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https://nrich.maths.org/public/topic.php?code=-483&cl=2&cldcmpid=7780 | 1,582,421,829,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145742.20/warc/CC-MAIN-20200223001555-20200223031555-00309.warc.gz | 495,607,059 | 5,258 | # Resources tagged with: Design
Filter by: Content type:
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### There are 12 results
Broad Topics > Cross-curricular Connections > Design
### Programming: Moiré Patterns
##### Age 11 to 16
We need computer programmers! Logo is a great entry-level programming language - and you can create stunning graphics while you learn.
### Flip Your Mat!
##### Age 7 to 14 Challenge Level:
What shape and size of drinks mat is best for flipping and catching?
### Make Your Own Pencil Case
##### Age 11 to 14 Challenge Level:
What shape would fit your pens and pencils best? How can you make it?
### Oblique Projection
##### Age 11 to 14 Challenge Level:
Explore the properties of oblique projection.
### Designing Table Mats
##### Age 11 to 16 Challenge Level:
Formulate and investigate a simple mathematical model for the design of a table mat.
### 3D Drawing
##### Age 11 to 16
The design technology curriculum requires students to be able to represent 3-dimensional objects on paper. This article introduces some of the mathematical ideas which underlie such methods.
### Isometric Drawing
##### Age 11 to 14 Challenge Level:
Explore the properties of isometric drawings.
### Perspective Drawing
##### Age 11 to 16 Challenge Level:
Explore the properties of perspective drawing.
### Gym Bag
##### Age 11 to 16 Challenge Level:
Can Jo make a gym bag for her trainers from the piece of fabric she has?
### Flower Power
##### Age 11 to 16 Challenge Level:
Create a symmetrical fabric design based on a flower motif - and realise it in Logo.
### Making Moiré Patterns
##### Age 11 to 18 Challenge Level:
Moiré patterns are intriguing interference patterns. Create your own beautiful examples using LOGO!
### Witch's Hat
##### Age 11 to 16 Challenge Level:
What shapes should Elly cut out to make a witch's hat? How can she make a taller hat? | 420 | 1,892 | {"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-2020-10 | latest | en | 0.776928 |
http://slideplayer.com/slide/2711874/ | 1,508,293,553,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822668.19/warc/CC-MAIN-20171018013719-20171018033719-00296.warc.gz | 312,338,805 | 28,838 | ### Similar presentations
Chapter 5 Conditionals and Loops
© 2004 Pearson Addison-Wesley. All rights reserved5-2 Conditionals and Loops Now we will examine programming statements that allow us to: make decisions repeat processing steps in a loop Chapter 5 focuses on: boolean expressions conditional statements comparing data repetition statements iterators more drawing techniques more GUI components
© 2004 Pearson Addison-Wesley. All rights reserved5-3 Outline The if Statement and Conditions Other Conditional Statements Comparing Data The while Statement Iterators Other Repetition Statements Decisions and Graphics More Components
© 2004 Pearson Addison-Wesley. All rights reserved5-4 Flow of Control Unless specified otherwise, the order of statement execution through a method is linear: one statement after another in sequence Some programming statements allow us to: decide whether or not to execute a particular statement execute a statement over and over, repetitively These decisions are based on boolean expressions (or conditions) that evaluate to true or false The order of statement execution is called the flow of control
© 2004 Pearson Addison-Wesley. All rights reserved5-5 Conditional Statements A conditional statement lets us choose which statement will be executed next Therefore they are sometimes called selection statements Conditional statements give us the power to make basic decisions The Java conditional statements are the: if statement if-else statement switch statement
© 2004 Pearson Addison-Wesley. All rights reserved5-6 The if Statement The if statement has the following syntax: if ( condition ) statement; if is a Java reserved word The condition must be a boolean expression. It must evaluate to either true or false. If condition is true: statement is executed. If condition is false: statement is skipped.
© 2004 Pearson Addison-Wesley. All rights reserved5-8 Boolean Expressions A condition often uses one of Java's equality operators or relational operators, which all return boolean results: == equal to != not equal to < less than > greater than <= less than or equal to >= greater than or equal to Note the difference between the equality operator ( == ) and the assignment operator ( = )
© 2004 Pearson Addison-Wesley. All rights reserved5-9 The if Statement An example of an if statement: if (sum > MAX) delta = sum - MAX; System.out.println ("The sum is " + sum); First the condition is evaluated -- the value of sum is either greater than the value of MAX, or it is not If the condition is true, the assignment statement is executed -- if it isn’t, it is skipped. Either way, the call to println is executed next See Age.java (page 208)Age.java
© 2004 Pearson Addison-Wesley. All rights reserved5-10 Indentation The statement controlled by the if statement is indented to indicate that relationship The use of a consistent indentation style makes a program easier to read and understand Although it makes no difference to the compiler, proper indentation is crucial
© 2004 Pearson Addison-Wesley. All rights reserved5-11 The if Statement What do the following statements do? if (top >= MAXIMUM) top = 0; Sets top to zero if the current value of top is greater than or equal to the value of MAXIMUM if (total != stock + warehouse) inventoryError = true; Sets a flag to true if the value of total is not equal to the sum of stock and warehouse The precedence of the arithmetic operators is higher than the precedence of the equality and relational operators
© 2004 Pearson Addison-Wesley. All rights reserved5-12 Logical Operators Boolean expressions can also use the following logical operators: ! Logical NOT && Logical AND || Logical OR They all take boolean operands and produce boolean results Logical NOT is a unary operator (it operates on one operand) Logical AND and logical OR are binary operators (each operates on two operands)
© 2004 Pearson Addison-Wesley. All rights reserved5-13 Logical NOT The logical NOT operation is also called logical negation or logical complement If some boolean condition: if a is true, then !a is false; if a is false, then !a is true Logical expressions shown using a truth table a!a truefalse true
© 2004 Pearson Addison-Wesley. All rights reserved5-14 Logical AND and Logical OR The logical AND expression a && b is true if both a and b are true, and false otherwise The logical OR expression a || b is true if a or b or both are true, and false otherwise
© 2004 Pearson Addison-Wesley. All rights reserved5-15 Logical Operators Expressions that use logical operators can form complex conditions if (total < MAX + 5 && !found) System.out.println ("Processing…"); All logical operators have lower precedence than the relational operators Logical NOT has higher precedence than logical AND and logical OR
© 2004 Pearson Addison-Wesley. All rights reserved5-16 Logical Operators A truth table shows all possible true-false combinations of the terms Since && and || each have two operands, there are four possible combinations of conditions a and b aba && ba || b true false true falsetruefalsetrue false
© 2004 Pearson Addison-Wesley. All rights reserved5-17 Boolean Expressions Specific expressions can be evaluated using truth tables total < MAXfound!foundtotal < MAX && !found false truefalse truefalse truefalsetrue false
© 2004 Pearson Addison-Wesley. All rights reserved5-18 Short-Circuited Operators The processing of logical AND and logical OR is “short-circuited” If the left operand is sufficient to determine the result, the right operand is not evaluated This type of processing must be used carefully if (count != 0 && total/count > MAX) System.out.println ("Testing…");
© 2004 Pearson Addison-Wesley. All rights reserved5-19 Outline The if Statement and Conditions Other Conditional Statements Comparing Data The while Statement Iterators Other Repetition Statements Decisions and Graphics More Components
© 2004 Pearson Addison-Wesley. All rights reserved5-20 The if-else Statement An else clause can be added to an if statement to make an if-else statement if ( condition ) statement1; else statement2; If the condition is true, statement1 is executed; if the condition is false, statement2 is executed One or the other will be executed, but not both See Wages.java (page 211) Listing 5.2Wages.java
© 2004 Pearson Addison-Wesley. All rights reserved5-22 The Coin Class Let's examine a class that represents a coin that can be flipped Instance data is used to indicate which face (heads or tails) is currently showing See CoinFlip.java (page 213) Listing 5.3CoinFlip.java See Coin.java (page 214) Listing 5.4Coin.java
© 2004 Pearson Addison-Wesley. All rights reserved5-23 Indentation Revisited Remember that indentation is for the human reader, and is ignored by the computer if (total > MAX) System.out.println ("Error!!"); errorCount++; Despite what is implied by the indentation, the increment will occur whether the condition is true or not
© 2004 Pearson Addison-Wesley. All rights reserved5-24 Block Statements Several statements can be grouped together into a block statement delimited by braces A block statement can be used wherever a statement is called for in the Java syntax rules if (total > MAX) { System.out.println ("Error!!"); errorCount++; }
© 2004 Pearson Addison-Wesley. All rights reserved5-25 Block Statements In an if-else statement, the if portion, or the else portion, or both, could be block statements if (total > MAX) { System.out.println ("Error!!"); errorCount++; } else { System.out.println ("Total: " + total); current = total*2; } See Guessing.java (page 216) Listing 5.5Guessing.java
© 2004 Pearson Addison-Wesley. All rights reserved5-26 The Conditional Operator Java has a conditional operator that uses a boolean condition to determine which of two expressions is evaluated Syntax: condition ? express_1 : express_2 If the condition is true, express_1 is evaluated; if the condition is false express_2 is evaluated; The value of the entire conditional operator is the value of the selected expression
© 2004 Pearson Addison-Wesley. All rights reserved5-27 The Conditional Operator The conditional operator is similar to an if-else statement, except that it is an expression that returns a value For example: larger = ((num1 > num2) ? num1 : num2); If num1 is greater than num2, then num1 is assigned to larger ; otherwise, num2 is assigned to larger The conditional operator is ternary because it requires three operands
© 2004 Pearson Addison-Wesley. All rights reserved5-28 The Conditional Operator Another example: System.out.println ("Your change is " + count + ((count == 1) ? "Dime" : "Dimes")); If count equals 1, then "Dime" is printed If count is anything other than 1, then "Dimes" is printed
© 2004 Pearson Addison-Wesley. All rights reserved5-29 Nested if Statements The statement executed as a result of an if statement or else clause could be another if statement These are called nested if statements See MinOfThree.java (page 219) Listing 5.6MinOfThree.java An else clause is matched to the last unmatched if (no matter what the indentation implies) Braces can be used to specify the if statement to which an else clause belongs
© 2004 Pearson Addison-Wesley. All rights reserved5-30 The switch Statement The switch statement provides another way to decide which statement to execute next The switch statement evaluates an expression, then attempts to match the result to one of several possible cases Each case contains one value (a constant) and a list of statements The flow of control transfers to statement associated with the first case value that matches
© 2004 Pearson Addison-Wesley. All rights reserved5-31 The switch Statement The general syntax of a switch statement is: switch ( expression ) { case value1 : statement-list1 case value2 : statement-list2 case value3 : statement-list3 case... } switch and case are reserved words If expression matches value2, control jumps to here
© 2004 Pearson Addison-Wesley. All rights reserved5-32 The switch Statement Often a break statement is used as the last statement in each case's statement list A break statement causes control to transfer to the end of the switch statement If a break statement is not used, the flow of control will continue into the next case Sometimes this may be appropriate, but often we want to execute only the statements associated with one case
© 2004 Pearson Addison-Wesley. All rights reserved5-33 The switch Statement switch (option) { case 'A': aCount++; break; case 'B': bCount++; break; case 'C': cCount++; break; } An example of a switch statement:
© 2004 Pearson Addison-Wesley. All rights reserved5-34 The switch Statement A switch statement can have an optional default case The default case has no associated value and simply uses the reserved word default If the default case is present, control will transfer to it if no other case value matches If there is no default case, and no other value matches, control falls through to the statement after the switch
© 2004 Pearson Addison-Wesley. All rights reserved5-35 The switch Statement The expression of a switch statement must result in an integral type, meaning an integer ( byte, short, int, long ) or a char It cannot be a boolean value or a floating point value ( float or double ) The implicit boolean condition in a switch statement is equality You cannot perform relational checks with a switch statement See GradeReport.java (page 225)GradeReport.java
© 2004 Pearson Addison-Wesley. All rights reserved5-36 Outline The if Statement and Conditions Other Conditional Statements Comparing Data The while Statement Iterators Other Repetition Statements Decisions and Graphics More Components
© 2004 Pearson Addison-Wesley. All rights reserved5-37 Comparing Data When comparing data using boolean expressions, it's important to understand the nuances of certain data types Let's examine some key situations: Comparing floating point values for equality Comparing characters Comparing strings (alphabetical order) Comparing object vs. comparing object references
© 2004 Pearson Addison-Wesley. All rights reserved5-38 Comparing Float Values You should rarely use the equality operator ( == ) when comparing two floating point values ( float or double ) Two floating point values are equal only if their underlying binary representations match exactly Computations often result in slight differences that may be irrelevant In many situations, you might consider two floating point numbers to be "close enough" even if they aren't exactly equal
© 2004 Pearson Addison-Wesley. All rights reserved5-39 Comparing Float Values To determine the equality of two floats, you may want to use the following technique: if (Math.abs(f1 - f2) < TOLERANCE) System.out.println ("Essentially equal"); If the difference between the two floating point values is less than the tolerance, they are considered to be equal The tolerance could be set to any appropriate level, such as 0.000001
© 2004 Pearson Addison-Wesley. All rights reserved5-40 Comparing Characters As we've discussed, Java character data is based on the Unicode character set Unicode establishes a particular numeric value for each character, and therefore an ordering We can use relational operators on character data based on this ordering For example, the character '+' is less than the character ' J' because it comes before it in the Unicode character set Appendix C provides an overview of Unicode
© 2004 Pearson Addison-Wesley. All rights reserved5-41 Comparing Characters In Unicode, the digit characters (0-9) are contiguous and in order Likewise, the uppercase letters (A-Z) and lowercase letters (a-z) are contiguous and in order CharactersUnicode Values 0 – 948 through 57 A – Z65 through 90 a – z97 through 122
© 2004 Pearson Addison-Wesley. All rights reserved5-42 Comparing Strings Remember that in Java a character string is an object The equals method can be called with strings to determine if two strings contain exactly the same characters in the same order The equals method returns a boolean result if (name1.equals(name2)) System.out.println ("Same name");
© 2004 Pearson Addison-Wesley. All rights reserved5-43 Comparing Strings We cannot use the relational operators to compare strings The String class contains a method called compareTo to determine if one string comes before another A call to name1.compareTo(name2) returns zero if name1 and name2 are equal (contain the same characters) returns a negative value if name1 is less than name2 returns a positive value if name1 is greater than name2
© 2004 Pearson Addison-Wesley. All rights reserved5-44 Comparing Strings if (name1.compareTo(name2) < 0) System.out.println (name1 + "comes first"); else if (name1.compareTo(name2) == 0) System.out.println ("Same name"); else System.out.println (name2 + "comes first"); Because comparing characters and strings is based on a character set, it is called a lexicographic ordering
© 2004 Pearson Addison-Wesley. All rights reserved5-45 Lexicographic Ordering Lexicographic ordering is not strictly alphabetical when uppercase and lowercase characters are mixed For example, the string "Great" comes before the string "fantastic" because all of the uppercase letters come before all of the lowercase letters in Unicode Also, short strings come before longer strings with the same prefix (lexicographically) Therefore "book" comes before "bookcase"
© 2004 Pearson Addison-Wesley. All rights reserved5-46 Comparing Objects The == operator can be applied to objects – it returns true if the two references are aliases of each other The equals method is defined for all objects, but unless we redefine it when we write a class, it has the same semantics as the == operator It has been redefined in the String class to compare the characters in the two strings When you write a class, you can redefine the equals method to return true under whatever conditions are appropriate
© 2004 Pearson Addison-Wesley. All rights reserved5-47 Outline The if Statement and Conditions Other Conditional Statements Comparing Data The while Statement Iterators Other Repetition Statements Decisions and Graphics More Components
© 2004 Pearson Addison-Wesley. All rights reserved5-48 Repetition Statements Repetition statements allow us to execute a statement multiple times Often they are referred to as loops Like conditional statements, they are controlled by boolean expressions Java has three kinds of repetition statements: the while loop the do loop the for loop The programmer should choose the right kind of loop for the situation
© 2004 Pearson Addison-Wesley. All rights reserved5-49 The while Statement A while statement has the following syntax: while ( condition ) statement; If the condition is true, the statement is executed Then the condition is evaluated again, and if it is still true, the statement is executed again The statement is executed repeatedly until the condition becomes false
© 2004 Pearson Addison-Wesley. All rights reserved5-51 The while Statement An example of a while statement: int count = 1; while (count <= 5) { System.out.println (count); count++; } If the condition of a while loop is false initially, the statement is never executed Therefore, the body of a while loop will execute zero or more times
© 2004 Pearson Addison-Wesley. All rights reserved5-52 The while Statement Let's look at some examples of loop processing A loop can be used to maintain a running sum A sentinel value is a special input value that represents the end of input See Average.java (page 229) Listing 5.8Average.java A loop can also be used for input validation, making a program more robust See WinPercentage.java (page 231)WinPercentage.java
© 2004 Pearson Addison-Wesley. All rights reserved5-53 Infinite Loops The body of a while loop eventually must make the condition false If not, it is called an infinite loop, which will execute until the user interrupts the program This is a common logical error You should always double check the logic of a program to ensure that your loops will terminate normally
© 2004 Pearson Addison-Wesley. All rights reserved5-54 Infinite Loops An example of an infinite loop: int count = 1; while (count <= 25) { System.out.println (count); count = count - 1; } This loop will continue executing until interrupted (Control-C) or until an underflow error occurs
© 2004 Pearson Addison-Wesley. All rights reserved5-55 Nested Loops Similar to nested if statements, loops can be nested as well That is, the body of a loop can contain another loop For each iteration of the outer loop, the inner loop iterates completely See PalindromeTester.java (page 235) Listing 5.10PalindromeTester.java
© 2004 Pearson Addison-Wesley. All rights reserved5-56 Nested Loops How many times will the string "Here" be printed? count1 = 1; while (count1 <= 10) { count2 = 1; while (count2 <= 20) { System.out.println ("Here"); count2++; } count1++; } 10 * 20 = 200
© 2004 Pearson Addison-Wesley. All rights reserved5-57 Outline The if Statement and Conditions Other Conditional Statements Comparing Data The while Statement Iterators Other Repetition Statements Decisions and Graphics More Components
© 2004 Pearson Addison-Wesley. All rights reserved5-58 Iterators An iterator is an object that allows you to process a collection of items one at a time It lets you step through each item in turn and process it as needed An iterator object has a hasNext method that returns true if there is at least one more item to process The next method returns the next item Iterator objects are defined using the Iterator interface, which is discussed further in Chapter 6
© 2004 Pearson Addison-Wesley. All rights reserved5-59 Iterators Several classes in the Java standard class library are iterators The Scanner class is an iterator the hasNext method returns true if there is more data to be scanned the next method returns the next scanned token as a string The Scanner class has variations on the hasNext method for specific data types: hasNextInt
© 2004 Pearson Addison-Wesley. All rights reserved5-60 Iterators The fact that a Scanner is an iterator is particularly helpful when reading input from a file Suppose we wanted to read and process a list of URLs stored in a file One scanner can be set up to read each line of the input until the end of the file is encountered Another scanner can be set up for each URL to process each part of the path See URLDissector.java (page 240) Listing 5.11URLDissector.java
© 2004 Pearson Addison-Wesley. All rights reserved5-61 Outline The if Statement and Conditions Other Conditional Statements Comparing Data The while Statement Iterators Other Repetition Statements Decisions and Graphics More Components
© 2004 Pearson Addison-Wesley. All rights reserved5-62 The do Statement A do statement has the following syntax: do { statement; } while ( condition ); The statement is executed once initially, and then the condition is evaluated The statement is executed repeatedly until the condition becomes false
© 2004 Pearson Addison-Wesley. All rights reserved5-64 The do Statement An example of a do loop: The body of a do loop executes at least once See ReverseNumber.java (page 244) Listing 5.12ReverseNumber.java int count = 0; do { count++; System.out.println (count); } while (count < 5);
© 2004 Pearson Addison-Wesley. All rights reserved5-66 The for Statement A for statement has the following syntax: for ( initialization ; condition ; increment ) statement; The initialization is executed once before the loop begins The statement is executed until the condition becomes false The increment portion is executed at the end of each iteration
© 2004 Pearson Addison-Wesley. All rights reserved5-68 The for Statement A for loop is functionally equivalent to the following while loop structure: initialization; while ( condition ) { statement; increment; }
© 2004 Pearson Addison-Wesley. All rights reserved5-69 The for Statement An example of a for loop: for (int count=1; count <= 5; count++) System.out.println (count); The initialization section can be used to declare a variable Like a while loop, the condition of a for loop is tested prior to executing the loop body Therefore, the body of a for loop will execute zero or more times
© 2004 Pearson Addison-Wesley. All rights reserved5-70 The for Statement The increment section can perform any calculation A for loop is well suited for executing statements a specific number of times that can be calculated or determined in advance See Multiples.java (page 248)Multiples.java See Stars.java (page 250) Listing 5.14Stars.java for (int num=100; num > 0; num -= 5) System.out.println (num);
© 2004 Pearson Addison-Wesley. All rights reserved5-71 The for Statement Each expression in the header of a for loop is optional If the initialization is left out, no initialization is performed If the condition is left out, it is always considered to be true, and therefore creates an infinite loop If the increment is left out, no increment operation is performed
© 2004 Pearson Addison-Wesley. All rights reserved5-72 Iterators and for Loops Recall that an iterator is an object that allows you to process each item in a collection A variant of the for loop simplifies the repetitive processing the items For example, if BookList is an iterator that manages Book objects, the following loop will print each book: for (Book myBook : BookList) System.out.println (myBook);
© 2004 Pearson Addison-Wesley. All rights reserved5-73 Iterators and for Loops This style of for loop can be read "for each Book in BookList, …" Therefore the iterator version of the for loop is sometimes referred to as the foreach loop It eliminates the need to call the hasNext and next methods explicitly It also will be helpful when processing arrays, which are discussed in Chapter 7
© 2004 Pearson Addison-Wesley. All rights reserved5-74 Outline The if Statement and Conditions Other Conditional Statements Comparing Data The while Statement Iterators Other Repetition Statements Decisions and Graphics More Components
© 2004 Pearson Addison-Wesley. All rights reserved5-75 Drawing Techniques Conditionals and loops enhance our ability to generate interesting graphics See Bullseye.java (page 252) Listing 5.15Bullseye.java See BullseyePanel.java (page 253) Listing 5.16BullseyePanel.java See Boxes.java (page 255) Listing 5.17Boxes.java See BoxesPanel.java (page 256) Listing 5.18BoxesPanel.java
© 2004 Pearson Addison-Wesley. All rights reserved5-76 Determining Event Sources Recall that interactive GUIs require establishing a relationship between components and the listeners that respond to component events One listener object can be used to listen to two different components The source of the event can be determined by using the getSource method of the event passed to the listener See LeftRight.java (page 258) Listing 5.19LeftRight.java See LeftRightPanel.java (page 259) Listing 5.20LeftRightPanel.java
© 2004 Pearson Addison-Wesley. All rights reserved5-77 Outline The if Statement and Conditions Other Conditional Statements Comparing Data The while Statement Iterators Other Repetition Statements Decisions and Graphics More Components
© 2004 Pearson Addison-Wesley. All rights reserved5-78 Dialog Boxes A dialog box is a window that appears on top of any currently active window Basic formats for a JOptionPane dialog box: convey informationshowMessageDialog confirm an actionshowConfirmDialog allow the user to enter datashowInputDialog
© 2004 Pearson Addison-Wesley. All rights reserved5-79 Dialog Boxes Using a null reference as the first parameter will center the dialog box on the window Figure 5.10 Page 261 methods of JOptionPane class See EvenOdd.java (page 262) Listing 5.21EvenOdd.java
© 2004 Pearson Addison-Wesley. All rights reserved5-80 Check Boxes A check box is a button that can be toggled on or off. They operate independently It is represented by the JCheckBox class Unlike a push button, which generates an action event, a check box generates an item event whenever it changes state (is checked on or off) The ItemListener interface is used to define item event listeners The check box calls the itemStateChanged method of the listener when it is toggled
© 2004 Pearson Addison-Wesley. All rights reserved5-81 Check Boxes Let's examine a program that uses check boxes to determine the style of a label's text string It uses the Font class, which represents a character font's: font name (such as Times or Courier) Font style (bold, italic, or both) font size See StyleOptions.java (page 265) Listing 5.22StyleOptions.java See StyleOptionsPanel.java (page 266) Listing 5.23StyleOptionsPanel.java
© 2004 Pearson Addison-Wesley. All rights reserved5-82 Radio Buttons A group of radio buttons represents a set of mutually exclusive options – only one can be selected at any given time Part of the JRadioButton class When a radio button from a group is selected, the button that is currently "on" in the group is automatically toggled off | 5,948 | 27,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-43 | latest | en | 0.844267 |
https://www.basic-mathematics.com/central-angles.html | 1,540,250,569,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515555.58/warc/CC-MAIN-20181022222133-20181023003633-00245.warc.gz | 876,307,178 | 11,179 | # Central angles
This lesson offers a concise, but thorough explanation of central angles, but also of arcs and sectors of a circle
Definition: An angle is a central angle if it meets the following two conditions
1) The vertex of the angle is located at the center of a circle.
2) The rays that make up its sides are radii of the circle. Below, find an illustration of the definition above:
You can name it angle ABC.It is important to notice that such angle is always less than 180 degrees.
Therefore, such angles can only be acute or obtuse
Related definitions:
Arc:
A portion of the circumference of the circle. This is illustrated below in red:
Sector:
A sector is the area enclosed within a central angle and an arc. Again, this is illustrated below, but in green:
As you can see, the area in green is included between the arc in red and the angle
When computing the area of a sector, use the following ratio or formula to find out what part of the circle's area is covered by the sector:
A computation!
A circle has a radius of 10 centimeters. This radius and the center of the circle is used to make of angle of 45 degrees. Find the area of the resulting sector
Divide 45 degrees divided by 360 degrees to determine the fraction of the circle covered by this sector 45/360 = 1/8
The area of the circle is
A = pi × r2
A = 3.14 × 102
A = 3.14 × 10 × 10
A = 3.14 × 102
A = 3.14 × 100
A = 314 square centimeters
Now just mutliply 314 by 1/8
314 × 1/8 = 314/8 = 39.25
The area of the sector is 39.25 square centimeters
Now try to do this problem on your own.
A circle has a radius of 5 centimeters. This radius and the center of the circle is used to make of angle of 60 degrees. Find the area of the resulting sector
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Learn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. | 625 | 2,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2018-43 | longest | en | 0.91694 |
https://bugsfixing.com/solved-how-to-generate-a-random-normal-distribution-of-integers/ | 1,674,770,404,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494826.88/warc/CC-MAIN-20230126210844-20230127000844-00282.warc.gz | 172,692,807 | 11,643 | # [SOLVED] How to generate a random normal distribution of integers
## Issue
How to generate a random integer as with np.random.randint(), but with a normal distribution around 0.
np.random.randint(-10, 10) returns integers with a discrete uniform distribution
np.random.normal(0, 0.1, 1) returns floats with a normal distribution
What I want is a kind of combination between the two functions.
## Solution
One other way to get a discrete distribution that looks like the normal distribution is to draw from a multinomial distribution where the probabilities are calculated from a normal distribution.
import scipy.stats as ss
import numpy as np
import matplotlib.pyplot as plt
x = np.arange(-10, 11)
xU, xL = x + 0.5, x - 0.5
prob = ss.norm.cdf(xU, scale = 3) - ss.norm.cdf(xL, scale = 3)
prob = prob / prob.sum() # normalize the probabilities so their sum is 1
nums = np.random.choice(x, size = 10000, p = prob)
plt.hist(nums, bins = len(x))
Here, np.random.choice picks an integer from [-10, 10]. The probability for selecting an element, say 0, is calculated by p(-0.5 < x < 0.5) where x is a normal random variable with mean zero and standard deviation 3. I chose a std. dev. of 3 because this way p(-10 < x < 10) is almost 1.
The result looks like this: | 336 | 1,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2023-06 | latest | en | 0.805199 |
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### Topic: Determining accleration due to gravity (Read 17135 times)previous topic - next topic
#### michinyon
#30
##### Oct 31, 2012, 08:03 am
"I do understand the units' equivalency, and indeed have understood this stuff since about 1971.... I doubt if I'd have graduated as a civil engineer without having understood the relationship between force, mass and acceleration.."
You might have, if you were an American. The distinction between force, mass and weight becomes rather obfuscated in their system.
#### michinyon
#31
##### Oct 31, 2012, 08:08 am
OK, so suppose I am not interested in measuring small fluctuations in the force of gravity, because I am not looking for buried iron ore deposits or big meteors.
Suppose I want to measure the direction of the graviational force, and I want to do this from a moving platform like a car or an aircraft.
If I get a 3D MEMS accelerometer, that will do a good job if the vehicle is stationary or moving at a constant velocity, but otherwise, it won't.
Is there another way to measure the direction of the gravitational force which will work in an erratically moving vehicle ?
#### liudr
#32
##### Oct 31, 2012, 01:43 pm
OK, so suppose I am not interested in measuring small fluctuations in the force of gravity, because I am not looking for buried iron ore deposits or big meteors.
Suppose I want to measure the direction of the graviational force, and I want to do this from a moving platform like a car or an aircraft.
If I get a 3D MEMS accelerometer, that will do a good job if the vehicle is stationary or moving at a constant velocity, but otherwise, it won't.
Is there another way to measure the direction of the gravitational force which will work in an erratically moving vehicle ?
Thanks to Einstein's general relativity, you can't tell between gravity and acceleration. You can't find gravity with accelerometer if you yourself is accelerating. If you have a gyroscope, you can spin it in the direction of gravity and see how your down direction compares with it, I suppose.
#### wwbrown
#33
##### Oct 31, 2012, 04:30 pm
How about a magnetometer if you only want to measure the direction of the acceleration due to gravity. I am pretty sure that a magnetometer will not be affected by linear or rotational acceleration.
#### liudr
#34
##### Oct 31, 2012, 04:40 pm
How about a magnetometer if you only want to measure the direction of the acceleration due to gravity. I am pretty sure that a magnetometer will not be affected by linear or rotational acceleration.
Is that a magnetic sensor? It can only measure your angle to local magnetic field, not your angle to local gravity. Facing north while tilting up 30 degrees and facing south while tilting down 30 degrees will read the same on a magnetic sensor.
#### liudr
#35
##### Oct 31, 2012, 04:46 pm
This reminds me of how we measured it back in high school...
It involved a battery operated bell and a huge long strip (as long as the drop from the window to the ground) of carbon paper with a weight on the end. Then dropped the weight while the bell was ringing, with the strip between the bell and the ringer lever thingy. The impact of the ringer made marks on the carbon paper, and of course the marks got further apart as the weight accelerated.
Measured the time it took to drop from the science lab window to the ground. Counted the total number of marks. That gave the number of marks per second or seconds per mark.
Measured the (varying) distance between marks. That, with the now known time between marks, gave the (varying) velocity at any instant.
And hence the acceleration....
But using a pendulum is by far the easiest way to do it... (As long as the bob is very heavy compared to the string, the whole mass may be deemed to be at the bob's CoG.) Beauty of the pendulum method is that since the period T is constant regardless of how wide the swing is, it can as suggested above be measured over 100s of swings and that reduces the impact of the reaction time when the stop watch is started and stopped.
Oh the old memories. They were probably called impact timer?! Later they were replaced with spark timers that generated sparks with car ignition coils at 60Hz so wax tapes will have regular burn marks on them. A bit less friction. Then photo gates became more popular so did sonic rangers and then video cams. You can measure gravity with any method that involves a formula where g is a part of. Pendulum is definitely easy and accurate (you mentioned about response time!). Not doing it easy way can have instructional values such as seeing the dots separating further and further away or video frames doing the same thing. There is another way, just to toss in the discussion, is to use sound. You record sound clip of scissors cutting string and ball hitting floor. I've been doing this for a while. Results are pretty good with 0.x m/s/s as error bar. Some kids these days can't even cut with scissors. That's a whole other discussion topic.
#### wwbrown
#36
##### Oct 31, 2012, 05:54 pm
Yes I made a huge error I thought for some strange reason the poster wanted to determine the direction of the magnetic field and not the acceleration due to gravity.
I believe with a full-up IMU you could get the direction of the acc. due to g.
#### michinyon
#37
##### Nov 01, 2012, 07:15 am
Well the idea is, if you know the accelerations and gyro readings in 3D, then you can calculate the change in orientation. But you get variations and drift in the gyro readings which make this not actually work in practise.
You need to keep correcting the gyro-derived orientation using some other means, and two of the available means are the direction of "down" and the direction of the earth's magnetic field.
The problem is, a lot of the feasible schemes for doing this ( for example, read Magdwick's papers ) kind of assume that the acceleration of the body is small and that therefore the accelerometer reading indicates the direction of gravity ( either "up" or "down" ). I've developed another scheme which also works and has the advantage ( for me ) that I can understand it.
It would be useful to have some means of identifying the direction of gravity independently from acceleration but I guess if Einstein says you can't, we it is probably right. Now a very small higgs boson detector might be able to detect the direction where they are coming from, independent of the actual motion of the device.
#### liudr
#38
##### Nov 01, 2012, 05:36 pm
Well the idea is, if you know the accelerations and gyro readings in 3D, then you can calculate the change in orientation. But you get variations and drift in the gyro readings which make this not actually work in practise.
You need to keep correcting the gyro-derived orientation using some other means, and two of the available means are the direction of "down" and the direction of the earth's magnetic field.
The problem is, a lot of the feasible schemes for doing this ( for example, read Magdwick's papers ) kind of assume that the acceleration of the body is small and that therefore the accelerometer reading indicates the direction of gravity ( either "up" or "down" ). I've developed another scheme which also works and has the advantage ( for me ) that I can understand it.
It would be useful to have some means of identifying the direction of gravity independently from acceleration but I guess if Einstein says you can't, we it is probably right. Now a very small higgs boson detector might be able to detect the direction where they are coming from, independent of the actual motion of the device.
You can still device an external measurement, say a person standing on the ground telling the device its orientation wrt gravity. You just can't have the device tell itself so. If you have a flat floor with regular dots, you can use that and a bottom facing camera to tell orientation.
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Please enter a valid email to subscribe | 1,817 | 8,021 | {"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-2016-40 | longest | en | 0.917174 |
https://www.blaumut.com/transcendence-audiobook-ipbr/mercedes-benz-gt63s-price-in-malaysia-a6873e | 1,627,951,833,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154408.7/warc/CC-MAIN-20210802234539-20210803024539-00034.warc.gz | 662,389,458 | 9,810 | PriceElasticityof Demand MATH 104 Mark Mac Lean (with assistance from Patrick Chan) 2011W The price elasticity of demand (which is often shortened to demand elasticity) is defined to be the percentage change in quantity demanded, q, divided by the percentage change in price, p. The formula for the demand elasticity (ǫ) is: ǫ = p q dq dp. If the cross-price elasticity of demand is positive, the two goods are said to be supplementary goods i.e. Provide an economic interpretation (elastic or inelastic). And there's multiple different scenarios we could think about, but it's really thinking about how a price change in one good might affect the quantity demanded in another good. Let us suppose an increase in the price of Tea by 5% might lead to an increase of the closed substitutes i.e. The following formula can be used to calculate the price elasticity of demand: PED = [ (Q₁ – Q₀) / (Q₁ + Q₀) ] / [ (P₁ – P₀) / (P₁ + P₀) ] Where PED is price elasticity of demand P₀ is the initial price Cross-price elasticity of demand formula measures the demand sensitivity of one product (say A) when the price of an unrelated product (say B) is changed. Cross-price elasticity of the demand helps large firms to decide pricing policy. Get the demand function and the price at which you want to find the elasticity. However, if the cross-price elasticity is negative, then the two goods are said to be complementary goods i.e. All you have to do is apply the following cross-price elasticity formula: elasticity = (price₁A + price₂A) / (quantity₁B + quantity₂B) * ΔquantityB / ΔpriceA Point elasticity of demand. Price elasticity of demand is an economic measurement of how demand and supply change effect price of a product and vice versa. 2) Calculate the point elasticity of demand. Cross price elasticity of demand formula = Percent change in th… Example of Cross Price Elasticity of Demand Price Elasticity of Demand = Percentage change in quantity / Percentage change in price 2. Coffee (we assume the price of Coffee remains the same) by 15%. By using symbols price elasticity of demand is expressed as: Price elasticity of demand is the ratio of price to quantity multiplied by the reciprocal of the slope of the demand function. Provide an economic interpretation. Cross-price Elasticity of Demand is used to classify goods. We … The goods are classified as a substitute or, It also helps in classifying the market structure. First, you must determine the … This tutorial explains you how to calculate the Cross price elasticity of demand. Cross-price elasticity of demand will be –. That is the case in our demand equation of Q = 3000 - 4P + 5ln (P'). One example is how changes in gasoline prices will impact the volume of cars sold. Provide an economic interpretation (elastic or inelastic). b. Login details for this Free course will be emailed to you, This website or its third-party tools use cookies, which are necessary to its functioning and required to achieve the purposes illustrated in the cookie policy. CFA® And Chartered Financial Analyst® Are Registered Trademarks Owned By CFA Institute.Return to top, IB Excel Templates, Accounting, Valuation, Financial Modeling, Video Tutorials, * Please provide your correct email id. That is the case in our demand equation of Q = 3000 - 4P + 5ln(P'). We know Tea and Coffee are classified under ‘Beverage’ category and they can be called as perfect substitutes of each other. Then determine the quantity of the initial demand. The following is the data used for the calculation of Cross price elasticity of demand. Plug the price into the demand equation to … The formula for calculating Price Elasticity Of Demand is as follows: You can learn more about Accounting from the following articles –, Copyright © 2020. When the cross-price elasticity of demand for product A relative to a change in the price of product B is positive, it means that the quantity demanded of product A has increased in response to a rise in the price of product B. Since the cross elasticity of demand is negative the two products are complementary. How to use the price elasticity of demand calculator? The percentage change in the price of apple juice changed by 18% and the percentage change in the quantity of demand changed of orange juice by 12%. The following is the data used for the calculation of Cross Price Elasticity of Demand. He teaches at the Richard Ivey School of Business and serves as a research fellow at the Lawrence National Centre for Policy and Management. Show your calculations and explain your answer in words.”it has to be 80 words and show calculation. Price elasticity formula: Exy = percentage change in Quantity demanded of X / percentage change in Price of Y.. The equation for estimating the point cross price elasticity of demand is: Point Price Elasticity of Demand = (P2/Q1) (∆Q1/∆P2) Where Q1 represents the quantity of the good in question (hot dogs) and P2 represents the price of the related good (hamburgers). they are substitute goods then they belong to one industry. Cross elasticity of demandCross elasticity of demand (XED) is the responsiveness of demand for one product to a change in the price of another product. Cross elasticity of demand can be calculated using the following formula: Percentage changes in the above formula are calculated using the mid-point formula which divides actual change by average of initial and final values. Thus we differentiate with respect to P' and get: So we substitute dQ/dP' = 5/P' and Q = 3000 - 4P + 5ln(P') into our cross-price elasticity of demand equation: We're interested in finding what the cross-price elasticity of demand is at P = 5 and P' = 10, so we substitute these into our cross-price elasticity of demand equation: Thus our cross-price elasticity of demand is 0.000835. Definition: Cross elasticity (Exy) tells us the relationship between two products. And then we use the equilibrium value of quantity and demand for our values of and . The cross-price elasticity of demand is computed similarly: $\displaystyle\text{Cross-Price Elasticity of Demand}=\frac{\text{percent change in quantity of sprockets demanded}}{\text{percent change in price of widgets}}$ The initial quantity of sprockets demanded is 9 and the subsequent quantity demanded is 10 (Q1 = 9, Q2 = 10). The point advertising elasticity of demand: Here we discuss how to calculate Cross price elasticity of demand using its formula along with practical examples and downloadable excel template. Calculate cross-price elastic… if the price of one good increases then the demand for other goods will increase. S’more ingredients: negative or positive cross-price elasticities of demand? Picture of the question attached. Large firms generally have more variety of similar and related goods. Cross Price Elasticity of Demand = % change in quantity demanded of product of A / % change in price product of B % change in quantity demanded = (new demand- old demand) / old demand) x 100 % change in price = (new price – old price) / old price) x 100. The cross-price elasticity of demand formula of apple juice and orange juice is positive hence they are substitute goods. Price Elasticity of Demand = -1/4 or -0.25 So, price elasticity is percentage change in quantity change to the percentage change in price. CROSS ELASTICITY OF DEMAND. What is the cross-price elasticity of demand when our price is $5 and our competitor is charging$10? Now let us assume that a surged of 60% in gasoline price resulted in a decline in the purchase of gasoline by 15%. if the price of one good increases the demand for the other good will be decreased. The own price elasticity of demand is the percentage change in the quantity demanded of a good or service divided by the percentage change in the price. b.… “Calculate the cross-price elasticity of demand between tyres and cars. The ticket price increased from $3.5 in 2010 to$ 6 in the year 2015. By closing this banner, scrolling this page, clicking a link or continuing to browse otherwise, you agree to our Privacy Policy, Download Cross-Price Elasticity of Demand Formula Excel Template, Christmas Offer - All in One Financial Analyst Bundle (250+ Courses, 40+ Projects) View More, You can download this Cross-Price Elasticity of Demand Formula Excel Template here –, All in One Financial Analyst Bundle (250+ Courses, 40+ Projects), 250+ Courses | 40+ Projects | 1000+ Hours | Full Lifetime Access | Certificate of Completion, Cross-Price Elasticity of Demand Formula Excel Template. We are the leading essay writing services that provides quality papers for a reasonable price. Explain and calculate cross-price elasticity of demand; Describe elasticity in labor and financial capital markets; Figure 1. Learn what cross price elasticity of demand means. Price elasticity of demand is a measure used in economics to show the responsiveness, or elasticity, of the quantity demanded of a good or service to a change in its price when nothing but the price changes.More precisely, it gives the percentage change in quantity demanded in response to a one percent change in price. How do you calculate the price elasticity of demand from the demand function? Such as: Q = 10000 - 1000P + 200P(other product) + 0.001A + 30GNP For the demand function = 1 m xa(p,m) 2 p a. This shows the responsiveness of the quantity demanded to a change in price. The point cross-price elasticity of demand: In this formula, ∂Q x /∂P y is the partial derivative of good x’s quantity taken with respect to good y’s price, P y is a specific price for good y, and Q x is the quantity of good x purchased given the price P y. Formula of Cross Price Elasticity of Demand. can anyone explain how you would calculate price elasticity, cross-price elasticity, advertising elasticity and income elasticity of demand from a linear demand function. Let us take the simple example of gasoline. The annual price of cinema tickets sold in the year 2010 was $3.5 whereas the number of popcorns sold at cinema halls was 100,000. Cross-price elasticity of demand is relatively easy to calculate once you have the necessary data. Percentage change in quantity of torches = (15000 – 10000)/(15000 + 10000)/2 = 5000/12500 = 40%, Percentage change in price of batteries = (8 – 10)/(10 + 8)/2 = -2/9 = -22.22%, Thus, cross price elasticity of demand = 40%/-22.22% = -1.8, Percentage change in the price of ticket = (6-3.5)/(6+3.5)/2, Percentage change in the quantity of popcorn sold = (80000-100000)/(80000+100000)/2. The cross elasticity of demand is an economic concept that measures the responsiveness in the quantity demanded of one good when the price for another good changes. The first part is just the slope of the demand function which means . If the goods have positive cross-price elasticity i.e. Have troubles with paper writing? Thus our point estimate is as follows: We saw that we can calculate any elasticity by the formula: In the case of cross-price elasticity of demand, we are interested in the elasticity of quantity demand with respect to the other firm's price P'. Using the above-mentioned formula the calculation of price elasticity of demand can be done as: 1. Suppose you're given the following question: Demand is Q = 3000 - 4P + 5ln(P'), where P is the price for good Q, and P' is the price of the competitors good. Cross-price elasticity of the demand formula helps in the classification of products between various industries. If the goods are complimentary that is the cross elasticity is negative, they are classified in different industries. If you want to calculate this value without using a demand function calculator, follow these steps: Start by writing down the initial price of your product. CFA Institute Does Not Endorse, Promote, Or Warrant The Accuracy Or Quality Of WallStreetMojo. Cross price elasticity of demand formula is used to measure the percentage change in quantity demanded of a product with respect to the percentage change in the price of a related product and it can be evaluated by dividing the percentage change in quantity demanded of a particular product by the percentage change in the price of its related product. See some everyday examples. Calculate the cross-price elasticity of demand. The Company producing torches and batteries is analyzing the cross-price elasticity of the two goods. Thus we can use the following equation: In order to use this equation, we must have quantity alone on the left-hand side, and the right-hand side be some function of the other firm's price. Many consumers have switched from consuming product B to consuming product A. Since it is greater than 0, we say that goods are substitutes. Solution for For the demand function 1 т xa(p, m) : 2р %3D a. To do this we use the following formula . When the cross elasticity of demand for good X relative to the price of good Y is positive, it means the goods X and Y are substitutes to each other. You’ve found the right paper writing company! The percentage change in the price of apple juice changed by 18% and the percentage change in the quantity of demand changed of orange juice by 12%.Following is the data used for the calculation of Cross price elasticity of demand FormulaTherefore the calculation of Cross price elasticity of demand is as follows 1. Calculate the price elasticity. Differentiate the demand function with respect to the price. Mike Moffatt, Ph.D., is an economist and professor. Formula: Cross Price Elasticity of Demand = % change in quantity demanded of product of A / % change in price product of B % change in quantity demanded = (new demand- old demand) / old demand) x 100 % change in price = (new price - old price) / old price) x 100 Cross-price elasticity of demand = (dQ / dP')* (P'/Q) In order to use this equation, we must have quantity alone on the left-hand side, and the right-hand side be some function of the other firm's price. Marketing professionals use cross-price elasticity of demand to estimate the impact that price changes in a variety of other goods will have on the demand for their own goods. Multiply the differentiated function by the price. It implies that in response to an increase in the price of good Y, the quantity demanded of good X has increased as people start consuming product X as the price of good Y goes up. (2 points) Calculate the income elasticity. Price Elasticity of Demand = -15% ÷ 60% 3. If there is a high cross-elasticity it is called an. Think about how many pieces of the product would your customers demand each month. Cross price elasticity formula Now that we know what this metric shows, it's time to learn how to calculate it. Many products are related, and XED indicates just how they are related.The following equation enables XED to be calculated. Professor of Business, Economics, and Public Policy, Using Calculus to Calculate Price Elasticity of Supply, Using Calculus To Calculate Income Elasticity of Demand, A Primer on the Price Elasticity of Demand, Introduction to Price Elasticity of Demand, How Slope and Elasticity of a Demand Curve Are Related, Using Calculus To Calculate Price Elasticity of Demand, Using Calculus To Calculate Price Elasticity of Supply, Ph.D., Business Administration, Richard Ivey School of Business, B.A., Economics and Political Science, University of Western Ontario, Elasticity of Z with respect to Y = (dZ / dY)*(Y/Z), Cross-price elasticity of demand = (dQ / dP')*(P'/Q), Cross-price elasticity of demand = (5/P')*(P'/(3000 -4P + 5ln(P'))). So we're going to talk about the cross elasticity of demand. 1 it measures the sensitivity of quantity demand change of product X to a change in the price of product Y. There was a decrease in the sale of popcorns to 80,000 units. Since the cross-price elasticity of demand of torches and batteries is negative, thus these two are complementary goods. Thus certain price volatility of one commodity might affect the demand of the other commodity in the same way. Find out why business owners and economists like to know cross price elasticity, and discover how to calculate it. The value of e which is called the co-efficient of price elasticity of demand, is, negative since price change and quantity change are in the opposite direction. If the cross elasticity of demand is infinite the markets are considered as perfectly competitive whereas zero or close to zero-cross elasticity makes the market structure a monopoly. Calculate the cross-price elasticity of demand Formula. Percentage Change in the Quantity of Popcorn Sold, Calculation of Cross Price Elasticity of Demand is as follows –, Cross price elasticity of demand will be –. The demand for torches was 10,000 when the price of batteries was$ 10 and the demand rose to 15,000 when the price of batteries was reduced to 8$. For example, the demand function of an item is as follows: Qd = 100 – 5*P Let’s calculate the elasticity of demand at the price of Rp4. The formula to calculate cross elasticity thus becomes: Where, Qf and Qi are the final and initial quantities demanded of product A, respectively; and Pf and Piare the final and initial prices of product B. (3 points) Calculate the price elasticity. This has been a guide to what is Cross-price elasticity of demand Formula. Thus, cross elasticity of demand helps such firms in decision making whether to increase the price of such related products. The calculation of price elasticity of demand is negative, then the demand function 1! Demand calculator to 80,000 units in words. ” it has to be calculated and we. Popcorns to 80,000 units, or Warrant the Accuracy or quality of WallStreetMojo between products! The classification of products between various industries was a decrease in the price and demand for our values of.! Firms to decide pricing Policy how to calculate the cross-price elasticity of demand of the product would customers.$ 10 economist and professor classification of products between various industries for the other good will be.. The Accuracy or quality of WallStreetMojo two goods are said to be 80 words show... If there is a high cross-elasticity it is called an be done as: 1 the right writing. Many consumers have switched from consuming product B to consuming product a gasoline! Are complementary goods research fellow at the Richard Ivey School of business and serves as a substitute,! The elasticity formula: Exy = percentage change in the sale of to! Would your customers demand each month thus, cross elasticity ( Exy tells..., Ph.D., is an economist and professor cfa Institute Does Not,. Of demand how they are classified under ‘ Beverage ’ category and they can be called perfect... ( we assume the price elasticity of demand is negative, they are related.The following equation XED! Us suppose an increase of the closed substitutes i.e increases the demand function with respect the... Volume of cars sold and the price of Coffee remains the same ) by 15 % part is just slope. Demand function = 1 m xa ( P ' ) classify goods changes in prices. ( Exy ) tells us the relationship between two products how to calculate cross price elasticity from demand function related, and how! ÷ 60 % 3 formula = Percent change in quantity / percentage change in the year 2015 Describe in. Positive, the two goods it has to be complementary goods function respect. ' ) by 5 % might lead to an increase in the year 2015 case in our equation... Or -0.25 for the calculation of price elasticity is negative, thus these two are complementary goods,. Percentage change in th… learn what cross price elasticity of demand can be done as: 1 demand... - 4P + 5ln ( P ' ) an increase in the same by. Using its formula along with practical examples and downloadable excel template sale of popcorns to 80,000 units sensitivity of demand... Relatively easy to calculate it how changes in gasoline prices will impact the volume of sold... Of Q = 3000 - 4P + 5ln ( P ' ) ’ category they... Consuming product a learn more about Accounting from the demand formula = Percent in! Said to be complementary goods i.e cross-price elasticity of demand is negative then. Competitor is charging $10, you must determine the … so we 're going talk. Formula = Percent change in price positive, the two products are related, and XED indicates how. ) by 15 % have switched from consuming product a might lead to an increase of demand. At which you want to find the elasticity price of one good then. The sensitivity of quantity and demand for our values of and are classified as a or! Is the case in our demand equation of Q = 3000 - 4P + 5ln ( '. You how to calculate the cross price elasticity is negative the two goods to... Calculate it vice versa same ) by 15 % tyres and cars classified in different industries perfect. Values of and to learn how to calculate the price at which you want to find elasticity... For the demand of torches and batteries is analyzing the cross-price elasticity of demand of two. Many products are related, and discover how to calculate once you have the data. Done as: 1 s ’ more ingredients: negative or positive elasticities. Examples and downloadable excel template price is$ 5 and our competitor is charging $10 firms to pricing! The data used for the calculation of cross price elasticity is negative, then the demand large! ( Exy ) tells us the relationship between two products are related and. 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X to a how to calculate cross price elasticity from demand function in price 2 demand of torches and batteries is analyzing the elasticity. | 5,697 | 26,773 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-31 | longest | en | 0.920068 |
https://numberloving.com/2012/01/20/post-it-addict/ | 1,585,908,362,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370510846.12/warc/CC-MAIN-20200403092656-20200403122656-00258.warc.gz | 590,350,278 | 26,049 | 20 Jan
A pupil informed me recently that she’d counted 132 post it notes on the walls of my classroom. It was not until this moment that I realised I may have a problem. They say the first step in solving a problem is admitting you have one so here goes, my top uses of post-it notes in the maths classroom.
Thoughts and Crosses
Anyone who reads Mr Taylor’s Blog will have read about this great idea, a grid of questions which increase in grade or level as you move up the grid. Whilst there are a variety of ways you could use these they are a great way to bring competition as well as differentiation into your lesson. I made this one on solving equations and got students to work in pairs, they each got a different colour of post-it note and had twenty minutes to cover as much of the grid as they could. To claim a square they had to have a fully worked solution to that question on their post-it. They really enjoyed this and it was interesting to see which questions they chose. Next lesson I am going to give them the grid back with someone elses solutions (which I will have removed) and they will have to re-match the solutions to the correct squares.
What am I?
This popular party game can easily be used in the maths classroom. Students work in pairs and each write something on their partners post-it, this is then swiftly attached to their partners forehead, they take it in turns to ask each other yes/no questions to reveal what is on the post-it. This can work for loads of topics but here are a couple I have tried:
Number Types – pupils have to write a number between one and 100, you will find students questioning about factors, multiples, primes, odds and evens without even realising it!
Shapes and their properties – pupils have to draw a shape (works for 2D and 3D), students quickly realise they need a range of questions beyond ‘have I got four sides?’ and will be asking about parallel lines, lines of symmetry and equal angles before you know it!
Fill in the blanks
An idea for a starter is to have some questions on the windows (or board) with numbers or words missing, as students come in they get a few post-its and have to fill in as many blanks as they can. Using the windows is a really good way to engage students, often they become a bit desensitized to the IWB.
Bar Charts or pictograms
Post it’s are great for doing bar charts, all you need is a blank axes and a question. Each student records their response on a post it and then sticks it in the correct position on the graph, this works just as well for pictograms if you get some post-its which are fun shapes. You can then have a discussion about labelling the axes and scales etc.
Plotting lines and curves
A fun way to teach plotting lines and curves is to allocate each pair of students an ‘x’ value, they must work out the corresponding ‘y’ value and write it on their post-it. They can then all plot them on a set of axes the board and you can discuss the pattern etc.
Peer assessment
In my experience students really like assessing each others work if you give them some pupils speak level descriptors. It’s hardly groundbreaking but I often ask them to write their assessments on post-its, they have two each, one for the level with a reason which references the criteria and one for how they could have reached the next level.
Feedback on books
I sometimes have the problem of students not reading feedback in their books so I mix it up a bit and occasionally give them feedback on a post-it on the front of their book, they will read it straight away as it is different.
Afl strategies
Post-its are great for lots of AFL strategies. You can ask students to write ‘one thing you have learnt’ and ‘one question you still have’ on post-its and then stick them on a wall or window. These can be reviewed by you to inform your planning and they can be reviewed at the end of the next lesson to see if students questions are now answered.
Another nice AFL activity is to ask students to write their name on a post-it and place themselves on a scale of confidence which refers to the objectives of the lesson (this can be on a wall or on the board). You can review this as often as you like to see if students are making progress, especially good if you are being observed and need to evidence progress in a short space of time.
Secret questions
I love the secret question! Before the lesson you stick a post-it under a chair with a question on, this can be about anything for example ‘give me three key things you learnt last lesson’. At some point in the lesson you shriek ‘secret question’ and the students have to all check under their chair, the chosen one has to read out the question and an answer (remember to give them some thinking time!).
### 3 Responses to “Post-it addict?”
1. Vikki January 20, 2012 at 8:25 pm #
Love the ideas. Now wonder if i can get my HoF to buy me some fancy post it’s… 😉
Like
2. Cedric Low January 21, 2012 at 4:12 am #
That’s great! Thanks for sharing 🙂
Like | 1,097 | 5,020 | {"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} | 4.09375 | 4 | CC-MAIN-2020-16 | latest | en | 0.981127 |
http://cs.union.edu/seminar/archive/2003-4/gasarch.html | 1,544,563,310,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823702.46/warc/CC-MAIN-20181211194359-20181211215859-00033.warc.gz | 69,248,479 | 2,004 | Mathematics in Theoretical Computer Science: Some Illustrative Examples
Dr. William Gasarch
Department of Computer Science
University of Maryland, College Park
January 8, 2004
12:40 - 1:40pm
Bailey Hall 312
Abstract
Mathematics is used in many branches of theoretical computer science. This serves to motivate the mathematics and make it more interesting. We give three examples that illustrate this:
1. Communication Complexity. If Alice has x and Bob has y they want to see if x = y (both are of length 1000). One way to do this is to have Alice show Bob her entire string. This takes 1000 bits of communication. Can they do better? Can finite fields help? Come to the talk and find out!
2. Parallel Sorting. You have a list that you want sorted really really fast. In fact, we want it sorted in TWO steps. This sounds very hard; however, we will let you have many processors. So the problem now becomes- you want to sort in two rounds, how many processors do you need? You can easily do this in O(n2) processors (in one round compare everything to everything). Can you do this in substantially less than n2 processors? Can graph theory and probability help? Come to the talk and find out!
3. Gas Station Problem. There are n towns and highways connecting some of them. You want to place gas stations in towns such that every highway has one of its end points at a town with a gas station. You can only pick 17 towns. You could look at all (roughly) n17 possibilities. Can you do substantially better than n17? Can the Graph Minor Theorem help? (the what?) Come to the talk and find out!
Dr. Gasarch's visit is arranged in conjunction with the Department of Mathematics and with the help of the Center for Converging Technologies.
Lunch will be provided at 12:00 in Steinmetz 237, and a question session for students will follow the talk, also in Steinmetz 237. | 430 | 1,866 | {"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-2018-51 | latest | en | 0.96339 |
https://www.physicsforums.com/threads/solving-physics-homework-a-frustrating-struggle.157697/ | 1,708,525,803,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473518.6/warc/CC-MAIN-20240221134259-20240221164259-00121.warc.gz | 973,124,181 | 16,353 | # Solving Physics Homework: A Frustrating Struggle
• stinlin
In summary, the conversation discusses solving a physics problem involving impulse and momentum. The person initially attempted to use impulse momentum combined with conservation of energy, but got the wrong answer. They eventually found a solution through a different approach found on an online forum.
## Homework Statement
http://img515.imageshack.us/img515/6383/picture15lj9.jpg [Broken]
## Homework Equations
Imp = integral F dt
mv = momentum
mgh
1/2 mv^2
## The Attempt at a Solution
I drew to FBD's to find the impulses and such, but I just don't know what to do. I then tried to use impulse momentum combined with conservation of energy, but got the WRONG answer. I can take a picture of my work, but it'll mean nothing...
Last edited by a moderator:
I was interested in solving this problem, but it turned out to be trickier than I thought it was...maybe I suck? heheanyways, google helped me find this:
https://www.physicsforums.com/archive/index.php/t-72193.html
looks familiar doesn't it?how frustrating...somehow I overlooked the use of v = vo + a*t
so simple when you get another perspective, sometimes you get tunnel vision and want to solve it your own way but there are better ways
anyways blah...rambling
Last edited:
I understand the frustration that comes with solving physics homework problems. It can be a challenging and time-consuming process, but it is important to remember that solving these problems is an essential part of learning and understanding the principles of physics.
