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https://forums.codeguru.com/showthread.php?561323-RESOLVED-can-i-convert-a-string-into-a-math-expression&s=018621933292c4eef7a7a85aa6460096	[ "CodeGuru Home VC++ / MFC / C++ .NET / C# Visual Basic VB Forums Developer.com\n\n# Thread: [RESOLVED] can i convert a string into a math expression?\n\n1.", null, "Senior Member", null, "Join Date\nApr 2009\nPosts\n1,333\n\n## [RESOLVED] can i convert a string into a math expression?\n\nsee these string:\n2+3 + 3(5+5)\ncan i convert it to a math expression for calculate?", null, "", null, "Reply With Quote\n\n2.", null, "Member +", null, "", null, "", null, "Join Date\nFeb 2017\nPosts\n653\n\n## Re: can i convert a string into a math expression?", null, "Originally Posted by Cambalinho", null, "see these string:\n2+3 + 3(5+5)\ncan i convert it to a math expression for calculate?\nThere is no standard function in C++ that does that (if available in a language it's often called eval).\n\nThe most famous algorithm for this purpose is the Shunting-yard algorithm (by Dijkstra),\n\nhttps://en.wikipedia.org/wiki/Shunting-yard_algorithm\n\nI'm sure you can locate a suitable implementation on the net.\nLast edited by wolle; April 2nd, 2018 at 03:31 AM.", null, "", null, "Reply With Quote\n\n3. ## Re: can i convert a string into a math expression?\n\nI have written math expressions' parser and evaluator in C++. It's a C++ wrapper class.\nWorks with 1, 2 variable, and thus can handle customized functions from 1,2 arguments for 2d and 3d processing.\nLet me know if you need such functionality in your application and I'll send you the sources.", null, "", null, "Reply With Quote\n\n4. ## Re: can i convert a string into a math expression?", null, "Originally Posted by Cambalinho", null, "see these string:\n2+3 + 3(5+5)\ncan i convert it to a math expression for calculate?\nThese types of expressions are called infix and are not really suited for computer evaluation. To make them suitable, they are converted into postfix notation (also called reverse polish notation) which doesn't use brackets, the ordering is strict left to right and the operators come after the operands.\n\neg 3 + 4 5 infix becomes 3 4 5 * + postfix. (2 + 3) * (4 + 5) becomes 2 3 + 4 5 + * postfix. 2 + 3 + 3 * (5 + 5) infix becomes 2 3 + 3 5 5 + * + postfix. Postfix notation can be easily evaluated using a stack. Numbers are pushed onto the stack and when an operator is found, this is applied to the top 1 or 2 numbers (depending whether it is a unary or binary operator) which are popped from the stack and the result is pushed back to the stack. This is repeated until the end of the postfix expression. The result should be one number on the stack. If this is not true or there is stack underflow then there is an error in the postfix expression. To convert from infix to postfix, there are various methods of which Dijkstra's Shunting-yard is perhaps the most famous (see Wolle's post #2). Another common one is Recursive Descent parser (which is the method I prefer) see https://en.wikipedia.org/wiki/Recursive_descent_parser.\n\nPS if you use one you're found on the internet, make sure it's fit for your purpose - eg some don't handle unary minus correctly. eg -2 + -3 is a valid infix expression. Some only handle integers etc etc. Do you also need it to handle functions eg sin(), cos(), abs() etc etc.\nLast edited by 2kaud; March 31st, 2018 at 07:40 AM. Reason: PS", null, "", null, "Reply With Quote\n\n5.", null, "Senior Member", null, "Join Date\nApr 2009\nPosts\n1,333\n\n## Re: can i convert a string into a math expression?\n\nfinally i get results:\nCode:\n```#include <iostream>#include <string>\n#include <vector>\n#include <sstream>\n\nusing namespace std;\n\nclass MathOperators\n{\npublic:\n\nenum enumMathOperators\n{\nParenthesesOpen='(',\nParenthesesClose=')',\nMinus='-',\nMultiplication ='',\nSpace=' ',\nDivision ='/'\n};\n\nstatic bool IsMathOperator(char enMath)\n{\nswitch(enMath)\n{\ncase ParenthesesOpen:\ncase ParenthesesClose:\ncase Minus:\ncase Multiplication:\ncase Space:\ncase Division:\nreturn true;\ndefault:\nreturn false;\n}\n}\n};\n\nvoid CalculateExpression(string strMathExpression){\n//clean empty spaces\nfor(unsigned int i=0; i<strMathExpression.size(); i++)\n{\nif(strMathExpression[i]== ' ' \|\| strMathExpression[i]== '\\t')\n{\nstrMathExpression.erase(i,1);\ni=0;\n}\n}\n\n//spare the operator and numbers\n//for add them on a vector:\nstring strNumber=\"\";\nvector<string> vecOperator;\nint OperatorCount=0;\n\n//we must add a null string terminator\n//for we convert the char to string\n//or we can get unexpected results:\nchar chr[]={'0','\\0'};\nfor(unsigned int i=0; i<=strMathExpression.size() ; i++)\n{\nchr=strMathExpression[i];\nif(isdigit(chr))\n{\nstrNumber+=chr;\n}\nelse\n{\nif(isdigit(strNumber))\nvecOperator.push_back(strNumber);\nif( MathOperators::IsMathOperator(chr)==true)\n{\nvecOperator.push_back(chr);\nOperatorCount++;\n}\nOperatorCount++;\nstrNumber=\"\";\n\n}\n}\n\n//print actual vector:\nfor (unsigned int a=0; a<vecOperator.size();a++)\n{\ncout << vecOperator[a] << \" \";\n}\ncout << \"\\n\";\n\n//making the math:\nfor(unsigned int i=0;i<vecOperator.size(); i++)\n{\nif(vecOperator[i]==\"+\")\n{\n//get result in int:\nint result=stoi(vecOperator[i-1])+ stoi(vecOperator[i+1]);\n\n//get the result in int to string:\nvecOperator[i-1]= to_string(result);\n\n//erase the next 2 elements:\nvecOperator.erase(vecOperator.begin()+i,vecOperator.begin()+i+2);\n\n//print the elements after changes:\nfor (unsigned int a=0; a<vecOperator.size();a++)\n{\ncout << vecOperator[a] << \" \";\n}\ncout << \"\\n\";\n\n//before continue the i must be -1\n//when the for restart, the i will be 0:\ni=-1;\ncontinue;\n}\n\nelse if(vecOperator[i]==\"-\")\n{\nint result=stoi(vecOperator[i-1])- stoi(vecOperator[i+1]);\nvecOperator[i-1]= to_string(result);\nvecOperator.erase(vecOperator.begin()+i,vecOperator.begin()+i+2);\nfor (unsigned int a=0; a<vecOperator.size();a++)\n{\ncout << vecOperator[a] << \" \";\n}\ncout << \"\\n\";\ni=-1;\ncontinue;\n}\nelse if(vecOperator[i]==\"/\")\n{\nint result=stoi(vecOperator[i-1])/ stoi(vecOperator[i+1]);\nvecOperator[i-1]= to_string(result);\nvecOperator.erase(vecOperator.begin()+i,vecOperator.begin()+i+2);\nfor (unsigned int a=0; a<vecOperator.size();a++)\n{\ncout << vecOperator[a] << \" \";\n}\ncout << \"\\n\";\ni=-1;\ncontinue;\n}\nelse if(vecOperator[i]==\"\")\n{\nint result=stoi(vecOperator[i-1])* stoi(vecOperator[i+1]);\nvecOperator[i-1]= to_string(result);\nvecOperator.erase(vecOperator.begin()+i,vecOperator.begin()+i+2);\nfor (unsigned int a=0; a<vecOperator.size();a++)\n{\ncout << vecOperator[a] << \" \";\n}\ncout << \"\\n\";\ni=-1;\ncontinue;\n}\n}\n}\n\nint main()\n{\nstring strExpression;\ngetline(cin, strExpression);\nCalculateExpression(strExpression);\nreturn 0;\n}```\nthe math is done correctly. can anyone give me 1 advice: how can i control the priory math?", null, "readers: don't forget see the way that i do the conversion from char array to string... it's very important add the '\\0'(string null terminator) for avoid unexpected results", null, "", null, "Reply With Quote\n\n6. ## Re: can i convert a string into a math expression?\n\nThis code doesn't use mathematical precedence. This evaluates 1 + 2 * 3 as 9 whereas it should be 7 as 1+ 2 * 3 should be evaluated as 1 + (2 * 3). Also it doesn't recognise parenthesis even though they are specified as an operator. There is an error in the 'clean empty space code' which anyhow would usually be done as a single statement\n\nCode:\n`\tstrMathExpression.erase(remove_if(strMathExpression.begin(), strMathExpression.end(), isspace), strMathExpression.end());`\nhow can i control the priory math\nYou implement an infix to postfix parser as per posts #2 and #4 - which requires a complete re-design of this code so no further comments.", null, "", null, "Reply With Quote\n\n7.", null, "Senior Member", null, "Join Date\nApr 2009\nPosts\n1,333\n\n## Re: can i convert a string into a math expression?\n\ni'm sorry... that line give me several compiler errors, that's why i use the for loop.\nerror: \"no matching function for call to 'remove_if(std::__cxx11::basic_string<char>::iterator, std::__cxx11::basic_string<char>::iterator, <unresolved overloaded function type>)' \"", null, "", null, "Reply With Quote\n\n8. ## Re: can i convert a string into a math expression?\n\nHave you got the required #include statements\n\nCode:\n```#include <algorithm>\n#include <cctype>\n#include <string>```", null, "", null, "Reply With Quote\n\n9.", null, "Senior Member", null, "Join Date\nApr 2009\nPosts\n1,333\n\n## Re: can i convert a string into a math expression?\n\nyes. but i'm using code blocks.. maybe that's why i get the errors", null, "", null, "Reply With Quote\n\n10. ## Re: can i convert a string into a math expression?\n\nthen consider\n\nCode:\n```for(unsigned int i = 0; i < strMathExpression.size();)\nif (strMathExpression[i] == ' ' \|\| strMathExpression[i] == '\\t')\nstrMathExpression.erase(i,1);\nelse\n++i;```", null, "", null, "Reply With Quote\n\n11.", null, "Senior Member", null, "Join Date\nApr 2009\nPosts\n1,333\n\n## Re: can i convert a string into a math expression?\n\nbut can you give me an advice for control the operator priority?", null, "", null, "Reply With Quote\n\n12. ## Re: can i convert a string into a math expression?\n\nIt's beautiful code. Modern C++. Well commented. Tested. Use it and you may even learn something.\n...but doesn't work with negative numbers (unary minus). -1+2 gives -1946156963 !!", null, "PS. It also uses integer division. So 3/22 gives 2 but 23/2 gives 3 whereas mathematically these are the same and should give the same answer (3). Doh!!\nLast edited by 2kaud; April 1st, 2018 at 09:19 AM. Reason: PS", null, "", null, "Reply With Quote\n\n13.", null, "Member +", null, "", null, "", null, "Join Date\nFeb 2017\nPosts\n653\n\n## Re: can i convert a string into a math expression?", null, "Originally Posted by 2kaud", null, "...but doesn't work with negative numbers (unary minus). -1+2 gives -1946156963 !!", null, "PS. It also uses integer division. So 3/22 gives 2 but 23/2 gives 3 whereas mathematically these are the same and should give the same answer (3). Doh!!\nSorry, I remove this reference.", null, "", null, "Reply With Quote\n\n14. ## Re: can i convert a string into a math expression?\n\nThe original link for a shunting-yard c++ implementation was https://gist.github.com/t-mat/b9f681b7591cdae712f6 has issues including those discussed in post #12. There is also a fundamental implementation issue with the shunting for a right parenthesis which has been fixed below. Based upon this code, consider the following code which works with negative numbers and uses floating point rather than integer arithmetic. The main changes are in exprToTokens() which uses the type of the last token as a state indicator to determine whether the current token is valid or not. Also the evaluation code has been placed into eval() so that it is easier to use from within other code. Also eval() now returns a pair which indicates whether the eval() has been successful or not and if it has, the value. No error messages are now displayed. Note that the test code in main() requires c++17 - though it is fairly easy to convert it back to c++14 code.\n\nCode:\n```// Modified from https://gist.github.com/t-mat/b9f681b7591cdae712f6\n\n#include <algorithm>\n#include <iostream>\n#include <string>\n#include <vector>\n#include <deque>\n#include <cmath>\n#include <cctype>\n#include <utility>\n\nusing Mtype = long double;\n\nstruct Token {\nenum class Type {\nUnknown = 0,\nNumber,\nOperator,\nLeftParen,\nRightParen,\n};\n\nToken(Type t, const std::string& s, int prec = 1, bool ra = false)\n: type {t}, str(s), precedence {prec}, rightAssociative {ra}\n{}\n\nType type {Type::Unknown};\nstd::string str;\nint precedence = -1;\nbool rightAssociative = false;\n};\n\nstd::deque<Token> exprToTokens(const std::string& expr)\n{\nconst auto isnum = [](char ch) {return isdigit(ch) \|\| (ch == '.');};\n\nstd::deque<Token> tokens;\nToken::Type lstok = Token::Type::Operator;\nint nopar = 0;\n\nfor (const auto p = expr.c_str(); p;)\nif (((lstok == Token::Type::Operator) \|\| (lstok == Token::Type::LeftParen)) && (isnum(p) \|\| (((p == '-') \|\| (p == '+')) && isnum((p + 1))))) {\nconst char* const b = ((p == '-') \|\| (p == '+')) ? p++ : p;\n\nfor (; isnum(p); ++p);\nlstok = Token::Type::Number;\nconst auto s = std::string(b, p);\n\nif (std::count(s.begin(), s.end(), '.') < 2)\ntokens.push_back(Token {lstok, s});\nelse\nreturn {};\n} else {\nint pr = -1;\t\t// precedence\nbool ra = false;\t// rightAssociative\n\nif (((lstok == Token::Type::LeftParen) \|\| (lstok == Token::Type::Operator)) && (p == '(')) {\nlstok = Token::Type::LeftParen;\n++nopar;\n} else\nif (((lstok == Token::Type::RightParen) \|\| (lstok == Token::Type::Number)) && (p == ')')) {\nlstok = Token::Type::RightParen;\n--nopar;\n} else\nif ((lstok == Token::Type::Number) \|\| (lstok == Token::Type::RightParen)) {\nlstok = Token::Type::Operator;\nswitch (p) {\ncase '^': pr = 4; ra = true; break;\ncase '': pr = 3; break;\ncase '/': pr = 3; break;\ncase '+': pr = 2; break;\ncase '-': pr = 2; break;\ndefault: return {};\n}\n} else\nreturn {};\n\ntokens.push_back(Token {lstok, std::string(1, p), pr, ra});\n++p;\n}\n\nreturn ((nopar == 0) && ((lstok == Token::Type::RightParen) \|\| (lstok == Token::Type::Number))) ? tokens : std::deque<Token> {};\n}\n\nstd::deque<Token> shuntingYard(const std::deque<Token>& tokens)\n{\nstd::deque<Token> queue;\nstd::vector<Token> stack;\n\n// While there are tokens to be read:\nfor (auto token : tokens) {\nswitch (token.type) {\ncase Token::Type::Number:\n// If the token is a number, then add it to the output queue\nqueue.push_back(token);\nbreak;\n\ncase Token::Type::Operator:\n{\n// If the token is operator, o1, then:\nconst auto o1 = token;\n\n// while there is an operator token,\nwhile (!stack.empty()) {\n// o2, at the top of stack, and\nconst auto o2 = stack.back();\n\n// either o1 is left-associative and its precedence is\n// less than or equal to that of o2,\n// or o1 if right associative, and has precedence\n// less than that of o2,\nif (!((!o1.rightAssociative && o1.precedence <= o2.precedence) \|\| (o1.rightAssociative && o1.precedence < o2.precedence)))\nbreak;\n\n// then pop o2 off the stack,\nstack.pop_back();\n// onto the output queue;\nqueue.push_back(o2);\n}\n\n// push o1 onto the stack.\nstack.push_back(o1);\n}\nbreak;\n\ncase Token::Type::LeftParen:\n// If token is left parenthesis, then push it onto the stack\nstack.push_back(token);\nbreak;\n\ncase Token::Type::RightParen:\n// If token is right parenthesis:\n{\nbool match = false;\n// Until the token at the top of the stack\n// is a left parenthesis,\n// pop operators off the stack\n// onto the output queue.\nwhile (!stack.empty() && stack.back().type != Token::Type::LeftParen) {\nqueue.push_back(stack.back());\nstack.pop_back();\nmatch = true;\n}\n\n// Pop the left parenthesis from the stack,\n// but not onto the output queue.\nstack.pop_back();\n\nif (!match && stack.empty()) {\n// If the stack runs out without finding a left parenthesis,\n// then there are mismatched parentheses.\n//std::cout << \"RightParen error (\" << token.str << \")\\n\";\nreturn {};\n}\n}\nbreak;\n\ndefault:\n//std::cout << \"Error (\" << token.str << \")\\n\";\nreturn {};\n}\n}\n\n// When there are no more tokens to read:\n// While there are still operator tokens in the stack:\nwhile (!stack.empty()) {\n// If the operator token on the top of the stack is a parenthesis,\n// then there are mismatched parentheses.\n\nif (stack.back().type == Token::Type::LeftParen) {\n//std::cout << \"Mismatched parentheses error\\n\";\nreturn {};\n}\n\n// Pop the operator onto the output queue.