeys the logistic equation (see footnote to the page). The growth of the system starts out exponentially until it becomes so large that its growth is limited. This model describes flowers growing in a field of limited size (800 sq. meters). Many think that World population will follow this sort of growth. The point here is not this specific example but to emphasize the importance of understanding the expected behavior of a system over time.
Linear Systems & Growth We refer to systems without loop structure or feedbacks as linear in that a cause has an effect, but the effect does not feedback and alter the cause. An example is a simple bucket filling with water. In reality someone will be watching the bucket fill and when it gets to the top they will turn off the faucet. This "feedback between water content and flow rate is a loop which would make the more realistic situation a non-linear system. The graph shows linear growth with two scenarios for filling the bucket with constant flow. Which line (A or B) corresponds to the greatest flow rate? From the graph, estimate this largest flow rate in gallons per minute.
• Linear thought is useful at times and appropriate for some situations.
• What does it mean when one says that "linear thought is a no-Karma thought process"?
• Thinking in terms of interconnectedness and loop structure is very important for understanding systems.
• The famous naturalist John Muir once said-- "When one tugs at a single thing in nature he finds it hitched to the rest of the universe"
Time Delays Between Cause and Effect
Understanding the dynamic behavior of systems requires the realization that there may be a significant delay between cause and effect.
In the figure a graph is shown for two hypothetical containers of cold (0 °C)water set in a 100 °C oven.
Can you tell which container of water is largest?
Why or why not?
Perturbations are small changes that disturb a system from its equilibrium state.
There is an inherent time delay between the time the perturbation is applied and when the new state is achieved.
Assuming that the graph applies to the figure below, what happened to cause the change in temperature shown in the graph?
#### Footnote
Here A is a variable (like area) and r and K are constants. | 783 | 4,048 | {"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-2018-22 | latest | en | 0.917205 |
https://www.enotes.com/homework-help/find-area-tirnagle-length-sides-3-8-9-250300 | 1,516,526,493,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890394.46/warc/CC-MAIN-20180121080507-20180121100507-00282.warc.gz | 906,212,558 | 9,715 | Find the area of the triangle if the length of the sides are 3, 8 and 9?
hala718 | Certified Educator
Given a triangle with know 3 sides.
We will use the formula of the area of a triangle given the length of the sides.
==> A = sqrt(s*(s-a)(s-b)(s-c) such that s is the perimeter/2 and a, b, and c are the length of the sides.
Let us calculate the perimeter.
==> p = 3+8+9 = 20
==> s = p/2 = 20/2 = 10
Let us substitute.
==> A = sqrt( 10*(10-3)(10-8)(10-9)
==> A = sqrt( 10*7*2*1) = sqrt140 = 2sqrt35
Then the area of the triangle is 2sqrt35 = 11.83 square units.
justaguide | Certified Educator
As we have the length of the sides of the triangle, we can derive the area using the relation
Area = sqrt [s*(s - a)*(s - b)*(s - c)], where s = (a + b + c)/2
a = 3 , b = 8 and c = 9
=> s = 10
Area = sqrt [10*(10 - 3)*(10 - 8)*(10 - 9)]
=> sqrt [10*7*2*1]
=> sqrt 140
The required area of the triangle is sqrt 140. | 323 | 929 | {"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-2018-05 | latest | en | 0.775849 |
https://numbermatics.com/n/104553157/ | 1,702,051,204,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100762.64/warc/CC-MAIN-20231208144732-20231208174732-00233.warc.gz | 471,235,783 | 6,690 | # 104553157
## 104,553,157 is an odd composite number composed of two prime numbers multiplied together.
What does the number 104553157 look like?
This visualization shows the relationship between its 2 prime factors (large circles) and 10 divisors.
104553157 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of ten divisors.
## Prime factorization of 104553157:
### 37 × 414
(37 × 41 × 41 × 41 × 41)
See below for interesting mathematical facts about the number 104553157 from the Numbermatics database.
### Names of 104553157
• Cardinal: 104553157 can be written as One hundred four million, five hundred fifty-three thousand, one hundred fifty-seven.
### Scientific notation
• Scientific notation: 1.04553157 × 108
### Factors of 104553157
• Number of distinct prime factors ω(n): 2
• Total number of prime factors Ω(n): 5
• Sum of prime factors: 78
### Divisors of 104553157
• Number of divisors d(n): 10
• Complete list of divisors:
• Sum of all divisors σ(n): 110063390
• Sum of proper divisors (its aliquot sum) s(n): 5510233
• 104553157 is a deficient number, because the sum of its proper divisors (5510233) is less than itself. Its deficiency is 99042924
### Bases of 104553157
• Binary: 1100011101101011010110001012
• Base-36: 1Q8XQD
### Scales and comparisons
How big is 104553157?
• 104,553,157 seconds is equal to 3 years, 16 weeks, 6 days, 2 hours, 32 minutes, 37 seconds.
• To count from 1 to 104,553,157 would take you about four years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 104553157 cubic inches would be around 39.3 feet tall.
### Recreational maths with 104553157
• 104553157 backwards is 751355401
• The number of decimal digits it has is: 9
• The sum of 104553157's digits is 31
• More coming soon!
#### Copy this link to share with anyone:
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Numbermatics. (2023). Number 104553157 - Facts about the integer. Retrieved 8 December 2023, from https://numbermatics.com/n/104553157/
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The information we have on file for 104553157 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 104553157, math, Factors of 104553157, curriculum, school, college, exams, university, Prime factorization of 104553157, STEM, science, technology, engineering, physics, economics, calculator, one hundred four million, five hundred fifty-three thousand, one hundred fifty-seven.
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• Nukes should destroy item entities
The title says it all; I love nukes, but when I detonate them, thousands of entities are made and it lags my system out. If they had a 100% block item destroy rate (TNT has a partial chance to destroy blocks, so it's possible), it would help them not lag so much.
It would also work to make it a configurable setting. The way they work now makes a 5 minute lag area that is inaccessible to players. If this is intentional, then so be it; I still would like the change.
Thanks for reading, and I hope this is implemented!
• True V=IR power draw
Solving for load (machine) power, cable resistance, and source voltage, it's possible to solve for power draw.
The voltage breakdown would work as it already does, and there's the potential for a temperature-based breakdown system, albeit with more complexity.
The voltage can be solved with this formula:
P = (4*Pl)/(2+sqrt(1-(4*Pl*R/V**2)))
Where
P is the power expended by the source of current,
Pl is the power received by the machine,
R is the resistance of the cables connecting the machine to the source of current,
and V is the voltage of the system.
This would allow for the power system in IC^2 to accurately model ideal loads and ideal resistors in series. Of course, this system would put a greater computational drain on the system, but I think it would be insignificant compared to rendering or block updates.
• Mark 1 reactor, 3 efficiency < Nope, it's Mark III
Wow, I haven't been following this thoroughly enough. I haven't put any effort into this reactor, and it really shows. If I had been at the top of my game, I would have made Grincore's 4.44 efficiency reactor, but I guess that's the deal. In any case, I think that my reactor is not worth mentioning.
As an aside: I miss nukes. T_T
I made myself an endgame in IC 1 of creating and detonating 100 nukes, and it was quite satisfying (no TMI, of course). The explosion on a completely filled nuclear reactor is measly in comparison.
• Mark 1 reactor, 3 efficiency < Nope, it's Mark III
Don't believe me, eh? I don't blame you. I am cheating considerably, as I am using a redstone clock to make this reactor work.
3:1 clock (1 on pulse to 4 off pulses) connected to a reactor with 6 chambers and a 2x2 uranium setup in the center (or wherever, coincidentally).
Simple as that! 2x2 square means 2 connections each; 2 connections means 3 pulses each tick, 3 pulses means 30 heat per uranium, that equates to 120 heat/tick when the reactor's on. 120/4 = 30, which is the average heat per tick.
3x3 cube (wait, I'm starting to doubt this, for it makes it such that only 1 block of water is necessary outside of the reactor chambers...) of water surrounding means 3x3x3 cooling, +6 because of additional cooling due to chambers, -2 due to redstone setup (cobble + redstone above). This all evens out to 31 cooling, and therefore the total heat production is -1.
This is an easy setup, as it requires few starting materials, and the only tending one has to perform is on uranium, and of course to cut the power when logging in/out, as the redstone can bug, which could cause catastrophic nuclear meltdown.
• Suggestion: electric boat?
I am also in support of this idea.
One thing I discovered upon cutting down my trees; using the mining laser in conjunction with the chainsaw decreases the requisite time by a ton. I think that the difficulty in cutting down trees is fine, as rubber for wiring has always been something one must devote a lot of time to getting.
• Nukes?
If I am right (to be honest, I have not confirmed), nukes cost 2 raw uranium and 4 processed uranium apiece. Unless there is a way to produce more processed uranium than the starting raw uranium, this punishes processing the uranium. Is this intentional?
Edit: I have tried to use processed uranium to create a nuke, but there was nothing... Where be the nukes?
Edit 2: I apologise, this is in the wrong forum. This is supposed to be for IC^2. If a mod would relocate, that would be nice.
• Miner uses infinite energy
I'm not too clear on the details, but it seems that if the miner goes into the void it discharges the OV or OD scanner in the device. Because it is coded to recharge these, it will use infinite power.
Tested by mining a bunch of things; I would often get a miner with multiple bedrock and an empty OV device.
Edit: I apologise, this is in the wrong forum. This is supposed to be in IC^2 bugs. If a mod would relocate, that would be much appreciated.
• Nerdgasms, anyone?
Okay, so I was bored playing minecraft, looking for an SMP server to play bukkit on. All my attempts failed, and I thought: "shoot; I'm bored and don't know what to do!". I was wandering the mods, and thought of IC^2, and decided "to heck with it, there might be something there. I remember the good old days of IndustrialCraft. Al probably has a beta release out by now!" I looked around, and I saw "IC^2 1.00 release" and I enjoyed it. I installed it and went through the classic get all the metals, get the machines started. I was getting my solar early when I looked up switch cables. I couldn't find them, and I thought: "what the heck? How am I supposed to make this work? Use those expensive MFEs? There'd better be a good reason for this..." I then put 5 cables together and BAM! Nerdgasm.
Al, you have impressed me time and again, and I have this to say to you:
On the first day Al made IndustrialCraft, and he got +200 internets (seriously, you work so fast!).
On the second day, Al updated Industrialcraft with awesome all ways to Thursday, and the entirety of the Minecraft Mod Community saw that it was good.
On the Third day, Al made Midievalcraft, and made non-bukkit multiplayer good for once.
On the Fourth day, Al updated IndustrialCraft and MidievalCraft simultaneously, and we all were impressed.
On the Fifth day, Al made Mine 4 Dead, and it was cool (sadly I got no SMP action...).
On the Sixth day, Al announced IC^2, and all the boys did all they could to not pee themselves in excitement.
On the Seventh day, Al said: "I don't need no rest! I can release IC^2 before anyone can blink!" | 1,488 | 6,177 | {"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-2022-05 | longest | en | 0.971254 |
https://vdocuments.site/topographic-maps-topographic-maps-what-are-topographic-maps-topographic-maps-.html | 1,591,276,621,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347441088.63/warc/CC-MAIN-20200604125947-20200604155947-00315.warc.gz | 573,505,273 | 18,997 | # topographic maps topographic maps. what are topographic maps? topographic maps- maps that show...
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• Slide 1
• Topographic Maps TOPOGRAPHIC MAPS
• Slide 2
• What are topographic maps? Topographic maps- maps that show changes in elevation of Earths surface Elevation- distance above sea level. Also show mountains, rivers, forests and bridges
• Slide 3
• Contour Lines Contour lines- connects points of equal elevation.
• Slide 4
• One Island.Two Views
• Slide 5
• One IslandTwo Views
• Slide 6
• Identify the Elevation
• Slide 7
• What is a contour interval? Contour interval- the difference in elevation between two side by side contour lines What is the difference in elevation between these two lines?
• Slide 8
• What are index contours? Index Contours- contour lines that have marked elevations. Can use them to find the contour interval Index Contours
• Slide 9
• Finding the Contour Interval 1.Find the index contours. 2.Determine the elevation difference between the two index contours 800 m 700m = 100 m 3. Count the number of spaces between the 2 index contours. 5 4. Divide the elevation difference between the 2 index contours by the number of spaces. 100 m / 5 = 20 m.
• Slide 10
• Scale Scale is the relationship between horizontal distance on the map and distance on the ground.
• Slide 11
• Three types of scale
• Slide 12
• Slide 13
• Slide 14
• CONTOUR MAPPING RULES Do Worksheet 1-6
• Slide 15
• 1. A contour line represents a single equal elevation, all points on the contour line have the same elevation.
• Slide 16
• 2. Contour lines do not cross other contour lines.
• Slide 17
• 3. Where one closed contour line surrounds another, the inner contour line represents the higher elevation. Which line has the higher elevation?
• Slide 18
• 4. Every 5 th contour line is darker and has the elevation marked. These are index contours.
• Slide 19
• What is slope? Slope- change in elevation over a given horizontal distance Formula for slope Slope = change in elevation horizontal distance Which hill has a steep slope? Gentle slope?
• Slide 20
• 5a. On a contour map, closely spaced lines indicate a steep slope.
• Slide 21
• 5b. Widely spaced lines indicate a gentle slope
• Slide 22
• 5c. Uniformly spaced lines indicate a uniform (constant) slope. Notice that the contour lines for each are the same distance apart in each picture. There is the same change in elevation for each given distance.
• Slide 23
• Comparing Slope
• Slide 24
• 6. A contour line that closes within the limits of a map indicates a hill, ridge or plateau.
• Slide 25
• 6a. Highest possible elevation of a hill is just below the value of the next contour line that would be drawn. The last contour line drawn is 220 m. The next line would be 240 m. Since there is no 240 m line, the hill cannot be higher than 239 m.
• Slide 26
• 7. Depressions are shown by small marks (hachures) pointing inward off the contour line. Elevation of the depression would be less than 140 m.
• Slide 27
• 8. Where a contour line crosses a stream or valley, the contour bends to form a V to point upstream.
• Slide 28
• Exit Card Without looking at your notes, write two things you learned today. Write one question you have about topographic maps. It can be something that confused you or something youd like more information about.
• Slide 29
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# confusing for Loop.
Ranch Hand
Posts: 435
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What will the following code print?
Ans: It will print 10
My question is that the if statement satisfies in the 2 checking loop itself.first
c=0;
i=0,j=0,k=0;--->c=1
i=0,j=0,k=1;---> c=2
here itself k>j i.e 1>0
so the value of c must be 2..
please explain where am i wrong?
Sonir
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the break statement in the most inner loop only stops the most inner loop.
the 2 outer ones will continue again so at the end c is 10.
Ranch Hand
Posts: 287
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If you put println's inside each for loop & one right after c is incremented you will see this result:
F:\JavaProgs>java Test
i is: 0
j is: 0
k is: 0
c is: 1
k is: 1
c is: 2
j is: 1
k is: 0
c is: 3
k is: 1
c is: 4
k is: 2
c is: 5
i is: 1
j is: 0
k is: 0
c is: 6
k is: 1
c is: 7
j is: 1
k is: 0
c is: 8
k is: 1
c is: 9
k is: 2
c is: 10
10
If you use this to follow the loops thru I think you will get it.....you have to remember that when you break out of a loop to another loop the inner loop variable will be re-initialized when it is hit again. Look at the code & these results & try writing it down in a small chart.....I thikn this will help you see whats up.
[ January 13, 2002: Message edited by: DC Dalton ]
Did you see how Paul cut 87% off of his electric heat bill with 82 watts of micro heaters? | 656 | 2,040 | {"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} | 2.765625 | 3 | CC-MAIN-2024-22 | latest | en | 0.803982 |
http://www.physicsforums.com/showthread.php?t=39248 | 1,369,397,538,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704655626/warc/CC-MAIN-20130516114415-00004-ip-10-60-113-184.ec2.internal.warc.gz | 654,281,747 | 27,828 | ## [SOLVED] probability (was Re: EEQT)
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nOn 2004-08-12, Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:\n> Probabilities of single events are meaningless.\n\nOh boy. This is a really blatant misconception of probability. I can\'t\ntell whether the context rescues this (it seems not, though you are\narguing against another \'interpretation\' that I also find odious), but\nthe words themselves make probability a useless concept.\n\nOne can try to rescue it, e.g. by looking at ensembles, assuming\nexchangeability & independence, and trying to identify frequencies with\nprobabilities, but this gets incoherent, circular, or irrelevant to\nexperiment very quickly.\n\nIndependence is a statement about non-correlation between probabilities\nof single events, which you claim are meaningless.\n\nExchangeability is a statement that certain classes of possible\nrealizations of ensembles are equally likely -- in other words, it\ndepends on having a concept of probability for single events, where\nthese single events are realizations of ensemble measurements.\n\nIf you want to use finite ensembles, all probabilities must now be\nrationals, which seems a big limitation. We\'re also not guaranteed that\nthe final frequency really has any connection to the probability as we\nwant it -- instead we have to bring up "for all practical purposes"\narguments.\n\nIf you want to use infinite ensembles (the only case where we can\n_really_ say that the limiting frequencies approach the probability),\nwe have another problem: infinite ensembles\' limiting frequencies are a\ntail property. We can only measure the head, which can be arbitrarily\ndifferent, no matter how far out you measure it. There\'s no grounding\nthat lets us connect the rest of the ensemble to what we actually\nmeasure.\n\nBoth types of ensembles are imaginary and have nothing to do with\nany data we\'ve actually measured.\n\n--\nAaron Denney\n-><-\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>On $2004-08-12,$ Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
> Probabilities of single events are meaningless.
Oh boy. This is a really blatant misconception of probability. I can't
tell whether the context rescues this (it seems not, though you are
arguing against another 'interpretation' that I also find odious), but
the words themselves make probability a useless concept.
One can try to rescue it, e.g. by looking at ensembles, assuming
exchangeability & independence, and trying to identify frequencies with
probabilities, but this gets incoherent, circular, or irrelevant to
experiment very quickly.
Independence is a statement about non-correlation between probabilities
of single events, which you claim are meaningless.
Exchangeability is a statement that certain classes of possible
realizations of ensembles are equally likely -- in other words, it
depends on having a concept of probability for single events, where
these single events are realizations of ensemble measurements.
If you want to use finite ensembles, all probabilities must now be
rationals, which seems a big limitation. We're also not guaranteed that
the final frequency really has any connection to the probability as we
want it -- instead we have to bring up "for all practical purposes"
arguments.
If you want to use infinite ensembles (the only case where we can
_really_ say that the limiting frequencies approach the probability),
we have another problem: infinite ensembles' limiting frequencies are a
tail property. We can only measure the head, which can be arbitrarily
different, no matter how far out you measure it. There's no grounding
that lets us connect the rest of the ensemble to what we actually
measure.
Both types of ensembles are imaginary and have nothing to do with
any data we've actually measured.
--
Aaron Denney
$-><-$
PhysOrg.com physics news on PhysOrg.com >> A quantum simulator for magnetic materials>> Atomic-scale investigations solve key puzzle of LED efficiency>> Error sought & found: State-of-the-art measurement technique optimised
Aaron Denney wrote: > On $2004-08-12,$ Arnold Neumaier wrote: > >>Probabilities of single events are meaningless. > > Oh boy. This is a really blatant misconception of probability. The misconception seems to be completely on your side. What is the probability that 'I will die of cancer'? This is a single event that either will happen, or will not happen. If you consider this single event only, the probability is 1 or depending on what will actually happen. (But this sort of probability is not what we talk about in physics.) On the other hand one may assign a probability based on some facts about me. These facts determine an ensemble of people, from which one can form a statistical estimate of the probability. It clearly depends on which sort of ensemble one regarde me to belong to, what probability you will assign. I belong to many ensembles, and the answer is different for each of these. Thus probabilities are meaningful not for the single event but only as a property of the ensemble under consideration. This can also be seen from the mathematical foundations. Probabilities are determined by measures on the set of elementary events. All statements in measure theory are _only_ about expectations and probabilities of all possible realizations simultaneously, and say nothing at all about any particular realization. For a finite binary random sequence with independent bits, the sequence 111111111 has exactly the same status and probability as the sequence 101001101 or 000000000, although only the second looks random. Arnold Neumaier
Arnold Neumaier says... >What is the probability that 'I will die of cancer'? >This is a single event that either will happen, or will not happen. >If you consider this single event only, the probability is 1 or >depending on what will actually happen. (But this sort of probability >is not what we talk about in physics.) > >On the other hand one may assign a probability based on some facts >about me. These facts determine an ensemble of people, from which >one can form a statistical estimate of the probability. I don't see how ensembles help give any more precise meaning to probability. Suppose you say that "The probability that someone in risk group A will die of cancer is 1/3". That doesn't mean that for any 3 people in group A, 1 of them will die of cancer. It doesn't mean that for any 30 people, 10 of them will die of cancer. It doesn't even mean that in the limit as N as goes to infinity, the ratio $f_N =$ #who die of cancer/N $= 1/3$. What is true is that almost surely $f_N$ goes to 1/3 as N goes to infinity, but that "almost surely" is a probabilistic concept, as well. So you can't define probabilities completely in terms of ensembles. Saying that there is $a 1/3$ chance that I will die of cancer is meaningful without ensembles if you interpret that as a measure of my *belief* that I will die of cancer (1 meaning that I'm certain I will, meaning that I'm certain I won't). Of course, that's unsatisfying because we feel that quantum mechanical probabilities are revealing something objective, rather than subjective. So I don't know what the resolution is, but I don't think ensembles are on any firmer ground than any other interpretation of probability. -- Daryl McCullough Ithaca, NY
## [SOLVED] probability (was Re: EEQT)
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nOn 2004-08-13, Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:\n> Aaron Denney wrote:\n>> On 2004-08-12, Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:\n>>\n>>>Probabilities of single events are meaningless.\n>>\n>> Oh boy. This is a really blatant misconception of probability.\n>\n> The misconception seems to be completely on your side.\n>\n> What is the probability that \'I will die of cancer\'?\n> This is a single event that either will happen, or will not happen.\n\nYep. Events don\'t have to "half-happen" to have a probability of 0.5.\n\n> If you consider this single event only, the probability is 1 or 0\n> depending on what will actually happen.\n\nAfter the fact, yes. Before no, unless you know enough to time\nevolve the system.\n\nThis same argument would also lead to the probability of heads on a coin\nflip being 0 or 1, depending on exactly how one flips it. Actually,\nif you know enough about the flip beforehand, these would be the proper\nones, no matter how fair the coin is.\n\n> (But this sort of probability is not what we talk about in physics.)\n> On the other hand one may assign a probability based on some facts\n> about me.\n\nIt isn\'t? It sounds like we would all the time, and it is just a\nlimiting case of your example.\n\nIf we have all the facts we can push the probabilities to 0 or 1.\n\n(speaking classically. A full QM treatmnet of your life\nwould involve interactions with radioactive particles, which\nwill undoubtedly keep it from being exactly 0 or 1, though\nlaws of large numbers will probably push near 0 or 1 -- either\nyou got enough or you didn\'t.)\n\n> These facts determine an ensemble of people, from which\n> one can form a statistical estimate of the probability.\n\nBut this step is unnecessary.\n\n> It clearly depends on which sort of ensemble one regarde me to belong\n> to, what probability you will assign. I belong to many ensembles, and\n> the answer is different for each of these.\n\nCan you tell me which one is the right one to use for this question?\nWhy or why not? Since you said it is either 0 or 1, which ensemble\ngives that answer?\n\nYou don\'t belong to _any_ ensemble. They\'re sometimes useful for\nconstructing models, but they\'re not the only way to do it.\n\nIf you want to assign something a probability of 1/3, that doesn\'t\nrequire the construction of an ensemble of size 3N, which includes N\nways that event can happen.\n\n> Thus probabilities are meaningful not for the single event but only\n> as a property of the ensemble under consideration.\n\nAnd yet people bet on individual events all the time.\n\n> This can also be seen from the mathematical foundations. Probabilities\n> are determined by measures on the set of elementary events.\n\nThat\'s one way of defining the axioms of probability theory and getting\nthe standard results for manipulating probabilities in various self\nconsistent and maximally useful ways. It\'s not the only way, though\nit can give a nice intuitive understanding (assuming one knows\nmeasure theory better than probability theory, which seems unlikely).\n\nIf you want to invoke measure theory, but the measures are now the\nprobabilities, and the measures are not determined by the ensembles, the\nset you choose for the elementary events.\n\n> All statements in measure theory are _only_ about expectations and\n> probabilities of all possible realizations simultaneously, and say\n> nothing at all about any particular realization.\n\nAll statements in measure theory are about, well, measures over\nsets and subsets. If you\'re modeling probability with it, then\nthe measure over a subset _is_ supposed to be the probability\nof that subset occuring, or the probability of those particular\nrealizations. Of course they don\'t say that the event will or\nwill not happen, unless the probability is zero or one.\n\n> For a finite binary random sequence with independent bits,\n> the sequence 111111111 has exactly the same status and probability\n> as the sequence 101001101 or 000000000, although only the second\n> looks random.\n\nI\'m not sure what you\'re getting at here. Were this phrased\na bit more precisely, I wouldn\'t disagree.\n\nHow do you define "random" and "independent" for each bit, if single\nevents don\'t have probabilities? If 1 is more or less probable than\n0 for each digit, they will indeed have different statuses and\nprobabilities.\n\n--\nAaron Denney\n-><-\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>On $2004-08-13,$ Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
> Aaron Denney wrote:
>> On $2004-08-12,$ Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
>>
>>>Probabilities of single events are meaningless.
>>
>> Oh boy. This is a really blatant misconception of probability.
>
> The misconception seems to be completely on your side.
>
> What is the probability that 'I will die of cancer'?
> This is a single event that either will happen, or will not happen.
Yep. Events don't have to "half-happen" to have a probability of .5.
> If you consider this single event only, the probability is 1 or
> depending on what will actually happen.
After the fact, yes. Before no, unless you know enough to time
evolve the system.
This same argument would also lead to the probability of heads on a coin
flip being or 1, depending on exactly how one flips it. Actually,
if you know enough about the flip beforehand, these would be the proper
ones, no matter how fair the coin is.
> (But this sort of probability is not what we talk about in physics.)
> On the other hand one may assign a probability based on some facts
It isn't? It sounds like we would all the time, and it is just a
limiting case of your example.
If we have all the facts we can push the probabilities to or 1.
(speaking classically. A full QM treatmnet of your life
would involve interactions with radioactive particles, which
will undoubtedly keep it from being exactly or 1, though
laws of large numbers will probably push near or 1 -- either
you got enough or you didn't.)
> These facts determine an ensemble of people, from which
> one can form a statistical estimate of the probability.
But this step is unnecessary.
> It clearly depends on which sort of ensemble one regarde me to belong
> to, what probability you will assign. I belong to many ensembles, and
> the answer is different for each of these.
Can you tell me which one is the right one to use for this question?
Why or why not? Since you said it is either or 1, which ensemble
You don't belong to _any_ ensemble. They're sometimes useful for
constructing models, but they're not the only way to do it.
If you want to assign something a probability of $1/3,$ that doesn't
require the construction of an ensemble of size 3N, which includes N
ways that event can happen.
> Thus probabilities are meaningful not for the single event but only
> as a property of the ensemble under consideration.
And yet people bet on individual events all the time.
> This can also be seen from the mathematical foundations. Probabilities
> are determined by measures on the set of elementary events.
That's one way of defining the axioms of probability theory and getting
the standard results for manipulating probabilities in various self
consistent and maximally useful ways. It's not the only way, though
it can give a nice intuitive understanding (assuming one knows
measure theory better than probability theory, which seems unlikely).
If you want to invoke measure theory, but the measures are now the
probabilities, and the measures are not determined by the ensembles, the
set you choose for the elementary events.
> All statements in measure theory are _only_ about expectations and
> probabilities of all possible realizations simultaneously, and say
> nothing at all about any particular realization.
All statements in measure theory are about, well, measures over
sets and subsets. If you're modeling probability with it, then
the measure over a subset _is_ supposed to be the probability
of that subset occuring, or the probability of those particular
realizations. Of course they don't say that the event will or
will not happen, unless the probability is zero or one.
> For a finite binary random sequence with independent bits,
> the sequence 111111111 has exactly the same status and probability
> as the sequence 101001101 or 000000000, although only the second
> looks random.
I'm not sure what you're getting at here. Were this phrased
a bit more precisely, I wouldn't disagree.
How do you define "random" and "independent" for each bit, if single
events don't have probabilities? If 1 is more or less probable than
for each digit, they will indeed have different statuses and
probabilities.