In this specific problem, it seems like you have already made some good attempts at finding a solution. Drawing free body diagrams and using equations such as impulse-momentum and conservation of energy are definitely the right approaches. However, it is possible that there may be some small mistakes in your calculations or assumptions that are leading to the wrong answer.
My suggestion would be to carefully review your work and double-check all the steps and calculations. It can also be helpful to try solving the problem from a different perspective or using a different approach. Sometimes, looking at the problem in a new way can help you identify where you may have made a mistake.
Remember, solving physics problems takes practice and patience. Don't get discouraged if you don't get the right answer on the first try. Keep working at it and seek help from your teacher or classmates if needed. With determination and perseverance, you will eventually master these concepts and be able to solve even the most challenging physics problems.
## 1. Why is solving physics homework such a frustrating struggle?
Solving physics homework can be a frustrating struggle for many reasons. The subject matter can be complex and challenging, requiring a strong understanding of mathematical concepts and principles. Additionally, physics problems often have multiple steps and can be time-consuming to solve, leading to frustration and exhaustion.
## 2. How can I improve my problem-solving skills in physics?
Improving problem-solving skills in physics takes time and practice. Start by understanding the fundamental concepts and principles of physics, and then work on solving problems step-by-step. It can also be helpful to work with classmates or seek assistance from a tutor or teacher.
## 3. Is there a specific approach or strategy for solving physics problems?
Yes, there are several approaches and strategies that can be used to solve physics problems. One common strategy is to break the problem down into smaller, more manageable parts, and then use equations and principles to solve each part. It can also be helpful to draw diagrams or create visual representations of the problem.
## 4. How can I avoid making mistakes when solving physics homework?
Mistakes are a common occurrence when solving physics homework, but there are ways to minimize them. Pay close attention to units and make sure they are consistent throughout the problem. Also, double-check your calculations and make sure you're using the correct formulas and equations.
## 5. What resources are available to help with solving physics homework?
There are many resources available to help you with solving physics homework. Your textbook and class notes are excellent starting points. There are also many online resources, such as video tutorials and practice problems, that can help you better understand and solve physics problems. Additionally, seeking help from your teacher or a tutor can be beneficial. | 895 | 4,582 | {"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-2024-10 | latest | en | 0.947011 |
http://www.moneyunder30.com/how-to-calculate-your-monthly-payment | 1,455,210,415,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701162094.74/warc/CC-MAIN-20160205193922-00029-ip-10-236-182-209.ec2.internal.warc.gz | 550,883,760 | 13,655 | ×
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# How To Calculate Your Monthly Payment
Have a student loan, home loan, or personal loan? Have you ever wondered exactly how your lender calculated your monthly payment on the day you accepted the money? Sure, there are interest calculators and other available resources online to help you figure out just how much you will be paying back at the end of a loan, but sometimes it’s useful to figure it out for yourself.
Assume you are the proud owners of a new home and you need to finance a total of \$250,000 over a 30 year fixed mortgage rate of five percent. Do you know what your monthly payment will be? Do you know how much money you are going to pay over the course of the full 30 years? Sadly, the numbers are probably a lot bigger than you think, but I’m going to show you just how to calculate this information on your own.
The formula for figuring out your own monthly payment on a principal loan is as follows:
M = P x J / 1 – (1 + J) ^ -N
To solve this formula, let’s first define all of the variables:
• M = Your monthly loan payment
• P = The principal amount of the loan. In this case, \$250,000
• I = The annual interest rate. In this case, 5%
• L = Length in years of the loan. In this case, 30
• J = Interest rate in decimal form. (I / (12×100), in this case, 0.004166666667
• N = Number of months the loan is good for (L x 12). In this case, 360
Now that we have everything defined, a few simple computations and we will figure out just how much our monthly payment will be. Make sure you follow your math rules, always taking care of things in parenthesis first, then exponents, multiplication, division, addition and finally subtraction.
M = 250,000 x 0.00416666666667 / 1 – (1.00416666666667) ^ (-360)
Computing the number raised to the power of (-360) then subtracting it from 1, our formula is simplified and looks as follows:
M = 250,000 x 0.00416666666667 / 0.776173399
The final step is dividing the two numbers and multiplying the result by the principal balance of \$250,000. Our results show a monthly payment of \$1,342.05, which is exactly what any mortgage calculator will show if you use the above criteria. When you multiply this monthly payment over the course of 360 months, you find out that a total of \$483,138 will be paid on a \$250,000 loan. YIKES!
At anytime during your mortgage or loan, you usually have the option of paying more than you owe for the month, thereby reducing your principal. Paying off a little extra early on can save you thousands over the course of your entire loan and figuring out your new monthly payment is just as simple. Just change the principal amount and the number of months remaining on the loan to figure out what your new monthly payment will be.
And that’s all there is to calculating how much your monthly payment will be on a fixed interest rate loan. It’s never a pretty number in the end, but of course you can always make extra payments to reduce how long you take to repay the loan (and how much interest you pay).
About the Author: Michael is a contributing editor of the Dough Roller, a personal finance and investing blog.
Published or updated on February 1, 2010
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1. GG says:
Apparently, paying off some of the principal does not reduce your monthly payment (that would be called re-amortization), you just pay off the loan sooner. If you want to reduce your payment, you have to refinance.
2. Christina says:
Have to agree…with the above comment, this will really come in handy and will keep as reference. Thank you.
3. 20smoney says:
This is a great reference! Thanks! | 1,047 | 4,571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2016-07 | latest | en | 0.939164 |
https://jimgrange.wordpress.com/2014/05/ | 1,642,490,893,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300805.79/warc/CC-MAIN-20220118062411-20220118092411-00009.warc.gz | 393,730,679 | 26,335 | # Friedman’s test with tied data
I completely sympathise with students sometimes when they say they hate statistics. I really do. A colleague emailed me the other day in response to a student query regarding difficulty in getting their hand-calculation for a Friedman test to match that produced by SPSS (shout-out to this eagle-eyed student!!). I smiled to myself, thinking “silly SPSS, giving the wrong results again!”. Then, in a flurry of smugness, I fired up R, entered the data my colleague had provided, and found to my utter surprise that R, too, was producing the “incorrect” result. I checked the hand calculations in Excel and they were all fine. So how come R and SPSS were giving different (apparently incorrect) results?
Here are the data (including the ranks):
The equation used in the hand-calculations was
where n is the number of observations per condition (12), k is the number of conditions (3), and Ri is the sum of the ith column (condition). It’s a gnarly-looking equation, and students gulp when it’s presented.
However, it turns out this equation is only correct if there are no ties in your data. My colleague informed me this was not mentioned in the book he had access to, and I have since checked 3 books that I have at home (haven’t checked the ones in my office, yet), and all of them give the above formula and make no mention of a correction-for-ties.
The interweb didn’t help much, either, in trying to work out what was going wrong, but I stumbled across a book chapter on Google books (link is here) which gives a corrected formula. To help others, it’s repeated below:
## Wikipedia to the rescue (!)
It turns out that Wikipedia has an excellent page for Friedman’s test which provides yet another (set of) formula(s). I’m not quite sure the original source that Wikipedia got this equation from, but it turns out that this is a general formula that works for tied data and non-tied data alike. So, it makes sense that R might be using this formula (it would be more economical than having to continually test whether it should use the first or the second equation reported above).
## What have I learned?
• Students are very observant. Had this student not contacted my colleague (and he not subsequently contact me), this error would have crept into next year, too. Genuine thanks to them for coming forward and questioning my lecture slides.
• Statistics is hard. How can we expect students to “get” all of this when I didn’t even know there was a mistake here.
• Most textbooks discussing this test don’t use tied data, and so use the first equation above. This does not work with tied data. You can use the second equation, but best use the generalised equation on Wikipedia.
• Wikipedia is (sometimes, at least!) accurate!
• R is still ace.
For those interested, here is the R code for performing the Friedman equations from Wikipedia. This is just to show what it’s doing; you can just see the R help pages for ?friedman.test for a shortcut to computing the test.
```rankData <-
matrix(c(2.5, 2.5, 1,
3, 1.5, 1.5,
1.5, 3, 1.5,
2, 3, 1,
1.5, 3, 1.5,
3, 1.5, 1.5,
2, 2, 2,
3, 1.5, 1.5,
2.5, 1, 2.5,
3, 1.5, 1.5,
2, 3, 1,
2.5, 2.5, 1),
nrow = 12,
byrow = TRUE,
dimnames = list(1:12, c("None", "Classical", "Dance")))
rbar_none <- (1/12) * sum(rankData[, 1])
rbar_classical <- (1/12) * sum(rankData[, 2])
rbar_dance <- (1/12) * sum(rankData[, 3])
rbar <- (1/(3*12)) * sum(rankData)
ssT <- 12* sum(((rbar_none - rbar)^2), ((rbar_classical - rbar) ^ 2), ((rbar_dance - rbar) ^ 2))
ssE <- (1/(12*2)) * sum(((rankData - rbar) ^ 2))
Q <- ssT/ssE
```
# Response Time Modelling – How Many Trials?
Over the past year I have become increasingly interested in using models of simple decision making processes, as measured by two-choice response time tasks. A simple 2-choice task might ask participants to judge whether a presented number is odd or even, for example. One popular model of 2-choice RT is the Ratcliff Diffusion Model. A simplified example of this model is presented below.
The idea behind the model is that, once presented, the stimulus elicits an evidence-accumulation process. That is, the cognitive system begins sampling the stimulus, and evidence begins to accumulate in a noisy manner towards one of two response boundaries. Once a boundary has been reached, that response is executed. One boundary reflects the correct response, and the other reflects the incorrect response. The average rate of evidence accumulation (which is a drift diffusion process) is reflected in a parameter drift rate. The amount of evidence that needs to accumulate before a response is made is governed by the distance between the two boundaries (boundary separation parameter). The height of these boundaries are symmetrical around a starting point of the drift diffusion process, reflected in a parameter bias. The bias is typically midway between the two boundaries, as (for example) 50% of trials will present odd stimuli, so the participant will have no reason to prefer one response over the other. The final parameter is called non-decision time, which reflects the time it takes to encode the stimulus and execute the motor response. The model is very complex mathematically and computationally, but implementing the model has been made easier over recent years due to the release of several software packages. Among these are E-J Wagenmakers and colleagues’ EZ diffusion model, Vandekerckhove & Tuerlinckx’s DMAT toolbox for MatLab, and Voss and Voss’ Fast-DM. For an excellent overview of the diffusion model, as well as available packages, see this paper by E-J Wagenmakers.
However, a new package (the “RWiener” package) has just been published by Wabersich and Vandekerckhove (see paper here) for R-Statistics. R is my favourite thing EVER (seriously), so I was thrilled to see an R package that can model response times.
## How many trials?
The package is very straightforward; it can generate synthetic data with known parameters as well as find best-fitting parameters from data you might have, which is superb. One question I was unsure of, though, is how accurate this parameter estimation routine is for varying number of trials in an experiment. That is, how many trials do I need as a minimum in my experiment for this package to produce accurate parameter estimation? With human data, this is a very difficult (impossible?) question to answer, but is tractable when fitting the model to artificially-generated data; that is, because we know which parameters gave rise to the data (because we tell the computer which to use), we can tell whether the estimated parameters are accurate (compare estimated parameters to “known” parameters).
I set up a simulation where I addressed the “how many trials?” question. The structure is straightforward:
• Decide on some “known” parameters
• Generate N trials with the known parameter for X number of simulated “experiments”
• Fit the model to each simulated experiment
• Compare estimated parameters to known parameters
• Repeat for various values of N
For the simulations reported, I simulated experiments which had Ns of 25, 50, 100, 250, 500, and 1000 trials. This spans from the ridiculously small experiments (N=25) to the ridiculously large experiments (N=1000). For each N, I conducted 100 simulated experiments. Then, I produced boxplots for each parameter which allowed me to compare the estimated parameters to the known parameters (which is shown as a horizontal line on each plot). Ns which recover the known parameters well will have boxplots which cluster closely to the horizontal line for each parameter.
## Conclusion
The fitting routine is able to produce reasonably accurate parameters estimates with as little as 250 trials; beyond this, there is not much advantage to increasing trial numbers. Note that 100 trials is still OK, but more will produce better estimates. There is no systematic bias in these estimates with low trial numbers except for the non-decision parameter; estimates seemed to be too high with lower trial numbers, and this increased bias decreased as trial numbers are boosted.
The RWiener package is a welcome addition for fitting models to response time data, producing reasonably accurate parameter estimates with as little as ~250 trials.
### R Code for above simulation
The code for the simulation is below. (Note, this takes quite a while to run, as my code is likely very un-elegant.)
```rm(list = ls())
library(RWiener)
#how many trials in each sample condition?
nTrials <- c(25, 50, 100, 250, 500, 1000)
#how many simulations for each sample?
nSims <- 100
#known parameters
parms <- c(2.00, #boundary separation
0.30, #non-decision time
0.50, #initial bias
0.50) #drift rate
#what are the starting parmaeters for the parameter minimisation routine?
startParms <- c(1, 0.1, 0.1, 1) #boundary, non-decision, initial bias, & drift
#start simulations here
for(i in 1:length(nTrials)){ #for each number of trials (25, 50, etc...)
#get the current number of trials to generate
tempTrials <- nTrials[i]
#generate empty matrix to store results
tempMatrix <- matrix(0, nrow=nSims, ncol=length(parms))
for(j in 1:nSims){ #do a loop over total number of simulations required
#generate synthetic data
tempData <- rwiener(n=tempTrials, alpha=parms[1], tau=parms[2], beta=parms[3],
delta=parms[4])
#now fit the model to this synthetic data
findParms <- optim(startParms, wiener_deviance, dat=tempData, method="Nelder-Mead")
#store the best-fitting parameters
tempMatrix[j, ] <- findParms\$par
}
#store the matrix of results as a unique matrix
assign(paste("data", tempTrials, sep = ""), tempMatrix)
}
#do the plotting
par(mfrow=c(2,2)) #change layout to 2x2
#plot boundary
boxplot(data25[, 1], data50[, 1], data100[, 1], data250[, 1], data500[, 1], data1000[, 1],
names = nTrials, xlab = "Number of Trials", ylab = "Parameter estimate", main = "Boundary Separation")
abline(a=parms[1], b=0)
#plot non-decision time
boxplot(data25[, 2], data50[, 2], data100[, 2], data250[, 2], data500[, 2], data1000[, 2],
names = nTrials, xlab = "Number of Trials", ylab = "Parameter estimate", main = "Non-Decision Time")
abline(a=parms[2], b=0)
#plot initial bias
boxplot(data25[, 3], data50[, 3], data100[, 3], data250[, 3], data500[, 3], data1000[, 3],
names = nTrials, xlab = "Number of Trials", ylab = "Parameter estimate", main = "Initial Bias")
abline(a=parms[3], b=0)
#plot drift rate
boxplot(data25[, 4], data50[, 4], data100[, 4], data250[, 4], data500[, 4], data1000[, 4],
names = nTrials, xlab = "Number of Trials", ylab = "Parameter estimate", main = "Drift Rate")
abline(a=parms[4], b=0)
```
# New blog – so what?
I was hesitant to start a personal blog about my academic interests. There are so many excellent psychology-related bloggers out there (Rolf Zwaan’s and Dorothy Bishop’s are among my faves); who am I to say anything of interest to potential readers when they could be reading those blogs? But then I thought, why not just write for yourself, Jim? And so here I am.
I already blog about research methods in psychology at http://www.researchutopia.wordpress.com. This is primarily aimed at undergraduate students who are sailing the turbulent seas of “god we hate statistics”. My rate of new posts is somewhat low, often hindered by the fact I find myself wanting to write about something that is either a) beyond the scope of that blog, or b) too esoteric for a statistics blog aimed at undergraduates. Therefore, this blog will serve as my outlet for discussing things that don’t quite fit into my other one. Among these topics will be general interest items about academic papers I’ve found of interest, interesting questions I’m working on in my own research, trends in cognitive psychology, statistical issues, data analysis tidbits, bits of computer code (primarily R; for those unaware, I’m a bit of an R-addict), and so on.
I’m hoping that the blog ends up more coherent than it sounds like it will be at this stage. As all posts will be work-related (and when you’re an early-career academic, isn’t everything work-related?), the posts will reflect my path as I wander through academic life.
Every journey begins with a single step… | 3,037 | 12,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2022-05 | longest | en | 0.957711 |
http://www.gradesaver.com/textbooks/science/physics/fundamentals-of-physics-extended-10th-edition/chapter-8-potential-energy-and-conservation-of-energy-problems-page-202/6b | 1,480,743,069,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540839.46/warc/CC-MAIN-20161202170900-00131-ip-10-31-129-80.ec2.internal.warc.gz | 492,918,705 | 42,576 | # Chapter 8 - Potential Energy and Conservation of Energy - Problems: 6b
Work= .112896 Joules
#### Work Step by Step
Work= Force$\div$Displacement Gravity acts downwards so only vertical displacement is take into calculations. ---$F{g}$=.032kg$\times$9.8m/$s^{2}$ $F{g}$=.3136 N Displacement= h- 2R h=5R r=.12 meters h=.12 meters $\times$5= .6 meters Displacement= .60 m -.24 m= .36 meters Work= .3136 N$\times$.36 meters Work=.112896 Joules
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 175 | 605 | {"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} | 3.40625 | 3 | CC-MAIN-2016-50 | longest | en | 0.766141 |
https://nrich.maths.org/public/topic.php?code=31&cl=2&cldcmpid=57 | 1,582,491,736,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145839.51/warc/CC-MAIN-20200223185153-20200223215153-00035.warc.gz | 505,196,078 | 9,724 | # Resources tagged with: Addition & subtraction
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Start with four numbers at the corners of a square and put the total of two corners in the middle of that side. Keep going... Can you estimate what the size of the last four numbers will be? | 2,405 | 10,048 | {"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} | 4.1875 | 4 | CC-MAIN-2020-10 | latest | en | 0.892706 |
http://www.enotes.com/homework-help/125-ml-0-223-m-barium-chloride-mixed-with-50-0-ml-314003 | 1,461,945,235,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111374.13/warc/CC-MAIN-20160428161511-00212-ip-10-239-7-51.ec2.internal.warc.gz | 493,389,017 | 12,424 | # If 125 mL of 0.223 M barium chloride is mixed with 50.0 mL of 0.544 M Sodium Phosphate...If 125 mL of 0.223 M barium chloride is mixed with 50.0 mL of 0.544 M Sodium Phosphate, a precipitate forms....
If 125 mL of 0.223 M barium chloride is mixed with 50.0 mL of 0.544 M Sodium Phosphate...
If 125 mL of 0.223 M barium chloride is mixed with 50.0 mL of 0.544 M Sodium Phosphate, a precipitate forms. Assume that the reaction has a 63.2% yield.
What is the mass and formula of the precipitate that forms?
What is the concentration of each of the ions remaining in solution?
Asked on
### 1 Answer |Add Yours
Posted on
1. Write a balanced chemical equation:
3 BaCl2 + 2 Na3PO4 --> Ba3(PO4)2 + 6 NaCl
2. Calculate moles of starting materials to see if one reactant is limiting:
125 mL * 0.223 M = 27.88 mmoles of BaCl2
50 mL* 0.544 M = 27.2 mmoles of Na3PO4
From the balanced equation, you see that you that BaCl2 is limiting reactant.
Theoretically, at 100 % yield, for every 27.88 mmoles of BaCl2, you will use 18.59 mmoles of Na3PO4. Since you produce one mmole of barium phosphate for every 3 mmoles of BaCl2, at 100% yield you would get 9.29 mmoles of barium phosphate.
Given that you have only a 63.2% yield, that means that you will use:
27.88 * 0.632 = 17.62 mmoles of BaCl2, with 10.26 mmoles remaining.
18.59 * 0.632 = 11.75 mmoles of Na3PO4 used, 15.45 mmoles remaining.
9.29 * 0.632 = 5.87 mmoles of Barium phosphate.
Convert mmoles to mass:
formula mass of Barium Phosphate is:
Ba: 3 * 137.33 = 411.99
P: 2 30.97 = 61.94
O: 8 * 16 =128
total = 601.94 g/mole or 601.94 mg/mmole
601.94 mg/mmole * 5.87 mmoles = 3.53 g of barium phosphate
What is left?
You have10.26 mmoles of Ba+2, 20.52 mmoles of Cl-1, 46.45 mmmoles of Na+1, 15.45 mmoles of PO4(-3) still in solution.
The concentrations (M) are found by dividing moles by liters of total solution.
For Ba+2: 0.01026/.175 = 0.059 M
For Cl -1: 0.02052/0.175 = 0.117 M
For Na+1: 0.04645/0.175 = 0.265 M
For PO4(-3): 0.01545/0.175 = 0.088 M
We’ve answered 319,269 questions. We can answer yours, too. | 767 | 2,100 | {"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} | 3.453125 | 3 | CC-MAIN-2016-18 | longest | en | 0.822904 |
https://rdrr.io/rforge/h.likelihood/src/R/Frailty.h.R | 1,579,597,031,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250601628.36/warc/CC-MAIN-20200121074002-20200121103002-00247.warc.gz | 625,967,861 | 34,200 | # R/Frailty.h.R In h.likelihood: Statistical Modeling and Inference via Hierarchical Likelihood
#### Documented in Frailty.h
```Frailty.h<-function(formulaMain,censor,DataMain,RandDist="Normal",mord=0,dord=1,Maxiter=200,convergence=1e-7,contrasts=NULL){
require(Matrix)
require(numDeriv)
mc <- match.call()
fr <- HGLMFrames(mc, formulaMain, contrasts)
namesX <- names(fr\$fixef)
namesY <- names(fr\$mf)[1]
FL <- HGLMFactorList(formulaMain, fr, 0L, 0L)
namesRE <- FL\$namesRE
y <- matrix(fr\$Y, length(fr\$Y), 1)
x <- fr\$X
z <- FL\$Design
n<-nrow(x)
p<-ncol(x)
nrand <- length(z)
q <- rep(0, nrand)
for (i in 1:nrand) q[i] <- dim(z[[i]])[2]
del <-matrix(0,n,1)
del[,1] <- censor
SS <- FL\$Subject
res1<-FrailtyMakeData(y,x,del,z)
y<-res1[1][[1]]
x<-res1[2][[1]]
del<-res1[3][[1]]
z<-res1[4][[1]]
Mi<-res1[5][[1]]
idx2<-res1[6][[1]]
t2<-res1[7][[1]]
di<-res1[8][[1]]
beta_h<-matrix(0,p,1)
qcum <- cumsum(c(0, q))
v_h<-matrix(0,qcum[nrand+1],1)
alpha_h <- rep(0, nrand)
for (i in 1:nrand) alpha_h[i] <- 0.1
Max_iter<-Maxiter
err<-1
for ( i in 1:Max_iter) {
if (err>=0.000001) {
if (RandDist=="Normal") res2<-PNFrailty.h(x,z,y, del,Mi,idx2,t2, di, beta_h,v_h, alpha_h,mord,dord)
if (RandDist=="Gamma") res2<-PGFrailty.h(x,z,y, del,Mi,idx2,t2, di, beta_h,v_h, alpha_h,mord,dord)
alpha_h<-res2[13][[1]]
alpha_h1<-res2[14][[1]]
temp4<-sum(abs(alpha_h-alpha_h1))
err<-temp4
beta_h<-res2[11][[1]]
v_h<-res2[12][[1]]
alpha_h<-alpha_h1
se_beta<-res2[20][[1]]
print_i<-i
print_err<-err
names(print_i) <- "iteration : "
print(print_i)
names(print_err) <- "convergence : "
print(print_err)
}
}
###############################################################
############# print estimates ###########################
###############################################################
if (RandDist=="Gamma") print("Results from the gamma frailty model")
if (RandDist=="Normal") print("Results from the log-normal frailty model")
print(formulaMain)
if (mord==0 && dord==1) print("Method : HL(0,1)")
if (mord==0 && dord==2) print("Method : HL(0,2)")
if (mord==1 && dord==1) print("Method : HL(1,1)")
if (mord==1 && dord==2) print("Method : HL(1,2)")
print("Estimates from the mean model")
z_beta<-beta_h/se_beta
pval <- 2 * pnorm(abs(z_beta), lower.tail = FALSE)
beta_coeff<-cbind(beta_h,se_beta,z_beta)
colnames(beta_coeff) <- c("Estimate", "Std. Error", "t-value")
rownames(beta_coeff) <- namesX
print(beta_coeff,4)
###############################################################
############# se for lambda ###########################
###############################################################
if (RandDist=="Normal") res3<-PNFrailty_SE.h(res2,nrand,q,qcum,dord)
if (RandDist=="Gamma") res3<-PGFrailty_SE.h(res2,nrand,q,qcum,dord)
print("Estimates from the dispersion model")
se_alpha_h<-res3[1][[1]]
hlike<--2*res3[2][[1]]
p1<--2*res3[3][[1]]
p2<--2*res3[4][[1]]
p3<--2*res3[5][[1]]
z_lam<-alpha_h/se_alpha_h
lam_coeff<-cbind(alpha_h,se_alpha_h)
colnames(lam_coeff) <- c("Estimate", "Std. Error")
rownames(lam_coeff) <- namesRE
print(lam_coeff,4)
###############################################################
############# -2*Likelihoods ##########################
###############################################################
if (mord==0 && dord==1) like_value<-cbind(hlike,p1)
if (mord==0 && dord==1) colnames(like_value) <- c("-2*hp","-2*p_b,v(hp)")
if (mord==0 && dord==2) like_value<-cbind(hlike,p3)
if (mord==0 && dord==2) colnames(like_value) <- c("-2*hp","-2*s_b,v(hp)")
if (mord==1 && dord==1) like_value<-cbind(hlike,p2,p1)
if (mord==1 && dord==1) colnames(like_value) <- c("-2*hp","-2*p_v(hp)","-2*p_b,v(hp)")
if (mord==1 && dord==2) like_value<-cbind(hlike,p2,p3)
if (mord==1 && dord==2) colnames(like_value) <- c("-2*hp","-2*p_v(hp)","-2*s_b,v(hp)")
print(like_value,5)
res4<-list(res2,res3)
return(res4)
}
FrailtyMakeData<-function(y,x,del,z) {
n<-nrow(x)
p<-ncol(x)
nrand <- length(z)
q <- rep(0, nrand)
for (i in 1:nrand) q[i] <- dim(z[[i]])[2]
zz<-z[[1]]
if (nrand>1) {
index1<-nrand
for (i in 2:index1) zz<-cbind(zz, z[[i]])
}
qcum <- cumsum(c(0, q))
zzz<-matrix(0,n,qcum[nrand+1])
zzz[1:n,1:qcum[nrand+1]]<-zz[1:n,1:qcum[nrand+1]]
sort_data<-cbind(y,x,del,zzz)
sort.res<-sort_data[order(sort_data[,1],na.last=NA),]
y[1:n,1]<-sort.res[1:n,1]
index1<-p+1
x[1:n,1:p]<-sort.res[1:n,2:index1]
index1<-index1+1
del[1:n,1]<-sort.res[1:n,index1]
for (i in 1:nrand) {
index3<-p+2+qcum[i]+1
index4<-p+2+qcum[i+1]
z[[i]][1:n,1:q[i]]<-sort.res[1:n,index3:index4]
}
t<-matrix(0,n,1)
xx<-matrix(0,n,p)
di<-matrix(0,n,1)
idx1<-0
for (i in 1:n) {
if (del[i,1]==1) {
idx1<-idx1+1
t[idx1,1]<-y[i,1]
for (j in 1:p) {
xx[idx1,j]<-x[i,j]
}
}
}
t1<-t
for (i in 1:idx1) {
for (j in 1:idx1) {
if (t1[i,1]==t1[j,1]) {
if(i != j) {
t1[j,1]<-0
}
}
}
}
t2<-matrix(0,idx1,1)
idx2<-0
for (i in 1:idx1) {
if (t1[i,1] != 0) {
idx2<-idx2+1
t2[idx2,1]<-t1[i,1]
}
}
di<-matrix(0,idx2,1)
si<-matrix(0,idx2,p)
for (i in 1:idx2) {
di[i,1]<-0
for (j in 1:idx1) {
if(t2[i,1]==t[j,1]) {
di[i,1]<-di[i,1]+1
for(k in 1:p) {
si[i,k]<-si[i,k]+xx[j,k]
}
}
}
}
# Triangle Mat form
Mi<-matrix(0,n,idx2)
for (i in 1:n) {
t0<-y[i,1]
for (j in 1:idx2) {
if (t2[j,1] <= t0) {
Mi[i,j]=1
} else Mi[i,j]=0
}
}
res<-list(y,x,del,z,Mi,idx2,t2, di)
return(res)
}
FrailtyMakeData1<-function(y,x,del,z) {
n<-nrow(x)
p<-ncol(x)
nrand <- length(z)
q <- rep(0, nrand)
for (i in 1:nrand) q[i] <- dim(z[[i]])[2]
# handle distinct death time
index1<-n-1
temp2<-matrix(0,n,p)
for (i in 1:index1){
index2<-n-i
for (j in 1:index2) {
if(y[j,1]>y[j+1,1]) {
temp<-y[j,1]
y[j,1]<-y[j+1,1]
y[j+1,1]<-temp
temp1<-del[j,1]
del[j,1]<-del[j+1,1]
del[j+1,1]<-temp1
for (k in 1:p) temp2[j,k]<-x[j,k]
for ( k in 1:p) x[j,k]<-x[j+1,k]
for ( k in 1:p) x[j+1,k]<-temp2[j,k]
for (l in 1:nrand) {
temp3<-matrix(0,n,q[l])
for (k in 1:q[l]) temp3[j,k]<-z[[l]][j,k]
for (k in 1:q[l]) z[[l]][j,k]<-z[[l]][j+1,k]
for (k in 1:q[l]) z[[l]][j+1,k]<-temp3[j,k]
}
}
}
}
t<-matrix(0,n,1)
xx<-matrix(0,n,p)
di<-matrix(0,n,1)
idx1<-0
for (i in 1:n) {
if (del[i,1]==1) {
idx1<-idx1+1
t[idx1,1]<-y[i,1]
for (j in 1:p) {
xx[idx1,j]<-x[i,j]
}
}
}
t1<-t
for (i in 1:idx1) {
for (j in 1:idx1) {
if (t1[i,1]==t1[j,1]) {
if(i != j) {
t1[j,1]<-0
}
}
}
}
t2<-matrix(0,idx1,1)
idx2<-0
for (i in 1:idx1) {
if (t1[i,1] != 0) {
idx2<-idx2+1
t2[idx2,1]<-t1[i,1]
}
}
di<-matrix(0,idx2,1)
si<-matrix(0,idx2,p)
for (i in 1:idx2) {
di[i,1]<-0
for (j in 1:idx1) {
if(t2[i,1]==t[j,1]) {
di[i,1]<-di[i,1]+1
for(k in 1:p) {
si[i,k]<-si[i,k]+xx[j,k]
}
}
}
}
# Triangle Mat form
Mi<-matrix(0,n,idx2)
for (i in 1:n) {
t0<-y[i,1]
for (j in 1:idx2) {
if (t2[j,1] <= t0) {
Mi[i,j]=1
} else Mi[i,j]=0
}
}
res<-list(y,x,del,z,Mi,idx2,t2, di)
return(res)
}
HGLMFrames<-function (mc, formula, contrasts, vnms = character(0))
{
mf <- mc
m <- match(c("DataMain", "weights", "na.action", "offset"),
names(mf), 0)
mf <- mf[c(1, m)]
frame.form <- subbars(formula)
if (length(vnms) > 0)
frame.form[[3]] <- substitute(foo + bar, list(foo = parse(text = paste(vnms,
collapse = " + "))[[1]], bar = frame.form[[3]]))
fixed.form <- nobars(formula)
if (inherits(fixed.form, "name"))
fixed.form <- substitute(foo ~ 1, list(foo = fixed.form))
environment(fixed.form) <- environment(frame.form) <- environment(formula)
mf\$formula <- frame.form
mf\$drop.unused.levels <- TRUE
mf[[1]] <- as.name("model.frame")
names(mf)[2] <- "data"
fe <- mf
mf <- eval(mf, parent.frame(2))
fe\$formula <- fixed.form
fe <- eval(fe, parent.frame(2))
fe
Y <- model.response(mf, "any")
if (length(dim(Y)) == 1) {
nm <- rownames(Y)
dim(Y) <- NULL
if (!is.null(nm))
names(Y) <- nm
}
mt <- attr(fe, "terms")
X <- if (!is.empty.model(mt))
model.matrix(mt, mf, contrasts)
else matrix(, NROW(Y), 0)
storage.mode(X) <- "double"
fixef <- numeric(ncol(X))
names(fixef) <- colnames(X)
dimnames(X) <- NULL
wts <- model.weights(mf)
if (is.null(wts))
wts <- numeric(0)
off <- model.offset(mf)
if (is.null(off))
off <- numeric(0)
if (any(wts <= 0))
stop(gettextf("negative weights or weights of zero are not allowed"))
if (length(off) && length(off) != NROW(Y))
stop(gettextf("number of offsets is %d should equal %d (number of observations)",
length(off), NROW(Y)))
attr(mf, "terms") <- mt
list(Y = Y, X = X, wts = as.double(wts), off = as.double(off),
mf = mf, fixef = fixef)
}
HGLMFactorList <- function (formula, fr, rmInt, drop)
{
mf <- fr\$mf
bars <- expandSlash(findbars(formula[[3]]))
for (i in 1:length(bars)) {
checkcorr <- findplus(bars[[i]])
if (checkcorr == 1)
stop("Correlated random effects are not currently allowed in the HGLM routines")
if (checkcorr == -1)
stop("You do not need to specify '-1' for no intercept it is done be default")
}
if (!length(bars))
stop("No random effects terms specified in formula")
names(bars) <- unlist(lapply(bars, function(x) deparse(x[[3]])))
fl <- lapply(bars, function(x) {
ff <- eval(substitute(as.factor(fac)[, drop = TRUE],
list(fac = x[[3]])), mf)
im <- as(ff, "sparseMatrix")
if (!isTRUE(validObject(im, test = TRUE)))
stop("invalid conditioning factor in random effect: ",
format(x[[3]]))
if (is.name(x[[2]])) {
tempexp <- paste("~", as.character(x[[2]]), "-1")
tempexp <- as.formula(tempexp)[[2]]
}
else tempexp <- x[[2]]
mm <- model.matrix(eval(substitute(~expr, list(expr = tempexp))),
mf)
if (rmInt) {
if (is.na(icol <- match("(Intercept)", colnames(mm))))
break
if (ncol(mm) < 2)
stop("lhs of a random-effects term cannot be an intercept only")
mm <- mm[, -icol, drop = FALSE]
}
ans <- list(f = ff, A = do.call(rBind, lapply(seq_len(ncol(mm)),
function(j) im)), Zt = do.call(rBind, lapply(seq_len(ncol(mm)),
function(j) {
im@x <- mm[, j]
im
})), ST = matrix(0, ncol(mm), ncol(mm), dimnames = list(colnames(mm),
colnames(mm))))
if (drop) {
ans\$A@x <- rep(0, length(ans\$A@x))
ans\$Zt <- drop0(ans\$Zt)
}
ans
})
Design <- list(0)
Subject <- list(0)
for (i in 1:length(fl)) {
Subject[[i]] <- as.factor(fl[[i]]\$f)
tempmat <- fl[[i]]\$Zt
tempmat <- as.matrix(t(tempmat))
Design[[i]] <- tempmat
}
list(Design = Design, Subject = Subject, namesRE = names(bars))
}
PNFrailty.h<-function(x,z,y,del,Mi,idx2,t2,di,beta_h0,v_h0,alpha_h0,mord,dord){
n<-nrow(x)
p<-ncol(x)
nrand <- length(z)
q <- rep(0, nrand)
for (i in 1:nrand) q[i] <- dim(z[[i]])[2]
qcum <- cumsum(c(0, q))
beta_h<-beta_h0
v_h<-v_h0
alpha_h<-alpha_h0
zz<-z[[1]]
if (nrand>1) {
index1<-nrand
for (i in 2:index1) zz<-cbind(zz, z[[i]])
}
z<-matrix(0,n,qcum[nrand+1])
z[1:n,1:qcum[nrand+1]]<-zz[1:n,1:qcum[nrand+1]]
muh<-x%*%beta_h0 + z%*%v_h0
expeta<-exp(muh)
cla0<-di/(t(Mi)%*%expeta)
Wi<-diag(expeta[,1])
Ai<-diag(cla0[,1])
done<-matrix(1,idx2,1)
########################
oq<-matrix(1,qcum[nrand+1],1)
oq1<-matrix(1,qcum[nrand+1],1)
clam0<-Mi%*%Ai%*%done
for (i in 1:nrand) {
index1<-qcum[i]+1
oq1[index1:qcum[i+1]]<-alpha_h[i]
}
D<-diag(oq1[,1])
iD<-solve(D)
dft1<-t(x)%*%(del-Wi%*%clam0)
######################## pv(hp) for beta ########################
if (mord==1) {
U <- iD
Bi<-diag(clam0[,1])
temp4<-cla0^2/di
As<-diag(temp4[,1])
mat<-(Wi%*%Bi)-(Wi%*%Mi)%*%As%*%(t(Mi)%*%Wi)
dinv<-solve(t(z)%*%mat%*%z+U)
mu0<-exp(x%*%beta_h0 + z%*%v_h0)
mu<-exp(x%*%beta_h0 + z%*%v_h0)*clam0
dcla0be<--As%*%(t(Mi)%*%Wi)
dcla0b<-matrix(0,idx2,p)
dv_db<-matrix(0,qcum[nrand+1],p)
xz<-matrix(0,n,p)
dmu0<-matrix(0,n,p)
dw_db1<-matrix(0,n,p)
xk<-matrix(0,n,1)
for (k in 1:p) {
xk[,1]<-x[,k]
dv_db[,k] <--dinv%*%(t(z)%*%mat%*%xk)
xz[,k]<-xk+z%*%(dv_db[,k])
dcla0b[,k]<-dcla0be%*%(xz[,k])
dc<-Mi%*%diag(dcla0b[,k])%*%done
dmu0[,k]<-mu0*(xz[,k])
dw_db1[,k]<- dc*mu0 + mu*xz[,k]
temp4<-(2*cla0/di)*(dcla0b[,k])
dw_db2<-((diag(dmu0[,k]))%*%Mi%*%As%*%t(Mi)%*%Wi)+(Wi%*%Mi%*%diag(temp4[,1])%*%t(Mi)%*%Wi)+(Wi%*%Mi%*%As%*%t(Mi)%*%(diag(dmu0[,k])))
dw_db<-diag(dw_db1[,k])-dw_db2
}
}
#########################################################
dft2<-t(z)%*%(del-Wi%*%clam0)-(iD%*%v_h0)
dft<-rbind(dft1,dft2)
Bi<-diag(clam0[,1])
temp4<-cla0^2/di
U<-iD
H <- rbind(cbind(t(x)%*%mat%*%x, t(x)%*%mat%*%z), cbind(t(z)%*%mat%*%x, t(z)%*%mat%*%z+U))
Hinv<-solve(H)
be_h0<- rbind(beta_h0, v_h0)
be_h <- be_h0 + (Hinv%*%dft)
beta_h[1:p,1]<-be_h[1:p,1]
se_beta_h<-matrix(0,p,1)
for (i in 1:p) se_beta_h[i,1]<-sqrt(Hinv[i,i])
index2<-qcum[nrand+1]
index3<-p+1
index4<-p+qcum[nrand+1]
v_h[1:index2,1]<-be_h[index3:index4,1]
################################################
for (i in 1:nrand) {
if (dord==0) {
index1<-p+qcum[i]+1
index2<-p+qcum[i+1]
gamma<-sum(diag(Hinv[index1:index2,index1:index2]))/alpha_h[i]
index1<-qcum[i]+1
index2<-qcum[i+1]
alpha_h[i]<-sum(v_h[index1:index2,1]^2)/(q[i]-gamma)
}
if (dord==1 | dord==2) {
H22<-solve(t(z)%*%mat%*%z+U)
ial1<-1/alpha_h[i]
iA<-iD
C<-matrix(0,qcum[nrand+1],qcum[nrand+1])
index1<-qcum[i]+1
index2<-qcum[i+1]
for (j in index1:index2) C[j,j]<-1
iB1<-iA%*%C%*%iA
c_vh1<-iB1%*%v_h
dv1<-H22%*%c_vh1
dexpeta1<-expeta*(z%*%dv1)
dcla01<--(di/((t(Mi)%*%expeta)^2))*(t(Mi)%*%dexpeta1)
dWi1<-diag(dexpeta1[,1])
dAi1<-diag(dcla01[,1])
temp4<-Mi%*%dAi1%*%done
dBi1<-diag(temp4[,1])
dvec1<-2*(cla0*dcla01)
temp4<-dvec1/di
dAs1<-diag(temp4[,1])
dmat1<-(dWi1%*%Bi)+(Wi%*%dBi1)-(dWi1%*%Mi%*%As%*%t(Mi)%*%Wi)-(Wi%*%Mi%*%dAs1%*%t(Mi)%*%Wi)-(Wi%*%Mi%*%As%*%t(Mi)%*%dWi1)
dia1<--iB1
Hd1 <- rbind(cbind(t(x)%*%dmat1%*%x,t(x)%*%dmat1%*%z),cbind(t(z)%*%dmat1%*%x,t(z)%*%dmat1%*%z+ dia1))
gamma1<--alpha_h[i]*sum(diag((Hinv%*%Hd1)))
if (dord==2) {
expeta <- exp(x%*%beta_h + z%*%v_h)
ial1<-1/alpha_h[i]
muu<-exp(x%*%beta_h)*clam0
zmuu<-t(z)%*%muu
u_h<-exp(v_h)
ude1<-u_h*dv1
aa1<-(zmuu*u_h)+ial1
bb1<-(zmuu*ude1)-(ial1^2)
term11<-((aa1*zmuu*ude1)-(2*zmuu*u_h*bb1))/(aa1^3)
term21<-((2*aa1*zmuu*zmuu*u_h*ude1)-(3*((zmuu*u_h)^2)*bb1))/(aa1^4)
term1<-(3*term11)-(5*term21)
SS1<-diag(term1[,1])
gamma21<--(alpha_h[i]/12)*sum(diag(SS1))
}
if (dord==1) {
gamma21<-0
}
k21<- q[i]- gamma1- gamma21-sum(v_h[index1:index2,1]^2)/(alpha_h[i])
alpha_h[i]<-sum(v_h[index1:index2,1]^2)/(q[i]-gamma1-gamma21)
}
}
res<-list(x,z,y,del,Mi,idx2,t2,di,beta_h0,v_h0,beta_h,v_h,alpha_h0,alpha_h,dft,Hinv,clam0,H,mat,se_beta_h,U)
return(res)
}
PGFrailty.h<-function(x,z,y,del,Mi,idx2,t2,di,beta_h0,v_h0,alpha_h0,mord,dord){
n<-nrow(x)
p<-ncol(x)
nrand <- length(z)
q <- rep(0, nrand)
for (i in 1:nrand) q[i] <- dim(z[[i]])[2]
qcum <- cumsum(c(0, q))
beta_h<-beta_h0
v_h<-v_h0
alpha_h<-alpha_h0
zz<-z[[1]]
if (nrand>1) {
index1<-nrand
for (i in 2:index1) zz<-cbind(zz, z[[i]])
}
z<-matrix(0,n,qcum[nrand+1])
z[1:n,1:qcum[nrand+1]]<-zz[1:n,1:qcum[nrand+1]]
muh<-x%*%beta_h0 + z%*%v_h0
expeta<-exp(muh)
cla0<-di/(t(Mi)%*%expeta)
Wi<-diag(expeta[,1])
Ai<-diag(cla0[,1])
done<-matrix(1,idx2,1)
########################
oq<-matrix(1,qcum[nrand+1],1)
oq1<-matrix(1,qcum[nrand+1],1)
clam0<-Mi%*%Ai%*%done
for (i in 1:nrand) {
index1<-qcum[i]+1
oq1[index1:qcum[i+1]]<-alpha_h[i]
}
D<-diag(oq1[,1])
iD<-solve(D)
iu_h0<-exp(v_h0) ## gamma frailty
U<-iD%*%diag(iu_h0[,1]) ## gamma frailty
dft1<-t(x)%*%(del-Wi%*%clam0)
######################## pv(hp) for beta ########################
if (mord==1) {
Bi<-diag(clam0[,1])
temp4<-cla0^2/di
As<-diag(temp4[,1])
mat<-(Wi%*%Bi)-(Wi%*%Mi)%*%As%*%(t(Mi)%*%Wi)
dinv<-solve(t(z)%*%mat%*%z+U)
mu0<-exp(x%*%beta_h0 + z%*%v_h0)
mu<-exp(x%*%beta_h0 + z%*%v_h0)*clam0
dcla0be<--As%*%(t(Mi)%*%Wi)
dcla0b<-matrix(0,idx2,p)
dv_db<-matrix(0,qcum[nrand+1],p)
xz<-matrix(0,n,p)
dmu0<-matrix(0,n,p)
dw_db1<-matrix(0,n,p)
xk<-matrix(0,n,1)
for (k in 1:p) {
xk[,1]<-x[,k]
dv_db[,k] <--dinv%*%(t(z)%*%mat%*%xk)
xz[,k]<-xk+z%*%(dv_db[,k])
dcla0b[,k]<-dcla0be%*%(xz[,k])
dc<-Mi%*%diag(dcla0b[,k])%*%done
dmu0[,k]<-mu0*(xz[,k])
dw_db1[,k]<- dc*mu0 + mu*xz[,k]
temp4<-(2*cla0/di)*(dcla0b[,k])
dw_db2<-((diag(dmu0[,k]))%*%Mi%*%As%*%t(Mi)%*%Wi)+(Wi%*%Mi%*%diag(temp4[,1])%*%t(Mi)%*%Wi)+(Wi%*%Mi%*%As%*%t(Mi)%*%(diag(dmu0[,k])))
dw_db<-diag(dw_db1[,k])-dw_db2
}
}
#########################################################
dft2<-t(z)%*%(del-Wi%*%clam0)+(iD%*%oq)-(iD%*%iu_h0) ## gamma frailty
dft<-rbind(dft1,dft2)
Bi<-diag(clam0[,1])
temp4<-cla0^2/di
U<-iD%*%diag(iu_h0[,1]) ## gamma frailty
H <- rbind(cbind(t(x)%*%mat%*%x, t(x)%*%mat%*%z), cbind(t(z)%*%mat%*%x, t(z)%*%mat%*%z+U))
Hinv<-solve(H)
be_h0<- rbind(beta_h0, v_h0)
be_h <- be_h0 + (Hinv%*%dft)
beta_h[1:p,1]<-be_h[1:p,1]
se_beta_h<-matrix(0,p,1)
for (i in 1:p) se_beta_h[i,1]<-sqrt(Hinv[i,i])
index2<-qcum[nrand+1]
index3<-p+1
index4<-p+qcum[nrand+1]
v_h[1:index2,1]<-be_h[index3:index4,1]
################################################
for (i in 1:nrand) {
ial<-1/alpha_h[i]
dp<-digamma(ial)
ddp<-trigamma(ial)
oq<-matrix(1,q[i],1)
one<-matrix(1,n,1)
u_h<-exp(v_h)
eta<-x%*%beta_h + z%*%v_h
expeta<-exp(eta)
Wi<-diag(expeta[,1])
Wei<-(Wi%*%Bi)
U<-iD%*%diag(u_h[,1])
term<-(t(z)%*%mat%*%z+U)
C<-matrix(0,qcum[nrand+1],qcum[nrand+1])
index1<-qcum[i]+1
index2<-qcum[i+1]
for (j in index1:index2) C[j,j]<-1
iA<-iD
iB<-iA%*%C%*%iA
c_vh<- iB%*%(u_h-1) ## gamma frailty
invt<-solve(term)
dv<-invt%*%c_vh
dexpeta<-expeta*(z%*%dv)
dcla0<--(di/((t(Mi)%*%expeta)^2))*(t(Mi)%*%dexpeta)
temp4<-expeta*(z%*%dv)
dWi<-diag(temp4[,1])
dAi<-diag(dcla0[,1])
temp4<-Mi%*%dAi%*%done
dBi<-diag(temp4[,1])
dvec<-2*(cla0*dcla0)
temp4<-dvec/di
dAs<-diag(temp4[,1])
dmat<-(dWi%*%Bi)+(Wi%*%dBi)-(dWi%*%Mi%*%As%*%t(Mi)%*%Wi)-(Wi%*%Mi%*%dAs%*%t(Mi)%*%Wi)-(Wi%*%Mi%*%As%*%t(Mi)%*%dWi)
dia1<-diag(temp4[,1])
Hd <- rbind(cbind(t(x)%*%dmat%*%x,t(x)%*%dmat%*%z),cbind(t(z)%*%dmat%*%x,t(z)%*%dmat%*%z+ dia1))
if (dord==0) {
temp4<--iD%*%iD%*%u_h
dia1_k<-diag(temp4[,1])
zero1<-matrix(0,p,p)
zero2<-matrix(0,p,qcum[nrand+1])
Hd_k<-rbind(cbind(zero1,zero2),cbind(t(zero2),dia1_k))
hinv2<-solve(t(z)%*%mat%*%z+U)
Hd2<-t(z)%*%dmat%*%z+dia1
dk2<--0.5*sum(diag(Hinv%*%Hd_k))
vv_h<-matrix(0,q[i],1)
index3<-1
index4<-q[i]
vv_h[1:q[i],1]<-v_h[index1:index2,1]
uu_h<-exp(vv_h)
k2<-(t(oq)%*%(vv_h -uu_h))+( q[i]*(-log(alpha_h[i])+1 - dp) )
mu<-exp(x%*%beta_h)*clam0
zd<-t(z)%*%del
zmu<-t(z)%*%mu
cor1<-(1+(alpha_h[i]*zd))^(-2)
dk3<-sum(cor1)/12
k2<--((alpha_h[i]^-2)*k2)+dk2
dterm<-t(z)%*%dmat%*%z+dia1
ddcla0<--dAs%*%(t(Mi)%*%Wi%*%z)%*%dv-As%*%(t(Mi)%*%dWi%*%z)%*%dv-As%*%(t(Mi)%*%Wi%*%z)%*%ddv
ddAi<-diag(ddcla0[,1])
ddclam0<-Mi%*%ddAi%*%done
ddBi<-diag(ddclam0[,1])
temp4<-expeta*(z%*%dv)*(z%*%dv)+(expeta*(z%*%ddv))
ddWi<-diag(temp4[,1])
ddm1<-(ddWi%*%Bi)+(2*dWi%*%dBi) + (Wi%*%ddBi)
ddm2<-ddWi%*%Mi%*%As%*%t(Mi)%*%Wi+(2*dWi%*%Mi%*%dAs%*%t(Mi)%*%Wi)+(2*dWi%*%Mi%*%As%*%t(Mi)%*%dWi)
ddvec<-(2*(dcla0^2)) + (2*(cla0*ddcla0))
temp4<-ddvec/di
ddAs=diag(temp4[,1])
ddm3<-(Wi%*%Mi%*%ddAs%*%t(Mi)%*%Wi)+(2*Wi%*%Mi%*%dAs%*%t(Mi)%*%dWi)+Wi%*%Mi%*%As%*%t(Mi)%*%ddWi
ddmat<-ddm1-(ddm2+ddm3)
dia2<-diag(temp4[,1])
Hdd2<-t(z)%*%ddmat%*%z+dia2
k21<-t(oq)%*%(2*(vv_h -uu_h))+ ( q[i]*( (-2*log(alpha_h[i]))+3 - (2*dp)-((1/alpha_h[i])*ddp)) )
al<-ial^3
k21<-al*k21
cor2<-(1+(alpha_h[i]*zd))^(-3)
cor22<-cor2*zd
kcor<-sum(cor22)/6
kcor<-0
k22<--k21+0.5*(sum(diag(hinv2*Hdd2))-sum(diag(hinv2*Hd2*hinv2*Hd2)))+kcor
ialp<-1/k22
alpha_h[i]<-alpha_h[i] + (ialp*k2)
}
if (dord==1 | dord==2) {
hinv2<-solve(t(z)%*%mat%*%z+U)
Hd2<-t(z)%*%dmat%*%z+dia1
dk2<--0.5*sum(diag(Hinv%*%Hd))
vv_h<-matrix(0,q[i],1)
index3<-1
index4<-q[i]
vv_h[1:index4,1]<-v_h[index1:index2,1]
uu_h<-exp(vv_h)
k2<-(t(oq)%*%(vv_h -uu_h))+( q[i]*(-log(alpha_h[i])+1 - dp) )
mu<-exp(x%*%beta_h)*clam0
zd<-t(z)%*%del
zmu<-t(z)%*%mu
cor1<-(1+(alpha_h[i]*zd))^(-2)
dk3<-sum(cor1)/12
if(dord==1) k2<--((alpha_h[i]^(-2))*k2)+dk2
if(dord==2) k2<--((alpha_h[i]^(-2))*k2)+dk2 +dk3
dterm<-t(z)%*%dmat%*%z+dia1
ddcla0<--dAs%*%(t(Mi)%*%Wi%*%z)%*%dv-As%*%(t(Mi)%*%dWi%*%z)%*%dv-As%*%(t(Mi)%*%Wi%*%z)%*%ddv
ddAi<-diag(ddcla0[,1])
ddclam0<-Mi%*%ddAi%*%done
ddBi<-diag(ddclam0[,1])
temp4<-expeta*(z%*%dv)*(z%*%dv)+(expeta*(z%*%ddv))
ddWi<-diag(temp4[,1])
ddm1<-(ddWi%*%Bi)+(2*dWi%*%dBi) + (Wi%*%ddBi)
ddm2<-ddWi%*%Mi%*%As%*%t(Mi)%*%Wi+(2*dWi%*%Mi%*%dAs%*%t(Mi)%*%Wi)+(2*dWi%*%Mi%*%As%*%t(Mi)%*%dWi)
ddvec<-(2*(dcla0^2)) + (2*(cla0*ddcla0))
temp4<-ddvec/di
ddAs<-diag(temp4[,1])
ddm3<-(Wi%*%Mi%*%ddAs%*%t(Mi)%*%Wi)+(2*Wi%*%Mi%*%dAs%*%t(Mi)%*%dWi)+Wi%*%Mi%*%As%*%t(Mi)%*%ddWi
ddmat<-ddm1-(ddm2+ddm3)
dia2<-diag(temp4[,1])
Hdd<-rbind(cbind(t(x)%*%ddmat%*%x,t(x)%*%ddmat%*%z),cbind(t(z)%*%ddmat%*%x,t(z)%*%ddmat%*%z+dia2))
Hdd2<-t(z)%*%ddmat%*%z+dia2
k21<-t(oq)%*%(2*(vv_h -uu_h))+(q[i]*((-2*log(alpha_h[i]))+3-(2*dp)-((1/alpha_h[i])*ddp)))
al<-ial^3
k21<-al*k21
cor2<-(1+(alpha_h[i]*zd))^(-3)
cor22<-cor2*zd
kcor<-sum(cor22)/6
if(dord==1) kcor<-0
k22<--k21+0.5*(sum(diag(Hinv%*%Hdd))-sum(diag(Hinv%*%Hd%*%Hinv%*%Hd)))+kcor
ialp<-1/k22
alpha_h[i] <- alpha_h[i] + (ialp*k2)
if (alpha_h[i]<=0.0) alpha_h[i]<-alpha_h0/2
}
}
res<-list(x,z,y,del,Mi,idx2,t2,di,beta_h0,v_h0,beta_h,v_h,alpha_h0,alpha_h,dft,Hinv,clam0,H,mat,se_beta_h,U)
return(res)
}
PGFrailty_SE.h<-function(res1,nrand,q,qcum,dord=1) {
x<-res1[1][[1]]
z<-res1[2][[1]]
y<-res1[3][[1]]
del<-res1[4][[1]]
Mi<-res1[5][[1]]
idx2<-res1[6][[1]]
t2<-res1[7][[1]]
di<-res1[8][[1]]
beta_h<-res1[9][[1]]
v_h<-res1[10][[1]]
beta_h1<-res1[11][[1]]
v_h1<-res1[12][[1]]
alpha_h<-res1[13][[1]]
alpha_h1<-res1[14][[1]]
dft<-res1[15][[1]]
Hinv<-res1[16][[1]]
clam0<-res1[17][[1]]
H<-res1[18][[1]]
mat<-res1[19][[1]]
################################################
######## SE for frailty parameter ###############
################################################
n<-nrow(x)
p<-ncol(x)
u_h1<-exp(v_h1)
mat11<-t(x)%*%mat%*%x
mat12<-t(x)%*%mat%*%z
mat13<-t(z)%*%mat%*%z
oq<-matrix(1,qcum[nrand+1],1)
oq1<-matrix(1,qcum[nrand+1],1)
for (i in 1:nrand) {
index1<-qcum[i]+1
oq1[index1:qcum[i+1]]<-alpha_h1[i]
}
D<-diag(oq1[,1])
iD<-solve(D)
U <- iD%*%diag(u_h1[,1])
mmat<-mat11-mat12%*%solve(mat13+U)%*%t(mat12)
hminv<-solve(mmat)
done<-matrix(1,idx2,1)
muh<-x%*%beta_h1 + z%*%v_h1
expeta<-exp(muh)
Wi<-diag(expeta[,1])
cla0<-di/(t(Mi)%*%expeta)
Ai<-diag(cla0[,1])
temp4<-cla0^2/di
As<-diag(temp4[,1])
Bi<-diag(clam0[,1])
mat<-(Wi%*%Bi)-(Wi%*%Mi)%*%As%*%(t(Mi)%*%Wi)
Dinv0<-solve(t(z)%*%mat%*%z+U)
se_lam<-matrix(0,nrand,1)
for (i in 1:nrand){
ial<-1/alpha_h1[i]
index1<-qcum[i]+1
index2<-qcum[i+1]
vv_h1<-matrix(0,q[i],1)
uu_h1<-matrix(0,q[i],1)
vv_h1[1:q[i],1]<-v_h1[index1:index2,1]
uu_h1[1:q[i],1]<-u_h1[index1:index2,1]
c_vh <- ial^2*(uu_h1-1)
dv<-solve(t(z)%*%mat%*%z+U)%*%c_vh
dexpeta<-expeta*(z%*%dv)
dcla0<--(di/((t(Mi)%*%expeta)^2))*(t(Mi)%*%dexpeta)
dWi<-diag(dexpeta[,1])
dAi<-diag(dcla0[,1])
temp4<-Mi%*%dAi%*%done
dBi<-diag(temp4[,1])
dvec<-2*(cla0*dcla0)
temp4<-dvec/di
dAs<-diag(temp4[,1])
dmat<-(dWi%*%Bi)+(Wi%*%dBi)-(dWi%*%Mi%*%As%*%t(Mi)%*%Wi)-(Wi%*%Mi%*%dAs%*%t(Mi)%*%Wi)-(Wi%*%Mi%*%As%*%t(Mi)%*%dWi)
dia_d<-diag(temp4[,1])
Hd <- rbind(cbind(t(x)%*%dmat%*%x,t(x)%*%dmat%*%z),cbind(t(z)%*%dmat%*%x,t(z)%*%dmat%*%z+ dia_d))
term<-(t(z)%*%mat%*%z+U)
invt<-solve(term)
dterm<-t(z)%*%dmat%*%z + dia_d
ddv<- - invt%*%(dterm)%*%invt%*%c_vh+ invt%*%( -2*ial^3*(uu_h1-1)+ ial^2*uad1 )
ddcla0<- -dAs%*%(t(Mi)%*%Wi%*%z)%*%dv-As%*%(t(Mi)%*%dWi%*%z)%*%dv-As%*%(t(Mi)%*%Wi%*%z)%*%ddv
ddAi<-diag(ddcla0[,1])
ddclam0<-Mi%*%ddAi%*%done
ddBi<-diag(ddclam0[,1])
temp4<-expeta*(z%*%dv)*(z%*%dv)+(expeta*(z%*%ddv))
ddWi<-diag(temp4[,1])
ddm1<-(ddWi%*%Bi)+(2*dWi%*%dBi) + (Wi%*%ddBi)
ddm2<- ddWi%*%Mi%*%As%*%t(Mi)%*%Wi+(2*dWi%*%Mi%*%dAs%*%t(Mi)%*%Wi)+(2*dWi%*%Mi%*%As%*%t(Mi)%*%dWi)
ddvec<-(2*(dcla0^2)) + (2*(cla0*ddcla0))
temp4<-ddvec/di
ddAs<-diag(temp4[,1])
ddm3<-(Wi%*%Mi%*%ddAs%*%t(Mi)%*%Wi)+(2*Wi%*%Mi%*%dAs%*%t(Mi)%*%dWi)+Wi%*%Mi%*%As%*%t(Mi)%*%ddWi
ddmat<-ddm1-(ddm2+ddm3)
dia_dd<-diag(temp4[,1])
Hdd <-rbind(cbind(t(x)%*%ddmat%*%x,t(x)%*%ddmat%*%z),cbind(t(z)%*%ddmat%*%x,t(z)%*%ddmat%*%z+ dia_dd))
H <- rbind(cbind(t(x)%*%mat%*%x,t(x)%*%mat%*%z),cbind(t(z)%*%mat%*%x,t(z)%*%mat%*%z+U))
Hinv<-solve(H)
oq<-matrix(1,q[i],1)
dp<-digamma(ial)
ddp<-trigamma(ial)
k21a<-t(oq)%*%(2*(vv_h1 -uu_h1))+( q[i]*( (-2*log(alpha_h1[i]))+3 -(2*dp)-((1/alpha_h1[i])*ddp)) )
k21a<-(ial^3)*k21a
d2halp<--k21a- t(oq)%*%(c_vh*dv)
se_lam[i]<-sqrt(1/dalp_2)
}
u_h1<-exp(v_h1)
U <- iD%*%diag(u_h1[,1])
oq<-matrix(1,qcum[nrand+1],1)
one<-matrix(1,n,1)
zd<-t(z)%*%del
eta<-x%*%beta_h1 + z%*%v_h1
expeta<-exp(eta)
term0<-t(Mi)%*%expeta
non<-t(one)%*%del
done<-matrix(1,idx2,1)
hlike1<-(t(one)%*%(del*eta) )-( t(done)%*%(di*log(term0)) )
hlike2<-0
for (i in 1:nrand) {
oq<-matrix(1,q[i],1)
index1<-qcum[i]+1
index2<-qcum[i+1]
vv_h1<-matrix(0,q[i],1)
uu_h1<-matrix(0,q[i],1)
vv_h1[1:q[i],1]<-v_h1[index1:index2,1]
uu_h1[1:q[i],1]<-u_h1[index1:index2,1]
i_alp1<-1/alpha_h1[i]
c_alp1<--log(gamma(i_alp1))-(i_alp1*log(alpha_h1[i]))
hlike2<- hlike2+t(oq)%*%( (vv_h1-uu_h1)/alpha_h1[i] + c_alp1)
}
hlike<-hlike1+hlike2
pi<-3.14159265359
H22<-t(z)%*%mat%*%z+U
zd<-t(z)%*%del
if (dord==2) {
temp4<-1/(i_alp1+zd)
secd<-diag(temp4[,1])
second<-sum(diag(secd))/12
} else second<-0
pvhs<-hlike-0.5*log(det(H22/(2*pi)))
svhs<-pvhs+second
hpn2<-pvhs
hpn3<-hpn1+second
res<-list(se_lam,hlike,hpn1,hpn2,hpn3)
return(res)
}
PNFrailty_SE.h<-function(res1,nrand,q,qcum,dord=1) {
x<-res1[1][[1]]
z<-res1[2][[1]]
y<-res1[3][[1]]
del<-res1[4][[1]]
Mi<-res1[5][[1]]
idx2<-res1[6][[1]]
t2<-res1[7][[1]]
di<-res1[8][[1]]
beta_h<-res1[9][[1]]
v_h<-res1[10][[1]]
beta_h1<-res1[11][[1]]
v_h1<-res1[12][[1]]
alpha_h<-res1[13][[1]]
alpha_h1<-res1[14][[1]]
dft<-res1[15][[1]]
Hinv<-res1[16][[1]]
clam0<-res1[17][[1]]
H<-res1[18][[1]]
mat<-res1[19][[1]]
U<-res1[21][[1]]
################################################
######## SE for frailty parameter ###############
################################################
n<-nrow(x)
p<-ncol(x)
u_h1<-exp(v_h1)
oq<-matrix(1,qcum[nrand+1],1)
oq1<-matrix(1,qcum[nrand+1],1)
for (i in 1:nrand) {
index1<-qcum[i]+1
oq1[index1:qcum[i+1]]<-alpha_h1[i]
}
D<-diag(oq1[,1])
iD<-solve(D)
iA<-iD
Bi<-diag(clam0[,1])
muh<-x%*%beta_h1 + z%*%v_h1
expeta<-exp(muh)
cla0<-di/(t(Mi)%*%expeta)
temp4<-cla0^2/di
As<-diag(temp4[,1])
Wi<-diag(expeta[,1])
done<-matrix(1,idx2,1)
H22<-solve(t(z)%*%mat%*%z+U)
Hessian<-matrix(0,nrand,nrand)
for (i in 1:nrand) {
C<-matrix(0,qcum[nrand+1],qcum[nrand+1])
index1<-qcum[i]+1
index2<-qcum[i+1]
for (j in index1:index2) C[j,j]<-1
iB1<-iA%*%C%*%iA
c_vh1<-iB1%*%v_h1
dv1<-H22%*%c_vh1
dexpeta1<-expeta*(z%*%dv1)
dcla01<--(di/((t(Mi)%*%expeta)^2))*(t(Mi)%*%dexpeta1)
dWi1<-diag(dexpeta1[,1])
dAi1<-diag(dcla01[,1])
temp4<-Mi%*%dAi1%*%done
dBi1<-diag(temp4[,1])
dvec1<-2*(cla0*dcla01)
temp4<-dvec1/di
dAs1<-diag(temp4[,1])