\nqueue.push_back(std::move(stack.back()));\nstack.pop_back();\n}\n\nreturn queue;\n}\n\nstd::pair<bool, Mtype> eval(const std::string& expr)\n{\nconst auto tokens = exprToTokens(expr);\nauto queue = shuntingYard(tokens);\nstd::vector<Mtype> stack;\n\nwhile (!queue.empty()) {\nconst auto token = queue.front();\nqueue.pop_front();\nswitch (token.type) {\ncase Token::Type::Number:\nstack.push_back(std::stold(token.str));\nbreak;\n\ncase Token::Type::Operator:\n{\nconst auto rhs = stack.back();\nstack.pop_back();\nconst auto lhs = stack.back();\nstack.pop_back();\n\nswitch (token.str) {\ndefault:\n//std::cout << \"Operator error [\" << token.str << \"]\\n\";\nreturn {false, 0};\ncase '^':\nstack.push_back(static_cast<Mtype>(pow(lhs, rhs)));\nbreak;\ncase '':\nstack.push_back(lhs rhs);\nbreak;\ncase '/':\nstack.push_back(lhs / rhs);\nbreak;\ncase '+':\nstack.push_back(lhs + rhs);\nbreak;\ncase '-':\nstack.push_back(lhs - rhs);\nbreak;\n}\n}\nbreak;\n\ndefault:\n//std::cout << \"Token error \" << (int)token.type << std::endl;\nreturn {false, 0};\n}\n}\n\nreturn (stack.size() != 1) ? std::make_pair(false, 0.0L) : std::make_pair(true, stack.back());\n}\n\nint main() {\n//std::string expr = \"20-30/3+42^3\";\t\t// Should give 42\n\nstd::cout << \"Enter expression :\";\nstd::string expr;\n\nstd::getline(std::cin, expr);\nexpr.erase(std::remove_if(expr.begin(), expr.end(), std::isspace), expr.end());\n\nif (auto [ok, res] = eval(expr); ok)\nstd::cout << \"Result : \" << res << std::endl;\nelse\nstd::cout << \"Error in expression\" << std::endl;\n\nreturn 0;\n}```\nExamples\n\nCode:\n```Enter expression :-1+-2-3\nResult : 5\n\nEnter expression :20-30/3+42^3\nResult : 42\n\nEnter expression :3/22\nResult : 3\n\nEnter expression :(-7+(7^2-4*12)^.5)/2\nResult : -3```\nLast edited by 2kaud; April 7th, 2018 at 05:04 AM. Reason: Fixed issue with illegal terminating token", null, "", null, "Reply With Quote\n\n15.", null, "Member +", null, "", null, "", null, "Join Date\nFeb 2017\nPosts\n653\n\n## Re: can i convert a string into a math expression?", null, "Originally Posted by 2kaud", null, "The original link for a shunting-yard c++ implementation was https://gist.github.com/t-mat/b9f681b7591cdae712f6 has issues including those discussed in post #12.\nWell, I made a mistake to uncritically recommended the GitHub implementation but with your fixes it should be okay now so I recommend your version in #14.", null, "My advice to the OP really was that if this isn't a exercise it's much better to use an existing implementation. Expression parsing in practice quickly becomes very complicated as soon as you have to deal with more than just the very basic operators,\n\nhttp://en.cppreference.com/w/c/langu...tor_precedence\nLast edited by wolle; April 3rd, 2018 at 02:01 AM.", null, "", null, "Reply With Quote\n\n####", null, "Posting Permissions\n\n• You may not post new threads\n• You may not post replies\n• You may not post attachments\n• You may not edit your posts\n•" ]	[ null, "https://forums.codeguru.com/images/statusicon/user-offline.png", null, "https://forums.codeguru.com/images/reputation/reputation_pos.png", null, "https://forums.codeguru.com/images/misc/progress.gif", null, "https://forums.codeguru.com/clear.gif", null, "https://forums.codeguru.com/images/statusicon/user-offline.png", null, "https://forums.codeguru.com/images/reputation/reputation_pos.png", null, "https://forums.codeguru.com/images/reputation/reputation_pos.png", null, 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http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blob_plain;f=topics/week7_environments_and_closures.mdwn;hb=7f98aa5b7f740662593c87e0fe138c95a62ede85	[ "One strategy for evaluating or interpreting terms avoids the tedious and inefficient _substituting_ we've become familiar with. Instead, when a variable becomes bound to a result value, we'll keep track of that binding in a separate scorecard, that we'll call an \"environment.\" This strategy has to introduce a new kind of structure, called a `Closure`, to handle binding to terms that have locally free variables in them. Let's see how this works. For exposition, I'll pretend that the code you're working with has primitive numbers in it, though [[the homework assignment\|/exercises/assignment7]] doesn't. (But [[the fuller code\|/topics/week7_untyped_evaluator]] it's simplified from does.) Now consider: term environment ---- ----------- (\\w.(\\y.y) w) 2 [] (\\y.y) w [w->2] y [y->w, w->2] In the first step, we bind `w` to the term `2`, by saving this association in our environment. In the second step, we bind `y` to the term `w`. In the third step, we would like to replace `y` with whatever its current value is according to our scorecard/environment. But a naive handling of this would replace `y` with `w`; and that's not the right result, because `w` should itself be mapped to `2`. On the other hand, in: term environment ---- ----------- (\\x w. x) w 2 [] (\\w. x) 2 [x->w] x [w->2, x->w] Now our final term _should_ be replaced with `w`, not with `2`. So evidently it's not so simple as just recursively re-looking up variables in the environment. A good strategy for handling these cases would be not to bind `y` to the term `w`, but rather to bind it to _what the term `w` then fully evaluates to_. Then in the first case we'd get:\n```term environment\n---- -----------\n(\\w.(\\y.y) w) 2 []\n(\\y.y) w [w->2]\ny [y->2, w->2]\n```\nAnd at the next step, `y` will evaluate directly to `2`, as desired. In the other example, though, `x` gets bound to `w` when `w` is already free. (In fact the code skeleton we gave you won't permit that to happen; it refuses to perform any applications except when the arguments are \"result values\", and it doesn't count free variables as such. As a result, other variables can never get bound to free variables.) So far, so good. But now consider the term: (\\f. x) ((\\x y. y x) 0) Since the outermost head `(\\f. x)` is already a `Lambda`, we begin by evaluating the argument `((\\x y. y x) 0)`. This results in: term environment ---- ----------- (\\f. x) (\\y. y x) [x->0] But that's not right, since it will result in the variable `x` in the head also being associated with the argument `0`. Instead, we want the binding of `x` to `0` to be local to the argument term `(\\y. y x)`. For the moment, let's notate that like this:\n```term environment\n---- -----------\n(\\f. x)1 (\\y. y x)2 []1 [x->0]2\n```\nNow, let's suppose the head is more complex, like so: (\\f. (\\x. f x) I) ((\\x y. y x) 0) That might be easier to understand if you transform it to: let f = (let x = 0 in \\y. y x) in let x = I in f x Since the outermost head `(\\f. (\\x. f x) I)` is already a `Lambda`, we begin by evaluating the argument `((\\x y. y x) 0)`. This results in:\n```term environment\n---- -----------\n(\\f. (\\x. f x) I)1 (\\y. y x)2 []1 [x->0]2\n```\nNow the argument is not itself any longer an `App`, and so we are ready to evaluate the outermost application, binding `f` to the argument term. So we get:\n```term environment\n---- -----------\n((\\x. f x) I)1 [f->(\\y. y x)2]1 [x->0]2\n```\nNext we bind `x` to the argument `I`, getting:\n```term environment\n---- -----------\n(f x)1 [x->I, f->(\\y. y x)2]1 [x->0]2\n```\nNow we have to apply the value that `f` is bound to to the value that `x` is bound to. But notice there is a free variable `x` in the term that `f` is bound to. How should we interpret that term? Should the evaluation proceed:\n```(\\y. y x1) x1 [x->I, ...]1 [x->0]2\ny x1 [y->I, x->I, ...]1 [x->0]2\nI I\nI\n```\nusing the value that `x` is bound to in context1, _where the `f` value is applied_? Or should it proceed:\n```(\\y. y x2) x1 [x->I, ...]1 [x->0]2\ny x2 [y->I, x->I, ...]1 [x->0]2\nI 0\n0\n```\nusing the value that `x` was bound to in context2, _where `f` was bound_? In fact, when we specified rules for the Lambda Calculus, we committed ourselves to taking the second of these strategies. But both of the kinds of binding described here are perfectly coherent. The first is called \"dynamic binding\" or \"dynamic scoping\" and the second is called \"lexical or static binding/scoping\". Neither is intrinsically more correct or appropriate than the other. The first is somewhat easier for the people who write implementations of programming languages to handle; so historically it used to predominate. But the second is easier for programmers to reason about. > In Scheme, variables are bound in the lexical/static way by default, just as in the Lambda Calculus; but there is special vocabulary for dealing with dynamic binding too, which is useful in some situations. As far as I'm aware, Haskell and OCaml only provide the lexical/static binding. Shell scripts, on the other hand, only use dynamic binding. If you type this at a shell prompt: `( x=0; foo() { echo \\$x; }; bar() { local x=1; foo; }; bar )`, it will print `1` not `0`. In any case, if we're going to have the same semantics as the untyped Lambda Calculus, we're going to have to make sure that when we bind the variable `f` to the value `\\y. y x`, that (locally) free variable `x` remains associated with the value `0` that `x` was bound to in the context where `f` is bound, not the possibly different value that `x` may be bound to later, when `f` is applied. One thing we might consider doing is _evaluating the body_ of the abstract that we want to bind `f` to, using the then-current environment to evaluate the variables that the abstract doesn't itself bind. But that's not a good idea. What if the body of that abstract never terminates? The whole program might be OK, because it might never go on to apply `f`. But we'll be stuck trying to evaluate `f`'s body anyway, and will never get to the rest of the program. Another thing we could consider doing is to substitute the `2` in for the variable `x`, and then bind `f` to `\\y. y 2`. That would work, but the whole point of this evaluation strategy is to avoid doing those complicated (and inefficient) substitutions. Can you think of a third idea? What we will do is have our environment associate `f` not just with a `Lambda` term, but also with the environment that was in place _when_ `f` was bound. Then later, if we ever do use `f`, we have that saved environment around to look up any free variables in. More specifically, we'll associate `f` not with the _term_: Lambda(\"y\", BODY) but instead with the `Closure` structure: Closure(\"y\", BODY, SAVED_ENV) where `BODY` is the term that constituted the body of the corresponding `Lambda`, and `SAVED_ENV` is the environment in place when `f` is being bound. (In this simple implementation, we will just save the _whole environment_ then in place. But in a more efficient implementation, we'd sift through it and only keep those bindings for variables that are free in `BODY`. That would take up less space.)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90596336,"math_prob":0.94650275,"size":7045,"snap":"2019-43-2019-47","text_gpt3_token_len":1904,"char_repetition_ratio":0.13662831,"word_repetition_ratio":0.060606062,"special_character_ratio":0.30092263,"punctuation_ratio":0.13253012,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98030365,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-18T08:12:17Z\",\"WARC-Record-ID\":\"<urn:uuid:8043e272-05fc-4c7f-b5c6-60e167e30966>\",\"Content-Length\":\"8729\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6a0667e4-e723-46ac-b8e9-683ae352901d>\",\"WARC-Concurrent-To\":\"<urn:uuid:4d2361c0-6a3d-4907-8bef-e0e467c11c63>\",\"WARC-IP-Address\":\"45.79.164.50\",\"WARC-Target-URI\":\"http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blob_plain;f=topics/week7_environments_and_closures.mdwn;hb=7f98aa5b7f740662593c87e0fe138c95a62ede85\",\"WARC-Payload-Digest\":\"sha1:DQISDTLRHU5756VBUK33DDNYKVWF73D2\",\"WARC-Block-Digest\":\"sha1:GH74WP3NT7HZEJLD5MJ4AFUWBZP7OMTG\",\"WARC-Identified-Payload-Type\":\"text/plain\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669730.38_warc_CC-MAIN-20191118080848-20191118104848-00539.warc.gz\"}"}
https://coursepivot.com/tutor-answers/why-is-the-adjusted-present-value-approach-appropriate-for-situations-with-a-changing-capital-structure/	[ "# Why is the adjusted present value approach appropriate for situations with a changing capital structure?\n\nCategory:\n ▲ 0 ▼ ♥ 0 Why is the adjusted present value approach appropriate for situations with a changing capital structure?\nAnswer and Explanation", null, "Explanation The adjusted present value approach in valuation is appropriate in situations where the capital structure is not constant. In this method, the value of operations is calculated using t...\n\nExplanation\n\nThe adjusted present value approach in valuation is appropriate in situations where the capital structure is not constant. In this method, the value of operations is calculated using the unlevered cost of equity, which eliminates the effect of debt in the capital structure. Using the unlevered cost of equity helps in eliminating the effect of debt on the capital structure. By eliminating the impact of debt on the target company's shareholders, the acquiring company can easily capture the target company's risk.Since the effect of both equity and debt on the valuation is calculated separately, this method is appropriate for varied capital structure. First, the value of operations is calculated assuming that it is an all-equity firm. The effect of financial risk(tax shield) is then added to calculate the final value of operations.\n\nThe adjusted present value approach is calculated using the unlevered cost of equity, which eliminates the effect of debt in the capital structure. The total value of the tax shield is calculated separately. The total value of operations is the sum of unlevered value of operations and tax shield. Therefore, the adjusted present value method is suitable for variable capital structure.", null, "" ]	[ null, "https://coursepivot.com/wp-content/uploads/Review-q.png", null, "https://coursepivot.com/wp-content/uploads/course_pivot-t1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9135208,"math_prob":0.9830157,"size":1471,"snap":"2023-40-2023-50","text_gpt3_token_len":269,"char_repetition_ratio":0.15541922,"word_repetition_ratio":0.2488889,"special_character_ratio":0.17471108,"punctuation_ratio":0.08,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9928629,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-02T08:13:34Z\",\"WARC-Record-ID\":\"<urn:uuid:e093352c-9018-46e9-a9b3-6b183db1960e>\",\"Content-Length\":\"63232\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:322fb427-8139-4573-88b6-06fd8cd2c00b>\",\"WARC-Concurrent-To\":\"<urn:uuid:8fad0532-e46b-4fa0-9119-5f4b53a25657>\",\"WARC-IP-Address\":\"104.21.67.118\",\"WARC-Target-URI\":\"https://coursepivot.com/tutor-answers/why-is-the-adjusted-present-value-approach-appropriate-for-situations-with-a-changing-capital-structure/\",\"WARC-Payload-Digest\":\"sha1:GMNA7CHC2YWDP5LP4LD5TDEKSHHGJHYR\",\"WARC-Block-Digest\":\"sha1:UXISJUCP3RDYXYSBGCVACMCIHHFVZQYD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510983.45_warc_CC-MAIN-20231002064957-20231002094957-00070.warc.gz\"}"}
https://thedebateinitiative.com/2012/04/20/speed-of-light-accurate-value-inspired-by-a-single-verse-from-the-quran/	[ "Islam\n\n# Speed of Light; Accurate value inspired by a single verse from the Quran!\n\nA little bit far from the traditional “Tafseer”, when a cosmologist finishes reading the verse Quran 32:5 “He [Allah] arranges [each] matter from the heaven to the earth; then it will ascend to Him in a Day, the extent of which is a thousand years of those which you count.” He wouldn’t miss the fabric of a physical model consisting of: Objects (Allah’s matters), Motion (from heaven to the earth; then it will ascend to him), Time/Duration (in a Day, the extent of which is a thousand years of those which you count), Distance and ultimately a derivable Velocity imbedded within the Holy verse!\n\nMoon and it’s orbiting motion around the earth being the instrument Allah had created to help mankind count the years and time, Quran 10:5 “It is He [Allah] who made the sun a shining light and the moon a derived light and determined for it phases – that you may know the number of years and account [of time]. Allah has not created this except in truth. He details the signs for a people who know” …One might suggest the following physical motion model/problem: What’s the velocity of an object that travels in ONE day a same distance the Moon travels in 1000 years?\n\nThis would immediately recall the simple formula (we learned in year 5 science) … Velocity = Distance / Time … In other words:\n\nV (km/s) = 1000 (years said by the verse) * 12 (laps/year) * Lunar Orbit Circumference (km/lap) / 1 day (said by the verse, in seconds, sidereal day)\n\nCalculating the Lunar Orbit Circumference:\n\nDistance = Velocity * Time\n\nThus; Lunar Orbit Circumference (km/lap) = Lunar Velocity (Relative to Earth) (km/hr) * Sidereal Month (day/lap) * 24 (hr/day)\n\nThe Moon is running in two courses at the same time. The first is around the Earth which is noticed by us. The second is around the Sun (following the Earth) which is not seen by us. Verse 32:5 is referring to 1000 Years relative to us (of those which you count) which are basically 12,000 Moon cycles around the Earth.\n\nIf you’re on board of an air plane you’ll walk 5 meters to the Toilet but to an observer on the ground you’ve already walked 20 km’s! What matters us is what relative to us, the 5 meters, that’s what we count while on board!