--
Aaron Denney
$-><-$
In article <411CA50D.8080100@univie.ac.at>, Arnold Neumaier writes: |> Aaron Denney wrote: $|> >$ On $2004-08-12,$ Arnold Neumaier wrote: $|> >$ |> >>Probabilities of single events are meaningless. $|> >|> > Oh$ boy. This is a really blatant misconception of probability. |> |> The misconception seems to be completely on your side. |> |> What is the probability that 'I will die of cancer'? |> This is a single event that either will happen, or will not happen. |> If you consider this single event only, the probability is 1 or |> depending on what will actually happen. (But this sort of probability |> is not what we talk about in physics.) |> |> On the other hand one may assign a probability based on some facts |> about me. These facts determine an ensemble of people, from which |> one can form a statistical estimate of the probability. I am sorry, but you are wrong on both counts. There are several concepts of probability, and mathematical statisticians work with all except the erroneous one: The purest mathematical one is a specialisation of measure theory, and probabilities are simply positive measures of $\sigma$ one over some Borel set. Nice and easy, but PURE mathematics. The simplest semi-physical approximation is a 'repeatable' experiment (let's leave the philosophy of repeatability out of it), which is what most people are taught at a naive level. There is a very common error where excessive simplification leads people to confuse the concepts of a distribution of a measurement over a sample and a probability. You have made that error, I am afraid, though not in its simple form. There is, however, also the concept of the probability of a non-repeatable event, which can be handled mathematically just as easily as a repeatable experiment. What you can't do is to MEASURE such probabilities, though you can do some measurement with ensembles (as you correctly state). Note, however, that this all depends on the existence of time's arrow (causality again!), because the probability has a meaning only up until the time that the event takes place (and it may change as time progresses, too). Whereafter, it is either or 1, true. So, if you are working with a model of the universe that does not have such a concept, your first statement is correct. But few physicists do. Regards, Nick Maclaren.
In article , Aaron Denney writes: |> On $2004-08-12,$ Arnold Neumaier wrote: $|> >$ Probabilities of single events are meaningless. |> $|> Oh$ boy. This is a really blatant misconception of probability. I can't |> tell whether the context rescues this (it seems not, though you are |> arguing against another 'interpretation' that I also find odious), but |> the words themselves make probability a useless concept. That is true :-( But see below for a possible cause of confusion. It is also claimed that talking about the probability of an inherently non-repeatable event is meaningless, but even that is based on a naive and mistaken view of probability. What is true is that it is quite hard to do much with such probabilities. |> Independence is a statement about non-correlation between probabilities |> of single events, which you claim are meaningless. Er, that is NOT well-phrased! There is also the serious problem that many people may be using "event" to mean what a probabilist would call "outcome". The word "event" is a common and fruitful source of confusion. Regards, Nick Maclaren.
Daryl McCullough wrote: > Arnold Neumaier says... > >>What is the probability that 'I will die of cancer'? >>This is a single event that either will happen, or will not happen. >>If you consider this single event only, the probability is 1 or >>depending on what will actually happen. (But this sort of probability >>is not what we talk about in physics.) >> >>On the other hand one may assign a probability based on some facts >>about me. These facts determine an ensemble of people, from which >>one can form a statistical estimate of the probability. > > I don't see how ensembles help give any more precise meaning to > probability. Suppose you say that "The probability that someone > in risk group A will die of cancer is 1/3". That doesn't mean > that for any 3 people in group A, 1 of them will die of cancer. > It doesn't mean that for any 30 people, 10 of them will die of > cancer. I claimed neither of these. To say that "The probability that someone in risk group A will die of cancer is 1/3" means nothing more or less than that exactly 1/3 of _all_ people in risk group A will die of cancer. Of course, we cannot check this before we have information about how all people in risk group A died, but once we have this information, we know. Usually we only have incomplete knowledge about the ensemble. This is why statisticians say that they _estimate_ probabilities based on _incomplete_ knowledge, collected from a sample. Whereas they _compute_ probabilities from $_assumed_complete$ knowledge about the ensemble, namely the theoretical probability distribution. Estimates are usually inaccurate but useful; this reconciles the two approaches; hence one finds no difficulties at all in actual practice. A more extensive discussion can be found in my theoretical physics FAQ at http://www.mat.univie.ac.at/~neum/physics-faq.txt It currently has the following entries about classical probability: 5a. Random numbers in probability theory 5b. How meaningful are probabilities of single events? 5c. What about the subjective interpretation of probabilities? 5d. What is the meaning of probabilities? 5e. How do probabilities apply in practice? 5f. Priors and entropy in probability theory > Saying that there is $a 1/3$ chance that I will die of cancer is > meaningful without ensembles if you interpret that as a measure > of my *belief* that I will die of cancer (1 meaning that I'm > certain I will, meaning that I'm certain I won't). Of course, > that's unsatisfying because we feel that quantum mechanical > probabilities are revealing something objective, rather than > subjective. This is also unsatisfactory because "belief" is an even poorer defined concept than probability, that depends on psychology of human beings, which is not a sound foundation for physics. On the other hands, "ensemble" is a precise concept with far-reaching applications, independent of any beliefs, and hence adequate for use in quantum mechanics. Arnold Neumaier
Nick Maclaren wrote: > In article <411CA50D.8080100@univie.ac.at>, > Arnold Neumaier writes: > |> > |> What is the probability that 'I will die of cancer'? > |> This is a single event that either will happen, or will not happen. > |> If you consider this single event only, the probability is 1 or > |> depending on what will actually happen. (But this sort of probability > |> is not what we talk about in physics.) > |> > |> On the other hand one may assign a probability based on some facts > |> about me. These facts determine an ensemble of people, from which > |> one can form a statistical estimate of the probability. > > I am sorry, but you are wrong on both counts. There are several > concepts of probability, and mathematical statisticians work with > all except the erroneous one: > > The purest mathematical one is a specialisation of measure > theory, and probabilities are simply positive measures of $\sigma$ > one over some Borel set. Nice and easy, but PURE mathematics. This case conforms to my statements. A discrete measure with probabilities that are integral multiples of N can be viewed as an ensemble of N experiments. All statements that make sense for discrete measures make corresponding assertions about such an ensemble. From a practical point of view, all other measures are just useful approximations to summarise the information in ensembles with very large N. > The simplest semi-physical approximation is a 'repeatable' > experiment (let's leave the philosophy of repeatability out of > it), which is what most people are taught at a naive level. Here the concept of probzbility is already dubious, unless you specify which set of repetitions (i.e., the ensemble) you are applying it to. > There is a very common error where excessive simplification > leads people to confuse the concepts of a distribution of a > measurement over a sample and a probability. You have made that > error, I am afraid, though not in its simple form. No. There is a sample distribution, and there is a theoretical distribution. Both assign probabilities to events. The sample distribution is the right one if the ensemble is taken as the sample you know, the theoretical distribution is the right one if the ensemble is taken as the set of all element in the set upon which the $\sigma$ algebra is based. > There is, however, also the concept of the probability of a > non-repeatable event, which can be handled mathematically just > as easily as a repeatable experiment. If it is as easy, please give the mathematical formulation of the probability that the earth will be hit by an asteroid in the year 9999? > What you can't do is to > MEASURE such probabilities, though you can do some measurement > with ensembles (as you correctly state). As I mentioned in another post, probability assignments to single events can be neither verified nor falsified. Thus they are meaningless. Arnold Neumaier
Aaron Denney wrote: > On $2004-08-13,$ Arnold Neumaier wrote: > >>What is the probability that 'I will die of cancer'? >>This is a single event that either will happen, or will not happen. > > Yep. Events don't have to "half-happen" to have a probability of .5. Probability assignments to single events can be neither verified nor falsified. Indeed, suppose we intend to throw a coin exactly once. Person A claims 'the probability of the coin coming out head is 50%'. Person B claims 'the probability of the coin coming out head is 20%'. Person C claims 'the probability of the coin coming out head is 80%'. Now we throw the coin and find 'head'. Who was right? It is undecidable. Thus there cannot be objective content in the statement 'the probability of the coin coming out head is p', when applied to a single case. Subjectively, of course, every person may feel (and is entitled to feel) right about their probability assignment. But for use in science, such a subjective view (where everyone is right, no matter which statement was made) is completely useless. >>If you consider this single event only, the probability is 1 or >>depending on what will actually happen. > > After the fact, yes. Before no, unless you know enough to time > evolve the system. What if someone knows the fact and someone else doesn't??? I am discussing objective probability, since physics is an objective science. >>It clearly depends on which sort of ensemble one regarde me to belong >>to, what probability you will assign. I belong to many ensembles, and >>the answer is different for each of these. > > Can you tell me which one is the right one to use for this question? > Why or why not? Since you said it is either or 1, which ensemble > gives that answer? The ensemble consisting of me only, as appropriate for a single case. In other ensembles, the probability is just the proportion of people in the ensemble dying of cancer; of course this probability, though it is a well-defined number, can be estimated only approximately $- at$ least until I am dead ;-) >>Thus probabilities are meaningful not for the single event but only >>as a property of the ensemble under consideration. > > And yet people bet on individual events all the time. Oh yes. They estimate probabilities, based on their faforite ensemble. But as you know, people often lose their bets! >>This can also be seen from the mathematical foundations. Probabilities >>are determined by measures on the set of elementary events. > > That's one way of defining the axioms of probability theory and getting > the standard results for manipulating probabilities in various self > consistent and maximally useful ways. It's not the only way, But it is the only consistent way. >>All statements in measure theory are _only_ about expectations and >>probabilities of all possible realizations simultaneously, and say >>nothing at all about any particular realization. > > All statements in measure theory are about, well, measures over > sets and subsets. If you're modeling probability with it, then > the measure over a subset _is_ supposed to be the probability "_is_", not "_is_ supposed to be". > of that subset occuring, or the probability of those particular > realizations. Of course they don't say that the event will or > will not happen, unless the probability is zero or one. Yes; this is why they say nothing at all about the single case. >>For a finite binary random sequence with independent bits, >>the sequence 111111111 has exactly the same status and probability >>as the sequence 101001101 or 000000000, although only the second >>looks random. > > I'm not sure what you're getting at here. Were this phrased > a bit more precisely, I wouldn't disagree. > > How do you define "random" and "independent" for each bit, if single > events don't have probabilities? If 1 is more or less probable than > for each digit, they will indeed have different statuses and > probabilities. Here I meant 'random bit' to say both probabilities 1/2. A random sequence is $_not_ a$ sequence of numbers but a sequence of random numbers. Only the realizations are sequences of ordinary numbers. Sequences of ordinary numbers are _never_ random, but they can 'look random' (in a subjective sense). Arnold Neumaier
In article <411F45F1.10704@univie.ac.at>, Arnold Neumaier wrote: >Daryl McCullough wrote: > >To say that "The probability that someone in risk group A will die >of cancer is 1/3" means nothing more or less than that exactly 1/3 >of _all_ people in risk group A will die of cancer. That is completely and utterly wrong. You can see that by taking a nice, simple example (i.e. not cancer). Fair coins have a probability .5 of coming up heads. If you toss 10 fair coins, it is NOT necessarily the case that exactly 5 will show heads. You get exactly the same situation with a fixed number of electrons in quantum mechanics. >Of course, we cannot check this before we have information about >how all people in risk group A died, but once we have this information, >we know. Let us say 7 coins out of 10 show heads. Would that mean that the probability of a fair coin showing heads was 70%? Oh, come now. If you toss them again (which is where repeatable experiments come in), you might well have 4 come up heads. And so on. >Usually we only have incomplete knowledge about the ensemble. >This is why statisticians say that they _estimate_ probabilities >based on _incomplete_ knowledge, collected from a sample. That is true, but it is only ONE of the things that statisticians do. And not the most important one, either. >Whereas they _compute_ probabilities from $_assumed_complete$ knowledge >about the ensemble, namely the theoretical probability distribution. Grrk. We also compute confidence intervals on probabilities based on data, compute probabilities based on a known mathematical model and values of its parameters, compute various forms of best estimates of probabilities based on data, and so on. >Estimates are usually inaccurate but useful; this reconciles the two >approaches; hence one finds no difficulties at all in actual practice. Hmm. I am glad that you find the problems so simple. Not all leading statisticians do. >> Saying that there is $a 1/3$ chance that I will die of cancer is >> meaningful without ensembles if you interpret that as a measure >> of my *belief* that I will die of cancer (1 meaning that I'm >> certain I will, meaning that I'm certain I won't). Of course, >> that's unsatisfying because we feel that quantum mechanical >> probabilities are revealing something objective, rather than >> subjective. > >This is also unsatisfactory because "belief" is an even poorer defined >concept than probability, that depends on psychology of human beings, >which is not a sound foundation for physics. Yes and no. That is not true if he defines the mathematical model he is using and the data and methods he is using to estimate the parameters of that model. That is, after all, precisely the statistical analogue of measuring a physical constant! >On the other hands, "ensemble" is a precise concept with far-reaching >applications, independent of any beliefs, and hence adequate for use >in quantum mechanics. Well, it wasn't a standard term when I did my (masters equivalent) course in mathematical statistics, though that was some 30+ years back. I can guess what it means, but I don't think that "a precise concept" is what a mathematical statistician would call it. Please note that I am replying to this posting and not yours to mine, because it gives a clearer example of some of the issues. Regards, Nick Maclaren.
Nick Maclaren wrote: > In article <411F45F1.10704@univie.ac.at>, > Arnold Neumaier wrote: > >>Daryl McCullough wrote: >> >>To say that "The probability that someone in risk group A will die >>of cancer is 1/3" means nothing more or less than that exactly 1/3 >>of _all_ people in risk group A will die of cancer. > > > That is completely and utterly wrong. You can see that by taking > a nice, simple example (i.e. not cancer). > > Fair coins have a probability .5 of coming up heads. If you toss > 10 fair coins, it is NOT necessarily the case that exactly 5 will > show heads. This is not a correct translation of my claim. If you take any finite $\sigma$ algebra representing a fair coin, one has a finite ensemble of elementary events, and exactly half of them come out heads. If you take an infinite $\sigma$ algebra, the ensemble is infinite, but with the natural weighting, again exactly half of them come out head. This is precisely what I claimed. 'Tossing 10 fair coins' is just a sloppy way of saying 'Selecting a sample of size 10 from the total ensemble', and it is obvious that here the number of heads is 5 only on average over many random samples, again as I had claimed in an unquoted part of the post to which you replied. >>On the other hands, "ensemble" is a precise concept with far-reaching >>applications, independent of any beliefs, and hence adequate for use >>in quantum mechanics. > > Well, it wasn't a standard term when I did my (masters equivalent) > course in mathematical statistics, though that was some 30+ years > back. I can guess what it means, but I don't think that "a precise > concept" is what a mathematical statistician would call it. I am talking here (s.p.r.) physics language. In mathematical terms, a classical ensemble is the set of elementary events underlying the $\sigma$ algebra over which the measure is defined. Arnold Neumaier
Arnold Neumaier says... >Probability assignments to single events can be neither verified nor >falsified. Probabilistic predictions can *never* be verified or falsified by any (finite) number of observations. If the prediction is that half of all particles of type X decay within T seconds, how many measurements does it take to prove the prediction is true? How many measurements does it take to prove the prediction is false? The answer is that there is no number that is sufficient. It's just that as we make more and more observations, we become and more confident that the prediction is true (or that it's false). There is never a point where it is absolutely verified or absolutely falsified, although we can get to a point where we are as good as certain one way or the other. But for any finite number of observations, we don't know whether the probabilistic prediction is true or not. We're in the same boat whether we are talking about 1 observation or 1000. -- Daryl McCullough Ithaca, NY
Arnold Neumaier says... >To say that "The probability that someone in risk group A will die >of cancer is 1/3" means nothing more or less than that exactly 1/3 >of _all_ people in risk group A will die of cancer. That is not true. That's the *frequency* with which people in group A die of cancer. It is *not* the probability. The frequency is supposed to *approach* the probability in some sense, but they aren't the same thing. Think about it. If you have a coin with a 50% chance of heads and 50% chance of tails, then the frequency jumps around as time goes on. With the first coin toss, the frequency of heads is either or 1. With the second coin toss, the frequency is either $0, 1/2,$ or 1. With the third toss, the frequency is either $0, 1/3, 2/3,$ or 1. Nobody would say that the *probability* jumps around like that. The probability is always the same (well, unless there is some time-dependent effect). Probability is *not* the same thing as relative frequency, and ensembles don't help to define probability. They can help define relative frequency, but only in the case of *finite* ensembles (frequency is not well-defined for an infinite ensemble). But it is exactly in the case of finite ensembles that the difference between probability and relativity frequency is the most pronounced. -- Daryl McCullough Ithaca, NY
On $2004-08-16,$ Arnold Neumaier wrote: > To say that "The probability that someone in risk group A will die > of cancer is 1/3" means nothing more or less than that exactly 1/3 > of _all_ people in risk group A will die of cancer. > Of course, we cannot check this before we have information about > how all people in risk group A died, but once we have this information, > we know. Okay, you have two choices for defining this ensemble. Either (a) it is actual people in this ensemble, and it's a finite set, or (b) it is imaginary people similar to the ones you actually care about. For (b), the probability is not objective, not checkable by anyone else. For (a), well, suppose I flip a coin ten times, and get 6 heads and 4 tails. I really, really, hope that you don't think that the coin has a probability of exactly .6 of coming up heads. If you flip it again twice and get two tails, is the probability for heads now .5? -- Aaron Denney $-><-$
On $2004-08-16,$ Arnold Neumaier wrote: > > > > > Aaron Denney wrote: >> On $2004-08-13,$ Arnold Neumaier wrote: >> >>>What is the probability that 'I will die of cancer'? >>>This is a single event that either will happen, or will not happen. >> >> Yep. Events don't have to "half-happen" to have a probability of .5. > > Probability assignments to single events can be neither verified nor > falsified. Right. Probability assignments inherently have some subjectivity -- what someone knows determines the assignment. > Indeed, suppose we intend to throw a coin exactly once. > Person A claims 'the probability of the coin coming out head is 50%'. > Person B claims 'the probability of the coin coming out head is 20%'. > Person C claims 'the probability of the coin coming out head is 80%'. > Now we throw the coin and find 'head'. Who was right? It is undecidable. Any of them, or none of them, depending on what they knew about the prior conditions of tossing. "Appropriate" probability assignment would be better language than "correct". All of them could be correct if A knows only that it has heads and tails, and that both can come up, if B knows that the coin is heavier on the heads side by a certain amount, and C knows that the tosser is extremely practiced and can make it come out heads 80% of time. > Thus there cannot be objective content in the statement > 'the probability of the coin coming out head is p', > when applied to a single case. Subjectively, of course, every person > may feel (and is entitled to feel) right about their probability assignment. > But for use in science, such a subjective view (where everyone is right, > no matter which statement was made) is completely useless. Not at all. Suppose someone you trust completely assures you that a coin is biased so that during tests it comes up one way 80% of the time, and the other 20% of the time, but refuses to tell you which is which. What is your probability assignment that the coin comes up heads on one toss? I claim this is the same case as flipping any coin. Deterministically it will come up whatever it comes up as. $P = 0,$ or 1, if you knew everything. Still, even in this case, where the "objective" probability is .8 or .2, the best representation of the information available to you for a single toss is .5 to heads and .5 to tails. If you know it will be flipped twice, you should assign .32 to HH and TT and .18 to HT and TH. >>>If you consider this single event only, the probability is 1 or >>>depending on what will actually happen. >> >> After the fact, yes. Before no, unless you know enough to time >> evolve the system. > > What if someone knows the fact and someone else doesn't??? > I am discussing objective probability, since physics is an objective > science. Then someone gets a better estimate. Probability theory is a way of reasoning about uncertainty. If someone knows that they know enough information, they get what will actually happen with probability 1. Someone who doesn't know enough will get a whole range of possible outcomes with different probabilities. These different probabilities will still be useful information about what choices to make. It's okay for the results to be subjective because different people start with different information. If someone starts with incorrect, rather than incomplete information, they'll get incorrect results. This is expected. Probability is _not_ figuring out how often something happens in repeatable experiments. It can be applied to that, yielding the well known de Finetti exchangeability results linking long-run frequency with probability, but limiting it to that case is perverse. >>>It clearly depends on which sort of ensemble one regarde me to belong >>>to, what probability you will assign. I belong to many ensembles, and >>>the answer is different for each of these. >> >> Can you tell me which one is the right one to use for this question? >> Why or why not? Since you said it is either or 1, which ensemble >> gives that answer? > > The ensemble consisting of me only, as appropriate for a single case. > > In other ensembles, the probability is just the proportion of > people in the ensemble dying of cancer; of course this probability, > though it is a well-defined number, can be estimated only approximately > $- at$ least until I am dead ;-) Do you allow infinite ensembles, or are only rational numbers acceptable probabilities? >>>Thus probabilities are meaningful not for the single event but only >>>as a property of the ensemble under consideration. >> >> And yet people bet on individual events all the time. > > Oh yes. They estimate probabilities, based on their favorite ensemble. > But as you know, people often lose their bets! Sure. That doesn't mean they estimated the probability wrong, or are misusing probability theory. Refusing to let probability theory deal with single events reduces its applicability to almost nothing, and people do successfully use it for single events. >>>This can also be seen from the mathematical foundations. Probabilities >>>are determined by measures on the set of elementary events. >> >> That's one way of defining the axioms of probability theory and getting >> the standard results for manipulating probabilities in various self >> consistent and maximally useful ways. It's not the only way, > > But it is the only consistent way. There are at least three or four different ways. They all give the same answers in the areas where they all apply. >> of that subset occuring, or the probability of those particular >> realizations. Of course they don't say that the event will or >> will not happen, unless the probability is zero or one. > > Yes; this is why they say nothing at all about the single case. Sure they, just not something definite. That's why there probabilities instead of certainties. Now, you can do most of this reasoning about uncertainty with ensembles rather than states of knowledge, but it's much harder and more complicated. You have to make sure that the ensembles you come up with are not only consistent with your state of knowledge, but also don't tell you anything more -- that they aren't biased. The choice of ensemble to use is just as subjective as the choices of A, B, and C in your example above. -- Aaron Denney $-><-$
In article <4121F056.7090303@univie.ac.at>, Arnold Neumaier >> >>>To say that "The probability that someone in risk group A will die >>>of cancer is 1/3" means nothing more or less than that exactly 1/3 >>>of _all_ people in risk group A will die of cancer. >> >That is completely and utterly wrong. You can see that by taking >a nice, simple example (i.e. not cancer). >> >Fair coins have a probability .5 of coming up heads. If you toss >10 fair coins, it is NOT necessarily the case that exactly 5 will >show heads. > >This is not a correct translation of my claim. Good, Because it is completely wrong. It IS a correct example of what you posted, so I am glad that you didn't mean it. > >If you take any finite $\sigma$ algebra representing a > >fair coin, one has a finite ensemble of elementary events, > >and exactly half of them come out heads. > > If you take an infinite $>\sigma$ algebra, the ensemble is infinite, but with the natural >weighting, again exactly half of them come out head. The very concept of "with the natural weighting, again exactly half of them come out head" is misleading to a degree when applied to a limit process (which this is). See below. >'Tossing 10 fair coins' is just a sloppy way of saying >'Selecting a sample of size 10 from the total ensemble', >and it is obvious that here the number of heads is 5 only on >average over many random samples, again as I had claimed in an >unquoted part of the post to which you replied. > >>>On the other hands, "ensemble" is a precise concept with far-reaching >>>applications, independent of any beliefs, and hence adequate for use >>>in quantum mechanics. >> >Well, it wasn't a standard term when I did my (masters equivalent) >course in mathematical statistics, though that was some 30+ years >back. I can guess what it means, but I don't think that "a precise >concept" is what a mathematical statistician would call it. > >I am talking here (s.p.r.) physics language. >In mathematical terms, a classical ensemble is the set of elementary >events underlying the $\sigma$ algebra over which the measure is defined. I am afraid that this shows a SERIOUS misunderstanding of measure theory (i.e. Borel sets and Lebesgue measure). Yes, discrete measures (i.e. over countable sets) have such a basis, but that does NOT extend to the general case. And using that 'simplification' vastly complicates the theory. In general, there IS no set of elementary events underlying the Borel set. Even when that is defined on top of another set that does have a concept of basic elements (which is not necessarily the case), it isn't rare for the measure of all such elements to be zero. The simple and classic example is the real interval [0,1] with the uniform measure. Regards, Nick Maclaren.
Daryl McCullough wrote: > Arnold Neumaier says... > > >>Probability assignments to single events can be neither verified nor >>falsified. > > Probabilistic predictions can *never* be verified or falsified by any > (finite) number of observations. If the prediction is that half of all > particles of type X decay within T seconds, how many measurements does > it take to prove the prediction is true? How many measurements does > it take to prove the prediction is false? All those in the defining ensemble. You seem to be thinking of an infinite ensemble; then your statement is true. But if the ensemble is finite, one knows the probability of any statement about a random variable x once all realizations $x(\omega)$ and their weight are known. This completely characterizes the ensemble. Arnold Neumaier | 11,347 | 49,706 | {"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} | 2.90625 | 3 | CC-MAIN-2013-20 | latest | en | 0.817129 |
https://math-frolic.blogspot.com/2014/12/an-epiphany.html | 1,501,144,029,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549427750.52/warc/CC-MAIN-20170727082427-20170727102427-00069.warc.gz | 668,957,061 | 21,483 | ## Pages
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## Sunday, December 14, 2014
### An Epiphany
Sunday reflection via Steven Strogatz....
"The teacher, Mr. diCurcio, said, 'I want you to figure out a rule about this pendulum.' He handed each of us a little toy pendulum with a retractable bob. You could make it a little bit longer or shorter in clicks in discrete steps. We were each handed a stopwatch and told to let the pendulum swing ten times, and then click, measure how long it takes for ten swings, and then click again, repeating the measurement after making the pendulum a little bit longer. The point was to see how the length of the pendulum determines how long it takes to make ten swings. The experiment was supposed to teach us about graph paper and how to make a relationship between one variable and another, but as I was dutifully plotting the length of time the pendulum took to swing ten times versus its length it occurred to me, after about the fourth or fifth dot, that a pattern was starting to emerge. These dots were falling on a particular curve I recognized because I'd seen it in my algebra class. It was a parabola, the same shape that water makes coming out of a fountain.
"I remember having an enveloping sensation of fear. It was not a happy feeling but an awestruck feeling. It was as if this pendulum knew algebra. What was the connection between the parabolas in algebra class and the motion of this pendulum? There it was on the graph paper. It was a moment that struck me, and was my first sense that the phrase 'law of nature' meant something. I suddenly knew what people were talking about when they said there could be order in the universe and that, more to the point, you couldn't see it unless you knew math. It was an epiphany that I've never really recovered from.
"
-- Steven Strogatz from "Who Cares About Fireflies"
[…If you have a favorite math-related passage that might make a nice Sunday morning reflection here let me know (SheckyR@gmail.com). If I use one submitted by a reader, I'll cite the contributor.] | 471 | 2,085 | {"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} | 3.328125 | 3 | CC-MAIN-2017-30 | longest | en | 0.976358 |
https://republicofsouthossetia.org/question/if-you-have-2-6-3-what-s-the-operation-you-should-perform-first-15145574-47/ | 1,632,712,624,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058263.20/warc/CC-MAIN-20210927030035-20210927060035-00594.warc.gz | 526,053,795 | 13,863 | ## If you have 2×6-3 what’s the operation you should perform first
Question
If you have 2×6-3 what’s the operation you should perform first
in progress 0
2 weeks 2021-09-13T10:33:52+00:00 2 Answers 0
see bold
Step-by-step explanation:
using P.E.M.D.A.S
( parenthisis, exponent, mulitiplication, division, adding, subtracting)
use multiplication ( 2×6) first
Step-by-step explanation:
Using the BODMAS rule
2×6-3
It was observed that the multiplication comes first before substration
Therefore to solve the question
2×6=12
Then 12-3=9 | 164 | 548 | {"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} | 3.828125 | 4 | CC-MAIN-2021-39 | latest | en | 0.839497 |
https://homesearchend.com/recommendations/how-to-find-the-y-intercept-of-a-function.html | 1,618,178,535,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038065492.15/warc/CC-MAIN-20210411204008-20210411234008-00006.warc.gz | 408,767,016 | 15,056 | # How To Find The Y Intercept Of A Function?
## How do you find the Y intercept in an equation?
To find the y intercept using the equation of the line, plug in 0 for the x variable and solve for y.
If the equation is written in the slope-intercept form, plug in the slope and the x and y coordinates for a point on the line to solve for y.
## What is the Y intercept of the function?
The intercepts of a graph are points at which the graph crosses the axes. The x-intercept is the point at which the graph crosses the x-axis. At this point, the y-coordinate is zero. The y-intercept is the point at which the graph crosses the y-axis.
## How do you find the Y intercept and zeros of a function?
Find the zeros and y intercept of the quadratic –
## How do you find the Y intercept of a polynomial?
Finding the y intercept of a Polynomial –
## What is the Y intercept formula?
The equation of any straight line, called a linear equation, can be written as: y = mx + b, where m is the slope of the line and b is the y-intercept. The y-intercept of this line is the value of y at the point where the line crosses the y axis.
## What is an example of Y intercept?
y -Intercepts. The y -intercept of a graph is the point where the graph crosses the y -axis. For example, we say that the y -intercept of the line shown in the graph below is 3.5 . When the equation of a line is written in slope-intercept form ( y=mx+b ), the y -intercept b can be read immediately from the equation.
## How do you do you find the Y intercept?
To find the y intercept using the equation of the line, plug in 0 for the x variable and solve for y. If the equation is written in the slope-intercept form, plug in the slope and the x and y coordinates for a point on the line to solve for y.
We recommend reading: How To Find The Surface Area Of A Sphere?