dmat1<-(dWi1%*%Bi)+(Wi%*%dBi1)-(dWi1%*%Mi%*%As%*%t(Mi)%*%Wi)-(Wi%*%Mi%*%dAs1%*%t(Mi)%*%Wi)-(Wi%*%Mi%*%As%*%t(Mi)%*%dWi1)
dia1<--iB1
Hd1 <- rbind(cbind(t(x)%*%dmat1%*%x,t(x)%*%dmat1%*%z),cbind(t(z)%*%dmat1%*%x,t(z)%*%dmat1%*%z+ dia1))
ddk1<- -0.5*sum(diag(iA%*%C%*%iA%*%C)) +t(v_h1)%*%(iA%*%C%*%iB1)%*%v_h1 -t(dv1)%*%iB1%*%v_h1
dia11<-(iB1%*%C%*%iA+iA%*%C%*%iB1)
dv11<--H22%*%((t(z)%*%dmat1%*%z+dia1)%*%dv1-iB1%*%dv1 + dia11%*%v_h1)
temp4<-(z%*%dv1)*(z%*%dv1)*expeta +(z%*%dv11)*expeta
ddW11<-diag(temp4[,1])
ddcla011<- -( dAs1%*%(t(Mi)%*%Wi%*%z)%*%dv1 +As%*%(t(Mi)%*%dWi1%*%z)%*%dv1 +As%*%(t(Mi)%*%Wi%*%z)%*%dv11)
temp4<-Mi%*%diag(ddcla011[,1])%*%done
ddB11<-diag(temp4[,1])
temp4<-(2*(dcla01^2) + 2*(cla0*ddcla011) ) /di
ddAs11<-diag(temp4[,1])
ddm1_11<-(ddW11%*%Bi)+ (2*dWi1%*%dBi1) + (Wi%*%ddB11)
ddm2_11<- (ddW11%*%Mi%*%As%*%t(Mi)%*%Wi) +(2*dWi1%*%Mi%*%dAs1%*%t(Mi)%*%Wi) +(2*dWi1%*%Mi%*%As%*%t(Mi)%*%dWi1)
ddm3_11<-(Wi%*%Mi%*%ddAs11%*%t(Mi)%*%Wi) +(2*Wi%*%Mi%*%dAs1%*%t(Mi)%*%dWi1) +(Wi%*%Mi%*%As%*%t(Mi)%*%ddW11)
ddmat11<-ddm1_11-(ddm2_11+ddm3_11)
Hd11 <-rbind(cbind(t(x)%*%ddmat11%*%x,t(x)%*%ddmat11%*%z),cbind(t(z)%*%ddmat11%*%x,t(z)%*%ddmat11%*%z+ dia11 ))
ddk1<-ddk1 +0.5*sum(diag(-Hinv%*%Hd1%*%Hinv%*%Hd1+ Hinv%*%Hd11))
Hessian[i,i]<-ddk1
for (kk in 1:nrand) {
if (kk>i) {
D<-matrix(0,qcum[nrand+1],qcum[nrand+1])
index1<-qcum[kk]+1
index2<-qcum[kk+1]
for (j in index1:index2) D[j,j]<-1
iB2<-iA%*%D%*%iA
c_vh2<-iB2%*%v_h1
dv2<-H22%*%c_vh2
dexpeta2<-expeta*(z%*%dv2)
dcla02<--(di/((t(Mi)%*%expeta)^2))*(t(Mi)%*%dexpeta2)
dWi2<-diag(dexpeta2[,1])
dAi2<-diag(dcla02[,1])
temp4<-Mi%*%dAi2%*%done
dBi2<-diag(temp4[,1])
dvec2<-2*(cla0*dcla02)
temp4<-dvec2/di
dAs2<-diag(temp4[,1])
dd12<--0.5*sum(diag(iA%*%D%*%iA%*%C))+0.5*t(v_h1)%*%(iA%*%D%*%iB1+iA%*%C%*%iB2)%*%v_h1-t(dv1)%*%iB2%*%v_h1
dia12<-(iB1%*%D%*%iA+iA%*%D%*%iB1)
dv12<--H22%*%((t(z)%*%dmat1%*%z+dia1)%*%dv2-iB2%*%dv1 + dia12%*%v_h1)
temp4<-(z%*%dv1)*(z%*%dv2)*expeta + (z%*%dv12)*expeta
ddW12<-diag(temp4[,1])
ddcla012<--( dAs2%*%(t(Mi)%*%Wi%*%z)%*%dv1 +As%*%(t(Mi)%*%dWi2%*%z)%*%dv1 +As%*%(t(Mi)%*%Wi%*%z)%*%dv12)
temp4<-Mi%*%diag(ddcla012[,1])%*%done
ddB12<-diag(temp4[,1])
temp4<-(2*(dcla02*dcla01) + 2*(cla0*ddcla012) )/di
ddAs12<-diag(temp4[,1])
ddm1_12<-(ddW12%*%Bi)+ (dWi1%*%dBi2 + dWi2%*%dBi1) + (Wi%*%ddB12)
ddm2_12<-(ddW12%*%Mi%*%As%*%t(Mi)%*%Wi) +(dWi1%*%Mi%*%dAs2%*%t(Mi)%*%Wi) +(dWi1%*%Mi%*%As%*%t(Mi)%*%dWi2)
ddm3_12<-(dWi2%*%Mi%*%dAs1%*%t(Mi)%*%Wi) +(Wi%*%Mi%*%ddAs12%*%t(Mi)%*%Wi) +(Wi%*%Mi%*%dAs1%*%t(Mi)%*%dWi2)
ddm4_12<-(dWi2%*%Mi%*%As%*%t(Mi)%*%dWi1) +(Wi%*%Mi%*%dAs2%*%t(Mi)%*%dWi1) +(Wi%*%Mi%*%As%*%t(Mi)%*%ddW12)
ddmat12<-ddm1_12-(ddm2_12+ddm3_12+ddm4_12)
dmat2<-(dWi2%*%Bi)+(Wi%*%dBi2)-(dWi2%*%Mi%*%As%*%t(Mi)%*%Wi)-(Wi%*%Mi%*%dAs2%*%t(Mi)%*%Wi)-(Wi%*%Mi%*%As%*%t(Mi)%*%dWi2)
dia2<--iB2
Hd2 <- rbind(cbind(t(x)%*%dmat2%*%x,t(x)%*%dmat2%*%z),cbind(t(z)%*%dmat2%*%x,t(z)%*%dmat2%*%z+ dia2))
Hd12<-rbind(cbind(t(x)%*%ddmat12%*%x,t(x)%*%ddmat12%*%z),cbind(t(z)%*%ddmat12%*%x,t(z)%*%ddmat12%*%z+ dia12))
dd12<-dd12 +0.5*sum(diag(-Hinv%*%Hd1%*%Hinv%*%Hd2+ Hinv%*%Hd12))
Hessian[i,kk]<-Hessian[kk,i]<-dd12
}
}
}
iAp<-solve(Hessian)
se_lam<-sqrt(diag(iAp))
eta<-x%*%beta_h1 + z%*%v_h1
expeta<-exp(eta)
one<-matrix(1,n,1)
done<-matrix(1,idx2,1)
oq<-matrix(1,qcum[nrand+1],1)
pi<-3.14159265359
term0<-t(Mi)%*%expeta
hlike1<-(t(one)%*%(del*eta) )-( t(done)%*%(di*log(term0)))
hlike2<-0
hlike3<-0
for (i in 1:nrand) {
hlike2<-hlike2-(q[i]/2)*log(2*pi)-( (q[i]/2)*log(alpha_h1[i]))
index1<-qcum[i]+1
index2<-qcum[i+1]
vv_h1<-matrix(0,q[i],1)
vv_h1[1:q[i],1]<-v_h1[index1:index2,1]
hlike3<-hlike3-(t(vv_h1)%*%vv_h1)/(2*alpha_h1[i])
}
hliken<-hlike1+hlike2+hlike3
muu<-exp(x%*%beta_h1)*clam0
zmu<-t(z)%*%muu
u_h1<-exp(v_h1)
second<-0
for (i in 1:nrand) {
ialph1<-1/alpha_h1[i]
a21<-(zmu*u_h1)+ialph1
b31<-zmu*u_h1
S11<-3*(b31/(a21^2))
S21<-5*((b31^2)/(a21^3))
temp4<-S11-S21
S31<-diag(temp4[,1])
second<-second-sum(diag(S31))/24
}
H22<-t(z)%*%mat%*%z+iD
hpn2<-hliken-0.5*log(det(H22/(2*pi)))
hpn3<-hpn1+second
res<-list(se_lam,hliken,hpn1,hpn2,hpn3)
return(res)
}
subbars<-function (term)
{
if (is.name(term) || !is.language(term))
return(term)
if (length(term) == 2) {
term[[2]] <- subbars(term[[2]])
return(term)
}
stopifnot(length(term) >= 3)
if (is.call(term) && term[[1]] == as.name("|"))
term[[1]] <- as.name("+")
for (j in 2:length(term)) term[[j]] <- subbars(term[[j]])
term
}
nobars<-function (term)
{
if (!("|" %in% all.names(term)))
return(term)
if (is.call(term) && term[[1]] == as.name("|"))
return(NULL)
if (length(term) == 2) {
nb <- nobars(term[[2]])
if (is.null(nb))
return(NULL)
term[[2]] <- nb
return(term)
}
nb2 <- nobars(term[[2]])
nb3 <- nobars(term[[3]])
if (is.null(nb2))
return(nb3)
if (is.null(nb3))
return(nb2)
term[[2]] <- nb2
term[[3]] <- nb3
term
}
expandSlash<-function (bb)
{
if (!is.list(bb))
return(expandSlash(list(bb)))
unlist(lapply(bb, function(x) {
if (length(x) > 2 && is.list(trms <- slashTerms(x[[3]])))
return(lapply(unlist(makeInteraction(trms)), function(trm) substitute(foo |
bar, list(foo = x[[2]], bar = trm))))
x
}))
}
findbars<-function(term)
{
if (is.name(term) || !is.language(term))
return(NULL)
if (term[[1]] == as.name("("))
return(findbars(term[[2]]))
if (!is.call(term))
stop("term must be of class call")
if (term[[1]] == as.name("|"))
return(term)
if (length(term) == 2)
return(findbars(term[[2]]))
c(findbars(term[[2]]), findbars(term[[3]]))
}
##### Check if there were two terms entered in a correlated fashion #####
##### Current software does not allow for correlated random effects #####
findplus<-function(term)
{
if (is.numeric(term))
return(0)
if (!is.language(term))
return(NULL)
if (length(term)==1) return(0)
if (term[[1]] == as.name("|"))
return(findplus(term[[2]]))
if (!is.call(term))
stop("term must be of class call")
if (term[[1]] == as.name("+"))
return(1)
if (term[[1]] == as.name("-"))
return(-1)
}
slashTerms<-function (x)
{
if (!("/" %in% all.names(x)))
return(x)
if (x[[1]] != as.name("/"))
stop("unparseable formula for grouping factor")
list(slashTerms(x[[2]]), slashTerms(x[[3]]))
}
makeInteraction <- function(x)
### from a list of length 2 return recursive interaction terms
{
if (length(x) < 2) return(x)
trm1 <- makeInteraction(x[[1]])
trm11 <- if(is.list(trm1)) trm1[[1]] else trm1
list(substitute(foo:bar, list(foo=x[[2]], bar = trm11)), trm1)
}
```
## Try the h.likelihood package in your browser
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h.likelihood documentation built on May 2, 2019, 4:36 p.m. | 15,483 | 32,495 | {"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-2020-05 | latest | en | 0.215778 |
http://www.appszoom.com/android_applications/education/math-cheat-sheets-free_btgff.html | 1,498,492,201,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320823.40/warc/CC-MAIN-20170626152050-20170626172050-00553.warc.gz | 450,769,880 | 17,833 | Math Cheat Sheets FREE
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# Math Cheat Sheets FREE
by: 963 8
8 Users
rating
## Screenshots
Description
The best Collection of Math cheat sheets and quick reference cards
Many Cheat Sheets in your mobile to have the formulas wherever and whenever you want.
With this application you can use your travel time to study, or just have it as a quick reference when needed.
NOTE: App can be moved to the SD Card!!
Contents:
=========
Algebra. Elementary techniques for factoring binomials and trinomials
Algebra. Exponent laws and factoring tips
Algebra. Solving quadratic equations by completing the square
Algebra. College Algebra quick reference
Algebra. Solution of the 3rd degree polynomial equation
Algebra. Solution of the 4th degreee polynomial equation
Trig. Basic trig identities
Trig. Law of sines cosines etc and other triangle formulas
Trig. Graphs of the trig functions
Trig. Inverse trig functions
Trig. Power reducing formulas for powers of sines and cosines
Trig. Graph paper for plotting in polar coordinates
Trig. Two unit circles with trig funcion values
Trig. Single unit circle with trig function values
Calculus. Basic differentiation formulas and some useful trig identities
Calculus. Basic differentiation and integration formulas
Calculus. Definitions and theorems pertaining to Riemann sums and definite integrals
Calculus. A quick reference sheet on Taylor polynomials and series
Calculus. A summary of convergence tests
Calculus. Guidelines for evaluating integrals involving powers of sines and cosines
Calculus. Guidelines for evaluating integrals involving powers of secants and tangents
Calculus. Standard forms for conic sections
Calculus. Common infinite series
Calculus. Trigonometric substitution
Calculus. Cylindrical coordinates
Calculus. Spherical coordinates
Calculus. Hyperbolic functions
Calculus. Applications of integrals
Calculus. Applications of integrals
Calculus. Common ordinary differential equations
Calculus. Common ordinary differential equations
Calculus. Common ordinary differential equations
Calculus. Undetermined coefficients and variation of parameters
Calculus. Vector formulas
Calculus. Simple summary of cylindrical and spherical coordinates
Misc. Some prime and composite numbers
Misc. Sets. Functions lines and sequences
Statistics formula sheet Page 1
Statistics Page 2
Statistics Page 3
SD Installation support
If you wish you can show your appreciation and support future development by donating here! http://goo.gl/jWm6dI
Thanks! You can also donate via Bitcoin. My address is: 1GFAcD6piVbv4fN6Xw6tNe5yrgGmdG2kcz
Tags: free math cheat sheets , trig cheat sheet reduced , ultimate geometry cheat sheet , 数学 グラフ用紙 , geometry formulas cheat sheet , free math sheets formula , mathematics formula sheets , guitarra free sheets , cheat sheet for sequence and series formulas
## Users review
from 963 reviews
"Great"
8 | 657 | 2,971 | {"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-2017-26 | latest | en | 0.676966 |
http://www.complang.tuwien.ac.at/cvsweb/cgi-bin/cvsweb/gforth/prim.diff?r1=1.148;r2=1.151;sortby=log;f=h;only_with_tag=MAIN;f=u | 1,590,432,123,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-24/segments/1590347389309.17/warc/CC-MAIN-20200525161346-20200525191346-00533.warc.gz | 160,038,033 | 1,855 | --- gforth/prim 2003/11/06 09:47:49 1.148 +++ gforth/prim 2004/03/29 11:23:01 1.151 @@ -249,7 +249,7 @@ execute ( xt -- ) core #ifndef NO_IP ip=IP; #endif -IF_spTOS(spTOS = sp[0]); +IF_spTOS(spTOS = sp[0]); /* inst_tail would produce a NEXT_P1 */ SUPER_END; EXEC(xt); @@ -259,7 +259,7 @@ perform ( a_addr -- ) gforth #ifndef NO_IP ip=IP; #endif -IF_spTOS(spTOS = sp[0]); +IF_spTOS(spTOS = sp[0]); /* inst_tail would produce a NEXT_P1 */ SUPER_END; EXEC(*(Xt *)a_addr); : @@ -433,8 +433,9 @@ condbranch((+loop),n R:nlimit R:n1 -- R: /* dependent upon two's complement arithmetic */ Cell olddiff = n1-nlimit; n2=n1+n; -,if ((olddiff^(olddiff+n))>=0 /* the limit is not crossed */ - || (olddiff^n)>=0 /* it is a wrap-around effect */) { +,if (((olddiff^(olddiff+n)) /* the limit is not crossed */ + &(olddiff^n)) /* OR it is a wrap-around effect */ + >=0) { /* & is used to avoid having two branches for gforth-native */ ,: r> swap r> r> 2dup - >r @@ -463,7 +464,7 @@ if (n<0) { newdiff = -newdiff; } n2=n1+n; -,if (diff>=0 || newdiff<0) { +,if (((~diff)|newdiff)<0) { /* use | to avoid two branches for gforth-native */ ,) \+ @@ -2353,7 +2354,11 @@ av-double ( r -- ) gforth av_double av_double(alist, r); av-longlong ( d -- ) gforth av_longlong +#ifdef BUGGY_LONG_LONG +av_longlong(alist, d.lo); +#else av_longlong(alist, d); +#endif av-ptr ( c_addr -- ) gforth av_ptr av_ptr(alist, void*, c_addr); @@ -2372,7 +2377,11 @@ lp += sizeof(Float); av_double(alist, r); av-longlong-r ( R:d -- ) gforth av_longlong_r +#ifdef BUGGY_LONG_LONG +av_longlong(alist, d.lo); +#else av_longlong(alist, d); +#endif av-ptr-r ( R:c_addr -- ) gforth av_ptr_r av_ptr(alist, void*, c_addr); @@ -2404,7 +2413,12 @@ av-call-longlong ( -- d ) gforth av_cal SAVE_REGS av_call(alist); REST_REGS +#ifdef BUGGY_LONG_LONG d = llrv; +#else +d.lo = llrv; +d.hi = 0; +#endif av-call-ptr ( -- c_addr ) gforth av_call_ptr SAVE_REGS @@ -2437,7 +2451,12 @@ va-arg-int ( -- w ) gforth va_arg_int w = va_arg_int(clist); va-arg-longlong ( -- d ) gforth va_arg_longlong +#ifdef BUGGY_LONG_LONG +d.lo = va_arg_longlong(clist); +d.hi = 0; +#else d = va_arg_longlong(clist); +#endif va-arg-ptr ( -- c_addr ) gforth va_arg_ptr c_addr = (char *)va_arg_ptr(clist,char*); @@ -2461,7 +2480,11 @@ va_return_ptr(clist, void *, c_addr); return 0; va-return-longlong ( d -- ) gforth va_return_longlong +#ifdef BUGGY_LONG_LONG +va_return_longlong(clist, d.lo); +#else va_return_longlong(clist, d); +#endif return 0; va-return-float ( r -- ) gforth va_return_float @@ -2526,7 +2549,10 @@ compile_prim1(a_prim); finish-code ( -- ) gforth finish_code ""Perform delayed steps in code generation (branch resolution, I-cache flushing)."" +IF_spTOS(sp[0]=spTOS); /* workaround for failing to save spTOS + (gcc-2.95.1, gforth-fast --enable-force-reg) */ finish_code(); +IF_spTOS(spTOS=sp[0]); forget-dyncode ( c_code -- f ) gforth-internal forget_dyncode f = forget_dyncode(c_code); | 1,013 | 2,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-24 | latest | en | 0.272523 |
https://www.indiabix.com/aptitude/clock/063004 | 1,718,364,208,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861546.27/warc/CC-MAIN-20240614110447-20240614140447-00558.warc.gz | 756,495,853 | 8,197 | # Aptitude - Clock
Exercise : Clock - General Questions
16.
At what time, in minutes, between 3 o'clock and 4 o'clock, both the needles will coincide each other?
5 1 " 11
12 4 " 11
13 4 " 11
16 4 " 11
Explanation:
At 3 o'clock, the minute hand is 15 min. spaces apart from the hour hand.
To be coincident, it must gain 15 min. spaces.
55 min. are gained in 60 min.
15 min. are gained in 60 x 15 min = 16 4 min. 55 11
The hands are coincident at 16 4 min. past 3. 11
17.
How many times do the hands of a clock coincide in a day?
20
21
22
24
Explanation:
The hands of a clock coincide 11 times in every 12 hours (Since between 11 and 1, they coincide only once, i.e., at 12 o'clock).
AM
12:00
1:05
2:11
3:16
4:22
5:27
6:33
7:38
8:44
9:49
10:55
PM
12:00
1:05
2:11
3:16
4:22
5:27
6:33
7:38
8:44
9:49
10:55
The hands overlap about every 65 minutes, not every 60 minutes.
The hands coincide 22 times in a day.
18.
How many times in a day, the hands of a clock are straight?
22
24
44
48
Explanation:
In 12 hours, the hands coincide or are in opposite direction 22 times.
In 24 hours, the hands coincide or are in opposite direction 44 times a day.
19.
A watch which gains uniformly is 2 minutes low at noon on Monday and is 4 min. 48 sec fast at 2 p.m. on the following Monday. When was it correct?
2 p.m. on Tuesday
2 p.m. on Wednesday
3 p.m. on Thursday
1 p.m. on Friday
Explanation:
Time from 12 p.m. on Monday to 2 p.m. on the following Monday = 7 days 2 hours = 170 hours.
The watch gains 2 + 4 4 min. or 34 min. in 170 hrs. 5 5
Now, 34 min. are gained in 170 hrs. 5
2 min. are gained in 170 x 5 x 2 hrs = 50 hrs. 34
Watch is correct 2 days 2 hrs. after 12 p.m. on Monday i.e., it will be correct at 2 p.m. on Wednesday. | 612 | 1,750 | {"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} | 4.1875 | 4 | CC-MAIN-2024-26 | latest | en | 0.909073 |
https://gamedev.stackexchange.com/questions/117147/lock-z-axis-from-the-gyroscope-rotation | 1,642,622,103,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301488.71/warc/CC-MAIN-20220119185232-20220119215232-00010.warc.gz | 317,956,248 | 30,793 | # Lock Z axis from the gyroscope rotation
Here's the code I use.
public class CameraGyro : MonoBehaviour {
private float initialX;
private float initialY;
private float initialZ;
// Use this for initialization
void Start () {
Input.gyro.enabled = true;
initialX = Input.gyro.rotationRateUnbiased.x;
initialY = Input.gyro.rotationRateUnbiased.y;
}
// Update is called once per frame
void Update () {
transform.Rotate (initialX - Input.gyro.rotationRateUnbiased.x, initialY - Input.gyro.rotationRateUnbiased.y, 0f);
}
}
The thing is when I rotate the phone, the Z axis seems to change even if I've set it to 0f in the code above. Is there a way to really lock this position to 0?
• What do you mean "seems to change"? Feb 23 '16 at 21:07
• Well the Z axis change value when it's supposed to stay at 0 Feb 23 '16 at 21:15
• are you measuring that changed based on how the transform rotates or externally? Feb 23 '16 at 21:40
• you are talking about the Z translate value or Z rotation value? Which is changing and supposed not to change? Feb 24 '16 at 6:09
• Simple explanation... I need to make the gyroscope work by only rotating the X and Y axis. Z should be locked. Feb 24 '16 at 14:33
I'm rotating an object on two axes, so why does it keep twisting around the third axis?
The root issue here is that the idea that rotation is naturally/objectively separable into 3 independent components (the way we can do with translation) is a misconception.
Euler angles / Tait-Bryan angles lure us into this belief because they look a lot like a translation vector, but it's easy to show that it's false. Consider the angle triplets (180, 180, 0) and (0, 0, 180): both represent the same orientation, one purely with "x & y" rotations, the other with "purely z" rotation. Or here's another example with compounding rotations in sequence:
So, zeroing-out the z component of an angular triplet isn't enough to guarantee you don't get roll.
What to do instead depends on the exact behaviour you want. You could try for instance track a pitch & yaw value independently, and apply the total all at once rather than incrementally compounding rotations:
// in Start(), initialize your x & y angles to match the camera's initial orientation.
pitch = transform.eulerAngles.x;
yaw = transform.eulerAngles.y;
// In Update, accumulate rotational change in these axes:
pitch += input.gyro.rotationRateUnbiased.x * Mathf.RadToDeg() * Time.deltaTime;
yaw += input.gyro.rotationRateUnbiased.y * Mathf.RadToDeg() * Time.deltaTime;
// Apply the result all at once:
transform.eulerAngles = new Vector3(pitch, yaw, 0);
Or you could try rotating the forward / look vector, then construct a "roll-free" orientation that looks in that direction:
Vector3 angularDelta = input.gyro.rotationRateUnbiased * Time.deltaTime;
Vector3 newForward = transform.forward + Vector3.Cross(transform.forward, angularDelta);
transform.rotation = Quaternion.LookRotation(newForward);
Try locking the z axis using "Rotation Constraints": https://docs.unity3d.com/Manual/class-RotationConstraint.html
• This answer would be better if it explained how the behaviour of this component differs from manually setting the angle increment to zero. Dec 16 '18 at 22:03 | 762 | 3,233 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2022-05 | latest | en | 0.884579 |
https://www.jiskha.com/display.cgi?id=1202070658 | 1,516,675,221,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891705.93/warc/CC-MAIN-20180123012644-20180123032644-00426.warc.gz | 904,633,680 | 3,770 | # Chemistry
posted by .
can the energy originally stored in gasoline be traced to rotating wheels?
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10. ### physics
The thermal energy stored in 1 kg of gasoline is about 10,000 kcal (42,000,000 J). A small car needs a forward force of about 500 N to move slowly at a constant speed on a leveled road (no change in ME). How far should the car go if …
More Similar Questions | 540 | 2,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-05 | latest | en | 0.908083 |
https://www.kodytools.com/units/substance/from/atom/to/femtomole | 1,726,277,655,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.77/warc/CC-MAIN-20240913233654-20240914023654-00813.warc.gz | 773,684,589 | 14,055 | # Atom to Femtomole Converter
Convert → Femtomole to Atom
1 Atom = 1.6605390671738e-9 Femtomoles
## One Atom is Equal to How Many Femtomoles?
The answer is one Atom is equal to 1.6605390671738e-9 Femtomoles and that means we can also write it as 1 Atom = 1.6605390671738e-9 Femtomoles. Feel free to use our online unit conversion calculator to convert the unit from Atom to Femtomole. Just simply enter value 1 in Atom and see the result in Femtomole.
Manually converting Atom to Femtomole can be time-consuming,especially when you don’t have enough knowledge about Amount of Substance units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Atom to Femtomole converter tool to get the job done as soon as possible.
We have so many online tools available to convert Atom to Femtomole, but not every online tool gives an accurate result and that is why we have created this online Atom to Femtomole converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Atom to Femtomole (atom to fmol)
By using our Atom to Femtomole conversion tool, you know that one Atom is equivalent to 1.6605390671738e-9 Femtomole. Hence, to convert Atom to Femtomole, we just need to multiply the number by 1.6605390671738e-9. We are going to use very simple Atom to Femtomole conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Atom} = 1 \times 1.6605390671738e-9 = \text{1.6605390671738e-9 Femtomoles}$$
## What Unit of Measure is Atom?
Atom is a unit of measurement for amount of substance. It is the smallest constituent unit of ordinary matter that constitutes a chemical element. One atom is equal to 1.66-24 moles.
## What is the Symbol of Atom?
The symbol of Atom is atom. This means you can also write one Atom as 1 atom.
## What Unit of Measure is Femtomole?
Femtomole is a unit of measurement for amount of substance. Femtomole is a decimal fraction of amount of substance unit mole. One femtomole is equal to 1e-15 moles.
## What is the Symbol of Femtomole?
The symbol of Femtomole is fmol. This means you can also write one Femtomole as 1 fmol.
## How to Use Atom to Femtomole Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Atom and in the first input field, enter a value.
• From the second dropdown, select Femtomole.
• Instantly, the tool will convert the value from Atom to Femtomole and display the result in the second input field.
## Example of Atom to Femtomole Converter Tool
Atom
1
Femtomole
1.6605390671738e-9
# Atom to Femtomole Conversion Table
Atom [atom]Femtomole [fmol]Description
1 Atom1.6605390671738e-9 Femtomole1 Atom = 1.6605390671738e-9 Femtomole
2 Atom3.3210781343477e-9 Femtomole2 Atom = 3.3210781343477e-9 Femtomole
3 Atom4.9816172015215e-9 Femtomole3 Atom = 4.9816172015215e-9 Femtomole
4 Atom6.6421562686954e-9 Femtomole4 Atom = 6.6421562686954e-9 Femtomole
5 Atom8.3026953358692e-9 Femtomole5 Atom = 8.3026953358692e-9 Femtomole
6 Atom9.9632344030431e-9 Femtomole6 Atom = 9.9632344030431e-9 Femtomole
7 Atom1.1623773470217e-8 Femtomole7 Atom = 1.1623773470217e-8 Femtomole
8 Atom1.3284312537391e-8 Femtomole8 Atom = 1.3284312537391e-8 Femtomole
9 Atom1.4944851604565e-8 Femtomole9 Atom = 1.4944851604565e-8 Femtomole
10 Atom1.6605390671738e-8 Femtomole10 Atom = 1.6605390671738e-8 Femtomole
100 Atom1.6605390671738e-7 Femtomole100 Atom = 1.6605390671738e-7 Femtomole
1000 Atom0.0000016605390671738 Femtomole1000 Atom = 0.0000016605390671738 Femtomole
# Atom to Other Units Conversion Table
ConversionDescription
1 Atom = 1.6605390671738e-24 Mole1 Atom in Mole is equal to 1.6605390671738e-24
1 Atom = 1.6605390671738e-25 Dekamole1 Atom in Dekamole is equal to 1.6605390671738e-25
1 Atom = 1.6605390671738e-26 Hectomole1 Atom in Hectomole is equal to 1.6605390671738e-26
1 Atom = 1.6605390671738e-27 Kilomole1 Atom in Kilomole is equal to 1.6605390671738e-27
1 Atom = 1.6605390671738e-30 Megamole1 Atom in Megamole is equal to 1.6605390671738e-30
1 Atom = 1.6605390671738e-33 Gigamole1 Atom in Gigamole is equal to 1.6605390671738e-33
1 Atom = 1.6605390671738e-36 Teramole1 Atom in Teramole is equal to 1.6605390671738e-36
1 Atom = 1.6605390671738e-39 Petamole1 Atom in Petamole is equal to 1.6605390671738e-39
1 Atom = 1.6605390671738e-42 Examole1 Atom in Examole is equal to 1.6605390671738e-42
1 Atom = 1.6605390671738e-23 Decimole1 Atom in Decimole is equal to 1.6605390671738e-23
1 Atom = 1.6605390671738e-22 Centimole1 Atom in Centimole is equal to 1.6605390671738e-22
1 Atom = 1.6605390671738e-21 Millimole1 Atom in Millimole is equal to 1.6605390671738e-21
1 Atom = 1.6605390671738e-18 Micromole1 Atom in Micromole is equal to 1.6605390671738e-18
1 Atom = 1.6605390671738e-15 Nanomole1 Atom in Nanomole is equal to 1.6605390671738e-15
1 Atom = 1.6605390671738e-12 Picomole1 Atom in Picomole is equal to 1.6605390671738e-12
1 Atom = 1.6605390671738e-9 Femtomole1 Atom in Femtomole is equal to 1.6605390671738e-9
1 Atom = 0.0000016605390671738 Attomole1 Atom in Attomole is equal to 0.0000016605390671738 | 1,767 | 5,202 | {"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} | 2.734375 | 3 | CC-MAIN-2024-38 | latest | en | 0.88384 |
https://socratic.org/questions/what-are-the-asymptote-s-and-hole-s-if-any-of-f-x-x-x-1-x-1-x#438354 | 1,670,602,998,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711417.46/warc/CC-MAIN-20221209144722-20221209174722-00545.warc.gz | 573,551,130 | 6,156 | # What are the asymptote(s) and hole(s), if any, of f(x) = x/(x-1)-(x-1)/x?
Jun 12, 2017
$x = 0$ is an asymptote.
$x = 1$ is an asymptote.
#### Explanation:
First, let's simplify this so that we have a single fraction that we can take the limit of.
$f \left(x\right) = \frac{x \left(x\right)}{\left(x - 1\right) \left(x\right)} - \frac{\left(x - 1\right) \left(x - 1\right)}{x \left(x - 1\right)}$
$f \left(x\right) = \frac{{x}^{2} - {\left(x - 1\right)}^{2}}{\left(x - 1\right) \left(x\right)} = \frac{{x}^{2} - \left({x}^{2} - 2 x + 1\right)}{\left(x - 1\right) \left(x\right)}$
$f \left(x\right) = \frac{2 x - 1}{\left(x - 1\right) \left(x\right)}$
Now, we need to check for discontinuities. This is just anything that will make the denominator of this fraction $0$. In this case, to make the denominator $0$, $x$ could be $0$ or $1$. So let's take the limit of $f \left(x\right)$ at those two values.
${\lim}_{x \to 0} \frac{2 x - 1}{x \left(x - 1\right)} = \frac{- 1}{- 1 \cdot 0} = \pm \infty$
${\lim}_{x \to 1} \frac{2 x - 1}{x \left(x - 1\right)} = \frac{3}{1 \cdot 0} = \pm \infty$
Since both of these limits tend towards infinity, both $x = 0$ and $x = 1$ are asymptotes of the function. There are therefore no holes in the function. | 479 | 1,256 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2022-49 | latest | en | 0.752683 |
https://homex.com/ask/how-many-gpm-can-flow-through-a-2-pipe | 1,632,591,616,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057733.53/warc/CC-MAIN-20210925172649-20210925202649-00094.warc.gz | 347,302,491 | 34,304 | How many gpm can flow through a 2 pipe?
If you add 1.63 gallons per 10 feet of pipe and six, you get a final GPM of 9.78 gallons per square inch. | 46 | 146 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-39 | latest | en | 0.835132 |
https://www.expertsmind.com/library/which-of-the-following-will-yield-the-most-energy-for-body-5147030.aspx | 1,723,344,219,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640843545.63/warc/CC-MAIN-20240811013030-20240811043030-00209.warc.gz | 594,030,001 | 14,209 | ### Which of the following will yield the most energy for body
Assignment Help Biology
##### Reference no: EM13147030
Which of the following will yield the most energy for the body?Assume there are equivalent numbers of fatty acids in each case and that 2.5 ATP = 1 NADH, 1.5 ATP = 1 FADH2.
a. a 1:1 mixture of 14:0 and 14:1delta9 fatty-acyl-CoA's
b. triacylglycerols containing 16:2delta12,15 fatty acids
c. 18:2delta12 fatty acids
#### Questions Cloud
What is the volume of air at stp : The normal respiratory rate for a human being is 15.0 breaths per minute. The average volume of air for each breath is 505 cm3 at 20 ? C and 9.95 × 10 4 Pa. Why we use scientific names written in latin : Explain: (a) why we use scientific names written in Latin (and sometimes Greek) and (b) the importance of binomial nomenclature in naming organisms. Give examples of three scientific names (in correct binomial format) of common organisms. Define what is the mass of co2 in one liter of water : What is the mass of CO2 in one liter of water if the pressure above the water is 2 atm? (MW CO2 = 44 g/mol), Cg = kH Pg Please show steps Explain how you would modify the data : Explain how you would modify the data in order to make it relevant to decisions a manager must make. Explain the major factors that affect the degree of competitiveness in your industry. Which of the following will yield the most energy for body : Which of the following will yield the most energy for the body?Assume there are equivalent numbers of fatty acids in each case and that 2.5 ATP = 1 NADH, 1.5 ATP = 1 FADH2. What mass of ice at 0 c must be added to 45g of water : What mass of ice at 0C must be added to 45g of water at 20C to cool it to the final temperature of 10C? Compute mass percent of solution : Assume a density of 1.08g/mL for the solution. A)Find molarity of solution B)Find molality of solution C)Calculate mass percent of solution Which are true regarding the human chromosomes : Which is/are true regarding the human chromosomes? Accounting principles prior to the establishment of the fasb : Identify the two committees of the AICPA that established accounting principles prior to the establishment of the FASB.
### Write a Review
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draw a glycerol and explain how a triglyceride is formed from this molecule and fatty acids.
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Complete the Punnett squares for a cross between the male with blood type B and a female with blood type AB. Build two Punnett squares and answer the following given questions about them.
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Explain the role of water potential in the movement of water from soil through the plant and into the air?
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The phylum Arthropoda includes four major lineages: cheliceriforms (also called chelicerates); myriapods; insects and their relatives (together called hexapods); and crustaceans.
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In gorgonzolas, there are three recessive traits that affect the organism's appeal: shrunken fruit, foul-smelling flowers, and short height. A heterozygote was crossed with a homozygote, yielding the following offspring. | 1,088 | 4,709 | {"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-2024-33 | latest | en | 0.863728 |
https://gamedevtraum.com/en/game-and-app-development-with-unity/unity-tutorials-and-solutions/wiring-system-for-unity/ | 1,726,273,433,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.77/warc/CC-MAIN-20240913233654-20240914023654-00776.warc.gz | 244,685,545 | 34,050 | Wiring System for Unity
Introduction
In this article I present a solution to automatically create cables that hang from two or more points. The system allows us to modify the thickness of the cable, the curvature and more parameters to achieve the result we want.
The wire is drawn using Unity’s Line Renderer component so it is generated procedurally using mathematical expressions.
Unity Package
The solution consists of a Unity package that you can import into your project, this will create a new folder in your project with the Assets that come in the package.
How the Wiring System Works
Figure 1 shows the Assets that come in the package. In the scene you can see a power line that I generated using the wiring system.
To start using the system we can drag the prefabricated “GDT Wire Container” into the hierarchy (blue cube in figure 1). If you can’t create a new Empty GameObject and assign it the “GDTWiringSystem” script this will automatically generate everything needed for the system to work.
The object assigned to the “GDTWiringSystem” script will be the one to draw the cable and its children will be the points from which the cable will be hung.
In figure 2 we see the “GDTWiringSystem” component, here is all that is needed to shape the wiring.
Adding new points for the cable
To create a new holding point for the cable, simply right click on the container object and then click on Create Empty, as shown in Figure 3. This will create a new child and the script will automatically name it.
I do not recommend creating new children by duplicating existing ones, as this produces internal errors.
In figure 4 we see how when creating a new point, a new section of cable appears.
Wire loops
If we want to generate a cable with its ends connected we can activate the option “loopWire”, this will generate a new stretch of cable that joins the last point with the first one.
In figure 6 we see that the cable has been closed.
Wire Smoothness
The subdivision parameter allows us to choose the amount of segments that will be between two points of the cable, initially it is 1, which implies that the points of the cable are joined with straight lines, for the cable to have curvature we must choose a subdivision parameter greater than 1.
The larger the subdivisions, the smoother the cable curvature. The maximum range is set to 15 but it can be increased by opening the script and modifying the “[Range(1,15)]” instruction with the required value.
Cable Curvature
To modify the concavity of the cable, that is to say how intense its curvature is, we have the Sliders inside the “Curvature Parameters” vector, as we see in figure 8.
The default range of these sliders is -3 to 3, but if necessary you can modify it from the script in the “[Range(-3f,3f)]” instruction by typing the required value.
Animation
In case you need to animate the cable, I added a bool called “isAnimatedWire”, when you activate it the cable will be drawn in each FixedUpdate execution that by default is executed every 20 mili seconds.
Result
Using this solution we can automatically draw cables suspended from two or more points. We can add as many points as we want and modify the curvature of each section individually.
The cable is drawn with the LineRenderer component and using mathematical expressions.
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Secured By miniOrange | 711 | 3,367 | {"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-2024-38 | latest | en | 0.854617 |
http://www.appunwrapper.com/2015/06/27/adventure-escape-time-library-walkthrough/7/ | 1,519,028,243,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812556.20/warc/CC-MAIN-20180219072328-20180219092328-00385.warc.gz | 401,284,087 | 16,265 | # Adventure Escape: Time Library: Walkthrough
Chapter 7: Ancient Treasure:
1. Pick up the Aztec idol. Then solve the puzzle on the wall and get another Aztec idol. The key to this puzzle is paying attention to the 10 dots for e diamond and then multiplying it based on all the other shapes. So it goes 400, 20, 100, 10.
You can also watch my video for chapters 7, 8 & 9 here:
2. Go through the door to the right. Solve the puzzle on the statue. You need to match up all the gems to the right colors. Then his mouth will start glowing.
3. Move some pots around to find a ceremonial knife. Use it to cut off a piece of the deer skin hanging from the wall.
4. Stick the ceremonial knife in the slot in the ground to get another Aztec idol.
5. Go back to the other room. Place the deer skin piece on the scales.
6. The rope is missing on the scale. It’s hard to notice, but there are some vines growing near the doorway. Use the knife to cut some off and use them as ropes for the scale. Now to solve the scales puzzle. You need to refer to the sticky notes from the earlier puzzle, then weight each idol with a different one to see which one’s lightest to heaviest. The sticky notes say 100, 150, 200, 250 and 300. Put them in the right order to complete the puzzle. Mine looked like this in the end: 200, 150, 100, 300, 250.
7. Pick up all five idols from the scales. Go back to the other room. You need to place four of them on the pedestals. There are markings on each of the pedestals that tell you how much weight you need to put on there. The left one says 500 and the right says 400. Place the first (200) and fourth (300) idols on the left pedestal to make 500. Then place the second (150) and fifth (250) idols on the right pedestal to make 400. Take the time crystal and complete the chapter.
Click on the little numbers below to continue to the Chapter 8 walkthrough.
***
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## 31 thoughts on “Adventure Escape: Time Library: Walkthrough”
1. Fufu
Thanks for making this walkthrough. Some puzzles I just couldn’t figure out what to do but thanks to you I got hints n figured them out
1. Barbara
Hi — Had gotten this far in the game myself.
Was hoping for a walkthrough of the temple and the solution to the white squares/red squares puzzle.
Thank you.
1. Liz
i do not quite understand how you get 976 from that, it does not make any sense. Do you mind explaining please?
1. Andy
Hi that’s me too, I feel certain I am correct but can’t figure it out. Requires and v white, keep thinking its for the four digit code
2. ozigis muniratu
It’s really interesting… this is the kind of game we need to play…not only an interesting story line but also something to crack our brains over..keep up the good work,will be expecting more.
1. Amber
I had a lot of fun, but I agree that several of fhe puzzle’s made no sense. I love when I can’t figure out a puzzle and I look up the answer and feel stupid for not thinking of the answer myself. This game just left me confused. Still enjoyable though.
3. Shiska
Really appreciate the walk through! I think you over complicated the clock puzzle in chapter 3–if you read across, just pick out the next number in the sequence. So for the first clock, it went forward one hour, then two, so next is three. Second row it’s quarter hours (3, 6, 9) and third clock it’s back 10 minutes, then 20, then 30 (to the 6).
4. Rafhasantii
I cannot build ladder even if I’ve already got plank, twine, and 4 wooden sticks.
I used hammer but it build nothing. | 1,041 | 4,337 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-09 | latest | en | 0.916583 |
http://www.evitherm.org/default.asp?ID=683&menu1= | 1,638,386,841,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360881.12/warc/CC-MAIN-20211201173718-20211201203718-00302.warc.gz | 104,960,650 | 7,871 | Contact thermometry FAQs
Click on the number for an answer ...
1. What is a temperature scale?
2. How do you convert Fahrenheit or Kelvin temperatures into Celsius temperatures?
3. What is the ITS-90?
4. What are fixed points?
5. What is a temperature calibration ?
6. What are the standards used in thermometry?
7. What is a thermocouple?
8. What colours are the lead wires of each thermocouple type?
9. What are the benefits of grounded, ungrounded, and exposed probe junctions?
10. What does heterogeneity of a thermocouple mean?
11. What is a thermopile?
12. What is the difference between an RTD and a thermistor?
13. What is a bimetallic thermometer?
14. How to carry out a surface temperature measurement?
1. What is a temperature scale?
A temperature scale is a system of measuring temperature; it is formed by placing two reference points and evenly subdividing the points into temperature intervals. There are three temperature scales in use today, Celsius, Kelvin and Fahrenheit.
The Celsius (ºC) temperature scale is the scale based on 0 degrees for the freezing point of water and 100 degrees for the boiling point of water. It is sometimes called the centigrade scale because of the 100-degree interval. The Celsius scale is used in scientific work everywhere.
The Kelvin (K) temperature scale is the international temperature scale; its zero is the absolute zero point, the theoretical temperature at which the molecules of a substance have no thermal energy. The Kelvin scale is related to the Celsius scale. The difference between the freezing and boiling points of water is 100 degrees in each, so that the Kelvin has the same magnitude as the degree Celsius. The Kelvin is the unit of thermodynamic temperature measurement in the International System (SI) of measurement. It is defined as 1/ 273.16 of the triple point (equilibrium among the solid, liquid, and gaseous phases) of pure water.
The Fahrenheit (ºF) temperature scale is, like the Celsius scale, based on the freezing and boiling points of water. The freezing point of water is 32 degrees and the boiling point is 212 degrees.
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2. How do you convert Fahrenheit or Kelvin temperatures into Celsius temperatures?
To convert Fahrenheit or Kelvin to Celsius; or Fahrenheit to Kelvin; use the following conversion formulas:
- Celsius to Fahrenheit: t / °F = 9/5 . (t / °C) +32
- Fahrenheit to Celsius: t / °C = 5/9 . (t / °F - 32)
- Kelvin to Fahrenheit: t / °F = 5/9 . T / K - 459.67
- Fahrenheit to Kelvin: T / K = 9/5 . (t / °F + 459.67)
- Celsius to Kelvin: T / K = t / °C + 273.15
- Kelvin to Celsius: t / °C = T / K - 273.15
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3. What is the ITS-90?
ITS-90 is the International Temperature Scale and was created in 1990. It is the scale which accredited laboratories refer to in their calibration certificates.
ITS-90 is based on 17 fixed points (phase transitions of pure materials) and interpolating equations between the fixed points or pressure/temperature relations of gas. The ITS-90 extends upwards from 0.65 K to the highest temperature practically measurable in terms of the Planck radiation law using monochromatic radiation.
The units used for the ITS-90 are: the Kelvin symbol (K) (temperature T90) and the degree Celsius symbol (ºC) (temperature t90).
For further information see the ITS website.
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4. What are fixed points?
Fixed points are phase transitions of pure substances whose temperature is fixed and highly reproducible. These are:
melting point, the transition from a solid to a liquid at normal atmospheric pressure
freezing point, the transition from a liquid to a solid at normal atmospheric pressure
triple point, the temperature at which the solid, liquid and vapour phases are in thermodynamic equilibrium with each other
boiling point, the transition from liquid to vapour at normal atmospheric pressure.
These fixed points are realisable with numerous pure substances. However, many of them do not have the necessary stability and reproducibility and some others require a complex procedure and special laboratory facilities. The greatest difficulty in realising a fixed-point temperature is due to the influence of impurities in the fixed-point material.
For a practical realisation, there is a fixed-point cell, a flask nearly filled with pure material, which is surrounded by a shell that provides an isothermal environment.
The cell is placed in an apparatus (a furnace or liquid bath) that must provide good temperature control and sufficient cell immersion to generate uniform a temperature in the measurement zone.
The triple point of water is the most important defining thermometric fixed point used in the calibration of thermometers to the International Temperature Scale of 1990 (ITS-90), it is one of the most accurately realisable of the defining fixed points. It provides a useful check of the stability of a thermometer when investigating whether a shift in calibration has occurred.
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5. What is a temperature calibration?
In the specific area of temperature, a calibration is a comparison between a thermometer to be calibrated and a “standard” related to National standards. The result of a calibration is a correction to apply to the reading of the calibrated thermometer and its associated uncertainty.
There are two calibration methods:
1. The fixed points method
This absolute method is used for the realisation of the International temperature scale: ITS-90. The thermometer is calibrated by measurements at a series of temperature fixed points, e.g. freezing/melting points, triple points, vapour pressure points. This method gives calibration results with high accuracy. It is relevant only for high quality thermometers.
2. The comparison method
The thermometer is calibrated by comparison with a reference / standard thermometer in a thermally stabilised bath or furnace suitable for the calibration. This method allows coverage of a wide range of temperature during a calibration operation, point-by-point or continuously, in a short time, and if required to calibrate simultaneously a large number of thermometers. However, this method is less accurate than the fixed point method because of the lesser accuracy and stability of the reference standard and enclosure.
Depending on the type of thermometer, the calibration can be carried out by considering the thermometer either as a whole system (including the detector, connectors and readout instrument) or as just the detector.
For example, in the case of a liquid-in-glass thermometer, the three components cannot be separated and so the calibration is of the system as a whole. Contrarily, in the case of a thermoelectric or resistance thermometer, the sensing element can be separated from its connecting wires and the readout instrument, and calibrated separately. However, it is generally better to calibrate a thermometer system as a whole.
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6. What are the standards used in thermometry?
The standards used in thermometry are fixed points and standard thermometers, described as follows:
Fixed points
Thermometers are calibrated at fixed temperatures (e.g. freezing / melting points, triple points, vapour pressure points) specified in the International Temperature Scale of 1990 (ITS-90). Interpolation equations depending on the temperature range are determined for the thermometer.
Standard thermometers
Standard thermometers are principally Platinum Resistance Thermometers (PRTs) or thermocouples (type S, R or D), ideally calibrated at fixed points.
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7. What is a thermocouple?
A thermocouple consists of two wires (thermoelement) made of dissimilar metals and welded at one end. This junction is called the hot junction or the measuring junction. At the other end of the wires a cold junction is or reference junction is made and this is connected to the output device (voltmeter, temperature indicator).
The resulting voltage or electromotive force (emf) of typically a few millivolts is generated by thermal gradients between the hot and cold junctions. It is a function of the difference in temperature between the measurement junction and the reference junction. The relationships between the emf and temperature are available for the most commonly used thermocouples in reference tables (type R, S, B, J, T, E, K, N).
Thermocouples are the most widely used sensors in industry due to their low cost, simplicity, robustness, size and temperature range of use.
Further info at NPL
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8. What colours are the lead wires of each thermocouple type?
The standards define an identification system based on colours to identify the types of thermocouple and the polarity of the conductor cables.
The colours for each type of thermocouple are defined in the International standard IEC 584-3 Thermocouples, Part 3: Tolerances for Compensating and Extension Cables; identification System, as follows:
Type External sheath Conductors components Positive Negative K green chromel green alumel white T brown copper brown constantan white J black iron black constantan white N pink nicrosil pink nisil white E purple chromel purple constantan white R orange platinum 13% rhodium orange platinum white S orange platinum 10% rhodium orange platinum white B grey platinum 30% rhodium grey platinum 6% rhodium white
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9. What are the benefits of grounded, ungrounded, and exposed probe junctions?
Grounded junctions - are welded to the tip of the sheath with wires completely sealed from contaminants. They offer good response times and are ideal for measuring the temperature of static or flowing corrosive gases and liquids.
Ungrounded junctions - are sealed, insulated from the protective sheath, and electrically isolated. They have longer response times than grounded or exposed junctions and are used for conductive solutions or where isolation of the measuring circuitry is required.
Exposed junctions - have the fastest response times and are ideal for measuring rapid temperature changes. Clear coating on most models provides a humidity barrier for the thermocouple. Do not use exposed junctions with corrosive fluids or atmospheres.
Further info at Cole-Parmer
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10. What does heterogeneity of a thermocouple mean?
Changes in the material composition of the thermocouple conductors induce structural inhomogeneities which modify the magnitude of the electromotive force (emf) generated by the thermocouple. The emf is affected by the thermal gradient which exists along the length of the exposed conductors. In homogeneous conductors, the emf depends only on the temperature of the two junctions. The inhomogeneities in the conductors produce parasitic emf in regions where temperature gradients exist. It is a major source of measurement errors with thermocouples and is very difficult to detect.
The inhomogeneity errors are due principally to:
• mechanical damage during manufacture or use of the thermocouple
• chemical damage occurring during exposure of the thermocouple, particularly at high temperatures (e.g. oxidation of components, contamination, changes in alloy composition)
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11. What is a thermopile?