\n\nTherefore; The Lunar Velocity (Relative to Sun) must be multiplied by the cosine of the Tangent Angle in order to eliminate the Earth orbiting the Sun effect and reduce the Lunar Velocity to what relative to us on Earth, just as inspired by verse 32:5\n\nTherefore; Lunar Velocity (Relative to Earth) = Lunar Velocity (Relative to Sun) * cos (Tangent Angle)\n\nWhere the Tangent Angle = 360 * (Sidereal Month/Sidereal Year)\n\nKnowing that:\n\nSidereal Month = 27.321661 (day/lap) given by NASA\n\nSidereal Year = 365.25636 (days/lap) given by NASA\n\nThen; cos(Tangent Angle) = 360 * (Sidereal Month/Sidereal Year) = cos(360 * (27.321661 / 365.25636)) = 0.89157\n\nRe-constructing all variables in one equation:\n\nLunar Orbit Circumference (km/lap) = Lunar Velocity (Relative to Sun) (km/hr) * cos (Tangent Angle) * Sidereal Month (day/lap) * 24 (hr/day)\n\nKnowing that:\n\nLunar Velocity (relative to Sun) = 3682.07 (km/hr) given by NASA\n\nAnd by substituting all terms in the equation:\n\nLunar Orbit Circumference (km/lap) = 3682.07 (km/hr) * 0.89157 * 27.321661 (day/lap) * 24 (hr/day)\n\nResulting; Lunar Orbit Circumference = 2,152,612.349 (km/lap)\n\nComing back to the main equation inspired by the verse 32:5;\n\nV (km/s) = 1000 (years said by the verse) * 12 (laps/year) * Lunar Orbit Circumference (km/lap) / 1 day (said by the verse, in seconds, sidereal day)\n\nKnowing that:\n\n1 Sidereal Day = 86164.0906 (seconds/lap) given by NASA\n\nAnd The Lunar Orbit Circumference = 2,152,612.349 (km/lap) as derived above\n\nAnd by substituting all terms in the equation:\n\nV (km/s) = 1000 * 12 * 2,152,612.349 (km/lap) / 86164.0906 (sec/lap) = 299,792.5 km/s = Speed of Light!!!!!!\n\nCategories: Islam\n\n### 3 replies »\n\n1.", null, "Farrukh says:\n\nit is amazing work Mashallah, it made me smile, i’ll verify it Inshallah.\n\n2.", null, "Farrukh says:\n\nhard to understand how you calculated circumference of moon, why not to do it with 2pir formula? circumference given on websites is different from what you have calculated.\n\n3.", null, "Farrukh says:\n\nAfter reading similar things on other websites, I think this calculation and method that is being used here is totally wrong. if our 1000 year is equal to one day, then what about our 50,000 years equal to one day mentioned in another verse??\nfrom 1000 year verse, it seems like angels are much slower than speed of light. and from 50,000 years verse, it seems like angels are slightly slower than speed of light.\nand if they approach speed of light, then time will stop for them and it will be 9999999…. years for us." ]	[ null, "https://1.gravatar.com/avatar/aa9bf7e63e100794061f32503936be1e", null, "https://1.gravatar.com/avatar/aa9bf7e63e100794061f32503936be1e", null, "https://1.gravatar.com/avatar/aa9bf7e63e100794061f32503936be1e", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9146733,"math_prob":0.9301763,"size":4604,"snap":"2021-04-2021-17","text_gpt3_token_len":1228,"char_repetition_ratio":0.12021739,"word_repetition_ratio":0.15974844,"special_character_ratio":0.28670722,"punctuation_ratio":0.105945945,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9786625,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-25T22:25:52Z\",\"WARC-Record-ID\":\"<urn:uuid:a9f1a833-09ce-44dd-8857-6fd8768e673e>\",\"Content-Length\":\"100626\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:65675b2a-d481-42be-9b3c-1cfb62dc3e36>\",\"WARC-Concurrent-To\":\"<urn:uuid:83dee3c7-b6de-4c2d-9095-98f581f8e27a>\",\"WARC-IP-Address\":\"192.0.78.25\",\"WARC-Target-URI\":\"https://thedebateinitiative.com/2012/04/20/speed-of-light-accurate-value-inspired-by-a-single-verse-from-the-quran/\",\"WARC-Payload-Digest\":\"sha1:A52A3W3CVYEOC3GP5O2VEO2RSZ4GTF6T\",\"WARC-Block-Digest\":\"sha1:Z4J3T7TXUYZJUPCM5S5FRHZW3T6MVDVX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610704792131.69_warc_CC-MAIN-20210125220722-20210126010722-00187.warc.gz\"}"}
https://www.jviotti.com/notes/sequences	[ "# Sequences\n\nA sequence is a mathematical collection of objects in a particular order\n\nFor example: , in that order. A sequence can be seen as the total bijective function . Following this definition, we can access the element of a sequence using function application notation. Given , .\n\nThe empty sequence is defined as . Sequences are homogeneous. They must contain elements of the same type.\n\nIf is a set, then represents all the finite sequences of elements of , and its defined as: .\n\n## Basic Operations\n\n### Concatenation\n\nGiven sequences and , their concatenation is denoted as . Notice that . Of course, , and concatenation is an associative operation. It also follows that if , then and .\n\n### Filtering (or Restriction)\n\nGiven a sequence , the sequence represents all the elements of that are included in , preserving the ordering. For example, . Notice that given set , .\n\nFiltering the empty sequence always yields back the empty sequence: , and filtering any sequence by the empty set also yields the empty sequence: .\n\nApplying multiple restrictions over the same sequence is the same as restricting by the intersection of those sets. Given sequence and sets and , then .\n\nThe first element of a sequence is called its head. For example, given , , or alternatively . The head of the empty sequence is the empty sequence. Notice that , and that given a non empty sequence , then .\n\n### Tail\n\nThe tail of a sequence is a sequence containing all the original elements except for the first one. For example, given , then , or alternatively . Notice that for any sequence , , and .\n\n### Indexing\n\nSequences are indexed from 0 using square bracket notation. For example .\n\n### Prefix\n\nA sequence is a prefix of sequence if the elements if there is a sequence such that . For example: but . Notice that given any set , . Also, given a sequence , .\n\nIf follows that given any sequence , , and that . This operation is asymmetric: , and transitive: . Of course, if , then .\n\nAn easy way to check if a prefix expression holds is by checking that .\n\nNotice that if two sequences are prefixes of the same sequence, then one is the prefix of the other or viceversa: .\n\nThe operator may have an exponent, in which case is an n-prefix of if starts with and it has up to elements removed. For example: , , but . This operator is defined as . It follows that and that . Also notice that .\n\n### Cardinality\n\nSince a sequence is a relation from natural numbers to the sequence elements, we can re-use the cardinality notation from sets. Thus, , or .\n\n### Flattening\n\nA sequence containing other sequences might be flattened to a single sequence by using a ditributed version of the concatenation operator. Given , then .\n\n### Repetition\n\nA sequence to the power of is equal to the sequence concatenated to itself times. For example: .\n\n### Reverse\n\nGiven a sequence , its inverse is the same sequence from right to left: . Of course, , and . Also, . The reverse of a concatenation is equal to the reverse concatenation of the inverse: .\n\n### Star\n\nThe star sequence of a set is the infinite set of finite sequences made of elements from . More formally: . It follows that for any set , .\n\nMembership on a set can be expressed in terms of membership to the star sequence of the set: . Also, means that both and are members of the star of .\n\n### Count\n\nGiven a sequence , represents the amount of time is inside the sequence . For example: . Formally, is defined as .\n\n### Includes\n\nThe operation represents whether a sequence is contained within a sequence . For example: . Notice that not .\n\n### Interleaves\n\nA sequence is an interleave of sequences and if can be constructed by arbitrarily extracting elements from and in order. For example , and . Of course, it holds that .\n\nThis operation is defined like this:\n\n## Properties\n\n### Injection\n\nAn injective sequence is one where its elements don’t appear more than once. Remember that sequences are defined as functions whose range is the set of natural numbers, and therefore the functional definition of injection (one-to-one) holds: a sequence without duplicates is one where every natural number from the domain points to a different element." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9082973,"math_prob":0.9698912,"size":4125,"snap":"2023-40-2023-50","text_gpt3_token_len":865,"char_repetition_ratio":0.19631158,"word_repetition_ratio":0.010738255,"special_character_ratio":0.21745455,"punctuation_ratio":0.16997519,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9797606,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-22T19:34:36Z\",\"WARC-Record-ID\":\"<urn:uuid:7c0289ea-fc52-4ed8-9890-197e9508be6b>\",\"Content-Length\":\"435747\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:300b21d2-0762-4991-b8fb-810dc4d22c5a>\",\"WARC-Concurrent-To\":\"<urn:uuid:0bebe878-69a6-4440-9fbb-ad22a3deae68>\",\"WARC-IP-Address\":\"185.199.111.153\",\"WARC-Target-URI\":\"https://www.jviotti.com/notes/sequences\",\"WARC-Payload-Digest\":\"sha1:5LIL7WVNNGGS236SRSVUXLRGTPTSNEJY\",\"WARC-Block-Digest\":\"sha1:G24OTRPECMONSLLLJP4ID2DPUL4ACAMH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506421.14_warc_CC-MAIN-20230922170343-20230922200343-00792.warc.gz\"}"}
https://xianblog.wordpress.com/2013/06/24/vanilla-is-russian/	[ "## vanilla is Russian!", null, "Mark Girolami kindly sent me his Russian Roulette paper on handling intractable normalising constants toward exact-approximation MCMC schemes: “Playing Russian Roulette with Intractable Likelihoods”… I had heard Mark’s talk in Warwick last month but reading the paper (written jointly with Anne-Marie Lyne, Heiko Strathmann, Dan Simpson and Yves Atchadé) made me understand the method(s) much better. The idea is to use a series representation (that they call MacLaurin and that I would call Taylor, but this may be related to MacLaurin being Scottish, a good enough reason!) to get from an approximation of the normalising constant to an unbiased estimation of the inverse of this constant, via either a geometric or an exponential version. The infinite series is then truncated by the Russian roulette, a physics trick that relies on a stopping rule to ensure both unbiasedness and an almost surely finite number of terms.\n\nNow, some wee bit of whining and an explanation for the (apparently insane) post title: in our vanilla Rao-Blackwellisation paper, Randal Douc and I used fairly similar techniques to find an unbiased estimate of an inverse probability to Rao-Blackwellise Metropolis-Hastings outcomes. Namely, a geometric series representation of this inverse probability identical to (7) in the current paper and a truncation of the infinite summation of products using uniforms and indicators as in Russian Roulette… So, although we were unaware of this at the time, we were essenially using Russian Roulette! I am thus a tad disappointed to see our work not referenced in this paper and a wee more disappointed that it got under the radar in an important segment of the community! (Besides this omission, the many refernces to my (other) papers are much appreciated!)\n\nBack to the current paper, a first item of interest is that the geometrically tilted estimator can rely on a biased estimator of the intractable constant, still resulting in an unbiased estimator in the end. (See below about this!) Although this solution requires the availability of upper bounds on the ratio between the estimator and the missing constant, which obviously is a drag, this is a good incentive for the method. (This and the fact that it is always possible to turn it into a positive estimate.) The boundedness also allows for geometric Russian Roulette as in our vanilla Rao-Blackwellisation paper. As an aside, I find the presentation in Section 3.1 a mite too convoluted: using the identity", null, "$\\dfrac{f(y,\\theta)}{\\mathcal{Z}(\\theta)}=\\dfrac{f(y,\\theta)}{\\tilde{\\mathcal{Z}}(\\theta)}\\times c(\\theta)\\times\\dfrac{\\tilde{\\mathcal{Z}}(\\theta)}{c(\\theta)\\mathcal{Z}(\\theta)}$\n\nseems to me to be sufficient to justify the estimation via the geometric series without a call to tilted exponential families. The alternative based on an exponential auxiliary variable seems slightly less appealing to me. First because it requires an unbiased estimator of the intractable constant. Second, I do not see how this auxiliary variable can be generated, given that the rate parameter in the exponential is the intractable constant. Maybe I am missing the points by thousands of miles…. In an email exchange, Mark pointed out that both methods use unbiased estimates of the constant. However, the paper mentions twice that its estimators can start from some approximation of Z(θ) that is not necessarily unbiased. Thus, I wonder whether or not having an infinite sequence of independent estimators with the same distribution as the initial approximation could be enough.\n\nThe solution proposed in the paper (to deal with negative density estimates) is quite clever, turning the negatives into positives thanks to the sign function. I however wonder about the overall efficiency of the trick for two (minor?) reasons. The first reason for worry is that, when the density estimate is very negative, the exact value is likely to be close to zero, rather than very large. Turning the estimate into its positive counterpart puts undue weight on this value. The second reason for worry is that the final estimate, being a ratio of estimates, will be biased. Obviously, this is not damning as this happens at the estimation stage rather than during the MCMC stage…\n\nA mathematical worry about the series convergence proof in the Appendix: it seems to me that the telescopic argument in the proof only applies to convergent series as, otherwise, we could be considering a difference of two infinities… Another worry: I find the explanation at the top of page 5 somehow confusing: unless the notation is generic, the fact that p can be evaluated in", null, "$\\hat{\\pi}(\\theta,\\xi\|y)=\\hat{p}(\\theta,\\xi\|y)/Z(y)$\n\ndoes not make sense if the lhs is the estimate found in the previous page.\n\n### One Response to “vanilla is Russian!”\n\n1.", null, "Intractable likelihoods, unbiased estimators and sign problem \| Statisfaction Says:\n\n[…] by Mark Girolami, Anne-Marie Lyne, Heiko Strathmann, Daniel Simpson, Yves Atchade, see also Xian’s comments on it. The motivation of the paper is slightly different than the one described above but the techniques […]\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]	[ null, "https://bearheartbakingco.files.wordpress.com/2010/02/cakeweb2wheel1.jpg", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://secure.gravatar.com/blavatar/c97e6139e1af9c4d7e7fb0cad771e3eb", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92714816,"math_prob":0.897895,"size":4777,"snap":"2023-14-2023-23","text_gpt3_token_len":984,"char_repetition_ratio":0.12319296,"word_repetition_ratio":0.0,"special_character_ratio":0.1909148,"punctuation_ratio":0.087759815,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9722505,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,6,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-05T16:26:40Z\",\"WARC-Record-ID\":\"<urn:uuid:35e88a64-e2d4-4237-bb19-2f11c14aef7b>\",\"Content-Length\":\"67275\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ac1bd6d9-43bf-4bdd-8430-b83b0d222d79>\",\"WARC-Concurrent-To\":\"<urn:uuid:f261cbc7-8a56-427a-9b77-4b1d7fdad76a>\",\"WARC-IP-Address\":\"192.0.78.13\",\"WARC-Target-URI\":\"https://xianblog.wordpress.com/2013/06/24/vanilla-is-russian/\",\"WARC-Payload-Digest\":\"sha1:7VUSK4D6U2KKUIAV7LYVLCUGPNWEN6WM\",\"WARC-Block-Digest\":\"sha1:HIUPNM327JU3DU3CNDJ6YLWDZFVCNIV3\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652149.61_warc_CC-MAIN-20230605153700-20230605183700-00219.warc.gz\"}"}
https://clientebancario.bde.es/pcb/en/menu-horizontal/podemosayudarte/simuladores/calculo-de-la-tae-de-un-prestamo-hipotecario.html	[ "# Calculate the APR of a mortgage\n\nCalculate the APR of your mortgage\n\nThis calculator will let you calculate the APR of a mortgage based on the amount of the loan, its nominal interest rate, repayment term, fees, and, if it were compulsory, the payment protection insurance premium.\n\nYou can make various calculations and compare the APR for each of them.", null, "Abre en ventana nueva\n\n• ### Calculation assumptions\n\nThis calculator can be used for loans incurring fees and expenses which have to be included in the calculation of the annual percentage rate (APR) and which are paid when the loan is arranged and on a regular basis.\n\nThis calculator is based on the assumption that the loan is repaid in equal monthly instalments and that the proportion of the monthly payment corresponding to interest is calculated using the following formula:\n\n(Outstanding capital x nominal annual rate) / 1200.\n\nCalculations have been simplified as follows, so that the calculator can be used with in as many cases as possible:\n\n• Customers pay all the loan assessment, evaluation and arrangement costs on the date the loan agreement is signed\n• Instalments are paid on a monthly basis\n• Every month of the year has the same number of days\n• There are no payment holidays and the first payment is made exactly one month after the loan has been arranged\n\nThe instructions for calculate the annual percentage rate can be found in Anexo II of Ley 5/2019, de 15 de marzo, reguladora de los contratos de crédito inmobiliario.Abre en ventana nueva.\n\nDid you find this information useful?" ]	[ null, "https://clientebancario.bde.es/f/webcb/RCL/img/ir_simulador_en.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9436039,"math_prob":0.96415585,"size":1462,"snap":"2021-04-2021-17","text_gpt3_token_len":303,"char_repetition_ratio":0.14677641,"word_repetition_ratio":0.008064516,"special_character_ratio":0.19904241,"punctuation_ratio":0.067669176,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9866128,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-17T00:37:21Z\",\"WARC-Record-ID\":\"<urn:uuid:853b2120-aa47-405c-9840-b8c4fbcfb793>\",\"Content-Length\":\"33003\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6029a4ab-cbbe-4c16-b245-e825c8948f55>\",\"WARC-Concurrent-To\":\"<urn:uuid:2b6c36c8-1a5b-44cd-bbba-ab81c0763ea4>\",\"WARC-IP-Address\":\"77.73.203.99\",\"WARC-Target-URI\":\"https://clientebancario.bde.es/pcb/en/menu-horizontal/podemosayudarte/simuladores/calculo-de-la-tae-de-un-prestamo-hipotecario.html\",\"WARC-Payload-Digest\":\"sha1:JDLFPHFLNW45K2QUEGFUMII5X6VWMQ3B\",\"WARC-Block-Digest\":\"sha1:UGJIMMGVV6L5WYU37VJJDWHZD3WY63IB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703507971.27_warc_CC-MAIN-20210116225820-20210117015820-00419.warc.gz\"}"}