## How do you find the Y intercept of a graph?
Finding y intercept from a graph –
## What is Y intercept in an equation?
Every straight line can be represented by an equation: y = mx + b. The equation of any straight line, called a linear equation, can be written as: y = mx + b, where m is the slope of the line and b is the y-intercept. The y-intercept of this line is the value of y at the point where the line crosses the y axis.
## How do you find the zero of a function?
Finding the zero of a function means to find the point (a,0) where the graph of the function and the y-intercept intersect. To find the value of a from the point (a,0) set the function equal to zero and then solve for x. This involves using different techniques depending on the type of function that you have.
## How do you find the real zeros of a function?
How To Find the Zeros of The Function –
eight zeroes | 651 | 2,743 | {"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} | 4.6875 | 5 | CC-MAIN-2021-17 | latest | en | 0.912312 |
https://www.convertunits.com/from/milligram/hour/to/milligram/minute | 1,679,852,885,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946445.46/warc/CC-MAIN-20230326173112-20230326203112-00575.warc.gz | 791,222,793 | 12,495 | ## Convert milligram/hour to milligram/minute
milligram/hour milligram/minute
How many milligram/hour in 1 milligram/minute? The answer is 60.
We assume you are converting between milligram/hour and milligram/minute.
You can view more details on each measurement unit:
milligram/hour or milligram/minute
The SI derived unit for mass flow rate is the kilogram/second.
1 kilogram/second is equal to 3600000000 milligram/hour, or 60000000 milligram/minute.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between milligrams/hour and milligrams/minute.
Type in your own numbers in the form to convert the units!
## Quick conversion chart of milligram/hour to milligram/minute
1 milligram/hour to milligram/minute = 0.01667 milligram/minute
10 milligram/hour to milligram/minute = 0.16667 milligram/minute
20 milligram/hour to milligram/minute = 0.33333 milligram/minute
30 milligram/hour to milligram/minute = 0.5 milligram/minute
40 milligram/hour to milligram/minute = 0.66667 milligram/minute
50 milligram/hour to milligram/minute = 0.83333 milligram/minute
100 milligram/hour to milligram/minute = 1.66667 milligram/minute
200 milligram/hour to milligram/minute = 3.33333 milligram/minute
## Want other units?
You can do the reverse unit conversion from milligram/minute to milligram/hour, or enter any two units below:
## Enter two units to convert
From: To:
## Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 502 | 1,915 | {"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-2023-14 | latest | en | 0.777533 |
http://www.learnquebec.ca/en_US/math-bridge-resources | 1,566,183,053,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314638.49/warc/CC-MAIN-20190819011034-20190819033034-00397.warc.gz | 288,645,417 | 18,996 | This collection of lessons was developed to bridge the learning gap for students who have successfully completed Math CST 4 and would like to pursue the Math Science (SN) stream. These resources include interactive videos, as well as supporting documents that identify and explore topics that are not covered in the CST 4 program. (Additional lessons are in development.)
PLEASE NOTE: No sign-in is required to view these lessons. Simply click the arrow on the lower left-hand side of the slide to play. Use the arrows on the lower right-hand side of the slide to advance the slides.
LEARN Design and Development Team: Author/Lead Teacher: Peggy Drolet, BEd; Editor: Tasha Ausman, PhD; and Instructional Designer: Kristine Thibeault, MEd.
# Factoring 1: Introduction and review of Greatest Common Factors (GCFs)
This video introduces students to the building blocks of polynomials for the Math SN4 option. Geared toward students who already have a solid grasp of polynomials and exponents from Secondary 3, this video reviews Greatest Common Factor (GCF).
Learning Goal:
-To be able to factor polynomials by finding the common factor
Related Textbook Pages:
Visions, Science, Student Book Volume 1, Secondary Cycle Two, Year Two, Les Editions CEC Inc, 2009, pp 148-177
Khan Academy: Factoring Polynomials - Common Factor 1
Khan Academy: Factoring Polynomials - Common Factor 2
# Factoring 2: Factoring by differences of two perfect squares
This video continues the methods of factoring necessary for the Math SN4 option: factoring the difference of two perfect squares.
Learning Goal:
-To be able to factor polynomials by using second degree algebraic identity: the difference of two perfect squares.
Related Textbook Pages:
VisionsScience, Student Book Volume 1, Secondary Cycle TwoYear Two, Les Editions CEC Inc, 2009, pp 148-177
Khan Academy: Factoring Differences of Squares with Two Variables
# Factoring 3: Factoring by grouping
This video continues the methods of factoring necessary for the Math SN4 option: factoring by grouping.
Learning Goal:
-To be able to factor polynomials by grouping terms
Related Textbook Pages:
Visions, Science, Student Book Volume 1, Secondary Cycle Two, Year Two, Les Editions CEC Inc, 2009, pp 148-177
# Factoring 4: Factoring trinomials with a value of 'a’ equal to 1 (“easy” trinomials)
In this lesson, students are introduced to “easy” trinomials. Building upon students’ knowledge of the process of FOIL from Secondary 3, students will learn how to work backwards to find two binomials leading to a simple trinomial with an a value of 1. As well, this lesson reminds students to remove any GCFs first, so that factoring might take place.
Learning Goal:
-To be able to factor second degree trinomials
Related Textbook Pages:
Visions, Science, Student Book Volume 1, Secondary Cycle Two, Year Two, Les Editions CEC Inc, 2009, pp 148-177
# Factoring 5: Factoring trinomials with a value of ‘a’ that is greater than 1
In this lesson, the decomposition method is introduced whereby trinomials in general form with an a value greater than 1 can be factored. The approach is to first find any GCFs and remove them, multiply the first and last terms to find a desired target value, and then determine which combination of integers results in the value of b using addition or subtraction. This video relies on prior knowledge of factoring by grouping.
Learning Goal:
-To be able to factor polynomials that contain decomposable second degree trinomials
Related Textbook Pages:
Visions, Science, Student Book Volume 1, Secondary Cycle Two, Year Two, Les Editions CEC Inc, 2009, pp 148-177
# Long Division: Dividing polynomials by monomials and binomials
The purpose of this video is to learn how to employ the principles of long division, originally taught in elementary school, to polynomials. This lesson first reminds you how to divide a polynomial by a monomial, and then draws upon the steps in long division to divide a binomial into larger polynomials (second and third degree).
Learning Goal:
-To be able to divide a polynomial by another polynomial (with or without a remainder)
Related Textbook Pages:
Visions, Science, Student Book Volume 1, Secondary Cycle Two, Year Two, Les Editions CEC Inc, 2009, pp 148-177
Dividing a Polynomial by a Binomial
# Rational Expressions 1: Simplifying rational expressions
In this introductory lesson to Rational Expressions (REs), students will build upon their knowledge of factoring to, firstly, define rational expressions, and secondly, understand basic simplification. Using the concept of Greatest Common Factor (GCF), several examples are worked through, emphasizing how a common term can be removed, leaving a final answer that cannot be reduced further.
Learning Goal:
-To be able to manipulate rational expressions
Related Textbook Pages:
Visions, Science, Student Book Volume 1, Secondary Cycle Two, Year Two, Les Editions CEC Inc, 2009, pp 148-177
Khan Academy: Intro to Rational Expression Simplification
Khan Academy: Simplifying Rational Expressions: Common Binomial Factors
Khan Academy: Simplifying Rational Expressions: Grouping
MathHelp: Simplifying Rational Expressions
# Rational Expressions 2: Multiplying and dividing rational expressions
Beginning with an introduction on multiplying fractions, this lesson first focuses on factoring out terms in the numerators and denominators of polynomial fractions. Once again, reminding ourselves that common factors can be cancelled top-to-bottom, these are removed, leaving us with a final, non-reducible solution. In division, the same method is employed, but with a poignant reminder that when dividing fractions, including polynomial ones, one must reciprocate the second term.
Learning Goal:
-To be able to manipulate rational expressions
Related Textbook Pages:
Visions, Science, Student Book Volume 1, Secondary Cycle Two, Year Two, Les Editions CEC Inc, 2009, pp 148-177
Khan Academy: Multipying and dividing rational expressions (Video Series)
# Rational Expressions 3: Adding and subtracting rational expressions
In this lesson, we learn about adding and subtracting rational expressions (polynomials in fraction form). Students are reminded about the basics of arithmetic, whereby a common denominator must be found before any two fractions can be added or subtracted. Special emphasis is placed upon distributing the negative (subtraction sign) in the numerator when subtracting rational expressions.
Learning Goal:
-To be able to manipulate rational expressions
Related Textbook Pages:
Visions, Science, Student Book Volume 1, Secondary Cycle Two, Year Two, Les Editions CEC Inc, 2009, pp 148-177
# Rational Expressions 4: Adding and subtracting rational expressions
In this lesson, the steps for adding and subtracting rational expressions are reviewed and emphasized, and several difficult examples are worked through. Try these examples alongside the step-by-step instructions, and check your work as you proceed. If you can do these questions, you have an excellent grasp of adding and subtracting rational expressions.
Learning Goal:
-To be able to manipulate rational expressions
Related Textbook Pages:
Visions, Science, Student Book Volume 1, Secondary Cycle Two, Year Two, Les Editions CEC Inc, 2009, pp 148-177 | 1,670 | 7,354 | {"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} | 4.03125 | 4 | CC-MAIN-2019-35 | latest | en | 0.809271 |
https://www.convertunits.com/from/exabar/to/foot+of+air | 1,620,599,900,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989018.90/warc/CC-MAIN-20210509213453-20210510003453-00605.warc.gz | 717,767,344 | 16,930 | ## ››Convert exabar to foot of air [0 °C]
exabar foot of air
Did you mean to convert exabar to foot of air [0 °C] foot of air [15 °C]
How many exabar in 1 foot of air? The answer is 3.8640888E-23.
We assume you are converting between exabar and foot of air [0 °C].
You can view more details on each measurement unit:
exabar or foot of air
The SI derived unit for pressure is the pascal.
1 pascal is equal to 1.0E-23 exabar, or 0.25879322442072 foot of air.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between exabars and feet of air.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of exabar to foot of air
1 exabar to foot of air = 2.5879322442072E+22 foot of air
2 exabar to foot of air = 5.1758644884144E+22 foot of air
3 exabar to foot of air = 7.7637967326217E+22 foot of air
4 exabar to foot of air = 1.0351728976829E+23 foot of air
5 exabar to foot of air = 1.2939661221036E+23 foot of air
6 exabar to foot of air = 1.5527593465243E+23 foot of air
7 exabar to foot of air = 1.8115525709451E+23 foot of air
8 exabar to foot of air = 2.0703457953658E+23 foot of air
9 exabar to foot of air = 2.3291390197865E+23 foot of air
10 exabar to foot of air = 2.5879322442072E+23 foot of air
## ››Want other units?
You can do the reverse unit conversion from foot of air to exabar, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Exabar
The SI prefix "exa" represents a factor of 1018, or in exponential notation, 1E18.
So 1 exabar = 1018 bars.
The definition of a bar is as follows:
The bar is a measurement unit of pressure, equal to 1,000,000 dynes per square centimetre (baryes), or 100,000 newtons per square metre (pascals). The word bar is of Greek origin, báros meaning weight. Its official symbol is "bar"; the earlier "b" is now deprecated, but still often seen especially as "mb" rather than the proper "mbar" for millibars.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 735 | 2,480 | {"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-2021-21 | latest | en | 0.836627 |
https://meta.m.wikimedia.org/wiki/Abstract_Wikipedia/Updates/2021-05-28/uk | 1,627,923,493,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154321.31/warc/CC-MAIN-20210802141221-20210802171221-00349.warc.gz | 411,731,138 | 12,065 | # Абстрактна Вікіпедія/Оновлення/2021-01-21
This page is a translated version of the page Abstract Wikipedia/Updates/2021-05-28 and the translation is 12% complete.
Other languages:
English • français • українська • 한국어
◀ Оновлення Абстрактної Вікіпедії ▶
Чим не є Вікіфункції.
Раніше ми багато говорили про мету Вікіфункцій: "Проект Вікімедіа для всіх, щоб спільно створити та підтримувати бібліотеку функцій коду для підтримки проектів Вікімедіа, а також, щоб усі могли викликати та повторно використовувати природні мови та мови програмування у світі".
Today, in the tradition of the influential WP:NOT policy on English Wikipedia, we publish an essay on what Wikifunctions aims not to be. WP:NOT was started back in 2001, and was an important influence on the early development of the English Wikipedia — evidenced by the fact that more than 2 million links to that page exist within the English Wikipedia.
Отже, без зайвих передмов — чим не є Вікіфункції:
Wikifunctions is not an encyclopædia of algorithms in the sense that we will have pages for famous and not-so famous algorithms such as Euclid’s, Newton’s, or Dijkstra’s algorithm, aiming to represent all existing algorithms faithfully and in their historical context.
Yes, we expect to have a function for the greatest common divisor (GCD) of two integers. And there might or might not be one or more implementations which are based on Euclid’s algorithm to calculate the GCD. But Wikifunctions would not be incomplete if it didn’t, and if, instead, we had alternative algorithms to calculate the GCD. If you are looking for that, many Wikipedias are actually great resources.
Unlike an encyclopedic overview of existing algorithms, Wikifunctions will also invite original work. We will not be restricted to functions that have been published elsewhere first, and we do not require for every function and implementation to be based on previously published work. Wikifunctions, much like Wikibooks and very unlike Wikipedia, will be open to novel contributions. The main criteria for implementations will be: under which conditions can we run a given implementation, and what resources is it expected to take?
Wikifunctions is not an app development site.
We do not expect to make it possible to create full-fledged, stand-alone apps within Wikifunctions — there will be no place to store state, we don’t aim to allow calling external APIs or directly cause changes to other sites, and we don’t aim to package up apps with icons and UX, etc. We absolutely expect Wikifunctions to be a very useful resource for app developers, and I can very much imagine apps that are basically wrappers around one or more functions from Wikifunctions, but these would still need code and other assets which wouldn’t be part of Wikifunctions. We are not competing in the area of no-code or low-code development sites.
Wikifunctions is not a code hosting service.
Yes, sure, Wikifunctions will host code, but not for whole projects, merely for individual functions. There won’t be libraries, apps, or services developed on Wikifunctions with bug-trackers, forums, etc. There won’t be a Web-based version control system such as mercurial or git running against Wikifunctions. Again, we hope that there will be libraries, apps, and services that will rely on functions available in Wikifunctions, but they would be developed on a different site, such as Gerrit, GitHub, or GitLab.
Wikifunctions is not a programming language, nor trying to evangelise a particular language.
In fact, Wikifunctions will allow for functions to be implemented in a multitude of programming languages. The possibility to compose functions together to create higher level functions may look a little bit like a new programming language, but it will be extremely limited compared to most other programming languages, since we only allow for nested function calls and that’s it.
Wikifunctions is not an Integrated Development Environment (IDE).
We won't provide you with an interface for creating and developing software projects, interfacing with build, testing, and source control systems.
Wikifunctions is not a question-and-answer website.
We are not competing with StackOverflow and similar websites, where a developer would ask how to achieve a certain task and have community members discuss and answer the question. We won’t contain code snippets to help answer the question, but we will organize code within our website to enable the evaluation of functions within a library of functions.
Wikifunctions is not a cloud computing platform.
We do not provide computing resources and access to services and APIs so that you can run your computational needs on our platform, either for money or for free. Use of Wikifunctions's evaluation platform is to improve access to knowledge for everyone.
Wikifunctions is not a code snippet website.
We are not competing with sites such as gist, or sites such as rosettacode.org, esolangs.org, or helloworldcollection.de, where code snippets are collected either to share them quickly with others or around a specific theme in different programming languages. The reason for having functions be implemented in multiple programming languages is not to contrast them and compare them for the education of the users of Wikifunctions, but in order to be able to efficiently and effectively evaluate functions in different environments and to improve the reliability of Wikifunctions as a whole.
Wikifunctions is not a code education platform.
We are not in the business of teaching people how to code, the material in Wikifunctions will not be laid out in a pedagogical order, and we also won’t make sure to comprehensively cover all topics important for coding. In fact, we aim for Wikifunctions to be usable for people who don’t know how to code and who don’t need to learn how to code to use most of Wikifunctions effectively. Though the Wikifunctions community may well help each other in sharing best practices, style guides, and tips on how to use the site in different languages, these will be aimed at the purpose of serving the world's knowledge.
Wikifunctions is, as far as we can tell, a new kind of website, aiming for a new community. We very much hope to work together with many of the tools, sites, communities, and kind of systems we have mentioned above: we want to play together with IDEs, with cloud computing platforms, with app development sites, and many more of the systems and tools we mentioned. But we aim to be a novel thing and we hope to shape a new unique space out for us: "Проект Вікімедіа для всіх, щоб спільно створити та підтримувати бібліотеку функцій коду для підтримки проектів Вікімедіа, а також, щоб усі могли викликати та повторно використовувати природні мови та мови програмування у світі".
In related news, the video recording of the keynote about Abstract Wikipedia and Wikifunctions at this year’s Web Conference is now available online:
http://videolectures.net/www2021_vrandecic_knowledge_equity/ | 1,565 | 7,030 | {"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} | 2.90625 | 3 | CC-MAIN-2021-31 | latest | en | 0.536824 |
https://www.unitconverters.net/volume/milliliter-to-teaspoon-metric.htm | 1,721,844,788,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518427.68/warc/CC-MAIN-20240724171328-20240724201328-00812.warc.gz | 889,666,203 | 3,903 | Home / Volume Conversion / Convert Milliliter to Teaspoon (metric)
# Convert Milliliter to Teaspoon (metric)
Please provide values below to convert milliliter [mL] to teaspoon (metric), or vice versa.
From: milliliter To: teaspoon (metric)
### Milliliter to Teaspoon (metric) Conversion Table
Milliliter [mL]Teaspoon (metric)
0.01 mL0.002 teaspoon (metric)
0.1 mL0.02 teaspoon (metric)
1 mL0.2 teaspoon (metric)
2 mL0.4 teaspoon (metric)
3 mL0.6 teaspoon (metric)
5 mL1 teaspoon (metric)
10 mL2 teaspoon (metric)
20 mL4 teaspoon (metric)
50 mL10 teaspoon (metric)
100 mL20 teaspoon (metric)
1000 mL200 teaspoon (metric)
### How to Convert Milliliter to Teaspoon (metric)
1 mL = 0.2 teaspoon (metric)
1 teaspoon (metric) = 5 mL
Example: convert 15 mL to teaspoon (metric):
15 mL = 15 × 0.2 teaspoon (metric) = 3 teaspoon (metric) | 248 | 837 | {"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} | 2.796875 | 3 | CC-MAIN-2024-30 | latest | en | 0.363757 |
https://starfishconfidential.com/toddlers/your-question-how-much-breastmilk-should-a-13-pound-baby-eat.html | 1,627,411,717,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153474.19/warc/CC-MAIN-20210727170836-20210727200836-00115.warc.gz | 543,142,951 | 19,812 | # Your question: How much breastmilk should a 13 pound baby eat?
Contents
Baby weight (lbs) Breast milk needed (oz)
10 lbs 24 oz
11 lbs 26 oz
12 lbs 29 oz
13 lbs 31 oz
## How much milk should a 13 pound baby drink?
According to the American Academy of Pediatrics (AAP), a baby should consume, on average, about 2.5 ounces of formula a day for every pound of their body weight. For example, a 12-pound baby would likely need 30 ounces in a 24-hour period.
By Weight.
Baby Weight in Pounds Ounces of Formula per Day
12 30
13 32.5
## How many ounces of breastmilk should a baby eat chart?
Formula Feeding Amounts by Age
Age # of feedings per day / 24 hours Average Bottle Size
0-4 weeks on-demand ~2-4 ounces / 60-120 ml
5-8 weeks 6-7 ~4 ounces / 120 ml
9-12 weeks/3 months 5 4-6 ounces / 120-180 ml
13-16 weeks/4 months 5 4-6 ounces / 120-180 ml
IT IS INTERESTING: You asked: Is 9 months considered a toddler?
## How much breastmilk should a 12 pound baby eat?
Amount of Breast Milk Needed By Baby as per their Weight
Baby weight (lbs) Breast milk needed (oz) Breast milk needed (ml)
11 lbs 26 oz 782 ml
12 lbs 29 oz 861 ml
13 lbs 31 oz 939 ml
14 lbs 34 oz 1000 ml
## How many ounces of breastmilk should a 15 pound baby eat?
Baby’s Weight Oz’s per Feeding (8 per day)
14 lbs 6,400 gr 4.7 oz
14 lbs 8 oz 6,704 gr 4.8 oz
15 lbs 6,795 gr 4.9 oz
15 lbs 8 oz 7,021 gr 5.0 oz
## How do I know if I’m overfeeding my baby?
Signs of overfeeding
1. Baby gains average or greater than average weight.
2. Eight or more heavily wet nappies per day.
3. Frequent sloppy, foul-smelling bowel motions.
4. Extreme flatulence.
5. Large belching.
6. Milk regurgitation.
7. Irritability.
8. Sleep disturbance.
4 июл. 2017 г.
## How much does a 3 month old weigh?
Parents want to know: How much should a 3-month-old weigh and measure? The average weight of a 3-month-old baby is 12.9 pounds for girls and 14.1 pounds for boys; average length is 23.5 inches for girls and 24.2 inches for boys.
## Is pumped breast milk still good for baby?
Babies who feed exclusively on pumped milk do not get the benefit of a feedback loop between their body and the breast milk. However, they do still gain access to a well-designed food that is rich in healthful fats and antibodies.
## How many ounces of breastmilk should a 5 month old eat?
Typically, a baby needs about 25 ounces of breast milk per day. So you’ll need to divide that by how many feedings your baby typically has. So if you feed baby about eight times per day, he should get about 3.1 ounces of breast milk at each feeding. That’s how much milk a 5-month-old should drink.
IT IS INTERESTING: Question: Do babies drink the same amount of formula as breast milk?
## Is breastmilk more filling than formula?
Simply put, yes, formula can be more filling. The answer is not what you would imagine. The reason why baby formulas are more filling than breastmilk is because babies can drink MORE of formulas. … Give them formula second, so they can still receive all the antibodies from the breastmilk and get filled up on the formula.
## How do I know my baby is getting enough milk?
If your baby isn’t getting enough milk, you may see him sucking rapidly but not swallowing slowly and rhythmically; he may also take long pauses while nursing or repeatedly fall asleep at your breast. 2. He’s satisfied. If your baby seems content and well fed after feeding sessions, all is likely going well.
## Can a 2 week old baby eat 4 oz?
During the first 2 weeks, babies will eat on average 1 – 2 oz at a time. By the end of the first month they eat about 4 oz at a time. By 2 months, increase to 6 oz per feed, and by 4 months, about 6-8 oz per feed.
## How much milk can a breast hold?
Studies show some women have as few as 3 milk lobules/ducts and others as many as 15. As a result the amount of milk that can fit in a woman’s breasts varies – anywhere from 2oz to 5oz combined is average but some women can store as much as 10 oz in one breast (this is very unusual).
## Is 4 oz every 3 hours too much for a newborn?
On average, a newborn drinks about 1.5-3 ounces (45-90 milliliters) every 2-3 hours. This amount increases as your baby grows and is able to take more at each feeding. At about 2 months, your baby may be taking 4-5 ounces (120-150 milliliters) at each feeding and the feedings may be every 3-4 hours.
IT IS INTERESTING: What happens if you get caught without a child seat?
## Is 4 oz of breastmilk too much for a newborn?
This is compared to a breastfed baby, who will usually eat every two to three hours. By the time your baby is 1 month old, they should be eating around 4 ounces every four hours.
## Is 5 oz of breastmilk too much for a newborn?
Most babies will eat 2-3.5 oz per feed. If your baby is eating more than 5 oz per feed, they are most likely eating too much at a time. | 1,328 | 4,865 | {"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-2021-31 | latest | en | 0.863188 |
http://book.caltech.edu/bookforum/showthread.php?s=2b74ea7df75df7ede5dc308aca552803&p=12191&mode=threaded | 1,571,363,149,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677412.35/warc/CC-MAIN-20191018005539-20191018033039-00510.warc.gz | 32,739,523 | 9,383 | LFD Book Forum Lecture 7: Shattering for d+2 points
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#1
11-01-2015, 07:02 PM
sandeeps Junior Member Join Date: Sep 2015 Posts: 5
Lecture 7: Shattering for d+2 points
In the argument,
After we arrive at the result that
but also state that
it is concluded that H does not shatter the given d+2 points. (I understood this).
However you then stated:
"You cannot shatter your set for any set you choose. Therefore you cannot shatter any set of d+2 points"
How did we go from not shattering a given set of d+2 points to not being to shatter any set of d+2 points? | 168 | 604 | {"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} | 2.640625 | 3 | CC-MAIN-2019-43 | latest | en | 0.927184 |
http://math.stackexchange.com/questions/62258/does-continuous-only-at-countable-many-points-not-differentiable-with-bounde | 1,469,365,634,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824037.46/warc/CC-MAIN-20160723071024-00289-ip-10-185-27-174.ec2.internal.warc.gz | 161,261,103 | 19,682 | # Does “continuous + only at countable many points not differentiable (with bounded derivative)” imply Lipschitz-continuity?
Let $f$ be continuous on $\mathbb R$ and differentiable with derivative $f'$ on $\mathbb R \setminus \{t_0, t_1, \dots \}$. Let $\sup | f'(t) | < \infty$, then $f$ is Lipschitz continuous with $L=\sup |f'(t)|$.
Does this hold? How could one prove it?
-
@Theo Oh. You're right, so I should rephrase like "at only countable points not differentiable", it's a bit due to translating to English while writing.. And thanks for correcting my title – Johannes L Sep 6 '11 at 11:01
You should also change the definition of L to $L = \sup |f'(t)|$. – Ragib Zaman Sep 6 '11 at 11:03
@all: I cast a vote to close this question because of my misreading it. Please ignore that vote. Johannes: sorry about that. – t.b. Sep 6 '11 at 11:19
@Joh The title still says differentiable at only countably many points. – Srivatsan Sep 6 '11 at 12:31
@Srivatsan I want to expres that f is differentiable on $\mathbb{R} \setminus \{ t_0, t_1, \dots\}$ where $\{t_0, t_1, \dots \}$ is countable, so this should be okay. Before it said "almost everywhere differentiable" which Theo pointed out to not to be what I meant. – Johannes L Sep 6 '11 at 12:44
Since $-L \leq f'(x) \leq L$ except on a set of only measure zero, we may integrate this between $x_1$ and $x_2$ and the desired $-L(x_2-x_1) \leq f(x_2) - f(x_1) \leq L(x_2-x_1)$ pops right out.
Note to all: The following is what was my intial answer, but it is faulty.
First do the problem for each segment that the function is differentiable (so $(-\infty,t_0), (t_0,t_1)$ etc): By the mean value theorem, for $x_1 , x_2 \in (t_k, t_{k+1})$ we get $$\frac{f(x_1)-f(x_2)}{x_1-x_2} = f'(c)$$ for some $c\in (x_1,x_2)$.
This means for any $x_1,x_2 \in \mathbb{R}$ we have in $(t_k, t_{k+1})$ that $$|f(x_1)-f(x_2) | \leq |f'(c)||x_1-x_2| \leq L|x_1-x_2|.$$
which makes it Lipschitz continuous in each segment with Lipschitz constant $L$. Can you see how to prove that if we stitch together Lipschitz continuous functions like this, it remains so?
-
From the notion I think I can really imagine it, $|\frac{f(x_1)-f(x_2)}{x_1-x_2}|$ as the slope of a secant, and this can be at tops as big as the maximum of the derivates between $x_1$ and $x_2$, and the non-differentiable points $t_0, t_1, \dots$ don't do any harm because with the continuity the function doesn't "rise" (or fall)... But in the moment I don't manage to make a straight argument out of this idea * ($x_1, x_2$ in different intervals) – Johannes L Sep 6 '11 at 11:19
Wonderful! This is such a nice and clean argument, it's perfect. (( and now I could even go back to "almost everywhere, because for this proof we actually use $f'$ almost everywhere defined.)) So, if $f$ has almost everywhere defined and bounded derivate the statement also holds. I think the continuity is still needed, when asserting $\int_{x_1}^{x_2} f'(x)=f(x_2)-f(x_1)$ when $x_2=t_k$ for example. – Johannes L Sep 6 '11 at 12:54
@Johannes L, I thought of that as well, and it worried me as my entire argument fails if the fundamental theorem of calculus required continuity. Luckily, a less well known version does not require it: See en.wikipedia.org/wiki/… – Ragib Zaman Sep 6 '11 at 13:23
@Ragib : There is another problem with your first argument, when the set $\{t_0, t_1, \dots \}$ is dense in $\mathbb{R}$. The second argument (proof) got rid of this. – Rajesh Dachiraju Sep 6 '11 at 13:26
@all I'm unsure of the etiquette in this situation? It turns out the first part of my answer is quite useless, but the 2nd proof is fine. Should I just delete the first part? – Ragib Zaman Sep 6 '11 at 13:28
I try a correct fix of @Ragib's answer, then we can continue to discuss here
Let $A=\{t_0, t_1, \dots \}$. Define
$$\tilde{f}'(x) = \begin{cases} 0 & x \in A\\ f'(x) & \text{else} \end{cases}$$ $A$ has measure 0 (in other words $f$ almost everywhere differentiable). Then $\sup|\tilde{f}'(t)| \equiv \sup |f(t)| =L$.