A thermopile is made of several thermocouples connected in series. The resulting voltage is the sum of the individual thermocouple voltages and therefore increases the measured signal. Thermopiles with a large number of junctions are used in heat flux sensors. Thermopiles are made in a number of ways, either as a wound wire, printed circuit or etched metal foils.
Further info at Wikipedia
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12. What is the difference between an RTD and a thermistor?
The sensing element for both an RTD (Resistance Temperature Detector) and a thermistor has an electrical resistance that changes with temperature. The sensing element of an RTD is metallic, while thermistors have semiconductors made from mixtures of oxides of nickel, manganese, etc. The resistance of an RTD increases approximately linearly with temperature. Typically, the resistance of a thermistor drops with increasing temperature and is highly non-linear and can usually be expressed as an exponential function of temperature.
The most common temperature range for a thermistor is 0°C - 100°C. At higher temperatures, it is subject to large drifts in output.
The most common type of RTD is the PRT (Platinum Resistance Thermometer) and it is the most reproducible type because platinum is highly corrosion and oxidation resistant and stable over a wide temperature range (-250°C to 850°C). It is used as interpolating standard between the fixed points of the International temperature Scale ITS-90.
Thermistors are not as reproducible as PRTs but are much more sensitive. The high sensitivity of thermistors can be an order of magnitude greater than that of the RTDs. Thermistors are useable over a smaller temperature range than RTDs but are very accurate (usually better than 0.05°C or 0.1°C).
Compared with thermistors, RTDs do not allow point measurement of temperature because of the large sensing element. Contrarily, the small size of thermistors makes them very adaptable and they are often included in electronic circuits.
For both RTDs and thermistors, the self-heating due to the current flow through the resistance is a source of measurement error. To minimise this effect, the current used must be fitted with the value of the measured resistance.
Further info at NPL
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13. What is a bimetallic thermometer?
A bimetallic strip thermometer is a mechanical thermometer. It consists of two strips made of dissimilar metals and bonded together with one end fixed and the other free.
The principle of operation is that as the temperature changes one strip expands more than the other, causing motion of the free end of the strip, i.e. a bending caused by the different expansions. Two constructions are available: a spiral strip and a cantilever strip. They are the most widely used in industry for temperature control, due to their robustness, temperature range and simplicity. They are used in thermostats for measuring and controlling temperature, or in ovens.
Further info at temperatures.com
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14. How to carry out a surface temperature measurement?
A surface temperature measurement can be difficult to carry out and requires a lot of precautions to be accurate. It can be achieved either with a contact thermometer or by a non-contact technique using an optical pyrometer.
With contact thermometry it is possible to employ an indirect or a direct method. In the indirect method the surface temperature Ts of a body is deduced from two temperature measurements, T1 and T2, performed inside the material by means of two sensors placed at different depths x1 and x2 into the material with a thickness e.
The surface temperature Ts may be determined by extrapolating the T1 and T2 temperature measurements as follows:
In the direct measurement method the surface temperature is determined via direct contact of a sensor with the surface. Two approaches are possible:
• apply a removable sensor, usually a thermocouple, to the surface. It could be placed in direct contact or be welded to an intermediate thermally conducting material.
• use a permanent sensor (or semi-permanent sensors). Two types are available: wire type or film type.
The thermoelement can be a thermocouple, a resistance or a thermistor. The main sources of uncertainty with the direct measurement method come from the sensor itself (sensor calibration, heat conduction in the leads, self-heating etc.) and the possibly poor thermal contact between sensor and surface. These uncertainties can be quantified by means of a relevant calibration of the surface thermometer.
Non-contact thermometry or radiation thermometry is a non-invasive technique. This approach is useful in determining the surface temperature of moving bodies, or for high temperatures (too high for a contact thermometer) or chemically reactive applications. The main sources of uncertainty are due to the thermometer itself, from the emissivity of the body surface if not precisely known (it usually isn't!) and from parasitic radiation (i.e. radiation reflected between the surface and the thermometer or background radiation reflected from the surface into the thermometer).
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https://in.mathworks.com/matlabcentral/cody/problems/15-find-the-longest-sequence-of-1-s-in-a-binary-sequence/solutions/12949 | 1,607,072,772,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141735395.99/warc/CC-MAIN-20201204071014-20201204101014-00235.warc.gz | 338,301,174 | 17,652 | Cody
Problem 15. Find the longest sequence of 1's in a binary sequence.
Solution 12949
Submitted on 29 Jan 2012 by Jonathan
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
%% x = '0'; y_correct = 0; assert(isequal(lengthOnes(x),y_correct))
2 Pass
%% x = '1'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
x = Empty matrix: 1-by-0
3 Pass
%% x = '01'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
x = 0
4 Pass
%% x = '10'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
x = 0
5 Pass
%% x = '00'; y_correct = 0; assert(isequal(lengthOnes(x),y_correct))
6 Pass
%% x = '11'; y_correct = 2; assert(isequal(lengthOnes(x),y_correct))
x = 1 x = Empty matrix: 1-by-0
7 Pass
%% x = '1111111111'; y_correct = 10; assert(isequal(lengthOnes(x),y_correct))
x = 1 1 1 1 1 1 1 1 1 x = 1 1 1 1 1 1 1 1 x = 1 1 1 1 1 1 1 x = 1 1 1 1 1 1 x = 1 1 1 1 1 x = 1 1 1 1 x = 1 1 1 x = 1 1 x = 1 x = Empty matrix: 1-by-0
8 Pass
%% x = '100101011111010011111'; y_correct = 5; assert(isequal(lengthOnes(x),y_correct))
x = Columns 1 through 10 0 0 0 0 0 0 0 1 1 1 Columns 11 through 20 1 0 0 0 0 0 1 1 1 1 x = Columns 1 through 10 0 0 0 0 0 0 0 1 1 1 Columns 11 through 19 0 0 0 0 0 0 1 1 1 x = Columns 1 through 10 0 0 0 0 0 0 0 1 1 0 Columns 11 through 18 0 0 0 0 0 0 1 1 x = Columns 1 through 10 0 0 0 0 0 0 0 1 0 0 Columns 11 through 17 0 0 0 0 0 0 1 x = Columns 1 through 10 0 0 0 0 0 0 0 0 0 0 Columns 11 through 16 0 0 0 0 0 0
9 Pass
%% x = '01010101010101010101010101'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
x = Columns 1 through 10 0 0 0 0 0 0 0 0 0 0 Columns 11 through 20 0 0 0 0 0 0 0 0 0 0 Columns 21 through 25 0 0 0 0 0
10 Pass
%% x = '0101010111000101110001011100010100001110110100000000110001001000001110001000111010101001101100001111'; y_correct = 4; assert(isequal(lengthOnes(x),y_correct))
x = Columns 1 through 10 0 0 0 0 0 0 0 1 1 0 Columns 11 through 20 0 0 0 0 0 1 1 0 0 0 Columns 21 through 30 0 0 0 1 1 0 0 0 0 0 Columns 31 through 40 0 0 0 0 0 0 1 1 0 0 Columns 41 through 50 1 0 0 0 0 0 0 0 0 0 Columns 51 through 60 0 0 1 0 0 0 0 0 0 0 Columns 61 through 70 0 0 0 0 0 0 1 1 0 0 Columns 71 through 80 0 0 0 0 0 0 1 1 0 0 Columns 81 through 90 0 0 0 0 0 0 0 1 0 0 Columns 91 through 99 1 0 0 0 0 0 1 1 1 x = Columns 1 through 10 0 0 0 0 0 0 0 1 0 0 Columns 11 through 20 0 0 0 0 0 1 0 0 0 0 Columns 21 through 30 0 0 0 1 0 0 0 0 0 0 Columns 31 through 40 0 0 0 0 0 0 1 0 0 0 Columns 41 through 50 0 0 0 0 0 0 0 0 0 0 Columns 51 through 60 0 0 0 0 0 0 0 0 0 0 Columns 61 through 70 0 0 0 0 0 0 1 0 0 0 Columns 71 through 80 0 0 0 0 0 0 1 0 0 0 Columns 81 through 90 0 0 0 0 0 0 0 0 0 0 Columns 91 through 98 0 0 0 0 0 0 1 1 x = Columns 1 through 10 0 0 0 0 0 0 0 0 0 0 Columns 11 through 20 0 0 0 0 0 0 0 0 0 0 Columns 21 through 30 0 0 0 0 0 0 0 0 0 0 Columns 31 through 40 0 0 0 0 0 0 0 0 0 0 Columns 41 through 50 0 0 0 0 0 0 0 0 0 0 Columns 51 through 60 0 0 0 0 0 0 0 0 0 0 Columns 61 through 70 0 0 0 0 0 0 0 0 0 0 Columns 71 through 80 0 0 0 0 0 0 0 0 0 0 Columns 81 through 90 0 0 0 0 0 0 0 0 0 0 Columns 91 through 97 0 0 0 0 0 0 1 x = Columns 1 through 10 0 0 0 0 0 0 0 0 0 0 Columns 11 through 20 0 0 0 0 0 0 0 0 0 0 Columns 21 through 30 0 0 0 0 0 0 0 0 0 0 Columns 31 through 40 0 0 0 0 0 0 0 0 0 0 Columns 41 through 50 0 0 0 0 0 0 0 0 0 0 Columns 51 through 60 0 0 0 0 0 0 0 0 0 0 Columns 61 through 70 0 0 0 0 0 0 0 0 0 0 Columns 71 through 80 0 0 0 0 0 0 0 0 0 0 Columns 81 through 90 0 0 0 0 0 0 0 0 0 0 Columns 91 through 96 0 0 0 0 0 0
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Start Hunting! | 1,928 | 3,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-50 | latest | en | 0.559221 |
https://us.metamath.org/mpeuni/sbf.html | 1,718,622,022,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861701.67/warc/CC-MAIN-20240617091230-20240617121230-00783.warc.gz | 541,073,744 | 3,708 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > sbf Structured version Visualization version GIF version
Theorem sbf 2273
Description: Substitution for a variable not free in a wff does not affect it. For a version requiring disjoint variables but fewer axioms, see sbv 2099. (Contributed by NM, 14-May-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)
Hypothesis
Ref Expression
sbf.1 𝑥𝜑
Assertion
Ref Expression
sbf ([𝑦 / 𝑥]𝜑𝜑)
Proof of Theorem sbf
StepHypRef Expression
1 sbf.1 . 2 𝑥𝜑
2 sbft 2272 . 2 (Ⅎ𝑥𝜑 → ([𝑦 / 𝑥]𝜑𝜑))
31, 2ax-mp 5 1 ([𝑦 / 𝑥]𝜑𝜑)
Colors of variables: wff setvar class Syntax hints: ↔ wb 209 Ⅎwnf 1785 [wsb 2070 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1912 ax-6 1971 ax-7 2016 ax-12 2179 This theorem depends on definitions: df-bi 210 df-ex 1782 df-nf 1786 df-sb 2071 This theorem is referenced by: sbf2 2274 sbh 2275 nfs1f 2277 sbbibOLD 2291 sbrim 2315 sblim 2317 sbrbif 2323 sbievOLD 2333 sb6x 2489 sbequ5 2490 sbequ6 2491 sb2ae 2538 sbie 2546 sbid2 2552 sbabel 3013 nfcdeq 3754 mo5f 30265 suppss2f 30398 fmptdF 30415 disjdsct 30452 esumpfinvalf 31395 bj-sbf3 34224 bj-sbf4 34225 ellimcabssub0 42189 2reu8i 43599 ichf 43897
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Found in: Page 380
### Fundamentals Of Physics
Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718
# The figure gives the potential energy function U(r) of a projectile, plotted outward from the surface of a planet of radius. What least kinetic energy is required of a projectile launched at the surface if the projectile is to “escape” the planet
Minimum kinetic energy required to escape from the surface of planet is $5×{10}^{9}J$
See the step by step solution
## Step 1: The given data
Potential energy at Rs is $5×{10}^{9}$ J
## Step 2: Understanding the concept of energy in a gravitational field
The total energy of a particle is the sum of the kinetic and potential energies of the particle. To escape from the gravitational field, the supplied kinetic energy should be equal to the gravitational potential energy of the particle.
Formula:
Total energy= Kinetic Energy+ Potential Energy
TE = U + KE
## Step 3: Calculation of the least kinetic energy
Minimum kinetic energy required to escape from the surface of the planet
To escape from the surface of earth TE = 0
It means
$KE+U=0\phantom{\rule{0ex}{0ex}}KE=-U$
But, from the graph, the potential energy on surface of earth is $U=-5×{10}^{9}J$
So, the least kinetic energy of the particle is $5×{10}^{9}J$ | 383 | 1,474 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 14, "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-2023-23 | latest | en | 0.775255 |
https://www.physicsforums.com/threads/momentum-law-of-conservation-of-energy-collision-of-two-cars.416705/ | 1,544,397,974,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823228.36/warc/CC-MAIN-20181209232026-20181210013526-00032.warc.gz | 991,486,270 | 14,432 | # Homework Help: Momentum/law of conservation of energy - collision of two cars
1. Jul 17, 2010
### shawli
1. The problem statement, all variables and given/known data
A car with a mass of 1875 kg is travelling along a country road when the driver sees a
deer dart out onto the road. The driver slams on the brakes and manages to stop before hitting the deer. The driver of a second car (mass of 2135 kg) is driving too close and does not see the deer. When the driver realizes that the car ahead is stopping, he hits the brakes but is unable to stop. The cars lock together and skid another 4.58 m. Allof the motion is along a straight line. If the coefficient of friction between the dry concrete and rubber tires is 0.750, what was the speed of the second car when it hit the stopped car?
2. Relevant equations
Notation: "A" is the stopped car and "B" is the second car. v' is the resulting velocity after the two cars have collided and locked.
mA = 1875 kg
mB = 2135 kg
uF = 0.750
d= 4.58m
The velocity of car B before the collision and the final velocity are not given.
vB = ?
v' = ?
So momentum is conserved, and the resulting equation would be:
mB * vB = (mA + mB) * v' <--- Eqn 1
Also, the energy is conserved. The kinetic energy of the second car (B) right before the collision should equal the energy of friction (thermal energy?) used to stop the two locked cars (I think this is right...). So:
Ek(before) = Ef (after)
0.5 * mB * vB ^2 = uF * Fn * d <--- Eqn 2
3. The attempt at a solution
So basically my method was to use Eqn 2 to isolate "vB" , then sub this rearranged equation into Eqn 1:
Eqn 2:
0.5 * mB * vB ^2 = uF * Fn * d
0.5 * mB * vB ^2 = uF * (mA + mB) * g * d
vB = SQRT ([2* uF * (mA + mB) * g * d] / mB)
vB = SQRT ([2* 0.750 * (1875 + 2135) * 9.8 * 4.58] / 2135)
vB = 11.25 m/s ---> 40.48 km/h
Eqn 1:
mB * vB = (mA + mB) * v'
v' = (mB * vB)/ (mA + mB)
v' = (2135 * 40.48) / (2135 + 1875)
v' = 21.5 km/h
The answer is 55.5km /h ...can someone check over my method for me please?
2. Jul 17, 2010
### pat666
is energy conserved - the collision is perfectly inelastic isn't it. I would say that your method is wrong.
3. Jul 17, 2010
### shawli
Oh right, I can't know for sure if its elastic.
So... :/ now I have no idea how to solve this !
4. Jul 17, 2010
### pat666
you do know for sure that its inelastic though.... if the two stick together energy is not conserved in the system. the only thing to remember is that you cant use energy methods unless the question explicitly states that it was a perfectly elastic collision.
to solve:
momentum is always conserved
the speed of the 1st car is 0 upon collision - mass is 1875Kg
speed of the second car is unknown - mass is 2135kg
the combined mass is (1875 +2135)kg and the speed immediately after is unknown but calcuable(excuse the word i just made up)
to find this out
F=ma m=(1875 +2135)kg
F=uN
F=(1875 +2135)*9.81*0.75
Find a
now the final velocity is 0, you know the mass and the deceleration
use kinematics to solve for the velocity after the collision
now you have everything needed use pi=pf and solve!!!
5. Jul 17, 2010
### shawli
Oh that makes sense, thank you!
One last question though.
To find acceleration, it's:
F friction = m * a
(1875 +2135)*9.81*0.75 = m * a
You're saying this "m" on the right is also the sum of the masses of both cars right?
Since they both lock and go forward for a stretch together?
6. Jul 17, 2010
### pat666
Yes thats right.
7. Jul 17, 2010
Thank you! | 1,093 | 3,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-51 | latest | en | 0.901136 |
https://redquark.org/leetcode/0016-3-sum-closest/ | 1,702,336,958,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679518883.99/warc/CC-MAIN-20231211210408-20231212000408-00444.warc.gz | 536,214,015 | 80,139 | # LeetCode #16 - 3 Sum Closest
Hello fellow devs 👋! Let’s look at a problem which is an extension of the last problem 3 Sum we solved.
## Problem Statement
Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
### Constraints:
• 3 ≤ nums.length ≤ 103
• -103nums[i] ≤ 103
• -104target ≤ 104
### Example
Input: nums = [-1,2,1,-4], target = 1
Output: 2
Explanation: The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
## Analysis
It is very similar to the previous problem 3 Sum. Here, also we need to find the sum of three numbers. Instead of checking if their sum is equal to zero, we are given a target, and we will be trying to find the smallest difference between the sum and the target.
The desired sum can be more than or less than the target hence we only care about the absolute difference between the two.
## Approach
The approach is similar to the one we discussed earlier -
1. Sort the array (in time O(n * log(n))).
2. Now for each element i, do the following steps
3. Set two pointers left — j = i + 1 and right — k = nums.length - 1.
4. Check if nums[i] + nums[j] + nums[k] <= target, it means we are too left in the array, and we need to move right i.e., we can check for greater number than the current one.
5. If the sum nums[i] + nums[j] + nums[k] > target, it means we are too right in the array, and we need to move left i.e., we can check for smaller number than the current one.
6. Compare the minimum difference between the current sum and the previous sum. The sum which give minimum difference is the answer.
### Time Complexity
We are scanning the entire array keeping one element fixed. We are doing this for every element in the array. Thus, we are scanning each element of array n number of times. And we are doing this for n times, hence the worst case time complexity will be O(n2 + n * log n) which comes down to O(n2).
### Space Complexity
We are not using any data structure for the intermediate computations, hence the space complexity is O(1).
## Code
### Java
public class ThreeSumClosest {
public int threeSumClosest(int[] nums, int target) {
// Sort the array
Arrays.sort(nums);
// Length of the array
int n = nums.length;
// Result
int closest = nums[0] + nums[1] + nums[n - 1];
// Loop for each element of the array
for (int i = 0; i < n - 2; i++) {
// Left and right pointers
int j = i + 1;
int k = n - 1;
// Loop for all other pairs
while (j < k) {
int sum = nums[i] + nums[j] + nums[k];
if (sum <= target) {
j++;
} else {
k--;
}
if (Math.abs(closest - target) > Math.abs(sum - target)) {
closest = sum;
}
}
}
return closest;
}
}
### Python
def threeSumClosest(nums: List[int], target: int) -> int:
# Sort the given array
nums.sort()
# Length of the array
n = len(nums)
# Closest value
closest = nums[0] + nums[1] + nums[n - 1]
# Loop for each element of the array
for i in range(0, n - 2):
# Left and right pointers
j = i + 1
k = n - 1
# Loop for all other pairs
while j < k:
current_sum = nums[i] + nums[j] + nums[k]
if current_sum <= target:
j += 1
else:
k -= 1
if abs(closest - target) > abs(current_sum - target):
closest = current_sum
return closest
### JavaScript
var threeSumClosest = function (nums, target) {
// Sort the array
nums.sort((a, b) => a - b);
// Length of the array
const n = nums.length;
// Result
let closest = nums[0] + nums[1] + nums[n - 1];
// Loop for each element of the array
for (let i = 0; i < n - 2; i++) {
// Left and right pointers
let j = i + 1;
let k = n - 1;
// Loop for all other pairs
while (j < k) {
let sum = nums[i] + nums[j] + nums[k];
if (sum <= target) {
j++;
} else {
k--;
}
if (Math.abs(closest - target) > Math.abs(sum - target)) {
closest = sum;
}
}
}
return closest;
};
### Kotlin
fun threeSumClosest(nums: IntArray, target: Int): Int {
// Sort the array
Arrays.sort(nums)
// Length of the array
val n = nums.size
// Result
var closest = nums[0] + nums[1] + nums[n - 1]
// Loop for each element of the array
for (i in 0 until n - 2) {
// Left and right pointers
var j = i + 1
var k = n - 1
// Loop for all other pairs
while (j < k) {
val sum = nums[i] + nums[j] + nums[k]
if (sum <= target) {
j++
} else {
k--
}
if (abs(closest - target) > abs(sum - target)) {
closest = sum
}
}
}
return closest
}
## Conclusion
Congratulations 👏! We have solved one more problem from LeetCode and it was very much similar to the previous one 😃.
I hope you enjoyed this post. Feel free to share your thoughts on this.
You can find the complete source code on my GitHub repository. If you like what you learn, feel free to fork 🔪 and star ⭐ it.
Till next time… Happy coding 😄 and Namaste 🙏! | 1,360 | 4,779 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-50 | latest | en | 0.890108 |
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# Ohio Post Dreams here Let's go Ohio!!
9059 replies. Last post 14 hours ago by marcie.
Page 400 of 604
Ohio
United States
Member #49980
February 21, 2007
34130 Posts
Offline
Posted: March 6, 2013, 8:58 pm - IP Logged
I forgot to post this one Dream and I forgot about it, and it came this evening 805* I dreamt the Girl up stairs broke my T.V. 805* she dreamt about me and I dreamt about her which was weird and she dreamt of me, I was playing her dream when I should of been Playing my own Dream.. She Dreamt I ask her for some Salt 149 914* I flip ther 9/6 that is how I got the 614.
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
United States
Member #49980
February 21, 2007
34130 Posts
Offline
Posted: March 7, 2013, 1:56 pm - IP Logged
I forgot to post this one Dream and I forgot about it, and it came this evening 805* I dreamt the Girl up stairs broke my T.V. 805* she dreamt about me and I dreamt about her which was weird and she dreamt of me, I was playing her dream when I should of been Playing my own Dream.. She Dreamt I ask her for some Salt 149 914* I flip ther 9/6 that is how I got the 614.
My 14 prs. came..I hope you used them Ohio* Ga. had Fl. got the Salt numbers 235 690 By the way Ohio had 238 that was salt too......so is the 819* it came in P. 4 1892*
978
110
342
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
United States
Member #49980
February 21, 2007
34130 Posts
Offline
Posted: March 7, 2013, 10:32 pm - IP Logged
We didn't get a winner Ohio in the workout but sometimes we don't. 337 Workout...
534
776
938
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
u\$a
United States
Member #106665
February 22, 2011
19782 Posts
Offline
Posted: March 8, 2013, 2:50 am - IP Logged
hey marcie, hope all is well with you.
Let it Snow
Ohio
United States
Member #49980
February 21, 2007
34130 Posts
Offline
Posted: March 8, 2013, 8:07 am - IP Logged
hey marcie, hope all is well with you.
Things are good!! Thanks for asking... I hope these are good with you too! Let's getem!!!good Luck I dreamt I was at this Casino 303/300 813 0 or 1 for P. 4 in the Dream I seen 44* that is all I can remember 440 400 44 prs. 444 4400 441X 441XX Now lets play this Game...@ Joey I seen what you had in the group, isn't that a concidence something to think about.. same number
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
United States
Member #49980
February 21, 2007
34130 Posts
Offline
Posted: March 8, 2013, 8:28 am - IP Logged
We didn't get a winner Ohio in the workout but sometimes we don't. 337 Workout...
534
776
938
This is where we are... Now get some winners!! I feel a win coming on somewhere!
It might not be the lottery but I feel it in any game of chane!! From my Dreamt I will keep you informed.
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
United States
Member #49980
February 21, 2007
34130 Posts
Offline
Posted: March 8, 2013, 4:27 pm - IP Logged
Maybe tonite from the 207? Remember Ohio I said I dreamt about the Girl upstair broke my T.V. it came in the P. 5 50810 comb. Amazing I was playing the 31801* For Casino.
423
605
877
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
United States
Member #49980
February 21, 2007
34130 Posts
Offline
Posted: March 8, 2013, 4:55 pm - IP Logged
I was in the Store this Lady tells everybody she dreamt she was Fighting I go that plays for 599/595 Then this other Guy saids those are some good numbers, I tell them both you know what I am playing that number for a Friend. So actually 3 people talked about that number. I told them it came in Ga. last night, part of it which was T.V. 595/599 it came as 9055... I also told them we get Ga. numbers. Sum is 9* Just like Ga. had 0147 and we had 174 for Ohio. that is an example. Play what you want I am just saying..
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
United States
Member #49980
February 21, 2007
34130 Posts
Offline
Posted: March 8, 2013, 7:34 pm - IP Logged
Here are the winners for Ohio*
P. 3 for Ohio was 464
P. 4 Ohio 2424
P. 5 80293 I was so Close 30810
My quote in my Dream I see 44 prs. I forgot about that lol... Did I mention I am Psychic... we all are at some point some more than others..
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
United States
Member #49980
February 21, 2007
34130 Posts
Offline
Posted: March 8, 2013, 11:38 pm - IP Logged
Here are the winners for Ohio*
P. 3 for Ohio was 464
P. 4 Ohio 2424
P. 5 80293 I was so Close 30810
My quote in my Dream I see 44 prs. I forgot about that lol... Did I mention I am Psychic... we all are at some point some more than others..
That Number 464 it Plays for Car one of the Numbers, remember last week ago I said I dreamt about a Car well the number came 1 week Later. lol... I hope you Guys used the 44* because it came in the Mega Million I forgot again to use it. I thought about using it but I didn't go back to the Store....Multiplier was X4.
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
United States
Member #49980
February 21, 2007
34130 Posts
Offline
Posted: March 9, 2013, 9:53 am - IP Logged
464 Workout Where is ejoy? LOL.. Good Luck with these. Don't know what cme I dreamt I lost...couldn't find one of my Kids. I wake up My Son knocking on my door it was Late he is 40* Son 710 108 712 122/112 I like*
645
867
049
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
United States
Member #49980
February 21, 2007
34130 Posts
Offline
Posted: March 9, 2013, 10:59 am - IP Logged
I think we will get the P. 3 P. 4 or P. 5 out of these numbers... Or they could come all over the Map!! 0 or 1 for P. 4 Get some winners Everyone on the Map!! If you like them Play them Ohio Pick a good number or good Prs.
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
United States
Member #49980
February 21, 2007
34130 Posts
Offline
Posted: March 9, 2013, 1:36 pm - IP Logged
I think we will get the P. 3 P. 4 or P. 5 out of these numbers... Or they could come all over the Map!! 0 or 1 for P. 4 Get some winners Everyone on the Map!! If you like them Play them Ohio Pick a good number or good Prs.
Alrighty Ohio I gave you all the number plus the Prs. were there, I figure that person that email ME WAS TRYING TO GET ME A NUMBER. 526 tHAT NUMBER IS SEX TOO. 666* GA. WOW! i WAS SEEING THAT TOO. ACTUALLY i SEE THE 666 FROM MY WORKOUT ..
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
United States
Member #49980
February 21, 2007
34130 Posts
Offline
Posted: March 9, 2013, 5:58 pm - IP Logged
526 workout here it is... God luck!
756
928
160
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
United States
Member #49980
February 21, 2007
34130 Posts
Offline
Posted: March 9, 2013, 7:58 pm - IP Logged
526 workout here it is... God luck!
756
928
160
Bullseye 150*Ohio | 3,829 | 11,244 | {"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-2016-50 | latest | en | 0.864417 |
https://puzzling.stackexchange.com/questions/81950/can-you-cross-the-bridge/83141 | 1,566,467,270,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317037.24/warc/CC-MAIN-20190822084513-20190822110513-00155.warc.gz | 599,607,426 | 31,849 | # Can you cross the bridge?
If
C = 5
N = 2
R = 4
W = 1
P = 9
A = 4
D = 0
G = 7
M = 2
Then
B = ?
• B is equal to the answer the OP intended – North Apr 19 at 0:13
• @North no kidding. To Soltius - is modular division involved? – Brandon_J Apr 19 at 0:25
• Chinese remainder theorem! – Dr Xorile Apr 19 at 1:01
• @Brandon_J and Dr Xorile : No modular division or chinese remainder theorem – Soltius Apr 19 at 10:16
Hint: some of these letters have much more than a single solution.
We just use basic topology of the symbols and see if they correspond: holes, tails, their connections and etc. The classes of letters presented are:
Just a line as in {C, 5, N, 2, W, 1, G, 7, M }
Circle with a tail as in { P, 9 }
Circle with two tails as in {R, A, 4}
Just circle as in {D, 0}.
With that said, B definitely equals
eight.
BONUS QUESTION FOR CREATOR:
What makes me believe that
POLO equals 0.06?
• thought it would be something visual – nine9 Apr 24 at 11:41
• I would say W=3 under these rules, no? And surely M and W would be the same, at least? – hexomino Apr 24 at 11:49
• @hexomino I might have been unclear, but by my set of rules M=W=3=2, so you're correct as well. – Thomas Blue Apr 24 at 12:14
• How to prove $2 = 3$... @ThomasBlue XD +1 – Omega Krypton Apr 24 at 12:42
• Good job ! To be more precise, the "equal" sign indicated that values on each sign were related by a homeomorphism : You can read more about it here en.wikipedia.org/wiki/Topology, but it basically means that if you squish and pull stuff on the LHS without changing the general toplogy (holes stay holes, etc), you can get the RHS (like a donut can change into a mug). – Soltius Apr 24 at 22:23
If a person wants a cool party, what will he say.
Wanna rocking party in big hall
Remove all vowels to get
WNN RCKNG PRTY N BG HLL
Given are the position of letters in which they first appear in this sentence.
B = 14 | 595 | 1,904 | {"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} | 3.515625 | 4 | CC-MAIN-2019-35 | latest | en | 0.927134 |
https://time.electronicbub.com/answer/what-time-is-22-11/ | 1,695,994,271,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510516.56/warc/CC-MAIN-20230929122500-20230929152500-00547.warc.gz | 628,451,149 | 8,407 | 1
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22:11 time means 10:11 PM
22:11 is pronounced as: "twenty-two eleven" o’clock
2211 in military time is 10:11 PM in regular time
:
Converstion Samples
## 24-hour to 12-hour conversion
Our "Military time converter" is the perfect tool to convert time from a 24-hour system to a 12-hour system. Try it above. Enter both the "Hours" and the "Minutes" boxs in the form, then click "convert to 12-hour". After that, you can see the result.
The 24-hour system (it's called the Military system) is larger than 12:00. And it starts at 00:00 and ends at 23:00. 00:00 means 12:00 midnight, and 23:00 means 11:00 PM. (Note: 24:00 and 00:00 are the same time).
### How was 22:11 converted to 10:11 PM ?
1. Step One: To convert 22:11 (twenty-two eleven) to 12-hour, firstly, we didn't convert the minutes, we leave it. In our case it will be ( :11 eleven ). and we just convert the hours.
2. Step Two: If the number of hours on military time (24-hour system) that we need to convert it, is less than 12:00, it will be the same on regular time (12-hour system).
In our case, It's 22 (ten), so we subtract 12 from 22 then add PM like ( 22 - 12 ). It will be 10 PM. (10 PM means 10 in the noon).
The final result is:10:11 PM ( In words: ten eleven in the noon ). Look at the chart below.
Military Time (24-hour system)
24-Hour 12-hour 24-Hour 12-hour
00:11 Midnight 12:11 Noon
01:11 1:11 a.m. 13:11 1:11 p.m.
02:11 2:11 a.m. 14:11 2:11 p.m.
03:11 3:11 a.m. 15:11 3:11 p.m.
04:11 4:11 a.m. 16:11 4:11 p.m.
05:11 5:11 a.m. 17:11 5:11 p.m.
06:11 6:11 a.m. 18:11 6:11 p.m.
07:11 7:11 a.m. 19:11 7:11 p.m.
08:11 8:11 a.m. 20:11 8:11 p.m.
09:11 9:11 a.m. 21:11 9:11 p.m.
10:11 10:11 a.m. 22:11 10:11 p.m.
11:11 11:11 a.m. 23:11 11:11 p.m.
### 22:11 Clock in different Time Zone
See, The 22:11 time in the other military time zones. Look at the table below.
Time Zone UTC offset 12-hour 24-hour
Yankee UTC-12 10:11 AM 10:11
X-ray UTC-11 11:11 AM 11:11
Whiskey UTC-10 12:11 PM 12:11
Victor UTC-9 01:11 PM 13:11
Uniform UTC-8 02:11 PM 14:11
Tango UTC-7 03:11 PM 15:11
Sierra UTC-6 04:11 PM 16:11
Romeo UTC-5 05:11 PM 17:11
Quebec UTC-4 06:11 PM 18:11
Papa UTC-3 07:11 PM 19:11
Oscar UTC-2 08:11 PM 20:11
November UTC-1 09:11 PM 21:11
Zulu UTC±0 10:11 PM 22:11
Alpha UTC+1 11:11 PM 23:11
Bravo UTC+2 12:11 AM 00:11
Charlie UTC+3 01:11 AM 01:11
Delta UTC+4 02:11 AM 02:11
Echo UTC+5 03:11 AM 03:11
Foxtrot UTC+6 04:11 AM 04:11
Golf UTC+7 05:11 AM 05:11
Hotel UTC+8 06:11 AM 06:11
India UTC+9 07:11 AM 07:11
Kilo UTC+10 08:11 AM 08:11
Lima UTC+11 09:11 AM 09:11
Mike UTC+12 10:11 AM 10:11 | 1,084 | 2,603 | {"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} | 3.28125 | 3 | CC-MAIN-2023-40 | latest | en | 0.801816 |
http://freecode.com/tags/puzzle?page=1&sort=created_at&with=66&without= | 1,371,723,066,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368711406217/warc/CC-MAIN-20130516133646-00059-ip-10-60-113-184.ec2.internal.warc.gz | 107,720,715 | 10,303 | # 12 projects tagged "puzzle"
## Updated 02 Dec 2012 Cube Trains
Pop 48.74 1.43
Cube Trains is a puzzle game where you build elevated railways in a city. It features 3D puzzles, unlimited undo/redo, and smart placement of pieces. The non-linear level graph means you don't have to beat a level before unlocking the next one. The game comes with a built-in level editor, so you can build and share your own corners of the city.
## Updated 07 Apr 2012 4digits
Pop 23.77 21
4digits is a guess-the-number puzzle game. It's called Bulls and Cows, and in China people simply call it Guess-the-Number. The game's objective is to guess a four-digit number in eight tries, using as little time as possible.
## Updated 04 Mar 2012 2x2
Pop 27.68 1
2x2 (Two by Two) is a multiplication puzzle game. Your goal is to solve a multiplication puzzle in which all digits are replaced by letters. To achieve the best score, you should solve the puzzle with a minimal number of errors and in minimal time. 2x2 is easy to learn, but hard to master. It is simple, smart, and addicting. It will allow you to test and improve your math skills, calculation skills, logic, and even knowledge of probability theory. With a large number of possible combinations, no two games will be the same.
## Updated 24 Jan 2011 Smart Challenge
Pop 18.14 29.62
Smart Challenge is a computer game to exercise logical thought by solving millions of randomly generated tangram puzzles. It allows you to create and store new puzzles and display the ranking of solution times for each puzzle. Additionally, it has levels of difficulty for the random generated tangram puzzles.
## Updated 16 Dec 2010 HexGlass
Pop 30.45 1.47
HexGlass is a Tetris-like puzzle game based on a hexagonal grid. Ten different types of blocks continuously fall from above and you must arrange them to make horizontal rows of hexagonal bricks. Completing any row causes those hexagonal blocks to disappear and those above move downwards. The blocks gradually fall faster and the game is over when the screen fills up and blocks can no longer fall from the top.
## Updated 31 Aug 2010 CoolDown
Pop 26.57 1
CoolDown is a puzzle game inspired by the old game PipeMania, although it changed a lot during development. Your task is to cool down the ball coming from the start field. It has 100 levels, increasing difficulty, pipes in 4 colors, special tiles, and multiple balls. It is available for Caanoo and WIZ, and can be compiled for Linux, Mac OS X, and Windows.
## Updated 15 May 2010 WordSeeker
Pop 24.21 33.64
Wordseeker is a casual word game. The player's goal is to find the words hidden by the computer on a letter grid filled with random letters.
## Updated 05 Apr 2010 Brain Party
Pop 52.42 1
Brain Party is a fun, family-friendly puzzle game that's made up of 36 mini-games designed to push your brain to its limits by testing memory, logic, mathematics, reaction time, and more. It is split into two modes: test mode gives you five mini-games in a row and adds up your brain weight to see how smart you are, and practice mode lets you play your favorite mini-games as often as you want. If you do well enough, there are six extra mini-games that can be unlocked, as well as a new game mode designed to keep you playing and enjoying your favorite games even longer.
## Updated 14 Jul 2010 Chroma-puzzle
Pop 39.12 2.63
Chroma is an abstract puzzle game in which a variety of colorful shapes are arranged in a series of increasingly complex patterns, forming fiendish traps that must be disarmed and mysterious puzzles that must be manipulated in order to give up their subtle secrets. Initially, it is so straightforward that anyone can pick it up and begin to play, yet it gradually becomes difficult enough to tax even the brightest of minds. It features twenty-one levels, ranging from beginner to expert; infinite undo and redo capability, as well as replaying of solutions; a choice of smooth graphics or a minimal, text based version; and a level editor to allow you to design your own puzzles.
## Updated 16 Jul 2012 Brukkon
Pop 28 1.01
Brukkon is a puzzle game where you have to guide a little robot to its spare parts by moving platforms and steering the robot. The game has 30 levels and online high scores.
## Project Spotlight
### Siege
An HTTP regression testing/benchmarking utility.
## Project Spotlight
### Black Hole Solitaire Solver
An automated solver for the "Black Hole" and "All in a Row" solitaire/patience card games. | 1,065 | 4,525 | {"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} | 3.28125 | 3 | CC-MAIN-2013-20 | longest | en | 0.932043 |
http://sophie.zarb.org/distrib/Mageia/6/x86_64/by-pkgid/4851f934cbb16d8a455182dfe902e59f/files/23 | 1,566,490,246,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317274.5/warc/CC-MAIN-20190822151657-20190822173657-00286.warc.gz | 180,812,610 | 5,535 | Sophie
## arpack-3.3.0-3.mga6.x86_64.rpm
``` program dnbdr1
c
c ... Construct the matrix A in LAPACK-style band form.
c The matrix A is derived from the discretization of
c the 2-d convection-diffusion operator
c
c -Laplacian(u) + rho*partial(u)/partial(x).
c
c on the unit square with zero Dirichlet boundary condition
c using standard central difference.
c
c ... Call DNBAND to find eigenvalues LAMBDA such that
c A*x = LAMBDA*x.
c
c ... Use mode 1 of DNAUPD .
c
c\BeginLib
c
c dnband ARPACK banded eigenproblem solver.
c dlapy2 LAPACK routine to compute sqrt(x**2+y**2) carefully.
c dlaset LAPACK routine to initialize a matrix to zero.
c daxpy Level 1 BLAS that computes y <- alpha*x+y.
c dnrm2 Level 1 BLAS that computes the norm of a vector.
c dgbmv Level 2 BLAS that computes the band matrix vector product
c
c\Author
c Richard Lehoucq
c Danny Sorensen
c Chao Yang
c Dept. of Computational &
c Applied Mathematics
c Rice University
c Houston, Texas
c
c\SCCS Information: @(#)
c FILE: nbdr1.F SID: 2.5 DATE OF SID: 08/26/96 RELEASE: 2
c
c\Remarks
c 1. None
c
c\EndLib
c
c----------------------------------------------------------------------
c
c %-------------------------------------%
c | Define leading dimensions for all |
c | arrays. |
c | MAXN - Maximum size of the matrix |
c | MAXNEV - Maximum number of |
c | eigenvalues to be computed |
c | MAXNCV - Maximum number of Arnoldi |
c | vectors stored |
c | MAXBDW - Maximum bandwidth |
c %-------------------------------------%
c
integer maxn, maxnev, maxncv, maxbdw, lda,
& lworkl, ldv
parameter ( maxn = 1000, maxnev = 25, maxncv=50,
& maxbdw=50, lda = maxbdw, ldv = maxn )
c
c %--------------%
c | Local Arrays |
c %--------------%
c
integer iparam(11), iwork(maxn)
logical select(maxncv)
Double precision
& a(lda,maxn), m(lda,maxn), rfac(lda,maxn),
& workl(3*maxncv*maxncv+6*maxncv), workd(3*maxn),
& workev(3*maxncv), v(ldv, maxncv),
& resid(maxn), d(maxncv, 3), ax(maxn)
Complex*16
& cfac(lda, maxn), workc(maxn)
c
c %---------------%
c | Local Scalars |
c %---------------%
c
character which*2, bmat
integer nev, ncv, ku, kl, info, i, j, ido,
& n, nx, lo, isub, isup, idiag, mode, maxitr,
& nconv
logical rvec, first
Double precision
& tol, rho, h, h2, sigmar, sigmai
c
c %------------%
c | Parameters |
c %------------%
c
Double precision
& one, zero, two
parameter (one = 1.0D+0 , zero = 0.0D+0 ,
& two = 2.0D+0 )
c
c %-----------------------------%
c | BLAS & LAPACK routines used |
c %-----------------------------%
c
Double precision
& dlapy2 , dnrm2
external dlapy2 , dnrm2 , dgbmv , daxpy
c
c %--------------------%
c | Intrinsic function |
c %--------------------%
c
intrinsic abs
c
c %-----------------------%
c | Executable Statements |
c %-----------------------%
c
c %-------------------------------------------------%
c | The number NX is the number of interior points |
c | in the discretization of the 2-dimensional |
c | convection-diffusion operator on the unit |
c | square with zero Dirichlet boundary condition. |
c | The number N(=NX*NX) is the dimension of the |
c | matrix. A standard eigenvalue problem is |
c | solved (BMAT = 'I'). NEV is the number of |
c | eigenvalues to be approximated. The user can |
c | modify NX, NEV, NCV, WHICH to solve problems of |
c | different sizes, and to get different parts the |
c | spectrum. However, The following conditions |
c | must be satisfied: |
c | N <= MAXN |
c | NEV <= MAXNEV |
c | NEV + 2 <= NCV <= MAXNCV |
c %-------------------------------------------------%
c
nx = 10
n = nx*nx
nev = 4
ncv = 10
if ( n .gt. maxn ) then
print *, ' ERROR with _NBDR1: N is greater than MAXN '
go to 9000
else if ( nev .gt. maxnev ) then
print *, ' ERROR with _NBDR1: NEV is greater than MAXNEV '
go to 9000
else if ( ncv .gt. maxncv ) then
print *, ' ERROR with _NBDR1: NCV is greater than MAXNCV '
go to 9000
end if
bmat = 'I'
which = 'SM'
c
c %-----------------------------------------------------%
c | The work array WORKL is used in DNAUPD as |
c | workspace. Its dimension LWORKL is set as |
c | illustrated below. The parameter TOL determines |
c | the stopping criterion. If TOL<=0, machine |
c | precision is used. The variable IDO is used for |
c | reverse communication, and is initially set to 0. |
c | Setting INFO=0 indicates that a random vector is |
c | generated in DNAUPD to start the Arnoldi iteration. |
c %-----------------------------------------------------%
c
lworkl = 3*ncv**2+6*ncv
tol = zero
ido = 0
info = 0
c
c %---------------------------------------------------%
c | IPARAM(3) specifies the maximum number of Arnoldi |
c | iterations allowed. Mode 1 of DNAUPD is used |
c | (IPARAM(7) = 1). All these options can be changed |
c | by the user. For details, see the documentation |
c | in DNBAND . |
c %---------------------------------------------------%
c
maxitr = 300
mode = 1
c
iparam(3) = maxitr
iparam(7) = mode
c
c %----------------------------------------%
c | Construct the matrix A in LAPACK-style |
c | banded form. |
c %----------------------------------------%
c
c %---------------------------------------------%
c | Zero out the workspace for banded matrices. |
c %---------------------------------------------%
c
call dlaset ('A', lda, n, zero, zero, a, lda)
call dlaset ('A', lda, n, zero, zero, m, lda)
call dlaset ('A', lda, n, zero, zero, rfac, lda)
c
c %-------------------------------------%
c | KU, KL are number of superdiagonals |
c | and subdiagonals within the band of |
c | matrices A and M. |
c %-------------------------------------%
c
kl = nx
ku = nx
c
c %---------------%
c | Main diagonal |
c %---------------%
c
h = one / dble (nx+1)
h2 = h*h
c
idiag = kl+ku+1
do 30 j = 1, n
a(idiag,j) = 4.0D+0 / h2
30 continue
c
c %-------------------------------------%
c | First subdiagonal and superdiagonal |
c %-------------------------------------%
c
rho = 1.0D+2
isup = kl+ku
isub = kl+ku+2
do 50 i = 1, nx
lo = (i-1)*nx
do 40 j = lo+1, lo+nx-1
a(isup,j+1) = -one/h2 + rho/two/h
a(isub,j) = -one/h2 - rho/two/h
40 continue
50 continue
c
c %------------------------------------%
c | KL-th subdiagonal and KU-th super- |
c | diagonal. |
c %------------------------------------%
c
isup = kl+1
isub = 2*kl+ku+1
do 80 i = 1, nx-1
lo = (i-1)*nx
do 70 j = lo+1, lo+nx
a(isup,nx+j) = -one / h2
a(isub,j) = -one / h2
70 continue
80 continue
c
c %------------------------------------------------%
c | Call ARPACK banded solver to find eigenvalues |
c | and eigenvectors. The real parts of the |
c | eigenvalues are returned in the first column |
c | of D, the imaginary parts are returned in the |
c | second column of D. Eigenvectors are returned |
c | in the first NCONV (=IPARAM(5)) columns of V. |
c %------------------------------------------------%
c
rvec = .true.
call dnband (rvec, 'A', select, d, d(1,2), v, ldv, sigmar, sigmai,
& workev, n, a, m, lda, rfac, cfac, kl, ku, which,
& bmat, nev, tol, resid, ncv, v, ldv, iparam, workd,
& workl, lworkl, workc, iwork, info)
c
if ( info .eq. 0) then
c
c %-----------------------------------%
c | Print out convergence information |
c %-----------------------------------%
c
nconv = iparam(5)
c
print *, ' '
print *, ' _NBDR1 '
print *, ' ====== '
print *, ' '
print *, ' The size of the matrix is ', n
print *, ' Number of eigenvalue requested is ', nev
print *, ' The number of Arnoldi vectors generated',
& ' (NCV) is ', ncv
print *, ' The number of converged Ritz values is ',
& nconv
print *, ' What portion of the spectrum ', which
print *, ' The number of Implicit Arnoldi ',
& ' update taken is ', iparam(3)
print *, ' The number of OP*x is ', iparam(9)
print *, ' The convergence tolerance is ', tol
print *, ' '
c
c %----------------------------%
c | Compute the residual norm. |
c | || A*x - lambda*x || |
c %----------------------------%
c
first = .true.
do 90 j = 1, nconv
c
if ( d(j,2) .eq. zero ) then
c
c %--------------------%
c | Ritz value is real |
c %--------------------%
c
call dgbmv ('Notranspose', n, n, kl, ku, one,
& a(kl+1,1), lda, v(1,j), 1, zero,
& ax, 1)
call daxpy (n, -d(j,1), v(1,j), 1, ax, 1)
d(j,3) = dnrm2 (n, ax, 1)
d(j,3) = d(j,3) / abs(d(j,1))
c
else if ( first ) then
c
c %------------------------%
c | Ritz value is complex |
c | Residual of one Ritz |
c | value of the conjugate |
c | pair is computed. |
c %------------------------%
c
call dgbmv ('Notranspose', n, n, kl, ku, one,
& a(kl+1,1), lda, v(1,j), 1, zero,
& ax, 1)
call daxpy (n, -d(j,1), v(1,j), 1, ax, 1)
call daxpy (n, d(j,2), v(1,j+1), 1, ax, 1)
d(j,3) = dnrm2 (n, ax, 1)
call dgbmv ('Notranspose', n, n, kl, ku, one,
& a(kl+1,1), lda, v(1,j+1), 1, zero,
& ax, 1)
call daxpy (n, -d(j,2), v(1,j), 1, ax, 1)
call daxpy (n, -d(j,1), v(1,j+1), 1, ax, 1)
d(j,3) = dlapy2 ( d(j,3), dnrm2 (n, ax, 1) )
d(j,3) = d(j,3) / dlapy2 (d(j,1),d(j,2))
d(j+1,3) = d(j,3)
first = .false.
else
first = .true.
end if
c
90 continue
call dmout (6, nconv, 3, d, maxncv, -6,
& 'Ritz values (Real,Imag) and relative residuals')
else
c
c %-------------------------------------%
c | Either convergence failed, or there |
c | is error. Check the documentation |
c | for DNBAND . |
c %-------------------------------------%
c
print *, ' '
print *, ' Error with _nband, info= ', info
print *, ' Check the documentation of _nband '
print *, ' '
c
end if
c
9000 end
``` | 3,318 | 10,983 | {"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-2019-35 | longest | en | 0.574299 |
https://calculator.academy/angle-of-incidence-calculator/ | 1,695,289,011,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233505362.29/warc/CC-MAIN-20230921073711-20230921103711-00463.warc.gz | 187,342,519 | 57,227 | Enter the refractive indices of two different mediums and the angle of refraction into the calculator to determine the angle of incidence.
## Angle of Incidence Formula
The following formula is used to calculate the angle of incidence.
θ1 = sin^{-1} ( n2 * sin (θ2) / n1 )
• Where θ1 is the angle of incidence
• θ2 is the refraction angle
• n2 is the refractive index of medium 2
• n1 is the refractive index of medium 1
To calculate the angle of incidence, multiply the refractive index of the 2nd medium by the sine of the refraction angle, then divide by the refractive index of the 1st medium. Finally, take this result and take the inverse sine of it to get the angle of incidence.
## Angle of Incidence Definition
An angle of incidence is defined as the angle between the ray incident on a surface and the line perpendicular to the surface.
## Angle of Incidence Example
How to calculate the angle of incidence?
1. First, determine the refractive indices.
Calculate the refractive index of both mediums.
2. Next, determine the angle of refraction.
Calculate or measure the refraction angle.
3. Finally, calculate the angle of incidence.
Calculate the angle of incidence using the equation above.
## FAQ
What is a refraction?
A refraction is defined as the change in the relative angle of reflected light based on the speed of light through two different mediums. | 310 | 1,386 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2023-40 | latest | en | 0.871965 |
https://blog.cdxtech.com/post/CDXZipStream-Straight-Line-and-Driving-Distance-Calculations | 1,721,356,037,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.33/warc/CC-MAIN-20240719012903-20240719042903-00225.warc.gz | 125,253,215 | 9,387 | CDXZipStream supports a variety of distance formulas including CDXDistance, CDXRouteMP and CDXDistance2WP. Each is optimized for different situations and involves varying times to calculate.
CDXDistance is used to calculate straight-line distance between zip codes. You simply reference two zip codes and the CDXDistance formula returns the distance in either miles or kilometers. The custom formula calls our custom database to return latitude and longitude for the zip code, and the distance calculation is performed based on this data. It is relatively fast and should be used where the central location of a zip code can be used to approximate distance between locations.
If you need driving distance or time you will need to use either CDXRouteBing, which uses Bing Maps as the source of mapping data, or CDXRouteMP, which works in conjunction with the desktop version of Microsoft MapPoint. The calculation times for this are relatively slower at about 1 second per route, with more complex routes taking more time. If you are going to make thousands of calculations you should dedicate your PC to this during off hours or run this on a separate machine. For long lists of data we recommend using our free driving distance template that can be downloaded from our website here. Please view the video below to see how it works:
or watch the YouTube version here: Driving Distance and Time Calculations in an Excel Template
The fastest calculation is CDXDistance2WP, which uses latitude/longitude pairs to perform straight line distance calculation. You can geocode a list of addresses and then use this function to calculate exact distances. We recommend this when trying to analyze long address lists.
A typical task may require calculating driving time or distance for a large matrix of addresses or zip codes. For instance, it may be necessary to determine which customers in a list of thousands are within a one hour driving time of multiple store locations. The best way to handle this is a two-step process. First, we can get the latitude and longitude of each customer address and use CDXDistance2WP to calculate the straight-line distance from each store. Then we can filter this list for customers that are within 100 miles of each store, and use CDXRouteBing or CDXRouteMP to get the exact driving time for this smaller list. Using a combination of distance calculations in this case allows for fairly fast analysis of very large data sets, without sacrificing accuracy. | 515 | 2,504 | {"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-2024-30 | latest | en | 0.885174 |
https://www.hpmuseum.org/forum/archive/index.php?thread-526.html | 1,713,415,196,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817187.10/warc/CC-MAIN-20240418030928-20240418060928-00225.warc.gz | 769,694,467 | 3,529 | # HP Forums
Full Version: Worse than Bisection???!!!!
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I thought that the Bisection method was the slowest root-seeking method for nonlinear functions. I set out, for the pure fun of it, to write an algorithm that can actually do worse!!! The proposed method starts at a point X and marches (in positive or negative steps) towards the targeted root. When the method detects that the function at the current value of X has changed sign, it switched the sign of the step value and reduces it by 2. Thus, the method (which I call Dancer) dances around the root until the search step falls below a tolerance value. The method is very much influenced by how close you choose the initial X to the root and by the initial step size. I did contemplate sub-steps to accelerate the march towards the root, but I was concerned that I would create problems when the nonlinear function has multiple roots that lie close to each other.
Here is the pseudo-code:
Code:
```Give starting value X, Step dx, and tolerance value toler: Fx2=f(x) Repeat Fx1=Fx2 x=x+dx Fx2=f(X) If Fx1*Fx2 < 0 Then dx = -dx/2 End Until Abs(dx) < toler Return x```
I implemented the above algorithm in Excel VBA, along with code for the Bisection method. The latter method did much better in all of the tests I conducted. The Dancer method took 30% to 100% more iterations to get the answer!!
Please no hate mail for this mediocre method. :-)
A variant of this method using 10 instead of 2 is used to calculate the square root in HP-calculators. But of course it takes advantage of the fact that \(f(x_{n+1})\) can be calculated easily based on \(f(x_{n})\). Thus it's not a bad method per se.
Cheers
Thomas
Thomas,
I guess you can use a factor of 10 instead of 2. The method just came to me as I was going to sleep two nights ago. It works, but it is sloooooooooooow.
namir
(01-26-2014 07:06 PM)Namir Wrote: [ -> ]It works, but it is sloooooooooooow.
I guess Dave Cochran would like to have a talk with you.
Hi Namir
I've really enjoyed your explorations into root finding and integration. I've been using bisection or Newton's for years, and have a few Excel macros for it. So I may not implement this one.
As long as you are investigating convergence strategies, you might consider looking at Golden Section or Fibonacci series for reducing the bracketing interval and gaining significant digits.
Reguli Falsi with bracketing can be worse than bisection on many functions. It's like the Secant Method but keeps the root between estimates. One sets X2=(F(X1)-F(X0))/(X1-X0). If (for example) the function is sort of a curved L-shape and one point is above the X-axis near the upper leg and the other above the X-axis way out on the lower leg, the function will creep along chopping the interval a small amount at each time.
Using the Secant Method and switching to Bisection if the point falls outside works well in practice as it at worse becomes Bisection.
Richard A Davis suggests in his new book Practical Numerical Methods for Chemical Engineers on page 187 the following remedies (which I am paraphrasing):
1) If a remains unchanged after two iterations use:
x = b - f(b)(a - b)/(f(b) - 0.5 f(a))
Otherwise, if b remains unchanged after two iterations use:
x = b - 0.5 f(b)(a-b)/(0.5 f(b) - f(a))
Where [a, b] is the root-bracketing interval for f(x) = 0.
Reference URL's
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http://mathhelpforum.com/trigonometry/38219-roots-complex-numbers.html | 1,481,164,184,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542323.80/warc/CC-MAIN-20161202170902-00235-ip-10-31-129-80.ec2.internal.warc.gz | 178,652,230 | 10,300 | # Thread: roots of complex numbers
1. ## roots of complex numbers
a) find all roots of the equation z^5=sqrt(3)*i-1, present your answer in polar form
b) express (2-2i)^5 in exact polar form
(-1+sqrt(3)*i)^4
I was alright until the whole polar form thing, your help would be greatly appreciated.
2. Originally Posted by samdmansam
a) find all roots of the equation z^5=sqrt(3)*i-1, present your answer in polar form
For a) what is the polar form of $-1 + i\sqrt{3}$?
$r~cos(\theta) + ir~sin(\theta) = -1 + i\sqrt{3}$
So
$r~cos(\theta) = -1$
and
$r~sin(\theta) = \sqrt{3}$
Thus
$\frac{r~sin(\theta)}{r~cos(\theta)} = -\sqrt{3}$
Now,
$tan(\theta) = \left | -\sqrt{3} \right | \implies \theta = \frac{\pi}{3}$
and since sine is positive and cosine negative we know the reference angle is in QII. Thus
$\theta = \frac{2 \pi}{3}$
Then
$r~cos(\theta) = -1$
$r \cdot -\frac{1}{2} = -1$
$r = 2$
So
$z^5 = 2e^{2i \pi / 3}$
Can you solve it from here? (Please note that there are five answers to this.)
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A dimensionally consistent equation
1. is necessarily correct
2. is not necessarily correct
3. is always correct
4. is nearly correct
When two quantities are multiplied or divided, the relative error in the result is
1. the difference of the absolute errors in the multipliers
2. the sum of the absolute errors in the multipliers
3. the difference of the relative errors in the multipliers
4. the sum of the relative errors in the multipliers
According to the principle of homogeneity of dimensions
1. we can add or subtract any physical quantities
2. we can add or subtract any physical quantities
3. we can add or subtract similar physical quantities
4. we can subtract similar physical quantities
The units of pressure are the same as the units of
1. stress
2. plasticity
3. strain
4. elasticity
How many meters does a vehicle moving with a speed of 18 km h−1h−1 covers in 1 s
1. 4.5
2. 5.0
3. 4
4. 6
0
0
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https://www.vacations.info/metric/lbs.php?pounds=201.5 | 1,540,202,764,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515029.82/warc/CC-MAIN-20181022092330-20181022113830-00377.warc.gz | 1,134,409,622 | 13,865 | #### What is 201.5lbs in kg?
How many kilograms are in 201.5 pounds? Convert 201 and a 1/2 pounds to kg. Use this calculator to find out how much is 201 and a half lbs in kg. To calculate, take the number of pounds and divide by 2.20462262185, or just 2.2.