http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=commitdiff;h=94af2dbc84f38e23ae79e560300923f458615aeb	[ "author jim Fri, 20 Mar 2015 12:42:29 +0000 (08:42 -0400) committer Linux User Fri, 20 Mar 2015 12:42:29 +0000 (08:42 -0400)\n\nindex 23f4ff6..2403da7 100644 (file)\n@@ -116,12 +116,21 @@ Here are the types of our crucial functions, together with our pronunciation, an\n<code>join: <span class=\"box2\">P</span> -> <u>P</u></code>\n\n+In the class handout, we gave the types for `>=>` twice, and once was correct but the other was a typo. The above is the correct typing.\n+\n+Also, in the handout we called `mid` `𝟭`. But now we've decided that `mid`\n+is better. (Think of it as \"m\" plus \"identity\", not as the start of \"midway\".)\n+\nThe menagerie isn't quite as bewildering as you might suppose. Many of these will\nbe interdefinable. For example, here is how `mcomp` and `mbind` are related: <code>k <=< j ≡\n\\a. (j a >>= k)</code>.\n\n-In most cases of interest, instances of these systems of functions will provide\n-certain useful guarantees.\n+We will move freely back and forth between using `>=>` and using `<=<` (aka `mcomp`), which\n+is just `>=>` with its arguments flipped. `<=<` has the virtue that it corresponds more\n+closely to the ordinary mathematical symbol `○`. But `>=>` has the virtue\n+that its types flow more naturally from left to right.\n+\n+These functions come together in several systems, and have to be defined in a way that coheres with the other functions in the system:\n\n* *Mappable* (in Haskelese, \"Functors\") At the most general level, box types are Mappable\nif there is a `map` function defined for that box type with the type given above. This\n@@ -130,7 +139,9 @@ has to obey the following Map Laws:\n<code>map (id : α -> α) = (id : <u>α</u> -> <u>α</u>)</code>\n<code>map (g ○ f) = (map g) ○ (map f)</code>\n\n- Essentially these say that `map` is a homomorphism from `(α -> β, ○, id)` to <code>(<u>α</u> -> <u>β</u>, ○', id')</code>, where `○'` and `id'` are `○` and `id` restricted to arguments of type <code><u>_</u></code>. That might be hard to digest because it's so abstract. Think of the following concrete example: if you take a `α list` (that's our <code><u>α</u></code>), and apply `id` to each of its elements, that's the same as applying `id` to the list itself. That's the first law. And if you apply the composition of functions `g ○ f` to each of the list's elements, that's the same as first applying `f` to each of the elements, and then going through the elements of the resulting list and applying `g` to each of those elements. That's the second law. These laws obviously hold for our familiar notion of `map` in relation to lists.\n+ Essentially these say that `map` is a homomorphism from the algebra of `(universe α -> β, operation ○, elsment id)` to that of <code>(<u>α</u> -> <u>β</u>, ○', id')</code>, where `○'` and `id'` are `○` and `id` restricted to arguments of type <code><u>_</u></code>. That might be hard to digest because it's so abstract. Think of the following concrete example: if you take a `α list` (that's our <code><u>α</u></code>), and apply `id` to each of its elements, that's the same as applying `id` to the list itself. That's the first law. And if you apply the composition of functions `g ○ f` to each of the list's elements, that's the same as first applying `f` to each of the elements, and then going through the elements of the resulting list and applying `g` to each of those elements. That's the second law. These laws obviously hold for our familiar notion of `map` in relation to lists.\n+\n+ > <small>As mentioned at the top of the page, in Category Theory presentations of monads they usually talk about \"endofunctors\", which are mappings from a Category to itself. In the uses they make of this notion, the endofunctors combine the role of a box type <code><u>_</u></code> and of the `map` that goes together with it.\n\n* *MapNable* (in Haskelese, \"Applicatives\") A Mappable box type is MapNable" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9095665,"math_prob":0.61912507,"size":3554,"snap":"2019-35-2019-39","text_gpt3_token_len":1025,"char_repetition_ratio":0.13239437,"word_repetition_ratio":0.43535188,"special_character_ratio":0.29234666,"punctuation_ratio":0.09928057,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9765193,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-20T08:22:58Z\",\"WARC-Record-ID\":\"<urn:uuid:5b5c6df4-7f9d-459c-b516-5e578df842b0>\",\"Content-Length\":\"17244\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:635749b9-755b-4f43-9768-3d53ce212f1d>\",\"WARC-Concurrent-To\":\"<urn:uuid:599ed552-6b3b-4b06-82b5-7b2f0b7c6a02>\",\"WARC-IP-Address\":\"45.79.164.50\",\"WARC-Target-URI\":\"http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=commitdiff;h=94af2dbc84f38e23ae79e560300923f458615aeb\",\"WARC-Payload-Digest\":\"sha1:DAG5TSGKOTCQUNKN6GYCOSPU6FXXBDFK\",\"WARC-Block-Digest\":\"sha1:4D4OEIFDTB6M7VTNSIAICDHQBA5UN7XW\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573908.70_warc_CC-MAIN-20190920071824-20190920093824-00262.warc.gz\"}"}
https://mran.revolutionanalytics.com/snapshot/2022-05-15/web/packages/err/readme/README.html	[ "# err", null, "## Introduction\n\nTo err is human - Alexander Pope (1711)\n\n`err` is a light-weight R package to produce customizable number and object sensitive error and warning messages.\n\n## Demonstration\n\n### Object Sensitive\n\nThe `co` functions produce object sensitive strings.\n\n``````library(err)\n\nfox <- c(\"The\", \"quick\", \"brown\", \"fox\", \"jumps\", \"over\", \"the\", \"lazy\", \"dog\")\nco(fox)\n#> \"fox has 9 values: 'The', 'quick', 'brown', 'fox', 'jumps', 'over', 'the', 'lazy', 'dog'\"\nco(fox)\n#> \"fox has 1 value: 'The'\"\nco(fox)\n#> \"fox has 0 values\"\nco(fox, nlots = 5)\n#> \"fox has 9 values: 'The', 'quick', 'brown', ..., 'dog'\"``````\n\n### Customizable\n\nThe object sensitive strings are fully customized.\n\n``````one <- \"darn! the vector %o of length %n has the following value: %c\"\nnone <- \"phew! vector %o is empty\"\nsome <- \"rats! vector %o has the following %n element%s: %c\"\nlots <- \"really?! the %n elements of vector %o are too numerous to print\"\n\nco(fox, one = one, none = none, some = some, lots = lots, nlots = 5)\n#> \"phew! vector fox is empty\"\nco(fox, one = one, none = none, some = some, lots = lots, nlots = 5)\n#> \"darn! the vector fox of length 1 has the following value: 'The'\"\nco(fox[1:3], one = one, none = none, some = some, lots = lots, nlots = 5)\n#> \"rats! vector fox[1:3] has the following 3 elements: 'The', 'quick', 'brown'\"\nco(fox[1:5], one = one, none = none, some = some, lots = lots, nlots = 5)\n#> \"really?! the 5 elements of vector fox[1:5] are too numerous to print\"``````\n\nThe following `sprintf`-like types can be used in the custom messages:\n\n• `%c`: the object as a comma separated list (produced by a `cc` function)\n• `%n`: the length of the object\n• `%o`: the name of the object\n• `%s`: ‘s’ if n != 1 otherwise ’’\n• `%r`: ‘are’ if n != 1 otherwise ‘is’\n\nAnd there are various formatting options\n\n``````co(fox[1:6], conjunction = \"or\", bracket = \"\|\", oxford = TRUE, ellipsis = 5)\n#> \"fox[1:6] has 6 values: \|The\|, \|quick\|, \|brown\|, ..., or \|over\|\"``````\n\n### Data Frames\n\nThere is also a method for data frames.\n\n``````cat(co(datasets::mtcars, conjunction = \"and\", oxford = TRUE, ellipsis = 5))\n#> datasets::mtcars has 11 columns\n#> mpg: 21, 21, 22.8, ..., and 21.4\n#> cyl: 6, 6, 4, ..., and 4\n#> disp: 160, 160, 108, ..., and 121\n#> ...\n#> and carb: 4, 4, 1, ..., and 2``````\n\n### Number Sensitive\n\nThe `cn` function produces number sensitive customizable messages\n\n``````cn(0)\n#> \"there are 0 values\"\ncn(1)\n#> \"there is 1 value\"\ncn(2)\n#> \"there are 2 values\"\ncn(100, lots = \"there %r %n value%s - this is a lot\")\n#> \"there are 100 values - this is a lot\"``````\n\n### Warning and Error Messages\n\nThe `co` and `cn` functions can be combined with the wrappers `msg`, `wrn` and `err` to produce a message, warning and error (without the call as part of the warning/error message).\n\n``````msg(cn(2))\n#> there are 2 values\nwrn(cn(2))\n#> Warning: there are 2 values\nerr(cn(2))\n#> Error: there are 2 values``````\n\n## Installation\n\nTo install the latest release version from CRAN\n\n``install.packages(\"err\")``\n\nTo install the latest development version from GitHub\n\n``````if(!\"devtools\" %in% installed.packages()[,1])\ninstall.packages(\"devtools\")\ndevtools::install_github(\"poissonconsulting/err\")``````\n\nTo install the latest development version from the Poisson drat repository\n\n``````if(!\"drat\" %in% installed.packages()[,1])\ninstall.packages(\"drat\")" ]	[ null, "https://mran.revolutionanalytics.com/snapshot/2022-05-15/web/packages/err/readme/man/figures/logo.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.77317727,"math_prob":0.9303563,"size":3405,"snap":"2022-40-2023-06","text_gpt3_token_len":1107,"char_repetition_ratio":0.12731549,"word_repetition_ratio":0.13111888,"special_character_ratio":0.35712188,"punctuation_ratio":0.21815719,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9630069,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-30T11:49:52Z\",\"WARC-Record-ID\":\"<urn:uuid:36c87302-e09a-4d30-bd4c-af58fafc1536>\",\"Content-Length\":\"16917\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aa8406f7-18d1-464b-97e8-9b21b9c951b9>\",\"WARC-Concurrent-To\":\"<urn:uuid:3f876bd2-5fac-46ef-b6b6-e592b38eadcc>\",\"WARC-IP-Address\":\"40.118.246.51\",\"WARC-Target-URI\":\"https://mran.revolutionanalytics.com/snapshot/2022-05-15/web/packages/err/readme/README.html\",\"WARC-Payload-Digest\":\"sha1:YMUPNWJ7MPCJSKJJYK53OUSRIDJQC6SM\",\"WARC-Block-Digest\":\"sha1:OH67RTJKD6BDRNGLT5PZ74ZQCB63ON55\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499816.79_warc_CC-MAIN-20230130101912-20230130131912-00009.warc.gz\"}"}
https://answers.everydaycalculation.com/compare-fractions/2-4-and-3-9	[ "Solutions by everydaycalculation.com\n\n## Compare 2/4 and 3/9\n\n2/4 is greater than 3/9\n\n#### Steps for comparing fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 4 and 9 is 36\n\nNext, find the equivalent fraction of both fractional numbers with denominator 36\n2. For the 1st fraction, since 4 × 9 = 36,\n2/4 = 2 × 9/4 × 9 = 18/36\n3. Likewise, for the 2nd fraction, since 9 × 4 = 36,\n3/9 = 3 × 4/9 × 4 = 12/36\n4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction\n5. 18/36 > 12/36 or 2/4 > 3/9\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]	[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87953645,"math_prob":0.9955734,"size":397,"snap":"2021-31-2021-39","text_gpt3_token_len":176,"char_repetition_ratio":0.2977099,"word_repetition_ratio":0.0,"special_character_ratio":0.4534005,"punctuation_ratio":0.037383176,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9956261,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-01T18:03:57Z\",\"WARC-Record-ID\":\"<urn:uuid:70cd81b8-d0fb-428f-8988-e71c030e5ff6>\",\"Content-Length\":\"8214\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6eccaaf0-b9d0-4187-b6fa-9ea6dfa26220>\",\"WARC-Concurrent-To\":\"<urn:uuid:5a9228ec-8070-4285-8eb2-fa13ed9ffc05>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/compare-fractions/2-4-and-3-9\",\"WARC-Payload-Digest\":\"sha1:GUL6KOCSLISOQZCGFEG6XJSFYJUKY26A\",\"WARC-Block-Digest\":\"sha1:OMHP5NOUXCVS3UFGA64VSDTIZ572SYGT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154214.63_warc_CC-MAIN-20210801154943-20210801184943-00530.warc.gz\"}"}
https://atomscientific.com/category/chemicals/volumetric-solutions/page/2	[ "phone 0161 366 5123 mail [email protected]", null, "A highly accurate and precisely manufactured range of Volumetric Solutions, Buffer Solutions, Calibration Buffers and Conductivity Standards\n\n## P\n\nPotassium Chloride Solution 4.0M (4.0N)\nPotassium Hydroxide Solution 0.075M (0.075N)\nPotassium Hydroxide Solution 0.1M (0.1N)\nPotassium Hydroxide Solution 0.5M (0.5N)\nPotassium Hydroxide Solution 1.0M (1.0N)\nPotassium Hydroxide Solution 5.0M (5.0N)\nPotassium Permanganate Solution 0.02M (0.1N)\nPotassium Permanganate Solution 0.1M (0.5N)\nPotassium Permanganate Solution 0.2M (1.0N)\nPotassium Thiocyanate Solution 0.05M (0.05N)\nPotassium Thiocyanate Solution 0.1M (0.1N)\n\n## S\n\nSilver Nitrate Solution 0.02M (0.02N)\nSilver Nitrate Solution 0.05N (0.05M)\nSilver Nitrate Solution 0.085M (0.085N)\nSilver Nitrate Solution 0.1709M (0.1709N)\nSilver Nitrate Solution 0.1M (0.1N)\nSilver Nitrate Solution 0.5M (0.5N)\nSilver Nitrate Solution 1.0M (1.0N)\nSodium Azide 0.1M Solution\nSodium Carbonate Solution 0.5M\nSodium Chloride Solution 0.1M (0.1N)\nSodium Chloride Solution 5.0M\nSodium Hydroxide Solution 2.5M (2.5N)\nSodium Hydroxide Solution 0.01M (0.01N)\nSodium Hydroxide Solution 0.02M (0.02N)\nSodium Hydroxide Solution 0.111M (0.111N)\nSodium Hydroxide Solution 0.14M (0.14N)\nSodium Hydroxide Solution 0.156M (0.156N)\nSodium Hydroxide Solution 0.1M (0.1N)\nSodium Hydroxide Solution 0.25M (0.25N)\nSodium Hydroxide Solution 0.2M (0.2N)\nSodium Hydroxide Solution 0.313M (0.313N)\nSodium Hydroxide Solution 0.33M (0.33N)\nSodium Hydroxide Solution 0.5M (0.5N)\nSodium Hydroxide Solution 1.0M (1.0N)\nSodium Hydroxide Solution 10.0M (10N)\nSodium Hydroxide Solution 12.5M (12.5N)\nSodium Hydroxide Solution 2.0M (2.0N)\nSodium Hydroxide Solution 3.0M (3N)\nSodium Hydroxide Solution 4.0M (4.0N)\nSodium Hydroxide Solution 5.0M (5N)\nSodium Hydroxide Solution 8.0M (8N)\nSodium Thiosulphate Solution 0.1M (0.1N)\nSodium Thiosulphate Solution 1.0M (1.0N)\nSulphosalicylic Acid Solution 0.8M\nSulphuric Acid Solution 0.05M (0.1N)\nSulphuric Acid Solution 0.1M (0.2N)\nSulphuric Acid Solution 0.25M (0.5N)" ]	[ null, "https://atomscientific.com/images/categorybanners/volumetric-solutions-2022-category-banner-png.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5921933,"math_prob":0.9408465,"size":2062,"snap":"2023-14-2023-23","text_gpt3_token_len":860,"char_repetition_ratio":0.3765792,"word_repetition_ratio":0.0,"special_character_ratio":0.3457808,"punctuation_ratio":0.2078522,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9905249,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-09T07:12:17Z\",\"WARC-Record-ID\":\"<urn:uuid:464d9407-fc7e-4a72-a97b-f5678baef39b>\",\"Content-Length\":\"27276\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5afc7337-12a8-4751-98d4-cf5b8ae457bd>\",\"WARC-Concurrent-To\":\"<urn:uuid:2331b5cd-e4ba-4b98-be7f-b1edc60c482a>\",\"WARC-IP-Address\":\"178.159.12.141\",\"WARC-Target-URI\":\"https://atomscientific.com/category/chemicals/volumetric-solutions/page/2\",\"WARC-Payload-Digest\":\"sha1:PIUFR3WVRAN2IU4HLN4LOZDW73LSKLHW\",\"WARC-Block-Digest\":\"sha1:IWSOXWGEEVUO7HKY74RFSCOKJC5WGF6V\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224655446.86_warc_CC-MAIN-20230609064417-20230609094417-00405.warc.gz\"}"}