$\tilde{f}'(t)$ has an anti-derivative $\forall x \in R$ and is Lebesgue-integrable, hence with the Fundamental Theorem of Analysis for Lebesgue-integral: $\int_{x_1}^{x_2} \tilde{f}'(x)=f(x_2)-f(x_1)$
In Wikipedia under generalizations it says "Part II of the theorem is true for any Lebesgue integrable function ƒ which has an antiderivative F (not all integrable functions do, though)."
- | 1,439 | 4,418 | {"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} | 3.578125 | 4 | CC-MAIN-2016-30 | latest | en | 0.880301 |
http://www.numbersaplenty.com/138900 | 1,586,184,697,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371637684.76/warc/CC-MAIN-20200406133533-20200406164033-00518.warc.gz | 274,099,473 | 3,514 | Search a number
138900 = 22352463
BaseRepresentation
bin100001111010010100
321001112110
4201322110
513421100
62551020
71115646
oct417224
9231473
10138900
11953a3
1268470
134b2b8
1438896
152b250
hex21e94
138900 has 36 divisors (see below), whose sum is σ = 402752. Its totient is φ = 36960.
The previous prime is 138899. The next prime is 138917. The reversal of 138900 is 9831.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 11 ways as a sum of consecutive naturals, for example, 69 + ... + 531.
2138900 is an apocalyptic number.
138900 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 138900, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (201376).
138900 is an abundant number, since it is smaller than the sum of its proper divisors (263852).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
138900 is a wasteful number, since it uses less digits than its factorization.
138900 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 480 (or 473 counting only the distinct ones).
The product of its (nonzero) digits is 216, while the sum is 21.
The square root of 138900 is about 372.6929030717. The cubic root of 138900 is about 51.7885893880.
The spelling of 138900 in words is "one hundred thirty-eight thousand, nine hundred". | 454 | 1,603 | {"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} | 3.03125 | 3 | CC-MAIN-2020-16 | latest | en | 0.898317 |
https://www.devx.com/terms/fuzzy-logic/ | 1,718,288,036,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00278.warc.gz | 679,412,373 | 50,486 | # Fuzzy Logic
## Definition
Fuzzy Logic is a computing approach based on “degrees of truth” rather than the usual true or false binary logic (0 or 1) on which modern computers are based. This system is designed to mimic human decision making processes and deal with problems that are uncertain or vague. In fuzzy logic, a statement can be both true and false at the same time, to varying degrees.
## Key Takeaways
1. Fuzziness: Unlike traditional boolean logic that deals in absolutes of ‘true’ or ‘false’, Fuzzy Logic deals with levels of truth. These are represented in a continuum of values ranging between absolute truth and absolute falsehood, enabling the modeling of complex systems with a high level of ambiguity or uncertainty.
2. Flexibility: Fuzzy Logic allows for the representation of human cognitive processes in a more flexible, intuitive manner. It mimics human reasoning by incorporating a degree of uncertainty, and can effectively handle complex models and operations that cannot easily be defined by traditional mathematical or logical constructs.
3. Applications: Fuzzy Logic is widely used in numerous applications such as control systems, artificial intelligence, decision-making processes, pattern recognition, and data analysis due to its ability to handle imprecise and inaccurate data.
## Importance
Fuzzy Logic is an essential concept in technology because it provides a method to process complex and ambiguous problems that traditional binary logic cannot address adequately.
As opposed to binary logic’s precise ‘true’ or ‘false’ outcomes, fuzzy logic introduces the concept of partial truth, reflecting how people make decisions in real life, which often involve factors that are somewhat true or partially false. This logic offers an invaluable tool in areas such as artificial intelligence, automation, control systems, and decision-making processes, allowing these systems to behave more like humans in their problem-solving approaches.
The importance of fuzzy logic lies in its ability to enhance the sophistication, accuracy, and efficiency of computing systems, leading to more effective technological applications.
## Explanation
Fuzzy Logic is essentially a system of rules that allows us to make decisions based on ‘degrees of truth’ rather than the traditional binary ‘true or false’ (1 or 0) used in standard logic. It fills the gap where ambiguity or vagueness is present, effectively providing an algorithm to handle subjective and uncertain scenarios.
This technology serves to mimic human reasoning and decision-making, as it operates on if-then logic, similarly to how an individual processes thoughts. It’s named ‘fuzzy’ to represent the concept of partial truth, where something can be both true and false to an extent simultaneously.In terms of its uses, Fuzzy Logic is immensely valuable in various industries and applications. It’s widely applied in control systems such as household appliances (like washing machines that adjust their cycle lengths and water consumption based on load weight and dirtiness), automated motor controls, and weather forecasting systems.
In artificial intelligence, fuzzy logic aids in making the system more human-like in processing data and making decisions. Its application can also be found in medical instrumentation and diagnostics, where it assists in interpreting uncertain human responses to medical questions and generating accurate diagnoses. Essentially, anywhere there is ambiguity or uncertainty in data, fuzzy logic can be utilized to draw meaningful and practical conclusions.
## Examples
1. Washing Machines: Modern washing machines use fuzzy logic control to automatically adjust settings like wash time, water level, and water temperature based on the load of laundry (weight and dirtiness). This not only reduces manual intervention but also efficiently uses water and electricity.2. Automated Vehicle Control Systems: To improve passenger safety, fuzzy logic is implemented in the vehicle control systems. For instance, anti-lock braking systems (ABS) use fuzzy logic to prevent wheels from locking up or skidding when braking. The system continuously monitors wheel speed and controls the brake fluid pressure, thus ensuring optimal braking.3. Smart Home Systems: Fuzzy logic is widely used in home automation systems. For example, it can control HVAC (Heating, Ventilation, and Air Conditioning) systems by monitoring and adjusting the indoor temperature based on different factors like outdoor weather conditions and the number of people present inside the room to maintain optimal comfort and energy efficiency.
### Q: What is Fuzzy Logic?
A: Fuzzy Logic is a mathematical approach to problem-solving that deals with uncertainties by simulating human decision-making process. It acknowledges that things can be partially true or false, and not just absolutely true or false, mimicking the way human beings make decisions.
### Q: Where is Fuzzy Logic used?
A: Fuzzy Logic is widely used in various industries and applications such as automotive systems (like in automatic gearboxes), household appliances (like washing machines, microwaves), environmental control systems, stock market analysis, medical diagnosis systems and much more.
### Q: Who developed Fuzzy Logic?
A: Fuzzy Logic was proposed by Lotfi Zadeh, a professor at the University of California, in the mid 1960s.
### Q: How does Fuzzy Logic differ from traditional Boolean Logic?
A: Unlike Boolean Logic which only allows values of 0 (false) and 1 (true), Fuzzy Logic allows values in between 0 and 1, representing degrees of truth. This provides a way to deal with uncertainties and vagueness present in real-world problems.
### Q: What is a Fuzzy Logic system?
A: A Fuzzy Logic system is a control system that uses Fuzzy Logic to analyze input and make decisions, often used in appliances, vehicles, and other systems. It mimics human control intuition and can handle imprecise inputs.
### Q: What are the advantages of Fuzzy Logic?
A: Fuzzy Logic allows for flexible input, has a simple design, and can model complex systems. As it is built to mimic human decision making, it can handle unanticipated events and unknowns better than some other mathematical approaches. It’s also relatively easy to adjust and optimize.
### Q: Are there any disadvantages to Fuzzy Logic?
A: Some disadvantages include the fact that Fuzzy Logic sometimes can be too general and can overcomplicate solutions that could have been solved using simpler logic or empirical models. Also, there might be a challenge to use fuzzy logic when precision and exactitude are highly required.
### Q: Is Fuzzy Logic used in AI (Artificial Intelligence)?
A: Yes, Fuzzy Logic is used in AI due to its ability to mimic human decision-making, which makes it useful in building intelligent systems. It allows AI systems to reason in a way that is not just black and white but also considers the grey areas.
## Related Finance Terms
• Fuzzy Sets
• Fuzzy Systems
• Fuzzy Control Systems
• Fuzzy Inference System
• Fuzzy Logic Algorithms
The DevX Technology Glossary is reviewed by technology experts and writers from our community. Terms and definitions continue to go under updates to stay relevant and up-to-date. These experts help us maintain the almost 10,000+ technology terms on DevX. Our reviewers have a strong technical background in software development, engineering, and startup businesses. They are experts with real-world experience working in the tech industry and academia.
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See our full editorial policy. | 1,522 | 7,957 | {"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} | 2.703125 | 3 | CC-MAIN-2024-26 | latest | en | 0.930678 |
https://www.enotes.com/homework-help/solve-x-16-2-7-x-2-22-2x-5-2-245721 | 1,484,571,233,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279176.20/warc/CC-MAIN-20170116095119-00426-ip-10-171-10-70.ec2.internal.warc.gz | 899,492,925 | 12,451 | # Solve for x: 16^2 + (7 - x)^2 = 22- (2x/5)^2
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
The equation you gave is : 16^2 + (7=x)^2 = (22- (2x/5)^2
I think it should be 16^2 + (7 - x)^2 = 22 - (2x/5)^2, the appropriate changes have been made.
16^2 + (7 - x)^2 = 22 - (2x/5)^2
=> 16^2 + 49 + x^2 - 14x = 22 - 4x^2 / 25
=> (305 + x^2 - 14x)* 25 = 22*25 - 4x^2
=> 305*25 + 25x^2 - 350x = 550 - 4x^2
=> 7625 + 25x^2 - 350x = 550 - 4x^2
=> 29x^2 - 350x + 7075 = 0
x1 = [-b + sqrt (b^2 - 4ac)] / 2a
=> x1 = [ 350 + sqrt ( 350^2 - 820700)]/58
=> x1 = [ 350 + sqrt ( -698200)]/58
=> x1 = 350 / 58 + i* (10/58)*sqrt 6982
x2 = 350 / 58 - i* (10/58)*sqrt 6982
The required solution of x is 350 / 58 + i* (10/58)*sqrt 6982 and 350 / 58 - i* (10/58)*sqrt 6982 | 400 | 793 | {"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-2017-04 | longest | en | 0.614137 |
https://de.mathworks.com/help/sps/powersys/ug/simulating-transients.html | 1,675,909,872,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501066.53/warc/CC-MAIN-20230209014102-20230209044102-00498.warc.gz | 219,187,981 | 19,502 | ## Simulating Transients
### Introduction
In this section, you
• Learn how to create an electrical subsystem
• Simulate transients with a circuit breaker
• Compare time domain simulation results with different line models
• Learn how to discretize a circuit and compare results thus obtained with results from a continuous, variable time step algorithm
### Simulating Transients with a Circuit Breaker
One of the main uses of Simscape™ Electrical™ Specialized Power Systems software is to simulate transients in electrical circuits. This can be done with either mechanical switches (circuit breakers) or switches using power electronic devices.
1. Open the power_gui example model.
2. To simulate a line energization, add a Breaker block from the Simscape > Electrical > Specialized Power Systems > Passives library to the model and set the parameters as follows:.
Breaker resistance Ron (Ohm) `0.001 `Ω Initial status `0 (open)` Snubber resistance Rs (Ohm) `inf` Snubber capacitance Cs (F) `0` Switching times `[(1/60)/4]`
3. Insert the Breaker block in series with the sending end of the line, then rearrange the circuit as shown in the figure.
4. Right-click the two lines that connect to the Scope 1 block and select Properties. In the dialog box, select Log signal data for the two signals and click . From the Simulation Data Inspector, select Send Logged Workspace Data to Data Inspector.
5. Open the 150 km Line block dialog box and set the number of sections to `1`. Start the simulation.
6. Open the 150 km Line block dialog box and change the number of sections from `1` to `10`. Start the simulation.
7. Add a Distributed Parameters Line block from the Simscape > Electrical > Specialized Power Systems > Passives library. Set the number of phases to `1` and use the same R, L, C, and length parameters as the 150 km Line block. Delete the 150 km Line block and replace it with the Distributed Parameters Line block. Start the simulation.
8. Compare the three waveforms obtained from the three line models. Open the Simulation Data Inspector. Select the Ub1 and Ub2 signals of Run 1, the Ub2 signals of Run 2, and Run 3.
A zoom on these waveforms is shown in the next figure. As expected from the frequency analysis performed during Analyze a Simple Circuit, the single PI model does not respond to frequencies higher than 229 Hz. The 10 PI section model gives better accuracy, although high-frequency oscillations are introduced by the discretization of the line. You can clearly see the propagation time delay of 1.03 ms associated with the distributed parameter line.
Receiving End Voltage Obtained with Three Different Line Models
### Discretizing the Electrical System
An important product feature is its ability to simulate either with continuous, variable step integration algorithms or with discrete solvers. For small systems, variable time step algorithms are usually faster than fixed step methods, because the number of integration steps is lower. For large systems that contain many states or many nonlinear blocks such as power electronic switches, however, it is advantageous to discretize the electrical system.
When you discretize your system, the precision of the simulation is controlled by the time step. If you use too large a time step, the precision might not be sufficient. The only way to know if it is acceptable is to repeat the simulation with different time steps and find a compromise for the largest acceptable time step. Usually time steps of 20 µs to 50 µs give good results for simulation of switching transients on 50-Hz or 60-Hz power systems or on systems using line-commutated power electronic devices such as diodes and thyristors. You must reduce the time step for systems using forced-commutated power electronic switches. These devices, the insulated-gate bipolar transistor (IGBT), the field-effect transistor (FET), and the gate-turnoff thyristor (GTO) are operating at high switching frequencies.
For example, simulating a pulse-width-modulated (PWM) inverter operating at 8 kHz would require a time step of at most 1 µs.
1. Open the power_3level example. In the Powergui block, note that the Simulation type is set to `Discrete` and the sample time is set to 1e-6 s by the Ts variable defined in the Model Properties. Run a first simulation.
2. To perform a continuous simulation, open the Powergui block parameters dialog box and set Simulation type to `Continuous`. Select the ```ode23tb variable-step solver``` in the Configuration Parameters dialog box. Simulate the model.
3. In the Powergui block, set Simulation type to `Discrete`. In the InitFcn model callback section of the Model Properties, specify `Ts = 20e-6`. Run the simulation.
4. Open the Simulation Data Inspector and compare the differences on the high-frequency transients.
The 1 µs compares reasonably well with the continuous simulation. However, increasing the time step to 20 µs produces appreciable errors. The 1 µs time step would therefore be acceptable for this circuit, while obtaining a gain on simulation speed. | 1,092 | 5,079 | {"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-2023-06 | latest | en | 0.78819 |
https://developerpublish.com/xnpv-function-in-excel/ | 1,675,139,607,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499842.81/warc/CC-MAIN-20230131023947-20230131053947-00021.warc.gz | 224,462,188 | 33,369 | # XNPV Function in Excel
In this article, you will learn about the XNPV function, the formula syntax and usage of the function in Microsoft Excel
## XNPV Function in Excel
The XNPV function in Excel is a financial function. It calculates the net present value (NPV) of an investment using a discount rate and a series of cash flows that occur at irregular intervals.
#### Syntax
=XNPV (rate, values, dates)
Arguments
• rate – Discount rate to apply to the cash flows.
• values – Values representing cash flows.
• dates – Dates that correspond to cash flows.
#### Usage Notes and Possible Errors
• As we know Excel stores dates as sequential series of numbers, i.e., January 1, 1900 is serial number 1, and January 1, 2008 is serial number 39448 as it is 39,448 days after January 1, 1900.
• Numbers in dates are truncated to integers.
• When a nonnumeric argument is entered, XNPV returns the #VALUE! error value.
• In case of the date is not a valid date, XNPV returns the #VALUE! error value.
• If the given date precedes the starting date, XNPV returns the #NUM! error value.
• If values and dates contain a different number of values, XNPV returns the #NUM! error value.
• XNPV is calculated as follows:
where:
• di = the ith, or last, payment date.
• d1 = the 0th payment date.
• Pi = the ith, or last, payment.
## How to use the XNPV function in Excel?
Using this function in a WS is simple; all you need to do is enter the function as a formula of the cell in the formula bar.
Take a look at the given example
To find out the net present values, Enter the given values in column A, enter the dates in Column B, and in Column C, enter the following formula
Formula: =XNPV (.15, A2:A6, B2:B6)
Here, the rate of discount is given as 15% or 0.15 and A2 refers to the cell name or the cell address. | 472 | 1,817 | {"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} | 3.515625 | 4 | CC-MAIN-2023-06 | longest | en | 0.809926 |
https://physics-network.org/what-are-the-concepts-of-fluid-statics/ | 1,669,661,326,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710534.53/warc/CC-MAIN-20221128171516-20221128201516-00862.warc.gz | 489,601,099 | 17,418 | # What are the concepts of fluid statics?
Fluid statics is the branch of fluid mechanics that studies incompressible fluids at rest. It encompasses the study of the conditions under which fluids are at rest in stable equilibrium as opposed to fluid dynamics, the study of fluids in motion.
## What are the 3 fluid principles?
The basic fluid mechanics principles are the continuity equation (i.e. conservation of mass), the momentum principle (or conservation of momentum) and the energy equation.
## What is the concept of fluid mechanics?
Fluid mechanics is the study of fluid behavior (liquids, gases, blood, and plasmas) at rest and in motion. Fluid mechanics has a wide range of applications in mechanical and chemical engineering, in biological systems, and in astrophysics.
## What is a fluids in physics?
In physics, a fluid is a liquid, gas, or other material that continuously deforms (flows) under an applied shear stress, or external force. They have zero shear modulus, or, in simpler terms, are substances which cannot resist any shear force applied to them.
## What are the two main classes of fluids?
Viscous or Non-viscous Flow Liquid flow can be viscous or non-viscous. Viscosity is a measure of the thickness of a fluid, and very gloppy fluids such as motor oil or shampoo are called viscous fluids.
## What is Newton law of viscosity?
Newton’s law of viscosity defines the relationship between the shear stress and shear rate of a fluid subjected to a mechanical stress. The ratio of shear stress to shear rate is a constant, for a given temperature and pressure, and is defined as the viscosity or coefficient of viscosity.
## Why is fluid dynamics so hard?
Fluid mechanics is difficult indeed. The primary reason is there seems to be more exceptions than rules. This subject evolves from observing behaviour of fluids and trying to put them in the context of mathematical formulation. Many phenomena are still not accurately explained.
## What is Bernoulli’s equation in fluid mechanics?
Bernoulli’s equation formula is a relation between pressure, kinetic energy, and gravitational potential energy of a fluid in a container. The formula for Bernoulli’s principle is given as follows: p + 1 2 ρ v 2 + ρ g h = c o n s t a n t.
## Why is it important to study fluid mechanics?
Fluid mechanics helps us understand the behavior of fluid under various forces and at different atmospheric conditions, and to select the proper fluid for various applications. This field is studied in detail within Civil Engineering and also to great extent in Mechanical Engineering and Chemical Engineering.
## What are the different types of fluids?
• Ideal fluid.
• Real fluid.
• Newtonian fluid.
• Non-Newtonian fluid.
• Ideal plastic fluid.
• Incompressible fluid.
• Compressible fluid.
## How many types of fluid are classified?
Fluids are separated in five basic types: Real Fluid. Newtonian Fluid. Non-Newtonian Fluid. Ideal Plastic Fluid.
## What are the 5 main characteristics of fluids?
All fluids, whether liquid or gas, have the same five properties: compressibility, pressure, buoyancy, viscosity, and surface tension. If a fluid is not compressible and it has zero viscosity it is considered an ideal fluid.
## What are 2 characteristics of all fluids?
• COMPRESSIBILITY. Compressibility is one of the characteristics where gases and liquids vary.
• SHAPE AND VOLUME. Unlike solids, fluids take the shape of the container they are stored in.
• SHEAR RESISTANCE.
• VISCOSITY.
• MOLECULAR SPACING.
## What are 3 characteristics of a liquid?
Liquids have the following characteristics: No definite shape (takes the shape of its container). Has definite volume. Particles are free to move over each other, but are still attracted to each other.
## What are 5 examples of fluids?
• Water.
• Air.
• Blood.
• Mercury.
• Honey.
• Gasoline.
• Any other gas or liquid.
## What is difference between Newtonian and non Newtonian fluid?
Newtonian fluids are those that obey Newton’s law of constant viscosity. These fluids have constant viscosity and zero shear rate at shear stress. Non-Newtonian fluids are fluids that do not have constant viscosity and have a variable relationship with shear stress.
## What is the SI unit of viscosity?
What is the unit of viscosity? The unit of viscosity is newton-second per square metre, which is usually expressed as pascal-second in SI units.
## Is blood a Newtonian fluid?
While the plasma is essentially a Newtonian fluid, the blood as a whole behaves as a non-Newtonian fluid showing all signs of non- Newtonian rheology which includes deformation rate dependency, viscoelasticity, yield stress and thixotropy.
## What is shear rate in fluids?
The shear rate is defined as the gradient in velocity, that is, the difference in velocity between the two surfaces containing the fluid, divided by the distance between them.
## What is the hardest physics formula?
Yet only one set of equations is considered so mathematically challenging that it’s been chosen as one of seven “Millennium Prize Problems” endowed by the Clay Mathematics Institute with a \$1 million reward: the Navier-Stokes equations, which describe how fluids flow.
## Which is easier fluid mechanics or thermodynamics?
Which is harder, fluid mechanics or thermodynamics? Fluid mechanics, by orders of magnitude.
## What is the most complicated physics equation?
The longest math equation contains around 200 terabytes of text called the Boolean Pythagorean Triples problem. It was first proposed by California-based mathematician Ronald Graham, back in the 1980s.
## What does Pascal’s principle state?
Pascal’s law states that when there is an increase in pressure at any point in a confined fluid, there is an equal increase at every other point in the container.
## What is kinetic energy in fluid flow?
In fluid dynamics, the kinetic energy per unit volume at each point in an incompressible fluid flow field is called the dynamic pressure at that point. is the dynamic pressure, and ρ is the density of the incompressible fluid. | 1,275 | 6,093 | {"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-2022-49 | latest | en | 0.946788 |
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# Cost of pumping well water to irigate 6 ac
## Cost of pumping well water to irigate 6 ac
(OP)
Hi, I am trying to find out a approximate cost off irrigating 6 acres with well water. I got a 5g/min pump however i do not know the specs of how much electricity it pulls and i am just looking for a average figure.
Say I need 1 inch/acre = 27154 gal. How much electricity would a average submersible pump use to pump 27154 gallons and how much would that cost?
Any one that has real life experience doing this that has a similar figure like I spend \$x amount of dollars per year
irrigating a 2acre pasture or so pleas feel free to inform me.
### RE: Cost of pumping well water to irigate 6 ac
How deep is the well?
What is the draw down under steady use?
What does electricity cost at your location?
It depends.
yours
### RE: Cost of pumping well water to irigate 6 ac
I would need to know a wee bit more about your particular circumstances to give you anything like a sensible answer.
What is the deapth from the surface of the water in the well to the rim of the well? What height above the well dose the water have to be pumped? What distance dose it have to be pumped.What size pipe are you using and how many fittings/elbows etc are you using?And finaly what is the means of distribution at the end of the line, is it sprinkelers,trickle,seep hose capilary or channel? All these things have a bearing on the final cost.
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Register now while it's still free! | 609 | 2,626 | {"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-2018-17 | latest | en | 0.938445 |
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# Standard deviation of the population (rather than sample)?
Hi JASP people,
A question about the descriptive statistics given by JASP. I have a small data set for which I'd like to calculate the standard deviation. JASP seems to treat the data as a sample and gives the sample standard deviation.
But my data set is not a sample, it's the whole population. How can I get the standard deviation of this population? Something like the STDEV.P function in Excel is what I'm looking for.
Any ideas?
Beginner
• So you want to divide by N instead of N-1, is that what you mean?
If I'm not mistaken you can take the SD from JASP, multiply by [sqrt(1/N) / sqrt(1/(N-1))] and you should get the result you want. I did this very quickly though, so best to check!
Actually this works out for your examples:
(sqrt(1/5) / sqrt(1/(5-1))) * 1.58 = 1.413195
and
(sqrt(1/5) / sqrt(1/(5-1))) * 2 = 1.788854
Cheers,
E.J. | 279 | 1,041 | {"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-2022-27 | latest | en | 0.942087 |
https://gitlab.psi.ch/mcmule/handyG/-/raw/1be3dd9bfee3064a6040367bf2cd239d7d203853/utils.f90 | 1,620,472,796,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-21/segments/1620243988858.72/warc/CC-MAIN-20210508091446-20210508121446-00294.warc.gz | 301,424,010 | 2,270 | ! Contains some functions that might be useful later ! Write your own print function with ability to suppress print ! Muss immer alle prints und warnings ausschalten können ! Test Programm schreiben mit exit codes -> gfortran 'test.f90' und dann 'echo \$?' ! Define GPL infinity ! Mach n optional ! Kommentar schreiben zu anderer Notation ! Funktion überprüfen! Tests schreiben! MODULE utils implicit none integer, parameter :: prec = selected_real_kind(15,32) ! logical :: print_enabled = .true. ! logical :: warnings_enabled = .true. CONTAINS FUNCTION get_condensed_m(z) result(m) ! returns condensed m where the ones not needed are filled with 0 complex(kind=prec) :: z(:), m(size(z)) integer :: pos = 1, i m = 1 do i = 1,size(z) if(z(i) == 0) then if(i == size(z)) then pos = pos + 1 else m(pos) = m(pos) + 1 end if else pos = pos + 1 end if end do m(pos:) = 0 END FUNCTION get_condensed_m FUNCTION get_condensed_z(m, z_in) result(z_out) ! returns condensed z vector integer :: m(:), i, pos complex(kind=prec) :: z_in(:), z_out(size(m)) pos = 0 do i=1,size(m) pos = pos + m(i) z_out(i) = z_in(pos) end do END FUNCTION get_condensed_z FUNCTION get_flattened_z(m,z_in) result(z_out) ! returns flattened version of z based on m and z integer :: m(:), i, pos complex(kind=prec) :: z_in(:), z_out(sum(m)) z_out = 0 pos = 0 do i=1,size(m) pos = pos + m(i) z_out(pos) = z_in(i) end do END FUNCTION get_flattened_z FUNCTION find_amount_trailing_zeros(z) result(res) complex(kind=prec) :: z(:) integer :: res, i res = 0 do i = size(z), 1, -1 if( z(i) == 0 ) then res = res + 1 else exit end if end do END FUNCTION find_amount_trailing_zeros FUNCTION find_first_zero(v) result(res) ! returns index of first zero, or -1 if there is no zero integer :: v(:), i, res res = -1 do i = 1,size(v) if(v(i) == 0) then res = i return end if end do END FUNCTION find_first_zero SUBROUTINE print_as_matrix(m) ! prints a 2d array as a matrix complex :: m(:,:) integer :: s(2), i s = shape(m) do i = 1,s(1) print*, m(i,:) end do END SUBROUTINE print_as_matrix FUNCTION shuffle_with_zero(a) result(res) ! rows of result are shuffles of a with 0 complex :: a(:) complex :: res(size(a)+1,size(a)+1) integer :: i,j, N N = size(a)+1 do i = 1,N ! i is the index of the row ! j is the index of the zero j = N+1-i res(i,j) = 0 res(i,1:j-1) = a(1:j-1) res(i,j+1:N) = a(j:) end do END FUNCTION shuffle_with_zero ! subroutine print(s1,s2,s3,s4,s5) ! character(len = *), intent(in), optional :: s1, s2, s3, s4, s5 ! if(print_enabled) then ! print*, s1, s2, s3, s4, s5 ! end if ! end subroutine print ! subroutine warn(s1,s2,s3,s4,s5) ! character(len = *), intent(in), optional :: s1, s2, s3, s4, s5 ! if(warnings_enabled) then ! print*, 'Warning: ', s1, s2, s3, s4, s5 ! end if ! end subroutine warn END MODULE utils ! PROGRAM test ! use utils ! implicit none ! ! complex(kind=prec), dimension(5) :: a = cmplx((/1,2,3/)) ! ! complex(kind=prec) :: z_flat(2) ! ! complex(kind=prec), allocatable :: z(:) ! ! integer :: m_prime(2), condensed_size ! ! integer, allocatable :: m(:) ! ! complex(kind=prec) :: b(size(a)+1,size(a)+1) ! ! ! ! test shuffling ! ! ! b = 1 ! ! ! b = shuffle_with_zero(a) ! ! ! call print_as_matrix(b) ! ! ! test condensing ! ! z_flat = cmplx((/4,0/)) ! ! m_prime = get_condensed_m(z_flat) ! ! if(find_first_zero(m_prime) == -1) then ! ! condensed_size = size(m_prime) ! ! else ! ! condensed_size = find_first_zero(m_prime)-1 ! ! end if ! ! allocate(m(condensed_size)) ! ! allocate(z(condensed_size)) ! ! m = m_prime(1:condensed_size) ! ! z = get_condensed_z(m,z_flat) ! ! z_flat = get_flattened_z(m,z) ! ! deallocate(m) ! ! deallocate(z) ! END PROGRAM test | 1,136 | 3,643 | {"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} | 2.984375 | 3 | CC-MAIN-2021-21 | latest | en | 0.348796 |
https://www.vadepares.cat/adding-and-subtraction-worksheets/ | 1,627,736,560,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154089.6/warc/CC-MAIN-20210731105716-20210731135716-00394.warc.gz | 1,121,277,328 | 14,482 | This page has lots of worksheets and activities on money addition. Use the numbers to make fact families.