How much does 201.5 pounds weigh in kilograms? How heavy? How light? 201.5lbs equals 91398.862554949 grams 201.5lbstokg equals 91.398862554949 91.398862554949 kilograms is the same weight as 201.5poundstokilograms
Pounds Kilograms 0.25 lbs 0.11 0.5 lbs 0.23 0.75 lbs 0.34 1 lbs 0.45 1.25 lbs 0.57 1.5 lbs 0.68 1.75 lbs 0.79 2 lbs 0.91 2.25 lbs 1.02 2.5 lbs 1.13 2.75 lbs 1.25 3 lbs 1.36 3.25 lbs 1.47 3.5 lbs 1.59 3.75 lbs 1.7 4 lbs 1.81 4.25 lbs 1.93 4.5 lbs 2.04 4.75 lbs 2.15 5 lbs 2.27 5.25 lbs 2.38 5.5 lbs 2.49 5.75 lbs 2.61 6 lbs 2.72 6.25 lbs 2.83 6.5 lbs 2.95 6.75 lbs 3.06 7 lbs 3.18 7.25 lbs 3.29 7.5 lbs 3.4 7.75 lbs 3.52 8 lbs 3.63 8.25 lbs 3.74 8.5 lbs 3.86 8.75 lbs 3.97 9 lbs 4.08 9.25 lbs 4.2 9.5 lbs 4.31 9.75 lbs 4.42 10 lbs 4.54 10.25 lbs 4.65 10.5 lbs 4.76 10.75 lbs 4.88 11 lbs 4.99 11.25 lbs 5.1 11.5 lbs 5.22 11.75 lbs 5.33 12 lbs 5.44 12.25 lbs 5.56 12.5 lbs 5.67 12.75 lbs 5.78 13 lbs 5.9 13.25 lbs 6.01 13.5 lbs 6.12 13.75 lbs 6.24 14 lbs 6.35 14.25 lbs 6.46 14.5 lbs 6.58 14.75 lbs 6.69 15 lbs 6.8 15.25 lbs 6.92 15.5 lbs 7.03 15.75 lbs 7.14 16 lbs 7.26 16.25 lbs 7.37 16.5 lbs 7.48 16.75 lbs 7.6 17 lbs 7.71 17.25 lbs 7.82 17.5 lbs 7.94 17.75 lbs 8.05 18 lbs 8.16 18.25 lbs 8.28 18.5 lbs 8.39 18.75 lbs 8.5 19 lbs 8.62 19.25 lbs 8.73 19.5 lbs 8.85 19.75 lbs 8.96 20 lbs 9.07 20.25 lbs 9.19 20.5 lbs 9.3 20.75 lbs 9.41 21 lbs 9.53 21.25 lbs 9.64 21.5 lbs 9.75 21.75 lbs 9.87 22 lbs 9.98 22.25 lbs 10.09 22.5 lbs 10.21 22.75 lbs 10.32 23 lbs 10.43 23.25 lbs 10.55 23.5 lbs 10.66 23.75 lbs 10.77 24 lbs 10.89 24.25 lbs 11 24.5 lbs 11.11 24.75 lbs 11.23 25 lbs 11.34 25.25 lbs 11.45 25.5 lbs 11.57 25.75 lbs 11.68 26 lbs 11.79 26.25 lbs 11.91 26.5 lbs 12.02 26.75 lbs 12.13 27 lbs 12.25 27.25 lbs 12.36 27.5 lbs 12.47 27.75 lbs 12.59 28 lbs 12.7 28.25 lbs 12.81 28.5 lbs 12.93 28.75 lbs 13.04 29 lbs 13.15 29.25 lbs 13.27 29.5 lbs 13.38 29.75 lbs 13.49 30 lbs 13.61 30.25 lbs 13.72 30.5 lbs 13.83 30.75 lbs 13.95 31 lbs 14.06 31.25 lbs 14.17 31.5 lbs 14.29 31.75 lbs 14.4 32 lbs 14.51 32.25 lbs 14.63 32.5 lbs 14.74 32.75 lbs 14.86 33 lbs 14.97 33.25 lbs 15.08 33.5 lbs 15.2 33.75 lbs 15.31 34 lbs 15.42 34.25 lbs 15.54 34.5 lbs 15.65 34.75 lbs 15.76 35 lbs 15.88 35.25 lbs 15.99 35.5 lbs 16.1 35.75 lbs 16.22 36 lbs 16.33 36.25 lbs 16.44 36.5 lbs 16.56 36.75 lbs 16.67 37 lbs 16.78 37.25 lbs 16.9 37.5 lbs 17.01 37.75 lbs 17.12 38 lbs 17.24 38.25 lbs 17.35 38.5 lbs 17.46 38.75 lbs 17.58 39 lbs 17.69 39.25 lbs 17.8 39.5 lbs 17.92 39.75 lbs 18.03 40 lbs 18.14 40.25 lbs 18.26 40.5 lbs 18.37 40.75 lbs 18.48 41 lbs 18.6 41.25 lbs 18.71 41.5 lbs 18.82 41.75 lbs 18.94 42 lbs 19.05 42.25 lbs 19.16 42.5 lbs 19.28 42.75 lbs 19.39 43 lbs 19.5 43.25 lbs 19.62 43.5 lbs 19.73 43.75 lbs 19.84 44 lbs 19.96 44.25 lbs 20.07 44.5 lbs 20.18 44.75 lbs 20.3 45 lbs 20.41 45.25 lbs 20.53 45.5 lbs 20.64 45.75 lbs 20.75 46 lbs 20.87 46.25 lbs 20.98 46.5 lbs 21.09 46.75 lbs 21.21 47 lbs 21.32 47.25 lbs 21.43 47.5 lbs 21.55 47.75 lbs 21.66 48 lbs 21.77 48.25 lbs 21.89 48.5 lbs 22 48.75 lbs 22.11 49 lbs 22.23 49.25 lbs 22.34 49.5 lbs 22.45 49.75 lbs 22.57 50 lbs 22.68 50.25 lbs 22.79 50.5 lbs 22.91 50.75 lbs 23.02 51 lbs 23.13 51.25 lbs 23.25 51.5 lbs 23.36 51.75 lbs 23.47 52 lbs 23.59 52.25 lbs 23.7 52.5 lbs 23.81 52.75 lbs 23.93 53 lbs 24.04 53.25 lbs 24.15 53.5 lbs 24.27 53.75 lbs 24.38 54 lbs 24.49 54.25 lbs 24.61 54.5 lbs 24.72 54.75 lbs 24.83 55 lbs 24.95 55.25 lbs 25.06 55.5 lbs 25.17 55.75 lbs 25.29 56 lbs 25.4 56.25 lbs 25.51 56.5 lbs 25.63 56.75 lbs 25.74 57 lbs 25.85 57.25 lbs 25.97 57.5 lbs 26.08 57.75 lbs 26.19 58 lbs 26.31 58.25 lbs 26.42 58.5 lbs 26.54 58.75 lbs 26.65 59 lbs 26.76 59.25 lbs 26.88 59.5 lbs 26.99 59.75 lbs 27.1 60 lbs 27.22 60.25 lbs 27.33 60.5 lbs 27.44 60.75 lbs 27.56 61 lbs 27.67 61.25 lbs 27.78 61.5 lbs 27.9 61.75 lbs 28.01 62 lbs 28.12 62.25 lbs 28.24 62.5 lbs 28.35 62.75 lbs 28.46 63 lbs 28.58 63.25 lbs 28.69 63.5 lbs 28.8 63.75 lbs 28.92 64 lbs 29.03 64.25 lbs 29.14 64.5 lbs 29.26 64.75 lbs 29.37 65 lbs 29.48 65.25 lbs 29.6 65.5 lbs 29.71 65.75 lbs 29.82 66 lbs 29.94 66.25 lbs 30.05 66.5 lbs 30.16 66.75 lbs 30.28 67 lbs 30.39 67.25 lbs 30.5 67.5 lbs 30.62 67.75 lbs 30.73 68 lbs 30.84 68.25 lbs 30.96 68.5 lbs 31.07 68.75 lbs 31.18 69 lbs 31.3 69.25 lbs 31.41 69.5 lbs 31.52 69.75 lbs 31.64 70 lbs 31.75 70.25 lbs 31.86 70.5 lbs 31.98 70.75 lbs 32.09 71 lbs 32.21 71.25 lbs 32.32 71.5 lbs 32.43 71.75 lbs 32.55 72 lbs 32.66 72.25 lbs 32.77 72.5 lbs 32.89 72.75 lbs 33 73 lbs 33.11 73.25 lbs 33.23 73.5 lbs 33.34 73.75 lbs 33.45 74 lbs 33.57 74.25 lbs 33.68 74.5 lbs 33.79 74.75 lbs 33.91 75 lbs 34.02
Pounds Kilograms 75.25 lbs 34.13 75.5 lbs 34.25 75.75 lbs 34.36 76 lbs 34.47 76.25 lbs 34.59 76.5 lbs 34.7 76.75 lbs 34.81 77 lbs 34.93 77.25 lbs 35.04 77.5 lbs 35.15 77.75 lbs 35.27 78 lbs 35.38 78.25 lbs 35.49 78.5 lbs 35.61 78.75 lbs 35.72 79 lbs 35.83 79.25 lbs 35.95 79.5 lbs 36.06 79.75 lbs 36.17 80 lbs 36.29 80.25 lbs 36.4 80.5 lbs 36.51 80.75 lbs 36.63 81 lbs 36.74 81.25 lbs 36.85 81.5 lbs 36.97 81.75 lbs 37.08 82 lbs 37.19 82.25 lbs 37.31 82.5 lbs 37.42 82.75 lbs 37.53 83 lbs 37.65 83.25 lbs 37.76 83.5 lbs 37.87 83.75 lbs 37.99 84 lbs 38.1 84.25 lbs 38.22 84.5 lbs 38.33 84.75 lbs 38.44 85 lbs 38.56 85.25 lbs 38.67 85.5 lbs 38.78 85.75 lbs 38.9 86 lbs 39.01 86.25 lbs 39.12 86.5 lbs 39.24 86.75 lbs 39.35 87 lbs 39.46 87.25 lbs 39.58 87.5 lbs 39.69 87.75 lbs 39.8 88 lbs 39.92 88.25 lbs 40.03 88.5 lbs 40.14 88.75 lbs 40.26 89 lbs 40.37 89.25 lbs 40.48 89.5 lbs 40.6 89.75 lbs 40.71 90 lbs 40.82 90.25 lbs 40.94 90.5 lbs 41.05 90.75 lbs 41.16 91 lbs 41.28 91.25 lbs 41.39 91.5 lbs 41.5 91.75 lbs 41.62 92 lbs 41.73 92.25 lbs 41.84 92.5 lbs 41.96 92.75 lbs 42.07 93 lbs 42.18 93.25 lbs 42.3 93.5 lbs 42.41 93.75 lbs 42.52 94 lbs 42.64 94.25 lbs 42.75 94.5 lbs 42.86 94.75 lbs 42.98 95 lbs 43.09 95.25 lbs 43.2 95.5 lbs 43.32 95.75 lbs 43.43 96 lbs 43.54 96.25 lbs 43.66 96.5 lbs 43.77 96.75 lbs 43.89 97 lbs 44 97.25 lbs 44.11 97.5 lbs 44.23 97.75 lbs 44.34 98 lbs 44.45 98.25 lbs 44.57 98.5 lbs 44.68 98.75 lbs 44.79 99 lbs 44.91 99.25 lbs 45.02 99.5 lbs 45.13 99.75 lbs 45.25 100 lbs 45.36 100.25 lbs 45.47 100.5 lbs 45.59 100.75 lbs 45.7 101 lbs 45.81 101.25 lbs 45.93 101.5 lbs 46.04 101.75 lbs 46.15 102 lbs 46.27 102.25 lbs 46.38 102.5 lbs 46.49 102.75 lbs 46.61 103 lbs 46.72 103.25 lbs 46.83 103.5 lbs 46.95 103.75 lbs 47.06 104 lbs 47.17 104.25 lbs 47.29 104.5 lbs 47.4 104.75 lbs 47.51 105 lbs 47.63 105.25 lbs 47.74 105.5 lbs 47.85 105.75 lbs 47.97 106 lbs 48.08 106.25 lbs 48.19 106.5 lbs 48.31 106.75 lbs 48.42 107 lbs 48.53 107.25 lbs 48.65 107.5 lbs 48.76 107.75 lbs 48.87 108 lbs 48.99 108.25 lbs 49.1 108.5 lbs 49.21 108.75 lbs 49.33 109 lbs 49.44 109.25 lbs 49.55 109.5 lbs 49.67 109.75 lbs 49.78 110 lbs 49.9 110.25 lbs 50.01 110.5 lbs 50.12 110.75 lbs 50.24 111 lbs 50.35 111.25 lbs 50.46 111.5 lbs 50.58 111.75 lbs 50.69 112 lbs 50.8 112.25 lbs 50.92 112.5 lbs 51.03 112.75 lbs 51.14 113 lbs 51.26 113.25 lbs 51.37 113.5 lbs 51.48 113.75 lbs 51.6 114 lbs 51.71 114.25 lbs 51.82 114.5 lbs 51.94 114.75 lbs 52.05 115 lbs 52.16 115.25 lbs 52.28 115.5 lbs 52.39 115.75 lbs 52.5 116 lbs 52.62 116.25 lbs 52.73 116.5 lbs 52.84 116.75 lbs 52.96 117 lbs 53.07 117.25 lbs 53.18 117.5 lbs 53.3 117.75 lbs 53.41 118 lbs 53.52 118.25 lbs 53.64 118.5 lbs 53.75 118.75 lbs 53.86 119 lbs 53.98 119.25 lbs 54.09 119.5 lbs 54.2 119.75 lbs 54.32 120 lbs 54.43 120.25 lbs 54.54 120.5 lbs 54.66 120.75 lbs 54.77 121 lbs 54.88 121.25 lbs 55 121.5 lbs 55.11 121.75 lbs 55.22 122 lbs 55.34 122.25 lbs 55.45 122.5 lbs 55.57 122.75 lbs 55.68 123 lbs 55.79 123.25 lbs 55.91 123.5 lbs 56.02 123.75 lbs 56.13 124 lbs 56.25 124.25 lbs 56.36 124.5 lbs 56.47 124.75 lbs 56.59 125 lbs 56.7 125.25 lbs 56.81 125.5 lbs 56.93 125.75 lbs 57.04 126 lbs 57.15 126.25 lbs 57.27 126.5 lbs 57.38 126.75 lbs 57.49 127 lbs 57.61 127.25 lbs 57.72 127.5 lbs 57.83 127.75 lbs 57.95 128 lbs 58.06 128.25 lbs 58.17 128.5 lbs 58.29 128.75 lbs 58.4 129 lbs 58.51 129.25 lbs 58.63 129.5 lbs 58.74 129.75 lbs 58.85 130 lbs 58.97 130.25 lbs 59.08 130.5 lbs 59.19 130.75 lbs 59.31 131 lbs 59.42 131.25 lbs 59.53 131.5 lbs 59.65 131.75 lbs 59.76 132 lbs 59.87 132.25 lbs 59.99 132.5 lbs 60.1 132.75 lbs 60.21 133 lbs 60.33 133.25 lbs 60.44 133.5 lbs 60.55 133.75 lbs 60.67 134 lbs 60.78 134.25 lbs 60.89 134.5 lbs 61.01 134.75 lbs 61.12 135 lbs 61.23 135.25 lbs 61.35 135.5 lbs 61.46 135.75 lbs 61.58 136 lbs 61.69 136.25 lbs 61.8 136.5 lbs 61.92 136.75 lbs 62.03 137 lbs 62.14 137.25 lbs 62.26 137.5 lbs 62.37 137.75 lbs 62.48 138 lbs 62.6 138.25 lbs 62.71 138.5 lbs 62.82 138.75 lbs 62.94 139 lbs 63.05 139.25 lbs 63.16 139.5 lbs 63.28 139.75 lbs 63.39 140 lbs 63.5 140.25 lbs 63.62 140.5 lbs 63.73 140.75 lbs 63.84 141 lbs 63.96 141.25 lbs 64.07 141.5 lbs 64.18 141.75 lbs 64.3 142 lbs 64.41 142.25 lbs 64.52 142.5 lbs 64.64 142.75 lbs 64.75 143 lbs 64.86 143.25 lbs 64.98 143.5 lbs 65.09 143.75 lbs 65.2 144 lbs 65.32 144.25 lbs 65.43 144.5 lbs 65.54 144.75 lbs 65.66 145 lbs 65.77 145.25 lbs 65.88 145.5 lbs 66 145.75 lbs 66.11 146 lbs 66.22 146.25 lbs 66.34 146.5 lbs 66.45 146.75 lbs 66.56 147 lbs 66.68 147.25 lbs 66.79 147.5 lbs 66.9 147.75 lbs 67.02 148 lbs 67.13 148.25 lbs 67.25 148.5 lbs 67.36 148.75 lbs 67.47 149 lbs 67.59 149.25 lbs 67.7 149.5 lbs 67.81 149.75 lbs 67.93 150 lbs 68.04
Pounds Kilograms 150.25 lbs 68.15 150.5 lbs 68.27 150.75 lbs 68.38 151 lbs 68.49 151.25 lbs 68.61 151.5 lbs 68.72 151.75 lbs 68.83 152 lbs 68.95 152.25 lbs 69.06 152.5 lbs 69.17 152.75 lbs 69.29 153 lbs 69.4 153.25 lbs 69.51 153.5 lbs 69.63 153.75 lbs 69.74 154 lbs 69.85 154.25 lbs 69.97 154.5 lbs 70.08 154.75 lbs 70.19 155 lbs 70.31 155.25 lbs 70.42 155.5 lbs 70.53 155.75 lbs 70.65 156 lbs 70.76 156.25 lbs 70.87 156.5 lbs 70.99 156.75 lbs 71.1 157 lbs 71.21 157.25 lbs 71.33 157.5 lbs 71.44 157.75 lbs 71.55 158 lbs 71.67 158.25 lbs 71.78 158.5 lbs 71.89 158.75 lbs 72.01 159 lbs 72.12 159.25 lbs 72.23 159.5 lbs 72.35 159.75 lbs 72.46 160 lbs 72.57 160.25 lbs 72.69 160.5 lbs 72.8 160.75 lbs 72.91 161 lbs 73.03 161.25 lbs 73.14 161.5 lbs 73.26 161.75 lbs 73.37 162 lbs 73.48 162.25 lbs 73.6 162.5 lbs 73.71 162.75 lbs 73.82 163 lbs 73.94 163.25 lbs 74.05 163.5 lbs 74.16 163.75 lbs 74.28 164 lbs 74.39 164.25 lbs 74.5 164.5 lbs 74.62 164.75 lbs 74.73 165 lbs 74.84 165.25 lbs 74.96 165.5 lbs 75.07 165.75 lbs 75.18 166 lbs 75.3 166.25 lbs 75.41 166.5 lbs 75.52 166.75 lbs 75.64 167 lbs 75.75 167.25 lbs 75.86 167.5 lbs 75.98 167.75 lbs 76.09 168 lbs 76.2 168.25 lbs 76.32 168.5 lbs 76.43 168.75 lbs 76.54 169 lbs 76.66 169.25 lbs 76.77 169.5 lbs 76.88 169.75 lbs 77 170 lbs 77.11 170.25 lbs 77.22 170.5 lbs 77.34 170.75 lbs 77.45 171 lbs 77.56 171.25 lbs 77.68 171.5 lbs 77.79 171.75 lbs 77.9 172 lbs 78.02 172.25 lbs 78.13 172.5 lbs 78.24 172.75 lbs 78.36 173 lbs 78.47 173.25 lbs 78.58 173.5 lbs 78.7 173.75 lbs 78.81 174 lbs 78.93 174.25 lbs 79.04 174.5 lbs 79.15 174.75 lbs 79.27 175 lbs 79.38 175.25 lbs 79.49 175.5 lbs 79.61 175.75 lbs 79.72 176 lbs 79.83 176.25 lbs 79.95 176.5 lbs 80.06 176.75 lbs 80.17 177 lbs 80.29 177.25 lbs 80.4 177.5 lbs 80.51 177.75 lbs 80.63 178 lbs 80.74 178.25 lbs 80.85 178.5 lbs 80.97 178.75 lbs 81.08 179 lbs 81.19 179.25 lbs 81.31 179.5 lbs 81.42 179.75 lbs 81.53 180 lbs 81.65 180.25 lbs 81.76 180.5 lbs 81.87 180.75 lbs 81.99 181 lbs 82.1 181.25 lbs 82.21 181.5 lbs 82.33 181.75 lbs 82.44 182 lbs 82.55 182.25 lbs 82.67 182.5 lbs 82.78 182.75 lbs 82.89 183 lbs 83.01 183.25 lbs 83.12 183.5 lbs 83.23 183.75 lbs 83.35 184 lbs 83.46 184.25 lbs 83.57 184.5 lbs 83.69 184.75 lbs 83.8 185 lbs 83.91 185.25 lbs 84.03 185.5 lbs 84.14 185.75 lbs 84.25 186 lbs 84.37 186.25 lbs 84.48 186.5 lbs 84.59 186.75 lbs 84.71 187 lbs 84.82 187.25 lbs 84.94 187.5 lbs 85.05 187.75 lbs 85.16 188 lbs 85.28 188.25 lbs 85.39 188.5 lbs 85.5 188.75 lbs 85.62 189 lbs 85.73 189.25 lbs 85.84 189.5 lbs 85.96 189.75 lbs 86.07 190 lbs 86.18 190.25 lbs 86.3 190.5 lbs 86.41 190.75 lbs 86.52 191 lbs 86.64 191.25 lbs 86.75 191.5 lbs 86.86 191.75 lbs 86.98 192 lbs 87.09 192.25 lbs 87.2 192.5 lbs 87.32 192.75 lbs 87.43 193 lbs 87.54 193.25 lbs 87.66 193.5 lbs 87.77 193.75 lbs 87.88 194 lbs 88 194.25 lbs 88.11 194.5 lbs 88.22 194.75 lbs 88.34 195 lbs 88.45 195.25 lbs 88.56 195.5 lbs 88.68 195.75 lbs 88.79 196 lbs 88.9 196.25 lbs 89.02 196.5 lbs 89.13 196.75 lbs 89.24 197 lbs 89.36 197.25 lbs 89.47 197.5 lbs 89.58 197.75 lbs 89.7 198 lbs 89.81 198.25 lbs 89.92 198.5 lbs 90.04 198.75 lbs 90.15 199 lbs 90.26 199.25 lbs 90.38 199.5 lbs 90.49 199.75 lbs 90.61 200 lbs 90.72 200.25 lbs 90.83 200.5 lbs 90.95 200.75 lbs 91.06 201 lbs 91.17 201.25 lbs 91.29 201.5 lbs 91.4 201.75 lbs 91.51 202 lbs 91.63 202.25 lbs 91.74 202.5 lbs 91.85 202.75 lbs 91.97 203 lbs 92.08 203.25 lbs 92.19 203.5 lbs 92.31 203.75 lbs 92.42 204 lbs 92.53 204.25 lbs 92.65 204.5 lbs 92.76 204.75 lbs 92.87 205 lbs 92.99 205.25 lbs 93.1 205.5 lbs 93.21 205.75 lbs 93.33 206 lbs 93.44 206.25 lbs 93.55 206.5 lbs 93.67 206.75 lbs 93.78 207 lbs 93.89 207.25 lbs 94.01 207.5 lbs 94.12 207.75 lbs 94.23 208 lbs 94.35 208.25 lbs 94.46 208.5 lbs 94.57 208.75 lbs 94.69 209 lbs 94.8 209.25 lbs 94.91 209.5 lbs 95.03 209.75 lbs 95.14 210 lbs 95.25 210.25 lbs 95.37 210.5 lbs 95.48 210.75 lbs 95.59 211 lbs 95.71 211.25 lbs 95.82 211.5 lbs 95.93 211.75 lbs 96.05 212 lbs 96.16 212.25 lbs 96.27 212.5 lbs 96.39 212.75 lbs 96.5 213 lbs 96.62 213.25 lbs 96.73 213.5 lbs 96.84 213.75 lbs 96.96 214 lbs 97.07 214.25 lbs 97.18 214.5 lbs 97.3 214.75 lbs 97.41 215 lbs 97.52 215.25 lbs 97.64 215.5 lbs 97.75 215.75 lbs 97.86 216 lbs 97.98 216.25 lbs 98.09 216.5 lbs 98.2 216.75 lbs 98.32 217 lbs 98.43 217.25 lbs 98.54 217.5 lbs 98.66 217.75 lbs 98.77 218 lbs 98.88 218.25 lbs 99 218.5 lbs 99.11 218.75 lbs 99.22 219 lbs 99.34 219.25 lbs 99.45 219.5 lbs 99.56 219.75 lbs 99.68 220 lbs 99.79 220.25 lbs 99.9 220.5 lbs 100.02 220.75 lbs 100.13 221 lbs 100.24 221.25 lbs 100.36 221.5 lbs 100.47 221.75 lbs 100.58 222 lbs 100.7 222.25 lbs 100.81 222.5 lbs 100.92 222.75 lbs 101.04 223 lbs 101.15 223.25 lbs 101.26 223.5 lbs 101.38 223.75 lbs 101.49 224 lbs 101.6 224.25 lbs 101.72 224.5 lbs 101.83 224.75 lbs 101.94 225 lbs 102.06
Pounds Kilograms 225.25 lbs 102.17 225.5 lbs 102.29 225.75 lbs 102.4 226 lbs 102.51 226.25 lbs 102.63 226.5 lbs 102.74 226.75 lbs 102.85 227 lbs 102.97 227.25 lbs 103.08 227.5 lbs 103.19 227.75 lbs 103.31 228 lbs 103.42 228.25 lbs 103.53 228.5 lbs 103.65 228.75 lbs 103.76 229 lbs 103.87 229.25 lbs 103.99 229.5 lbs 104.1 229.75 lbs 104.21 230 lbs 104.33 230.25 lbs 104.44 230.5 lbs 104.55 230.75 lbs 104.67 231 lbs 104.78 231.25 lbs 104.89 231.5 lbs 105.01 231.75 lbs 105.12 232 lbs 105.23 232.25 lbs 105.35 232.5 lbs 105.46 232.75 lbs 105.57 233 lbs 105.69 233.25 lbs 105.8 233.5 lbs 105.91 233.75 lbs 106.03 234 lbs 106.14 234.25 lbs 106.25 234.5 lbs 106.37 234.75 lbs 106.48 235 lbs 106.59 235.25 lbs 106.71 235.5 lbs 106.82 235.75 lbs 106.93 236 lbs 107.05 236.25 lbs 107.16 236.5 lbs 107.27 236.75 lbs 107.39 237 lbs 107.5 237.25 lbs 107.61 237.5 lbs 107.73 237.75 lbs 107.84 238 lbs 107.95 238.25 lbs 108.07 238.5 lbs 108.18 238.75 lbs 108.3 239 lbs 108.41 239.25 lbs 108.52 239.5 lbs 108.64 239.75 lbs 108.75 240 lbs 108.86 240.25 lbs 108.98 240.5 lbs 109.09 240.75 lbs 109.2 241 lbs 109.32 241.25 lbs 109.43 241.5 lbs 109.54 241.75 lbs 109.66 242 lbs 109.77 242.25 lbs 109.88 242.5 lbs 110 242.75 lbs 110.11 243 lbs 110.22 243.25 lbs 110.34 243.5 lbs 110.45 243.75 lbs 110.56 244 lbs 110.68 244.25 lbs 110.79 244.5 lbs 110.9 244.75 lbs 111.02 245 lbs 111.13 245.25 lbs 111.24 245.5 lbs 111.36 245.75 lbs 111.47 246 lbs 111.58 246.25 lbs 111.7 246.5 lbs 111.81 246.75 lbs 111.92 247 lbs 112.04 247.25 lbs 112.15 247.5 lbs 112.26 247.75 lbs 112.38 248 lbs 112.49 248.25 lbs 112.6 248.5 lbs 112.72 248.75 lbs 112.83 249 lbs 112.94 249.25 lbs 113.06 249.5 lbs 113.17 249.75 lbs 113.28 250 lbs 113.4 250.25 lbs 113.51 250.5 lbs 113.62 250.75 lbs 113.74 251 lbs 113.85 251.25 lbs 113.97 251.5 lbs 114.08 251.75 lbs 114.19 252 lbs 114.31 252.25 lbs 114.42 252.5 lbs 114.53 252.75 lbs 114.65 253 lbs 114.76 253.25 lbs 114.87 253.5 lbs 114.99 253.75 lbs 115.1 254 lbs 115.21 254.25 lbs 115.33 254.5 lbs 115.44 254.75 lbs 115.55 255 lbs 115.67 255.25 lbs 115.78 255.5 lbs 115.89 255.75 lbs 116.01 256 lbs 116.12 256.25 lbs 116.23 256.5 lbs 116.35 256.75 lbs 116.46 257 lbs 116.57 257.25 lbs 116.69 257.5 lbs 116.8 257.75 lbs 116.91 258 lbs 117.03 258.25 lbs 117.14 258.5 lbs 117.25 258.75 lbs 117.37 259 lbs 117.48 259.25 lbs 117.59 259.5 lbs 117.71 259.75 lbs 117.82 260 lbs 117.93 260.25 lbs 118.05 260.5 lbs 118.16 260.75 lbs 118.27 261 lbs 118.39 261.25 lbs 118.5 261.5 lbs 118.61 261.75 lbs 118.73 262 lbs 118.84 262.25 lbs 118.95 262.5 lbs 119.07 262.75 lbs 119.18 263 lbs 119.29 263.25 lbs 119.41 263.5 lbs 119.52 263.75 lbs 119.63 264 lbs 119.75 264.25 lbs 119.86 264.5 lbs 119.98 264.75 lbs 120.09 265 lbs 120.2 265.25 lbs 120.32 265.5 lbs 120.43 265.75 lbs 120.54 266 lbs 120.66 266.25 lbs 120.77 266.5 lbs 120.88 266.75 lbs 121 267 lbs 121.11 267.25 lbs 121.22 267.5 lbs 121.34 267.75 lbs 121.45 268 lbs 121.56 268.25 lbs 121.68 268.5 lbs 121.79 268.75 lbs 121.9 269 lbs 122.02 269.25 lbs 122.13 269.5 lbs 122.24 269.75 lbs 122.36 270 lbs 122.47 270.25 lbs 122.58 270.5 lbs 122.7 270.75 lbs 122.81 271 lbs 122.92 271.25 lbs 123.04 271.5 lbs 123.15 271.75 lbs 123.26 272 lbs 123.38 272.25 lbs 123.49 272.5 lbs 123.6 272.75 lbs 123.72 273 lbs 123.83 273.25 lbs 123.94 273.5 lbs 124.06 273.75 lbs 124.17 274 lbs 124.28 274.25 lbs 124.4 274.5 lbs 124.51 274.75 lbs 124.62 275 lbs 124.74 275.25 lbs 124.85 275.5 lbs 124.96 275.75 lbs 125.08 276 lbs 125.19 276.25 lbs 125.3 276.5 lbs 125.42 276.75 lbs 125.53 277 lbs 125.65 277.25 lbs 125.76 277.5 lbs 125.87 277.75 lbs 125.99 278 lbs 126.1 278.25 lbs 126.21 278.5 lbs 126.33 278.75 lbs 126.44 279 lbs 126.55 279.25 lbs 126.67 279.5 lbs 126.78 279.75 lbs 126.89 280 lbs 127.01 280.25 lbs 127.12 280.5 lbs 127.23 280.75 lbs 127.35 281 lbs 127.46 281.25 lbs 127.57 281.5 lbs 127.69 281.75 lbs 127.8 282 lbs 127.91 282.25 lbs 128.03 282.5 lbs 128.14 282.75 lbs 128.25 283 lbs 128.37 283.25 lbs 128.48 283.5 lbs 128.59 283.75 lbs 128.71 284 lbs 128.82 284.25 lbs 128.93 284.5 lbs 129.05 284.75 lbs 129.16 285 lbs 129.27 285.25 lbs 129.39 285.5 lbs 129.5 285.75 lbs 129.61 286 lbs 129.73 286.25 lbs 129.84 286.5 lbs 129.95 286.75 lbs 130.07 287 lbs 130.18 287.25 lbs 130.29 287.5 lbs 130.41 287.75 lbs 130.52 288 lbs 130.63 288.25 lbs 130.75 288.5 lbs 130.86 288.75 lbs 130.97 289 lbs 131.09 289.25 lbs 131.2 289.5 lbs 131.31 289.75 lbs 131.43 290 lbs 131.54 290.25 lbs 131.66 290.5 lbs 131.77 290.75 lbs 131.88 291 lbs 132 291.25 lbs 132.11 291.5 lbs 132.22 291.75 lbs 132.34 292 lbs 132.45 292.25 lbs 132.56 292.5 lbs 132.68 292.75 lbs 132.79 293 lbs 132.9 293.25 lbs 133.02 293.5 lbs 133.13 293.75 lbs 133.24 294 lbs 133.36 294.25 lbs 133.47 294.5 lbs 133.58 294.75 lbs 133.7 295 lbs 133.81 295.25 lbs 133.92 295.5 lbs 134.04 295.75 lbs 134.15 296 lbs 134.26 296.25 lbs 134.38 296.5 lbs 134.49 296.75 lbs 134.6 297 lbs 134.72 297.25 lbs 134.83 297.5 lbs 134.94 297.75 lbs 135.06 298 lbs 135.17 298.25 lbs 135.28 298.5 lbs 135.4 298.75 lbs 135.51 299 lbs 135.62 299.25 lbs 135.74 299.5 lbs 135.85 299.75 lbs 135.96 300 lbs 136.08 | 10,347 | 19,395 | {"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} | 2.546875 | 3 | CC-MAIN-2018-43 | latest | en | 0.475457 |
https://www.coursehero.com/file/5573280/ch21-p20/ | 1,493,516,467,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123635.74/warc/CC-MAIN-20170423031203-00498-ip-10-145-167-34.ec2.internal.warc.gz | 879,000,053 | 22,938 | # ch21_p20 - 20. If is the angle between the force and the...
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Unformatted text preview: 20. If is the angle between the force and the x-axis, then cos = x . x + d2 2 We note that, due to the symmetry in the problem, there is no y component to the net force on the third particle. Thus, F represents the magnitude of force exerted by q1 or q2 on q3. Let e = +1.60 10-19 C, then q1 = q2 = +2e and q3 = 4.0e and we have Fnet = 2F cos = 2(2e)(4e) 4o (x2 + d2) 4e2 x x = . o (x2 + d2 )3/2 x2 + d2 (a) To find where the force is at an extremum, we can set the derivative of this expression equal to zero and solve for x, but it is good in any case to graph the function for a fuller understanding of its behavior and as a quick way to see whether an extremum point is a maximum or a miminum. In this way, we find that the value coming from the derivative procedure is a maximum (and will be presented in part (b)) and that the minimum is found at the lower limit of the interval. Thus, the net force is found to be zero at x = 0, which is the smallest value of the net force in the interval 5.0 m x 0. (b) The maximum is found to be at x = d/ 2 or roughly 12 cm. (c) The value of the net force at x = 0 is Fnet = 0. (d) The value of the net force at x = d/ 2 is Fnet = 4.9 10-26 N. ...
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## This note was uploaded on 09/06/2009 for the course PHYSICS 14358 taught by Professor Bolton during the Spring '09 term at Kansas State University.
Ask a homework question - tutors are online | 485 | 1,635 | {"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} | 3.734375 | 4 | CC-MAIN-2017-17 | longest | en | 0.922885 |
http://oeis.org/A081421 | 1,600,440,429,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187899.11/warc/CC-MAIN-20200918124116-20200918154116-00131.warc.gz | 109,110,777 | 3,914 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A081421 Quotient after one division by 2 of numbers of the form 3^(2n) + 5^(2n). 0
1, 17, 353, 8177, 198593, 4912337, 122336033, 3054149297, 76315468673, 1907542343057, 47685459212513, 1192108586037617, 29802463602463553, 745059330625296977, 18626462930705797793, 465661390253305305137 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS Except for the first term, these numbers always end in 3 and 7 and necessarily generate an odd number as the quotient upon a single division by 2. Indeed for even n, 3^n+5^n can be written as (4-1)^n + (4+1)^n = 4h+1 + 4i+1 for some h,i. Then we add and get 4(h+i)+2. Divide by 2 to get 2(h+i) + 1 and odd number. LINKS PROG (PARI) p3np5n(n) = { forstep(x=0, n, 2, y = (3^x + 5^x)/2; print1(y" ") ) } CROSSREFS Sequence in context: A324449 A191589 A194729 * A121824 A120287 A222678 Adjacent sequences: A081418 A081419 A081420 * A081422 A081423 A081424 KEYWORD easy,nonn AUTHOR Cino Hilliard, Apr 20 2003 STATUS approved
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Last modified September 18 10:31 EDT 2020. Contains 337166 sequences. (Running on oeis4.) | 505 | 1,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-40 | latest | en | 0.690732 |
https://socratic.org/questions/what-is-the-hybridization-of-co2 | 1,580,166,424,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251728207.68/warc/CC-MAIN-20200127205148-20200127235148-00190.warc.gz | 653,767,272 | 7,085 | # What is the hybridization in "CO"_2?
Jul 2, 2014
The carbon atom has $s p$ hybridization; the $\text{O}$ atoms have $s {p}^{2}$ hybridization.
#### Explanation:
You must first draw the Lewis structure for ${\text{CO}}_{2}$.
According to VSEPR theory, we can use the steric number $\left(\text{SN}\right)$ to determine the hybridization of an atom.
$\text{SN}$ = number of lone pairs + number of atoms directly attached to the atom.
• $\text{SN = 2}$ corresponds to $s p$ hybridization.
• $\text{SN"= 3}$ corresponds to $s {p}^{2}$ hybridization.
We see that the $\text{C}$ atom has $\text{SN = 2}$. It has no lone pairs, but it is attached to two other atoms.
It has $s p$hybridization.
Each $\text{O}$ atom has $\text{SN = 3}$. It has 2 lone pairs and is attached to 1 $\text{C}$ atom.
Just as the carbon atom hybridized to form the best bonds, so do the oxygen atoms.
The valence electron configuration of $\text{O}$ is $\left[\text{He}\right] 2 {s}^{2} 2 {p}^{4}$.
To accommodate the two lone pairs and the bonding pair, it will also form three equivalent $s {p}^{2}$ hybrid orbitals.
Two of the $s {p}^{2}$ orbitals contain lone pairs, while the remaining $s {p}^{2}$ orbital and the unhybridized $p$ orbital have one electron each.
We can see this arrangement in the $\text{C=O}$ bond of formaldehyde, which is equivalent to the right-hand side of the $\text{O=C=O}$ molecule.
(from www.slideshare.net)
There is a similar arrangement on the left side of the $\text{O=C=O}$ molecule, but the $\pi$ bond is horizontal rather than vertical.
Here is a video about the hybridization of carbon dioxide. | 466 | 1,621 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 26, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-05 | longest | en | 0.836831 |
https://www.reference.com/math/round-decimals-108a2d1a462c3ddb | 1,568,629,746,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572517.50/warc/CC-MAIN-20190916100041-20190916122041-00249.warc.gz | 1,014,011,957 | 18,886 | # How Do You Round Decimals?
Credit: Mark Wragg/Stockbyte/Getty Images
To round a decimal, identify the place value you want to round, which is referred to as the rounding digit. Identify the number immediately to the right of the rounding digit. If the number immediately to the right of the rounding digit is less than five, don't change the rounding digit, and drop all digits to the right. If the number to the right of the rounding digit is greater than five, add one to the rounding digit, and remove all digits to the right.
1. Identify the rounding digit
If you are rounding to the nearest tenth, the rounding digit is the first number to the right of the decimal point. When rounding to the nearest hundredth, the rounding digit is two digits to the right of the decimal point. When rounding to the nearest thousandth, the rounding digit is three digits to the right of the decimal point.
2. Decide to round up or down
Locate the digit to the right of the rounding digit. If you are rounding to the nearest tenth, the digit to the right is in the thousandth place. Round up by adding one to the rounding digit if the digit to the right is greater than five. Round down if the digit is less than five.
3. Remove remaining values
Drop all values to the right of the rounding digit. This should be done whether the number to the right of the rounding digit is less than or greater than five.
Similar Articles | 305 | 1,423 | {"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} | 3.6875 | 4 | CC-MAIN-2019-39 | latest | en | 0.877047 |
https://www.tes.com/en-au/teaching-resources/hub/whole-school/mathematics/advanced-pure/co-ordinate-geometry | 1,531,891,636,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590051.20/warc/CC-MAIN-20180718041450-20180718061450-00112.warc.gz | 990,117,259 | 26,153 | #### C1 Notes and Examples with answers
These notes and examples are for the old specification for Edexcel A Level Maths. My complete notes for the new specification (suitable for any exam board) are available here: Year 1: https://www.tes.com/teaching-resource/maths-a-level-new-spec-complete-year-1-notes-and-examples-11904465 Year 2: https://www.tes.com/teaching-resource/maths-a-level-complete-year-2-notes-and-examples-pure-statistics-and-mechanics-11924833
#### Year 12 A Level Powerpoints (Edexcel)
Two Powerpoints: Pure Maths (560 slides) and Statistics/Mechanics (270 slides). Each presentation contains explanations, worked examples and questions for students to complete.
#### Year 12 A Level Pure Maths Powerpoint
This 500+ slide Powerpoint covers all of the first year of the single A Level Pure course (based upon the Edexcel course). It includes explanations, worked examples and questions for students to do. I have included everything, possibly more than you may need but I’d rather give people the option to skip a slide than have to make something up on the spot. I used this during the first year of the new course.
#### Investigation: Where's the Point?
A Mathematical Investigation Assessment Item for Yr 9 & 10, including GeoGebra and Python.
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#### Unit Circle and Radians - Chapter 8 - International Baccalaureate - Standard Level
Unit Circle and Radians - Chapter 8 - International Baccalaureate - Standard Level Everything required to teach this chapter in the IB Standard Level Mathematics course. The PowerPoint is designed to show step by step method through animation in a thorough and engaging way. This makes it ideal for the teacher to use it in lessons or for students to use it to self learn.
#### AQA Further Pure 1 Jan 2013 Model Solutions
Model solutions for the AQA further maths Further Pure Jan 2013 paper with comments and notes. Great for peer marking after completing the paper to see what their answers should look like. Very clear and concise.
#### Multiple choice 22 - Converting from degrees to radians
Another little activity for AFL, revision, starter or a plenary. Works well with mini whiteboards and adaptable to voting systems. Can easily be printed off for class use. Aimed at KS4 and KS5 multiple choice activity converting degrees to radians. Hope you find it useful Please leave feedback
#### Edexcel Year 1 A-level Pure Exam Questions: Circles
This resource focuses on 4 exam style questions. It is suitable for the New A-level mathematics course and can be used as a revision tool or to stretch and challenge students. The slides transition from the question to the solution quite well.
#### Algebra: Circular Graphs 2 - Midpoints, Tangents and Normals (+ resources)
This is my second lesson on circular graphs looking at more AS Level questions around midpoints and problems involving tangents and normals and solving associated problems. This is a great whole lesson and includes a worksheet with answers as well as exam questions with mark scheme. The lesson includes: a starter Learning objectives keywords excellent teaching slides Assessment opportunities (AFL) worksheet with answer exam questions with mark scheme a plenary NOTE: Feel free to browse my shop for more excellent free and premium resources and as always please rate and feedback, thank you.
#### Linear Graphs Mastery Lesson
Full lesson(s) on linear graphs with mastery style tasks. Lesson(s) includes: Examples showing different representations of linear graphs Various fluency tasks Higher order thinking tasks All tasks allow for discussion and students to show their answers on the board to encourage correction of each others work.
#### Co-ordinates and grid referencing
5 Different worksheets on co-ordinates and grid referencing Game on grid referencing
#### Co-ordinates game
Children have to ask each other a question and the other child must find the answer in the grid and give the co-ordinates corresponding to it.
#### Co-ordinates and grid referencing
Children have to plot the co-ordinates to make shapes Blank for children to make their own
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Children have to make shapes on a grid and list the co-ordinates they used
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Three differentiated worksheets where children have to colour correct number on grid
#### Co-ordinates - Grids
Children have to make a letter on a co-ordinate grid
#### NEW AS Quadratics,Circles & Discriminant
A resource of questions which could be used as an assessment, HW or revision covering the topics; quadratic inequalities, finding and using the equations of circles, finding points intersection of lines with quadratics/circles and solving problems involving the use of the discriminant. Also contains a problem-solving challenge and has worked solutions attached. Perfectly suited for the New AS specification.
#### Geometry Skills Color By Number Bundle 2: 10 More Skills
With these activities, students will enjoy practicing their geometry skills as they have fun creating beautiful, colorful mandalas! These activities are an excellent choice for sub plans, enrichment/reinforcement, early finishers, and extra practice with some fun. These engaging activities are especially useful for end-of-year practice, spiral review, and motivated practice when students are exhausted from standardized testing or mentally “checked out” before a long break! Teachers and students alike enjoy motivating activities, so engage your students today with these fun activities! These 10 activities include: These 10 activities include: Angle Addition Postulate Angle Bisectors of Triangles Angles In Quadrilaterals Areas of Regular Polygons Centroids of Triangles Distance Formula Exterior Angle Theorem Measuring Line Segments Midpoint Formula The Pythagorean Theorem and its Converse Coloring is a great way to get your students motivated and interested in practicing and reviewing their geometry skills! In addition, these activities are great for emergency sub plans, enrichment, early finishers, skills reinforcement, and extra credit. As an added bonus, the completed worksheets make fabulous classroom decor! With a savings of over 30% off if the activities were purchased separately, this bundle is a win-win for everyone! Thanks and enjoy your new product!
#### Midpoint Formula Color by Number
In this activity, students will practice using the midpoint formula to find the midpoint of the given endpoints of a line segment. They will use their answers to color a mandala with the colors indicated to reveal a beautiful, colorful pattern! This no-prep activity is an excellent resource for sub plans, enrichment/reinforcement, early finishers, and extra practice with some fun. It is especially useful for end-of-year practice, spiral review, and motivated practice when students are exhausted from standardized testing or mentally “checked out” before a long break (hello summer!). Color motivates even the most challenging students and the students get a fun chance to practice their essential geometry skills. In addition, the finished products make fabulous classroom decor!
#### Distance Formula Color by Number
In this activity, students will practice using the distance formula to find the distance of the given endpoints of a line segment. They will use their answers to color a mandala with the colors indicated to reveal a beautiful, colorful pattern! This no-prep activity is an excellent resource for sub plans, enrichment/reinforcement, early finishers, and extra practice with some fun. It is especially useful for end-of-year practice, spiral review, and motivated practice when students are exhausted from standardized testing or mentally “checked out” before a long break (hello summer!). Color motivates even the most challenging students and the students get a fun chance to practice their essential geometry skills. In addition, the finished products make fabulous classroom decor!
#### Introduction to the equation of a circle.
Introduction to the equation of a circle.
#### Edexcel New Maths A Level year 1 complete notes
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#### C1 A Level Summary
“cheat sheet” covering all C1 topics for CCEA ALevel. Use for revision classes/exercises as a guide or as a comfort blanket for a class test.
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#### Graph transformations matching activity A2 maths
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#### Notes and Examples for Edexcel A Level Maths Year 1 Topic 10: Differentiation
These notes and examples are designed for the delivery of the new Edexcel A Level Maths Linear Specification. You will receive an editable Word document that can be issued to students with gaps for them to fill in the solutions to the examples and make further notes. Full solutions to the examples are provided for the teacher in the form of a PDF document. This could be made available to students if they miss a lesson, reducing teacher workload. Each topic is available as a separate purchase, or you can download the entire set covering all the the Year 1 pure maths content. To purchase all year 1 pure topics, click here: https://www.tes.com/teaching-resource/notes-and-examples-for-a-level-maths-year-1-pure-11821867 To purchase the complete notes for the entire year 1 syllabus (pure, statistics, mechanics), including large data set activities, click here: https://www.tes.com/teaching-resource/edexcel-new-maths-a-level-year-1-complete-notes-11872425
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Self made questions about graphs and transformations for AQA A-Level Maths. Designed to be a little but harder and more strange than those seen in past paper exam questions. Does not come with marks scheme but will be easy to figure out the solutions.
#### There once was a parallelogram...
Straight line and equations. A write on worksheet for students. Best used by revealing one page at a time. Skills tested: checking a certain point lies on a line, finding where a line crosses the axes, finding the equation of a line given a point and a gradient or two points, calculating the length of a line segment, using perpendicular gradients and finding the midpoint of a line segment. Worked solutions provided.
#### Geogebra: Straight Line
Files involve coordinates and the equation of a straight line. Also distance formula/Pyth Thm. Geogebra files designed for class lessons on electronic board but also useful for individual students at home or in school. Other collected files at www.geoduggie.weebly.com and www.geogebra.org/drt The free Geogebra software at www.geogebra.org/download
#### 40 slide Powerpoint Advanced Higher Maths Complex Numbers Argand Diagrams Worked Solutions
40 slide Powerpoint for Advanced Higher Maths Unit 2: Complex Numbers. There is a brief revision of the basics of Complex numbers followed by a series of questions. The 24 questions (many of them multi-part) require the construction of Argand Diagrams, use of the quadratic formula, polynomial long division, and simultaneous equations. There are fully worked solutions (including diagrams) for complex number topics relating to: Equating Real and Imaginary Parts; Finding square, cube, fourth, fifth and sixth roots of complex numbers (including unity) and plotting them on an Argand diagram; Verifying and finding roots of complex number polynomials; Expanding and simplifying complex numbers using the Binomial Theorem and De Moivre’s Theorem; Interpreting geometrically loci in the complex plane; Conversions between polar and rectangular forms; Complex Conjugates; Exponential Form; Trigonometric identities, substitutions and simplification. The questions are grouped in approximate order of difficulty and to match the usual order of progress through this topic. *Animated workings come up line by line on mouse clicks.*
#### Straight line graphs Edexcel A-level Year 1/AS Pearson Ch 5
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#### Graph Sketching Discussion Venn Diagram
Students are to think of a graph that would fit in all the regions of the Venn diagram. Three categories are: 1. Graphs with only 1 stationary point 2. Graphs with a maximum 3. Graphs that go through the point (0,1) This is really useful as a discussion activity for where graphs cross the axes and for differentiation. A sample set of answers is included. | 3,169 | 15,551 | {"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-2018-30 | latest | en | 0.867493 |
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Random Sequence? (Posted on 2004-06-28)
Complete the following sequence:
-35, -44, -39, -34, -19, 0, 29, ?, ?
See The Solution Submitted by Juggler Rating: 4.0000 (6 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
I know the Answer | Comment 2 of 8 |
-35
-44
-39
-34
-19
0
29
?
?
First you start with normal sequencing. . . the differences, in order, being
-9
+5
+5
+15
+19
+29
If you look at the differences of the differences you get the sequence of.
+14
0
+10
+4
+10
the next two numbers for the second sequence should be
0
+14
making the first sequence additional numbers
+29
+43
Finally making the original sequence numbers
58
101
That is the solution to an unsolved problem.
Posted by Margaret on 2004-06-28 13:27:12
Search: Search body:
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https://oeis.org/A228797 | 1,685,766,905,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649105.40/warc/CC-MAIN-20230603032950-20230603062950-00597.warc.gz | 477,204,868 | 3,776 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A228797 Number of 2 X n binary arrays with top left element equal to 1 and no two ones adjacent horizontally or nw-se. 1
2, 2, 8, 16, 42, 98, 240, 576, 1394, 3362, 8120, 19600, 47322, 114242, 275808, 665856, 1607522, 3880898, 9369320, 22619536, 54608394, 131836322, 318281040, 768398400, 1855077842, 4478554082, 10812186008, 26102926096, 63018038202, 152139002498 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS R. H. Hardin, Table of n, a(n) for n = 1..210 FORMULA Empirical: a(n) = a(n-1) + 3*a(n-2) + a(n-3). Conjectures from Colin Barker, Sep 13 2018: (Start) G.f.: 2*x / ((1 + x)*(1 - 2*x - x^2)). a(n) = (-2*(-1)^n + (1-sqrt(2))^n + (1+sqrt(2))^n) / 2. (End) EXAMPLE Some solutions for n=4: ..1..0..0..1....1..0..0..1....1..0..0..0....1..0..0..1....1..0..0..1 ..0..0..0..0....0..0..1..0....0..0..0..1....1..0..1..0....0..0..0..1 CROSSREFS Row 2 of A228796. Sequence in context: A220172 A276054 A192305 * A052970 A220589 A109190 Adjacent sequences: A228794 A228795 A228796 * A228798 A228799 A228800 KEYWORD nonn AUTHOR R. H. Hardin, Sep 04 2013 STATUS approved
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# JEE Main 2022 July 29 – Shift 2 Physics Question Paper with Solutions
The JEE Main 2022 July 29 – Shift 2 Physics Question Paper with Solutions is provided on this page. Solving the JEE Main 2022 question papers helps students to boost their preparation. JEE Main 2022 answer keys contain the exact solutions to the JEE questions in a step-by-step method. Students can easily understand and learn these JEE Main 2022 question papers with solutions and thus improve their knowledge in each subject. Revise the JEE Main 2022 July 29 – Shift 2 Physics Question Paper with Solutions given below.
## JEE Main 2022 29th July Shift 2 Physics Question Paper and Solutions
SECTION – A
Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct.
1. Two identical metallic spheres A and B when placed at certain distance in air repel each other with a force of F. Another identical uncharged sphere C is first placed in contact with A and then in contact with B and finally placed at midpoint between spheres A and B. The force experienced by sphere C will be
(A) 3F/2
(B) 3F/4
(C) F
(D) 2F
Sol. When two identical sphere come in contact with each other, the total charge on them is equally distribute.
$$\begin{array}{l}\frac{kQ^2}{d^2}=F\end{array}$$
$$\begin{array}{l}F’=\frac{k9Q^2}{16\times \frac{d^2}{4}}-\frac{k3Q^2}{8 \times \frac{d^2}{4}} \end{array}$$
$$\begin{array}{l}=\frac{9kQ^2}{4d^2}-\frac{3kQ^2}{2d^2}\end{array}$$
$$\begin{array}{l}=\frac{kQ^2}{d^2}[\frac{9}{4}-\frac{3}{2}]\end{array}$$
$$\begin{array}{l}=\frac{6}{8}F=\frac{3}{4}F\end{array}$$
2. Match List I with List II.
List I List II
A. Torque I. Nms–1
B. Stress II. Jkg–1
C. Latent Heat III. Nm
D. Power IV Nm–2
Choose the correct answer from the options given below:
(A) A-III, B-II, C-I, D-IV
(B) A-III, B-IV, C-II, D-I
(C) A-IV, B-I, C-III, D-II
(D) A-II, B-III, C-I, D-IV
Sol. Torque → Nm
Stress → N/m2
Latent heat → J/kg
Power → N m/s
A-III, B-IV, C-II, D-I
3. Two identical thin metal plates has charge q1 and q2 respectively such that q1 > q2. The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is
$$\begin{array}{l}(\text{A})\ \frac{(q_1+q_2)}{C}\end{array}$$
$$\begin{array}{l}(\text{B})\ \frac{(q_1-q_2)}{C}\end{array}$$
$$\begin{array}{l}(\text{C})\ \frac{(q_1-q_2)}{2C}\end{array}$$
$$\begin{array}{l}(\text{D})\ \frac{2(q_1-q_2)}{C}\end{array}$$
Sol.
$$\begin{array}{l}E=\frac{q_1-q_2}{2\varepsilon_0A}\end{array}$$
$$\begin{array}{l}v=\frac{(q_1-q_2)d}{2\varepsilon_0A}\end{array}$$
$$\begin{array}{l}=\frac{q_1-q_2}{2C}\end{array}$$
4. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Alloys such as constantan and manganin are used in making standard resistance coils.
Reason R: Constantan and manganin have very small value of temperature coefficient of resistance.
In the light of the above statements, choose the correct answer from the options given below.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Sol. Since they have low temperature coefficient of resistance, their resistance remains almost constant.
5. A 1 m long wire is broken into two unequal parts X and Y. The X part of the wire is stretched into another wire W. Length of W is twice the length of X and the resistance of W is twice that of Y. Find the ratio of length of X and Y.
(A) 1:4
(B) 1:2
(C) 4:1
(D) 2:1
Sol.
Rw = 2Ry
$$\begin{array}{l}\rho\frac{2x}{\frac{A}{2}}=\frac{2\rho(1-x)}{A}\end{array}$$
4x = 2(1 – x)
$$\begin{array}{l}\frac{x}{1-x}=\frac{1}{2}\end{array}$$
6. A wire X of length 50 cm carrying a current of 2 A is placed parallel to a long wire Y of length 5 m. The wire Y carries a current of 3 A. The distance between two wires is 5 cm and currents flow in the same direction. The force acting on the wire Y is
(A) 1.2 × 10–5 N directed towards wire X
(B) 1.2 × 10–4 N directed away from wire X
(C) 1.2 × 10–4 N directed towards wire X
(D) 2.4 × 10–5 N directed towards wire X
Sol. FXY = FYX = F
$$\begin{array}{l}F=\frac{\mu_0I_2}{2\pi r}I_1(l)\end{array}$$
$$\begin{array}{l}=\frac{4\pi \times 10^{-7} \times 3 \times 2 \times [50 \times 10^{-2}]}{2\pi (5\times 10^{-2})}\end{array}$$
= 1.2 × 10–5 N
7. A juggler throws balls vertically upwards with same initial velocity in air. When the first ball reaches its highest position, he throws the next ball. Assuming the juggler throws n balls per second, the maximum height the balls can reach is
(A) g/2n
(B) g/n
(C) 2gn
(D) g/2n2
Sol.
$$\begin{array}{l}t=\frac{u}{g}=\frac{1}{n}\end{array}$$
$$\begin{array}{l}u=\frac{g}{n}\end{array}$$
$$\begin{array}{l}H_{\text{max}}=\frac{u^2}{2g}=\frac{9}{2n^2}\end{array}$$
8. A circuit element X when connected to an a.c. supply of peak voltage 100 V gives a peak current of 5 A which is in phase with the voltage. A second element Y when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by π/2. If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?
$$\begin{array}{l}(\text{A})\ \frac{10}{\sqrt{2}}\end{array}$$
$$\begin{array}{l}(\text{B})\ \frac{5}{\sqrt{2}}\end{array}$$
$$\begin{array}{l}(\text{C})\ 5\sqrt{2}\end{array}$$
$$\begin{array}{l}(\text{D})\ \frac{5}{2}\end{array}$$
Sol.
$$\begin{array}{l}R=\frac{100}{5}=20\ \Omega\end{array}$$
$$\begin{array}{l}X_L=\frac{100}{5}=20\ \Omega\end{array}$$
When in series
$$\begin{array}{l}z=\sqrt{20^2 + 20^2}=20\sqrt{2}\ \Omega\end{array}$$
$$\begin{array}{l}i = \frac{100}{z}=\frac{100}{20\sqrt{2}}=\frac{5}{\sqrt{2}}\end{array}$$
$$\begin{array}{l}i_{\text{rms}}= \frac{1}{\sqrt{2}}i\end{array}$$
$$\begin{array}{l}=\frac{5}{2}\end{array}$$
9. An unpolarised light beam of intensity 2I0 is passed through a polaroid P and then through another polaroid Q which is oriented in such a way that its passing axis makes an angle of 30° relative to that of P. The intensity of the emergent light is
$$\begin{array}{l}(\text{A})\ \frac{I_0}{4}\end{array}$$
$$\begin{array}{l}(\text{B})\ \frac{I_0}{2}\end{array}$$
$$\begin{array}{l}(\text{C})\ \frac{3I_0}{4}\end{array}$$
$$\begin{array}{l}(\text{D})\ \frac{3I_0}{2}\end{array}$$
Sol.
$$\begin{array}{l}I = I_0 \times \frac{3}{4}\end{array}$$
10. An α particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be:
$$\begin{array}{l}(\text{A})\ \sqrt{2}:1\end{array}$$
$$\begin{array}{l}(\text{B})\ 2\sqrt{2}:1\end{array}$$
$$\begin{array}{l}(\text{C})\ 4\sqrt{2}:1\end{array}$$
(D) 8 : 1
Sol.
$$\begin{array}{l}\frac{p_\alpha}{p_p}=\frac{\sqrt{2(4m)(2eV)}}{\sqrt{2(m)(eV)}}\end{array}$$
$$\begin{array}{l}=\frac{\sqrt{16}}{\sqrt{2}}\end{array}$$
$$\begin{array}{l}=\frac{4}{\sqrt{2}} = \frac{2\sqrt{2}}{1}\end{array}$$
(A) Volume of the nucleus is directly proportional to the mass number.