https://numbermatics.com/n/11163/	[ "# 11163\n\n## 11,163 is an odd composite number composed of two prime numbers multiplied together.\n\nWhat does the number 11163 look like?\n\nThis visualization shows the relationship between its 2 prime factors (large circles) and 6 divisors.\n\n11163 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of six divisors.\n\n## Prime factorization of 11163:\n\n### 3 × 612\n\n(3 × 61 × 61)\n\nSee below for interesting mathematical facts about the number 11163 from the Numbermatics database.\n\n### Names of 11163\n\n• Cardinal: 11163 can be written as Eleven thousand, one hundred sixty-three.\n\n### Scientific notation\n\n• Scientific notation: 1.1163 × 104\n\n### Factors of 11163\n\n• Number of distinct prime factors ω(n): 2\n• Total number of prime factors Ω(n): 3\n• Sum of prime factors: 64\n\n### Divisors of 11163\n\n• Number of divisors d(n): 6\n• Complete list of divisors:\n• Sum of all divisors σ(n): 15132\n• Sum of proper divisors (its aliquot sum) s(n): 3969\n• 11163 is a deficient number, because the sum of its proper divisors (3969) is less than itself. Its deficiency is 7194\n\n### Bases of 11163\n\n• Binary: 101011100110112\n• Base-36: 8M3\n\n### Squares and roots of 11163\n\n• 11163 squared (111632) is 124612569\n• 11163 cubed (111633) is 1391050107747\n• The square root of 11163 is 105.6550992619\n• The cube root of 11163 is 22.3491138687\n\n### Scales and comparisons\n\nHow big is 11163?\n• 11,163 seconds is equal to 3 hours, 6 minutes, 3 seconds.\n• To count from 1 to 11,163 would take you about three hours.\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 11163 cubic inches would be around 1.9 feet tall.\n\n### Recreational maths with 11163\n\n• 11163 backwards is 36111\n• The number of decimal digits it has is: 5\n• The sum of 11163's digits is 12\n• More coming soon!\n\n#### Copy this link to share with anyone:\n\nMLA style:\n\"Number 11163 - Facts about the integer\". Numbermatics.com. 2023. Web. 5 December 2023.\n\nAPA style:\nNumbermatics. (2023). Number 11163 - Facts about the integer. Retrieved 5 December 2023, from https://numbermatics.com/n/11163/\n\nChicago style:\nNumbermatics. 2023. \"Number 11163 - Facts about the integer\". https://numbermatics.com/n/11163/\n\nThe information we have on file for 11163 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!\n\nKeywords: Divisors of 11163, math, Factors of 11163, curriculum, school, college, exams, university, Prime factorization of 11163, STEM, science, technology, engineering, physics, economics, calculator, eleven thousand, one hundred sixty-three.\n\nOh no. Javascript is switched off in your browser.\nSome bits of this website may not work unless you switch it on." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8572975,"math_prob":0.9585965,"size":2662,"snap":"2023-40-2023-50","text_gpt3_token_len":707,"char_repetition_ratio":0.12151994,"word_repetition_ratio":0.025821596,"special_character_ratio":0.31442523,"punctuation_ratio":0.16602316,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98476386,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-05T19:47:54Z\",\"WARC-Record-ID\":\"<urn:uuid:61ef7aeb-ab30-4852-9c8a-532d34cd32c6>\",\"Content-Length\":\"19073\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b649a069-1b67-421c-a62e-11467594f514>\",\"WARC-Concurrent-To\":\"<urn:uuid:53886e14-96c1-4602-aff9-547d14a39f18>\",\"WARC-IP-Address\":\"72.44.94.106\",\"WARC-Target-URI\":\"https://numbermatics.com/n/11163/\",\"WARC-Payload-Digest\":\"sha1:QGKV4DQG7KQU2EBRIZJY52RPM2EMBZHJ\",\"WARC-Block-Digest\":\"sha1:WB7WLLBHW4HVTEZTV5VEDJHARDWK7KIB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100555.27_warc_CC-MAIN-20231205172745-20231205202745-00036.warc.gz\"}"}
https://math.stackexchange.com/questions/430543/how-do-i-understand-the-meaning-of-the-phrase-up-to-in-mathematics/430555	[ "# How do I understand the meaning of the phrase \"up to~\" in mathematics?\n\nI am reading a book that explains elementary number theory: Number Theory: A Lively Introduction with Proofs, Applications, and Stories by James Pommersheim, Tim Marks and Erica Flapan.\n\nThe authors say,\n\n\"We express this idea in the statement of the Fundamental of Arithmetic by saying that prime factorization are unique up to order. ... for example, 40 and -40 are equal up to sign, the functions $12x^2$ and $x^2$ are equal up to a constant factor, and so forth.\"\n\nBecause I am not an English native speaker, the phrase \"up to\" is ambiguous for me a little bit. I would like to figure out the meaning of the phrase and use it properly. How do I do?\n\n• Have you learned about equivalence relation? Think about the mean of \"up to\" comparing with equivalence relation. Jun 27, 2013 at 5:09\n• By the way, many feel that it is in poor taste to refer to only one of the authors of a book. Imagine how much time and energy it would take to write a math book, even with the help of some coauthors. Every author's name deserves to be given. Jun 27, 2013 at 5:43\n\nWhen one says \"$X$ is true up to $Y$\", then one means that it is not strictly speaking correct that $X$ is true, and the $Y$ which occurs after the up to clarifies the sense in which $X$ is not quite true.\n\nThus it really means \"Roughly speaking $X$ is true, except for $Y$\".\n\nAs you can see, this is a very informal construction. It could be used very abusively. In order to be acceptable, the sense in which $Y$ \"corrects\" $X$ needs to be very familiar to the reader, or the underlying logic should be made more explicit.\n\nSo when one says that factorizations in $\\mathbb{Z}$ are unique up to order, then it is understood that of course we could have $abc= bac = cab = \\ldots$ and we don't want to count these as being essentially different factorizations. In this case this sloppy language is probably a good expository choice, since most readers have an intuitive idea of what the \"exception\" is, whereas spelling it out explicitly would probably involve something like the following:\n\nUniqueness of Prime Factorizations: If $r$ and $s$ are positive integers, $p_1,\\ldots,p_r,q_1,\\ldots,q_s$ are prime numbers and $p_1 \\cdots p_r = q_1 \\cdots q_s$, then $r = s$ and there is a bijection $\\sigma: \\{1,\\ldots,r\\} \\rightarrow \\{1,\\ldots,s\\}$ such that $p_i = q_{\\sigma(i)}$ for all $1 \\leq i \\leq r$.\n\nThe point is that this more precise statement will be gibberish to someone who does not have a certain amount of mathematical sophistication. (On the other hand, at some point a student of mathematics should be able to supply the statement above or something equally explicit and logically equivalent. In fact I happen to recall one colleague of mine, a veteran research mathematician, who told me that he was always uncomfortable with the statement of uniqueness of factorization because of the vague phrase \"up to\". He has very high standards for clear exposition and thinking.) The book that you speak of is written for a much more general audience, so their expository choice is a good one.\n\n(As an aside, I was first introduced to number theory in a course taught by Marks and Pommersheim, a course they taught to talented high school students during the summer over a period of many years. I haven't read the book that you mention, coauthored with Flapan, but I'd have to think that it is largely based on these courses. This course was one of the great influences on my intellectual life -- not coincidentally I am now a number theorist! -- so I expect the book is pretty good.)\n\n• Good point. As an example of sloppy thinking I offer myself: if asked to make the statement more explicit without \"up to\", I'd probably have written something like \"If $r$ and $s$ are positive integers, and $p_1,\\dots,p_r$ and $q_1,\\dots,q_s$ are prime numbers satisfying $p_1\\le\\dots\\le p_r$ and $q_1\\le\\dots\\le q_s$, then $r=s$ and $p_i=q_i$ for all $1\\le i\\le r$.\" Now I think of it, this statement, while correct, is actually weaker than the original statement, and has essentially pushed the issue of order under the rug. (This picks one particular bijection $\\sigma$ as above of course.) Jun 27, 2013 at 6:21\n• @ShrreevatsaR: sure, the result is often stated that way, and it is more palatable than my version. I was really thinking of the statement about unique factorization into irreducibles in a general integral domain, in which there need not be a natural ordering on the irreducibles. (But then you need to say that $p_i$ and $q_{\\sigma(i)}$ are associate, not equal.) Jun 27, 2013 at 7:44\n\n\"$A$ and $B$ are equal up to $x$\" means that either $A = B$, or that $A$ can be obtained from $B$ by merely modifying $x$ or applying $x$.\n\nExamples:\n\n• $4$ and $-4$ are equal up to sign because you can obtain one from the other by modifying the sign.\n• $(2^3, 7^4)$ and $(7^4, 2^3)$ are equal up to order since you can obtain one from the other by modifying the order.\n• $S_2$ and $\\mathbb{Z}/2\\mathbb{Z}$ are equal up to isomorphism (as groups) since you can obtain one from the other by applying an isomorphism.\n\nIt's understood that when ever this terminology is used, the relation between $A$ and $B$ is an equivalence relation on objects of the type of $A$ and $B$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.95979583,"math_prob":0.9924771,"size":2488,"snap":"2022-27-2022-33","text_gpt3_token_len":605,"char_repetition_ratio":0.08494364,"word_repetition_ratio":0.0,"special_character_ratio":0.23794213,"punctuation_ratio":0.096114516,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99813133,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-08T22:33:54Z\",\"WARC-Record-ID\":\"<urn:uuid:68870463-45c4-4eff-935f-1dd305107c30>\",\"Content-Length\":\"238605\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2c093c2a-b2b1-47f8-898f-2746ec194bee>\",\"WARC-Concurrent-To\":\"<urn:uuid:e1143db1-96a5-4d72-a81e-019e9eb11e4c>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/430543/how-do-i-understand-the-meaning-of-the-phrase-up-to-in-mathematics/430555\",\"WARC-Payload-Digest\":\"sha1:GKQYXXAKD4ZZJOFRVWGEIBMK7NY75NOL\",\"WARC-Block-Digest\":\"sha1:35Z6JITGJ6OWCERG6QMTDRUNYQXNUEUW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570879.1_warc_CC-MAIN-20220808213349-20220809003349-00046.warc.gz\"}"}
http://www.cs.cmu.edu/~bmm/LeisersonM88.html	[ "• Communication-efficient parallel algorithms for distributed random-access machines. C. E. Leiserson and B. M. Maggs, Algorithmica, Vol. 3, 1988, pp. 53-77.\nThis paper introduces a model for parallel computation called the distributed random-access machine (DRAM). A DRAM is an abstraction of a parallel computer in which memory accesses are implemented by routing messages through a communication network. A DRAM explicitly models the congestion of messages across cuts of the network. We introduce the notion of a conservative algorithm as one whose communication requirements at each step can be bounded by the congestion of pointers of the input data structure across cuts of a DRAM. We give a simple lemma that shows how to ``shortcut'' pointers in a data structure so that remote processors can communicate without causing undue congestion. We give O(log n)-step, linear-processor, linear-space, conservative algorithms for a variety of problems on n-node trees, such as computing treewalk numberings and evaluating all subexpressions in an expression tree, and O(log^2 n)-step conservative algorithms for problems on graphs of size n, including finding a minimum-cost spanning forest, computing biconnected components, and constructing an Eulerian cycle.\n\nBack to other publications" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81363106,"math_prob":0.83802855,"size":1316,"snap":"2019-13-2019-22","text_gpt3_token_len":287,"char_repetition_ratio":0.094512194,"word_repetition_ratio":0.0,"special_character_ratio":0.19452888,"punctuation_ratio":0.11255411,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9745633,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-25T04:12:19Z\",\"WARC-Record-ID\":\"<urn:uuid:b937cc58-ba49-4410-a2a0-78ea41e545af>\",\"Content-Length\":\"2211\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e75cde0e-2476-4309-a704-504db25b28e9>\",\"WARC-Concurrent-To\":\"<urn:uuid:001ec5a2-5666-4b45-adb3-b705fd686c36>\",\"WARC-IP-Address\":\"128.2.42.95\",\"WARC-Target-URI\":\"http://www.cs.cmu.edu/~bmm/LeisersonM88.html\",\"WARC-Payload-Digest\":\"sha1:22MBCSHATYBEKJ5FUJF4XAHCSPDUCU33\",\"WARC-Block-Digest\":\"sha1:GHM7NWXY46O5TEERAW7IJVCDDF7FVKPK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912203548.81_warc_CC-MAIN-20190325031213-20190325053213-00245.warc.gz\"}"}
https://www.cheenta.com/sequence-and-integers-aime-i-2007-question-14/	[ "", null, "Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2007 based on Sequence and Integers.\n\n## Sequence and Integers – AIME I, 2007\n\nA sequence is defined over non negetive integral indexes in the following way $a_0=a_1=3$, $a_{n+1}a_{n-1}=a_n^{2}+2007$, find the greatest integer that does not exceed $\\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}$\n\n• is 107\n• is 224\n• is 840\n• cannot be determined from the given information\n\n### Key Concepts\n\nSequence\n\nInequalities\n\nIntegers\n\nBut try the problem first…\n\nSource\n\nAIME I, 2007, Question 14\n\nElementary Number Theory by David Burton\n\n## Try with Hints\n\nFirst hint\n\n$a_{n+1}a_{n-1}$=$a_{n}^{2}+2007$ then $a_{n-1}^{2} +2007 =a_{n}a_{n-2}$ adding these $\\frac{a_{n-1}+a_{n+1}}{a_{n}}$=$\\frac{a_{n}+a_{n-2}}{a_{n-1}}$, let $b_{j}$=$\\frac{a_{j}}{a_{j-1}}$ then $b_{n+1} + \\frac{1}{b_{n}}$=$b_{n}+\\frac{1}{b_{n-1}}$ then $b_{2007} + \\frac{1}{b_{2006}}$=$b_{3}+\\frac{1}{b_{2}}$=225\n\nSecond Hint\n\nhere $\\frac{a_{2007}a_{2005}}{a_{2006}a_{2005}}$=$\\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}$ then $b_{2007}$=$\\frac{a_{2007}}{a_{2006}}$=$\\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}$$\\gt$$\\frac{a_{2006}}{a_{2005}}$=$b_{2006}$\n\nFinal Step\n\nthen $b_{2007}+\\frac{1}{b_{2007}} \\lt b_{2007}+\\frac{1}{b_{2006}}$=225 which is small less such that all $b_{j}$ s are greater than 1 then $\\frac{a_{2006}^{2}+ a_{2007}^{2}}{a_{2006}a_{2007}}$=$b_{2007}+\\frac{1}{b_{2007}}$=224." ]	[ null, "https://www.facebook.com/tr", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5565305,"math_prob":0.99996746,"size":1657,"snap":"2020-24-2020-29","text_gpt3_token_len":714,"char_repetition_ratio":0.20871143,"word_repetition_ratio":0.0,"special_character_ratio":0.51840675,"punctuation_ratio":0.057228915,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000068,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-16T14:51:15Z\",\"WARC-Record-ID\":\"<urn:uuid:2b97b6f1-2f2e-4d3a-91ce-d43623e6550b>\",\"Content-Length\":\"149338\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:30040bd1-aa66-42cb-962d-2938028a2711>\",\"WARC-Concurrent-To\":\"<urn:uuid:04954495-98fe-4bb3-8178-9fb014a25038>\",\"WARC-IP-Address\":\"35.213.169.84\",\"WARC-Target-URI\":\"https://www.cheenta.com/sequence-and-integers-aime-i-2007-question-14/\",\"WARC-Payload-Digest\":\"sha1:V5LXAEK5VE3Q27626NYFR5ZWTFWM22RN\",\"WARC-Block-Digest\":\"sha1:5TXO4ATYRM5LWS23BFCZERPCZ34PM6QU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593657169226.65_warc_CC-MAIN-20200716122414-20200716152414-00591.warc.gz\"}"}
https://electronics.stackexchange.com/questions/90464/how-to-apply-fourier-transform-to-0-5n-un	[ "# How to apply fourier transform to $0.5^n u(n)$\n\nI'm working in a signals class for continuous signals, and we have this problem shown above. I have tried using this function $f_1f_2 = F_1 F_2$, where I'm assuming this means multiplication of two functions is equal to the convolution of their fourier transforms. I'm using $f_1 = 0.5^n$ and $f_2 = u(n)$.\n\nSo I can calculate the fourier transorm of $u(n)$ fine. It is $\\pi\\delta(\\omega) + 1/(j\\omega))$. However, I cannot for the life of me figure out $0.5^n$. I tried to put it into the fourier transform integral integral of$(0.5^t)/(e^{j \\omega t})dt$ from negative infinity to infinity, but I end up with $0.5t/e^{jw}$, and when evaluated from negative infinity to infinity, I end up with $\\infty$ as my answer, unless of course the integration is wrong.\n\nTherefore, either the answer is $\\infty * \\pi\\delta(w) + 1/(j\\omega)$, which when convoluted would equal just the second function..? OR am I going about this problem completely wrong?\n\n## 1 Answer\n\nI think you'd better not split the functions. If you want to calculate the fourier transform of $0.5^t u(t)$, then you might put all that into the integral. Therefore, your integral limits will become $0$ to $+\\infty$ because of $u(t)$. And your integral result should not be infinity because your function $0.5^t$ goes to zero.\n\nSolving the integral:\n\n$\\large\\int_0^\\infty e^{(-jwt)} . 0.5^t dt$\n\nthe inside part could be rewrite:\n\n$\\huge\\frac{e^{(-jwt)}}{2^t} => (\\frac{e^{(-jw)}}{2})^t$\n\nYou could name $\\large\\frac{e^{(-jw)}}{2} = u$ so you get integral of $\\large u^t dt = \\frac{u^t}{ln(u)} + C$\n\nThen you get (replacing u):\n\n$\\huge\\frac{(\\frac{e^{-jw}}{2})^t}{ ln(e^{-jw}/2)}$\n\nWhen t goes to $+\\infty$ you get zero. When it goes to zero you get:\n\n$\\large\\frac{1}{ln(e^{-jw}/2)}$\n\nwhich left us:\n\n$\\large\\frac{1}{-jw - ln(2)}$\n\nI'm not sure I did not get lost in calculations but I believe its correct.\n\n• You're completely right. I got mixed up after separating the functions. Thank you! – nathpilland Nov 25 '13 at 2:12\n• You are welcome. Good luck – Felipe_Ribas Nov 25 '13 at 3:26" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91709775,"math_prob":0.9991708,"size":970,"snap":"2021-21-2021-25","text_gpt3_token_len":275,"char_repetition_ratio":0.098343685,"word_repetition_ratio":0.025641026,"special_character_ratio":0.30515465,"punctuation_ratio":0.12182741,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99990404,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-16T10:56:20Z\",\"WARC-Record-ID\":\"<urn:uuid:e7d790c3-49f9-485b-92a6-6ea5d527f024>\",\"Content-Length\":\"166325\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a3a8b213-de94-4679-9a9a-06ac5f18c38b>\",\"WARC-Concurrent-To\":\"<urn:uuid:5aa28ee1-5d29-49ce-9602-2c4aedc0b1ac>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://electronics.stackexchange.com/questions/90464/how-to-apply-fourier-transform-to-0-5n-un\",\"WARC-Payload-Digest\":\"sha1:O7QTTV7W3OHPB53FWKBKQUD3YEUWEIHH\",\"WARC-Block-Digest\":\"sha1:UJSOFWNEE5OJ5PYYA2SNN4VEVVUNYTDM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487623596.16_warc_CC-MAIN-20210616093937-20210616123937-00322.warc.gz\"}"}