Single Digit Addition and Subtraction Worksheet Math
### Find the answers and name the animals.
Adding and subtraction worksheets. Writing the four addition and subtraction facts in the house models, dominoes, picture models and more. These subtraction worksheets include timed math fact tests, multiple digit problems, subtraction with and without regrouping and much more. Our grade 2 subtraction worksheets provide the practice needed to master basic subtraction skills.
Free interactive exercises to practice online or download as pdf to print. There is also a versatile worksheet generator that allows for a limitless number of addition and/ or subtraction questions. Choose a topic below and check back often for new topics and features!
Addition and subtraction with pictures, adding 3 numbers practice sheets. Ways to make 5, 10, and 15. Subtraction is the inverse operation of addition.it has two important property:
It may be printed, downloaded or saved and used in your classroom, home school, or other educational environment to help someone. Adding and subtracting on a number line worksheets a number line is a great tool that helps children to visualize how addition and subtraction works. Addition and subtraction worksheets for kindergarten free printable simple mixed adding and subtracting numbers up to 20 practice worksheets for kindergarten kids.
Read Coloring Book Pages Of Animals
Worksheets for introducing decimal concepts, comparing decimals, and ordering decimals. Addition and subtraction fact family worksheets. Free interactive exercises to practice online or download as pdf to print.
Mixed addition and subtraction worksheets for grade 1. This worksheet generator produces a variety of worksheets for the four basic operations (addition, subtraction, multiplication, and division) with fractions and mixed numbers, including with negative fractions. Subtraction up to subtraction with no regrouping subtraction with regrouping more worksheets
Each page includes 14 equations, a total of 28. Adding and subtracting decimals worksheets pdf for 6th grade 6th grade add and subtract decimals exercises with answers. Worksheets > math > grade 2 > subtraction.
The free printable integer subtraction worksheets are sure to go a long way in gaining speed, precision, and accuracy in subtracting pairs of integers. 200+ worksheets available here and free to be downloaded! Adding & subtracting three fractions worksheets these fractions worksheets are great for testing children in their adding and subtracting of three fractions.
Whether at home or in class, get engaged and have access to a random supply of qualitative mixed addition and subtraction worksheets for grade 1.these 1 st grade printable mixed operations will help to evaluate kid’s simultaneous progression in both addition and. Solve multiplication problems that have decimal factors. Listed below are all the addition and subtraction worksheets available on the site.there are fact family worksheets up to 9.
These particular worksheets include adding and taking away numbers within the first 20. Subtraction has been around for several years now. Fraction worksheets 1 fraction addition, subtraction, multiplication, and division.
Your first graders will get plenty of practice in addition and subtraction within 20 with these free worksheets. These free printable worksheets cater to the students of grade 6, grade 7, and grade 8. Up the skill levels of different children.
You can make the worksheets in both html and pdf. Build your own subtraction worksheets in seconds! After students understand how to do 3 digit addition without needing to regroup, they can begin to practice 3 digit addition with regrouping.
It is important when learning the basic math operations to develop the skill of looking at the operation itself on each problem. Often, when we focus on only a single type of math fact at a time, progressing our way through addition, subtraction, multiplication and division facts, we can find that students become 'modal' when looking at a worksheet. You will find two printable pages in this pdf.
These worksheets will generate 10 or 15 mixed number subtraction problems per worksheet. Adding and subtracting time worksheets whether it is calculating the time taken to do two different tasks, or finding the difference in time taken by two people doing the same task, these printable adding and subtracting time worksheets have them covered in a variety of exercises. Enjoy adding subtracting fractions worksheet to improve your subtraction skills developing worksheets for youngsters involves creativity to really make it look like an.
This page has worksheets with decimal long division problems. Thankfully, our subtraction worksheets eliminate both of those issues with a variety of lessons that teach this important skill in an entertaining way. Addition and subtraction worksheets and online activities.
Well maybe more than a few, so it's probably a good thing. Addition and subtraction are the simplest and most basic mathematical operations. Incorporate the addition and subtraction fact family worksheets comprising sorting the number sets, find the missing members in the triangles, circles, number bonds and bar models;
Subtraction worksheets and online activities. They cover 2nd grade topics ranging from basic subtraction facts to subtracting in columns with regrouping. The addition and subtraction worksheets included both numbers to 10 and numbers to 20.
A brief description of the worksheets is on each of the worksheet widgets. Each worksheet is interactive, with a timer and instant scoring.
Adding and Subtracting Integers educationlevel subject
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Addition and subtraction activities and worksheets
The Adding And Subtracting Three Digit Numbers A Mixed
Scholastic Free Winter Addition and Subtraction Worksheets | 1,182 | 6,429 | {"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} | 3.078125 | 3 | CC-MAIN-2021-31 | latest | en | 0.89164 |
https://bestbinlxbcrjz.netlify.app/mignot68899kito/calculate-sample-variance-online-qeda | 1,726,164,508,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651491.39/warc/CC-MAIN-20240912174615-20240912204615-00880.warc.gz | 107,129,677 | 11,341 | ## Calculate sample variance online
Variance calculator. For variance calculation, please enter numerical data separated with comma (or space, tab, semicolon, or newline). For example: - 720.9 The tool goes beyond serving as a mean, median, and mode calculator: it also calculates sample variance, standard deviation, and standard error. These are
Shows you how to find the standard deviation and variance on a TI83 or TI84 now, we won't concern ourselves with whether this is sample or population data. Standard Deviation, Standard Error, Variance, Mean and Median Calculator. Smaller SD value means samples are clustered tightly, vice versa. The formula of Some of the functions calculate the sample variance and some calculate the population variance. Some of the functions ignore text and logical values, while other The Variance Calculator will calculate variance quickly and easily. Just enter in a comma separated list of the numbers that you want the variance calculated for In probability theory and statistics, variance is used to describe the amount of variation in a population or sample of observations. The square root of Variance is Sample Variance. The equations given above show you how to calculate variance for an entire population. However, when doing science project, you will almost
## Variance calculator. Variance calculator and how to calculate. Population variance and sample variance calculator
Variance calculator and how to calculate. Population variance and sample variance calculator. Enter values: Data type: Whole population data Instructions: Use this Sample Variance Calculator to compute showing all the steps the sample variance s^2. All you need to do is to provide the sample data. Free online standard deviation calculator and variance calculator with steps. Hundreds of statistics articles and videos, help for every topic! For 2 ≤ k ≤ n, the kth estimate of the variance is s2 = Sk/(k – 1). The C++ class RunningStat given below uses this method to compute the mean, sample variance, The calculator above computes population standard deviation and sample can be measured, and is the square root of the variance of a given data set. Ruby will need to know how to find the population and sample variance of her data. Variance is how far a set of numbers are spread out. This is very different
### Find the online calculator here: http://www.statisticshowto.com/calculators/ variance-and-standard-deviation-calculator/, plus hundreds more articles and videos.
It is often useful to be able to compute the variance in a For such an online algorithm, a recurrence relation is Here, xn denotes the sample mean of the first n samples (x1,, xn), s 2 The sample variance, s², is used to calculate how varied a sample is. In statistics, a data sample is a set of data collected from a population. Typically, the Use this calculator to compute the variance from a set of numerical values. Data is from: Population Sample Enter comma separated data (numbers only): Note: If you want to calculate the pooled variance of two samples, then you can use our pooled variance calculator. Scores. Population or Sample. Population. Variance calculator and how to calculate. Population variance and sample variance calculator. Enter values: Data type: Whole population data Instructions: Use this Sample Variance Calculator to compute showing all the steps the sample variance s^2. All you need to do is to provide the sample data. Free online standard deviation calculator and variance calculator with steps. Hundreds of statistics articles and videos, help for every topic!
### Ruby will need to know how to find the population and sample variance of her data. Variance is how far a set of numbers are spread out. This is very different
Calculator with step by step explanations to find standard deviation, variance, skewness and kurtosis. We compute the sample mean by adding and dividing by the number of samples, 6. 34 + 43 + 81 + Variance, Standard Deviation and Coefficient of Variation. Note: var(y) instructs R to calculate the sample variance of Y. In other words it uses n-1 'degrees of freedom', where n is the number of observations in Y. 19 Mar 2009 But the sample variance is biased. The unbiased variance estimate is larger by a factor of n/(n-1):. \bar{\sigma}^2 = \frac{n}{n. In conclusion, the Variance calculator. For variance calculation, please enter numerical data separated with comma (or space, tab, semicolon, or newline). For example: - 720.9 The tool goes beyond serving as a mean, median, and mode calculator: it also calculates sample variance, standard deviation, and standard error. These are
## 31 Jan 2020 To calculate variance of a sample, add up the squares of the differences between the mean of the sample and the individual data points, and
The formula for the Standard Deviation is square root of the Variance. Here is a free online arithmetic standard deviation calculator to help you solve your statistical questions. This can also be used as a measure of variability or volatility for the given set of data. The Sample Variance. The sample variance $$s^2$$ is one of the most common ways of measuring dispersion for a distribution. When a sample of data $$X_1, X_2, ., X_n$$ is given, the sample variance measures the dispersion of the sample values with respect to the sample mean. This simple tool will calculate the variance and standard deviation of a set of data. Simply enter your data into the textbox below, either one score per line or as a comma delimited list, and then press "Calculate". Note: If you want to calculate the pooled variance of two samples, then you can use our pooled variance calculator. Using a sample variance is highly recommended when making calculations on population variance becomes too tedious. The slight difference is that the sample variance uses a sample mean and the deviations get added up over this. Then you divide the sum by (n – 1). If you’re solving for the sample variance, n refers to how many sample points. The answer is, you can use the variance to figure out the standard deviation — a much better measure of how spread out your weights are. In order to get the standard deviation, take the square root of the sample variance: √9801 = 99. The standard deviation, in combination with the mean,
In probability theory and statistics, variance is used to describe the amount of variation in a population or sample of observations. The square root of Variance is | 1,348 | 6,497 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-38 | latest | en | 0.836322 |
https://dssmanuals.mo.gov/family-mo-healthnet-magi/1805-000-00/1805-030-00/1805-030-15/1805-030-15-10/ | 1,718,296,866,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861480.87/warc/CC-MAIN-20240613154645-20240613184645-00180.warc.gz | 191,847,510 | 8,506 | # 1805.030.00 Modified Adjusted Gross Income (MAGI) Methodology
### 1805.030.15.10 Reasonable Compatibility (RC) With More Than One Income in the Household
To determine RC for households with more than one income:
1. Combine all of the self-attested income amounts in the MAGI household
2. Combine all of the electronically obtained income (EOI) in the MAGI household
3. Compare both amounts to the program threshold
1. If they are both under the program threshold, the income is reasonably compatible.
2. If they are not both under the program threshold, but are within 10%, the income is reasonably compatible. Refer to Appendix B – Reasonable Compatibility Calculator.
Example: Marilee and Jacob have applied for their two children. Jacob self-attested to \$1500 in wages and Marilee self-attested to \$1400 in wages. The combined income is \$2900.00 per month. Jacob’s EOI shows \$1700 and Marilee’s shows \$1550, or \$3250 combined. Their income is RC because the combined self-attested income and the combined EOI are both under the Title XIX program threshold.
If the combined MAGI household income is not RC, complete an RC determination using income of the individual participants:
1. Determine the self-attested income amount of the first participant
2. Determine the EOI amount of the same participant
3. Determine if the incomes are within 10% of each other (refer to Appendix B – Reasonable Compatibility Calculator)
1. If the incomes are within 10%, the income is RC.
2. If the incomes are not within 10%, contact the participant for more information.
4. Repeat steps 1-3 for each participant with income
Example: Marcy and Joe apply for AEG and self-attest income of \$2000, which is under the Title XIX program threshold for their household size. The combined EOI for both is \$2400 which is over the threshold. The amounts are also not within 10%. However, Marcy’s self-attested amount is \$1000 and her EOI shows \$1074 which is within 10% of her self-attested, so her income is RC. Joe’s self-attested is \$1000, but his EOI shows \$1326, which is not within 10% of his self-attested. Clarification is needed for Joe’s income. | 521 | 2,152 | {"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} | 3.796875 | 4 | CC-MAIN-2024-26 | latest | en | 0.930557 |
http://mathhelpforum.com/calculus/69807-variable-substitution-derivative.html | 1,481,283,621,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542695.22/warc/CC-MAIN-20161202170902-00402-ip-10-31-129-80.ec2.internal.warc.gz | 174,414,736 | 9,792 | Thread: Variable substitution in a derivative
1. Variable substitution in a derivative
f'(z) = du(x,y)/dx + i dv(x,y)/dx with z=x+iy
Hi,
What happens when I change for z=r*(cos(theta) + i*sin(theta))?
I get:
f'(z) = du(r*cos(theta),r*sin(theta))/dx + i dv(r*cos(theta),r*sin(theta))/dx
I can't derivate about dx anymore, as I don't have the variable x, it has been replaced by x=r*cos(theta) and y=r*sin(theta)... What happens to the derivative?
Btw, how to use the MATH tags??? I looked in the wiki and FAQ but found nothing...
Thanks!
2. http://www.mathhelpforum.com/math-he...-tutorial.html | 170 | 603 | {"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-2016-50 | longest | en | 0.824964 |
https://questions.examside.com/past-years/jee/question/pa-light-ray-emits-from-the-origin-making-an-angle-30c-jee-main-mathematics-trigonometric-functions-and-equations-qvdmlolgtse1vjpp | 1,720,844,748,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514484.89/warc/CC-MAIN-20240713020211-20240713050211-00517.warc.gz | 396,001,023 | 52,248 | 1
JEE Main 2023 (Online) 29th January Morning Shift
+4
-1
A light ray emits from the origin making an angle 30$$^\circ$$ with the positive $$x$$-axis. After getting reflected by the line $$x+y=1$$, if this ray intersects $$x$$-axis at Q, then the abscissa of Q is :
A
$${2 \over {\left( {\sqrt 3 - 1} \right)}}$$
B
$${2 \over {3 - \sqrt 3 }}$$
C
$${{\sqrt 3 } \over {2\left( {\sqrt 3 + 1} \right)}}$$
D
$${2 \over {3 + \sqrt 3 }}$$
2
JEE Main 2022 (Online) 29th July Evening Shift
+4
-1
Let $$m_{1}, m_{2}$$ be the slopes of two adjacent sides of a square of side a such that $$a^{2}+11 a+3\left(m_{1}^{2}+m_{2}^{2}\right)=220$$. If one vertex of the square is $$(10(\cos \alpha-\sin \alpha), 10(\sin \alpha+\cos \alpha))$$, where $$\alpha \in\left(0, \frac{\pi}{2}\right)$$ and the equation of one diagonal is $$(\cos \alpha-\sin \alpha) x+(\sin \alpha+\cos \alpha) y=10$$, then $$72\left(\sin ^{4} \alpha+\cos ^{4} \alpha\right)+a^{2}-3 a+13$$ is equal to :
A
119
B
128
C
145
D
155
3
JEE Main 2022 (Online) 29th July Evening Shift
+4
-1
Let $$\mathrm{A}(\alpha,-2), \mathrm{B}(\alpha, 6)$$ and $$\mathrm{C}\left(\frac{\alpha}{4},-2\right)$$ be vertices of a $$\triangle \mathrm{ABC}$$. If $$\left(5, \frac{\alpha}{4}\right)$$ is the circumcentre of $$\triangle \mathrm{ABC}$$, then which of the following is NOT correct about $$\triangle \mathrm{ABC}$$?
A
area is 24
B
perimeter is 25
C
D
4
JEE Main 2022 (Online) 29th July Morning Shift
+4
-1
Let the circumcentre of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab > 0 be P(1,1). If the line AP intersects the line BC at the point Q$$\left(k_{1}, k_{2}\right)$$, then $$k_{1}+k_{2}$$ is equal to :
A
2
B
$$\frac{4}{7}$$
C
$$\frac{2}{7}$$
D
4
EXAM MAP
Medical
NEET | 686 | 1,734 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-30 | latest | en | 0.674907 |
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# Anderson, Loren Runar et al "RING DEFORMATION"
## Structural Mechanics of Buried Pipes
Boca Raton: CRC Press LLC,2000
Figure 3-1 (top) Vertical compression (strain) in a medium transforms an imaginary circle into an ellipse with
decreases in circumference and area.
(bottom) Now if a flexible ring is inserted in place of the imaginary ellipse and then is allowed to expand such
that its circumference remains the same as the original imaginary circle, the medium in contact with the ring
is compressed as shown by infinitesimal cubes at the spring lines, crown and invert.
## CHAPTER 3 RING DEFORMATION
Deformation of the pipe ring occurs under any load.
For most buried pipe analyses, this deformation is
small enough that it can be neglected. For a few
analyses, however, deformation of the ring must be
considered. This is particularly true in the case of
instability of the ring, as, for example, the hydrostatic
collapse of a pipe due to internal vacuum or external
pressure. Collapse may occur even though stress
has not reached yield strength. But collapse can
occur only if the ring deforms. Analysis of failures
requires a knowledge of the shape of the deformed
ring.
For small ring deflection of a buried circular pipe, the
basic deflected cross section is an ellipse. Consider
the infinite medium with an imaginary circle shown
in Figure 3-1 (top). If the medium is compressed
(strained) uniformly in one direction, the circle
becomes an ellipse. This is easily demonstrated
mathematic ally. Now suppose the imaginary circle
is a flexible ring. When the medium is compressed,
the ring deflects into an approximate ellipse with
slight deviations. If the circumference of the ring
remains constant, the ellipse must expand out into
the medium, increasing compressive stresses
between ring and medium. See Figure 3-1 (bottom).
The ring becomes a hard spot in the medium. On
the other hand, if circumference of the ring is
reduced, the ring becomes a soft spot and pressure
is relieved between ring and medium. In either case,
the basic deformation of a buried ring is an ellipse
slightly modified by the relative decreases in areas
within the ring and without the ring. The shape is
also affected by non-uniformity of the medium. For
example, if a concentrated reaction develops on the
bottom of the ring, the ellipse is modified by a flat
spot. Nevertheless, for small soil strains, the basic
ring deflection of a flexible buried pipe is an ellipse.
Following are some pertinent approximate
geometrical properties of the ellipse that are
sufficiently accurate for most buried pipe analyses.
Greater accuracy would require solutions of infinite
series.
## Geometry of the Ellipse
The equation of an ellipse in cartesian coordinates,
x and y, is:
a2x2 + b2y2 = a2b2
where (See Figure 3-2):
a = minor semi-diameter (altitude)
b = major semi-diameter (base)
r = radius of a circle of equal circumference
The circumference of an ellipse is p(a+b) which
reduces to 2pr for a circle of equal circumference.
In this text a and b are not used because the pipe
industry is more familiar with ring deflection, d.
Ring deflection can be written in terms of semidiameters a and b as follows:
d = D/D = RING DEFLECTION
. . . . . (3.1)
where:
D
= decrease in vertical diameter of ellipse from
a circle of equal circumference,
= 2r = mean diameter of the circle
diameter to the centroid of wall crosssectional areas,
a = r(1-d) for small ring deflections (<10%),
b = r(1+d) for small ring deflections (<10%).
Assuming that circumferences are the same for
circle and ellipse, and that the vertical ring deflection
is equal to the horizontal ring deflection, area within
the ellipse is Ae = Bab; and
Ae = pr2 (1 - d2)
The ratio of areas within ellipse and circle is:
A r = A e / A o = ratio of areas.
See Figure 3-3.
Figure 3-2 Some approximate properties of an ellipse that are pertinent to ring analyses of pipes where d is
the ring deflection and ry and rx are the maximum and minimum radii of curvature, respectively.
## Figure 3-3 Ratio of areas, Ar = Ae /Ao
(Ae within an ellipse and Ao within a
circle of equal circumference) shown
plotted as a function of ring deflection.
## What of the assumption that the horizontal and
vertical ring deflections are equal if the
circumferences are equal for circle and ellipse? For
the circle, circumference is 2pr. For the ellipse,
circumference is (b+a)(64-3R4)/(64-16R2), where R
is approximately R = (b-a)/(b+a). Only the first
terms of an infinite series are included in this
approximate ellipse circumference. See texts on
analytical geometry. Equating circumferences of the
circle and the ellipse, and transforming the values of
a and b into vertical and horizontal values of ring
deflection, dy and dx, a few values of d y and the
corresponding dx are shown below for comparison.
dy (%)
______
0.
5.00
10.00
15.00
20.00
dx (%)
______
0.
4.88
9.522
13.95
18.116
Deviation
(dy - dx)/dy
__________
0.
0.024
0.048
0.070
0.094
## For ring deflections of d = dy = 10%, the
corresponding dx is less than 10% by only
4.8%(10%) = 0.48%. This is too small to be
significant in most calculations such as areas within
the ellipse and ratios of radii.
## An important property of the ellipse is the ratio of
radii rr = ry/rx, which is:
rr = (1 + d)3 / (1 - d)3
. . . . . (3.2)
where:
rr = ratio of the maximum to minimum radii of
curvature of the ellipse. See graph of Figure 3-4.
## Measurement of Radius of Curvature
In practice it is often necessary to measure the
radius of curvature of a deformed pipe. This can be
done from either inside or outside of the pipe. See
Figure 3-5. Inside, a straightedge of known length L
is laid as a cord. The offset e is measured to the
curved wall at the center of the cord. Outside, e can
be found by laying a tangent of known length L and
by measuring the offsets e to the pipe wall at each
end of the tangent. The average of these two
offsets is the value for e. Knowing the length of the
cord, L, and the offset, e, the radius of curvature of
the pipe wall can be calculated from the following
equation:
r = (4e2 + L2)/8e
. . . . . (3.3)
## Radii of curvature of the sides (spring lines) and the
top and bottom (crown and invert) of the ellipse are:
## It is assumed that radius of curvature is constant
within cord length L. The calculated radius is to the
surface from which e measurements are made.
Example
## An inspection reveals that a 72-inch corrugated
metal pipe culvert appears to be flattened somewhat
on top. From inside the pipe, a straightedge (cord)
12 inches long is placed against the top, and the midordinate offset is measured and found to be 11/32
inch. What is the radius of curvature of the pipe ring
at the top?
## For ring deflection less than d = 10%, and neglecting
higher orders of d,
rx = r (1 - 3d), and ry = r (1 + 3d).
However, more precise, and almost as easy to use,
are the approximate values:
rx = a2 / b = r (1 - d)2 / (1 + d)
ry = b2 / a = r (1 + d)2 / (1 - d)
2000 CRC Press LLC
## From Equation 3.3, r = (4e2 + L2)/8e. Substituting in
values and solving, ry = 52.5 inches which is the
average radius within the 12-inch cord on the inside
of the corrugated pipe. On the outside, the radius is
greater by the depth of the corrugations.
d (%)
rr
0
1
2
3
4
5
6
8
10
12
15
20
1.000
1.062
1.128
1.197
1.271
1.350
1.434
1.618
1.826
2.062
2.476
3.375
## Figure 3-4 Ratio of radii,
rr = ry/rx = (1+d)3/(1-d)3,
(ry and rx are maximum and
for ellipse) shown plotted
as a function of ring deflection d.
Figure 3-5 Procedure for calculating the radius of curvature of a ring from measurements of a cord of length
L and the middle ordinate e.
## When subjected to uniform internal pressure, the
pipe expands. The radius increases. Ring deflection
is equal to percent increase in radius;
## Computer software is available for evaluating the
deformation of a pipe ring due to any external
virtual work according to Castigliano. Analysis
provides a component of deflection of some point B
on a structure with respect to a fixed point A. It is
convenient to select point A as the origin of fixed
coordinate axes the axes are neither translated
nor rotated. See Appendix A.
## d = Dr/r = DD/D = 2pre /2pr = e
where:
d = ring deflection (percent),
Dr and DD are increases due to internal pressure,
D = mean diameter,
e = circumferential strain,
E = modulus of elasticity = s/e.
s = circumferential stress = Ee = Ed.
But s = P'(ID)/2A, from Equation 2.1,
where:
P' = uniform internal pressure,
ID = inside diameter,
A = cross sectional area of wall per unit length.
Equating the two values for s , and solving for d,
d = P'(ID)/2AE
. . . . . (3.4)
## Figure 3-6 Quadrant of a circular cylinder fixed at
the crown A-A-A with Q-load at the spring line, BB-B, showing a slice isolated for analysis.
## 2000 CRC Press LLC
Example
Consider the quadrant of a circular cylinder shown
in Figure 3-6. It is fixed along edge A-A-A, and is
loaded with vertical line load Q along free edge B-BB. What is the horizontal deflection of free edge B
with respect to fixed edge A? This is a twodimensional problem for which a slice of unit width
can be isolated for analysis. Because A is fixed,
the horizontal deflection of B with respect to A is xB
for which, according to Castigliano:
xB = f (M/EI)(dM/dp)ds
. . . . . (3.5)
where:
xB = displacement of point B in the x-direction,
EI = wall stiffness,
E = modulus of elasticity,
I = centroidal moment of inertia of the cross
section of the wall per unit length of cylinder,
M = moment of force about the neutral axis at C,
point B in the direction assumed for deflection,
ds = differential length along the slice, = rdq
r = mean radius of the circular cylinder.
It is assumed that deflection is so small that radius r
remains constant. It is also assumed that the
deflection is due to moment M, flexure not to
shear or axial loads. In Figure 3-6, consider arc CB
as a free-body-diagram. Apply the dummy load p at
B acting to the right assuming that deflection xB will
be in the x-direction. If the solution turns out to be
negative, then the deflection is reversed. From the
free-body-diagram CB,
M = Qr(1-cosq) + pr(sinq)
M/ p = r(sinq)
PROBLEMS
3-1 A plain polyethylene pipe of 16-inch outside
diameter and DR = 15 is subjected to internal
pressure of 50 psi. The surfaces are smooth and
cylindrical (not ribbed or corrugated).
DR
(dimension ratio) = (OD)/t where t = wall thickness.
Modulus of elasticity is 115 ksi. What is the ring
deflection? DR is dimension ratio = (OD)/t.
(d = 0.28%)
## 3-2 At ring deflection of 15%, and assuming the
pipe cross section is an ellipse, what is the percent
error in finding the ratio of maximum to minimum
radii of curvature by means of approximate
Equation 3.2,
rr = (1+d)3 / (1-d)3?
(0.066%)
## 3-3 A 36 OD PVC buried pipeline is uncovered at
one location. The top of the pipe appears to be
flattened. A straight edge 200 mm long is laid
horizontally across the top and the vertical distances
down to the pipe surface at each end of the straight
edge are measured and found to be 9.2 and 9.4 mm.
What is the radius of curvature of the outside
surface of the pipe at the crown?
Ry = 542 mm = 21.35 inches)
M = Qr(1-cosq)
ds = rdq
## 3-4 Assuming that the ring of problem 3-3 is
deflected into an ellipse, approximately what is the
ring deflection? Maximum ring deflection is usually
limited to 5% according to specifications.
(d = 5.74%)
## Substituting into Equation 3.5,
xB = (Qr/EI) (1-cosq) r(sinq) rdq
Integrating and substituting in limits of q from 0 to
p/2,
## 3-5 What is the percent decrease in cross-sectional
area inside the deflected pipe of problem 3-4 if the
ring deflection is d = 5.74%?
(0.33%)
xB = Qr3/2EI
This is one of a number of the most useful
deflections of rings recorded in Table A-1.
## 3-6 What is the approximate ratio of maximum to
minimum radii, rr, for an ellipse?
(rr = 1.8)
## 3-7 A horizontal, rectangular plate is a cantilever
beam loaded by a uniform vertical pressure, P, and
supported (fixed) along one edge. What is the
vertical deflection of the opposite edge? The
thickness of the plate is t, the length measured from
the fixed edge is L, and the modulus of elasticity is
E. Elastic limit is not exceeded. Use the Castigliano
equation.
(y = 3PL4 / 2Et 3)
3-8 A half of a circular ring is loaded at the crown
The reactions are rollers at the spring lines B, as
shown. If the wall stiffness is EI, what is the
vertical deflection of point A?
(yA = 0.1781 Fr3/EI)
## 3-9 What is the vertical ring deflection of the hinged
arch of problem 3-8 if it is loaded with a uniform
## 3-10 The top and bottom halves of the circular
cylinder of problem 3-8 are symmetrical. If the
spring lines of the two halves are hinged together,
what is the ring deflection due to the F-load and an
equal and opposite reaction at the bottom?
(d = 0.1781 Fr2/EI)
3-11 Sections of pipe are tested by applying an Fload. For flexible rings, the F-load test is called a
parallel plate test. What is the ring deflection if
elastic limit is not exceeded?
[d = 0.0186F/(EI/D3)D] | 3,562 | 13,165 | {"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-2019-35 | latest | en | 0.89637 |
http://bootmath.com/showing-one-point-compactification-is-unique-up-to-homeomorphism.html | 1,531,855,174,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589892.87/warc/CC-MAIN-20180717183929-20180717203929-00444.warc.gz | 50,678,892 | 6,677 | # Showing one point compactification is unique up to homeomorphism
First for clarity I’ll define things as I’m familiar with them:
1. A compactification of a non-compact topological space $X$
is a compact topological space $Y$
such that $X$
can be densley embedded in $Y$
.