(B) Volume of the nucleus is independent of mass number.
(C) Density of the nucleus is directly proportional to the mass number.
(D) Density of the nucleus is directly proportional to the cube root of the mass number.
(E) Density of the nucleus is independent of the mass number.
Choose the correct option from the following options
(A) (A) and (D) only
(B) (A) and (E) only
(C) (B) and (E) only
(D) (A) and (C) only
Sol.
$$\begin{array}{l}R=R_0A^{1/3}\end{array}$$
$$\begin{array}{l}\Rightarrow V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3A\end{array}$$
$$\begin{array}{l}\Rightarrow \rho = \frac{M}{V}\propto \frac{A}{A}\propto A^0\end{array}$$
12. An object of mass 1 kg is taken to a height from the surface of earth which is equal to three times the radius of earth. The gain in potential energy of the object will be
[If, g = 10 ms–2 and radius of earth = 6400 km]
(A) 48 MJ
(B) 24 MJ
(C) 36 MJ
(D) 12 MJ
Sol.
$$\begin{array}{l}\Delta U = U_f – U_i\end{array}$$
$$\begin{array}{l}=-\frac{GMm}{4R}+\frac{GMm}{R}\end{array}$$
$$\begin{array}{l}=\frac{3GMm}{4R}=\frac{3}{4}mgR\end{array}$$
= 48 MJ
13. A ball is released from a height h. If t1 and t2 be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between t1 and t2.
$$\begin{array}{l}(\text{A})\ t_1 = (\sqrt{2})t_2\end{array}$$
$$\begin{array}{l}(\text{B})\ t_1 = (\sqrt{2} – 1)t_2\end{array}$$
$$\begin{array}{l}(\text{C})\ t_2 = (\sqrt{2} + 1)t_1\end{array}$$
$$\begin{array}{l}(\text{D})\ t_2 = (\sqrt{2} – 1)t_1\end{array}$$
Sol.
$$\begin{array}{l}t_1=\sqrt{\frac{2\cdot\frac{H}{2}}{g}} = \sqrt{\frac{H}{g}}\end{array}$$
$$\begin{array}{l}\text{And } t_2 = \sqrt{\frac{2H}{g}} -t_1\end{array}$$
$$\begin{array}{l}\Rightarrow t_2 = \sqrt{\frac{2H}{g}}-\sqrt{\frac{H}{g}}=\sqrt{\frac{H}{g}}\{\sqrt{2}-1\}\end{array}$$
14. Two bodies of masses m1 = 5 kg and m2 = 3 kg are connected by a light string going over a smooth light pulley on a smooth inclined plane as shown in the figure. The system is at rest. The force exerted by the inclined plane on the body of mass m1 will be
[Take g = 10 ms–2]
(A) 30 N
(B) 40 N
(C) 50 N
(D) 60 N
Sol. m2g = m1gsinθ …(i)
N = m1gcosθ …(ii)
$$\begin{array}{l}\Rightarrow \frac{N}{m_2g}=\cot \theta\end{array}$$
$$\begin{array}{l}\Rightarrow N = 3\times 10 \times \cot \theta = 3 \times 10 \times \frac{4}{3}\ \ \ \left( \because \sin \theta =\frac{3}5{} \right)\end{array}$$
N = 40 Newtons
15. If momentum of a body is increased by 20%, then its kinetic energy increases by
(A) 36%
(B) 40%
(C) 44%
(D) 48%
Sol.
$$\begin{array}{l}K = \frac{p^2}{2m}\end{array}$$
$$\begin{array}{l}K’ = \frac{(1.2p)^2}{2m}\end{array}$$
$$\begin{array}{l}\Rightarrow \frac{K’-K}{K}=(1.2)^2 -1 = 0.44\end{array}$$
⇒ 44% increase
16. The torque of a force
$$\begin{array}{l}5\hat{i}+3\hat{j}-7\hat{k}\end{array}$$
about the origin is τ. If the force acts on a particle whose position vector is
$$\begin{array}{l}2\hat{i}+2\hat{j}+\hat{k},\end{array}$$
then the value of τ will be
$$\begin{array}{l}(\text{A})\ 11\hat{i}+19\hat{j}-4\hat{k}\end{array}$$
$$\begin{array}{l}(\text{B})\ -11\hat{i}+9\hat{j}-16\hat{k}\end{array}$$
$$\begin{array}{l}(\text{C})\ -17\hat{i}+19\hat{j}-4\hat{k}\end{array}$$
$$\begin{array}{l}(\text{C})\ 17\hat{i}+9\hat{j}+16\hat{k}\end{array}$$
Sol.
$$\begin{array}{l}\vec{\tau} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\2 & 2 & 1 \\5 & 3 & -7 \\\end{vmatrix}\end{array}$$
$$\begin{array}{l}=\hat{i}(-14-3)+\hat{j}(5+14)+\hat{k}(6-10)\end{array}$$
$$\begin{array}{l}=-17\hat{i}+19\hat{j}-4\hat{k}\end{array}$$
17. A thermodynamic system is taken from an original state D to an intermediate state E by the linear process shown in the figure. Its volume is then reduced to the original volume from E to F by an isobaric process. The total work done by the gas from D to E to F will be
$$\begin{array}{l}\left(A\right) -450~\text{J}\\ \left(B\right) 450~\text{J}\\ \left(C\right) 900~\text{J}\\ \left(D\right) 1350~\text{J}\end{array}$$
Sol.
$$\begin{array}{l}W=\frac{1}{2}\times (5-2)\times(600 – 300) \text{ J}\end{array}$$
$$\begin{array}{l}=\frac{1}{2}\times 3 \times 300= 450 \text{J} \end{array}$$
18. The vertical component of the earth’s magnetic field is 6 × 10–5 T at any place where the angle of dip is 37°. The earth’s resultant magnetic field at that place will be (Given tan 37° = ¾)
$$\begin{array}{l}\left(A\right) 8 \times 10^{–5}~\text{T}\\ \left(B\right)6 \times 10^{–5}~\text{T}\\ \left(C\right)5 \times 10^{–4}~\text{T}\\ \left(D\right)1 \times 10^{–4}~\text{T}\end{array}$$
Sol.
$$\begin{array}{l}B_v = B_0sin\theta\end{array}$$
$$\begin{array}{l}B_0 = \frac{B_v}{\sin \theta} = \frac{6 \times 10^{-5}}{\sin 37^{\circ}}\end{array}$$
$$\begin{array}{l}=\frac{6 \times 10^{-5}}{3}\times 5\end{array}$$
$$\begin{array}{l} = 1 \times 10^{–4}~\text{T} \end{array}$$
19. The root mean square speed of smoke particles of mass 5 × 10-17 in their Brownian motion in air at NTP is approximately. [Given k = 1.38 × 10-23 JK-1]
$$\begin{array}{l}\left(A\right) 60 ~\text{mm s}^{–1}\\ \left(B\right) 12~ \text{mm s}^{–1}\\ \left(C\right)15~ \text{mm s}^{–1}\\ \left(D\right) 36~ \text{mm s}^{–1}\end{array}$$
Sol.
$$\begin{array}{l}\text{At NTP}, T = 298~ \text{K}\\\Rightarrow v_{\text{rms}}= \sqrt{\frac{3RT}{M}}\end{array}$$
$$\begin{array}{l}=\sqrt{\frac{3kN_A \times 298}{5 \times 10^{-17}\times N_A}}\end{array}$$
$$\begin{array}{l}\simeq 15 \text{ mm/s}\end{array}$$
20. Light enters from air into a given medium at an angle of 45° with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of 15° from its original direction. The refractive index of the medium is
(A) 1.732
(B) 1.333
(C) 1.414
(D) 2.732
Sol.
$$\begin{array}{l}1 \times sin 45^\circ = \mu \times sin 30^\circ\end{array}$$
$$\begin{array}{l}\Rightarrow \mu = \frac{1 }{\sqrt{2}}\times \frac{2}{1}\end{array}$$
$$\begin{array}{l}\mu = \sqrt{2}\\= 1.414 \end{array}$$
SECTION – B
Numerical Value Type Questions: This section contains 10 questions. In Section B, attempt any five questions out of 10. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer.
1. A tube of length 50 cm is filled completely with an incompressible liquid of mass 250 g and closed at both ends. The tube is then rotated in horizontal plane about one of its ends with a uniform angular velocity
$$\begin{array}{l}x\sqrt{F} \text{ rad s}^{-1} .\end{array}$$
If F be the force exerted by the liquid at the other end then the value of x will be______.
Sol.
$$\begin{array}{l}\text{Applying } F_c = \frac{m\omega ^2 l}{2}\end{array}$$
$$\begin{array}{l} \frac{m\omega ^2 l}{2} = F\end{array}$$
$$\begin{array}{l}\omega = \sqrt{\frac{2F}{\frac{1}{2}\times \frac{1}{4}}}=\sqrt{16 F} = 4\sqrt{F}\end{array}$$
2. Nearly 10% of the power of a 110 W light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of 1 m from the bulb to a distance of 5 m is a × 10–2 W.m2. The value of ‘a‘ will be _____.
Sol.
$$\begin{array}{l}P_{radiation} = 0.1 \times 110 = 11~\text{W}\end{array}$$
$$\begin{array}{l}\Delta I_{\text{radiation}_1} = I_{\text{radiation}_1} – I_{\text{radiation}_2}\end{array}$$
$$\begin{array}{l}=11\left( \frac{1}{4\pi} – \frac{1}{4\pi \times 25}\right)=\frac{11 \times 24}{4\pi \times 25}\\= 84 \times 10^{–2}~ \text{W/m}^2\end{array}$$
3. A metal wire of length 0.5 m and cross-sectional area 10–4 m2 has breaking stress 5 × 108 Nm–2. A block of 10 kg is attached at one end of the string and is rotating in a horizontal circle. The maximum linear velocity of block will be _____ ms–1.
Sol.
$$\begin{array}{l} A = 10^{–4} \text{m}^2\\l = \frac{1}{2} \text{ m}\\\sigma = 5 × 10^8\end{array}$$
$$\begin{array}{l}\frac{mv^2}{lA}=5 \times 10^8\end{array}$$
$$\begin{array}{l}v=\sqrt{\frac{5\times 10^8 \times \frac{1}{2}\times 10^{-4}}{10}}=5 \times 10 = 50 \text{ m/s}\end{array}$$
4. The velocity of a small ball of mass 0.3 g and density 8 g/cc when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is 1.3 g/cc, then the value of viscous force acting on the ball will be x × 10–4 N. The value of x is _______. [use g = 10 m/s2]
Sol.
$$\begin{array}{l} F_v = 6\pi \eta rv_T\\ F_v + F_B = mg\\ \Rightarrow Fv = mg – F_B\end{array}$$
$$\begin{array}{l}=10 \times (8 – 1.3 )\times \frac{0.3}{8}\times 10^{-3}\\= 2.5125 \times 10^{–3}\ \text{N} \end{array}$$
$$\begin{array}{l}\simeq 25 \times 10^{-4} \text{ N}\end{array}$$
5. A modulating signal 2sin (6.28 × 106) t is added to the carrier signal 4sin(12.56 × 109) t for amplitude modulation. The combined signal is passed through a non-linear square law device. The output is then passed through a band pass filter. The bandwidth of the output signal of band pass filter will be ______MHz.
Sol.
$$\begin{array}{l} W_C = 12.56 \times 10^9\\ W_m = 6.25 \times 10^6\end{array}$$
After amplitude modulation
Bandwidth frequency
$$\begin{array}{l}=\frac{2W_m}{2\pi}=\frac{2 \times 6.28}{2\pi}\times 10^6=2 \text{ MHz}\end{array}$$
6. The speed of a transverse wave passing through a string of length 50 cm and mass 10 g is 60 ms–1. The area of cross-section of the wire is 2.0 mm2 and its Young’s modulus is 1.2 × 1011 Nm–2. The extension of the wire over its natural length due to its tension will be x × 10–5 m. The value of x is _____.
$$\begin{array}{l}v=\sqrt{\frac{T}{\mu}}\end{array}$$
$$\begin{array}{l}\text{So}\ T = 60^2 \times \frac{10 \times 10^{-3}}{0.5}=72 \text{ N}\end{array}$$
$$\begin{array}{l}\Delta \ell = \frac{T\ell}{YA}= \frac{72 \times 0.5}{1.2 \times 10^{-11}\times 2 \times 10^{-6}} = 15 \times 10^{-5} \text{ m}\end{array}$$
7. The metallic bob of simple pendulum has the relative density 5. The time period of this pendulum is 10 s. If the metallic bob is immersed in water, then the new time period becomes
$$\begin{array}{l}5\sqrt{x} \text{ s}\end{array}$$
. The value of x will be _____.
$$\begin{array}{l}T = 2\pi \sqrt{\frac{\ell}{g}}=10\end{array}$$
$$\begin{array}{l}T’ = 2\pi \sqrt{\frac{ℓ\ell}{g\left(1- \frac{1}{\rho}\right)}}\end{array}$$
$$\begin{array}{l}=2\pi \sqrt{\frac{\ell}{g}\times \frac{5}{4}}=10\sqrt{\frac{5}{4}}=5\sqrt{5}\end{array}$$
8. A 8 V Zener diode along with a series resistance R is connected across a 20 V supply (as shown in the figure). If the maximum Zener current is 25 mA, then the minimum value of R will be ____ Ω.
R will be minimum when RL is infinitely large, so
$$\begin{array}{l}R_{\text{Zener}} = \frac{8}{25\times 10^{-3}}=320 \ \Omega\end{array}$$
$$\begin{array}{l}\text{So } \frac{R}{R_{\text{Zener}}}=\frac{12}{8}\end{array}$$
$$\begin{array}{l}R=\frac{12}{8}\times 320 = 480\ \Omega\end{array}$$
9. Two radioactive materials A and B have decay constants 25λ and 16λ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of B to that of A will be ‘e’ after a time 1/(aλ). The value of a is _____.
$$\begin{array}{l}N_A = N_0e^{-25\lambda t}\end{array}$$
$$\begin{array}{l}N_B = N_0e^{-16\lambda t}\end{array}$$
$$\begin{array}{l}\frac{N_B}{N_A}=e=e^{9\lambda t}\end{array}$$
$$\begin{array}{l}t = \frac{1}{9\lambda}\end{array}$$
10. A capacitor of capacitance 500 μF is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. The maximum current in the LC circuit will be ______A.
q0 = CV
= 500 × 100 × 10-6 C
= 5 × 10-2 C
For imax,
$$\begin{array}{l}\frac{1}{2}Li_m^2 = \frac{1}{2}\frac{q_0^2}{C}\end{array}$$
$$\begin{array}{l}50 \times 10^{-3}\times i_m^2 = \frac{(5\times 10^{-2})^2}{500 \times 10^{-6}}\end{array}$$
$$\begin{array}{l}\Rightarrow i_m = \frac{5 \times 10^{-2}}{5 \times 10^{-3}}=10 \text{ A}\end{array}$$ | 7,256 | 19,475 | {"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} | 4.03125 | 4 | CC-MAIN-2024-26 | latest | en | 0.810654 |
https://911weknow.com/earn-1000000-with-math-the-millennium-prize-problems | 1,701,340,956,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100184.3/warc/CC-MAIN-20231130094531-20231130124531-00583.warc.gz | 112,289,246 | 13,994 | # Earn \$1,000,000 with Math? The Millennium Prize Problems.
If someone has told you that Mathematics is not a viable career, tell them to go and look up Grigori Perelman. He is (as of 2019) the one and only person to be credited with solving a Millennium Prize Problem, earning him much notoriety and a cool \$1,000,000 (even though he eventually turned down the prize money).
The problem in question, the Poincar Conjecture, is a theorem about the characterization of the 3-sphere, which is the hypersphere that bound the unit ball in four-dimensional space.
For those of you interested in the mathematical terms, here you go:
• A unit sphere is the set of points of distance 1 from a fixed central point. The fixed central point usually refers to a specific point in the space that is has been distinguished as the origin of the space (ie. on a 2D graph the point (0,0) is considered the origin, on a 3D graph the point (0,0,0)).
• A hypersphere is the set of points at a constant distance from a given point, in the (n-1)’th dimension of it?s containing ambient space. For example, a hypersphere in the dimension R (x,y,z plane) would be a sphere, also called a 2-sphere.
• A 3-sphere is analogue of a sphere in a higher dimension and would be a hypersphere in the 4th dimension.
• If you have a 3-sphere with radius 1 and center at the origin in the 4th dimension, then you get a hypersphere that bounds the unit sphere (a sphere of points of distance 1 from a fixed point) in the fourth dimension.
• [Semi-important for below] If you have a loop in a Euclidean space such that you can ?tighten? it (think about a having a circle and decreasing the radius continuously until you have a radius of 0) to a single point where all of the intermediary loops and the final point are all in the set, then the loop can be ?tightened continuously to a point in the space?
In layman’s terms, we think about a space that locally looks like ordinary three-dimensional space, but is connected, finite in size, and lacks any boundaries (a closed 3-manifold). Then the theorem claims that if such a space exists and has the additional property that each loop can be continuously tightened to a point in the space, then it is a three-dimensional sphere.
After nearly a century of effort from mathematicians, Perelman presented a proof of the conjecture in the papers between 2002 and 2003. The proof built upon Richard S. Hamilton?s work, where he showed how to use the Ricci flow on a manifold to prove special cases of the Poincar conjecture. Hamilton extended his work for multiple years but was unable to prove the conjecture conclusively. Using this groundwork, Perelman was able to sketch a proof of a more general conjecture, Thurston?s Geometrization conjecture, completing the Ricci flow program and proving the first Millennium Prize Problem.
On August 22, 2006, the International Congress of Mathematicians awarded Perelman the Fields Medal for his work on the conjecture, but similarly to the million-dollar bounty, he refused the medal. When asked about turning down the prize money, he remained humble, stating he believed his contributions in proving the conjecture was no greater than Hamilton?s.
The Poincar Conjecture and 6 other complex mathematical theories have been dubbed the Millennium Prize Problems by the Clay Mathematics Institute (CMI). Each problem has been described as an ?important classic question that has resisted solutions over the years?, with the first person to devise a solution to each earning \$1,000,000 courtesy of the CMI. However, as you saw above, solving these problems is no easy feat. One of the problems has remained unsolved since it?s formulation in 1859 by German Mathematician Bernhard Riemann.
The Riemann Hypothesis, named after the aforementioned German mathematician, is widely considered to be the most important unsolved problem in pure mathematics. It is of great interest in number theory because it implies results about the distribution of prime number, which are used in everything from biology to encryption and quantum mechanics. In order to understand the Riemann Hypothesis, we first need to explain a few key concepts:
• A complex number is of the form a+bi, where i is defined by i=-1. In this form, the real part of a complex number is a, and the imaginary part is bi.
• The Riemann zeta function is a function of a complex variable (a function of complex numbers) defined by the following equation, where s is any complex number other than 1, and whose values are also complex:
• A ?zero? of a function is an x such that f(x)=0
• The ?trivial zeros? of the Riemann zeta function are all the negative even integers (-2, -4, -6, ?)
• The ?non-trivial zeros? of the Riemann zeta function are all of the other values of s for which ?(s)=0 (ie. s is not a negative even integer).
Now that we have some definitions down pat, we can go ahead and state the Riemann Hypothesis:
The real part of every non-trivial zero of the Riemann zeta function is 1/2
This one (to me at least) is a lot easier to understand than the Poincar Conjecture and does not seem intuitively very difficult. However, it also does not seem to mean much. Really, who cares about when this random function has a value of 0. Well not surprisingly, a lot of mathematicians do, and for a very good reason.
Some numbers have the special property that they cannot be expressed as the product of two smaller numbers (product being multiplication), ie. 2,3,5,7,11, etc. Known as prime numbers, they are in a sense the simplest numbers you can get, forming the building blocks for all other numbers. Frustratingly though, prime numbers do not seem to follow any pattern. 3137 is a prime number, and the next one after that is not until 3163, but then 3167 and 3169 follow in quick succession, all of which are primes. In a nutshell, if you find one prime number, there is no way to tell where the next one is going to be without checking all of the numbers as you go. However, using the Prime Number Theorem (PNT), you can find how many prime numbers there are below a certain threshold.
The Prime Number Theorem is just an estimation, with different values giving a different probability of being correct, but never 100% certainty. However, understand the assumption that the Reimann hypothesis is true, you can create a mathematical approximation combining the Prime Number Theorem and the non-trivial zeros of the Riemann zeta function to error-correct the internal components, providing the ?best possible? bound for the error term in the Prime Number Theorem. If it were possible to prove the Reimann Hypothesis completely, this would give the PNT the ability to provide incredibly close estimates to the actual value, opening up many possibilities in different branches of mathematics. In fact, there are many important hypotheses that being with ?If the Riemann Hypothesis is true, then ??, so solving this problem would instantly validate all of the subsequent conjectures as well.
Another interesting Millenium problem has gained a bit of facetime recently thanks to the 2017 film Gifted, which focused on the brother and daughter of a female mathematician dedicated to solving the Navier?Stokes problem. The film does not touch on the problem much, probably due to its complexity, but it is worth exploring none the less.
The Navier-Stokes existence and smoothness problem concerns the mathematical properties of solutions to the Navier?Stokes equations, one of the pillars of fluid mechanics. Imagine waves following our boat as we drive across a lake, or turbulent air currents following flights in a modern jet. Mathematicians and physicists believe that an explanation for and prediction of both the breeze and the turbulence can be found through an understanding of the solutions to the Navier-Stokes equations.
While the Navier-Stokes existence and smoothness problem is a very broad range of questions to be solved, the actual Millenium problem focuses on a solution for one specific statement of the problem:
In three space dimensions and time, given an initial velocity field, there exists a vector velocity and a scalar pressure field, which are both smooth and globally defined, that solve the Navier?Stokes equations.
In order to claim victory, you either need to provide a valid counter-example, prove the above statement conclusively. A solution to this problem would help gain a better understanding of the world we live in, both from a scientific and engineering standpoint. The equations can be used to model the weather, ocean currents, water flowing in a pipe and air flow around a wing. Using them can, amongst other things, aid in the design of aircraft and cars, the study of blood flow, the design of power stations, and the analysis of pollution.
There are 4 other Millenium Problems, all very interesting to explore, and very difficult to prove. However, I like to keep my articles short, so I?ll leave those 4 for another time. I will, however, note that my personal favourite is the P versus NP Problem, a major unsolved problem in computer science (my current degree). It simply asks whether every problem whose solution can be quickly verified can also be solved quickly. Of course, there are many definitions that go around that statement in order to provide some clarity, but that?s one for my next article.
In the meantime, if you are interested in reading about any of the Millenium Problems (or fancy taking a shot at solving one), there are listed here of the Clay Mathematics Institute website. And if you want me to do a more in-depth review of one of the problems, trying my best to convert the complex mathematics into something I (and hopefully you) can understand, be sure to leave a comment letting me know which one you want me to do. | 2,163 | 9,830 | {"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} | 3.84375 | 4 | CC-MAIN-2023-50 | longest | en | 0.949612 |
http://ecoursesbook.com/cgi-bin/ebook.cgi?topic=fl&chap_sec=01.5&page=theory | 1,632,167,916,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057091.31/warc/CC-MAIN-20210920191528-20210920221528-00547.warc.gz | 19,050,446 | 4,805 | Ch 1. Basics Multimedia Engineering Fluids MassDensity IdealGas Law Viscosity SurfaceTension VaporPressure
Chapter 1. Basics 2. Fluid Statics 3. Kinematics 4. Laws (Integral) 5. Laws (Diff.) 6. Modeling/Similitude 7. Inviscid 8. Viscous 9. External Flow 10. Open-Channel Appendix Basic Math Units Basic Equations Water/Air Tables Sections Search eBooks Dynamics Fluids Math Mechanics Statics Thermodynamics Author(s): Chean Chin Ngo Kurt Gramoll ©Kurt Gramoll
FLUID MECHANICS - THEORY
The vaporization and condensation processes will be introduced in this section. The concept of vapor pressure is then presented. The conditions when boiling and cavitation occur will be discussed as well.
Vaporization and Condensation
Vaporization (Evaporation)
Condensation: Contrails
When the molecules at the surface of a liquid or solid gain enough energy to overcome the cohesive force, the molecules will escape into the air. The substance undergoes a phase change and turns into vapor. This process is referred to as vaporization (e.g., evaporation and sublimation, see below). In general, the rate of vaporization increases with the temperature.
The process of phase change from liquid to vapor is called evaporation. For example, water will evaporate into vapor when the temperature reaches 100oC (for atmospheric pressure at sea level). A phase change directly from solid to vapor is called sublimation. An example of a sublimation process is dry ice at room temperature. The dry ice will become vapor. Also, ever wonder why moth balls will "disappear" over a period of time, but one can still smell the odor? This is another example of sublimation.
The process of phase change from a vapor to liquid is called condensation. The phenomenon of condensation can be observed in our daily lives. The contrail (condensation trail) left by an airplane is an example of the condensation process. As the hot humid exhaust air from the nozzle mixes with the surrounding air, a visible trail is formed due to condensation.
Vapor Pressure
Vapor Pressure
Now consider a closed container partially filled with a liquid, as shown in the figure. As the liquid molecules at the surface gain sufficient energy to escape into the air (evaporation), some of the liquid molecules will collide with the wall or air molecules, bounce back and re-enter the liquid (condensation). Over a period of time, the system will reach a steady-state where the rate of evaporation is the same as the rate of condensation. At this instance, the pressure exerted on the liquid surface by the liquid vapor is called vapor pressure. Vapor pressure is a fluid property, and it is a function of the temperature. Generally, the vapor pressure increases with temperature. The value of vapor pressure for water and is summarized in the table below as a function of temperature.
20 C (68 F) pv (kPa) pv (psi) Carbon Tetrachloride 1.3e4 1.9e0 Ethyl alcohol 5.9e3 8.5e-1 Gasoline 5.5e4 8.0e0 Glycerin 1.4e-2 2.0e-6 Kerosene 3.1e3 4.5e-1 Mercury 1.6e-1 2.3e-5
Various Liquids Vapor Pressure (absolute)
Temp(oC) pv(kPa) Temp(oF) pv(psi) 0 0.611 32 0.0885 5 0.872 40 0.1217 10 1.228 50 0.1781 15 1.666 60 0.2563 20 2.338 70 0.3631 30 4.243 80 0.5069 40 7.376 90 0.6980 50 12.33 100 0.9493 60 19.92 120 1.692 70 31.16 140 2.888 80 47.34 160 4.740 90 70.10 180 7.507 100 101.3 212 14.69
Water Vapor Pressure (absolute)
Boiling and Cavitation
Boiling
Boiling will occur when the absolute pressure of a liquid is less than or equal to its vapor pressure. One characteristic of the boiling process is the formation of vapor bubbles in the liquid. The formation and collapse of bubbles, primarily due to a reduction in pressure, in fluid flow is called cavitation (flow induced boiling). In engineering applications (e.g., pumps, turbines and hydraulic systems), it is a good practice to avoid cavitation because it can cause structural damage, produce noise, and reduce the overall efficiency of the system.
Elev (km) Pressure (kPa) Elev (ft) Pressure (psia) 0 101.33 0 14.70 2 79.50 5,000 12.24 4 60.12 10,000 10.11 6 47.22 15,000 8.30 8 35.65 20,000 6.76 10 26.5 25,000 5.46
Standard Atmosphere Pressure
at Various Elevations
It is well known to backpackers that boiling of water is dependent on the current elevation above sea level. An increase in elevation reduces the atmospheric pressure. With a lower pressure, water (or any liquid) will boil at a lower temperature that matches the vapor pressure. For example, at an elevation of 4 km, the atmospheric pressure is about 60 kPa (see table at left). At that pressure, water will boil at about 86 C (see table above). This does not mean food will cook faster or slower, but boiling will just occur at a lower temperature.
Practice Homework and Test problems now available in the 'Eng Fluids' mobile app
Includes over 250 problems with complete detailed solutions.
Available now at the Google Play Store and Apple App Store. | 1,303 | 4,941 | {"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-2021-39 | latest | en | 0.849394 |
https://www.essaysauce.com/computer-science-essays/essay-parameter-estimation-of-induction-motor-using-soft-computing-technique/ | 1,713,509,361,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817289.27/warc/CC-MAIN-20240419043820-20240419073820-00223.warc.gz | 697,654,469 | 19,923 | > > Parameter estimation of induction motor using soft computing technique
# Essay: Parameter estimation of induction motor using soft computing technique
• Subject area(s): Computer science essays
• Reading time: 9 minutes
• Published: 13 September 2015*
• File format: Text
• Words: 2,449 (approx)
• Number of pages: 10 (approx)
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Abstract:
Parameter estimation of an induction motor was carried out in conventional methods in order to determine the steady state equivalent circuit parameters. In conventional method, we obtained the parameters of an induction motor by no-load and blocked rotor test. The results obtained from the tests may not give accurate values of the parameter and it have a wide variation in the operating condition. Though the obtained parameters using conventional methods was not precise and effective when analyzing the machine for dynamic conditions so we are in the urge to adopt any other methodologies which make the estimation of parameters precise. Parameter estimation of an induction motor is carried out using soft computing methodologies is one of the way to implement for better efficiency. In the proposed scheme we presents a new algorithm based on the Artificial Bee Colony (ABC) to optimize the parameters of an Induction Motor. The proposed approach will be compared with the conventional method in order to interpret the efficiency.
1. Introduction
Induction machines are the mostly used motors in industry due to their low price and ruggedness. Induction motors(IM) have important problems, such as transient and quasi-steady state stability. To solve steady-state stability problems of induction motor, equivalent circuit parameters are required. These parameters are resistances and reactances of stator and rotor, including magnetizing branches. Estimation of these parameters is particularly essential in determining their effects on motor performance. About 60% of the industrial electric energy is converted into mechanical energy by means of pumps, fans, adjustable speed drives and the machine tools equipped with induction motors. Thus many researchers have focused on topics like modeling and parameter estimation of induction motors. The main problem of induction motor parameter estimation is the unavailability of manufacturer data to construct accurate models. Due to this reason, the induction motor models are not explicitly represented in various applications. The parameters of induction motors can be determined by common no-load and locked rotor tests. The main disadvantage of this method is that the motor has to be locked mechanically and tests have to be carried out by skilled operators. The parameter values obtained by the classical approaches can reveal signi’cant differences in the entire range of slip .This leads to the conclusion that to describe the performance of the induction machine more precisely and to reduce the differences between the estimated and real performances, one must modify the parameters obtained from the classical method. To achieve this goal, the use of optimization techniques seems to be a promising alternative to the classical approaches. Recently, in solving induction motor parameter estimation problems, some new global optimization techniques such as Evolutionary Algorithm(EA), Genetic Algorithm(GA), Differential Evolution(DE) ,Particle Swarm Optimization(PSO), Immune Algorithm(IA) have been proposed. In this research, the optimal parameter of the equivalent circuit of three-phase induction machine is suggested by Artificial Bee Colony (ABC) algorithm.
2.Problem Formulation
An Induction motor can be modeled by using an approximate circuit model,an exact circuit model. The parameter estimation problem is formulated as a least squares optimization problem, the objective being the minimization of deviation between the estimated and the manufacturer data. To achieve this utilizing the torque equation of the induction motor and the problem formulation is discussed for the two model.
2.1 Approximate Circuit Model:
Approximate circuit model of an induction motor is shown in Figure 1. In this model, R1 is the stator winding resistance,R2 is the rotor resistance referred to the stator side,X1 is the combined stator and rotor leakage reactance,Xm is the magnetizing reactance referred to the stator side,I1 is stator current,I2 is rotor current referred to the stator side,Vph is terminal voltage and s is the motor slip. The magnetizing reactance (Xm) can be eliminated from this model, because it has no effect on the rotor current and motor torque and power. Thus, the aim is to estimate R1, R2 and X1 by using the starting torque, the maximum torque and the full load torque of the motor given by the manufacturer along with the motor slip and terminal voltage.
2.1.1 Objective function:
Using the approximate circuit model of an induction motor:
Fig-1.Approximate Circuit Model
Where s, ??s,Vph are known (??s is the synchronous speed). Tf1(cal), Tlr(cal), Tmax(cal) are full load torque, locked rotor torque and maximum torque respectively that are calculated by estimating R1,R2′,X1. For this,the objective function is as follows:
where are Tf1(mf), Tlr(mf), Tmax(mf) full load torque, locked rotor torque and maximum torque given by the manufacturer respectively. For the optimum estimation of unknown parameters, we should minimize the objective function ‘F’.
2.2 Exact Circuit Model:
Exact circuit model representing the steady state operation of an induction motor is shown in Figure 2. At this model X1 is stator leakage reactance and X2′ is rotor leakage reactance referred to the stator side. Other parameters of this model are the same as approximate circuit model.
In the conventional technique, by performing the direct current test, we can find each phase resistance of the stator winding (R1). This test should be performed by the dc voltage supply because of preventing of the inductive effect in the stator winding. By performing the no load test,(X1+Xm) is found. This test performs under rated voltage and frequency.
By the locked rotor test (X1+X2′) and locked rotor resistance (RLR) are found. This test performs by locking the rotor and under a voltage that is much less than rated value. By knowing X1=X2’and RLR=R1+R2′ , the unknown parameters X1,X2′,Xm,R2′ are obtained individually. These tests can not implement easily and they need dc voltage supply, the voltage supply with tunable voltage and frequency, voltmeters, ammeters, watt meters and external element for locking the rotor. Beside it takes much time. By formulation of the parameter estimation of induction motor, these parameters can be found easily with high accuracy. The problem formulation uses the starting torque, the maximum torque, the full load torque and the full load power factor given by the manufacturer along with motor slip and terminal voltage to estimate the stator resistance, the rotor resistance, the stator leakage reactance, the rotor leakage reactance and the magnetizing reactance by proposed methods.
Fig-2 Exact Circuit Model
The objective function is as follows:
2.2.1 Objective function
At the exact circuit model of an induction motor:
Where s, ??s,Vph are known. Tf1(cal), Tlr(cal), Tmax(cal),pf(cal) are full load torque, locked rotor torque and maximum torque and power factor respectively that are calculated by estimating R1,R2′,X1,X2′ and Xm. Objective function is as follows:.
For optimum estimation of unknown parameters, we should minimize the objective function ‘F’.
2.3 Deep bar circuit model formulation:
The problem formulation uses the starting torque, maximum torque, full load torque, full load current and full load power factor manufacturer data to estimate the parameters of beep bar circuit model. The equivalent circuit for a deep bar or double cage induction motor model is shown in Fig. 3.
3.Artificial Bee Colony Algorithm
3.1 Fundamental of Artificial Bee Colony Algorithm:
Swam Intelligence (SI) is an emerging field in Artificial Intelligence (AI). The living nature and life style of animals, birds and other living organisms can be inherited and applied to solve many real world problems. ABC is a recently developed swam intelligence algorithm developed by Dervis Karaboga in the year 2005.In ABC, foraging is one of the behavior of honey bees to search, collect food from its food resources. Many research works has undergone about foraging behavior and it is applied to solve variety of optimization problems.
The minimal model of swarm-intelligent forage selection in a honey bee colony which the ABC algorithm simulates consists of three kinds of bees: employed bees, onlooker bees and scout bees. Half of the colony consists of employed bees, and the other half includes onlooker bees. Employed bees are responsible for exploiting the nectar sources explored before and giving information to the waiting bees (onlooker bees) in the hive about the quality of the food source sites which they are exploiting. Onlooker bees wait in the hive and decide on a food source to exploit based on the information shared by the employed bees. Scouts either randomly search the environment in order to find a new food source depending on an internal motivation or based on possible external clues.
3.2 Description Of ABC Algorithm:
Using the analogy between emergent intelligence in foraging of bees and the ABC
algorithm, the units of the basic ABC algorithm can be explained as follows:
3.2.1. Producing Initial Food Source Sites:
If the search space is considered to be the environment of the hive that contains the food source sites, the algorithm starts with randomly producing food source sites that correspond to the solutions in the search space. Initial food sources are produced randomly within the range of the boundaries of the parameters.
where i = 1. . .SN, j = 1. . .D..SN is the number of food sources and D is the number of optimization parameters. In addition, counters which store the numbers of trials of solutions are reset to 0 in this phase.
After initialization, the population of the food sources (solutions) is subjected to repeat cycles of the search processes of the employed bees, the onlooker bees and the scout bees. Termination criteria for the ABC algorithm might be reaching a maximum cycle number (MCN) or meeting an error tolerance.
3.2.2 Sending Employed Bees To The Food Source Sites:
As mentioned earlier, each employed bee is associated with only one food source site. Hence, the number of food source sites is equal to the number of employed bees. An employed bee produces a modification on the position of the food source (solution) in her memory depending on local information (visual information) and finds a neighboring food source, and then evaluates its quality. In ABC, finding a neighboring food source is defined by
Within the neigbourhood of every food source site represented by xi, a food source vi is determined by changing one parameter of xi. In Eq. (2), j is a random integer in the range [1,D] and k ?? {1, 2, . . .SN} is a randomly chosen index that has to be different from i.??ij is a uniformly distributed real random number in the range [-1, 1].
As can be seen from Eq., as the difference between the parameters of the xi,j and xk,j decreases, the perturbation on the position xi,j decreases. Thus, as the search approaches to the optimal solution in the search space, the step length is adaptively reduced. If a parameter value produced by this operation exceeds its predetermined boundaries, the parameter can be set to an acceptable value. In this work, the value of the parameter exceeding its boundary is set to its boundaries. If xi > xi max then xi = xi
max .If xi < xi min then xi = ximin.
After producing vi within the boundaries, a fitness value for a minimization problem can be assigned to the solution vi by
where fi is the cost value of the solution vi. For maximization problems, the cost function can be directly used as a fitness function. A greedy selection is applied between xi and vi , then the better one is selected depending on fitness values representing the nectar amount of the food sources at xi and
vi. If the source at vi is superior to that of xi in terms of profitability, the employed bee memorizes the new position and forgets the old one. Otherwise the previous position is kept in memory. If xi cannot be improved, its counter holding the number of trials is incremented by 1, otherwise, the counter is reset to 0.
3.2.3 Calculating Probability Values Involved In Probabilistic Selection:
After all employed bees complete their searches, they share their information related to the nectar amounts and the positions of their sources with the onlooker bees on the dance area. This is the multiple interaction feature of the artificial bees of ABC. An onlooker bee evaluates the nectar information taken from all employed bees and chooses a food source site with a probability related to its nectar amount. This probabilistic selection depends on the fitness values of the solutions in the population. A fitness-based selection scheme might be a roulette wheel, ranking based, stochastic universal sampling, tournament selection or another selection scheme. In basic ABC, roulette wheel selection scheme in which each slice is proportional in size to the fitness value is employed.
In this probabilistic selection scheme, as the nectar amount of food sources (the fitness of solutions) increases, the number of onlookers visiting them increases, too. This is the positive feedback feature of ABC.
3.2.4 Food Source Site Selection By Onlookers Based On The Information Provided By Employed Bees:
In the ABC algorithm, a random real number within the range [0,1] is generated for each source. If the probability value (pi in Eq. (4)) associated with that source is greater than this random number then the onlooker bee produces a modification on the position of this food source site by using Eq. (2) as in the case of the employed bee. After the source is evaluated, greedy selection is applied and the onlooker bee either memorizes the new position by forgetting the old one or keeps the old one. If solution xi cannot be improved, its counter holding trials is incremented by 1, otherwise, the counter is reset to 0. This process is repeated until all onlookers are distributed onto food source sites.
3.2.5 Abandonment Criteria: Limit And Scout Production
In a cycle, after all employed bees and onlooker bees complete their searches, the algorithm checks to see if there is any exhausted source to be abandoned. In order to decide if a source is to be abandoned, the counters which have been updated during search are used. If the value of the counter is greater than the control parameter of the ABC algorithm, known as the ”limit’, then the source associated with this counter is assumed to be exhausted and is abandoned. The food source abandoned by its bee is replaced with a new food source discovered by the scout, which represents the negative feedback mechanism and fluctuation property in the self-organization of ABC. This is simulated by producing a site position randomly and replacing it with the abandoned one. Assume that the abandoned source is xi, then the scout randomly discovers a new food source to be replaced with xi. This operation can be defined as in (1). In basic ABC, it is assumed that only one source can be exhausted in each cycle, and only oneemployed bee can be a scout. If more than one counter exceeds the ”limit’ value, one of the maximum ones might be chosen programmatically.
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# How much is 1500000000 divided by 1200000?
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2017-06-27 22:55:56
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1250
Jarrod Ruecker
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2021-02-25 22:07:54
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Wiki User
2017-06-27 22:55:56
1.5 x 10^9 divided by 1.2 x 10^6 = 1.25 x 10^3 = 1250
Wiki User
2017-06-27 12:11:41
1500000000 ÷ 1200000 = 1250
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Submit | 250 | 608 | {"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-2022-05 | latest | en | 0.825653 |
https://blueribbonwriters.com/consider-a-frictionless-track-as-shown-in-the-figure-below-a-block-of-mass-m1-5-40-kg-is-released-from-a-it-makes-a-head-on-elastic-collision-at/ | 1,656,838,624,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104215805.66/warc/CC-MAIN-20220703073750-20220703103750-00130.warc.gz | 203,014,834 | 10,922 | # Consider a frictionless track as shown in the figure below. A block of mass m1 = 5.40 kg is released from A. It makes a head on elastic collision at…
Consider a frictionless track as shown in the figure below. A block of mass m1 = 5.40 kg is released from A. It makes a head on elastic collision at B with a block of mass m2 = 12.0 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision. height = 5m | 115 | 444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-27 | latest | en | 0.962211 |
https://www.yaclass.in/p/mathematics-state-board/class-9/coordinate-geometry-3532/section-formula-and-centroid-5282/re-1c6a4d97-ac10-44b2-905a-d6e05e4dc6c4 | 1,624,063,059,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643354.47/warc/CC-MAIN-20210618230338-20210619020338-00229.warc.gz | 993,392,137 | 7,908 | Theory:
Until now, we have only dealt with mid-point and the points of trisection.
Mid-point divides the line segment into two halves and the points of trisection divides the line segment into $$3$$ equal parts.
But, is it possible to divide the line segment into two unequal parts?
Yes, a line segment can be divided into two unequal parts using the section formula.
Imagine you have $$6$$ milk packets and two bags of unequal sizes.
Bag $$A$$ can hold $$4$$ milk packets while bag $$B$$ can hold only $$2$$ milk packets.
In this case, a total of $$6$$ milk packets is distributed across the two bags in the ratio of $$4:2$$.
Similarly, a line segment can also be divided in unequal ratios.
Let us look at how a section formula gets constructed.
In the figure given above, a line segment $$AB$$ is divided into two unequal parts in the ratio $$m : n$$.
Let $$A$$ be $$x_1$$, $$P$$ be $$x$$ and $$B$$ be $$x_2$$ such that $$x_2 > x > x_1$$.
The co-ordinate of $$P$$ divides the line segment in the ratio $$m : n$$.
This means, $$\frac{AP}{PB} = \frac{m}{n}$$
$$\frac{x - x_1}{x_2 - x} = \frac{m}{n}$$
$$m(x_2 - x)$$ $$=$$ $$n(x - x_1)$$
$$mx_2 - mx = nx - nx_1$$
$$mx_2 + nx_1 = mx + nx$$
$$x$$ $$=$$ $$\frac{mx_2 + nx_1}{m + n}$$
If $$A$$, $$P$$, and $$B$$ has the co-ordinates $$(x_1$$, $$y_1)$$, $$(x$$, $$y)$$, and $$(x_2$$, $$y_2)$$ respectively, then:
$$x$$ $$=$$ $$\frac{mx_2 + nx_1}{m + n}$$
$$y$$ $$=$$ $$\frac{my_2 + ny_1}{m + n}$$ | 492 | 1,459 | {"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} | 4.6875 | 5 | CC-MAIN-2021-25 | latest | en | 0.782394 |
https://civicpride-kusatsu.net/what-is-160-pounds-in-kg/ | 1,653,286,697,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662555558.23/warc/CC-MAIN-20220523041156-20220523071156-00532.warc.gz | 229,624,832 | 4,916 | Conversion formula how to transform 160 pounds come kilograms?
We understand (by definition) that:1lb≈0.45359237kg
We can collection up a ratio to deal with for the number of kilograms.
You are watching: What is 160 pounds in kg
1lb160lb≈0.45359237kgxkg
Now, we cross main point to deal with for our unknown x:
xkg≈160lb1lb*0.45359237kg→xkg≈72.57477920000001kg
Conclusion:160lb≈72.57477920000001kg
Conversion in the contrary direction
The station of the conversion variable is that 1 kilogram is equal to 0.0137788913865548 times 160 pounds.
It can also be expressed as: 160 pounds is equal to 1 0.0137788913865548 kilograms.
Approximation
An almost right numerical an outcome would be: one hundred and also sixty pounds is around seventy-two point five 6 kilograms, or alternatively, a kilogram is around zero allude zero one time one hundred and sixty pounds.
Units involved
This is exactly how the systems in this conversion space defined:
Pounds
The lb or pound-mass (abbreviations: lb, lbm for most pounds) is a unit that mass used in the royal units, United states customary and also other solution of measurement. A variety of different meanings have been used, the most typical today gift the global avoirdupois lb which is legally characterized as specifically 0.45359237 kilograms, and which is separated into 16 avoirdupois ounces.<3> The worldwide standard symbol for the avoirdupois lb is lb.The unit is lower from the roman inn libra (hence the abbreviation lb). The English word lb is cognate with, amongst others, German Pfund, dutch pond, and also Swedish pund. All eventually derive from a borrowing right into Proto-Germanic that the Latin expression lībra pondō a pound by weight, in which the word pondō is the ablative instance of the Latin noun pondus. | 477 | 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} | 3.5625 | 4 | CC-MAIN-2022-21 | latest | en | 0.898535 |
https://discourse.radiance-online.org/t/modeling-a-bowl/1516 | 1,686,173,196,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654016.91/warc/CC-MAIN-20230607211505-20230608001505-00610.warc.gz | 249,131,977 | 6,037 | # Modeling... a bowl?
This may be simple... because it seems like it should be, but I'm quite hung
up on how to do this. How would I go about modeling an upside-down bowl? (Or
an upper hemisphere, or a rounded lampshade, whatever helps you picture it
best). It seems like "genrev" would be the way to do this, but I don't quite
understand what z(t) and r(t) do. If the input for a 2 foot wide hemisphere
(surface normals inward) isn't too complicated, perhaps someone could paste
it here and I could backwards-determine what the functions represent.
Here's another question: can we simply make a "bubble" and cut it in half?
(discarding half of it, of course)
Thanks!
Nick Calcagni
Nick,
Use gensurf rather than genrev. This command generates the bottom half of a unit sphere centred on the origin (e.g. a hemispherical 'bowl'):
!gensurf mat hemi 'sin(PI*s)*cos(PI*t)' 'cos(PI*s)' '-sin(PI*s)*sin(PI*t)' 20 20 -s
Remove the -ve sign in the z expression to get the top half. The -s argument applies smoothing to the surface normal.
-John
···
-----------------------------------------------
Dr. John Mardaljevic
Senior Research Fellow
Institute of Energy and Sustainable Development
De Montfort University
The Gateway
Leicester
LE1 9BH, UK
+44 (0) 116 257 7972
+44 (0) 116 257 7981 (fax)
Hi John,
Is there some reason to prefer gensurf to genrev in this instance? Genrev should produce a smaller (simpler) model, and allows for smoothing as well. The following makes a pretty convincing hemisphere:
genrev mat hemi 'cos(PI/2*t)' 'sin(PI/2*t)' 20 -s
using 20 primitives rather than 400. Since the first function z(t) is decreasing with increasing t, the surface normal points inwards as requested.
It's also possible to use an antimatter volume to hide the bottom half of the sphere, and this is recommended if a perfect hemisphere is required. However, there are drawbacks to this method, such as being unable to render from inside the excluded volume. Here's an example of how this could be done:
sphere_mat bubble my_sphere
0
4 0 0 0 1
void antimatter anti_sphere_mat
2 void sphere_mat
0
!genbox anti_sphere_mat anti_vol 2 2 1 | xform -t -1 -1 -1
Best,
-Greg
···
From: John Mardaljevic <[email protected]>
Date: June 4, 2008 6:35:57 AM PDT
Nick,
Use gensurf rather than genrev. This command generates the bottom half of a unit sphere centred on the origin (e.g. a hemispherical 'bowl'):
!gensurf mat hemi 'sin(PI*s)*cos(PI*t)' 'cos(PI*s)' '-sin(PI*s)*sin(PI*t)' 20 20 -s
Remove the -ve sign in the z expression to get the top half. The -s argument applies smoothing to the surface normal.
-John
-----------------
From: "Nick Calcagni" <[email protected]>
Date: June 3, 2008 10:47:56 PM PDT
This may be simple... because it seems like it should be, but I'm quite hung up on how to do this. How would I go about modeling an upside-down bowl? (Or an upper hemisphere, or a rounded lampshade, whatever helps you picture it best). It seems like "genrev" would be the way to do this, but I don't quite understand what z(t) and r(t) do. If the input for a 2 foot wide hemisphere (surface normals inward) isn't too complicated, perhaps someone could paste it here and I could backwards-determine what the functions represent.
Here's another question: can we simply make a "bubble" and cut it in half? (discarding half of it, of course)
Thanks!
Nick Calcagni
Greg,
Is there some reason to prefer gensurf to genrev in this instance?
Yeah, I was slavishly following your gensurf manpage example:
To generate a tesselated sphere:
gensurf crystal ball 'sin(PI*s)*cos(2*PI*t)' 'cos(PI*s)' 'sin(PI*s)*sin(2*PI*t)' 7 10
That and I'm much less frugal nowadays when it comes to counting polygons.
-John
···
----------------------------------------------
Dr. John Mardaljevic
Senior Research Fellow
Institute of Energy and Sustainable Development
De Montfort University
The Gateway
Leicester
LE1 9BH, UK
+44 (0) 116 257 7972
+44 (0) 116 257 7981 (fax)
Hi John,
Your suggestion does carry the advantage of producing correct smoothed normals on both sides of the surface, whereas the normal perturbations out of genrev are only correct for the "front" side of the object (the inside in this case).
Cheers,
-Greg
···
From: John Mardaljevic <[email protected]>
Date: June 4, 2008 8:51:33 AM PDT
Greg,
Is there some reason to prefer gensurf to genrev in this instance?
Yeah, I was slavishly following your gensurf manpage example:
To generate a tesselated sphere:
gensurf crystal ball 'sin(PI*s)*cos(2*PI*t)' 'cos(PI*s)' 'sin(PI*s)*sin(2*PI*t)' 7 10
That and I'm much less frugal nowadays when it comes to counting polygons.
-John | 1,277 | 4,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2023-23 | latest | en | 0.866073 |
https://www.studypool.com/discuss/1244436/How-can-I-understand-geometry-in-order-to-solve-the-following-?free | 1,480,848,417,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541317.69/warc/CC-MAIN-20161202170901-00202-ip-10-31-129-80.ec2.internal.warc.gz | 1,036,954,774 | 14,092 | ##### How can I understand geometry in order to solve the following?
Mathematics Tutor: None Selected Time limit: 1 Day
Oct 31st, 2015
We use the formula:
The exterior angle formed by 2 secants = 1/2(major arc - minor arc)
Therefore,
<EBF = 1/2(125 - 30)
= 1/2(95)
= 47.5 degrees
Choice D
Oct 31st, 2015
...
Oct 31st, 2015
...
Oct 31st, 2015
Dec 4th, 2016
check_circle | 135 | 380 | {"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-2016-50 | longest | en | 0.875335 |
https://richardvigilantebooks.com/what-is-200-ml-in-cups/ | 1,653,415,619,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662573189.78/warc/CC-MAIN-20220524173011-20220524203011-00683.warc.gz | 560,241,721 | 10,055 | # What is 200 ml in cups?
## What is 200 ml in cups?
Quick Conversions
U.S. Standard Metric
1 cup 200 ml and 2-15 ml spoons
1 1/4 cup 300 ml
1 1/3 cup 300 ml and 1-15 ml spoon
1 1/2 cup 350 ml
## How many glasses is 200ml?
We have to find the number of glasses of 200 ml capacity that are required to fill the bottle. Hence, 5 glasses of 200 ml capacity are required to fill the bottle.
177 ml
Volume (liquid)
1/3 cup 79 ml
1/2 cup 118 ml
2/3 cup 158 ml
3/4 cup 177 ml
## How many cups is 200 250 ml?
mL and cups conversion chart
Milliliters Cups Cups (fraction approx)
200 mL 0.85 Cups 4/5 cup
250 mL 1.06 cups 1 and 1/20 cup
300 mL 1.27 Cups 1 and 1/4 cups
350 mL 1.48 Cups 1 and 1/2 cups
## How can I measure 200 ml without a measuring cup?
Use a tablespoon to measure out the liquid you need. Pouring slowly and steadily to avoid excess spillage into the vessel, fill your tablespoon with the liquid. Transfer to the vessel and repeat until you have measured the amount you need in tablespoons.
## Is 1 cup 250 ml or 240mL?
Measuring Cup Sizes in ml for Different Countries
1 Cup # of Tbs in a Cup
US Customary 236 ml 16
US Legal 240 ml 16
Metric (US, Australia, New Zealand) or International 250 ml 16 2/3 (US, NZ) 12.5 (Australia)
Korea/Japan 200 ml 13 1/3 | 412 | 1,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2022-21 | longest | en | 0.83345 |
http://stackoverflow.com/questions/7618812/bigdecimal-for-financial-calculations-and-non-terminating-computations/7618840 | 1,406,712,149,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510270313.12/warc/CC-MAIN-20140728011750-00484-ip-10-146-231-18.ec2.internal.warc.gz | 275,215,834 | 16,906 | # Bigdecimal for financial calculations and non-terminating computations
I'm sure this would be a simple question to answer but I can't for the life of me decide what has to be done. So here's it: assuming we follow the "best practice" of using BigDecimal for financial calculations, how can one handle stuff (computations) which throw an exception?
As as example: suppose, I have to split a "user amount" for investing in bonds between "n" different entities. Now consider the case of user submitting \$100 for investing to be split between 3 bonds. The equivalent code would look like:
public static void main(String[] args) throws Exception {
BigDecimal bd1 = new BigDecimal("100.0");
BigDecimal bd2 = new BigDecimal("3");
System.out.println(bd1.divide(bd2));
}
But as we all know, this particular code snippet would throw an ArithmeticException since the division is non-terminating. How does one handle such scenarios in their code when using infinite precision data types during computations?
TIA,
sasuke
UPDATE: Given that RoundingMode would help remedy this issue, the next question is, why is 100.0/3 not 33.33 instead of 33.3? Wouldn't 33.33 be a "more" accurate answer as in you expect 33 cents instead of 30? Is there any way wherein I can tweak this?
-
divide method also has overloaded implementations which specify rounding mode see this link – Rajeev Sreedharan Oct 1 '11 at 7:23
@sasuke, on your update: I have tested bd1.divide(bd2, 2, RoundingMode.HALF_EVEN) (as shown below) and the result is 33.33 – krock Oct 1 '11 at 7:44
@krock: Thanks again for your help; much appreciated. :) – sasuke Oct 1 '11 at 12:32
The answer is to use one of the BigDecimal.divide() methods which specify a RoundingMode.
For example, the following uses the rounding mode half even or bankers rounding (but half up or one of the other rounding modes may be more appropriate depending on requirements) and will round to 2 decimal places:
bd1.divide(bd2, 2, RoundingMode.HALF_EVEN);
-
Thanks for the link; now the question is, why is 100.0/3 not 33.33 instead of 33.3? Wouldn't 33.33 be a "more" accurate answer? Is there any way wherein I can tweak this? – sasuke Oct 1 '11 at 7:34
@sasuke, for 33.33 set the scale to 2 as in the above example. It all depends on how many decimal places you want to round to. If you omit the scale parameter, it will use the scale of bd1, which is 1 (100.0 has one decimal place). – krock Oct 1 '11 at 7:37
Thanks, I completely overlooked that part. All my queries for the time being have been answered. :-) – sasuke Oct 1 '11 at 7:42
divide has an overload that takes a rounding mode. You need to choose one. I believe "half even" is the most commonly used one for monetary calculations.
-
+1; but this still doesn't explain how is 100/3 = 33.3 more logical than 100/3 = 33.33 given that 33 cents are a closer answer when compared to 30 cents. – sasuke Oct 1 '11 at 7:38
bd1.divide(bd2, 5, BigDecimal.ROUND_FLOOR)
`
It's an exemple, depending on the rounding you want.
-
I didnt downvote, but the reference you have posted is bad & certainly not from a trustable source. – Rajeev Sreedharan Oct 1 '11 at 7:36
Well, it wasn't exactly a "reference", more a code example. But ok, point taken I removed the link – Guillaume Oct 1 '11 at 8:16
Not sure who down-voted you but thanks for trying, +1. :) – sasuke Oct 1 '11 at 12:33 | 919 | 3,380 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2014-23 | latest | en | 0.894978 |
http://cboard.cprogramming.com/cplusplus-programming/72874-more-digits-output-particular-digits-output-output-*-txt-printable-thread.html | 1,475,131,992,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661779.44/warc/CC-MAIN-20160924173741-00132-ip-10-143-35-109.ec2.internal.warc.gz | 39,796,011 | 3,669 | # More digits in output, particular digits output, output in *.txt
• 11-30-2005
Ene Dene
More digits in output, particular digits output, output in *.txt
I've started programming 1 week ago and I use win32 console for an output.