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https://numbermatics.com/n/1468/	[ "# 1468\n\n## 1,468 is an even composite number composed of two prime numbers multiplied together.\n\nWhat does the number 1468 look like?\n\nThis visualization shows the relationship between its 2 prime factors (large circles) and 6 divisors.\n\n1468 is an even composite number. It is composed of two distinct prime numbers multiplied together. It has a total of six divisors.\n\n## Prime factorization of 1468:\n\n### 22 × 367\n\n(2 × 2 × 367)\n\nSee below for interesting mathematical facts about the number 1468 from the Numbermatics database.\n\n### Names of 1468\n\n• Cardinal: 1468 can be written as One thousand, four hundred sixty-eight.\n\n### Scientific notation\n\n• Scientific notation: 1.468 × 103\n\n### Factors of 1468\n\n• Number of distinct prime factors ω(n): 2\n• Total number of prime factors Ω(n): 3\n• Sum of prime factors: 369\n\n### Divisors of 1468\n\n• Number of divisors d(n): 6\n• Complete list of divisors:\n• Sum of all divisors σ(n): 2576\n• Sum of proper divisors (its aliquot sum) s(n): 1108\n• 1468 is a deficient number, because the sum of its proper divisors (1108) is less than itself. Its deficiency is 360\n\n### Bases of 1468\n\n• Binary: 101101111002\n• Base-36: 14S\n\n### Squares and roots of 1468\n\n• 1468 squared (14682) is 2155024\n• 1468 cubed (14683) is 3163575232\n• The square root of 1468 is 38.3144881213\n• The cube root of 1468 is 11.3651547063\n\n### Scales and comparisons\n\nHow big is 1468?\n• 1,468 seconds is equal to 24 minutes, 28 seconds.\n• To count from 1 to 1,468 would take you about twenty-four minutes.\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 1468 cubic inches would be around 0.9 feet tall.\n\n### Recreational maths with 1468\n\n• 1468 backwards is 8641\n• The number of decimal digits it has is: 4\n• The sum of 1468's digits is 19\n• More coming soon!\n\nMLA style:\n\"Number 1468 - Facts about the integer\". Numbermatics.com. 2023. Web. 30 March 2023.\n\nAPA style:\nNumbermatics. (2023). Number 1468 - Facts about the integer. Retrieved 30 March 2023, from https://numbermatics.com/n/1468/\n\nChicago style:\nNumbermatics. 2023. \"Number 1468 - Facts about the integer\". https://numbermatics.com/n/1468/\n\nThe information we have on file for 1468 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!\n\nKeywords: Divisors of 1468, math, Factors of 1468, curriculum, school, college, exams, university, Prime factorization of 1468, STEM, science, technology, engineering, physics, economics, calculator, one thousand, four hundred sixty-eight.\n\nOh no. Javascript is switched off in your browser.\nSome bits of this website may not work unless you switch it on." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86197686,"math_prob":0.9468554,"size":2616,"snap":"2023-14-2023-23","text_gpt3_token_len":698,"char_repetition_ratio":0.112940274,"word_repetition_ratio":0.025943397,"special_character_ratio":0.3046636,"punctuation_ratio":0.16472869,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98571074,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-30T14:47:29Z\",\"WARC-Record-ID\":\"<urn:uuid:2a9db348-64c1-469d-b63c-336657d5a0b4>\",\"Content-Length\":\"17164\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d2ee5c17-e7a6-42a5-a301-6db50c7f86d0>\",\"WARC-Concurrent-To\":\"<urn:uuid:d1e04a41-c7e8-4e84-ab8a-1bdd208486f4>\",\"WARC-IP-Address\":\"72.44.94.106\",\"WARC-Target-URI\":\"https://numbermatics.com/n/1468/\",\"WARC-Payload-Digest\":\"sha1:CDACMA2OSYACXWWE3GHDNGVGZXS65IYZ\",\"WARC-Block-Digest\":\"sha1:CYX6S25JDMSWFGZAZ4GWJVD6SBC3IFMH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949331.26_warc_CC-MAIN-20230330132508-20230330162508-00735.warc.gz\"}"}
https://proxies-free.com/nt-number-theory-a-novel-identity-connecting-permanents-to-bernoulli-numbers/	[ "# nt.number theory – A novel identity connecting permanents to Bernoulli numbers\n\nFor a matrix $$(a_{j,k})_{1le j,kle n}$$ over a field, its permanent is defined by\n$$mathrm{per}(a_{j,k})_{1le j,kle n}:=sum_{piin S_n}prod_{j=1}^n a_{j,pi(j)}.$$\nIn a recent preprint of mine, I investigated arithmetic properties of some permanents.\n\nLet $$n$$ be a positive integer. I have proved that\n$$mathrm{per}left(leftlfloorfrac{j+k}nrightrfloorright)_{1le j,kle n}=2^{n-1}+1$$\nand that\n$$detleft(leftlfloorfrac{j+k}nrightrfloorright)_{1le j,kle n}=(-1)^{n(n+1)/2-1} qquadtext{if} n>1.$$\nwhere $$lfloor cdotrfloor$$ is the floor function. The proofs are relatively easy.\n\nRecall that the Bernoulli numbers $$B_0,B_1,ldots$$ are given by\n$$frac x{e^x-1}=sum_{n=0}^infty B_nfrac{x^n}{n!} (\|x\|<2pi).$$\nThose $$G_n=2(1-2^n)B_n (n=1,2,3,ldots)$$ are sometimes called Genocchi numbers.\n\nBased on my numerical computation, here I pose the following two conjectures.\n\nConjecture 1. For any positive integer $$n$$, we have\n$$mathrm{per}left(leftlfloorfrac{2j-k}nrightrfloorright)_{1le j,kle n}=2(2^{n+1}-1)B_{n+1}.tag{1}$$\n\nConjecture 2. For any positive integer $$n$$, we have\n$$detleft(leftlfloorfrac{2j-k}nrightrfloorright)_{1le j,kle n}=begin{cases}(-1)^{(n^2-1)/8}&text{if} 2nmid n,\\0&text{if} 2mid n.end{cases}tag{2}$$\n\nConjecture 2 seems easier than Conjecture 1.\n\nQUESTION. Are the identities $$(1)$$ and $$(2)$$ correct? How to prove them?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5455444,"math_prob":0.99993694,"size":1348,"snap":"2021-31-2021-39","text_gpt3_token_len":512,"char_repetition_ratio":0.1235119,"word_repetition_ratio":0.041666668,"special_character_ratio":0.328635,"punctuation_ratio":0.13289036,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999944,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-20T17:31:36Z\",\"WARC-Record-ID\":\"<urn:uuid:c9ef2e59-f8de-48db-94e3-1e1a7d64a4a6>\",\"Content-Length\":\"27875\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9d7cdb37-c175-4eb5-9aaa-997a22fdd7f7>\",\"WARC-Concurrent-To\":\"<urn:uuid:e1ca32c6-12f2-48a2-a36e-922b654020b4>\",\"WARC-IP-Address\":\"173.212.203.156\",\"WARC-Target-URI\":\"https://proxies-free.com/nt-number-theory-a-novel-identity-connecting-permanents-to-bernoulli-numbers/\",\"WARC-Payload-Digest\":\"sha1:LCGZJFA7RXEMVDPMY4WJY7WC25CHJHE7\",\"WARC-Block-Digest\":\"sha1:YNXVPKSXNAF3FB5OC6QQJ3MRPPFSICHY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057083.64_warc_CC-MAIN-20210920161518-20210920191518-00295.warc.gz\"}"}
https://www.jiskha.com/search?query=The+relative+amount+of+gold+in+an+alloy+is+commonly+expressed+in+units+of+carats.+Pure+gold+is+24+carats%2C+and+the+percentage+of+gold+in+an+alloy+is+given+as+a+percentage+of+this+value.+For+example%2C+an+alloy	[ "# The relative amount of gold in an alloy is commonly expressed in units of carats. Pure gold is 24 carats, and the percentage of gold in an alloy is given as a percentage of this value. For example, an alloy\n\n33,851 results\n1. ## Chemistry\n\nPure platinum is too soft to be used in jewelry because it scratches easily. To increase the hardness so that it can be used in jewelry, platinum is combined with other metals to form an alloy. To determine the amount of platinum in an alloy, a 7.804 g\n\n2. ## Chemistry\n\nDuralin is an alloy of aluminium containing 4.5% manganese, 3.5% copper, and 0.45% magnesium. How much magnesium is present in a 32 g sample of this alloy? Answer in units of mol\n\n3. ## math\n\nshanon needs 20g of 80% of gold to make a pendan. she has some 85% gold and some 70% gold, from broken jewelry and wants to know how much she should use. determine the quantity of each alloy that she should use.\n\n4. ## Physiology\n\nPure gold has a density of 19 g/ml. If you bought a gold ring and found it had a volume of 0.3 ml and it weighed 5.7 grams, is it real gold? Show me your calculation\n\n5. ## math\n\na metalworker has a metal alloy thay is 25% copper and another alloy that is 60% copper. how many kilograms of each alloy should the metalworker combine to create 90kg of a 53% coppet alloy\n\n6. ## Algebra\n\nA goldsmith has two gold alloys. The first alloy is 20% gold; the second alloy is 60% gold. How many grams of each should be mixed to produce 40 grams of an alloy that is 30% gold? amount of 20% gold_____g Amount of 60% gold_____g\n\n7. ## Chemistry\n\nA 2.35 g sample of a substance suspected of being pure gold is warmed to 72.2 ∘C and submerged into 15.5 g of water initially at 24.1 ∘C. The final temperature of the mixture is 26.5 ∘C. Part A What is the heat capacity of the unknown substance? Part\n\n8. ## algebra\n\nA goldsmith has two gold alloys. The first alloy is 30% gold; the second alloy is 80% gold. How many grams of each should be mixed to produce 10 grams of an alloy that is 40% gold? amount of 30% gold ___ g amount of 80% gold ___ g\n\n9. ## Physics\n\nArchimedes purportedly used his principle to verify that the king's crown was pure gold, by weighing the crown while it was submerged in water. Suppose the crown's actual weight was 14.0 N. The densities of gold, silver, and water are 19.3 g/cm^3, 10.5\n\n10. ## chemistry\n\nDuralin is an alloy of aluminium containing 4.5% manganese, 3.5% copper, and 0.45% magnesium. How much magnesium is present in a 90 g sample of this alloy? Answer in units of mol\n\nThe \"karat\" is a dimensionless unit that is used to indicate the proportion of gold in a gold-containing alloy. An alloy that is one karat gold contains a weight of pure gold that is one part in twenty-four. What is the volume of gold in a 20.0 karat gold\n\n12. ## Physics, still don't get it!\n\nThe \"karat\" is a dimensionless unit that is used to indicate the proportion of gold in a gold-containing alloy. An alloy that is one karat gold contains a weight of pure gold that is one part in twenty-four. What is the volume of gold in a 20.0 karat gold\n\n13. ## Science\n\nHow many number of gold atom are present in 0.6 gm of 18 carat gold.The 24 carat gold is taken as 100% pure gold.(atomic mass of gold=197a.m.u)\n\n14. ## physics\n\nShow that for 1 kg of pure gold the volume of water displaced is 51.8 cm^3 I know that the density of pure gold is 19.32 gm/cm^3 so 1 kg/19.32 gm/cm^3= 0.05175 ? I have come along way with this problem and feel like I am close, but now I need help. How do\n\n15. ## math\n\nAnne and Nancy use a metal alloy that is 20.45% copper to make jewelry. How many ounces of a 12% alloy must be mixed with a 25% alloy to form 100 ounces of the desired alloy?\n\n16. ## Algebra\n\nsolve linear equation-an alloy of metal is 25% copper. Another alloy is 50% copper. how much of each alloy should be used to make 1000 grams of an alloy that is 45% copper\n\n17. ## College Algebra\n\nA goldsmith has two gold alloys. The first alloy is 30% gold; the second alloy is 80% gold. How many grams of each should be mixed to produce 50 grams of an alloy that is 68% gold? amount of 30% gold ______grams amount of 80% gold ______grams\n\n18. ## Physics\n\nA piece of gold-aluminium alloy weighs 49N when suspended from a spring balance and submerged in water it weighs 39.2N what is the weight of Gold in the alloy if the specific gravity of gold is 19.3 and aluminium is 2.5?\n\n19. ## physics\n\na piece of gold aluminuim alloy weighs 100g in air and 80g in water.what is the weight of the gold in the alloy if the relative density of gold is 19.3 and that of aluminium is 2.5\n\n20. ## physics\n\nAccording to legend, to determine whether the king's crown was made of pure gold, Archimedes measured the crown's volume by determining how much water it displaced. The density of gold is 19.3 g/cm^3. If the crown's mass was 6.00 * 10^2g, what volume of\n\n21. ## math\n\nThe amount of gold jewelry and other products is measured in karats (K), where 24K represents pure gold. The mark 14K on a chain indicates that the ratio between the mass of the gold in the chain and the mass of the chain is . If a gold ring is marked 18K\n\n22. ## physics2 lab\n\na 50 g bracelet is suspected of being gold plated lead instead of pure gold.when it is dropped into a full glass of water,4cm of water overflows.is the bracelet pure gold?if not,what proportions of its mass is gold?determine also the amount of its\n\n23. ## Maths Percentage\n\nFind the percentage of pure gold in 22-carat gold, if 24-carat gold is 100% pure gold.\n\n24. ## Science\n\nCompounds with high melting points have ___ a. Covalent bonds b. Metallic Bonds c. Ionic bonds* d. No chemical bonds How could you determine if a sample of gold is pure? Compare the sample's density with that of pure gold.* Compare the sample's color\n\n25. ## Algebra/Math\n\nCan you show work por favor? :) A metalworker has a metal alloy that is 20% copper and another alloy that is 60% copper. How many kilograms of each alloy should be combined to create 80 kg of a 53% copper alloy?\n\n26. ## Chemistry\n\nThe relative amount of gold in an alloy is commonly expressed in units of carats. Pure gold is 24 carats, and the percentage of gold in an alloy is given as a percentage of this value. For example, an alloy that is 50 percent gold is 12 carats. State the\n\n27. ## Physics\n\nThe \"karat\" is a dimensionless unit that is used to indicate the proportion of gold in a gold-containing alloy. An alloy that is one karat gold contains a weight of pure gold that is one part in twenty-four. What is the volume of gold in a 20.0 karat gold\n\n28. ## physical science\n\nA 50.g bracelet is suspected pf being gold plated lead instead of pure gold. It was dropped into a full to the rim glass of water and 4cm to the third power overflowed. Is the the bracelet pure gold. Explain. (Density of gold=19 g/cm to the third power and\n\n29. ## ALGEBRA\n\nIf a jeweller wishes to form 12 ounces oif 75%pure gold from substances that arev 60% AND 80% pure gold .How much of each substances must be mixed together to produce this?\n\n30. ## Algebra 1 (Ms. Sue)\n\n1). Solve each equation. 17 = w - 4 2/3s = -6 1/3y + 1/4 = 5/12 x/8 - 1/2 = 6 5/6x - 1/3 = 5/2 2). In 1668, the Hope diamond was reduced from its original weight by about 45 carats, resulting in a diamond weighing about 67 carats. Write and solve an\n\n31. ## Chemistry\n\nFrom the jewelry descriptions given, select the homogeneous mixture or solution. Answer?? 10 carat (K) electroplated gold jewelry 24 carat (K) gold jewelry (i know this is pure gold) 18 carat (K) gold jewelry Are these right? CaCl2>>calcium chloride\n\n32. ## science\n\nPure gold has a density of 19.32 g/cm3. How large would a piece of gold be if it had a mass of 483 g?\n\n33. ## chemistry\n\nA 9.35 gram of ring with it's volume of 0.654 cm3 is an alloy of gold and with density of gold and silver respectively 19.3 g/cm3 and 10.5 g/cm3 assume there is no change in volume when pure metals are mixed. calculate volume of gold?\n\n34. ## Algebra 2\n\nThe purity of gold is measured in karats. Twenty-four karat gold is pure gold and k karat gold is k/24 gold by mass. One piece of 16-karat gld jewelry has a mass of 30 grams and another piece of 10-karat gold jewelry has a mass of 40 grams. Find the number\n\n35. ## Math (Word Problem into formula)\n\nA chemist is studying the properties of a bronze alloy (mixture) of copper and tin. She begins with 6 kg of an alloy that is one sixth tin. Keeping the amount of copper constant, she adds small amounts of tin to the alloy. Letting x be the total amount of\n\n36. ## Chemistry\n\nA mettallurgist decided to make a special alloy. He melted 30 grams of Au(Gold) with 30 grams of Pb. This melting process resulted to 42 grams of an Au-Pb alloy and some leftover Pb. This scientist wanted to make 420 grams of this alloy with no leftovers\n\n37. ## science\n\nhowmany number of gold atoms are present in 0.6 of 18 carat gold. 24 carat gold is taken as 100 % pure gold\n\n38. ## physics\n\nThe British Sovereign coin is an alloy of gold and copper having a total mass of 7.988 g and is 22-karat gold. (a) Find the mass of gold in kg using that # of karats = 24 x (mass of gold)/(total mass). (b) Calculate the volumes of gold and copper used to\n\n39. ## difference between 10k, 14k, 18k gold\n\nI am in the process of ordering a class ring and the options for the gold are 10K, 14K, and 18K. I know the price is different, but I was just wondering if anybody knows if you can actually tell a difference between them when you look at them or do they\n\n40. ## maths word problem help!