2. In particular a compacitifaction is said to be a one-point compactification if $\left|Y\backslash X\right|=1$
3. The Alexandroff one-point compactification of a a topological space $\left(X,\mathcal{T}_{X}\right)$
is the set $X^{*}=X\cup\left\{ \infty\right\}$
for some element $\infty\notin X$
given the topology $$\mathcal{T}^{*}:=\mathcal{T}_{X}\cup\left\{ U\subseteq X^{*}\,|\,\infty\in U\,\wedge\, X\backslash U\,\mbox{is compact and closed in }\left(X,\mathcal{T}_{X}\right)\right\}$$
If $\left(X,\mathcal{T}_{X}\right)$
is a Hausdorff space one can omit the requirement that $X\backslash U$
is closed.
It is easy to show that given two choices of elements $\infty_{1},\infty_{2}\notin X$
the one-point compactifications $X\cup\left\{ \infty_{1}\right\}$
and $X\cup\left\{ \infty_{2}\right\}$
with the topology defined as that of the Alexandroff one-point compactification are homeomorphic. What I’m wondering is why isn’t there another possible way to define the topology on $X^{*}$
that would also yield a compactification (which is in particular not homeomorphic to the Alexandroff one-point topology)
As far as I see it there are two approaches to answering this question:
1. Show that any topology on $X^{*}$
that yields a compact space in which $X$
is dense is homeomorphic to $\mathcal{T}^{*}$.
2. Show it’s not possible to consturct any other topology on $X^{*}$ that results in a compactification.
I’m quite interested in seeing the reasoning to both approaches if possible.
#### Solutions Collecting From Web of "Showing one point compactification is unique up to homeomorphism"
You get the uniqueness result if the space is Hausdorff.
Let $\langle X,\tau\rangle$ be a compact space. Suppose that $p\in X$ is in the closure of $Y=X\setminus\{p\}$, and let $\tau_Y$ be the associated subspace topology on $Y$; $\langle X,\tau\rangle$ is then a compactification of $\langle Y,\tau_Y\rangle$.
Suppose that $p\in U\in\tau$, and let $V=U\cap Y$. Then $\varnothing\ne V\in\tau_Y$, so $Y\setminus V$ is closed in $Y$. Moreover, $Y\setminus V=X\setminus U$ is also closed in $X$, which is compact, so $Y\setminus V$ is compact. That is, every open nbhd of $p$ in $X$ is the complement of a compact, closed subset of $Y$. Thus, if $\tau’$ is the topology on $X$ that makes it a copy of the Alexandroff compactification of $Y$, then $\tau\subseteq\tau’$.
Now let $K\subseteq Y$ be compact and closed in $Y$, and let $V=Y\setminus K\in\tau_Y$. If $X\setminus K=V\cup\{p\}\notin\tau$, then $p\in\operatorname{cl}_XK$. If $X$ is Hausdorff, this is impossible: in that case $K$ is a compact subset of the Hausdorff space $X$ and is therefore closed in $X$. Thus, if $X$ is Hausdorff we must have $\tau=\tau’$, and $X$ is (homeomorphic to) the Alexandroff compactification of $Y$.
If $X$ is not Hausdorff, however, we can have $\tau\subsetneqq\tau’$. A simple example is the sequence with two limits. Let $D$ be a countably infinite set, let $p$ and $q$ be distinct points not in $D$, and let $X=D\cup\{p,q\}$. Points of $D$ are isolated. Basic open nbhds of $p$ are the sets of the form $\{p\}\cup(D\setminus F)$ for finite $F\subseteq D$, and basic open nbhds of $q$ are the sets of the form $\{q\}\cup(D\setminus F)$ for finite $F\subseteq D$. Let $Y=D\cup\{q\}$. Then $Y$ is dense in $X$, and $X$ is compact, and $Y$ itself is a closed, compact subset of $Y$ whose complement is not open in $X$.
Improved example (1 June 2015): Let $D$ and $E$ be disjoint countably infinite sets, let $p$ and $q$ be distinct points not in $D\cup E$, let $X=D\cup E\cup\{p,q\}$, and let $Y=D\cup E\cup\{q\}$. Points of $D\cup E$ are isolated. Basic open nbhds of $q$ are the sets of the form $\{q\}\cup D\cup (E\setminus F)$ for finite $F\subseteq E$, and basic open nbhds of $p$ are the sets of the form $\{p\}\cup\big((D\cup E)\setminus F\big)$ for finite $F\subseteq D\cup E$. Then $Y$ is a non-compact dense subspace of the compact space $X$, so $X$ is a (non-Hausdorff) compactification of $Y$. Let $K=\{q\}\cup E$. Then $K$ is a compact closed subset of $Y$, but $X\setminus K=\{p\}\cup E$ is not open in $X$.
(This avoids the question of whether it’s legitimate to look at the Alexandrov compactification of a compact space.) | 1,415 | 4,514 | {"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} | 2.515625 | 3 | CC-MAIN-2018-30 | latest | en | 0.837119 |
http://badcreditlineofcredit.tk/multiples-of-three-work-sheet.html | 1,566,770,872,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330907.46/warc/CC-MAIN-20190825215958-20190826001958-00534.warc.gz | 23,862,133 | 3,873 | Multiples of three work sheet
Work sheet
Multiples of three work sheet
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Work sheet
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``multiples of three work sheet``
4th through 6th Grades. Least Common Multiple: 3 Numbers. Find the LCM for each set of three of numbers. | 861 | 3,884 | {"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} | 2.703125 | 3 | CC-MAIN-2019-35 | latest | en | 0.898308 |
http://www.tech-archive.net/Archive/Word/microsoft.public.word.mailmerge.fields/2006-02/msg00322.html | 1,369,416,918,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704933573/warc/CC-MAIN-20130516114853-00068-ip-10-60-113-184.ec2.internal.warc.gz | 666,856,672 | 4,114 | # Re: Skip Record if Comparison of Several Fields Fails
You can look up the available functions and operators in the Word help for
the = field. However, that can be hard to find as the field isn't listed in
the help index - you can search for "formula field" in the "Search for" box
in Word Help.
+ is just the usual addition operator and * is the usual multiplication
operator. it is sometimes easier to use these rather than the or and and
functions which are also available because or and and only take two
parameters, i.e. you can have
{ =or(param1,param2) }
but for three parameters I think you need
{ =or(param1,or(param1,param2)) }
The thing to bear in mind is that within the formula field "True" is
represented as "1" and "False" as "0", so in the first example above, if
param1 or param2 is true or 1, then the result is "1".
However, the fields as suggested here obviously won't work because your
structures are actually slightly different.
Peter Jamieson
"Nick" <nking@xxxxxxxxxxxxxxx> wrote in message
news:1139950494.967560.128130@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Peter -
In the solution below:
{ SKIPIF
{ =
{ IF { MERGEFIELD target } = { MERGEFIELD choice_1 } "1" "0" }
+{ IF { MERGEFIELD target } = { MERGEFIELD choice_2 } "1" "0" }
+{ IF { MERGEFIELD target } = { MERGEFIELD choice_3 } "1" "0" }
*{ IF { MERGEFIELD rating_1 } = { MERGEFIELD rating_2 }
"{ IF { MERGEFIELD rating_2 } = { MERGEFIELD rating_3 } "0" "1" }"
"1" }
there are some operators (+ and *). Can you tell me about other
available operators? Is there an "or" operator?
Nick King
.
## Relevant Pages
• Re: Inserting "tomorrows" date
... Formula Field in the Word Help. ... I have a template which I use often. ... Please reply to the newsgroup to maintain the thread. ...
(microsoft.public.mac.office.word) | 497 | 1,805 | {"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-2013-20 | latest | en | 0.895201 |
http://www.ibpsguide.com/best-5-shortcuts-on-boats-and-stream-problems-for-sbi-clerk-ibps-exams-2017?q=/2017/06/best-5-shortcuts-on-boats-and-stream-problems-for-sbi-ibps-exams-.html | 1,503,080,098,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105086.81/warc/CC-MAIN-20170818175604-20170818195604-00243.warc.gz | 570,628,437 | 32,430 | # BEST 5 Shortcuts on Boats and Stream Problems for SBI Clerk / IBPS Exams 2017
BEST 5 Shortcuts on Boats and Stream Problems for SBI Clerk / IBPS Exams 2017:
Dear Readers, Here we have given the BEST 5 Shortcuts on Boats and Stream Problems for SBI Clerk / IBPS Exams 2017, candidates those who are preparing for the upcoming SBI Clerk/IBPS Exams 2017 can make use of it.
BASIC CONCEPT OF BOATS AND STREAM
Still water:
If the water is not moving then it is called still water.
Speed of boat in still water is= ½ (Downstream Speed + Upstream Speed)
Stream:
Moving water of the river is called stream
Upstream:
If a boat or a swimmer moves in the opposite direction of the stream then it is called upstream movement. In this case motion of boat or swimmer isopposedby the motion of stream.
Downstream:
If a boat or a swimmer moves in the same direction of the stream then it is called downstream movement. In this case motion of boat or swimmer issupported by the motion of stream.
When speed of boat or a swimmer is given then it normally means speed in still water.
Concepts and formulas of boats and streams are also applicable to problems involving:
Cyclist and wind:
Cyclist analogous to boat and wind analogous to stream.
Swimmer and stream:
Swimmer analogous to boat
If speed of boat in still water is ‘b’ km/hr and speed of stream is ‘s’ km/hr,
Speed of boat downstream = (b+s) km/hr, since the boat goes with the stream of water hence its speed increase.
Speed of boat upstream = (b – s) km/hr.the boat goes against the stream of water and hence its speed gets reduced.
Distance = Speed × Time
D = ST
Type 1:
If Ratioof downstream and upstream speeds of a boat is a : b. then ratio of time taken = b : a
Speed of stream = ((a – b)/ ( a + b))× speed in still water.
Speed in still water = ((a + b) /(a – b)) × speed of stream.
Example:
1) A man can row a boat to a certain distanced upstream in 4 hours and takes 3 hours to row downstream the same distance. What is the speed of boat in still water, if the speed of the stream is 2 km/hr?
a) 9km/hr b) 10 km/hr c) 12 km/hr d) 14 km/hr
Solution:
Ratio of speed downstream and upstream = 3 : 4
Speed of boat in still water = ((4+3) / (4-3)) × 2 = 14 km/hr
2) A man can row a boat 120 km with stream in 5 hours. If speed of the boat is double the speed of the stream, find the speed stream?
a) 6 km/hr b) 8 km/hr c) 9 km/hr d) 12 km/hr
Solution:
Speed of the boat downstream = 120/5 = 24 km/hr
Ratio of speed of boat and stream = 2 : 1àspeed of stream = 1/3 *24 = 8 km/hr
Type 2:
1) If a steam boat goes 8 km upstream in 40 minutes and the speed of stream is 5 km/hr, In still water what would be the speed of the boat?
Solution:
Rate of stream = 8 x 60 / 40 = 12 km/h. [convert minutes to hour we multiplying by 60] the speed of stream is 5 km/h
Let speed in still water be x km / hr.
If stream speed 5 km then speed of upstream would be = (x – 5) km/hr
Then, speed upstream = (x – 5) km / hr.
So, x – 5 = 12
x = 12 + 5 = 17 km/hr.
Type 3:
3) Pavi rows in still water with a speed of 4.5 kmph to go to a certain place and to come back. Find his average speed for the whole journey, if the river is flowing with a speed of 1.5 kmph ?
Solution:
Pavi’s speed upstream = 4.5 – 1.5 = 3 kmph
Pavi’s speed downstream = 4.5+1.5 = 6 kmph
Distance = X km
Time Taken in upstream = X/3
Time Taken in downstream = X/6
Average Speed = 2X/(X/3 + X/6 ) = 4 kmph
Type 4:
When a boat’s speed in still water is ‘a’ km/h and river is flowing with a speed of ‘b’ km/h an time taken to cover the same distance downstream, then
Example:
1)A boat’s speed in still water is 10 km/h while river is flowing with speed of 2 km/h and time to cover a certain distance upstream is 4 h more than time taken to cover the same distance downstream. Find the distance.
Solution:
Here, a =10 km/h, b = 2 km/h, T = 4 h
By using formula:
Required distance = [(10^2 – 2^2)/(2×2)]x4
= 96 km
Type 5:
Speed of boat in still water& speed of stream are given. Find average speed of boat that covers certain distance& returns through the same path.
Example:
1) If the speed of boat in still water is 10 km/hr& the speed of stream is 3 km/hr, the boat rows to a place which is 50 km far& returns through the same path. What would be the average speed of boat during the journey?
a) 2 km/hr b) 4.5 km/hr c) 9.1 km/hr d) 15 km/hr
Solution:
If a boat moves at ‘x’ km/hr speed and covers the same distance up and down in a stream at the speed of ‘y’ km/hr, then average speed is calculated by,
Speed of boat in still water = x = 10 km/hr
Speed of stream = y = 3 km/hr
We have,
If You Have Any Queries, Feel Free to Ask us in the below Comment Section. | 1,362 | 4,693 | {"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} | 4.21875 | 4 | CC-MAIN-2017-34 | latest | en | 0.938286 |
https://testbook.com/question-answer/in-the-given-figure-ac%E2%88%A5-oedoe--61153f852d4c34d25e15012c | 1,638,049,626,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358233.7/warc/CC-MAIN-20211127193525-20211127223525-00336.warc.gz | 647,058,495 | 30,065 | # In the given figure AC ∥ OE, ∠DOE = 50° and ∠ABF = 30°. Find the value of ∠FBD.
1. 110°
2. 100°
3. 90°
4. 105°
Option 2 : 100°
## Detailed Solution
Given:
∠DOE = 50° and ∠ABF = 30°
Concept:
Linear pair:- Sum of all the angles on the straight line is 180°
Parallel lines are lines that are lying on the same plane but will never meet
When a transversal line cuts two lines, the properties below will help us determine whether the lines are parallel.
The properties below will help us determine and show that two lines are parallel.
1. Corresponding angles are equal.
b = f and a = e
2. Alternate interior angles are equal.
c = f and d = e
3. Alternate exterior angles are equal.
a = h and b = g
4. Consecutive interior angles add up to 180°.
c + e = 180° and d + f = 180°
5. Consecutive exterior angles add up to 180°.
Calculation:
∠DOE = ∠OBC = 50° ----(Corresponding angles are equal)
By linear pair,
∠OBC + ∠FBD + ∠ABF = 180°
⇒ 50° + ∠FBD + 30° = 180°
⇒ ∠FBD = 100° | 340 | 1,003 | {"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} | 4.21875 | 4 | CC-MAIN-2021-49 | latest | en | 0.701415 |
https://www.mkamimura.com/2019/08/Mathematics-Python-Analytics-Series-Integral-Judgment-Logarithm-Function-Power-Power-Power-Power-Reciprocal-Convergence-Partial-Integration-Method.html | 1,695,768,636,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510225.44/warc/CC-MAIN-20230926211344-20230927001344-00845.warc.gz | 993,192,658 | 51,789 | ## 2019年8月28日水曜日
### 数学 - Python - 解析学 - 級数 - 積分による判定法 - 対数関数の累乗(べき乗)、累乗(べき乗)、逆数、収束、部分積分方
1. $\begin{array}{l}\int \frac{1}{x{\left(\mathrm{log}x\right)}^{1+\epsilon }}\mathrm{dx}\\ =\frac{\mathrm{log}x}{{\left(\mathrm{log}x\right)}^{1+\epsilon }}-\int \left(\mathrm{log}x\right)·\frac{-1}{{\left(\mathrm{log}x\right)}^{2\left(1+\epsilon \right)}}\left(1+\epsilon \right){\left(\mathrm{log}x\right)}^{\epsilon }·\frac{1}{x}\mathrm{dx}\\ =\frac{1}{{\left(\mathrm{log}x\right)}^{\epsilon }}+\left(1+\epsilon \right)\int \frac{1}{x{\left(\mathrm{log}x\right)}^{1+\epsilon }}\mathrm{dx}\\ -\epsilon \int \frac{1}{x{\left(\mathrm{log}x\right)}^{1+\epsilon }}\mathrm{dx}=\frac{1}{{\left(\mathrm{log}x\right)}^{\epsilon }}\\ \int \frac{1}{x{\left(\mathrm{log}x\right)}^{1+\epsilon }}=-\frac{1}{\epsilon {\left(\mathrm{log}x\right)}^{\epsilon }}\end{array}$
よって、
$\begin{array}{l}\underset{b\to \infty }{\mathrm{lim}}{\int }_{2}^{b}\frac{1}{x{\left(\mathrm{log}x\right)}^{1+\epsilon }}\mathrm{dx}\\ =\underset{b\to \infty }{\mathrm{lim}}\left(-\frac{1}{\epsilon }{\left[\frac{1}{{\left(\mathrm{log}x\right)}^{\epsilon }}\right]}_{2}^{b}\right)\\ =-\frac{1}{\epsilon }\underset{b\to \infty }{\mathrm{lim}}\left(\frac{1}{{\left(\mathrm{log}b\right)}^{\epsilon }}-\frac{1}{{\left(\mathrm{log}2\right)}^{\epsilon }}\right)\\ =\frac{1}{\epsilon {\left(\mathrm{log}2\right)}^{\epsilon }}\end{array}$
ゆえに、 問題の無限級数は収束する。
コード
Python 3
#!/usr/bin/env python3
from sympy import pprint, symbols, summation, oo, Integral, plot, log
import matplotlib.pyplot as plt
print('13.')
n = symbols('n')
epsilon = 2
f = 1 / (n * log(n) ** (1 + epsilon))
s = summation(f, (n, 2, oo))
I = Integral(f, (n, 2, oo))
for o in [s, I, I.doit()]:
pprint(o)
print()
p = plot(f,
(n, 2, 12),
legend=True,
show=False)
colors = ['red', 'green', 'blue', 'brown', 'orange',
'purple', 'pink', 'gray', 'skyblue', 'yellow']
for s, color in zip(p, colors):
s.line_color = color
p.show()
p.save('sample13.png')
def g(m):
return sum([f.subs({n: k}) for k in range(2, m)])
ms = range(2, 12)
plt.plot(ms, [g(m) for m in ms])
plt.legend(['Σ 1 / n(log n) ^ (1+ε)', '1 / n(log n)^(1+ε)'])
plt.savefig('sample13.png')
C:\Users\...>py sample12.py
13.
∞
____
╲
╲ 1
╲ ─────────
╱ 3
╱ n⋅log (n)
╱
‾‾‾‾
n = 2
∞
⌠
⎮ 1
⎮ ───────── dn
⎮ 3
⎮ n⋅log (n)
⌡
2
1
─────────
2
2⋅log (2)
c:\Users\...> | 1,023 | 2,370 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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} | 4.03125 | 4 | CC-MAIN-2023-40 | latest | en | 0.240129 |
http://qpr.ca/blogs/2011/05/ | 1,601,396,856,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400202418.22/warc/CC-MAIN-20200929154729-20200929184729-00636.warc.gz | 94,746,472 | 9,322 | ## Archive for May, 2011
### How to find the equation of a quadratic function from its graph :: squareCircleZ
Wednesday, May 18th, 2011
Murray Bourne of squareCircleZ has posted on ‘How to find the equation of a quadratic function from its graph‘. This is indeed the type of discussion and exercise that we need to see more of. Not only does it promote a deeper understanding of the mathematics than the reverse but it is also relevant to more practical applications. Occasionally we do come up with a formula and want to see what it looks like but, especially when it comes to specific examples as opposed to general patterns, it is more often that we have data and want to find or verify a formula. One of the activities in my own “Blue Meanies” game (at http://qpr.ca/math/applets/meanies/ )asks students to “guess” the equation of a parabola through three points by imagining the curve and using its geometry (in various ways) to determine the equation. Of course in such “modelling” problems, with limited data there will be many possible model types that can be used, and there is an interesting interplay between fitting with a particular class of functions (eg polynomial or exponential) and giving reasons why one or other such class might be more appropriate in a given situation.
Friday, May 13th, 2011
Howard Knopf doesn’t like the idea of extending the tax (or calling it one).
I didn’t like having to pay a tax, or “levy”, on the CDs I bought years ago to store photos and backup my HD, but I don’t see any difference between that bit of theft and this one. In fact, although I resent the presumption that the tax, or “levy”, is a fee for some service that I have no intention of using, I can live with the idea of a tax on media being used to support creative activities if that is the collective will of the nation.
Just don’t call it a “levy”, and interpret it as a fee for service, unless
(a)it entitles me to fill it with unlimited personal use copies of any works that I do buy, and
(b)there is some provision for levy-free media which are precluded from being used for copyright material (like the coloured tax-free fuel that is available in some places for farmers).
### Time for a Change
Tuesday, May 3rd, 2011
OK today must be the start of a four year campaign to reach agreement between the NDP, Greens, and remaining Liberals to:
1. Support an electoral reform which will provide for proportional representation – NOT based on party lists but on something more democratic like STV or some other preferential balloting system
2. Educate the Canadian public as to the acceptability, efficiency, and desirability of “coalition” governments – even (or perhaps especially) when not including the party which happens to have the most seats in parliament
3. Commit to endorsing strategic voting in the next Federal Election in order to achieve these ends
### The Escape of Osama Bin Laden
Tuesday, May 3rd, 2011
In retrospect it is obvious that the mission had no hope of success. Obviously anyone with the resources available to Osama bin Laden, when building a special-purpose hiding place, would have included an escape shaft from his apartment which led to a subterranean bomb shelter with a tunnel connecting it to the outside. At the first appearance of US helicopters OBL surely walked into the escape closet climbed down the shaft to his bomb shelter and by the time that the SEALs had fought their way up to his apartment he was well on the way down the half kilometer long tunnel from which he climbed out at an unknown location on the bank of a nearby gully or river. Good luck finding him now!
### Only Obama
Tuesday, May 3rd, 2011
Only Obama can deal with the financial collapse caused by the right wing libertarian attitude towards proper regulation of financial institutions
Only Obama could persuade us that the Lion King was born in Hawaii
But Only Limbaugh could come up with the slogan which will should drive the great Democratic election sweep of 2012 | 878 | 4,015 | {"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} | 2.765625 | 3 | CC-MAIN-2020-40 | latest | en | 0.953994 |
https://seekingalpha.com/article/4260779-eurozones-long-depression | 1,558,383,898,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256147.15/warc/CC-MAIN-20190520202108-20190520224108-00199.warc.gz | 634,329,841 | 99,235 | # The Eurozone's Long Depression
|
Includes: ADRU, BBEU, DBEU, DBEZ, DEUR, DEZU, DRR, EDOM, EEA, EPV, ERO, EUFL, EUFN, EUFX, EUO, EURL, EZU, FEEU, FEP, FEU, FEUZ, FEZ, FIEE, FIEU, FLEE, FXE, GSEU, HEDJ, HEZU, HFXE, IEUR, IEV, PTEU, RFEU, UEUR, ULE, UPV, URR, VGK
by: Frances Coppola
Summary
"Net saving" is what is left over after the private sector has paid its taxes, met its other obligations, fed and clothed itself, etc. It is a residual.
The public sector can also "net save." A government that is running a persistent budget surplus is "net saving."
Ever since the crisis, as net saving has increased in the eurozone, the rest of the world has benefited from its rising exports of capital.
The current account surplus, and associated capital exports, are protective.
Sectoral balances can tell us so much about what is going on in an economy. Especially when they are expressed as a time series, as in this remarkable chart from the ECB:
Although it is a time series, this is not a rate of change chart. The y axis is in billions of euros, not in percentage growth rates. But the chart nevertheless shows that eurozone net saving has risen steadily since the financial crisis, except during the eurozone crisis of 2011-12, when it dipped slightly.
What do we mean by "net saving"? The legend appears to conflate saving with investment, and the brief explanation at the bottom of the chart doesn't really help. So, here's some simple algebra to sort it out.
In national accounting, "saving" is the excess of income over desired consumption. For the private sector, it looks like this:
Sp = Y - T - C
where Y is the net income of the private sector from all sources, T is tax payments, and C is all other consumption.
Thus, "net saving" is what is left over after the private sector has paid its taxes, met its other obligations such as rent and debt service, fed and clothed itself, and bought the latest Mulberry handbag. It is a residual. So, when economists and politicians say "we need more saving in the economy," what they really mean is people and corporations should spend less.
The public sector can also "net save." It looks like this:
Sg = T - G
where T is tax revenue (the same T as in the private sector equation) and G is government spending. The OECD defines government net saving thus:
Net saving arises, and accrues over time, when revenues exceed expenditures without taking into account capital expenditures, such as public investment, transfers to publicly-owned enterprises or transfers to financial institutions (for instance, rescuing them during the financial crisis).
Thus, a government that is running a persistent budget surplus is "net saving."
Putting private and public "net saving" together gives us this:
Sp+g = Y - G - C
which is also equal to total net investment:
S = I
This identity is crucial. The chart above breaks down total net saving, S, into its investment (not saving) components:
S = Ih+ Ic + Ig + Ix
where h is households, c is corporations, g is government and x is the rest of the world (positive sign here indicates outward investment, i.e., capital exports).
The chart shows us that since 2009, with the exception of the two worst years of the eurozone crisis, the eurozone as a whole has been investing like crazy. But as Adam Tooze observed on Twitter, the investment has not gone into the eurozone:
It is clear from the chart that something fundamentally changed when the eurozone crisis struck. As Adam points out, government net investment completely vanished and has never returned. But it was not exactly large prior to the crisis. The chart shows that collapsing corporate investment is the real story of the eurozone crisis. European corporations all but stopped investing in the eurozone in 2012-13. Even now, corporate investment in the eurozone is barely higher than it was in 2011, in the wake of the Great Financial Crisis.
But European corporations didn't stop investing. They simply looked elsewhere. Net saving didn't dip much in the eurozone crisis - but the associated investment no longer went into the eurozone economy. It flowed out of the eurozone to the rest of the world. Ever since the crisis, as net saving has increased in the eurozone, the rest of the world has benefited from its rising exports of capital.
It wasn't only eurozone corporations that stopped investing in the eurozone. As Adam says, the growing external surplus is the flip side of the eurozone's current account balance. The chart shows that in 2011, the eurozone abruptly switched from net borrowing from the rest of the world to net lending to it. This is consistent with a "sudden stop," in which external investors abruptly pull their funds, causing severe economic damage. If so, then we would expect to see the current account switch from deficit to surplus at this time (net borrowing from the rest of the world indicates a current account deficit). And that is indeed what Eurostat shows:
The eurozone's "sudden stop" in the last quarter of 2011 is hardly news. Many people have commented on it. But the point is that it is driven by the behaviour of external investors. It was not just domestic investment that collapsed in the eurozone crisis. External investment fled too. And the persistently wide current account surplus since then tells us that it has never returned.
Economies don't always bounce back after a "sudden stop." If investment doesn't return, then the economy stays stuck in a slump. That is what has happened to the eurozone. It needs much, much more investment. Investment by domestic corporations is slowly growing, but the ECB says that households are still deleveraging. And government investment is all but absent, as balanced-budget rules proliferate across the eurozone and Brussels threatens to impose draconian sanctions on any country that dares run much of a deficit. Meanwhile, external investors stay away, put off by the eurozone's poor growth prospects and rising political risks.
But the eurozone authorities have very good reasons for pursuing policies that discourage external investment. The "sudden stop" in 2011 nearly destroyed the euro. Those in charge in the eurozone daren't risk another one. The current account surplus, and associated capital exports, are protective. After all, if you never borrow from the outside world, they can't hurt you. And if you are a net lender to the outside world, you have some leverage over them. Of course, they can still trash your currency, but as you don't need to service external debt, that doesn't cause you a major problem. It simply helps you to sell them more goods. Alternatively, they can refuse to buy your goods, in which case the loss of export income will make your population poorer. But hey, the people are already used to austerity, because that's how you maintain your current account surplus. They won't notice even more pain when the export income dries up.
In fact, the eurozone is behaving in much the same way as developing countries have since the Asian crisis of 1997, and for much the same reasons. Just as developing countries pursued export-led growth strategies and built up large FX reserves to protect themselves from sudden stops, so now the eurozone, scarred by its own "sudden stop," is doing the same. Perhaps eurozone leaders think the investment chill that is causing the eurozone's poor growth, high unemployment and stubbornly low inflation is a fair price to pay to keep the eurozone together.
But the eurozone is not a young, developing country whose population is generally living on a low income. It is a mature Western economy with a population that is used to a relatively high standard of living and whose net worth took an absolute beating not only in the 2007-08 financial crisis, but also in the eurozone crisis:
(Chart from the ECB)
Rising property prices seem to be the only means the people of the eurozone have of restoring their lost wealth. But for the young, already suffering from high unemployment, rising house prices mean an even more impoverished future. For how much longer will they tolerate this Long Depression?