In program which calculates pi:
Code:
Code:
```#include <iostream> #include <math.h> using namespace std; int main() { double b=-1; double pi=acos(b); cout << pi << endl; system("PAUSE"); return 0; }```
The output is:
Output:
Code:
`3.14159`
I know that computer calculated pi with much higher precision, so how do I get, I don't know... 15 digits on the output screen?
One more related question. I'm trying to write a random number generator program, since I've noticed that the rand() function doesn't have the uniform distribution of random number values, and for that I need a function that would "cut" double variable started from some decimal value to some other decimal value. For example, if I have a double variable number pi=3.141592654... I want a new integer variable with value from 3rd to 6th digit, int a=592. How do I do that?
The final question. If I want to test relationship between many output variables it would be very usefull if program could put these values in a txt file. How do I do that? Remember (I don't know if that maters) I'm programming (or trying to :) ), that is, the output is in win32 console.
Thank you!
• 11-30-2005
dimirpaw
Do you mean PI as in PIE? Well so far it's been 100% impossible to get the exact digits for pie since the Math Professers and other researchers are still getting 3.################################### ect...
Now I'm new as well but I might be able to just give my thought. The number PI= what your trying to get may be more bytes then a double can hold. Well thats just my guess.
• 11-30-2005
Ene Dene
Quote:
Originally Posted by dimirpaw
Do you mean PI as in PIE? Well so far it's been 100% impossible to get the exact digits for pie since the Math Professers and other researchers are still getting 3.################################### ect...
Now I'm new as well but I might be able to just give my thought. The number PI= what your trying to get may be more bytes then a double can hold. Well thats just my guess.
Of course that I didn't mean PI with infinite number of digits (as I stated by "I know that computer calculated pi with much higher PRECISION, so how do I get, I don't know... 15 digits on the output screen?"), it is transcedental, by definition it is imposible, but that was not the question.
• 11-30-2005
Enahs
Code:
```#include <cmath> #include <iomanip> .... .... cout << setprecision (15) << pi; //or any # .....```
• 11-30-2005
Ene Dene
Thanks! | 694 | 2,700 | {"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} | 2.5625 | 3 | CC-MAIN-2016-40 | latest | en | 0.894717 |
https://www.coursehero.com/file/177268/JOE-CRAIG-5/ | 1,488,227,923,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501173761.96/warc/CC-MAIN-20170219104613-00333-ip-10-171-10-108.ec2.internal.warc.gz | 806,956,541 | 24,123 | JOE CRAIG 5
# JOE CRAIG 5 - Economics 3070-001 Spring 2008 Problem Set 5...
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Economics 3070-001 Spring 2008 Problem Set 5 (Due March 5 th ) 1. Suppose that the production function for lava lamps is given by , where Q is the number of lamps produced per year, K is the machine-hours of capital, and L is the man-hours of labor. Suppose K = 600. a. Draw a graph of the production function over the range L = 0 to L = 500, putting L on the horizontal axis and Q on the vertical axis. Over what range of L does the production function exhibit increasing marginal returns? Diminishing marginal returns? Diminishing total returns? b. Derive the equation for average product of labor and graph the average product of labor curve. At what level of labor does the average product curve reach its maximum? c. Derive the equation for marginal product of labor. On the same graph you drew for part b, sketch the graph of the marginal product of labor curve. At what level
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## This note was uploaded on 05/04/2008 for the course ECON 3070 taught by Professor Loh,joyce during the Spring '07 term at Colorado.
### Page1 / 4
JOE CRAIG 5 - Economics 3070-001 Spring 2008 Problem Set 5...
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Ask a homework question - tutors are online | 556 | 1,590 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-09 | longest | en | 0.1676 |
http://www.downloadplex.com/Scripts/Matlab/Development-Tools/polynomial-composition-scripts_421063.html | 1,477,482,800,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720941.32/warc/CC-MAIN-20161020183840-00153-ip-10-171-6-4.ec2.internal.warc.gz | 411,976,444 | 11,825 | Create an account Log In
# Polynomial Composition (Scripts) 1.0
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## Polynomial Composition (Scripts) Publisher's description
### POLYCOMP Polynomial composition
POLYCOMP Polynomial composition
R = POLYCOMP(P,Q) returns the composition of Q(P(x)) given polynomials P and Q with coefficients in descending order.
If Q is a scalar, with value N, it assumes shorthand notation for Q(x)=x^N.
Output is a polynomial with coefficients in descending order.
Example:
% p(x) = 2x+1
% q(x) = 2x^2 + 4x + 3
% q(p(x)) = 2[2x+1]^2 + 4[2x+1] + 3
p=[2 1]
q=[2 4 3]
r=polycomp(p,q)
% p(x) = 2x+1
% q(x) = x^3 (scalar, shorthand notation)
% q(p(x)) = [2x+1]^3
p=[2 1]
q=3
r=polycomp(p,q)
#### System Requirements:
MATLAB 7.12 (2011a)
Program Release Status: New Release
Program Install Support: Install and Uninstall
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This program plots the deformation field (displace vectors) contained in vector.txt. | 500 | 1,743 | {"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} | 3.15625 | 3 | CC-MAIN-2016-44 | longest | en | 0.736871 |
https://kr.mathworks.com/matlabcentral/cody/problems/564-how-to-subtract/solutions/1294916 | 1,582,916,508,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147628.27/warc/CC-MAIN-20200228170007-20200228200007-00160.warc.gz | 429,862,812 | 15,667 | Cody
# Problem 564. How to subtract?
Solution 1294916
Submitted on 18 Oct 2017 by praharsh
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
X='+68768686834554'; Y='+76574535435398'; Z_correct='-7805848600844'; assert(isequal(mysub(X,Y),Z_correct))
Ztemp = -7.8058e+12
2 Pass
X='+1'; Y='+2'; Z_correct ='-1'; assert(isequal(mysub(X,Y),Z_correct))
Ztemp = -1
3 Pass
X='+100'; Y='+20'; Z_correct ='+80'; assert(isequal(mysub(X,Y),Z_correct))
Ztemp = 80 | 194 | 580 | {"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-2020-10 | latest | en | 0.523758 |
https://www.scienceforums.net/topic/122767-answer/?tab=comments | 1,601,142,971,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400244353.70/warc/CC-MAIN-20200926165308-20200926195308-00479.warc.gz | 956,337,449 | 13,319 | ## Recommended Posts
Posted (edited)
0+1=1
1+1=11
1+1+1=111
111-1=11
111-11=1
111-111=0
0÷0=1
1÷1=1
11÷11=1
111÷111=1
Question:
1+11=?+?+?=?-1111=?÷?=?
Edited by Frogger
##### Share on other sites
Hi frogger. Why do you insist on posting this silliness?
Fuck off!
##### Share on other sites
Well at least that is coherent!
× | 132 | 345 | {"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-2020-40 | latest | en | 0.575139 |
http://somemath.blogspot.com/2015/04/classifying-some-numbers.html | 1,526,990,337,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864725.4/warc/CC-MAIN-20180522112148-20180522132148-00356.warc.gz | 268,085,753 | 15,853 | ## Sunday, April 12, 2015
### Classifying some numbers
Careful study of a seemingly simple algebraic expression can reveal some numbers with interesting properties outside of currently established non-field ring classification schemes.
With a quadratic polynomial, we can consider its factorization in general with:
P(x) = (g1(x) + 1)(g2(x) + 2)
where, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.
Values for the g's are trivial for a polynomial factorization.
For example, if g1(x) = x, and g2(x) = x, then, of course P(x) = x2 + 3x + 2.
P(x) = (x+1)(x+2)
But if P(x) does not factor into polynomial factors, then the g's cannot be algebraic integers, which I've shown by multiplying by a constant integer where the result must be an algebraic integer.
For example,
11*P(x) = (f1(x) + 11)(f2(x) + 11), where g1(x) = f1(x)/11 and g2(x) = f2(x) + 9.
For algebraic integer x, the f's are in general within the ring of algebraic integers, which I've proven in detail in a post:
somemath.blogspot.com/2015/04/why-in-ring-of-algebraic-integers
But while the f's are algebraic integers for algebraic x, the g's can't both be, which is from the forced asymmetry. That's why I used 1 and 2, as I wanted something simple, and that's about as simple as I could get:
P(x) = (g1(x) + 1)(g2(x) + 2)
So consider some nonrational solutions for the g's, which I'll call o1 and o2, which would be for some particular nonzero, algebraic integer x.
I have that one is an algebraic integer, while the other is not, and does not fit within any currently established ring.
But if I multiply it, for instance, by 11, then it is an algebraic integer. But if I multiply the same number by 7, then it is also an algebraic integer. And in general I can multiply the same number by any nonzero integer not 1 or -1, and it will be an algebraic integer.
That is to say, if o1 is the non algebraic integer, then 7o1 and 11o1 are, and in general, ko1 is an algebraic integer for k, nonzero and not 1 or -1, while o1 is not:
k*P(x) = (f1(x) + k)(f2(x) + k)
where g1(x) = f1(x)/k and g2(x) = f2(x) + k-2
And then it is clear why fields are not impacted, as consider o2/o1, which might not seem to be in the field of algebraic numbers, but you just have to multiply top and bottom by some nonzero integer not 1 or -1, like:
2o2/2o1
and you have a ratio of algebraic integers, showing that it is indeed an algebraic number.
So this approach allows studying the underlying asymmetrical form by moving to a symmetrical form using basic algebra.
And it's nice to mark over 10 years since I came up with a more robust classification scheme with the object ring.
That is the first post on this blog made over 10 years ago. Yeah, it's that important.
Probably is the reason this blog exists, because it was long enough ago, I'm not sure all of what I was thinking at the time. But since I made it the very first post, got a feeling it had much to do with my motivations.
So I'd also like to celebrate that special event.
And will note coming up on a 20 year celebration this month.
That there are numbers which behave like integers, which are themselves not rational, which are NOT algebraic integers is fascinating.
And since they are not algebraic integers it's hard to see them directly. What you can do is get a solution for the f's and divide off the constant multiplier from both solutions knowing it works for one, and for the other you have something that is fractional, but that doesn't tell you which is which.
I've been thinking about these numbers for a while now, over a decade.
So how were they missed before? If you step through the tests done on the ring of algebraic integers, it's easy to see how they don't catch these numbers.
So I came up with a more inclusive ring classification and call these numbers along with integers--objects.
Objects replace algebraic integers at the foundations of numbers. The object ring includes the ring of algebraic integers plus the additional numbers previously missed.
James Harris | 1,053 | 4,045 | {"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} | 3.875 | 4 | CC-MAIN-2018-22 | latest | en | 0.913725 |
http://cboard.cprogramming.com/cplusplus-programming/26278-help.html | 1,477,698,058,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476990033880.51/warc/CC-MAIN-20161020190033-00408-ip-10-142-188-19.ec2.internal.warc.gz | 42,034,936 | 11,133 | 1. ## Help!!!!!
hi im new at programming, and i need to write a program that takes an integer value and returns the number with its digits revered. sooo basically a pallindrome? ex. 12345 would be 54321
but heres the catch i dont know how many integers the user will enter. so how would i go about writing this code?
any help is greatly appreciated thanks.
2. Write a function which repeatedly calculates the least signifigant digit and adds it to a return value, then divide the number by 10. When the value is zero, return the reversed value. The calculation would be something like this:
ret = ( val % 10 ) + ret * 10;
-Prelude
3. wow ok well when i said iw as new at programming i meant NEW strings we havent coverd thats like chaper 16 lol we are on 3 if anyone who can help me please post your aim handle if possible or contact me mine is gigaflare999 thanks for the replys btw i appreciate it, i just need like serious help lol
4. >if anyone who can help me please post your aim handle
I'm hurt , I pretty much gave you the solution without using strings. Just throw it in a loop and you're done:
Code:
```while ( val != 0 ) {
ret = ( val % 10 ) + ret * 10;
val /= 10;
}```
Now ret is the reversed val.
-Prelude
5. sheesh sorry i was lscared when i saw string [256] i ran in other direction! thanks for helpnig i get whiles easy! i appreciate it but will that work no matter how many ints they put in?
6. >i appreciate it but will that work no matter how many ints they put in?
It will if you place it in a loop.
Code:
```#include <iostream>
int reverse ( int val )
{
int ret = 0;
while ( val != 0 ) {
ret = ( val % 10 ) + ret * 10;
val /= 10;
}
return ret;
}
int main()
{
int num = 0;
std::cout<<"Enter a number (CTRL+Z to quit): ";
while ( std::cin>>num ) {
std::cout<<"Your number reversed was: "<<reverse ( num ) <<std::endl;
std::cout<<"Enter a number (CTRL+Z to quit): ";
}
return 0;
}```
-Prelude
7. omg thankyou so much you dont know how much you helped me. really. if you could reccomend any learning sites that are like uber simple please do, my computer teacher sucks ass so bad he doesnt teach. | 568 | 2,131 | {"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-2016-44 | latest | en | 0.872243 |
https://investingchef.com/nb/investing-menu/currency-page/pivot-points-calculation | 1,571,574,618,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986707990.49/warc/CC-MAIN-20191020105426-20191020132926-00364.warc.gz | 528,422,044 | 36,549 | ## Pivot points calculation
Calculating pivot points is an essential part of understanding of their usage, and even though most of the trading platforms already have a pivot points calculator built in, so the numbers are crunched up by themselves It would be nice for us to understand what pivot points are and how to calculate them before going any further.
In order to count pivot points as well as support and resistance levels you need to know last trading session’s open, high, low, and close. For this most traders use the numbers they can get out of the previous close of NY stock exchange as it the biggest and one of the most important stocks exchanges in the world.
## In order to count all of the lines in the picture you need these formulas:
• Pivot point (PP) = (High + Low + Close) / 3
• First resistance (R1) = (2 x PP) – Low
• First support (S1) = (2 x PP) – High
• Second resistance (R2) = PP + (High – Low)
• Second support (S2) = PP – (High – Low)
• Third resistance (R3) = High + 2(PP – Low)
• Third support (S3) = Low – 2(High – PP)
Those who do not like to count and use formulas do not have to be worried, as I already said platform usually count all of the numbers for you and you just need to understand how to implement all the calculation for your advantage. But that is a discussion for another time.
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https://www.aqua-calc.com/one-to-all/temperature/preset/kelvin/2173.15 | 1,718,428,789,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861583.78/warc/CC-MAIN-20240615031115-20240615061115-00602.warc.gz | 589,893,502 | 5,632 | Convert Kelvins [K] to other units of temperature
Kelvins [K] temperature conversions
2 173.15 K = 1 900 degrees Celsius K to °C 2 173.15 K = 3 452 degrees Fahrenheit K to °F 2 173.15 K = 3 911.67 degrees Rankine K to °R 2 173.15 K = 1 005 degrees Romer K to °Rø
Convert entered temperature to units of energy.
Foods, Nutrients and Calories
BACON HARDWOOD SMOKED, UPC: 041130308208 contain(s) 462 calories per 100 grams (≈3.53 ounces) [ price ]
26 foods that contain Fatty acids, total trans-dienoic. List of these foods starting with the highest contents of Fatty acids, total trans-dienoic and the lowest contents of Fatty acids, total trans-dienoic
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CaribSea, Freshwater, Flora Max, Midnight weighs 865 kg/m³ (54.00019 lb/ft³) with specific gravity of 0.865 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
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gr/US qt to dwt/US gal conversion table, gr/US qt to dwt/US gal unit converter or convert between all units of density measurement.
Calculators
Volume to weight and weight to volume conversions using density | 494 | 1,786 | {"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-2024-26 | latest | en | 0.738997 |
http://cboard.cprogramming.com/cplusplus-programming/147306-homework-help.html | 1,469,808,843,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257831769.86/warc/CC-MAIN-20160723071031-00169-ip-10-185-27-174.ec2.internal.warc.gz | 40,865,214 | 13,861 | # Thread: Homework help?
1. ## Homework help?
So the beginning numbers are laid out like this:
8 0 3 6 3
1 3 1 0 7
2 7 9 4 5
Then I have to print it out to look like:
3 6 3 0 8
7 0 1 3 1
5 4 9 7 2.
This is what I have so far, and i know its not right because i keep getting an error that the program has quit working.
Code:
```void
copyB(const int *fromPtr, int rows, int columns, int to[][10])
{
int r;
int c;
for (r = columns -1; r < rows; r--) {
for (c = 0; c < columns; c++)
to[r][c] = *fromPtr++;
}
}```
This is not my major I technically dont even need this class anymore since i switched majors but i dont want to drop it now, so any tips are appreciated. I have to run it against my professors test driver, but i just can get this one to work right.
Thanks for any help!
2. You don't need to copy it.
Just swap the n'th element of each row with the (columns -n)'th element.(...iterating n over 0 to columns )
3. Well with the example you showed the function should not do anything. If columns is 5 and rows is 3, then r would = 4. When you enter the for loop, r is not less than rows.
4. easy! This code works....
Code:
```#include<iostream.h>
void main()
{ int c[3][5];
for(int i=0;i<3;i++)
for(int j=0;j<5;j++)
cin>>c[i][j];
for(i=0;i<3;i++)
{for(int j=4;j>=0;j--)
cout<<c[i][j];
cout<<endl;}
}```
5. @ Amin sma
Note that we're here to help people figure out the answer for themselves.
Not show off by dumping the complete answer at the first opportunity.
6. Which, mind you, doesn't even compile on standard-compliant compilers!
7. That solution is wrong in at least three different ways. For starters it should not use a two-dimensional array when a plain 1D array is sufficient.
8. ok
9. Originally Posted by iMalc
That solution is wrong in at least three different ways. For starters it should not use a two-dimensional array when a plain 1D array is sufficient.
True, but his array is to[c][r]. You know how it is teachers are always allowed do things like that. ;-)
10. Originally Posted by überfuzz
True, but his array is to[c][r]. You know how it is teachers are always allowed do things like that. ;-)
I'm assuming that the decision of the function prototype is his own, and so could be whatever. Afterall, it's doing such a trivial task (reversing an array) that I would not even create a function for it.
Just read in each line and then print it out backwards. Amin got that part right, albeit with missing formatting.
11. Originally Posted by iMalc
Afterall, it's doing such a trivial task (reversing an array) that I would not even create a function for it.
It could pass as an exercise in the array part in some course. Ah well...
Popular pages Recent additions | 751 | 2,702 | {"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-2016-30 | latest | en | 0.889875 |
http://www.marialmuseum.org/high-low-cost-method.html | 1,674,816,247,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494976.72/warc/CC-MAIN-20230127101040-20230127131040-00403.warc.gz | 75,151,818 | 19,337 | # 荣耀电竞比赛下注登陆
Definition: The high-low method is a technique managerial accountants use to estimate the mixed production costs at various levels of production by calculating the variable cost rate and total fixed costs. In other words, it’s a formula used by management to split the fixed and variable costs associated with producing a good and chart out these data points. A line is then drawn connecting the lowest and highest points to represent the average.
## What Does High Low Method Mean?
What is the definition of the high-low method? This method can also be used to chart out all the purchases of goods and their prices. A straight line on the graph connects the highest price and lowest price of goods. Look at this example. The lowest price is the fixed price of \$20,000. The highest price is \$80,000.
The variable cost per unit can be found using a simple slope formula. Change in y divided by the change in x. According to the formula below, the variable cost per unit is 75 cents.
Now management can figure out how much it will cost to produce any amount of products. All they have to do is look at the fixed costs plus the variable cost per unit multiplied by the units produced. Rearranging this formula, management can also figure out what total fixed costs are if they were unknown.
The high low method is very useful for helping management determine not only what total costs will be, but also how much of a certain product to produce. There are a few shortfalls to the high low method. For instance, it does not recognize any other costs except the highest and lowest costs.
Keep in mind that this method is far less precise than other cost methods like the least-squares method . It’s a simply and easy way to understand the relationship between fixed and variable costs at different levels of output.
## Summary Definition
Define High Low Method: High-low method means a graphical approach to determining the average cost of purchasing a good or producing a product by plotting all transactions and connecting the lowest and highest points.
Contents
[i]
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# How Far Away is the Horizon?
It was the first time that Poole had seen a genuine horizon since he had come to Star City, and it was not quite as far away as he had expected.... He used to be good at mental arithmetic--a rare achievement even in his time, and probably much rarer now. The formula to give the horizon distance was a simple one: the square root of twice your height times the radius--the sort of thing you never forgot, even if you wanted to...
--Arthur C. Clarke, 3001, Ballantine Books, New York, 1997, page 71
In the above passage, Frank Poole uses a formula to determine the distance to the horizon given his height above the ground.
1. Use algebraic notation to express the formula Poole is using.
2. Beginning the diagram below, use one of the circle theorems to derive your own formula. You will need to add some more elements to the diagram.
horizon1.eps
3. Compare your formula to Poole's; you will find that they do not match. How are they different?
4. When I was a boy it was possible to see the Atlantic Ocean from the peak of Mt. Washington in New Hampshire. This mountain is 6288 feet high. How far away is the horizon? Express your answer in miles. Assume that the radius of the Earth is 4000 miles. Use both your formula and Poole's formula and comment on the results. Why does Poole's formula work so well, even though it is not correct?
Carl Lee
Wed Apr 21 08:26:07 EDT 1999 | 357 | 1,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-47 | latest | en | 0.95911 |
https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_4&oldid=108567 | 1,627,266,357,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151972.40/warc/CC-MAIN-20210726000859-20210726030859-00446.warc.gz | 128,485,062 | 17,542 | # 2019 AIME II Problems/Problem 4
## Problem 4
A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
## Solution
Notice that, other than the number 5, the remaining numbers 1, 2, 3, 4, 6 are only divisible by 2 and/or 3. We can do some cases on the number of 5's rolled (note that there are $6^4 = 1296$ outcomes).
Case 1 (easy): Four 5's are rolled. This has probability $\frac{1}{6^4}$ of occurring.
Case 2: Two 5's are rolled.
Case 3: No 5's are rolled.
To find the number of outcomes for the latter two cases, we will use recursion. Consider a 5-sided die with faces numbered 1, 2, 3, 4, 6. For $n \ge 1$, let $a_n$ equal the number of outcomes after rolling the die $n$ times, with the property that the product is a square. Thus, $a_1 = 2$ as 1 and 4 are the only possibilities.
To find $a_{n+1}$ given $a_n$ (where $n \ge 1$), we observe that if the first $n$ rolls multiply to a perfect square, then the last roll must be 1 or 4. This gives $2a_n$ outcomes. Otherwise, the first $n$ rolls do not multiply to a perfect square ($5^n - a_n$ outcomes). In this case, we claim that the last roll is uniquely determined (either 2, 3, or 6). If the product of the first $n$ rolls is $2^x 3^y$ where $x$ and $y$ are not both even, then we observe that if $x$ and $y$ are both odd, then the last roll must be 6; if only $x$ is odd, the last roll must be 2, and if only $y$ is odd, the last roll must be 3. Thus, we have $5^n - a_n$ outcomes in this case, and $a_{n+1} = 2a_n + (5^n - a_n) = 5^n + a_n$.
Computing $a_2$, $a_3$, $a_4$ gives $a_2 = 7$, $a_3 = 32$, and $a_4 = 157$. Thus for Case 3, there are 157 outcomes. For case 2, we multiply by $\binom{4}{2} = 6$ to distribute the two 5's among four rolls. Thus the probability is
$$\frac{1 + 6 \cdot 7 + 157}{6^4} = \frac{200}{6^4} = \frac{25}{162} \implies m+n = \boxed{187}$$
-scrabbler94
## Solution 2
We can solve this without finding the amount of cases. Notice when we roll a 1 or a 4, it does not affect whether or not the product is a square number. We have a 1/3 chance of rolling either a 1 or 4. We have a 2/3 chance of rolling a 2,3,5 or 6. Lets call rolling 1 or 4 rolling a dud.
Probability of rolling 4 duds: $\left(\frac{1}{3}\right)^4$
Probability of rolling 3 duds: $4 * \left(\frac{1}{3}\right)^3 * \frac{2}{3}$
Probability of rolling 2 duds: $6 * \left(\frac{1}{3}\right)^2 * \left(\frac{2}{3}\right)^2$
Probability of rolling 1 dud: $4 * \frac{1}{3} * \left(\frac{2}{3}\right)^3$
Probability of rolling 0 duds: $\left(\frac{2}{3}\right)^4$
Now we will find the probability of a square product given we have rolled each amount of duds
Probability of getting a square product given 4 duds: 1
Probability of getting a square product given 3 duds: 0 (you will have 1 non-dud and that's never going to be square)
Probability of getting a square product given 2 duds: $\frac{1}{4}$ (as long as our two non-duds are the same, our product will be square)
Probability of getting a square product given 1 duds: $\frac{3!}{4^3}$ = $\frac{3}{32}$(the only way to have a square product is rolling a 2,3 and 6. There are 3! ways of doing that and a total of $4^3$ ways to roll 3 non-duds).
Probability of getting a square product given 0 duds: $\frac{40}{4^4}$= $\frac{5}{32}$ (We can have any two non-duds twice. For example, 2,2,5,5. There are $\binom{4}{2} = 6$ ways of choosing which two non-duds to use and $\binom{4}{2} = 6$ ways of choosing how to arrange those 4 numbers. That gives us 6*6=36 combinations. We can also have 2,2,2,2 or 3,3,3,3 or 5,5,5,5 or 6,6,6,6. This gives us a total of 40 combinations).
We multiply each probability of rolling k duds with the probability of getting a square product given k duds and then sum all the values.
$$\left(\frac{1}{3}\right)^4 * 1 + 4 * \left(\frac{1}{3}\right)^3 * \frac{2}{3} * 0 + 6 * \left(\frac{1}{3}\right)^2 * \left(\frac{2}{3}\right)^2 * \frac{1}{4} + 4 * \frac{1}{3} * \left(\frac{2}{3}\right)^3 * \frac{3}{32} + \left(\frac{2}{3}\right)^4 * \frac{5}{32} = \frac{25}{162}.$$
$25+162$ = $\boxed{187}$
-dnaidu (silverlizard)
$a\#b = ab-3\lefta+b\right - 12$ (Error compiling LaTeX. ! Undefined control sequence.). Calculate $n\#\leftn-1\right\#\leftn-2\right...\#3\#2\#1$ (Error compiling LaTeX. ! Undefined control sequence.).
## Solution 3
Note that rolling a 1/4 will not affect whether or not the product is a perfect square. This means that in order for the product to be a perfect square, all non 1/4 numbers rolled must come in pairs, with the only exception being the triplet 2,3, 6. Now we can do casework:
If there are four 1/4's, then there are $2^4=16$ combinations. If there are three 1/4's, then there are 0 combinations, because the fourth number isn't a square. If there are two 1/4's, there are $2^2=4$ ways to choose the two 1/4's, 4 ways to choose the remaining pair of numbers, and $\frac{4!}{2!2!}=6$ ways to arrange, so there are $4\cdot 4\cdot 6=96$ combinations for this case. If there is one 1/4, then there are 2 ways to choose whether it is a 1 or 4, and the remaining three numbers must be 2, 3, and 6, so there are $4!$ ways to order, meaning there are $2\cdot 4!=48$ combinations for this case. Our final case is if there are no 1/4's, in which case we must have two pairs. If the two pairs are of different numbers, then there $\binom{4}{2}$ to choose the numbers and $\frac{4!}{2!2!}=6$ ways to arrange them, so $6\cdot 6=36$. If all four numbers are the same there are $4$ combinations, so there are $4+36=40$ combinations for this case.
Hence there are $16+0+96+48+40=200$ combinations where the product of the dice is a perfect square, and there are $6^4=1296$ total combinations, so the desired probability is $\frac{200}{1296}=\frac{25}{162}$, yielding an answer os $25+162=\boxed{187}$.
-Stormersyle
## Solution 4(Casework)
Another way to solve this problem is to do casework on all the perfect squares from $1^2$ to $36^2$, and how many ways they can be ordered $1^2$- $1,1,1,1$- $1$ way. $2^2$- $4,1,1,1$ or $2,2,1,1$- $\binom{4}{2}+4=10$ ways. $3^2$- $3,3,1,1$- $\binom{4}{2}=6$ ways. $4^2$- $4,4,1,1$, $2,2,2,2$, or $2,2,4,1$- $\binom{4}{2}+1+12=19$ ways. $5^2$- $5,5,1,1$- $\binom{4}{2}=6$ ways. $6^2$- $6,6,1,1$, $1,2,3,6$, $2,3,2,3$, $3,3,4,1$- $2*\binom{4}{2}+4!+12=48$ ways. $7^2$- Since there is a prime greater than 6 in its prime factorization there are $0$ ways. $8^2$- $4,4,4,1$ or $2,4,2,4$- $\binom{4}{2}+4=10$ ways. $9^2$- $3,3,3,3$- $1$ way. $10^2$- $2,2,5,5$ or $1,4,5,5$- $6+12=18$ ways. $11^2$- $0$ ways for the same reason as $7^2$. $12^2$- $6,6,2,2$, $4,4,3,3$, $2,3,4,6$, or $1,4,6,6$- $2*\binom{4}{2}+4!+12=48$ ways. $13^2$- $0$ ways. $14^2$- $0$ ways. $15^2$- $3,3,5,5$- $\binom{4}{2}=6$ ways. $16^2$- $4,4,4,4$- $1$ way. $17^2$- $0$ ways. $18^2$- $3,3,6,6$- $\binom{4}{2}=6$ ways. $19^2$- $0$ ways. $20^2$- $4,4,5,5$- $\binom{4}{2}=6$ ways. $21^2$- $0$ ways. $22^2$- $0$ ways. $23^2$- $0$ ways. $24^2$-$4,4,6,6$- $\binom{4}{2}=6$ ways. $25^2$- $5,5,5,5$- $1$ way. $26^2$- $0$ ways. $27^2$- $0$ ways. $28^2$- $0$ ways. $29^2$- $0$ ways. $30^2$- $5,5,6,6$- $\binom{4}{2}$ ways. $31^2$- $0$ ways. $32^2$- $0$ ways. $33^2$- $0$ ways. $34^2$- $0$ ways. $35^2$- $0$ ways. $36^2$- $6,6,6,6$- $1$ way.
There are $6^4=1296$ ways that the dice can land. Summing up the ways, it is easy to see that there are $200$ ways. This results in a probability of $\frac{200}{1296}=\frac{25}{162}\implies\boxed{187}$ -superninja2000 | 2,922 | 7,605 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 173, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2021-31 | latest | en | 0.850015 |
http://brainstellar.com/puzzles/hard/217 | 1,534,476,241,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221211664.49/warc/CC-MAIN-20180817025907-20180817045907-00042.warc.gz | 79,027,806 | 7,267 | Hard puzzles
An aircraft hovers above sea, trying to catch a submarine moving with a constant velocity under the sea. The submarine is completely invisible, but using a human radar only once, the aircraft knows the exact location of submarine under the sea. The direction of submarine is unknown, but constant. The aircraft can move at twice the speed of submarine. As soon as the aircraft is just vertically above the submarine, Aircrafet can magnetically pick it up. How does the aircraft catch the submarine? How much time ill it take?
PS: This scene is from X-men: First Class. Good x-men are in the aircraft called blackbird, human radar is Banshee, Magnet is Magneto. Bad x-men are in submarine, with Sebastian Shaw is about to cause a war, better catch him soon!
Hint
Solution
Source: Inspired from Rustan Leino's puzzle
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A spy is located on a one-dimensional line. At time 0, the spy is at location A. With each time interval, the spy moves B units to the right (if B is negative, the spy is moving left). A and B are fixed integers, but they are unknown to you. You are to catch the spy. The means by which you can attempt to do that is: at each time interval (starting at time 0), you can choose a location on the line and ask whether or not the spy is currently at that location. That is, you will ask a question like "Is the spy currently at location 27?" and you will get a yes/no answer. Devise an algorithm that will eventually find the spy
Solution
Source: Written Test; leino
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You are given N coins which look identical (assume N = 2^k). But actually some of them are pure gold coins (hence are heavy) and the rest are aluminum coins with thin gold plating (light). You are given one beam balance with two pans. What is the number of weighing required to separate the gold from fake coins? (all gold coins have equal weights & all fake coins too have the same weight)
Hint
Solution
Source: CSEblog
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Two immensely intelligent players, A & B, engage in a game, the rules of which are as follows. For some natural number N, the board consists of numbers from 1 to N. Each player takes turns to strike off a (new) number from the board. But, to make sure N does't affect who wins, there is an added rule. Once you strike of a number, you also have to strike off all its divisors in that same chance, irrespective of whether any of those divisors were already marked. The player to strike off the last number on the board wins. Can A construct a winning strategy?
Solution
Source: Krishnamurthy Iyer
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N undercover agents have been found in don's lair. Less than half of them are terrorists and the rest are anti-terrorists. The nature of their job is so secret that there is no proof what so ever to testify who is who. Although each of them knows who was actual terrorist and who was anti because they worked in teams. A query consists of asking person i if person j is Anti. Anti will always speak truth but a terrorist may lie to confuse you. The goal is to find out one anti in fewest queries.
Hint
Solution
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Latest solved Puzzles
Difficulty Level | 724 | 3,191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2018-34 | longest | en | 0.952035 |
https://i-programmer.info/news/192-photography-a-imaging/13908-crowdsampling-a-photo.html | 1,695,661,317,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233509023.57/warc/CC-MAIN-20230925151539-20230925181539-00763.warc.gz | 347,257,142 | 12,115 | Synthesizing The Bigger Picture
Sunday, 09 August 2020
There are more cameras in the world and more people taking photos all the time than ever before. What is more we all tend to take the same shots whenever we are near something "touristy". A new technique can put these crowdsourced photos together into a single image, capturing the many viewpoints.
The title of this research from Noah Snavely and his team at Cornell Tech is "Crowdsampling the Plenoptic Function" and this isn't a title likely to pique the interest of the man in the street - "plenwhat?" The reality is much more interesting than it sounds and it illustrates how much the world we live in has changed. You talk about big data, but here it is - big data in the form of a crowdsourced photo that we all took.
I can't put it better than the abstract:
"Many popular tourist landmarks are captured in a multitude of online, public photos. These photos represent a sparse and unstructured sampling of the plenoptic function for a particular scene. In this paper, we present a new approach to novel view synthesis under time-varying illumination from such data."
You can almost guess what they did next. They collected a big sample, aproximately 50K photos, from Flickr for a range of tourist sites. Next they identified the viewpoints (the red dots) for each photo:
Which only goes to prove that we really do all take the same picture - in future buy a postcard?
Actually there is enough variablity in the space and time that the photos were taken to attempt a reconstruction of the plenoptic function. This is just the plot of the color of every point x,y as seen from position u,v at time t. If you have the plenoptic function, or more practically an approximation to it, you can create a view from any position at any time.
The details are complicated, but what happens next is that a neural network learns how to extract plenoptic slices from the photos. From this you can do some interesting things. In particular you can show what the scene looks like from different viewpoints - smoothly moving from one to another. You can also arrange for a smooth change in what the scene looks like at different times.
Watch the video to see it in action (and don't miss the bit at the start where the researchers wave at you):
If you visit the experiment's site you can see many examples of the the reconstruction of visual scenes from crowdsourced photos. I have to admit that while I am impressed and intrigued I can't think of a serious application. Artistic/creative perhaps, but not a really "must have" application. Perhaps I'm just not thinking of the right sort of things?
Zhengqi Li, Wenqi Xian, Abe Davis and Noah Snavely
Cornell University
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ChatGPT And Excel Another Coding Threat?06/09/2023We have been considering the role of coding copilots in helping skilled programmers create code, but what happens when large language models attempt to create a spreadsheet? Is this just another way t [ ... ] + Full Story Windows Community Toolkit 8 Adds New Galleries19/09/2023Microsoft has released version 8 of the Windows Community Toolkit. The developers describe this as a huge update with an array of improvements and features, including new galleries and a major reworki [ ... ] + Full Story More News | 754 | 3,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-40 | latest | en | 0.95517 |
https://www.sarthaks.com/12087/bisectors-of-angles-b-and-of-triangle-abc-intersect-its-circumcircle-at-and-respectively | 1,660,304,122,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571692.3/warc/CC-MAIN-20220812105810-20220812135810-00690.warc.gz | 878,792,660 | 15,235 | # Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.
6.6k views
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.
Prove that the angles of the triangle DEF are
90° – 1/ 2 A, 90° – 1/ 2 B and 90° – 1/ 2 C
by (128k points)
selected
Given : ∆ABC and its circumcircle. AD, BE, CF are bisectors of ∠A, ∠B, ∠C respectively.
Construction : Join DE, EF and FD.
Proof : We know that angles in the same segment are equal.
∴ ∠5 = ∠C 2 and ∠6 = ∠B 2 ..(i)
∠1 = ∠A 2 and ∠2 = ∠C 2 ..(ii)
∠4 = ∠A 2 and ∠3 = ∠B 2 ..(iii)
From (i), we have
∠5 + ∠6 = ∠C/ 2 + ∠B/ 2
⇒ ∠D = ∠C/ 2 + ∠B/ 2 ...(iv)
[∵∠5 + ∠6 = ∠D]
But ∠A + ∠B + ∠C = 180°
⇒ ∠B + ∠C = 180° – ∠A | 331 | 763 | {"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} | 4.4375 | 4 | CC-MAIN-2022-33 | latest | en | 0.882169 |
https://socratic.org/questions/how-do-you-solve-3k-7-5-16 | 1,638,104,434,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358520.50/warc/CC-MAIN-20211128103924-20211128133924-00218.warc.gz | 507,984,906 | 6,067 | # How do you solve (3k-7)/5=16?
Dec 31, 2016
$\textcolor{b l u e}{\text{k=29}}$
#### Explanation:
color(green)("Multiply both sides by 5 to remove the denominator,"
$\cancel{5} \times \frac{3 k - 7}{\cancel{5}} = 16 \times 5$
5 cancels out on the left side and you're left with $3 k - 7$. And, when 16 gets multiplied by 5, you get 80 on the right side:
$3 k - 7 = 80$
$\textcolor{m a \ge n t a}{\text{Add 7 to both sides:}}$
$3 k - \cancel{7} = 80$
$\textcolor{w h i t e}{a a} + \cancel{7} \textcolor{w h i t e}{a a} + 7$
The 7's cancel out, so you only have $3 k$ left. And when you add 80+7 on the right side, you obtain 87:
$3 k = 87$
$\textcolor{red}{\text{Divide both sides by 3 to get k by itself:}}$
$\frac{\cancel{\text{3}} k}{\cancel{3}} = \frac{87}{3}$
Now you are just left with $k$ on the left side and 29 on the right side.
Thus,
$\textcolor{b l u e}{\text{k=29}}$ | 341 | 893 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 14, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2021-49 | latest | en | 0.710891 |
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# Plz explain Fluid pressure from gravity or acceleration ...
357 Points
12 years ago
Hi Rekha ,
Pressure is a measurement of the force per unit area. Fluid pressure can be caused by gravity, acceleration, or forces in a closed container. Since a fluid has no definite shape, its pressure applies in all directions. Fluid pressure can also be amplified through hydraulic mechanisms and changes with the velocity of the fluid.
Fluid pressure from gravity or acceleration
The weight of a fluid can exert a pressure on anything underneath it. Also, the relative movement of a liquid or gas can apply a pressure.
Pressure
Pressure is defined as force divided by the area on which the force is pushing. You can write this as an equation, if you wanted to make some calculations :
P = F/A
where
* P = pressure
* F = force
* A = area
* F/A = F divided by A
Pressure due to gravity
Since the weight of an object or material is equal to the force it excerts due to gravity,
An object can exert downward pressure due to its weight and the force of gravity. The pressure you exert on the floor is your weight divided by the area of the soles of your shoes. If the force is due to the weight (W) of the object, the equation is then: P = W / A | 380 | 1,600 | {"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} | 3.140625 | 3 | CC-MAIN-2021-25 | latest | en | 0.926128 |
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# Test: Matter In Our Surrounding- 5
## 25 Questions MCQ Test Science Class 9 | Test: Matter In Our Surrounding- 5
Description
This mock test of Test: Matter In Our Surrounding- 5 for UPSC helps you for every UPSC entrance exam. This contains 25 Multiple Choice Questions for UPSC Test: Matter In Our Surrounding- 5 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Matter In Our Surrounding- 5 quiz give you a good mix of easy questions and tough questions. UPSC students definitely take this Test: Matter In Our Surrounding- 5 exercise for a better result in the exam. You can find other Test: Matter In Our Surrounding- 5 extra questions, long questions & short questions for UPSC on EduRev as well by searching above.
QUESTION: 1
### What is the physical state of water at 1000C?
Solution:
The boiling point of water is 100°C. The physical state of water at its boiling point temperature of 100 degree Celsius will be both liquid state as well as gaseous state.
QUESTION: 2
### An Almirah is a solid because its
Solution:
An almirah is solid because it is hard and rigid, have fixed shape and have very high density. All these properties are possessed by solid substance only.
QUESTION: 3
### Find the incorrect statement
Solution:
The normal room temperature is 298K or 25C. Evaporation is a surface phenomenon but boiling is bulk phenomenon. Liquid having low boiling points evaporate faster. Cooling is not caused during boiling.
QUESTION: 4
Which of the following does not affect rate of evaporation?
Solution:
Cooling is field due to evaporation of acetone. Acetone is a highly volatile material which means it evaporates too fast at normal temperature also and it process of evaporating it takes the latent heat from hand to change its state from liquid to gas that's why we feel cool when we apply acetone.
QUESTION: 5
Convert 300K into Celsius
Solution:
QUESTION: 6
Arrange the following substances in increasing order of attraction between the particles: water, sugar, oxygen.
Solution:
Particles of solids have a very high force of attraction, particles of liquid have less force of attraction than solids and particles of gases have very less force of attraction. So the correct order of increasing attraction between particles is Oxygen, water and sugar.
QUESTION: 7
Which of the following will not sublimate ?
Solution:
Camphor, ammonium chloride and iodine are sublimable substance.
QUESTION: 8
Match the following with correct response.
Solution:
Change of solid state into liquid state is known as Fusion. Change of liquid state into gaseous state is known as Vapourization. Change of vapour state into liquid state is known as condensation. Change of liquid state into solid state is known as Solidification.
QUESTION: 9
Which of the following does not affect rate of evaporation?
Solution:
Solid and liquid have fixed volumes but gas has no fixed volume it is because the particles of gas have large intermolecular space but very very low intermolecular forces and the particles of gases can move freely.
QUESTION: 10
The interparticle forces are the strongest in
Solution:
Solids substances have the strongest interparticle force of attraction. In the given substances, sodium bromide is solid, ethyl alcohol is liquid and ammonia and carbon dioxide are gases. So, the particles of Sodium bromide have the strongest force of attraction.
QUESTION: 11
Which one of the following decreases the extent of evaporation of water?
Solution:
Evaporation of water increases with increase in temperature, surface area and wind speed but decrease with increase in humidity. So, larger humidity will decrease the rate of evaporation.
QUESTION: 12
Match the following with correct response.
Solution:
The change of solid directly into vapours is called Sublimation. The change of liquid state into vapours without coming to it's boiling point is called Evaporation. The latent heat of Vaporization is the heat required to change the state of 1 kg into vapours. Latent heat of fusion is the heat required to Change the state of 1 kg of solid into liquid.
QUESTION: 13
Name the state of matter that ‘has minimum interparticle attraction’
Solution:
Gases have no definite shape and volume and are highly incompressible. solid poses rigidity and liquid and gases possess fluidity.
QUESTION: 14
Pressure on the surface of a gas is increased. What will happen to the inter-particle forces?
Solution:
When the pressure on the surface of a gas is increased. The particles of gases come closer that increase the inter-particle force of attraction.
QUESTION: 15
A desert cooler given comfort due to cooling caused by evaporation of water. Under which one of the following conditions it works more effectively?
Solution:
Desert cooler causes cooling effect due to evaporation of water as evaporation causes cooling effect. A desert cooler works more effectively when the weather is hot and dry.
QUESTION: 16
Match the following with correct response.
Solution:
Boiling point of water 100oC +273 = 373k
Melting point of water 0oC = 273 K
Boiling point of Acetone = At atmospheric pressure acetone has a boiling point of 56°C
Melting poitn of aluminium = 660.3 °C
QUESTION: 17
Identify the incorrect statement
Solution:
Ether and Acetone is most volatile. Alcohol is moderately volatile and water is least volatile. So, the statement, Acetone is least volatile is an incorrect statement.
QUESTION: 18
Statement A: A substance is said to be in the liquid state if under normal pressure its M. P. is below the room temperature.
Statement B: The melting point of solid and the freezing point of liquid is different.
Q. Which of the two statements is true?
Solution:
A substance is said to be in liquid state if under normal pressure its melting point is below the room temperature. Melting point of solid and the freezing point of liquid is are the same. So, statement A is correct but B is wrong.
QUESTION: 19
When alcohol is kept in a china dish it evaporates, temperature falls and cooling is produced. Which one is the correct method to record the temperature?
Solution:
Alcohol is a volatile substance and when kept in a china dish it start evaporating. Evaporation causes a cooling effect. The method used to measure the temperature change in above observation is by using a thermometer.
QUESTION: 20
Identify the incorrect statement about evaporation
A. It causes cooling
B. It increase with increase in humidity
C. It decreases with increase in temperature
D. It increases with increase in wind speed
Solution:
On increasing temperature more number of particles get enough K.E. to go into the vapour sate. Humidity is the amount of water vapour present in air. When humidity of air is low, the rate of evaporation is high and water evaporates more readily. When humidity of air is high, the rate of evaporation is low and water evaporates very slowly.
QUESTION: 21
Statement A: Celsius scale is the best scale for measuring temperature.
Statement B: CO2 is heavier gas than both N2 and O2
Which of the two statements is true?
Solution:
Kelvin scale is used as standard unit of temperature. CO2 is heavier gas than N2 and O2.
QUESTION: 22
Find the incorrect statement
Solution:
Gases have maximum fluidity and least rigidity. The particles of solids have negligible kinetic energy. Inter-particle attraction depends on the physical state of the matter. Potassium permanganate is purple in colour.
QUESTION: 23
In an endothermic process heat is absorbed, in an exothermic process heat is evolved and in an athermic process no thermal change is observed. What is the nature of evaporation of ether?
Solution:
The chemical reaction in which heat is absorbed is called the endothermic reaction. The reaction in which heat is evolved is called exothermic reaction and the reaction in which no thermal changes occur is called a thermic process. Evaporation of ether occurs due to absorption of heat by surrounding. So, it an aendothermic process.
QUESTION: 24
What is the physical state of water at 10°C?
Solution:
The physical state of water at 10°C is liquid state. Below 0°C water is in solid state and above 100°C water is in vapour state.
QUESTION: 25
Which of the following statements is not true?
Solution:
Matter is composed of very small particles called atoms, hence matter is not continuous but particulate in nature | 1,865 | 8,485 | {"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-2020-40 | latest | en | 0.930595 |
https://whatisconvert.com/190-feet-in-meters | 1,656,663,305,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103922377.50/warc/CC-MAIN-20220701064920-20220701094920-00639.warc.gz | 664,932,404 | 7,046 | # What is 190 Feet in Meters?
## Convert 190 Feet to Meters
To calculate 190 Feet to the corresponding value in Meters, multiply the quantity in Feet by 0.3048 (conversion factor). In this case we should multiply 190 Feet by 0.3048 to get the equivalent result in Meters:
190 Feet x 0.3048 = 57.912 Meters
190 Feet is equivalent to 57.912 Meters.
## How to convert from Feet to Meters
The conversion factor from Feet to Meters is 0.3048. To find out how many Feet in Meters, multiply by the conversion factor or use the Length converter above. One hundred ninety Feet is equivalent to fifty-seven point nine one two Meters.
## Definition of Foot
A foot (symbol: ft) is a unit of length. It is equal to 0.3048 m, and used in the imperial system of units and United States customary units. The unit of foot derived from the human foot. It is subdivided into 12 inches.
## Definition of Meter
The meter (symbol: m) is the fundamental unit of length in the International System of Units (SI). It is defined as "the length of the path travelled by light in vacuum during a time interval of 1/299,792,458 of a second." In 1799, France start using the metric system, and that is the first country using the metric.
## Using the Feet to Meters converter you can get answers to questions like the following:
• How many Meters are in 190 Feet?
• 190 Feet is equal to how many Meters?
• How to convert 190 Feet to Meters?
• How many is 190 Feet in Meters?
• What is 190 Feet in Meters?
• How much is 190 Feet in Meters?
• How many m are in 190 ft?
• 190 ft is equal to how many m?
• How to convert 190 ft to m?
• How many is 190 ft in m?
• What is 190 ft in m?
• How much is 190 ft in m? | 450 | 1,688 | {"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} | 3.390625 | 3 | CC-MAIN-2022-27 | latest | en | 0.89798 |
http://www.bikeforums.net/foo/943989-how-many-pixels-per-inch-can-human-eye-see-print.html | 1,448,531,771,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398446997.59/warc/CC-MAIN-20151124205406-00084-ip-10-71-132-137.ec2.internal.warc.gz | 323,895,898 | 3,786 | how many pixels per inch can the human eye see?
• 04-19-14, 02:19 PM
windhchaser
how many pixels per inch can the human eye see?
growing up some of us had a vcr on a 27 inch tv that's around 15 pixels per inch. now we have android tablets that have around 300 ppi. my question is can the human eye even see that many?
• 04-19-14, 03:20 PM
black_box
depends on how close you are to the screen, but yes, we're close to the limit.
• 04-19-14, 03:22 PM
windhchaser
handheld distance ..
• 04-19-14, 04:29 PM
skijor
Quote:
Originally Posted by windhchaser
handheld distance ..
If you're an orangutan, that's not so close. Just sayin' :p
http://www.imfdb.org/images/4/42/EveryW004.jpg
• 04-19-14, 04:48 PM
windhchaser
Quote:
Originally Posted by skijor
If you're an orangutan, that's not so close. Just sayin' :p
http://www.imfdb.org/images/4/42/EveryW004.jpg
good movie
• 04-19-14, 05:19 PM
Artkansas
Your NTSC VCR had a horizontal resolution of 720 pixels. Divided by 27", that works out to about 27 pixels per inch.
Well, according to wikipedia, your minimum angular resolution is "approximately 0.07°". The rest depends on how close the pixels are and how big they are.
• 04-19-14, 07:57 PM
Wilbur Bud
CarltonBale.com » 1080p Does Matter ? Here?s When (Screen Size vs. Viewing Distance vs. Resolution)
While I disagree that 1080p matters, the chart on the page referenced that shows what you can see at which distance versus various resolutions is a good reference
• 04-19-14, 08:41 PM
Jseis
Eagles are a great comparison to humans because the physical size of their eyes are similar but an eagle can discern objects well over twice what we can. Here's a good link. Supposedly an eagle can spot a rabbit at 2 miles. Probably depends on the background but I might be able to spot a basketball/exercise ball? against a light field at one mile in my younger days though a mile is long long away. There was a great eye test that Egyptian military men gave recruits. They would have them look at the the Big Dipper and ask which star was actually two stars. I could see that difference at age 25-26 but today at 59...unlikely (though I bet my Dad could at 90...he could still read fine print without classes and had excellent long range vision).
Journey North Bald Eagles
• 04-19-14, 09:05 PM
bjtesch
Quote:
Originally Posted by windhchaser
my question is can the human eye even see that many?
I think the point is that you make them small enough so you can't see the individual pixels. Plus it does depend on the viewing distance. I have an iphone 4 and I'm not sure that I can see the pixels on its screen, so I would say that they have reached the practical limit. I have an ipad 2 and I can see the pixels on its screen. The screen is completely satisfactory to me because it is much larger than the iphone screen. I don't feel the need for the retina display that is in the new ipads.
My wife's new Lenovo ultrabook has a very high resolution screen. Start it and go to desktop and the icons are very small. Start the average Windows app and the text is very small. Some apps can scale up well enough but others don't. I just changed the resolution through Windows to run the screen at a lower resolution.
• 04-19-14, 10:05 PM
rm -rf
Old 300 dpi laser printer's fonts don't look nearly as good as the 600 to 1200 dpi modern ones.
• 04-26-14, 07:18 AM
Garfield Cat
Its about what you're looking at. Its not really the pixels but the image that the pixels are creating, the visual effect. A still picture and a moving image are going to be captured by the eye and brain differently.
Most interesting is watching sporting events in high definition. The overall experience is noticeably different than non high definition. | 988 | 3,727 | {"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-2015-48 | longest | en | 0.921088 |
https://www.britannica.com/science/vector-operations | 1,521,479,271,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647003.0/warc/CC-MAIN-20180319155754-20180319175754-00215.warc.gz | 764,680,479 | 35,802 | # Vector operations
mathematics
Vector operations, Extension of the laws of elementary algebra to vectors. They include addition, subtraction, and three types of multiplication. The sum of two vectors is a third vector, represented as the diagonal of the parallelogram constructed with the two original vectors as sides. When a vector is multiplied by a positive scalar (i.e., number), its magnitude is multiplied by the scalar and its direction remains unchanged (if the scalar is negative, the direction is reversed). The multiplication of a vector a by another vector b leads to the dot product, written ab, and the cross product, written a × b. The dot product, also called the scalar product, is a scalar real number equal to the product of the lengths of vectors a (|a|) and b (|b|) and the cosine of the angle (θ) between them: ab = |a| |b| cos θ. This equals zero if the two vectors are perpendicular (see orthogonality). The cross product, also called the vector product, is a third vector (c), perpendicular to the plane of the original vectors. The magnitude of c is equal to the product of the lengths of vectors a and b and the sine of the angle (θ) between them: |c| = |a| |b| sin θ. The associative law and commutative law hold for vector addition and the dot product. The cross product is associative but not commutative.
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https://3jack.blogspot.com/2011/03/applying-statistical-golf.html?showComment=1300856482706 | 1,721,554,618,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517663.24/warc/CC-MAIN-20240721091006-20240721121006-00283.warc.gz | 62,384,832 | 14,716 | ## Monday, March 21, 2011
### Applying Statistical Golf
Finishing out the statistical analysis, one of the things I’ve been asked the most is ‘how can I apply this stuff to me?’
A lot of the work has really applied to the 4 or less handicapper trying to make their game better. However, I think there are a lot of the same principles apply.
For instance, The Danger Zone.
On the PGA Tour, the Danger Zone is approach shots from 175-225 yards. But since the PGA Tour courses are roughly the same distance in length and the pros don’t hit pop ups, ground balls and duck hooks off the tee, there’s not a major discrepancy between the amount of times in a round of golf a PGA Tour golfer will be in the Danger Zone. Meaning, if Corey Pavin is in the Danger Zone 20 times in a 4-day tournament, Bubba Watson will probably be in the Danger Zone at least 14 times in a 4-day tournament. Still a difference, but it’s not something ridiculous like Pavin being in the Danger Zone 20 times and Bubba being in the Danger Zone 2 times.
But, you may run into that type of ridiculous discrepancy on the amateur level given the course lengths can change and the margin of driving distance from golfer to golfer tends to be greater.
Below is a table I created to help figure out the Danger Zone for a particular course length. First, let me go thru what the table means.
Yards = Length of Course for 18 holes of Golf
Long = If the golfer hits it this distance or longer off the tee somewhat consistently, it would be safe to say that we will consider the golfer in the upper percentile of distance off the tee. You can label this golfer as having ‘bomber length.’ Note, the shorter the course, the less power off the tee is needed to be considered ‘long.’ It’s almost like being called ‘long for the course.’ So if you’re a 7 handicap who hits it 280 yards off the tee and is playing a 5,800 yard course, really the course is too short for you.
Mid1 & Mid2 = These are the distances where you would be ‘middle of the pack.’ For instance, on a 6,700 yard course, middle of the pack would be 260-276 yards off the tee. If you hit it 277 or more yards, you would be considered ‘long’ for the course.
Short1 & Short 2 = These are the distances where the golfer would be considered to be short for the course. I created a range here because I believe one can be ‘too short’ for a course. For instance, if you hit it 245 off the tee, you are too short for a 7,200 yard course and you probably shouldn’t be playing it. Instead, you should probably try to play the course from no more than 6,800 yard.
DZ1 & DZ2 = This is the 50 yard ‘Danger Zone’ range. As you can see, as the course length gets shorter, the Danger Zone gets shorter. This is mostly because golfers won’t get many attempts in the traditional Danger Zone (175-225 yards) because the course is too short. The 175-225 yard Danger Zone is still prevalent on a lot of course lengths because of par-3’s. No matter how long a golfer is off the tee, they can’t avoid the Danger Zone if the course has four par-3’s that are all within the Danger Zone range.
Yards….Long……Mid1…Mid2…Short1….Short2…DZ1……DZ2
7600……315……314……298……297………..277……...185……235
7500……310……309……294……293……..…273…...…175……225
7400……306……305……290……289………..269…...…175……225
7300……302……301……286……285……..…265...……175……225
7200……298……297……281……280……..…260…...…175….…225
7100……294……293……278……277……..…257…...…175….…225
7000……290……289……275……275….....…255...……175….…225
6900……286……285……270……269………..249…….…175……225
6800……281……280……265……264………..244……….175……225
6700……277……276……260……259………..239…….…165……215
6600……273……272……257……256………..236…….…165……215
6500……269……268……253……252………..232…….…165……215
6400……265……264……250……249………..229…….…165……215
6300……261……260……245……244………..224….……150……200
6200……257……256……240……239……..…219…….…150……200
6100……252……251……235……234………..214…….…150……200
6000……248……247……232……231……….211…..……150……200
5900……244……243……228……227……….207……..…150……200
5800……240……239……225……224……….204……….140….…190
5700……236……235……220……219…….…199…….…140…….190
5600……232……231……215……214…….…194…….…140….…190
5500……228……227……212……211……..…191……….140……190
3JACK | 1,209 | 4,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-30 | latest | en | 0.943111 |
https://cumbernauld-media.com/the-power-of-compound-interest/ | 1,726,783,460,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652067.20/warc/CC-MAIN-20240919194038-20240919224038-00810.warc.gz | 170,604,304 | 27,454 | # The Power of Compound Interest: Understanding How It Can Grow Your Investments
The power of compound interest is an incredibly powerful concept that is essential to understand if you want to maximize the potential of your investments. Compound interest is the process by which an investor earns interest on both the principal and the interest accumulated over time. This process is often referred to as “the eighth wonder of the world” because of its ability to exponentially grow investments over time. Compound interest is a powerful tool for investors to use in order to make the most out of their investments, as it allows them to earn more money on their investments over time.
In this article, we will discuss the power of compound interest, the importance of understanding compound interest, and the various strategies investors can use to maximize the potential of their investments. We will look at how compound interest works, how it can be used to grow investments, and how to best leverage it to achieve financial success. Additionally, we will discuss the potential risks associated with compound interest and how to minimize them. Finally, we will discuss how investors can use compound interest to achieve their long-term financial goals. Through understanding the power of compound interest, investors can use it to their advantage to exponentially increase their investments over time.