\n\nA and B are two alloys of gold and copper,prepared by mixing metals in the ratio 7:2 and 7:11 respectively.if equal quantity of alloys are melted to form a third alloy C.find the ratio of gold and C show working thanks GOD bless\n\n41. ## math\n\njewelry marked 14 karat gold is 14/24 pure gold. what is the percent\n\n42. ## Physics\n\nThe karat is a dimensionless unit that is used to indicate the proportion of gold in a gold-containing alloy. An alloy that is one karat gold contains a weight of pure gold that is one part in twenty-four. What is the volume of gold in a 20.0-karat gold\n\n43. ## chemistry\n\nA 14 karat gold alloy contains 58% m/m gold. How much of this alloy can a jeweler make using 50.0 g of pure gold?\n\nThe amount of gold jewelry and other products is measured in karats (K), where 24K represents pure gold. The mark 14K on a chain indicates that the ratio between the mass of the gold in the chain and the mass of the chain is 14:24 If a gold ring is marked\n\n45. ## Chemistry 101\n\nA crown is determine to weigh 2.65 kg. When this crown is submerged in a full basin of water, the overflow is found to be 145 mL. Is the crown pure gold or a gold-silver alloy? The density of silver is 10.5 g/mL and gold that of gold is 19.3 g/mL. Show\n\n46. ## Chemistry 101\n\nA crown is determine to weigh 2.65 kg. When this crown is submerged in a full basin of water, the overflow is found to be 145 mL. Is the crown pure gold or a gold-silver alloy? The density of silver is 10.5 g/mL and gold that of gold is 19.3 g/mL\n\n47. ## Chemistry\n\nA homogenous mixture of metals is called an alloy. An alloy is found to be 30% by volume gold with the rest being aluminum. Aluminum has a density of 2.70 g/cm³. Gold has a density of 13.1 g/cm³. If a piece of alloy contains 3.70 lbs. of gold, how many\n\n48. ## chemistry\n\nA 9.35 gram of ring with it's volume of 0.654 cm3 is an alloy of gold and with density of gold and silver respectively 19.3 g/cm3 and 10.5 g/cm3 assume there is no change in volume when pure metals are mixed. The volume of gold is?\n\n49. ## Algebra\n\nThe purity of gold is measured in karats. Twenty-four karat gold is pure gold and k karat gold is k/24 gold by mass. One piece of 16-karat gold jewelry has a mass of 30 grams and another piece of 10-karat gold jewelry has a mass of 40 grams. Find the\n\n50. ## Chemistry\n\nA street merchant is trying to sell you a 14 karat (24 karat is 100% pure) bracelet for \\$450. The gold is alloyed with copper. Before buying the item, you weigh it to be 110 grams and you also measure its water displacement as 7.59 mL. What is the %\n\n51. ## college algebra word problem\n\nA goldsmith has two alloys that are different purities of gold. The first is 3/4 pure and the second is 5/12 pure. How many ounces of each should be melted and mixed in order to obtain a 6 oz mixture that is 2/3 pure gold?\n\n52. ## physics\n\nthe relative density of an alloy is 6.5.find the mass of the solid alloy cube of side 20cm.what volume of the alloy has a mass of 13kg\n\n53. ## CHEMISTRYYYY HELP\n\nA sample of gold ore contains 22.0 g of gold metal per 100kg of ore. If gold has a market value of \\$350.00 per oz., what mass of ore, in tons, must be refined to get one million dollars worth of pure gold?\n\n54. ## Physics\n\nThe relative density of an alloy is 6.5.a.find the mass of a solid cube of side 20cm.b.what volume of the alloy has a mass of 13kg\n\nSolve the problem. Anne and Nancy use a metal alloy that is 24.9% copper to make jewelry. How many ounces of a 20% alloy must be mixed with a 27% alloy to form 100 ounces of the desired alloy?\n\n56. ## College Pre Calculus\n\nanne and nancy use a metal alloy that is 19.3% copper to make jewelry. how many ounces of a 19% alloy must be mixed with a 20% alloy to form 110 ounces of the desired alloy?\n\n57. ## physics honors\n\nSuppose that pure gold was to be made into a very long, thin wire for high-end stereo systems. (Gold is often used because it does not oxidize so it makes very good electrical contacts). If the gold wire was to have a radius of 0.01 cm, how long of a wire\n\n58. ## Physics\n\nAn alloy of copper and gold displaces 1.935 cm^3 of water. It has a mass of 33.44g. If gold has a density of 19.3g/cm^3 and copper has a density of 8.9g/cm^3, what is the percent of gold by mass in the object?\n\n59. ## physics\n\nn alloy of copper and gold displaces 1.935 cm^3 of water. It has a mass of 33.44g. If gold has a density of 19.3g/cm^3 and copper has a density of 8.9g/cm^3, what is the percent of gold by mass in the objec\n\n60. ## college algebra\n\na chemist has two alloys one of which is 5% gold and 15% lead and the other of which is 25% gold and 30% lead how many grams of each of the two alloys should be used to make an alloy that contains 40g of gold and 93g of lead ?\n\n61. ## chemistry\n\nFrom the jewelry descriptions given, select the homogeneous mixture or solution. Answer?? 10 carat (K) electroplated gold jewelry 24 carat (K) gold jewelry (i know this is pure gold) 18 carat (K) gold jewelry Are these right? CaCl2>>calcium chloride\n\n62. ## Chemistry\n\nA 15g sample of what appears to be gold that was heated from 22 degrees C to 39 degrees C, required 26.3 J of energy. If the specific heat of gold is .129 J/(g x degrees C), is the metal pure gold? Defend your answer with calculations.\n\n63. ## science\n\npure gold is highly malleable. describe a use og gold when this property is not desirable\n\nHow to get pure gold from impure gold..please specify the chemical equation.. any example is fine,e.g. AuCl3 -> Au\n\n65. ## Chemistry\n\nA 2.6 g of a substance that looks like gold needed 9.4 J of heat energy to raise its temperature from 20 to 38 ºC. Is the metal pure gold?\n\n66. ## Chemistry\n\nDiamonds are measured in carats, and 1 carat=0.200g. The density of diamonds is 3.51g/cm3. what is the volume of a 5.0-carat diamond, expressed in liters?\n\n67. ## s.h.s chemistry\n\nwhich has a higher melting point pure gold or impure gold\n\n68. ## math\n\njewelry marked 14 karat gold is 14/25 pure gold. what is the percent\n\n69. ## Chemistry\n\nThe density of pure gold is 19.3 g/centimeter cube. A quantity of what appears to be gold has a mass of 465 grams and a volume of 26.5 mL. Is the substance likely to be pure gold? please tell me how you got your answer. Thanks! density = mass/volume You\n\n70. ## chem\n\ngold hasa density of 19 g/cm^3 if an 18 inch necklace is made of pure gold how much would it weigh in grams if it is 1 cm wide and 1 mm thick\n\n71. ## chemistry\n\nA 4.90 g nugget of pure gold absorbed 255 J of heat. What was the final temperature of the gold if the initial temperature was 24.0°C? The specific heat of gold is 0.129 J/(g·°C).\n\n72. ## Chem\n\nPure gold adopts a face-centered cubic crystal structure. It also has a density of 19.30g/mL. What is the edge length of gold's unit cell? What is the atomic radius of gold? I understand how to do this problem the other way: when you have the unit cell and\n\n73. ## chemistry\n\nIt is observed that when solid copper is added to a solution containing gold ions, the copper dissolves and the gold precipitates. Similarly, almost every other metal will displace gold from solutions of gold salts. a. Do these results mean that the gold\n\n74. ## algebra\n\nassume that a jeweler needs 80 grams of an alloy that is 30% gold. He has avaliable one alloy that is 5% gold and another that is 85% gold; he will mix some of each If x equal the number of grams of the first alloy (so the number of grams of the second is\n\n75. ## Chemistry\n\nA piece of metal alloy (a solution of metals) is made of gold and silver. It is 20% by mass silver. An irregularly shaped piece of this alloy contains 0.356 lb of gold. When the alloy is added to a graduated cylinder that contains 31.55 mL of alcohol, the\n\n76. ## math\n\nHow many ounces of an alloy containing 30% gold on the world market must be added to 10 ounces of an alloy containing 5% to produce an alloy containing 20% gold?\n\n77. ## maths\n\nIn 60 ounces of alloy for watch cases there are 20 ounces of gold.how much copper must be added to the alloy so that a watch case weighing 4 ounces,made from the new alloy,will contain 1 ounce of gold.......answer is 20 ounces\n\n78. ## College Algebra - Chemistry\n\nA goldsmith has two gold alloys. The first alloy is 25% gold and the second is 55% gold. How many grams of each should be mixed to produce 40 grams of an alloy that is 43%?\n\n79. ## Math\n\nThe amounts by weight of gold, silver and lead in three alloys of these metals are in the ratio 1:5:3 in the first alloy, 2:3:4 in the second and 5:2:2 in the third. How many kg of the first alloy must be used to obtain 10kg of an alloy containing equal\n\n80. ## algebra\n\na jeweler purchased 5 oz of a gold alloy and 20 oz of a silver alloy for a total cost of \\$540. The next day at the same price per ounce, the jeweler purchased 4 oz of th egold alloy and 25 oz of the silver alloy for a total cost of \\$450. Find the cost per\n\n81. ## chemistry\n\nArchimedes used density along with mass and volume of alloys (mixtures) of gold and silver to determine the percent of gold (and also silver) in these alloys by simple, non-destructive measurements. Prove that the mass of gold in a given mass of silver\n\n82. ## physics\n\nThe relative Density of an alloy is 65.(a) find the mass of a solid alloy cube of side 20cm.(b) what vol of the alloy has a mass of 13kg? Please explain in details\n\n83. ## physics\n\nthe relative density of an alloy is 6.5.find the mass of the solid alloy cube of side 20cm.what volume of the alloy has a mass of 13kg\n\n84. ## Physics\n\nThe relative density of an alloy is 6.5,calculate the mass of the solid alloy cube of sides 20cm.what volume of alloy has a mass of 3kg\n\n85. ## Chemistry\n\nNovember's birthstone is citrine, a yellow member of the quartz family. It is primarily silicon dioxide, but small amounts of iron(III) ions give it its yellow color. A high quality citrine containing about 0.040 moles of SiO2 costs around \\$225. If this\n\n86. ## algebra\n\nA chemist has x grams of an alloy that is 20% silver and 10% gold, and y grams of an alloy that is 25% silver and 30% gold. Express the number of grams of gold in the two allows in terms of x and y.\n\n87. ## Chemistry\n\nAn alloy contains 39.5 mass% silver (Ag) and has density og 7.86 g/ml What volume of alloy would be needed to contain 12.0g of pure Ag?\n\n88. ## problem solving\n\nYou purchased 18 karat gold chain that weights 15 grams. How much gold did you purchase in grams? In carats? (1 carat=0/2 grams)\n\n89. ## math\n\na diamond weighing 2.5 carats was cut into two pieces. The two pieces were put on a balance scale, one on each side. An additional weight of 0.9 carats on one side was required to maintain balance. Find the weight, in carats, of the larger piece\n\n90. ## Chemistry\n\nTo assay silver content in an alloy, a 13.38 g sample of alloy was dissolved in nitric acid. The addition of 0.53 molar sodium chloride resulted in the precipitation of silver chloride. If the alloy were pure silver, what volume of sodium chloride is\n\n91. ## algebra\n\nSolve the problem. Anne and Nancy use a metal alloy that is 24.9% copper to make jewelry. How many ounces of a 20% alloy must be mixed with a 27% alloy to form 100 ounces of the desired alloy?\n\n92. ## math\n\nAnne and Nancy use a metal alloy that is 23.5% copper to make jewelry. How many ounces of a 21% alloy must be mixed with a 30% alloy to form 108 ounces of the desired alloy?\n\n93. ## Physics\n\n9. A lump of gold sample is suspected to have a mass of 500g and is found to have a relative den of 5.2 Find what mass of gold is present in the sample If the relative den of gold and aluminium are 19.3 and 2.6\n\n94. ## Math\n\nThere are two alloys consisting of zinc, copper and tin. The first alloy contains 25% zinc and the second alloy contains 50% copper. The percentage of tin in the first alloy is twice that in the second alloy. 200 kg of the first alloy and 300 kg of the\n\n95. ## Definitions....\n\nWhat is pure liquid? What is impure liquid? Thx You are going to have to define these in context. \"Pure\" is a relative term. For instance, the term \"pure\" water...is it really pure? It depends on the definition of pure being used. Pure liquid could mean\n\n96. ## Chemistry\n\nHow many grams silver is added to 13 grams gold to produce 85% gold alloy?\n\n97. ## Chemistry\n\nHow many grams silver is added to 13 grams gold to produce 85% gold alloy?\n\n98. ## Chemistry\n\nHow many grams silver is added to 13 grams gold to produce 85% gold alloy?\n\n99. ## Honors chemistry\n\ni am doing a project for school and i was told to research the alloy of 16 carat gold. I have found all the other information that i need, but i cannot find the melting or boiling point of 16 carat gold! please help soon my project is due tomorrow.\n\n100. ## science\n\nis there any alloy which is equally dense with gold?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9243675,"math_prob":0.9182098,"size":21791,"snap":"2020-45-2020-50","text_gpt3_token_len":5889,"char_repetition_ratio":0.1889659,"word_repetition_ratio":0.28787163,"special_character_ratio":0.27483824,"punctuation_ratio":0.09462276,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9680105,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-04T11:47:10Z\",\"WARC-Record-ID\":\"<urn:uuid:bba96b99-f5d4-49fc-92cf-37d8ef721513>\",\"Content-Length\":\"66951\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f1209883-b68e-413e-ae6a-df01877edfa9>\",\"WARC-Concurrent-To\":\"<urn:uuid:5fde4f58-28aa-4bb7-8e82-e4f43f1ff9d0>\",\"WARC-IP-Address\":\"66.228.55.50\",\"WARC-Target-URI\":\"https://www.jiskha.com/search?query=The+relative+amount+of+gold+in+an+alloy+is+commonly+expressed+in+units+of+carats.+Pure+gold+is+24+carats%2C+and+the+percentage+of+gold+in+an+alloy+is+given+as+a+percentage+of+this+value.+For+example%2C+an+alloy\",\"WARC-Payload-Digest\":\"sha1:Y5LGETI7I5C3SLUZOS7ZDS3FM7IS7SQL\",\"WARC-Block-Digest\":\"sha1:4JMU2H7NNGKJATQZRKT2RDV2WAROAUOM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141735600.89_warc_CC-MAIN-20201204101314-20201204131314-00414.warc.gz\"}"}