Editor’s Note: The summary bullets for this article were chosen by Seeking Alpha editors. | 1,800 | 8,242 | {"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} | 2.53125 | 3 | CC-MAIN-2019-22 | latest | en | 0.949586 |
http://en.wikipedia.org/wiki/Homomorphic_secret_sharing | 1,406,851,570,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510273766.32/warc/CC-MAIN-20140728011753-00275-ip-10-146-231-18.ec2.internal.warc.gz | 98,361,551 | 12,734 | # Homomorphic secret sharing
In cryptography, homomorphic secret sharing is a type of secret sharing algorithm in which the secret is encrypted via homomorphic encryption. A homomorphism is a transformation from one algebraic structure into another of the same type so that the structure is preserved. Importantly, this means that for every kind of manipulation of the original data, there is a corresponding manipulation of the transformed data.[1]
## Technique
Homomorphic secret sharing is used to transmit a secret to several recipients as follows:
1. Transform the "secret" using a homomorphism. This often puts the secret into a form which is easy to manipulate or store. In particular, there may be a natural way to 'split' the new form as required by step (2).
2. Split the transformed secret into several parts, one for each recipient. The secret must be split in such a way that it can only be recovered when all or most of the parts are combined. (See secret sharing)
3. Distribute the parts of the secret to each of the recipients.
4. Combine each of the recipients' parts to recover the transformed secret, perhaps at a specified time.
5. Reverse the homomorphism to recover the original secret.
## Example: decentralized voting protocol
Suppose a community wants to perform an election, but they want to ensure that the vote-counters won't lie about the results. Using a kind of homomorphic secret sharing known as Shamir's secret sharing, each member of the community can put his vote into a form that can be split into pieces, then submit each piece to a different vote-counter. The pieces are designed so that the vote-counters can't predict how altering a piece of a vote will affect the whole vote; thus, vote-counters are discouraged from tampering with their pieces. When all votes have been received, the vote-counters combine all the pieces together, which allows them to reverse the alteration process and to recover the aggregate election results.
In detail, suppose we have an election with:
• Two possible outcomes, either yes or no. We'll represent those outcomes numerically by +1 and -1, respectively.
• A number of authorities, k, who will count the votes.
• A number of voters, n, who will submit votes.
Assume the election has two outcomes, so each member of the community can vote either yes or no. We'll represent those votes numerically by +1 and -1, respectively.
1. In advance, each authority generates a publicly available numerical key, xk.
2. Each voter encodes his vote in a polynomial pn according to the following rules: The polynomial should have degree k-1, its constant term should be either +1 or -1 (corresponding to voting "yes" or voting "no"), and its other coefficients should be randomly generated.
3. Each voter computes the value of his polynomial pn at each authority's public key xk.
• This produces k points, one for each authority.
• These k points are the "pieces" of the vote: If you know all of the points, you can figure out the polynomial pn (and hence you can figure out how the voter voted). However, if you know only some of the points, you can't figure out the polynomial. (This is because you need k points to determine a degree-k-1 polynomial. Two points determine a line, three points determine a parabola, etc.)
4. The voter sends each authority the value that was produced using the authority's key.
5. Each authority collects the values that he receives. Since each authority only gets one value from each voter, he can't discover any given voter's polynomial. Moreover, he can't predict how altering the submissions will affect the vote.
6. Once the voters have submitted their votes, each authority k computes and announces the sum Ak of all the values he's received.
7. There are k sums, Ak; when they are combined together, they determine a unique polynomial P(x)---specifically, the sum of all the voter polynomials: P(x) = p1(x) + p2(x) + … + pn(x).
• The constant term of P(x) is in fact the sum of all the votes, because the constant term of P(x) is the sum of the constant terms of the individual pn.
• Thus the constant term of P(x) provides the aggregate election result: if it's positive, more people voted for +1 than for -1; if it's negative, more people voted for -1 than for +1.
An illustration of the voting protocol. Each column represents the pieces of a particular voter's vote. Each row represents the pieces received by a particular authority.
### Features
This protocol works as long as not all of the $k$ authorities are corrupt — if they were, then they could collaborate to reconstruct $P(x)$ for each voter and also subsequently alter the votes.
The protocol requires t+1 authorities to be completed, therefore in case there are N>t+1 authorities, N-t-1 authorities can be corrupted, which gives the protocol a certain degree of robustness.
The protocol manages the IDs of the voters (the IDs were submitted with the ballots) and therefore can verify that only legitimate voters have voted.
Under the assumptions on t:
1. A ballot cannot be backtracked to the ID so the privacy of the voters is preserved.
2. A voter cannot prove how they voted.
3. It is impossible to verify a vote.
The protocol implicitly prevents corruption of ballots. This is because the authorities have no incentive to change the ballot since each authority has only a share of the ballot and has no knowledge how changing this share will affect the outcome.
### Vulnerabilities
• The voter cannot be certain that their vote has been recorded correctly.
• The authorities cannot be sure the votes were legal and equal, for example the voter can choose a value which is not a valid option (i.e. not in {-1, 1}) such as -20, 50 which will tilt the results in their favor.
## References
1. ^ Schoenmakers, Berry (1999). Advances in Cryptology 1666: 148–164. CiteSeerX: 10.1.1.102.9375. | 1,287 | 5,879 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "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-2014-23 | longest | en | 0.951127 |
https://www.my.freelancer.com/job-search/solve-httpd-problem-linux/ | 1,550,559,374,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489425.55/warc/CC-MAIN-20190219061432-20190219083432-00609.warc.gz | 910,526,755 | 38,370 | # Solve httpd problem linuxpekerjaan
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6,091 solve httpd problem linux tugasan ditemui, harga dalam USD
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My First Diagnostic Results – Need Help
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My First Diagnostic Results – Need Help [#permalink]
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18 Oct 2010, 07:46
Hello everyone,
This past Sunday I took my first GMAT practice test using the GMATPrep software. My total score was 480 Q: 27 V: 28. The schools I am targeting want to see 700+. I plan to take the GMAT April 2011. How can I go about making such a large jump in my performance on the test? Have you heard of someone improving this much before?
I am a great undergraduate student with a 3.7 GPA. This score definitely does not reflect the value I can add in an MBA program. I am open to any suggestions.
With appreciation,
zay1917
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18 Oct 2010, 07:52
Welcome to the group mate.
Build a plan and stick to that plan, you will rock on the test.
Have a look at these links gmat-study-plan-for-gmat-novices-start-your-gmat-journey-80727.html?highlight=ml301
gmat-prep-course-reviews-discount-codes-78451.html
best-gmat-stories-period-98512.html
_________________
http://drambedkarbooks.com/
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Re: My First Diagnostic Results – Need Help [#permalink]
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18 Oct 2010, 08:06
zay1917 wrote:
Hello everyone,
This past Sunday I took my first GMAT practice test using the GMATPrep software. My total score was 480 Q: 27 V: 28. The schools I am targeting want to see 700+. I plan to take the GMAT April 2011. How can I go about making such a large jump in my performance on the test? Have you heard of someone improving this much before?
I am a great undergraduate student with a 3.7 GPA. This score definitely does not reflect the value I can add in an MBA program. I am open to any suggestions.
With appreciation,
zay1917
Welcome zay1917 !
By the way what does your nickname mean?
Yes, you can achieve 700+ even with this score. Your verbal score is not so bad , it is in 47%. I personally struggled a lot to achieve 31, I have started my jorney from 19 in verbal.
Follow the link that kissthegmat posted above and get involved in forum this is a great community!
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Audaces fortuna juvat!
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Re: My First Diagnostic Results – Need Help [#permalink]
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18 Oct 2010, 10:41
Thanks to you both! Zay is my nickname and 1917 is the year my grandmother was born.
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Re: My First Diagnostic Results – Need Help [#permalink] 18 Oct 2010, 10:41
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My First Diagnostic Results – Need Help
Moderator: HiLine
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,036 | 3,828 | {"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} | 2.671875 | 3 | CC-MAIN-2017-43 | latest | en | 0.846786 |
http://www.listserv.uga.edu/cgi-bin/wa?A2=ind0908c&L=sas-l&D=0&F=P&P=2557 | 1,369,143,404,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700074077/warc/CC-MAIN-20130516102754-00059-ip-10-60-113-184.ec2.internal.warc.gz | 586,313,383 | 4,458 | Date: Sun, 16 Aug 2009 10:35:25 -0400 Joe Whitehurst "SAS(r) Discussion" Joe Whitehurst Re: sas gplot. aliitle basic question To: tal <33af868b-83ee-4696-9814-d01de1e80f0a@t13g2000yqt.googlegroups.com> text/plain; charset=ISO-8859-1
Here's an example program that you may find helpful for gaining an understanding of how to generate plots using PROC GLOT. This is a medium complexity example. You can generate a simple default plot with just this amount of code; Proc gplot date=mydata.sas_data_set; plot x*y; run;
Where x an y are variables on mydata.sas_data_set.
********************************************************************************************************************* Medium complexity Example **********************************************************************************************************************
%let begdate=%str('01sep99'd); %let enddate=%str('30sep99'd);
data trafsum2; set sub3.monthly; pgsviewd=pageview/100000; visitors=visitors/10000; run;
FILENAME trafsumm FTP '/sasweb/images/MTD_Traffic_Summary.gif' host='10.102.8.248' user='sascont1' pass='sas4auto99' recfm=s ;
goptions reset=all gaccess=gsasfile gsfmode=replace prompt dev=gif733 gsfname=trafsumm vpos=0 hpos=0 htext=1 ftext=swissb rotate=landscape border hby=1 fby=swissb cby=black ctext=black;; *goptions reset=all dev=win; *goptions reset=all rotate=landscape border dev=win targetdevice=gif733; title1 c=black f=swissb h=2 j=c ' '; title2 c=black f=swissb h=2 j=c 'TRAFFIC SUMMARY'; %annomac; data weekends; %system(2,1); %dclanno; text='WKND'; do day=&begdate to &enddate by 1; weekday=weekday(day); if weekday=7 then do; when='B'; %system(2,1); %bar(day,0,day+1,100,pay,0,s); text='WKND'; %label(day+.5,100,text,black,0,0,1,swissb,b); end; if weekday=1 then wkday='S'; else if weekday=2 then wkday='M'; else if weekday=3 then wkday='T'; else if weekday=4 then wkday='W'; else if weekday=5 then wkday='T'; else if weekday=6 then wkday='F'; else if weekday=7 then wkday='S'; %system(2,1); %label(day,97,wkday,black,0,0,1,swissb,b); end; run; symbol1 interpol=join line=1 value=dot h=.5 width=2 color=blue; symbol2 interpol=join line=1 value=dot h=.5 width=2 color=red; symbol3 interpol=join line=1 value=dot h=.1 width=1 color=red; symbol4 interpol=join line=1 value=dot h=.1 width=1 color=pink; symbol5 interpol=join line=1 value=dot h=.1 width=1 color=cyan; symbol6 interpol=join line=1 value=dot h=.1 width=1 color=white;
symbol7 interpol=join line=1 value=dot h=.1 width=1 color=red; symbol8 interpol=join line=1 value=dot h=.1 width=1 color=red; symbol9 interpol=join line=1 value=dot h=.1 width=1 color=blue; symbol10 interpol=join line=1 value=dot h=.1 width=1 color=pink; symbol11 interpol=join line=1 value=dot h=.1 width=1 color=cyan; symbol12 interpol=join line=1 value=dot h=.1 width=1 color=white;
axis1 order=(0 to 10 by .5) minor=(number=5) label=(rotate=0 angle=0 j=r h=1.25 f=swissb c=blue 'Daily' j=r 'Pages' j=r 'Viewed' j=r "(100k's)") value=(color=blue f=swissb) length=70pct offset=(0,2) width=3; axis2 order=('01sep99'd to '30sep99'd by day) minor=none /*(number=0) */ label=(color=black f=swissb h=1.25 'September 1999') value=(color=black f=swissb h=.8) length=80pct offset=(0,0) width=3; axis3 order=(0 to 14 by 1) minor=(number=3) label=(rotate=0 angle=0 j=l h=1.25 f=swissb c=red 'Daily ' j=l 'Visitors' j=l "(10K's)") value=(color=red f=swissb) length=70pct offset=(0,2) width=3; proc gplot data=trafsum2 anno=weekends; format date day2. pgsviewd mtdpages cvisitor 3.2; plot pgsviewd*date / vaxis=axis1 haxis=axis2 /*cframe=yellow*/; plot2 visitors*date /
vaxis=axis3 haxis=axis2; run; quit;
To control titles, plot axes, legends, etc., you use TITLE, AXIS, PATTERN, SYMBOL and LEGEND statements;
On Sun, Aug 16, 2009 at 9:23 AM, tal <talila.ar@gmail.com> wrote:
> Hi, > I'd like to do like a bar plot, but thinner (sorry for the stupid > question). > > My problem is that i don't know the term in english to look for this > types of plots in google. > > I'd like to draw something like what I draw below, only with vertical > lines in-between the stars ( the stars are not necessary- it's just > for explaining visually what I need) > > Thanks in advance > > > > * > * > * > * * > * * * > * * * > * * * > ____________________________________ >
Back to: Top of message | Previous page | Main SAS-L page | 1,403 | 4,346 | {"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-2013-20 | latest | en | 0.390224 |
https://statcompute.wordpress.com/2013/05/04/a-sas-macro-for-scorecard-evaluation-with-weights/ | 1,500,656,083,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423787.24/warc/CC-MAIN-20170721162430-20170721182430-00464.warc.gz | 726,396,051 | 33,498 | I can calculate the motion of heavenly bodies but not the madness of people. -Isaac Newton
## A SAS Macro for Scorecard Evaluation with Weights
On 09/28/2012, I posted a SAS macro evaluation the scorecard performance, e.g. KS / AUC statistics (https://statcompute.wordpress.com/2012/09/28/a-sas-macro-for-scorecard-performance-evaluation). However, this macro is not generic enough to handle cases with a weighting variable. In a recent project that I am working on, there is a weight variable attached to each credit applicant due to the reject inference. Therefore, there is no excuse for me to continue neglecting the necessity of developing another SAS macro that can take care of the weight variable with any positive values in the scorecard evaluation. Below is a quick draft of the macro. You might have to tweak it a little to suit your needs in the production.
```%macro wt_auc(data = , score = , y = , weight = );
***********************************************************;
* THE MACRO IS TO EVALUATE THE SEPARATION POWER OF A *;
* SCORECARD WITH WEIGHTS *;
* ------------------------------------------------------- *;
* PARAMETERS: *;
* DATA : INPUT DATASET *;
* SCORE : SCORECARD VARIABLE *;
* Y : RESPONSE VARIABLE IN (0, 1) *;
* WEIGHT: WEIGHT VARIABLE WITH POSITIVE VALUES *;
* ------------------------------------------------------- *;
* OUTPUTS: *;
* A SUMMARY TABLE WITH KS AND AUC STATISTICS *;
* ------------------------------------------------------- *;
* CONTACT: *;
* WENSUI.LIU@53.COM *;
***********************************************************;
options nocenter nonumber nodate mprint mlogic symbolgen
orientation = landscape ls = 125 formchar = "|----|+|---+=|-/\<>*";
data _tmp1 (keep = &score &y &weight);
set &data;
where &score ~= . and &y in (0, 1) and &weight >= 0;
run;
*** CAPTURE THE DIRECTION OF SCORE ***;
ods listing close;
ods output spearmancorr = _cor;
proc corr data = _tmp1 spearman;
var &y;
with &score;
run;
ods listing;
data _null_;
set _cor;
if &y >= 0 then do;
call symput('desc', 'descending');
end;
else do;
call symput('desc', ' ');
end;
run;
proc sql noprint;
create table
_tmp2 as
select
&score as _scr,
sum(&y) as _bcnt,
count(*) as _cnt,
sum(case when &y = 1 then &weight else 0 end) as _bwt,
sum(case when &y = 0 then &weight else 0 end) as _gwt
from
_tmp1
group by
&score;
select
sum(_bwt) into :bsum
from
_tmp2;
select
sum(_gwt) into :gsum
from
_tmp2;
select
sum(_cnt) into :cnt
from
_tmp2;
quit;
%put &cnt;
proc sort data = _tmp2;
by &desc _scr;
run;
data _tmp3;
set _tmp2;
by &desc _scr;
retain _gcum _bcum _cntcum;
_gcum + _gwt;
_bcum + _bwt;
_cntcum + _cnt;
_gpct = _gcum / &gsum;
_bpct = _bcum / &bsum;
_ks = abs(_gpct - _bpct) * 100;
_rank = int(_cntcum / ceil(&cnt / 10)) + 1;
run;
proc sort data = _tmp3 sortsize = max;
by _gpct _bpct;
run;
data _tmp4;
set _tmp3;
by _gpct _bpct;
if last._gpct then do;
_idx + 1;
output;
end;
run;
proc sql noprint;
create table
_tmp5 as
select
a._gcum as gcum,
(b._gpct - a._gpct) * (b._bpct + a._bpct) / 2 as dx
from
_tmp4 as a, _tmp4 as b
where
a._idx + 1 = b._idx;
select
sum(dx) format = 15.8 into :AUC
from
_tmp5;
select
max(_ks) format = 15.8 into :KS_STAT
from
_tmp3;
select
_scr format = 6.2 into :KS_SCORE
from
_tmp3
where
_ks = (select max(_ks) from _tmp3);
create table
_tmp6 as
select
_rank as rank,
min(_scr) as min_scr,
max(_scr) as max_scr,
sum(_cnt) as cnt,
sum(_gwt + _bwt) as wt,
sum(_gwt) as gwt,
sum(_bwt) as bwt,
sum(_bwt) / calculated wt as bad_rate
from
_tmp3
group by
_rank;
quit;
proc report data = _last_ spacing = 1 split = "/" headline nowd;
column("GOOD BAD SEPARATION REPORT FOR %upcase(%trim(&score)) IN DATA %upcase(%trim(&data))/
MAXIMUM KS = %trim(&ks_stat) AT SCORE POINT %trim(&ks_score) and AUC STATISTICS = %trim(&auc)/ /"
rank min_scr max_scr cnt wt gwt bwt bad_rate);
define rank / noprint order order = data;
define min_scr / "MIN/SCORE" width = 10 format = 9.2 analysis min center;
define max_scr / "MAX/SCORE" width = 10 format = 9.2 analysis max center;
define cnt / "RAW/COUNT" width = 10 format = comma9. analysis sum;
define wt / "WEIGHTED/SUM" width = 15 format = comma14.2 analysis sum;
define gwt / "WEIGHTED/GOODS" width = 15 format = comma14.2 analysis sum;
define bwt / "WEIGHTED/BADS" width = 15 format = comma14.2 analysis sum;
define bad_rate / "BAD/RATE" width = 10 format = percent9.2 order center;
rbreak after / summarize dol skip;
run;
proc datasets library = work nolist;
delete _tmp: / memtype = data;
run;
quit;
***********************************************************;
* END OF THE MACRO *;
***********************************************************;
%mend wt_auc;
```
In the first demo below, a weight variable with fractional values is tested.
```*** TEST CASE OF FRACTIONAL WEIGHTS ***;
data one;
set data.accepts;
weight = ranuni(1);
run;
%wt_auc(data = one, score = bureau_score, y = bad, weight = weight);
/*
GOOD BAD SEPARATION REPORT FOR BUREAU_SCORE IN DATA ONE
MAXIMUM KS = 34.89711721 AT SCORE POINT 678.00 and AUC STATISTICS = 0.73521009
MIN MAX RAW WEIGHTED WEIGHTED WEIGHTED BAD
SCORE SCORE COUNT SUM GOODS BADS RATE
-------------------------------------------------------------------------------------------
443.00 619.00 539 276.29 153.16 123.13 44.56%
620.00 644.00 551 273.89 175.00 98.89 36.11%
645.00 660.00 544 263.06 176.88 86.18 32.76%
661.00 676.00 555 277.26 219.88 57.38 20.70%
677.00 692.00 572 287.45 230.41 57.04 19.84%
693.00 707.00 510 251.51 208.25 43.26 17.20%
708.00 724.00 576 276.31 243.89 32.42 11.73%
725.00 746.00 566 285.53 262.73 22.80 7.98%
747.00 772.00 563 285.58 268.95 16.62 5.82%
773.00 848.00 546 272.40 264.34 8.06 2.96%
========== ========== ========== =============== =============== ===============
443.00 848.00 5,522 2,749.28 2,203.49 545.79
*/
```
In the second demo, a weight variable with positive integers is also tested.
```*** TEST CASE OF INTEGER WEIGHTS ***;
data two;
set data.accepts;
weight = rand("poisson", 20);
run;
%wt_auc(data = two, score = bureau_score, y = bad, weight = weight);
/*
GOOD BAD SEPARATION REPORT FOR BUREAU_SCORE IN DATA TWO
MAXIMUM KS = 35.58884479 AT SCORE POINT 679.00 and AUC STATISTICS = 0.73725030
MIN MAX RAW WEIGHTED WEIGHTED WEIGHTED BAD
SCORE SCORE COUNT SUM GOODS BADS RATE
-------------------------------------------------------------------------------------------
443.00 619.00 539 10,753.00 6,023.00 4,730.00 43.99%
620.00 644.00 551 11,019.00 6,897.00 4,122.00 37.41%
645.00 660.00 544 10,917.00 7,479.00 3,438.00 31.49%
661.00 676.00 555 11,168.00 8,664.00 2,504.00 22.42%
677.00 692.00 572 11,525.00 9,283.00 2,242.00 19.45%
693.00 707.00 510 10,226.00 8,594.00 1,632.00 15.96%
708.00 724.00 576 11,497.00 10,117.00 1,380.00 12.00%
725.00 746.00 566 11,331.00 10,453.00 878.00 7.75%
747.00 772.00 563 11,282.00 10,636.00 646.00 5.73%
773.00 848.00 546 10,893.00 10,598.00 295.00 2.71%
========== ========== ========== =============== =============== ===============
443.00 848.00 5,522 110,611.00 88,744.00 21,867.00
*/
```
As shown, the macro works well in both cases. Please feel free to let me know if it helps in your cases. | 2,543 | 8,893 | {"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-2017-30 | longest | en | 0.690119 |
https://www.physicsforums.com/threads/euler-spiral-forces.817011/ | 1,611,780,871,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704832583.88/warc/CC-MAIN-20210127183317-20210127213317-00063.warc.gz | 925,702,356 | 23,294 | # Euler spiral forces?
Trying to figure out direction of forces of an object traveling on an Euler spiral path. As an example if you had an astronaut with a jetpack and he wanted to change his direction 90 degrees he could aim his thrusters outward from the center of a circle and he would turn at a constant rate with a constant radius. But if he wanted to change his direction as quickly as possible and shrink the radius he was traveling on as quickly as possible in which direction would he aim his thrusters? I don't have a higher math understanding so please keep things intuitive if possible. Thanks
haruspex
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Trying to figure out direction of forces of an object traveling on an Euler spiral path. As an example if you had an astronaut with a jetpack and he wanted to change his direction 90 degrees he could aim his thrusters outward from the center of a circle and he would turn at a constant rate with a constant radius. But if he wanted to change his direction as quickly as possible and shrink the radius he was traveling on as quickly as possible in which direction would he aim his thrusters? I don't have a higher math understanding so please keep things intuitive if possible. Thanks
First, I need to be sure we have the problem properly defined. There is an astronaut travelling with velocity v. She has a thruster that can be pointed in any direction and delivers constant thrust. She wishes to travel at the same speed but at right angles to v, and do so in the least time. Right?
What is the overall momentum change? What is the relationship between force, momentum and time?
Thruster can be pointed in any direction and offers constant thrust. Speed can vary. Let's say she is starting out at 100 mph in a straight line and wants to change direction 90 degrees by changing her radius of travel as quickly as possible. I would assume the path taken would be a Euler spiral. The idea is to get through 90 degrees of the spiral as quickly as possible. Trying to figure out ideal direction for her to point the thruster in order to accomplish this.
haruspex
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Thruster can be pointed in any direction and offers constant thrust. Speed can vary. Let's say she is starting out at 100 mph in a straight line and wants to change direction 90 degrees by changing her radius of travel as quickly as possible. I would assume the path taken would be a Euler spiral. The idea is to get through 90 degrees of the spiral as quickly as possible. Trying to figure out ideal direction for her to point the thruster in order to accomplish this.
I don't know why you think the answer should be an Euler spiral. The application of those to transport is to avoid too rapid a change in acceleration. It sounds like that is not a constraint here.
I repeat my original questions. If the mass is m, the initial velocity is ##\vec v_i## and the final velocity is ##\vec v_f## :
What is the overall momentum change?
What is the relationship between force, momentum and time?
Thanks for your help. I"m actually not sure what you are asking, but let me try posing the question a different way with illustrations to see if that helps.
Let's say the astronaut is traveling along a line at 100 mph. They need to reach a parallel line as quickly as possible, but they need to have stopped all their velocity along the initial axis by the point they hit the second line. We can say the lines are 100 feet apart if that helps. We can also say their combined mass and thrust capabilities allowed 1 g of acceleration. One way of doing this would be to slow down to whatever speed was needed and then take a circular path with constant radius and velocity by pointing their thruster in the opposite direction to the center point of a circle they traveled on.
Alternatively I would think it would be faster if they used their thruster to slow down and turn at the same time. I would think that would give them a Euler spiral shaped path, but please let me know if I am wrong. I'm primarily wondering where the center of force they aim their thruster opposite of would be for this path. Would it move?
sophiecentaur
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If the thrusters deliver a fixed thrust, the best direction to fire them must (surely?) be towards the centre of a circular path with a radius, set by the equation
F = mv2/r
I can't think of a path that would achieve a 90° change of direction, any quicker, with the same final speed.
There are variations around this; for instance, when she/he doesn't insist on the speed being maintained. The quickest way to just get a 90° course change would be to fire them backwards and leave just enough 'squirt' to give a finite velocity at right angles to the original.
PS Are we discussing Sandra Bullock or George Clooney here?
Thanks for the input. We'll say it's Sandra:) The velocity doesn't have to be maintained. Not really concerned with what the final velocity is just the quickest way to get to 90 degrees on the second line. I'm more concerned with the "squirt". What angle would that need to be at? Would it change and how could it be calculated?
The circular path will only be the quickest if the thruster's maximum thrust is exactly what would be needed to make a turn with a radius equal to the distance between the lines. If the thrust is less then that, then the astronaut will still have forward velocity when they reach the 2'nd line. If the thruster is capable of higher thrust then a quarter circle turn would have to be followed by a straight path to reach the second line.
This is actually a very interesting problem. I'm not sure what the solution is yet, but its only a quarter circle in 1 very special case.
Thanks, yeah I was not thinking the circular path would be the ideal solution, I just put it there as an example of a possible solution to explain the problem better. I put the 100 mph, 100 feet, 1 g constraints because I would think that would require deceleration prior to the circular path. Or I was thinking ideally combining the deceleration with the turning in the Euler spiral example. Not sure if you can tell by the illustrations, but the blue line in the 2nd illustration is the 1st 90 degrees of a Euler spiral.
sophiecentaur
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2020 Award
The circular path will only be the quickest if the thruster's maximum thrust is exactly what would be needed to make a turn with a radius equal to the distance between the lines. If the thrust is less then that, then the astronaut will still have forward velocity when they reach the 2'nd line. If the thruster is capable of higher thrust then a quarter circle turn would have to be followed by a straight path to reach the second line.
This is actually a very interesting problem. I'm not sure what the solution is yet, but its only a quarter circle in 1 very special case.
I was assuming that the turn is to be as quick as possible and with no change of speed (and that gives you the radius). If you have more thrust, then the radius (and the time taken) can both be less.
I'm more concerned with the "squirt". What angle would that need to be at?
How long is a piece of string? If you are interested in a 'vestigial' final speed - just in the wanted direction - the direction would be in the reverse direction, with a squirt at right angles at the very end. The two could be combined into just one constant direction, of course and if the reverse thrust is much larger than the lateral thrust, you could simplify the situation into x and y co ordinates with the thrust being directed in a global sense, rather than relative to the craft. You would then just have motion under constant acceleration in both x and y directions, so the resultant would be the combination of two parabolas. I expect that curve will have been given a name by someone but it would be easy enough to plot it out with the help of Excel or equivalent. I don't think that would give an optimal course, though - that is if the circular path would be optimal.
haruspex
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Thanks, yeah I was not thinking the circular path would be the ideal solution, I just put it there as an example of a possible solution to explain the problem better. I put the 100 mph, 100 feet, 1 g constraints because I would think that would require deceleration prior to the circular path. Or I was thinking ideally combining the deceleration with the turning in the Euler spiral example. Not sure if you can tell by the illustrations, but the blue line in the 2nd illustration is the 1st 90 degrees of a Euler spiral.
If you only care that the final speed equals the initial speed then there is a quicker and simpler solution than a circular path. The questions I asked you lead to this.
If you don't even care about the final speed then the question ceases to make sense. You can drop the final speed to zero, so that there is not even a direction at the end. The problem then reduces to stopping as quickly as possible.
What would be the simpler solution than circular with constant speed? I don't actually care about final speed, but am curious what you mean here.
Not sure what you mean by question not making sense if I don't care about final speed. Maybe people are misunderstanding the problem. The colored paths just show the curved part in between the two lines, but the astronaut is assumed to keep going after the end of the path at whatever final speed they reach. If their speed needed to drop to zero that seems like it would take longer than if they were able to turn 90 degrees and cross the line while they were still going 50 mph or whatever speed is attainable on that radius at the end of the spiral.
sophiecentaur
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2020 Award
but the astronaut is assumed to keep going after the end of the path at whatever final speed they reach
That doesn't make sense to me. "whatever final speed" implies it can be arbitrarily low, as haruspex states, and the transition is accomplished in the shortest possible time.