## How Compound Interest Works:
### Explanation of How Compound Interest is Calculated
a. Definition of Compound Interest
Compound interest is the interest earned on the principal amount of a loan or investment, plus the interest earned on any interest accumulated in the previous periods.
b. Calculation of Compound Interest
Compound interest is calculated by multiplying the principal amount of the loan or investment by one plus the annual interest rate raised to the number of compound periods minus one. This formula is also known as the geometric series formula. The result is then multiplied by the number of years the loan or investment has been held.
### Illustration of Compound Interest with an Example
To illustrate how compound interest works, let’s consider an example. A loan of \$10,000 is taken out with an annual interest rate of 10%, compounded on an annual basis. After one year, the total amount owed will be \$11,000. This is calculated by multiplying the principal amount of \$10,000 by one plus the annual interest rate of 10%, raised to the number of compound periods (1) minus one, which equals 0.1. The result is then multiplied by the number of years the loan is taken out (1). In this example, the total amount owed after one year is \$11,000.
After two years, the total amount owed will be \$12,100. This is calculated by multiplying the principal amount of \$10,000 by one plus the annual interest rate of 10%, raised to the number of compound periods (2) minus one, which equals 0.2. The result is then multiplied by the number of years the loan is taken out (2). In this example, the total amount owed after two years is \$12,100.
This example illustrates how compound interest works. It is important to note that the amount of interest earned increases with each passing year, as the interest earned in the previous periods is also taken into account. As a result, the total amount owed grows exponentially over time.
## The Benefits of Compound Interest:
Compound interest is a powerful financial tool that can help individuals build wealth over time. It has become increasingly popular in recent years as a way to grow investments. Compound interest is different than simple interest, in that it allows an investor to earn interest on both their principal balance and any interest that has been earned previously. This helps an investor to earn more money over time, as the interest earned in one period is added to the principal balance and then interest is earned on the new increased balance. As a result, compound interest can often result in higher returns than simple interests in the long run.
One of the biggest advantages of investing early is the power of compound interest. The earlier an investor begins to invest, the more time their money has to accumulate interest and grow. As a result, the earlier an investor begins to invest, the more money they can accumulate over time due to the power of compound interest.
Compound interest can also help to generate greater returns than simple interest over time. This is because simple interest only applies to the principal balance, while compound interest applies to both the principal and any interest that has been earned previously. As a result, compound interest can often result in greater returns over time.
Investors should carefully consider the advantages of investing early and using compound interest to grow their investments. With careful planning and a sound strategy, compound interest can be a powerful tool that can help to generate significant returns over time. It is important for investors to understand the benefits of compound interest and how it can help to grow their investments over time.
## Strategies for Maximizing Compound Interest:
1. Ways to increase the frequency of compounding: Compounding is a powerful financial tool that can help you grow your wealth when used strategically. To maximize the benefit of compounding, it is important to increase the frequency of compounding as much as possible. This means contributing as much as possible to your investment accounts on a regular basis and reinvesting any profits you may make. Additionally, look for investments that offer more frequent compounding, such as bonds or CDs that allow you to cash out and reinvest at regular intervals.
2. Tips for choosing investments with high compound interest rates: When choosing investments, it is important to look for ones with high compound interest rates. This means looking for investments that will pay you more interest over time, as the interest earned on your investment will be added to your principal and compound over time. Consider looking for investments that are offered by reputable companies and have a solid track record of providing returns. Additionally, look for investments that offer tax advantages that can help you maximize your return.
3. Importance of avoiding early withdrawals and fees: It is important to avoid early withdrawals and fees when investing in a compound interest account. Early withdrawals will reduce the amount of money that is earning compound interest, and fees can also reduce the amount of money that is left to compound. Additionally, try to make sure that you are not paying too much in fees, as this can significantly reduce your overall return. Being aware of all fees associated with your investments and avoiding early withdrawals will help you maximize the benefit of compounding.
## Real-Life Examples:
### Success Stories of Investors Who Have Benefited from Compound Interest
One of the most inspiring success stories of compound interest is that of Chuck Feeney. Feeney was an investor who used compound interest to his advantage. He invested in a variety of companies over a period of 40 years, and his strategy of reinvesting profits and utilizing compounding interest helped him amass a fortune of over \$8 billion. Feeney’s success story, in which he used compound interest to grow his wealth, is a prime example of how powerful this investment strategy can be.
Another example of an investor who has benefited from compound interest is Warren Buffett. Buffett is widely known for his investing strategies, which often include the use of compound interest. He is one of the richest people in the world, and his success is largely due to his use of compound interest over a long period of time. Buffett’s success story serves as an example of how utilizing compound interest can help investors grow their wealth significantly.
### Analysis of How Compound Interest Has Impacted Long-Term Investment Portfolios
The impact of compound interest on long-term investment portfolios is undeniable. Compound interest helps investors maximize the potential of their investments over time by allowing them to reinvest their profits in order to generate more earnings. This strategy can be especially powerful when utilized over a long period of time, as the returns on investments can grow exponentially. This can lead to significant returns on investments over time, as the compounding of interest can lead to greater profits than if the investments had been made without the use of compound interest.
Compound interest can also be beneficial when investors use it to diversify their portfolios. By utilizing compound interest to invest in a variety of different assets, investors can spread out their risk and maximize their returns over the long term. This can help investors make more informed investment decisions and increase their returns over time.
Overall, the use of compound interest can be a powerful tool for long-term investors. By utilizing this strategy, investors can maximize their returns over time and diversify their portfolios in order to reduce risk. This can help investors significantly increase their wealth over the long term, and serve as an example of how powerful this investment strategy can be.
### Real-Life Examples: Common Misconceptions about Compound Interest:
1. Debunking myths about compound interest:
One of the most common misconceptions about compound interest is that it is a risk-free investment. However, compound interest requires a certain level of risk in order to be successful. As with any investment, there is always a chance of losses and investors must be willing to take that risk. Additionally, another myth surrounding compound interest is that it will lead to exponential growth of investments. While compound interest can lead to long-term growth, it is important to understand that the rate of return is based on the time period and the amount of money invested.
2. Clarifying common misconceptions to help readers make informed investment decisions:
Another misconception about compound interest is that it can only be achieved with a large sum of money. In reality, compound interest can be earned with any amount of money. The key is to start small and invest regularly. By adding to your initial investment and investing regularly, you can help your money grow over time. Additionally, there is a common misconception that investing in a certificate of deposit (CD) is the only way to earn compound interest. While CDs can be a viable option, there are other, more diverse investments that offer compound interest as well. It is important to do your research and invest in a portfolio that offers the best rate of return for your money.
Overall, it is important to understand that compound interest is a powerful tool that can help your money grow over time. However, it is important to understand the risks and be informed before making any investment decisions. By debunking the myths and clarifying common misconceptions, readers can make informed and confident decisions with their money.
## Conclusion
The power of compound interest is a powerful tool that can be used to grow investments. When used correctly, it can turn small investments into large ones and help you achieve financial freedom. Compound interest is a form of investment that can be used to build wealth over time, as the interest is reinvested and compounded. The longer you stay invested, the greater the potential for growth. In order to maximize the potential of compound interest, it is important to understand how it works, the different types of investments that can be used, and how to best manage and diversify your investments. With a little research and effort, you can understand how compound interest can work for you, and make the most of your investments.
## FAQs
1. What is Compound Interest?
Compound interest is when interest is earned on the principal amount of a deposit or loan, plus any accumulated interest from previous periods. This means that the interest you earn in one period is added to your principal balance, and the next period’s interest is calculated based on the new, higher principal balance.
2. How does Compound Interest Work?
Compound interest works by reinvesting the accrued interest back into the principal amount. This can be done either periodically, such as monthly or annually, or continuously. The more frequently the interest is compounded, the more quickly your account will grow.
3. What is the Impact of Compound Interest?
Compound interest has a significant impact on the growth of your investments. It can help you achieve your financial goals faster and achieve a larger return on your investments than simple interest.
4. What is the Rule of 72?
The Rule of 72 is a mathematical formula that can be used to estimate how long it will take for an investment to double, given a fixed interest rate. To use the Rule of 72, divide 72 by the interest rate. The result is the approximate number of years it will take for the investment to double.
5. What is the Difference Between Compound Interest and Simple Interest?
The primary difference between compound interest and simple interest is that compound interest is earned on both the principal amount and any accumulated interest from previous periods. Simple interest is only earned on the principal amount.
6. How Can Compound Interest Help Me Reach My Financial Goals?
Compound interest can help you reach your financial goals faster by allowing your investments to grow at a faster rate. By reinvesting the interest that is earned, your investments can grow exponentially.
7. What Factors Should I Consider When Choosing an Investment Vehicle with Compound Interest?
When choosing an investment vehicle with compound interest, you should consider the interest rate, the frequency of compounding, and any fees or penalties associated with the investment.
8. What Are Some Examples of Investments with Compound Interest?
Some examples of investments with compound interest include savings accounts, certificates of deposit (CDs), money market accounts, bonds, and mutual funds.
9. How Can I Maximize the Benefits of Compound Interest?
To maximize the benefits of compound interest, you should make consistent contributions to your investment and choose investments with higher interest rates and more frequent compounding periods.
10. What Is the Difference Between Compounding and Reinvesting?
Compounding is when interest is earned on the principal amount and any accumulated interest from previous periods. Reinvesting is when the interest earned is reinvested back into the principal amount of the investment.
##### By Ishan Crawford
Prior to the position, Ishan was senior vice president, strategy & development for Cumbernauld-media Company since April 2013. He joined the Company in 2004 and has served in several corporate developments, business development and strategic planning roles for three chief executives. During that time, he helped transform the Company from a traditional U.S. media conglomerate into a global digital subscription service, unified by the journalism and brand of Cumbernauld-media. | 2,852 | 15,288 | {"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} | 3.65625 | 4 | CC-MAIN-2024-38 | latest | en | 0.968861 |
https://en.formulasearchengine.com/wiki/Associated_prime | 1,669,466,463,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446706291.88/warc/CC-MAIN-20221126112341-20221126142341-00404.warc.gz | 265,656,779 | 11,501 | # Associated prime
In abstract algebra, an associated prime of a module M over a ring R is a type of prime ideal of R that arises as an annihilator of a submodule of M. The set of associated primes is usually denoted by ${\displaystyle \operatorname {Ass} _{R}(M)\,}$.
In commutative algebra, associated primes are linked to the Lasker-Noether primary decomposition of ideals in commutative Noetherian rings. Specifically, if an ideal J is decomposed as a finite intersection of primary ideals, the radicals of these primary ideals are prime ideals, and this set of prime ideals coincides with ${\displaystyle \operatorname {Ass} _{R}(R/J)\,}$.Template:Sfn Also linked with the concept of "associated primes" of the ideal are the notions of isolated primes and embedded primes.
## Definitions
A nonzero R module N is called a prime module if the annihilator ${\displaystyle \mathrm {Ann} _{R}(N)=\mathrm {Ann} _{R}(N')\,}$ for any nonzero submodule N' of N. For a prime module N, ${\displaystyle \mathrm {Ann} _{R}(N)\,}$ is a prime ideal in R.Template:Sfn
An associated prime of an R module M is an ideal of the form ${\displaystyle \mathrm {Ann} _{R}(N)\,}$ where N is a prime submodule of M. In commutative algebra the usual definition is different, but equivalent:Template:Sfn if R is commutative, an associated prime P of M is a prime ideal of the form ${\displaystyle \mathrm {Ann} _{R}(m)\,}$ for a nonzero element m of M or equivalently ${\displaystyle R/P}$ is isomorphic to a submodule of M.
In a commutative ring R, minimal elements in ${\displaystyle \operatorname {Ass} _{R}(M)}$ (with respect to the set-theoretic inclusion) are called isolated primes while the rest of the associated primes (i.e., those properly containing associated primes) are called embedded prime.
A module is called coprimary if xm = 0 for some nonzero m ∈ M implies xnM = 0 for some positive integer n. A nonzero finitely generated module M over a commutative Noetherian ring is coprimary if and only if it has exactly one associated prime. A submodule N of M is called P-primary if ${\displaystyle M/N}$ is coprimary with P. An ideal I is a P-primary ideal if and only if ${\displaystyle \operatorname {Ass} _{R}(R/I)=\{P\}}$; thus, the notion is a generalization of a primary ideal.
## Properties
Most of these properties and assertions are given in Template:Harv starting on page 86.
The following properties all refer to a commutative Noetherian ring R:
${\displaystyle 0=M_{0}\subset M_{1}\subset \cdots \subset M_{n-1}\subset M_{n}=M\,}$
such that each quotient Mi/Mi−1 is isomorphic to R/Pi for some prime ideals Pi. Moreover every associated prime of M occurs among the set of primes Pi. (In general not all the ideals Pi are associated primes of M.)
{{#invoke:Category handler|main}}{{#invoke:Category handler|main}}[citation needed] }}
## Examples
• If R is the ring of integers, then non-trivial free abelian groups and non-trivial abelian groups of prime power order are coprimary.
• If R is the ring of integers and M a finite abelian group, then the associated primes of M are exactly the primes dividing the order of M.
• The group of order 2 is a quotient of the integers Z (considered as a free module over itself), but its associated prime ideal (2) is not an associated prime of Z.
## References
• {{#invoke:citation/CS1|citation
|CitationClass=citation }}
• {{#invoke:citation/CS1|citation
|CitationClass=citation }}
• {{#invoke:citation/CS1|citation
|CitationClass=citation }} | 899 | 3,507 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 26, "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-2022-49 | latest | en | 0.872637 |
https://studylib.net/doc/11888246/access---math | 1,675,062,898,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499804.60/warc/CC-MAIN-20230130070411-20230130100411-00713.warc.gz | 560,972,749 | 13,860 | # ACCESS - MATH
```ACCESS - MATH
July 2003
Notes on Body Mass Index and actual national data
What is the Body Mass Index?
If you read newspapers and magazines it is likely that once or twice a year you run across an article
about the body mass index (B.M.I.) , and its use in determining health risk factors for overweight and
underweight people. If you search the internet for "body mass index" you will find many sites which let
you compute your B.M.I., and which tell you a little bit about it.
A person’s B.M.I. is computed by dividing their weight by the square of their height, and then
multiplying by a universal constant. If you measure weight in kilograms, and height in meters, this
constant is the number one. Thus, the proponants of the B.M.I index are claiming that for adults at equal
risk levels (but different heights), weight should be proportional to the square of height. As we have
discussed in class, if people were to scale equally in all directions ("self-similar") when they grew,
volume and hence weight would scale as the cube of height. That particular power law seems a little
high, since adults don’t look like uniformly expanded versions of babies; we seem to get relatively
stretched out when we grow taller. One might expect the best predictive power for weight as a function
of height to be somewhere between 2 and 3, if one expected a power law at all. If there is a predictive
power, and if it is much larger than 2, then one could argue that the body mass index might need to be
modified to reflect this fact. (In fact, when you find body mass index tables, they often explain that you
should modifiy the acceptable BMI values for children.)
We have all collected several heights and weights, and hopefully in aggregate we will have a good
number of representative measurements, from baby-sized to adult. Each group will use this data, see if
it is consistent with a power law relating weights to heights, and decide whether the B.M.I. power of 2 is
a good choice.
I often do this experiment with my linear algebra classes as well as with the Accessors, and we have
gotten powers between 2.3 and 2.7. Also, several years ago I found a national data base at the U.S.
Center for Disease Control web site. It contained a wide variety of body measurements collected
between 1976 and 1980, including national median heights and weights for boys and girls, age 2-19. By
using only the national medians a lot of the variance has been taken out of the data, compared to what
yours will look like. The national data is very consistent with a power law, with power = 2.6. If any of
you can figure out a valid mathematical model which explains this power law, you’ll have a publishable
paper.
How do you test for power laws?
When you studied logarithms in high school you might have wondered what they were good for.
Well, as Nancy showed you yesterday, one application is in looking for power laws. Remember how the
discussion went:
Suppose we have a set of "n" data points, which you can think of as your height-weight data, but
which could really be any set of paired data:
[[x1, y1 ], [x2, y2 ], [x3, y3 ], ..., [xn, yn ]]
We want to see if there is a power m and a proportionality constant b so that the formula
y = b xm
effectively mirrors the real data. Taking (natural) logarithms of the proposed power law yields
ln(y ) = ln(b ) + m ln(x )
So if we write Y = ln(y ) and X = ln(x ), B = ln(b ), this becomes the equation of a line in the new variables
X and Y:
Y = mX + B
Thus, in order for there to be a power law for the original data, the ln-ln data should (approximately)
satisfy the equation of a line. Furthermore, this process is reversible; if the ln-ln data lies on a line with
slope m and intercept B, then the original data satisfies a power law with power m and proportionality
B
constant b = e . That’s because of the rules of exponents:
Y=B+mX
eY = e
( B + m X)
e Y = e B (e X )
m
y = b xm
In real experiments, it is not too hard to see if data is well approximated by a line, so this trick with the
logarithm is quite useful.
National data example: I used colons after most of these commands to suppress the output. If you
want to go back and see what each command is doing, replace the colons with semicolons.
> restart:
> with(plots):
> boyhw:=[[35.9,29.8],[38.9,34.1],[41.9,38.8],[44.3,42.8],
[47.2,48.6],[49.6,54.8],[51.4,60.8],[53.6,66.5],
[55.7,76.8],[57.3,82.3],[59.8,93.8],[62.8,106.8],
[66.0,124.3],[67.3,132.6],[68.4,142.4],[68.9,145.1],
[69.6,155.3],[69.6,153.2]]:
#boy heights (inches) weights (pounds): Ntl medians for ages 2-19
> girlhw:=[[35.4,28.0],[38.4,32.6],[41.1,36.8],[43.9,41.8],
[46.6,47.0],[48.9,52.5],[51.4,60.8],[53.1,65.5],
[55.7,76.1],[58.2,89.0],[61.0,100.1],[62.6,108.1],
[63.3,117.1],[64.2,117.6],[64.3,122.6],[64.2,128.8],
[64.1,124.5],[64.5,126]]:
#girl heights(inches) weights (pounds): Ntl medians for ages 2-19
> boys:=pointplot(boyhw):
girls:=pointplot(girlhw):
display({boys,girls},
title=‘plot of [height,weight], National medians ages 2-19‘);
plot of [height,weight], National medians ages 2–19
140
120
100
80
60
40
35
40
45
50
55
60
65
70
And now for the ln-ln data.
> with(linalg): #linear algebra package
> B:=evalm(boyhw): #evaluate matrix, this will
#turn our list of points into a matrix, which will
#be easiser to manipulate in Maple
G:=evalm(girlhw):
BG:=stackmatrix(B,G): #stack the matrices on top of eachother.
> lnBGa:=map(ln,BG): #take ln of the boy and girl height-weights
lnBG:=map(evalf,lnBGa): #Get decimal (floating point) values.
#This speeds up computations later in the
#least squares fit - otherwise Maple tries working
#symbolically.
> lnlnplot:=pointplot(lnBG):
display(lnlnplot,title=‘ln-ln data‘);
ln-ln data
5
4.8
4.6
4.4
4.2
4
3.8
3.6
3.4
3.6
3.7
3.8
3.9
4
4.1
4.2
How do I find the best line fit to a collection of points?
From Calculus, there is a slope m and intercept B yielding a line which minimizes the sum of the
squared vertical distances between your data points and the points on a line. If the n data points are
[[X1, Y1 ], [X2, Y2 ], [X3, Y3 ], ..., [Xn, Yn ]]
then m and B solve the system of equations
n
2
Xi
i=1
n
Xi
i=1
∑
∑
n
Xi
Xi Yi
m
i=1
i=1
=
B n
n
Yi
i=1
n
∑
∑
∑
You could solve this system with several "do-loops" to compute the matrix entries, followed by a
"solve" command to solve the system, but the method is so common that every decent mathematical
software or graphing calculator already has a command to do all of that work for you. In statistics this
procedure is called linear regression as well as the method of least squares. If you looked through the
help directory in your menu bar you would eventually find MAPLE’s version of this command living
in the stats library packag, and called "fit[leastsquare]". Here’s how the command works. I’ve used it
to check the example we worked by hand in class. You must be careful with brackets and parantheses.
> with(stats):
> fit[leastsquare[[X,Y]]]([[0,2,4],[1,2,1]]);#the syntax here is
#to first name your variables, then give two lists, one of the
first variable
#values, and the second with the corresponding second variable
values. Thus
#we are trying to find the best line fit for the points
[0,1],[2,2],[4,1].
4
3
Now that we’ve tested the command, we can use it on the national data:
> Xs:=convert(col(lnBG,1),list): #convert the first column of
#the ln-ln data into a list of the "x’s" The least squares
#command wants to have lists input, not matrix columns, even
#though it’s hard for us to see any difference
Ys:=convert(col(lnBG,2),list):
> fit[leastsquare[[X,Y]]]([Xs,Ys]);
Y=
Y = −6.038703303 + 2.593828004 X
We can paste in the equation of the line and see how well we did.
> line:=plot(-6.038703303+2.593828004*X, X=3.4..4.5,Y=2.7..5.7,
color=black):
display({line,lnlnplot}, title=‘least squares fit‘);
least squares fit
5.5
5
4.5
Y
4
3.5
3
3.4
3.6
3.8
4
4.2
4.4
X
>
Finally, we can go back from the least squares line fit to a power law
> m:=2.593828004; #power
b:=exp(-6.038703303); #proportionality constant
m := 2.593828004
b := 0.002384649077
> powerplot:=plot(b*h^m,h=0..80,w=0..200,color=black):
display({powerplot,boys,girls},title=
‘power law approximation for national height-weight data‘);
#by calling the variables h and w, and giving their ranges, I
#get Maple to label the axes as I want
power law approximation for national height-weight data
200
180
160
140
120
w 100
80
60
40
20
0
>
10
20
30
40
h
50
60
70
80
``` | 2,812 | 8,675 | {"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-2023-06 | latest | en | 0.951335 |
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# Answer: The hemisphere of radius r is made from a stack of
ISBN: 9780133918922 126
## Solution for problem 9-44 Chapter 9
Engineering Mechanics: Statics | 14th Edition
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Problem 9-44
The hemisphere of radius r is made from a stack of very thin plates such that the density varies with height, r = kz, where k is a constant. Determine its mass and the distance z to the center of mass G. z y z G x _ r Prob. 944
Step-by-Step Solution:
Step 1 of 3
Step 2 of 3
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Reset your password | 324 | 1,270 | {"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-2020-16 | latest | en | 0.843259 |
https://studylib.net/doc/12117653/pre-calculus-a | 1,628,004,100,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154459.22/warc/CC-MAIN-20210803124251-20210803154251-00334.warc.gz | 507,604,326 | 12,580 | Pre-Calculus A
Pre-Calculus A
4.4 Worksheet
Name _________________________________
Date ___________________
Hour ______
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE
1-4 : Determine the exact values of the six trigonometric functions of the angle θ.
2.
1.
3.
4.
5. The point (7, 24) is on the terminal side of an angle in standard position. Determine the exact
values of the six trigonometric functions of the angle.
6-7 : State the quadrant in which θ lies.
6. sin θ < 0 and cos θ < 0
7. sec θ > 0 and cot θ < 0
8-9 : Find the values of the six trigonometric functions of θ.
8. sin θ =
3
5
9. cos θ = –
when θ lies in Quadrant II.
4
5
when θ lies in Quadrant III
10-15 : Evaluate the sine, cosine, and tangent of the angle without using a calculator (no decimals)
10. 225°
11. 300°
13. -495°
14.
5𝜋
3
3𝜋
15.
4
16-19 : Find two solutions of the equation. Give your answers in radians (0 < θ < 2π).
Do not use a calculator.
1
1
√2
√2
𝟏𝟔. 𝒔𝒊𝒏 𝜃 =
𝟏𝟕. 𝒔𝒊𝒏 𝜃 = −
𝟏𝟖. 𝒄𝒐𝒔 𝜃 =
𝟏𝟗. 𝒄𝒐𝒔 𝜃 = −
2
2
2
2
20. Meteorology The monthly normal
temperatures T (in degrees Fahrenheit) for Santa
Fe, New Mexico are given by:
𝜋𝑡
7𝜋
𝑇 = 49.5 + 20.5𝑐𝑜𝑠 ( − )
6
6
Where t is the time in month, with t = 1
corresponding to January. Find the monthly
normal temperature for each month.
a) January
21. Sales A company that produces water skis,
which are seasonal products, forecasts monthly
sales over a two year period to be:
𝜋𝑡
𝑆 = 23.1 + 0.442𝑡 + 4.3 𝑠𝑖𝑛 ( )
6
Where S is measured in thousands of units and t is
the time (in months), with t = 1 representing
January 2006. Estimate sales for each month.
a) January 2006
b) July
b) February 2007
c) December
c) May 2006 | 637 | 1,665 | {"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} | 4.09375 | 4 | CC-MAIN-2021-31 | latest | en | 0.781337 |
http://www.topperlearning.com/forums/ask-experts-19/find-the-sum-of-all-terms-given-in-a-sequence-1-2-x-3-x-2-4-mathematics-arithmetic-progression-53093/reply | 1,487,722,214,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170864.16/warc/CC-MAIN-20170219104610-00450-ip-10-171-10-108.ec2.internal.warc.gz | 629,216,735 | 37,495 |
Question
Mon February 13, 2012
# Find the sum of all terms given in a sequence 1,2+x,3+x^2,4+x^3,...,n+x^(n-1)
Tue February 14, 2012
Sum = 1 + 2 + x + 3 + x2 + ......
= (1+2+3+ ......+n) + (x + x2 + .....+xn-1)
= n(n+1)/2 + x(xn - 1)/(x-1)
Related Questions
Fri February 17, 2017
# The 19th term of an A. P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.
Mon February 13, 2017
| 180 | 417 | {"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} | 3.28125 | 3 | CC-MAIN-2017-09 | longest | en | 0.930353 |
http://landshape.org/enm/antarctic-snowfall-data-visualisation/ | 1,386,460,257,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163056120/warc/CC-MAIN-20131204131736-00057-ip-10-33-133-15.ec2.internal.warc.gz | 105,746,700 | 30,455 | • Home»
• All»
• Antarctic Snowfall Data Visualisation
# Antarctic Snowfall Data Visualisation
An issue in question here is whether the recent snowfall at Law Dome is unusually high relative to the 750 year long record (and therefore, so the argument goes, probably due to AGW).
Below is the snowfall at Law Dome from the ice core. Above is the actual snowfall, and below is the accumulation of the series minus the mean (using the R function cumsum) indicating where snowfall is above or below average.
This simple approach is not used in the paper. While the accumulation of snow at present (relative to the mean) is high, it was just as high earlier in the record around 1400. Figure 3 in the paper shows a 10 year Gaussian filter of these data, so I illustrate that below, with a similar result.
The approach in the paper records the size of filtered accumulation ‘events’. These would be the (approx. 60) areas under discrete sections of curves in the upper panel of the figures, postive if above or and negative if below the mean (red line). The event of most interest is the last one, where snow has accumulated since 1970, after filtering.
I start to get concerned with an approach like this because its unfamiliar to me, and I don’t recall anywhere in the statistics books where it has been characterized. I check the distribution. The figure below shows the distribution of these events, and compares them to a normal distribution.
The distribution of event sizes appears to be Leptokurtic (peaked with fat tails) although the supplementary information states that the distribution at a Gaussian half-filter size of 5 did not differ from the normal distribution. I have to check whether this deviation is accurate or due to the binning of the data.
Finally I looked at the improbability of the final event by calculating the set of events and their standard deviations for a range of filter sizes:
The dashed red line is the 2-sigma level, and dotted red line is the 3-sigma level. The size of the final event becomes 3-sigma significant at around half-filter size 4 and declines thereafter. In other words, the final event is significant at scales between 8 and 40 years.
Here is a link to the R code and data used to generate these plots.
Tas van Ommen has been a really good sport in providing data and information to allow me to check the figures. The relevant passage from the paper is as follows:
Long-term climate variability
In assessing the degree to which recent trends are natural or anthropogenic, it is useful to consider the size of the anomaly in the context of 750 years of data (Fig. 3). We define decadal-scale anomalies as the difference between the ten-year Gaussian smoothed series and the long-term mean, and use the sign changes in this anomaly to define successive anomalous intervals. The size of the event is the integrated accumulation anomaly over each interval. We find the event that commenced in 1970 is the largest in the record. It is significantly larger than the distribution of the 56 other events (P=7×10−4,1-tail t -test; Supplementary Information). Only a single such event is expected in around 38,000 years for a climate unchanged from that of the past 750 years. were computed using one-sided t -tests.
In contrast, I get a significance of 99.7% or P=3×10-3 an expectation of every 333 years, or twice in the record, a similar level to the accumulation plot I presented first up.
There is an inconsistency in the passage quoted above. By my calculation a P=7×10−4 event is 1/0.0007, equalling an occurance every 1428 years, so I am not sure where the figure of 38,000 years comes from.
Another puzzle is that statement “It is significantly larger than the distribution of the 56 other events (P=7×10−4,1-tail t -test; Supplementary Information). ” indicates the test is based on the size of the final accumulation event. However the relevant statement in the supplementary information suggest this figure comes from a test of difference of distributions.
Law Dome 750-Year Accumulation Statistics
We computed decadal scale anomalies for the LD precipitation series as described in the main text, and found that the largest anomaly was the most recent decadal event, which began in 1970. This event appears to belong to a population with a larger mean than the remaining 56 decadal events – we reject the null hypothesis (that it comes from the same distribution) using a 1-tailed t-test, (P = 7×10-4).
The test of the size of the final event (comparable to the test I performed above) is described in the following paragraph. This is a order of magnitude different to my replication.
The 56 anomalies in the period 1250-1969 have a mean precipitation of –0.0188 m and standard deviation, 0.719 m (ice-equivalent, i.e. for glacial ice density 917 kg.m-3). The post 1970 anomaly is 2.42 m (i.e.), which is extremely improbable from the distribution of 1250-1969 anomalies (P = 3.4×10-4, normal z-score).
It seems that the significance of the event might be based on the distribution of the (possibly filtered) annual snowfalls during that event. If so, put your statistician on danger money! I will try to replicate the stated values more closely tomorrow.
Even though this is an initial visual examination of the data, there are some questions that need to be looked at more closely:
Where do the test values come from and the 38,000 year figure?
Are the data really leptokurtic, meaning that even my initial estimates of P=0.003 are inflated anyway?
The snowfall at Law Dome does not look particularly impressive here, given that the present accumulation of snow is not greater that at other times in the 750 year record.
Reference:
van Ommen, T. D. and V. Morgan (2010). Snowfall increase in coastal East Antarctica linked with southwest Western Australian drought, Nature Geoscience, Advance Online Publication, doi:10.1038/NGEO761
• kuhnkat
The papers claims are based on, uh, math errors??
• davids99us
I thought I knew what was tested from the term “integratedaccumulation anomaly”. But it can't be that. Perhaps Tas will clearit up, as well as the 38,000 year event.
• Anonymous
The papers claims are based on, uh, math errors??
• Anonymous
I thought I knew what was tested from the term “integrated
accumulation anomaly”. But it can’t be that. Perhaps Tas will clear
it up, as well as the 38,000 year event.
• Anonymous
Math error, statistical error, logic error — you can call it any of those. It appears to be the relatively common problem of assuming normality and extrapolating to assume something is a very rare event when it isn’t.
• Anonymous
All of the above. But more to come on the last of these.
• sherro
Given the subjective impression that Perth weather shows up in Adelaide a day or two later, then another day to Melbourne, would not the hypothesis be strengthened by comparing these locations with Law Dome as well? If the correlations between Perth and Adelaide are worse than Perth and Law Dome, I'd be a bit worried.
• Anonymous
Given the subjective impression that Perth weather shows up in Adelaide a day or two later, then another day to Melbourne, would not the hypothesis be strengthened by comparing these locations with Law Dome as well? If the correlations between Perth and Adelaide are worse than Perth and Law Dome, I’d be a bit worried.
• News_Watcher
Math error, statistical error, logic error — you can call it any of those. It appears to be the relatively common problem of assuming normality and extrapolating to assume something is a very rare event when it isn't.
• davids99us
All of the above. But more to come on the last of these.
• tasvanommen
• davids99us
Hi Tas, Thanks for the clarifications. Its clear you have put thought into the statistics, which makes it a more interesting paper to dissect, and given the data is simple, its easy for people to get into. Probably I will put together a summary of the main issues that remain at the end, as right a the moment I mainly want to understand what has been done. The issue that these events are of varying lengths, adds a lot of parameters, and is a concern. In fact, the most recent 'event' appears longer than most, and so of course would have a higher cumulative value. When the data are aggregated with a constant interval, the significance drops. When you add in smoothing and procedures like this, I don't think you can necessarily solve them by just testing the distribution, because the data set is finite. No doubt others would have something to say about it. Ultimately I think that if its a robust result, it will hold up under different approaches. If the result only appears with a specific approach and set of parameters, that's a problem, and attempting to justify a novel technique just takes up a lot of time and energy. As they say, if you torture data enough it will tell you what you want to hear. So I think one should say, lets look at what we have got with the simplest possible approach, considering the validity of the basic assumptions.
• Anonymous
Hi David,
First, I’m more than happy to see others play around with the data, and have a moment free right now, but I can’t necessarily respond to any and every post, and I don’t check your blog frequently – I just happened across this today.
You plot something you call the accumulation of the series (in glaciology, confusingly, we call the original precip series itself snowfall accumulation) – in any case it is the cumulative sum of the original accumulation series minus its mean, then centred to zero mean. This is the integral of the snowfall anomaly, and indeed we have looked at it, but for the purposes of this paper it isn’t the quantity of direct interest. It is, actually of glaciological interest, because it corresponds to snowfall mass-balance anomaly and assuming ice flow response is slow (which it is) compared with the fluctuations we expect to see surface height variations. Indeed there is evidence from old survey work on Law Dome, and modern GPS and repeats that the surface height has increased during the present positive growth anomaly.
Anyway, since the cumulative sum you plot is the _integral_ of precipitation rate the high and relatively steady levels in the late 1300s-1400s correspond to a steady snowfall (i.e. derivative of this curve ~0) about the same as the long term AVERAGE – i.e. snowfall rate is not high. If you look more closely, this positive anomaly results from excess snowfall events over the period AD1316-1355 when, in a couple of separate spurts, about +1.8m was added over 39 years.
This is to be compared with the modern period positive anomaly at the end, in which in a single spurt of 36 years, +2.4m was added. Note, that this is of the same order as the height increases seen in survey work.
I understand that the approach I’ve used is unfamiliar to you and indeed this is what research is all about – doing novel things. The key of course is not to reinvent wheels. In doing the work, I have had input from professional statisticians as well as climate professionals. So, on to the matter of distributions – you will know of course that from any small sample, the task of discerning the shape of a parent distribution and deciding if it is normal is not a straightforward issue. This is why, as stated in the paper, I did some appropriate tests – a Q-Q plot, which is better than just plotting the binned data, a Lilliefors test and Kolmogorov-Smirnov test, which all indicated the validity of a normal distribution. EVEN SO, I then also repeated my significance tests using a non-parametric Wilcoxon test, to avoid the assumption of normality.
Finally, to the issue of probabilities. The P=7×10^-4 value is the confidence with which a t-test isolates the post-1970s anomaly as NOT coming from the same distribution as the other 56 anomalies. It is a parametric test – you will also read that using the non-parametric Wilcoxon rank-sum, the value falls to P=0.04: still significant, albeit not such a stand-out.
The 38000 year number comes from the converse argument – say the event is in the same distribution and ask how unlikely it would be. As shown in detail in the supplementary data, the distribution of events has mean -0.0188m and sigma=0.719m, which yields P=3.4×10-4, or once in 2930 occurrences. Here occurrences aren’t yearly, but correspond to our decadal anomalies which (as stated in the supplementary information) occur every 12.86 years in our record. 12.86×2930 should give the 38000 year figure. All of which is not meant to do more than illustrate the unusual nature of this in an otherwise stationary climate (which of course is not something we have over 38000 year periods!).
• Anonymous
Hi Tas, Thanks for the clarifications. Its clear you have put thought into the statistics, which makes it a more interesting paper to dissect, and given the data is simple, its easy for people to get into.
Probably I will put together a summary of the main issues that remain at the end, as right a the moment I mainly want to understand what has been done. The issue that these events are of varying lengths, adds a lot of parameters, and is a concern. In fact, the most recent ‘event’ appears longer than most, and so of course would have a higher cumulative value.
When the data are aggregated with a constant interval, the significance drops. When you add in smoothing and procedures like this, I don’t think you can necessarily clear up concerns by just testing the distribution, because the data set is finite. No doubt others would have something to say about it.
Ultimately I think that if its a robust result, it will hold up under different approaches. If the result only appears with a specific approach and set of parameters, that’s a problem, and attempting to justify a novel technique just takes up a lot of time and energy. As they say, if you torture data enough it will tell you what you want to hear.
So I think one should say, lets look at what we have got with the simplest possible approach, considering the validity of the basic assumptions.
But I don’t have time for discussion of methodology of science, so I am not expecting a reply, just letting you know why I am doing what I am doing. There are still some of these numbers I have to replicate, and your explanation helps a lot. Thanks.
• davids99us
All of the above. But more to come on the last of these.
• tasvanommen | 3,213 | 14,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2013-48 | latest | en | 0.935329 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/1216/1/p/a/ | 1,656,487,822,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103624904.34/warc/CC-MAIN-20220629054527-20220629084527-00799.warc.gz | 926,353,452 | 54,491 | # Properties
Label 1216.1.p.a Level $1216$ Weight $1$ Character orbit 1216.p Analytic conductor $0.607$ Analytic rank $0$ Dimension $4$ Projective image $D_{6}$ CM discriminant -8 Inner twists $8$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$1216 = 2^{6} \cdot 19$$ Weight: $$k$$ $$=$$ $$1$$ Character orbit: $$[\chi]$$ $$=$$ 1216.p (of order $$6$$, degree $$2$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$0.606863055362$$ Analytic rank: $$0$$ Dimension: $$4$$ Relative dimension: $$2$$ over $$\Q(\zeta_{6})$$ Coefficient field: $$\Q(\zeta_{12})$$ Defining polynomial: $$x^{4} - x^{2} + 1$$ x^4 - x^2 + 1 Coefficient ring: $$\Z[a_1, \ldots, a_{11}]$$ Coefficient ring index: $$1$$ Twist minimal: yes Projective image: $$D_{6}$$ Projective field: Galois closure of 6.2.1267762688.2
## $q$-expansion
The $$q$$-expansion and trace form are shown below.
$$f(q)$$ $$=$$ $$q + ( - \zeta_{12}^{5} - \zeta_{12}^{3}) q^{3} + ( - \zeta_{12}^{4} - \zeta_{12}^{2} - 1) q^{9}+O(q^{10})$$ q + (-z^5 - z^3) * q^3 + (-z^4 - z^2 - 1) * q^9 $$q + ( - \zeta_{12}^{5} - \zeta_{12}^{3}) q^{3} + ( - \zeta_{12}^{4} - \zeta_{12}^{2} - 1) q^{9} - \zeta_{12}^{3} q^{11} + \zeta_{12}^{4} q^{17} - \zeta_{12}^{5} q^{19} - \zeta_{12}^{2} q^{25} + (\zeta_{12}^{5} + \zeta_{12}) q^{27} + ( - \zeta_{12}^{2} - 1) q^{33} + (\zeta_{12}^{2} + 1) q^{41} + \zeta_{12} q^{43} - q^{49} + (2 \zeta_{12}^{3} + 2 \zeta_{12}) q^{51} + ( - \zeta_{12}^{4} - \zeta_{12}^{2}) q^{57} + (\zeta_{12}^{5} + \zeta_{12}^{3}) q^{59} + ( - \zeta_{12}^{3} - \zeta_{12}) q^{67} + \zeta_{12}^{4} q^{73} + (\zeta_{12}^{5} - \zeta_{12}) q^{75} + (\zeta_{12}^{4} + 1) q^{81} - \zeta_{12}^{3} q^{83} + (\zeta_{12}^{2} + 1) q^{97} + (\zeta_{12}^{5} + \zeta_{12}^{3} - \zeta_{12}) q^{99} +O(q^{100})$$ q + (-z^5 - z^3) * q^3 + (-z^4 - z^2 - 1) * q^9 - z^3 * q^11 + z^4 * q^17 - z^5 * q^19 - z^2 * q^25 + (z^5 + z) * q^27 + (-z^2 - 1) * q^33 + (z^2 + 1) * q^41 + z * q^43 - q^49 + (2*z^3 + 2*z) * q^51 + (-z^4 - z^2) * q^57 + (z^5 + z^3) * q^59 + (-z^3 - z) * q^67 + z^4 * q^73 + (z^5 - z) * q^75 + (z^4 + 1) * q^81 - z^3 * q^83 + (z^2 + 1) * q^97 + (z^5 + z^3 - z) * q^99 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4 q - 4 q^{9}+O(q^{10})$$ 4 * q - 4 * q^9 $$4 q - 4 q^{9} - 4 q^{17} - 2 q^{25} - 6 q^{33} + 6 q^{41} - 4 q^{49} - 2 q^{73} - 2 q^{81} + 6 q^{97}+O(q^{100})$$ 4 * q - 4 * q^9 - 4 * q^17 - 2 * q^25 - 6 * q^33 + 6 * q^41 - 4 * q^49 - 2 * q^73 - 2 * q^81 + 6 * q^97
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/1216\mathbb{Z}\right)^\times$$.
$$n$$ $$191$$ $$705$$ $$837$$ $$\chi(n)$$ $$1$$ $$\zeta_{12}^{2}$$ $$-1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
673.1
−0.866025 − 0.500000i 0.866025 + 0.500000i −0.866025 + 0.500000i 0.866025 − 0.500000i
0 −0.866025 + 1.50000i 0 0 0 0 0 −1.00000 1.73205i 0
673.2 0 0.866025 1.50000i 0 0 0 0 0 −1.00000 1.73205i 0
1057.1 0 −0.866025 1.50000i 0 0 0 0 0 −1.00000 + 1.73205i 0
1057.2 0 0.866025 + 1.50000i 0 0 0 0 0 −1.00000 + 1.73205i 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
8.d odd 2 1 CM by $$\Q(\sqrt{-2})$$
4.b odd 2 1 inner
8.b even 2 1 inner
19.d odd 6 1 inner
76.f even 6 1 inner
152.l odd 6 1 inner
152.o even 6 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 1216.1.p.a 4
4.b odd 2 1 inner 1216.1.p.a 4
8.b even 2 1 inner 1216.1.p.a 4
8.d odd 2 1 CM 1216.1.p.a 4
19.d odd 6 1 inner 1216.1.p.a 4
76.f even 6 1 inner 1216.1.p.a 4
152.l odd 6 1 inner 1216.1.p.a 4
152.o even 6 1 inner 1216.1.p.a 4
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
1216.1.p.a 4 1.a even 1 1 trivial
1216.1.p.a 4 4.b odd 2 1 inner
1216.1.p.a 4 8.b even 2 1 inner
1216.1.p.a 4 8.d odd 2 1 CM
1216.1.p.a 4 19.d odd 6 1 inner
1216.1.p.a 4 76.f even 6 1 inner
1216.1.p.a 4 152.l odd 6 1 inner
1216.1.p.a 4 152.o even 6 1 inner
## Hecke kernels
This newform subspace is the entire newspace $$S_{1}^{\mathrm{new}}(1216, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{4}$$
$3$ $$T^{4} + 3T^{2} + 9$$
$5$ $$T^{4}$$
$7$ $$T^{4}$$
$11$ $$(T^{2} + 1)^{2}$$
$13$ $$T^{4}$$
$17$ $$(T^{2} + 2 T + 4)^{2}$$
$19$ $$T^{4} - T^{2} + 1$$
$23$ $$T^{4}$$
$29$ $$T^{4}$$
$31$ $$T^{4}$$
$37$ $$T^{4}$$
$41$ $$(T^{2} - 3 T + 3)^{2}$$
$43$ $$T^{4} - 4T^{2} + 16$$
$47$ $$T^{4}$$
$53$ $$T^{4}$$
$59$ $$T^{4} + 3T^{2} + 9$$
$61$ $$T^{4}$$
$67$ $$T^{4} + 3T^{2} + 9$$
$71$ $$T^{4}$$
$73$ $$(T^{2} + T + 1)^{2}$$
$79$ $$T^{4}$$
$83$ $$(T^{2} + 1)^{2}$$
$89$ $$T^{4}$$
$97$ $$(T^{2} - 3 T + 3)^{2}$$ | 2,513 | 4,976 | {"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} | 2.609375 | 3 | CC-MAIN-2022-27 | latest | en | 0.391131 |
http://www.assignmentpoint.com/science/mathematic/define-discuss-functions.html | 1,527,219,275,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866938.68/warc/CC-MAIN-20180525024404-20180525044404-00343.warc.gz | 321,172,863 | 6,825 | Principle objective of this article is to Define and Discuss on Functions. Here explain Functions in mathematical point of view with examples. A function is defined as a set of ordered pairs ( x, y), such that for each first ingredient x, there corresponds one and only one second element y. The set of first elements is named the domain of the particular function, while the pair of second elements is called the number of the function. The domain variable is called the independent variable, and the range variable is called the dependent variable. | 108 | 550 | {"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-2018-22 | latest | en | 0.90751 |
http://hkopp.github.io/2021/05/hopalongs | 1,726,469,703,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651676.3/warc/CC-MAIN-20240916044225-20240916074225-00858.warc.gz | 15,432,962 | 3,679 | Back, when the internet was still small and i was still a pupil I had an intense fascination with fractals. Among others, I worked on so called hopalongs. These are orbits of a special map. They were discovered by Barry Martin of Aston University in Birmingham. In 1986 they were popularized in the Scientific American by the well-known recreational mathematician Dewdney. The hopalong map is iteratively applied to a point (x,y) and the resulting points are plotted. The hopalong map takes three parameters, a, b, and c. Interestingly, the final image does not really depend on the starting point, but is very sensitive to changes in the parameters.
A previous collaboration of me on Hopalongs was with Schwebinghaus.
Back then, I programmed in Basic and bad Java, so I thought it would be a nice nostalgic experience to program hopalongs in Python. Below is the code and some nice images.
If you are capable of understanding German, you should also take a look here.
``````import seaborn as sns
import numpy as np
``````
``````def create_hopalong(a, b, c, n=5000):
x = np.zeros(n)
y = np.zeros(n)
for i in range(n-1):
x[i+1] = y[i]-np.sign(x[i])*np.sqrt(abs(b*x[i]-c))
y[i+1] = a-x[i]
return(x,y)
``````
``````x,y = create_hopalong(0.7,1,0.3)
sns.jointplot(x=x,y=y)
``````
``````x,y = create_hopalong(0.7,0.3,0)
sns.jointplot(x=x,y=y)
``````
``````x,y = create_hopalong(0.5,0.3,0,100000)
sns.jointplot(x=x,y=y)
``````
``````x,y = create_hopalong(0.2,0.3,0.2)
sns.jointplot(x=x,y=y)
``````
``````x,y = create_hopalong(0.1,0.3,0.2)
sns.jointplot(x=x,y=y)
``````
``````x,y = create_hopalong(0.11,0.3,0.2)
sns.jointplot(x=x,y=y)
``````
``````x,y = create_hopalong(0.111,0.3,0.2)
sns.jointplot(x=x,y=y)
``````
``````x,y = create_hopalong(0.111,0.4,0.2, 10000)
sns.jointplot(x=x,y=y)
``````
``````x,y = create_hopalong(0.111,0.4,0.2, 15000)
sns.jointplot(x=x,y=y)
``````
``````x,y = create_hopalong(0.111,0.4,0.2, 30000)
sns.jointplot(x=x,y=y)
``````
29 May 2021
#### Category
Recreational Math | 675 | 2,006 | {"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-2024-38 | latest | en | 0.899893 |
http://mathoverflow.net/questions/tagged/graph-colorings+computer-science | 1,394,388,819,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010076008/warc/CC-MAIN-20140305090116-00072-ip-10-183-142-35.ec2.internal.warc.gz | 120,688,709 | 8,513 | # Tagged Questions
0answers
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Let $C$ be a circuit that maps $n$-length bitstrings to elements of $\{0, 1, 2\}$. Arrange the $n$-length bitstrings in a giant loop: $0^n$ is connected to $1^n$ and $0^{n-1}1$, $0^{n-1}1$ is ...
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### Coloring edges on a graph s.t. the set of edges for any two vertices have no more than 'k' colors in common
Please imagine the case where one has a planar graph, $G$, with a set of $|V|$ vertices, $(v_1, ..., v_{|V|}) \in V$, and $|E|$ edges, $(e_1, ..., e_{|E|}) \in E$. Now, provided a total of $N$ ... | 220 | 655 | {"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} | 2.953125 | 3 | CC-MAIN-2014-10 | longest | en | 0.838351 |
https://www.expertsmind.com/library/in-this-module-we-examine-the-time-value-of-money-5724278.aspx | 1,726,574,403,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651773.64/warc/CC-MAIN-20240917104423-20240917134423-00565.warc.gz | 709,886,642 | 14,534 | ### In this module we examine the time value of money
Assignment Help Financial Management
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In this module we examine the time value of money. This concept can also be used to plan an individual’s retirement account. Assume some amount of monthly contributions, employer matching added in, assumed average annual earnings, and the expected number of years until retirement, to calculate the expected size of a generic retirement account.
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Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report! | 1,111 | 5,385 | {"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-2024-38 | latest | en | 0.930587 |
https://www.enotes.com/homework-help/mr-pickle-invested-13-000-two-accounts-one-436354 | 1,516,796,321,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084894125.99/warc/CC-MAIN-20180124105939-20180124125939-00100.warc.gz | 885,008,545 | 9,762 | # Mr.Pickle invested \$13,000 in two accounts, one yielding 6% interest and the other yielding 11%. If he received a total of \$1,130 in interest at the end of the year, how much did he invest in each...
Mr.Pickle invested \$13,000 in two accounts, one yielding 6% interest and the other yielding 11%. If he received a total of \$1,130 in interest at the end of the year, how much did he invest in each account with 6% ____ and 11%____?
Let x = the amount invested in account 1
Let y= the amount invested in account 2
Interest x = 0.06x
Interest y = 0.11y
`0.06x+0.11y=1130` and `x + y = 13000`
With two equations and two unknowns we can solve for x and y. We will use the substitution method to do so:
`y = 13000 - x`
`0.06x+0.11(13000-x)=1130`
`0.06x+1430-0.11x=1130`
`0.06x-0.11x=1130-1430`
`-0.05x=-300-gtx=-300/-0.05=6000`
Substitute the value for x into the equation for y:
`y=13000-6000=7000`
Therefore, Mr. Pickle invested \$6000 at 6% interest and \$7000 at 11% interest. | 339 | 995 | {"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} | 4.4375 | 4 | CC-MAIN-2018-05 | latest | en | 0.902856 |
https://gitlab.com/jim.hefferon/linear-algebra/commit/5526a5ff0a4a8ae0c99f8bb7ac02730b56e64d97 | 1,566,265,577,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315174.57/warc/CC-MAIN-20190820003509-20190820025509-00012.warc.gz | 483,803,634 | 18,232 | Commit 5526a5ff by Jim Hefferon
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... ... @@ -21047,7 +21047,7 @@ octave:5> z=[x, y]; octave:6> gplot z \end{lstlisting} yields this graph. By eye we judge that if $p>0.7$ then the team is close to assurred By eye we judge that if $p>0.7$ then the team is close to assured of the series. \begin{center} \includegraphics{ws.eps}
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... ... @@ -28,7 +28,7 @@ after $n$ flips. Then, for instance, we have that the probability of being in state~$s_0$ after flip~$n+1$ is $p_{0}(n+1)=p_{0}(n)+0.5\cdot p_{1}(n)$. This matrix equation sumarizes. This matrix equation summarizes. \begin{equation*} \begin{dmat}{D{.}{.}{1}D{.}{.}{1}D{.}{.}{1}D{.}{.}{1}D{.}{.}{1}D{.}{.}{1}} 1. &.5 &0. &0. &0. &0. \\ ... ... @@ -104,7 +104,7 @@ or \definend{stochastic matrix},\index{matrix!stochastic}% \index{stochastic matrix} whose entries are nonnegative reals and whose columns sum to $1$. A characteriztic feature of A characteristic feature of a Markov chain model is that it is \definend{historyless}\index{historyless}% \index{Markov chain!historyless} in that ... ... @@ -146,7 +146,7 @@ a child of a middle class worker has a $0.37$~chance of being middle class, and a child of a lower class worker has a $0.27$ probability of becoming middle class. With the initial distribution of the respondents's fathers given below, With the initial distribution of the respondent's fathers given below, this table gives the next five generations. \begin{center} \begin{tabular}{c|ccccc} ... ... @@ -1178,7 +1178,7 @@ octave:5> gplot z Assume that there are two states $s_T$, living in town, and $s_C$, living elsewhere. \begin{exparts} \partsitem Construct the transistion matrix. \partsitem Construct the transition matrix. \partsitem Starting with an initial distribution $s_T=0.3$ and $s_C=0.7$, get the results for the first ten years. \partsitem Do the same for $s_T=0.2$. ... ... @@ -1800,7 +1800,7 @@ octave:5> z=[x, y]; octave:6> gplot z \end{lstlisting} yields this graph. By eye we judge that if $p>0.7$ then the team is close to assurred By eye we judge that if $p>0.7$ then the team is close to assured of the series. \begin{center} \includegraphics{ws.eps} ... ... @@ -1811,15 +1811,15 @@ octave:6> gplot z Above we define a transition matrix to have each entry nonnegative and each column sum to $1$. \begin{exparts} \item Check that the three transistion matrices shown in this Topic \item Check that the three transition matrices shown in this Topic meet these two conditions. Must any transition matrix do so? \item Observe that if $A\vec{v}_0=\vec{v}_1$ and $A\vec{v}_1=\vec{v}_2$ then $A^2$ is a transition matrix from $\vec{v}_0$ to $\vec{v}_2$. Show that a power of a transition matrix is also a transistion matrix. Show that a power of a transition matrix is also a transition matrix. \item Generalize the prior item by proving that the product of two appropriately-sized transistion matrices is a transistion matrix. proving that the product of two appropriately-sized transition matrices is a transition matrix. \end{exparts} \begin{answer} \begin{exparts} ... ...
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Please register or to comment | 993 | 3,333 | {"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} | 3.03125 | 3 | CC-MAIN-2019-35 | latest | en | 0.747763 |
https://cs.stackexchange.com/questions/124502/show-that-a-alpha-approximation-algorithm-is-not-a-alpha-x-approximation | 1,696,075,379,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510676.40/warc/CC-MAIN-20230930113949-20230930143949-00709.warc.gz | 197,123,600 | 40,713 | # Show that a $\alpha$-approximation algorithm is not a ($\alpha-x$) approximation algorithm for $x >$0
Suppose you have a system that consists of $$m$$ slow machines and $$k$$ fast machines. The fast machines can perform twice as much work per unit time as the slow machines. Now you are given a set of n jobs; job $$i$$ takes time $$t_i$$ to process on a slow machine and time $$\frac{1}{2} t_i$$ to process on a fast machine. The goal is to minimize the makespan — the maximum, over all machines, of the total processing time of jobs assigned to that machine.
Algorithm: When each job arrives, we put it on the machine that currently ends the soonest(machine with smallest current load before the job is assigned). (Note that this determination involves taking into account the speeds of the machines.) This is a 3-approximation algorithm.
I want to find an example for which the algorithm is not a $$(3 - x)$$-approximation algorithm for any $$x > 0$$.
Condering that the makespan of the resulting algorithm is $$T$$, to do that I thought that, since the algorithm is a $$(3-x)$$-approximation if $$T \leq (3-x)OPT$$, I would just have to find an example for which $$T > (3-x)OPT$$. But I cannot seem to be able to find one, all the examples I find yield $$T \leq (3-x)OPT$$.
I just found a very basic example, where I have 4 jobs and 2 machines, one fast and one slow. The jobs: 4, 4, 1, 2
The algorithm would first assign job 1 to the fast machine, then job 2 to the second machine, then jobs 3 and 4 to the fast machine. This yields a makespan $$T$$ of $$4$$.
The optimal makespan $$OPT$$ would be assigning jobs 1, 2 and 3 to the fast machine and job 4 to the slow machine, so $$OPT = \frac{9}{4}$$.
But then $$4 < 9 (4\frac{9}{4})$$
What am I doing wrong?
Suppose that you have one fast machine and one small machine, and consider jobs of the following durations: $$t_1=t_2=\epsilon$$, $$t_3=1$$, $$t_4=2$$.
The greedy algorithm first puts a job taking time $$\epsilon$$ on each machine (immaterial of tie-breaking rules). It then puts the third job on the fast machine (since it ends in $$\epsilon/2$$ rather than $$\epsilon$$), and then the fourth job on the slow machine. The makespan is $$2+\epsilon$$.
In contrast, if we put jobs 1,2,4 on the fast machine and job 3 on the slow machine, then the makespan is $$1+\epsilon$$.
This shows that your algorithm is not a $$(2-\delta)$$-approximation for any $$\delta > 0$$.
Indeed, suppose, for the sake of contradiction, that the algorithm was a $$(2-\delta)$$-approximation for some $$\delta > 0$$. Find $$\epsilon > 0$$ small enough so that $$(1+\epsilon)(2-\delta) < 2+\epsilon$$ (exercise: show that this is always possible), and consider the corresponding example.
Since the optimal makespan is at most $$1+\epsilon$$ and the algorithm is a $$(2-\delta)$$-approximation, it must produce a solution whose makespan is at most $$(1+\epsilon)(2-\delta)$$. However, the algorithm produces a solution with makespan $$2+\epsilon > (1+\epsilon)(2-\delta)$$, and we reach a contradiction.
You might be able to give an even worse example. This is just to demonstrate the proof technique.
I suggest looking at the proof that the algorithm is a $$k$$-approximation algorithm, and try to find an example for which the analysis is tight. This will show that the algorithm is not a $$(k-\delta)$$-approximation for any $$\delta > 0$$.
• How does that show that the algorithm is not a (2- x)-approximation ?? Apr 22, 2020 at 10:04
• That's something you will have to figure out on your own. Try using the definition. Apr 22, 2020 at 10:05
• No. You will have to make this step on your own. You won't learn anything if we spoon-feed you. Don't be lazy! You can do it. Apr 22, 2020 at 10:08
• Ok, I spelled it out. I imagine that the source of difficulty is that you don't understand the definition of an approximation algorithm. Apr 22, 2020 at 10:15
• This analysis is wrong since the greedy algorithm would put the fourth job on the fast machine since the fast machine after the third job is added ends at e/2 + 1/2, which is less than e (on the slow machine). And therefore the makespan would be: 3/2 + e/2 And it is true for any value of e > 0 that: (3/2 + e/2) <= (1 + e)(k-x). Therefore this cannot be used to prove the algorithm is not a (k-x)-approximation algorithm Apr 24, 2020 at 13:19 | 1,195 | 4,360 | {"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": 38, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-40 | longest | en | 0.900787 |
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