There is another point to consider and that is the total amount of energy available needs to be specified - otherwise, the final speed could be as high as you like.
I think it is necessary to restate the OP more precisely if we are to progress this further.
Thanks, for calculation purposes I had put that the mass and thrust of the astronaut allows 1 g of constant acceleration. Starting speed of 100 mph. The lines are 100 feet apart. Is there something else we need?
I would be interested in how to calculate the speed they are at as they reach the second line, but mainly interested in which direction their thruster would be pointed to accomplish this movement in the minimum time possible.. Intuitively I would think there is a certain amount of force needed to stop all movement along the original axis and then a certain amount needed to cross the 100 feet at the same time. I would think the thruster would want to be aimed primarily opposite their original direction of travel, but aimed away from the second line at some shallow angle. Maybe this angle would ideally change as they approached the second line. Hopefully this is making sense. Sorry, I don't have much formal physics/math training beyond high school.
haruspex
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What would be the simpler solution than circular with constant speed? I don't actually care about final speed, but am curious what you mean here.
Not sure what you mean by question not making sense if I don't care about final speed. Maybe people are misunderstanding the problem. The colored paths just show the curved part in between the two lines, but the astronaut is assumed to keep going after the end of the path at whatever final speed they reach. If their speed needed to drop to zero that seems like it would take longer than if they were able to turn 90 degrees and cross the line while they were still going 50 mph or whatever speed is attainable on that radius at the end of the spiral.
I mean there is a simpler solution in regards to what the astronaut has to do. (Remember, you have provided no means by which the astronaut will rotate on her own axis, so we can assume she faces in a constant direction. Arranging to move in a circular path could be tricky.)
If she does move in a circular path under constant thrust, the speed will be constant. You can deduce the radius from F=mv^2/r. How long will it take that way?
If you do not care about final speed, she could just use the thruster to come to a stop, then with one tiny blast at right angles move off in the desired direction. How long will that take? Which is quicker?
The interesting case is where the final speed is required to equal the original speed but you don't require it to be constant during the transition. For this, think in terms of the overall change in momentum required, and how that would be most obviously achieved.
Let's assume we can create thrust in any direction instantaneously so we don't have to worry about rotation. I would think the stopping and then right angle approach would be slower than even the circular method. I was thinking the combined turning/slowing of the euler spiral shaped path would be closer to ideal
haruspex
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Let's assume we can create thrust in any direction instantaneously so we don't have to worry about rotation. I would think the stopping and then right angle approach would be slower than even the circular method. I was thinking the combined turning/slowing of the euler spiral shaped path would be closer to ideal
You may think those things, but you would be wrong.
The reason you think stopping and then moving off at a right angle would be slow is, I suggest, that intuitively you are thinking in terms of then accelerating to regain the original speed. But you have said that is not required, so it is only the time taken to stop that counts.
You seem reluctant to do the calculations. They are not difficult. If you show some attempt I'm happy to assist.
Okay, just to make sure I understand, bringing your speed along the original line to zero and then turning the thruster 90 degrees and accelerating directly toward the second line is the fastest way to do it? Honestly I would have no clue where to even begin calculating these things.
Another idea is to work in a reference frame where the object would start out at rest. If you start out at rest and need to get to a speed (vx,vy,vz), I think it should be obvious in wich direction you need to thrust.
haruspex
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Okay, just to make sure I understand, bringing your speed along the original line to zero and then turning the thruster 90 degrees and accelerating directly toward the second line is the fastest way to do it?
Only if you really do not care what your new speed is. Applying the thrust for a tiny fraction of a second in the new direction is enough to get you moving, so does not effectively contribute to the total time.
How long does it take to come to a stop, starting at speed v and decelerating under a constant force F? Since the force is constant (in magnitude and direction) the acceleration is constant, F/m. Are you familiar with the SUVAT equations? If not, use your favourite search engine.
Okay so if we are starting at 100 mph then it would take 4.56 seconds to reach 0 mph if I'm doing that right. Then we need to go 100 feet so not sure how to calculate exactly how long that would take at 1 g, but I am estimating around 2 seconds if we can do 32 feet per second squared. That would be around 6.56 seconds total.
edit: I think I underestimated the time it would take to travel the 100 feet by a little bit.
Correct me if I'm wrong, but intuitively that doesn't seem like the fastest way because if we switched the order and did the 100 feet first we would continue traveling at approx. 64 fps past the line as we slowed down to 0 mph on the initial axis of travel and would be 291 feet past the second line by the time we finally stopped moving along our initial axis.
If we started our sideways thrust first and wanted to time it so that we hit the line at the same time we reached 0 on initial axis it seems we would need to create an initial sideways velocity of 22 feet per second which would take around 2/3 of a second of thrust. Then we turn our thrusters and slow down to 0 on the initial axis which would again take 4.56 seconds. Total time of a little over 5 seconds. Except for the initial little sideways thrust this seems like it would create a path at least similar to a Euler spiral. Would this be the quickest way? Or would combining the 2 different thrust periods into one single event at some angle in the middle be even faster than this?
Last edited:
haruspex
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if we are starting at 100 mph
It will be more educational if we stick to symbolic values and see how the formulae turn out.
we need to go 100 feet
What 100 feet? This seems like a new constraint you are adding.
1 g
1g? Is that the available thrust? Again, let's stick to symbols: Initial speed u, mass m, available thrust F, time t.
it would create a path at least similar to a Euler spiral
I don't know why you're so keen on Euler spirals in this context. As I mentioned, the benefit of Euler spirals is in minimising 'jerk'. There will definitely be solutions that take less time by not worrying about jerk.
A.T.
Okay, just to make sure I understand, bringing your speed along the original line to zero and ...
Since you don’t care about final speed, you can stop right here.
A.T.
If the thrusters deliver a fixed thrust, the best direction to fire them must (surely?) be towards the centre of a circular path with a radius, set by the equation
F = mv2/r
I can't think of a path that would achieve a 90° change of direction, any quicker, with the same final speed.
Firing the thruster at 45° to the original direction is faster.
sophiecentaur
Gold Member
2020 Award
Firing the thruster at 45° to the original direction is faster.
I guess someone needs to do the actual sums to prove or disprove that. It would be a matter of comparing the time to zero velocity with acceleration of -a√2 and the time for a quarter turn with a centrepetal acceleration a.
Volunteers please. I have to go shopping right now. | 3,991 | 18,335 | {"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} | 3.96875 | 4 | CC-MAIN-2021-04 | latest | en | 0.984013 |
http://www.jiskha.com/display.cgi?id=1217315474 | 1,462,410,046,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860125750.3/warc/CC-MAIN-20160428161525-00109-ip-10-239-7-51.ec2.internal.warc.gz | 609,492,557 | 3,670 | Wednesday
May 4, 2016
# Homework Help: Math
Posted by AK on Tuesday, July 29, 2008 at 3:11am.
If 31/2 pounds of bananas cost \$.98, how much would one pound cost?
A. \$.14
B. \$.18
C. \$.20
D. \$.28 | 80 | 202 | {"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-2016-18 | longest | en | 0.920633 |
https://uclocks.com/how-many-combinations-on-a-3-digit-lock-discover-the-possibilities/ | 1,685,691,086,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648465.70/warc/CC-MAIN-20230602072202-20230602102202-00134.warc.gz | 639,987,535 | 21,869 | # How Many Combinations on a 3-Digit Lock? Discover the Possibilities
Are you wondering how many combinations are possible on a 3-digit lock? Whether you’re considering purchasing a new lock or just curious about the math behind it, understanding the number of possible combinations can provide valuable insights into your security options.
## The Math Behind It
A three-digit lock has three columns. Each column can represent one of ten possibilities: the numbers 0 through 9. Therefore, for any single column, there are ten possible options. To calculate the total number of combinations possible with a three-digit combination lock, we need to multiply these individual probabilities together.
The calculation looks like this:
10 x 10 x 10 = 1000
Therefore, there are 1,000 possible combinations on a typical cheap three-digit combination luggage lock (or bike locks).
## Security Considerations
Knowing that there are only 1,000 potential combinations can put your mind at ease regarding an unsecured item such as gym locker or bicycle. However, when it comes to securing more high-value items such as safes or door locks and come in higher price ranges do use complicated algorithms and multi-point locking mechanisms providing increased levels of protection.
When choosing a combination lock for high-security needs their experience varies from user to user so it’s difficult to say what would be suit best for everyone but most high-end products provide over tens thousands even millions permutations allowing maximum level of safety which is virtually impossible crack in limited time according to statistics shown by manufacturer themselves.
But despite all technological wonders used by industry leaders figures still show us that successfully penetrating these types takes relentless effort and highly sophisticated knowledge therefore minimizing risk compared with common everyday locks with clear passwords like “123” ,”abcd”.
## Tips for Choosing Your Combination Lock
If you plan on using a combination lock for securing valuable possessions or accessing sensitive areas my recommendation is never underestimating how much harder a longer and more complex password will be to crack.
Here are some tips for choosing your combination lock:
• Look for locks with higher number codes. These locks offer a greater number of potential combinations, making them more secure.
• Choose passwords that are not easy to guess such as birthdays or popular phrases
• Avoid using common sequences such as 1234, 4321 or counting up and down from the middle point
By following these guidelines, you’ll increase the security level of your lock while also ensuring it’s easier for you to remember.
## Conclusion
In conclusion, a typical cheap three-digit combination lock provides only 1,000 possible permutations. For high-security applications like safes and door locks companies provide far more substantial locking mechanisms with millions of permutation combinations which reduces the risk drastically.
When purchasing any kind of lock make sure you consider security requirements appropriate for what you’re trying to protect – all expensive items should be protected better than their cheaper counterparts. By doing this along with selecting an effective code strategy can safeguard both personal possessions and sensitive areas against intruders who lack strong hacking abilities.
If you follow these recommendations when choosing your next combination lock — whether working out at gym or securing property–you’ll enhance your protection level considerably while enjoying peace of mind knowing that cracking it will require relentless effort accompanied by sophisticated knowledge.
## FAQs
Sure, here are three popular FAQs with answers for “How Many Combinations on a 3-Digit Lock? Discover the Possibilities”:
Q: How many possible combinations are there on a 3-digit lock?
A: There are 1,000 possible combinations on a 3-digit lock. This is because each digit can be one of ten numbers (0-9), and there are three digits in total. Therefore, you can calculate the number of possibilities using the formula n^r, where n = number of options per digit (10) and r = number of digits (3).
Q: How long does it take to crack a 3-digit combination lock?
A: The time it takes to crack a 3-digit combination lock depends on several factors such as the type of tools being used by an attacker, whether or not they have access to multiple attempts at guessing combinations, etc., but generally speaking with modern techniques this would be rather quick for an experienced locksmith. However, if you use unique codes that aren’t easily guessable like common sequences or repeating numbers it will offer much stronger protection.
Q: Can I change my own combination on a 3-digit lock?
A: It depends on the specific model and design of your lock; some models allow users to change their own combinations while others require specialized knowledge or tools to do so safely and correctly.\ For most locks that allow self-programming new codes involves resetting tumblers within metal shackle which can often include instructions included with packaging depending upon manufacturer & product details. | 968 | 5,211 | {"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} | 2.890625 | 3 | CC-MAIN-2023-23 | longest | en | 0.911312 |
https://www.jiskha.com/display.cgi?id=1268623939 | 1,516,425,469,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889325.32/warc/CC-MAIN-20180120043530-20180120063530-00482.warc.gz | 907,692,275 | 4,399 | # Chemistry
posted by .
A 25.00 ml sample of a standard solution containing 1 g of CaCO3/L required 25.20 ml of EDTA to fully complex the Ca present. Calculate the volume of EDTA stoichiometrically equivalent to 1.0 mg of CaCO3.
From the volume of EDTA required in titrating the sample, calculate the equivalent mass of CaCO3 in the aliquot of sample titrated.
Volume of sample: 100 ml
Volume of CaCO3 taken: 25.0 mL
Volume EDTA needed for 25 ml CaCO3 :25.20 mL
Volume EDTA needed for sample alone: 3.2225 mL
• Chemistry -
Do you have two problems or one problem here?
• Chemistry -
It's two problems.
The first one I got 1.01 mL EDTA.
Oh, it's 25.00 ml sample of a standard solution containing 1 mg of CaCO3/L.
The second problem, I'm not sure how to approach it.
• Chemistry -
I am totally confused about the concn. I don't know if it is 1 g/L as in the original post or if the 1 mg/L is a correction.
• Chemistry -
I'm sorry, it's a correction.
• Chemistry -
bump
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# The Druid
The Druid Army has only one piece of each type. The pawns were created by taking the 16 possible combinations of Ferz, Wazir, Alfil, and Dabaaba move and capture, and getting rid of the Alfil and Dabaaba, and the ones with move and capture that went the same directions with different lengths. Those that have a two-square move or capture were given a feline name, the others were given a canine name. The Rook, Knight, and Bishop are replaced with one identical piece and another piece of roughly the same value with a similar feel. Each of those pieces is used in one combination piece to get the replacement pieces for the Queen, Marshall, and Cardinal.
### Starting setup
The starting setup for black is the mirror image of that for white.
``` +---+---+---+---+---+---+---+---+---+---+
3 | Fo| J | Co| L | Pu| Ch| T | W | Pa| D |
+---+---+---+---+---+---+---+---+---+---+
2 | B | H | Ea| R | K | U | Pe| Fa| S | |
+---+---+---+---+---+---+---+---+---+---+
1 | | | | | | | | | | El|
+---+---+---+---+---+---+---+---+---+---+
a b c d e f g h i j
```
### Piece Listings
Each piece is listed with its abbreviation in parentheses after its name, followed by the notation for its move in Ralph Betza's notation in brackets, if its move can be expressed in that notation.
#### Fox (Fo) [F]
The Fox moves one square diagonally.
#### Dog (D) [W]
The Dog moves one square orthogonally.
#### Jaguar (J) [mAcW]
The Jaguar either jumps two squares diagonally without capturing or moves one square orthogonally to capture.
#### Panther (Pa) [mDcF]
The Panther either jumps two squares orthogonally without capturing or moves one square diagonally to capture.
#### Coyote (Co) [mFcW]
The Coyote moves either one square diagonally without capturing or one square orthogonally to capture.
#### Wolf (W) [mWcF]
The Wolf moves either one square orthogonally without capturing or one square diagonally to capture.
#### Lion (L) [mFcD]
The Lion either moves one square diagonally without capturing or jumps two squares orthogonally to capture.
#### Tiger (T) [mWcA]
The Tiger either moves one square orthogonally without capturing or jumps two squares diagonally to capture.
#### Puma (Pu) [mDcA]
The Puma jumps either two squares orthogonally without capturing or two squares diagonally to capture.
#### Cheetah (Ch) [mAcD]
The Cheetah jumps either two squares diagonally without capturing or two squares orthogonally to capture.
#### Bear (B) [R4F]
The Bear moves either up to four squares orthogonally or one square diagonally.
#### Elephant (El) [R]
The Elephant moves like a Rook.
#### Horse (H) [N]
The Horse moves like a Knight.
#### Stag (S) [WH]
The Stag either moves one square orthogonally or jumps three squares orthogonally.
#### Eagle (Ea)
The Eagle either moves two or more squares diagonally or jumps two squares orthogonally.
#### Falcon (Fa) [B]
The Falcon moves like a Bishop.
#### Pegasus (Pe)
The Pegasus has the combined moves of an Eagle and a Stag.
#### Roc (R) [Q]
The Roc moves like a Queen.
#### Unicorn (U) [R4FN]
The Unicorn has the combined moves of a Knight and a Bear.
#### Druid (K) [K]
The Druid is identical to a King in standard Chess, except that it may not castle.
Written by Peter Hatch.
WWW page created: November 26, 2000. | 936 | 3,493 | {"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} | 2.53125 | 3 | CC-MAIN-2023-06 | latest | en | 0.924178 |
http://www.jiskha.com/display.cgi?id=1285884146 | 1,498,615,775,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128322275.28/warc/CC-MAIN-20170628014207-20170628034207-00453.warc.gz | 559,577,062 | 3,489 | # math
posted by .
copy and place parentheses to make each statement true
22. 6*6+6*6=426
23. 6+6/6*6-6=0
24. 6-6*66+6/6=1
25. 6+6/6+6*6=6
26. 6-6/6*6+6=0 | 89 | 157 | {"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} | 3.03125 | 3 | CC-MAIN-2017-26 | latest | en | 0.269574 |
https://nrich.maths.org/public/topic.php?code=-40&cl=4&cldcmpid=2816 | 1,582,831,690,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146809.98/warc/CC-MAIN-20200227191150-20200227221150-00223.warc.gz | 487,050,601 | 6,583 | # Resources tagged with: Mathematical induction
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Broad Topics > Mathematical Thinking > Mathematical induction
### Fibonacci Fashion
##### Age 16 to 18 Challenge Level:
What have Fibonacci numbers to do with solutions of the quadratic equation x^2 - x - 1 = 0 ?
### Golden Powers
##### Age 16 to 18 Challenge Level:
You add 1 to the golden ratio to get its square. How do you find higher powers?
### Farey Fibonacci
##### Age 16 to 18 Short Challenge Level:
Investigate Farey sequences of ratios of Fibonacci numbers.
### OK! Now Prove It
##### Age 16 to 18 Challenge Level:
Make a conjecture about the sum of the squares of the odd positive integers. Can you prove it?
### Golden Fractions
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Find the link between a sequence of continued fractions and the ratio of succesive Fibonacci numbers.
### Binary Squares
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If a number N is expressed in binary by using only 'ones,' what can you say about its square (in binary)?
### Farey Neighbours
##### Age 16 to 18 Challenge Level:
Farey sequences are lists of fractions in ascending order of magnitude. Can you prove that in every Farey sequence there is a special relationship between Farey neighbours?
### An Introduction to Mathematical Induction
##### Age 16 to 18
This article gives an introduction to mathematical induction, a powerful method of mathematical proof.
### Gosh Cosh
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Explore the hyperbolic functions sinh and cosh using what you know about the exponential function.
### Particularly General
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By proving these particular identities, prove the existence of general cases.
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In the limit you get the sum of an infinite geometric series. What about an infinite product (1+x)(1+x^2)(1+x^4)... ?
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When is $7^n + 3^n$ a multiple of 10? Can you prove the result by two different methods?
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Which is larger: (a) 1.000001^{1000000} or 2? (b) 100^{300} or 300! (i.e.factorial 300)
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With n people anywhere in a field each shoots a water pistol at the nearest person. In general who gets wet? What difference does it make if n is odd or even?
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Find the values of n for which 1^n + 8^n - 3^n - 6^n is divisible by 6.
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Add powers of 3 and powers of 7 and get multiples of 11.
### Dirisibly Yours
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Find and explain a short and neat proof that 5^(2n+1) + 11^(2n+1) + 17^(2n+1) is divisible by 33 for every non negative integer n.
### One Basket or Group Photo
##### Age 7 to 18 Challenge Level:
Libby Jared helped to set up NRICH and this is one of her favourite problems. It's a problem suitable for a wide age range and best tackled practically.
### Counting Binary Ops
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How many ways can the terms in an ordered list be combined by repeating a single binary operation. Show that for 4 terms there are 5 cases and find the number of cases for 5 terms and 6 terms.
### Symmetric Tangles
##### Age 14 to 16
The tangles created by the twists and turns of the Conway rope trick are surprisingly symmetrical. Here's why!
##### Age 14 to 16 Challenge Level:
A walk is made up of diagonal steps from left to right, starting at the origin and ending on the x-axis. How many paths are there for 4 steps, for 6 steps, for 8 steps?
### Binomial Coefficients
##### Age 14 to 18
An introduction to the binomial coefficient, and exploration of some of the formulae it satisfies.
### Overarch 2
##### Age 16 to 18 Challenge Level:
Bricks are 20cm long and 10cm high. How high could an arch be built without mortar on a flat horizontal surface, to overhang by 1 metre? How big an overhang is it possible to make like this? | 1,020 | 4,112 | {"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-2020-10 | latest | en | 0.794772 |
https://www.math.arizona.edu/~models/Ruled_Surfaces/index.html | 1,610,822,700,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703506832.21/warc/CC-MAIN-20210116165621-20210116195621-00652.warc.gz | 900,570,846 | 2,207 | 51 Ruled hyperboloid 52 Quartic ruled surface 53 Intersection 54 Sextic ruled surface 55 Twisted cubic 56 Quartic scroll 57 Osculating hyperboloid 58 Ruled elliptic hyperboloid 59 Intersecting cones 60 Helicoid 61 Conoid 62 Double helicoid 63 Intersecting cylinders 65 Ruled cone 66 Ruled surface 67 Ruled hyperbolic paraboloid 68 Doubly ruled hyperboloid
A ruled surface (or scroll) is a surface swept out by a straight line as it moves through space. For example, a cylinder is formed by moving a straight line around a curve in a plane, keeping it perpendicular to the plane at all times; a cone is formed by moving a line so that it stays fixed at one point, but changes direction (65); and a helicoid is formed by moving a straight line along another straight line, keeping it perpendicular but rotating it as it moves (60).
These models show ruled surfaces using stretched string to represent the position of the straight line at various times. The straight lines in the ruling are called generators of the surface. Some quadratic surfaces are ruled: hyperboloids of one sheet, hyperbolic paraboloids, and quadratic cones and cylinders. The first two are doubly ruled, that is, they have two distinct ways of being generated by a moving straight line (67 and 68). A general way to form a ruled surface is to take three curves in space, and move a straight line so that it intersects all three curves at all times. This procedure, applied to the three lines in one of the rulings of a hyperboloid of one sheet or a hyperbolic paraboloid, will give the other ruling. This procedure is also illustrated in model 56 (Baker No. 84), with two straight lines and an ellipse, and in model 52, with two circles and a straight line. Models 53, 55, 59, and 63 show space curves obtained by taking the intersection of two ruled surfaces. | 428 | 1,833 | {"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-2021-04 | latest | en | 0.943666 |
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# Will O'Connell
228 total contributions since 2017
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9 months ago | 1,285 | 4,958 | {"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-2019-39 | latest | en | 0.800193 |
https://forum.codingame.com/t/stuck-on-bulls-and-cows/185933 | 1,696,091,810,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510697.51/warc/CC-MAIN-20230930145921-20230930175921-00851.warc.gz | 289,476,360 | 4,586 | # Stuck on "Bulls and Cows"
In the classic code-breaking game of Bulls and Cows, your opponent chooses a 4-digit secret number, and you need to guess it. After each guess, your opponent tells you how many “bulls” and how many “cows” are in your guess, interpreted as:
• Each bull indicates a digit in your guess that exactly matches the value and position of a digit in your opponent’s secret number.
• Each cow indicates a digit in your guess that matches the value of a digit in your opponent’s secret number, but is in the wrong position.
So for example, if the secret number is 1234 and you guess 5678, your guess has 0 bulls and 0 cows. However, if you guess 2324 then your guess has 1 bull (the 4) and 2 cows (one of the 2s, and the 3.)
You will be given a series of guesses along with the number of bulls and cows in each guess. Your job is to determine the secret number based on the given information.
NOTE: This version of the game deviates from the classic Bulls and Cows rules in that digits may be repeated any number of times in the secret number.
It will be easier to help you if you tell us what you understood, what you tried, etc.
General tips for this problem:
Bulls should be easy to count, just zip/compare.
Also, it’s easier to count both bulls/cows together than cows alone.
And then you substract bulls to get cows alone.
Write few examples and try to find a way to count bulls/cows together with basic observations.
1 Like
This is my approach:- I can find no of bulls & cows. But i am not able to figure out how to get the secret number
``````class Solution {
public String getHint(String secret, String guess) {
int[] map = new int[10];
int n = secret.length();
int bulls =0, cows =0;
for(char c : secret.toCharArray()){
map[c-'0']++;
}
for(char c : guess.toCharArray()){
if(map[c-'0']-- >0)
cows++;
}
for(int i=0; i<n; i++){
if(secret.charAt(i) == guess.charAt(i))
bulls++;
}
return String.valueOf(bulls)+"A"+String.valueOf(cows-bulls)+"B";
}
}``````
Ok so now all you have to do is to check all strings from 0000 to 9999 and find the one that matches all the bulls/cows given. | 534 | 2,119 | {"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} | 3.703125 | 4 | CC-MAIN-2023-40 | latest | en | 0.878648 |
https://mail.scipy.org/pipermail/scipy-user/2008-May/016686.html | 1,506,281,874,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690203.42/warc/CC-MAIN-20170924190521-20170924210521-00592.warc.gz | 692,529,909 | 3,263 | # [SciPy-user] scipy.stats rv objects from data
Erik Tollerud erik.tollerud@gmail....
Fri May 2 16:36:34 CDT 2008
```On Mon, Apr 28, 2008 at 12:29 AM, Stéfan van der Walt <stefan@sun.ac.za> wrote:
> That should be 1.0/len(x), otherwise all the probabilities are 0.
>
> Cheers
> Stéfan
I had >>>from __future__ import division above when I actually tested
this, so they aren't zero (although you're right they would be
otherwise), but you're right that they would be if this was run as-is.
I figured the pmf part out, though based on Michael's examples - the
pmf SHOULD be zero everywhere other then exactly at one of the
values... when I generate the cdf, it has non-zero values.
On Mon, Apr 28, 2008 at 2:02 AM, Michael <mnandris@btinternet.com> wrote:
> There are at least 2 ways of using rv_discrete
>
> e.g. 2 ways to calculate the next element of a simple Markov Chain with
> x(n+1)=Norm(0.5 x(n),1)
>
> from scipy.stats import rv_discrete
> from numpy.random import multinomial
>
> x = 3
> n1 = stats.rv_continuous.rvs( stats.norm, 0.5*x, 1.0 )[0]
> print n1
> n2 = stats.rv_discrete.rvs( stats.rv_discrete( name='sample',
> values=([0,1,2],[3/10.,5/10.,2/10.])), 0.5*x, 1.0 )[0]
> print n2
> print
> sample = stats.rv_discrete( name='sample',
> values=([0,1,2],[3/10.,5/10.,2/10.]) ).rvs( size=10 )
> print sample
>
> The multinomial distribution from numpy.random is somewhat faster (40
> times or so) but has a different idiom:
>
> SIZE = 100000
> VALUES = [0,1,2,3,4,5,6,7]
> PROBS = [1/8.,1/8.,1/8.,1/8.,1/8.,1/8.,1/8.,1/8.]
>
> The idiom for rv_discrete is
> rv_discrete( name='sample', values=(VALUES,PROBS) )
>
> The idiom for numpy.multinomial is different; if memory serves, you get
> frequencies as output instead of the actual values
> multinomial( SIZE, PROBS )
>
> >>> from numpy.random import multinomial
> >>> multinomial(100,[ 0.2, 0.4, 0.1, 0.3 ])
> array([12, 44, 10, 34])
> >>> multinomial( 100, [0.2, 0.0, 0.8, 0.0] ) <-- don't do this
> ...
> >>> multinomial( 100, [0.2, 1e-16, 0.8, 1e-16] ) <-- or this
> >>> multinomial( 100, [0.2-1e-16, 1e-16, 0.8-1e-16, 1e-16] ) <-- ok
> array([21, 0, 79, 0])
>
> the last one is ok since the probability adds up to 1... painful, but it
> works
As explained above, I think I got the discrete to work (spurred on by
your simpler example) - but what's the utility of using the
multinomial idiom over the rv_discrete syntax? Is it faster?.
> Continuous v's discrete: i found this in ./stats/scstats.py
>
> from scipy import stats, r_
> from pylab import show, plot
> import copy
>
> # SOURCE: ./stats/scstats.py
>
>
> class cdf( object ):
> """ Baseclass for the task of determining a sequence of numbers {vi}
> which is distributed as a random variable X
> """
> def integerDensityFunction( self ):
> """
> Outputs an integer density function: xs (ints) and ys (probabilities)
> which are the correspondence between the whole numbers on the x axis
> to the probabilities on the y axis, according to a normal distribution.
> """
> opt = []
> opt.append(( i, stats.norm.cdf(i) )) # ( int, P(int) )
> return zip(*opt) # [ (int...), (P...) ]
>
> def display( self ):
> xs, ys = self.integerDensityFunction()
> plot( xs, ys )
> show()
>
> if __name__=='__main__':
> d = cdf()
> d.display()
I don't really understand what you're suggesting, but anyway, I don't
see a stats/scstats.py file in the scipy directory (at least in
0.6)...
> Continuous: i can only suggest using rv_continuous
>
> stats.rv_continuous.rvs( stats.norm, 0.5*x, 1.0 ).whatever
>
> .rvs( shape, loc, scale ) is the random variates
> .pdf( x, shape, loc, scale ) is the probability density function which,
> i think, is or should be genuinely continuous
My interpretation of this is that it is using the normal distribution
- I want a distribution that is a smoothed/interpolated version of the
discrete distribution I generated above. I take this to mean there's
no built-in utility to do this, so I just have to make my own - this
seems like a useful thing for data analysis, though, so I may submit
it later to be added to SVN.
``` | 1,382 | 4,320 | {"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} | 2.90625 | 3 | CC-MAIN-2017-39 | latest | en | 0.699515 |
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