https://www.litscape.com/word_analysis/unsoiled	[ "# Definition of unsoiled\n\n## \"unsoiled\" in the adjective sense\n\n### 1. unsoiled, unspotted, unstained\n\nwithout soil or spot or stain\n\nSource: WordNet® (An amazing lexical database of English)\n\nPrinceton University \"About WordNet®.\"\nWordNet®. Princeton University. 2010.\n\nView WordNet® License\n\n# unsoiled in Scrabble®\n\nThe word unsoiled is playable in Scrabble®, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.\n\nUNSOILED\n(90)\nUNSOILED\n(90)\n\n## Seven Letter Word Alert: (4 words)\n\nelusion, indoles, nodules, unoiled\n\nUNSOILED\n(90)\nUNSOILED\n(90)\nUNSOILED\n(60)\nUNSOILED\n(60)\nUNSOILED\n(36)\nUNSOILED\n(36)\nUNSOILED\n(33)\nUNSOILED\n(30)\nUNSOILED\n(30)\nUNSOILED\n(30)\nUNSOILED\n(30)\nUNSOILED\n(30)\nUNSOILED\n(24)\nUNSOILED\n(24)\nUNSOILED\n(24)\nUNSOILED\n(22)\nUNSOILED\n(22)\nUNSOILED\n(22)\nUNSOILED\n(22)\nUNSOILED\n(22)\nUNSOILED\n(22)\nUNSOILED\n(22)\nUNSOILED\n(22)\nUNSOILED\n(20)\nUNSOILED\n(20)\nUNSOILED\n(20)\nUNSOILED\n(20)\nUNSOILED\n(20)\nUNSOILED\n(20)\nUNSOILED\n(20)\nUNSOILED\n(18)\nUNSOILED\n(18)\nUNSOILED\n(18)\nUNSOILED\n(18)\nUNSOILED\n(15)\nUNSOILED\n(13)\nUNSOILED\n(13)\nUNSOILED\n(13)\nUNSOILED\n(13)\nUNSOILED\n(13)\nUNSOILED\n(12)\nUNSOILED\n(12)\nUNSOILED\n(11)\nUNSOILED\n(11)\nUNSOILED\n(11)\nUNSOILED\n(11)\n\nUNSOILED\n(90)\nUNSOILED\n(90)\nINDOLES\n(82 = 32 + 50)\nUNOILED\n(82 = 32 + 50)\nNODULES\n(82 = 32 + 50)\nUNOILED\n(80 = 30 + 50)\nINDOLES\n(80 = 30 + 50)\nNODULES\n(80 = 30 + 50)\nELUSION\n(78 = 28 + 50)\nUNOILED\n(77 = 27 + 50)\nUNOILED\n(77 = 27 + 50)\nINDOLES\n(77 = 27 + 50)\nINDOLES\n(77 = 27 + 50)\nINDOLES\n(77 = 27 + 50)\nINDOLES\n(77 = 27 + 50)\nUNOILED\n(77 = 27 + 50)\nUNOILED\n(77 = 27 + 50)\nNODULES\n(77 = 27 + 50)\nUNOILED\n(77 = 27 + 50)\nINDOLES\n(77 = 27 + 50)\nUNOILED\n(77 = 27 + 50)\nINDOLES\n(77 = 27 + 50)\nNODULES\n(77 = 27 + 50)\nNODULES\n(77 = 27 + 50)\nNODULES\n(77 = 27 + 50)\nNODULES\n(77 = 27 + 50)\nINDOLES\n(77 = 27 + 50)\nNODULES\n(77 = 27 + 50)\nUNOILED\n(77 = 27 + 50)\nNODULES\n(77 = 27 + 50)\nUNOILED\n(74 = 24 + 50)\nELUSION\n(74 = 24 + 50)\nELUSION\n(74 = 24 + 50)\nINDOLES\n(74 = 24 + 50)\nELUSION\n(74 = 24 + 50)\nELUSION\n(74 = 24 + 50)\nELUSION\n(74 = 24 + 50)\nELUSION\n(74 = 24 + 50)\nNODULES\n(74 = 24 + 50)\nINDOLES\n(74 = 24 + 50)\nNODULES\n(74 = 24 + 50)\nELUSION\n(74 = 24 + 50)\nELUSION\n(74 = 24 + 50)\nINDOLES\n(72 = 22 + 50)\nUNOILED\n(72 = 22 + 50)\nNODULES\n(72 = 22 + 50)\nELUSION\n(71 = 21 + 50)\nINDOLES\n(70 = 20 + 50)\nUNOILED\n(70 = 20 + 50)\nINDOLES\n(70 = 20 + 50)\nINDOLES\n(70 = 20 + 50)\nUNOILED\n(70 = 20 + 50)\nUNOILED\n(70 = 20 + 50)\nUNOILED\n(70 = 20 + 50)\nUNOILED\n(70 = 20 + 50)\nUNOILED\n(70 = 20 + 50)\nNODULES\n(70 = 20 + 50)\nNODULES\n(70 = 20 + 50)\nUNOILED\n(70 = 20 + 50)\nNODULES\n(70 = 20 + 50)\nINDOLES\n(70 = 20 + 50)\nNODULES\n(70 = 20 + 50)\nINDOLES\n(70 = 20 + 50)\nNODULES\n(70 = 20 + 50)\nUNOILED\n(68 = 18 + 50)\nELUSION\n(68 = 18 + 50)\nINDOLES\n(68 = 18 + 50)\nNODULES\n(68 = 18 + 50)\nNODULES\n(68 = 18 + 50)\nNODULES\n(68 = 18 + 50)\nNODULES\n(68 = 18 + 50)\nNODULES\n(68 = 18 + 50)\nNODULES\n(68 = 18 + 50)\nELUSION\n(68 = 18 + 50)\nELUSION\n(68 = 18 + 50)\nELUSION\n(68 = 18 + 50)\nUNOILED\n(68 = 18 + 50)\nINDOLES\n(68 = 18 + 50)\nINDOLES\n(68 = 18 + 50)\nINDOLES\n(68 = 18 + 50)\nUNOILED\n(68 = 18 + 50)\nINDOLES\n(68 = 18 + 50)\nUNOILED\n(68 = 18 + 50)\nINDOLES\n(68 = 18 + 50)\nUNOILED\n(68 = 18 + 50)\nELUSION\n(68 = 18 + 50)\nINDOLES\n(68 = 18 + 50)\nUNOILED\n(68 = 18 + 50)\nELUSION\n(68 = 18 + 50)\nNODULES\n(68 = 18 + 50)\nNODULES\n(66 = 16 + 50)\nNODULES\n(66 = 16 + 50)\nINDOLES\n(66 = 16 + 50)\nINDOLES\n(66 = 16 + 50)\nINDOLES\n(66 = 16 + 50)\nINDOLES\n(66 = 16 + 50)\nINDOLES\n(66 = 16 + 50)\nNODULES\n(66 = 16 + 50)\nNODULES\n(66 = 16 + 50)\nNODULES\n(66 = 16 + 50)\nUNOILED\n(66 = 16 + 50)\nUNOILED\n(66 = 16 + 50)\nELUSION\n(66 = 16 + 50)\nELUSION\n(66 = 16 + 50)\nELUSION\n(66 = 16 + 50)\nUNOILED\n(66 = 16 + 50)\nELUSION\n(66 = 16 + 50)\nELUSION\n(66 = 16 + 50)\nELUSION\n(66 = 16 + 50)\nELUSION\n(66 = 16 + 50)\nUNOILED\n(66 = 16 + 50)\nELUSION\n(66 = 16 + 50)\nUNOILED\n(66 = 16 + 50)\nELUSION\n(64 = 14 + 50)\nELUSION\n(64 = 14 + 50)\nINDOLES\n(64 = 14 + 50)\nUNOILED\n(64 = 14 + 50)\nELUSION\n(64 = 14 + 50)\nELUSION\n(64 = 14 + 50)\nNODULES\n(64 = 14 + 50)\nELUSION\n(64 = 14 + 50)\nINDOLES\n(62 = 12 + 50)\nNODULES\n(62 = 12 + 50)\nUNOILED\n(62 = 12 + 50)\nNODULES\n(62 = 12 + 50)\nUNOILED\n(62 = 12 + 50)\nINDOLES\n(62 = 12 + 50)\nUNOILED\n(62 = 12 + 50)\nUNOILED\n(62 = 12 + 50)\nNODULES\n(62 = 12 + 50)\nINDOLES\n(62 = 12 + 50)\nUNOILED\n(61 = 11 + 50)\nNODULES\n(61 = 11 + 50)\nELUSION\n(61 = 11 + 50)\nINDOLES\n(61 = 11 + 50)\nELUSION\n(61 = 11 + 50)\nNODULES\n(61 = 11 + 50)\nELUSION\n(61 = 11 + 50)\nINDOLES\n(61 = 11 + 50)\nINDOLES\n(61 = 11 + 50)\nNODULES\n(61 = 11 + 50)\nNODULES\n(60 = 10 + 50)\nNODULES\n(60 = 10 + 50)\nELUSION\n(60 = 10 + 50)\nNODULES\n(60 = 10 + 50)\nUNSOILED\n(60)\nNODULES\n(60 = 10 + 50)\nNODULES\n(60 = 10 + 50)\nUNSOILED\n(60)\nUNOILED\n(60 = 10 + 50)\nUNOILED\n(60 = 10 + 50)\nUNOILED\n(60 = 10 + 50)\nINDOLES\n(60 = 10 + 50)\nUNOILED\n(60 = 10 + 50)\nINDOLES\n(60 = 10 + 50)\nINDOLES\n(60 = 10 + 50)\nUNOILED\n(60 = 10 + 50)\nINDOLES\n(60 = 10 + 50)\nELUSION\n(60 = 10 + 50)\nUNOILED\n(60 = 10 + 50)\nINDOLES\n(60 = 10 + 50)\nELUSION\n(59 = 9 + 50)\nUNOILED\n(59 = 9 + 50)\nNODULES\n(59 = 9 + 50)\nELUSION\n(59 = 9 + 50)\nELUSION\n(59 = 9 + 50)\nELUSION\n(59 = 9 + 50)\nELUSION\n(59 = 9 + 50)\nELUSION\n(59 = 9 + 50)\nELUSION\n(59 = 9 + 50)\nINDOLES\n(59 = 9 + 50)\nELUSION\n(58 = 8 + 50)\nUNSOILED\n(36)\nUNSOILED\n(36)\nUNSOILED\n(33)\nUNSOILED\n(30)\nUNSOILED\n(30)\nUNSOILED\n(30)\nUNSOILED\n(30)\nUNSOILED\n(30)\nUNSOLD\n(27)\nINDOLE\n(27)\nUNDOES\n(27)\nOLDIES\n(27)\nSOULED\n(27)\nSOILED\n(27)\nUNDIES\n(27)\nNODULE\n(27)\nLOUSED\n(27)\nINODES\n(27)\nNOISED\n(27)\nNUDIES\n(27)\nNUDIES\n(24)\nOILED\n(24)\nNOSED\n(24)\nNOISED\n(24)\nNOISED\n(24)\nDIONE\n(24)\nNUDIES\n(24)\nNUDIES\n(24)\n\n# unsoiled in Words With Friends™\n\nThe word unsoiled is playable in Words With Friends™, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.\n\nNODULES\n(92 = 57 + 35)\nUNOILED\n(92 = 57 + 35)\n\n## Seven Letter Word Alert: (4 words)\n\nelusion, indoles, nodules, unoiled\n\nUNSOILED\n(84)\nUNSOILED\n(84)\nUNSOILED\n(60)\nUNSOILED\n(56)\nUNSOILED\n(56)\nUNSOILED\n(54)\nUNSOILED\n(54)\nUNSOILED\n(54)\nUNSOILED\n(48)\nUNSOILED\n(48)\nUNSOILED\n(48)\nUNSOILED\n(48)\nUNSOILED\n(48)\nUNSOILED\n(42)\nUNSOILED\n(32)\nUNSOILED\n(32)\nUNSOILED\n(32)\nUNSOILED\n(32)\nUNSOILED\n(28)\nUNSOILED\n(28)\nUNSOILED\n(28)\nUNSOILED\n(28)\nUNSOILED\n(28)\nUNSOILED\n(28)\nUNSOILED\n(26)\nUNSOILED\n(26)\nUNSOILED\n(24)\nUNSOILED\n(24)\nUNSOILED\n(24)\nUNSOILED\n(24)\nUNSOILED\n(24)\nUNSOILED\n(24)\nUNSOILED\n(24)\nUNSOILED\n(24)\nUNSOILED\n(20)\nUNSOILED\n(18)\nUNSOILED\n(18)\nUNSOILED\n(18)\nUNSOILED\n(17)\nUNSOILED\n(17)\nUNSOILED\n(17)\nUNSOILED\n(16)\nUNSOILED\n(16)\nUNSOILED\n(16)\nUNSOILED\n(16)\nUNSOILED\n(16)\nUNSOILED\n(15)\nUNSOILED\n(15)\nUNSOILED\n(15)\nUNSOILED\n(15)\nUNSOILED\n(15)\nUNSOILED\n(14)\nUNSOILED\n(14)\nUNSOILED\n(14)\nUNSOILED\n(14)\nUNSOILED\n(13)\nUNSOILED\n(13)\nUNSOILED\n(13)\n\nNODULES\n(92 = 57 + 35)\nUNOILED\n(92 = 57 + 35)\nNODULES\n(86 = 51 + 35)\nNODULES\n(86 = 51 + 35)\nUNOILED\n(86 = 51 + 35)\nNODULES\n(86 = 51 + 35)\nUNOILED\n(86 = 51 + 35)\nUNSOILED\n(84)\nUNSOILED\n(84)\nELUSION\n(83 = 48 + 35)\nINDOLES\n(83 = 48 + 35)\nINDOLES\n(83 = 48 + 35)\nELUSION\n(83 = 48 + 35)\nINDOLES\n(83 = 48 + 35)\nELUSION\n(83 = 48 + 35)\nNODULES\n(80 = 45 + 35)\nUNOILED\n(80 = 45 + 35)\nUNOILED\n(80 = 45 + 35)\nUNOILED\n(80 = 45 + 35)\nNODULES\n(80 = 45 + 35)\nNODULES\n(80 = 45 + 35)\nUNOILED\n(80 = 45 + 35)\nUNOILED\n(79 = 44 + 35)\nNODULES\n(79 = 44 + 35)\nNODULES\n(79 = 44 + 35)\nUNOILED\n(79 = 44 + 35)\nNODULES\n(79 = 44 + 35)\nUNOILED\n(79 = 44 + 35)\nINDOLES\n(77 = 42 + 35)\nELUSION\n(77 = 42 + 35)\nINDOLES\n(77 = 42 + 35)\nELUSION\n(77 = 42 + 35)\nELUSION\n(77 = 42 + 35)\nELUSION\n(75 = 40 + 35)\nINDOLES\n(75 = 40 + 35)\nINDOLES\n(75 = 40 + 35)\nELUSION\n(75 = 40 + 35)\nINDOLES\n(75 = 40 + 35)\nELUSION\n(75 = 40 + 35)\nNODULES\n(74 = 39 + 35)\nNODULES\n(74 = 39 + 35)\nUNOILED\n(74 = 39 + 35)\nUNOILED\n(74 = 39 + 35)\nUNOILED\n(74 = 39 + 35)\nNODULES\n(74 = 39 + 35)\nINDOLES\n(71 = 36 + 35)\nINDOLES\n(71 = 36 + 35)\nELUSION\n(71 = 36 + 35)\nINDOLES\n(71 = 36 + 35)\nELUSION\n(71 = 36 + 35)\nINDOLES\n(71 = 36 + 35)\nELUSION\n(71 = 36 + 35)\nINDOLES\n(71 = 36 + 35)\nELUSION\n(71 = 36 + 35)\nUNOILED\n(65 = 30 + 35)\nNODULES\n(65 = 30 + 35)\nUNOILED\n(65 = 30 + 35)\nUNOILED\n(65 = 30 + 35)\nNODULES\n(65 = 30 + 35)\nNODULES\n(65 = 30 + 35)\nUNOILED\n(65 = 30 + 35)\nINDOLES\n(63 = 28 + 35)\nINDOLES\n(63 = 28 + 35)\nELUSION\n(63 = 28 + 35)\nELUSION\n(63 = 28 + 35)\nINDOLES\n(63 = 28 + 35)\nELUSION\n(63 = 28 + 35)\nUNOILED\n(61 = 26 + 35)\nNODULES\n(61 = 26 + 35)\nUNOILED\n(61 = 26 + 35)\nUNOILED\n(61 = 26 + 35)\nNODULES\n(61 = 26 + 35)\nNODULES\n(61 = 26 + 35)\nUNOILED\n(61 = 26 + 35)\nNODULES\n(61 = 26 + 35)\nUNOILED\n(61 = 26 + 35)\nUNOILED\n(61 = 26 + 35)\nNODULES\n(61 = 26 + 35)\nNODULES\n(61 = 26 + 35)\nUNSOILED\n(60)\nINDOLES\n(59 = 24 + 35)\nELUSION\n(59 = 24 + 35)\nINDOLES\n(59 = 24 + 35)\nUNOILED\n(59 = 24 + 35)\nELUSION\n(59 = 24 + 35)\nUNOILED\n(59 = 24 + 35)\nINDOLES\n(59 = 24 + 35)\nINDOLES\n(59 = 24 + 35)\nELUSION\n(59 = 24 + 35)\nNODULES\n(59 = 24 + 35)\nNODULES\n(59 = 24 + 35)\nELUSION\n(59 = 24 + 35)\nELUSION\n(59 = 24 + 35)\nINDOLES\n(59 = 24 + 35)\nNODULES\n(59 = 24 + 35)\nELUSION\n(59 = 24 + 35)\nINDOLES\n(59 = 24 + 35)\nELUSION\n(57 = 22 + 35)\nUNOILED\n(57 = 22 + 35)\nELUSION\n(57 = 22 + 35)\nUNOILED\n(57 = 22 + 35)\nUNOILED\n(57 = 22 + 35)\nINDOLES\n(57 = 22 + 35)\nUNOILED\n(57 = 22 + 35)\nNODULES\n(57 = 22 + 35)\nNODULES\n(57 = 22 + 35)\nNODULES\n(57 = 22 + 35)\nINDOLES\n(57 = 22 + 35)\nUNOILED\n(57 = 22 + 35)\nNODULES\n(57 = 22 + 35)\nNODULES\n(57 = 22 + 35)\nELUSION\n(57 = 22 + 35)\nINDOLES\n(57 = 22 + 35)\nUNOILED\n(57 = 22 + 35)\nNODULES\n(57 = 22 + 35)\nUNOILED\n(57 = 22 + 35)\nNODULES\n(57 = 22 + 35)\nUNSOILED\n(56)\nUNSOILED\n(56)\nINDOLES\n(55 = 20 + 35)\nELUSION\n(55 = 20 + 35)\nINDOLES\n(55 = 20 + 35)\nELUSION\n(55 = 20 + 35)\nELUSION\n(55 = 20 + 35)\nINDOLES\n(55 = 20 + 35)\nELUSION\n(55 = 20 + 35)\nELUSION\n(55 = 20 + 35)\nINDOLES\n(55 = 20 + 35)\nELUSION\n(55 = 20 + 35)\nINDOLES\n(55 = 20 + 35)\nINDOLES\n(55 = 20 + 35)\nELUSION\n(55 = 20 + 35)\nINDOLES\n(55 = 20 + 35)\nNODULES\n(54 = 19 + 35)\nNODULE\n(54)\nNODULES\n(54 = 19 + 35)\nUNSOILED\n(54)\nUNSOILED\n(54)\nUNSOILED\n(54)\nUNOILED\n(54 = 19 + 35)\nINDOLES\n(53 = 18 + 35)\nELUSION\n(53 = 18 + 35)\nNODULES\n(52 = 17 + 35)\nUNOILED\n(52 = 17 + 35)\nUNOILED\n(52 = 17 + 35)\nUNOILED\n(52 = 17 + 35)\nUNOILED\n(52 = 17 + 35)\nNODULES\n(52 = 17 + 35)\nUNOILED\n(52 = 17 + 35)\nSOULED\n(51)\nINDOLES\n(51 = 16 + 35)\nUNOILED\n(51 = 16 + 35)\nNODULES\n(51 = 16 + 35)\nLUNIES\n(51)\nNUDIES\n(51)\nELUSION\n(51 = 16 + 35)\nELUSION\n(51 = 16 + 35)\nUNDIES\n(51)\nINDOLES\n(51 = 16 + 35)\nELUSION\n(51 = 16 + 35)\nLOUSED\n(51)\nUNOILED\n(51 = 16 + 35)\nUNDOES\n(51)\nNODULES\n(51 = 16 + 35)\nINDOLES\n(51 = 16 + 35)\nNODULES\n(50 = 15 + 35)\nNODULES\n(50 = 15 + 35)\nNODULES\n(50 = 15 + 35)\nUNOILED\n(50 = 15 + 35)\nUNOILED\n(50 = 15 + 35)\nELUSION\n(50 = 15 + 35)\nUNOILED\n(50 = 15 + 35)\nNODULES\n(50 = 15 + 35)\nNODULES\n(50 = 15 + 35)\nNODULES\n(50 = 15 + 35)\nNODULES\n(49 = 14 + 35)\nNODULES\n(49 = 14 + 35)\nNODULES\n(49 = 14 + 35)\nNODULES\n(49 = 14 + 35)\nELUSION\n(49 = 14 + 35)\nNODULES\n(49 = 14 + 35)\nNODULES\n(49 = 14 + 35)\nINDOLES\n(49 = 14 + 35)\nNODULES\n(49 = 14 + 35)\nINDOLES\n(49 = 14 + 35)\nINDOLES\n(49 = 14 + 35)\nINDOLES\n(49 = 14 + 35)\nINDOLES\n(49 = 14 + 35)\nUNOILED\n(49 = 14 + 35)\nUNOILED\n(49 = 14 + 35)\nUNOILED\n(49 = 14 + 35)\nELUSION\n(49 = 14 + 35)\nINDOLES\n(49 = 14 + 35)\nUNOILED\n(49 = 14 + 35)\nELUSION\n(49 = 14 + 35)\nUNOILED\n(49 = 14 + 35)\nUNOILED\n(49 = 14 + 35)\nINDOLES\n(48 = 13 + 35)\nINDOLES\n(48 = 13 + 35)\nELUSION\n(48 = 13 + 35)\n\n# Word Growth involving unsoiled\n\nled oiled soiled\n\noil oiled soiled\n\noil soil soiled\n\nso soil soiled\n\nun\n\n## Longer words containing unsoiled\n\n(No longer words found)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.69259655,"math_prob":1.0000082,"size":254,"snap":"2021-04-2021-17","text_gpt3_token_len":68,"char_repetition_ratio":0.148,"word_repetition_ratio":0.0,"special_character_ratio":0.22047244,"punctuation_ratio":0.1904762,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999683,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-22T13:58:05Z\",\"WARC-Record-ID\":\"<urn:uuid:88d38679-e8d1-4e4a-b1f3-7875f1fbf9ba>\",\"Content-Length\":\"160883\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fbaa008b-0af6-4feb-803d-6cc5a73f64b4>\",\"WARC-Concurrent-To\":\"<urn:uuid:4f681c06-a4a0-49be-bb9b-8399983daa28>\",\"WARC-IP-Address\":\"104.21.48.109\",\"WARC-Target-URI\":\"https://www.litscape.com/word_analysis/unsoiled\",\"WARC-Payload-Digest\":\"sha1:DMYENJEIHEHFHKTZ3UY7VEPGZ3IEPFL2\",\"WARC-Block-Digest\":\"sha1:23VMLP5CGIGXRIYDJ4YLT3A63VEA2IH7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039610090.97_warc_CC-MAIN-20210422130245-20210422160245-00379.warc.gz\"}"}
https://tex.stackexchange.com/questions/226500/draw-a-dual-time-representation-diagram	[ "# draw a dual time representation diagram\n\nDoes anybody know how to draw a dual time representation in a professional way? Any suggestions for how to do it with LaTeX or other software?", null, "This is an attempt where two block styles are defined, one with sharp corners and the other with rounded corners. Here varwidth and enumerate environments from varwith and enumitem packages are used to provide some flexibility.", null, "Code\n\n\\documentclass{article}\n\\usepackage[margin=1cm]{geometry}\n\\usepackage[latin1]{inputenc}\n\\usepackage{tikz}\n\\usetikzlibrary{shapes,arrows,positioning}\n\\usepackage{varwidth,enumitem,amsmath}\n\\setlist[enumerate]{labelindent=2pt,leftmargin=}\n\n\\begin{document}\n\\begin{tikzpicture}[auto,\nblock/.style ={rectangle, draw=black, thick, fill=white,align=left,execute at begin node=\\footnotesize,outer sep=0pt,},\nblockrc/.style ={rectangle, draw=black, thick, fill=white, rounded corners,align=left,execute at begin node=\\footnotesize,outer sep=0pt},\nmyarr/.style={->,>=latex'}\n]\n\\node [block] (diag1) {\n\\begin{varwidth}{3cm}\n\\baselineskip=2pt\n\\begin{enumerate}\n\\item[1.] Attend to stimulus sorce, push the buttom marked with green tape and more and more\n\\end{enumerate}\n\\end{varwidth}\n};\n\\node [right = 0.5cm of diag1,blockrc,] (1a) {\\begin{varwidth}{2.5cm}\n\\baselineskip=2pt\n\\begin{enumerate}\n\\item[1a.] Be aware of locations. Green locations\n\\end{enumerate}\n\\end{varwidth}};\n\\node[block,anchor=north west] at (diag1.south east)(diag2)\n{\\begin{varwidth}{2cm}\n\\baselineskip=2pt\n\\begin{enumerate}\n\\item[2.] Intertrial internal\\\\ of \\\\duration\n\\end{enumerate}\n\\end{varwidth}\n};\n\\node [right= 0.5cm of diag2, blockrc] (2a) {\\begin{varwidth}{2cm}\n\\baselineskip=2pt\n\\begin{enumerate}\n\\item[2a.] Attend stimulus souce\n\\end{enumerate}\n\\end{varwidth}};\n\n%--- repeated diagonal entries\n\\node[block,anchor=north west] at (diag2.south east)(diag3)\n{\\begin{varwidth}{2cm}\n\\baselineskip=2pt\n\\begin{enumerate}\n\\item[3.] Stimulus light appears\n$i=1$ (green)\n$i=2$ (red)\n\\end{enumerate}\n\\end{varwidth}\n};\n\\node [right= 0.5cm of diag3, blockrc] (3a) {%\n\\begin{varwidth}{2cm}\n\\baselineskip=2pt\n\\begin{enumerate}\n\\item[3a.] Attend stimulus souce\n\\end{enumerate}\n\\end{varwidth}\n};\n\n\\node [below right= 0.2cm and 0.1cm of 3a, blockrc] (3b) {%\n\\begin{varwidth}{1.5cm}\n\\baselineskip=2pt\n\\begin{enumerate}\n\\item[3b.] Some text here.\n\\end{enumerate}\n\\end{varwidth}\n};\n\n\\node [below right= 0.2cm and 0.1cm of 3b, blockrc] (3c) {%\n\\begin{varwidth}{1.5cm}\n\\baselineskip=2pt\n\\begin{enumerate}\n\\item[3c.] Some text here\n\\end{enumerate}\n\\end{varwidth}\n};\n\n\\node [below right= 0.2cm and 0.1cm of 3c, blockrc] (3d) {%\n\\begin{varwidth}{1.5cm}\n\\baselineskip=2pt\n\\begin{enumerate}\n\\item[3d.] Some text here.\n\\end{enumerate}\n\\end{varwidth}\n};\n\n\\node [below right= 0.2cm and 0.1cm of 3d, blockrc] (3e) {%\n\\begin{varwidth}{1.5cm}\n\\baselineskip=2pt\n\\begin{enumerate}\n\\item[3e.] Some text here\n\\end{enumerate}\n\\end{varwidth}\n};\n\n% last diagonal entry\n\\node[block,anchor=north west] at ([shift={(7cm,-4.5cm)}]diag3.south east)(diag4)\n{\\begin{varwidth}{1.5cm}\n\\baselineskip=2pt\n\\begin{enumerate}[leftmargin=]\n\\item[4.] Push buttom $j$\n\\end{enumerate}\n\\end{varwidth}\n};\n\n% draw lines\n\n\\draw[myarr] (diag1.east)--(1a.west);\n\\draw[myarr] (diag2.east)--(2a.west);\n\\draw[myarr] (diag3.east)--(3a.west);\n\\draw[dashed,myarr] (diag3.south east)--(diag4.north west);\n\n\\draw[myarr] ([xshift=1cm]1a.south)--([xshift=1cm]1a.south \|- 2a.north);\n\\draw[myarr] (diag2.east)--(2a.west);\n\\draw[myarr] (diag3.east)--(3a.west);\n\\draw[myarr] (3e.south)--(3e.south \|- diag4.north);\n\\foreach \\t in{a,b,c,d,e}{\n\\draw[myarr,rounded corners] (1a.east)-\|(3\\t.north);\n}\n\\foreach \\f/\\t in{a/b,b/c,c/d,d/e}{\n\\draw[myarr,rounded corners] (3\\f.east)-\|(3\\t.north);\n}\n\n\\draw[-] (diag1.south east) -- ++ (0,-8.5cm)coordinate(a)\n(diag2.south east) -- ++ (0,-8cm)coordinate(b)\n(diag4.south west) -- ++ (0,-1cm)coordinate(c);\n\\draw[<->,>=latex'] (a)--node[midway]{Choice of RT}([yshift=0.5cm]c);\n\\node[block,draw=none] [below right = 0.2cm and -1.7cm of diag4]{\n\\left . \\begin{aligned} &\\mathtt{Correct:}\\, i=j\\\\ &\\mathtt{Error:}\\, i \\neq j \\end{aligned} \\right \\}\\Rightarrow \\mathtt{\\%Errors}};\n\\node[block,draw=none] [above=1ex of diag1]{INSTRUCTIONS};\n\\node[block,draw=none] [below=1ex of diag1]{\\small Hand locations is\\\\ counterbalanced};\n\\node[block,draw=none] [above=1ex of 1a]{ Reinforced by \\\\ practice trials};\n\\node[block,draw=none] [below=1ex of diag2]{ \\small Varies \\\\ randomly\\\\ over 4,6,8};\n\\end{tikzpicture}\n\\end{document}\n\n• Wow, thank you. I should discover it line by line :-) – Ehsan Masoudi Feb 5 '15 at 12:49\n\nHave a look at tikz, e.g. these examples:\n\nEdit: ok,hope this helps:\n\n\\begin{tikzpicture}\n\\node[rounded corners=3pt, draw](n1) at (0,0) {Main Node};\n\\node[rounded corners=3pt, draw](n2) at (2,-1){Node x};\n\\node[rounded corners=3pt, draw](n3) at (3,-2) {Node y};\n\\node[rounded corners=3pt, draw](n4) at (4,-3) {Node z};\n\n\\draw[rounded corners] (n1) -- (2,0) -- (n2);\n\\draw[rounded corners] (n1) -- (3,0) -- (n3);\n\\draw[rounded corners] (n1) -- (4,0) -- (n4);\n\\end{tikzpicture}" ]	[ null, "https://i.stack.imgur.com/EOhod.png", null, "https://i.stack.imgur.com/M1sbi.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5047161,"math_prob":0.99062884,"size":5425,"snap":"2021-04-2021-17","text_gpt3_token_len":1940,"char_repetition_ratio":0.16380742,"word_repetition_ratio":0.15714286,"special_character_ratio":0.3212903,"punctuation_ratio":0.19465649,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99513036,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-23T05:41:55Z\",\"WARC-Record-ID\":\"<urn:uuid:5c8811a5-4eab-4ef3-a7c2-6d2a4a7738db>\",\"Content-Length\":\"161181\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c901bdef-4f29-46f9-9bba-f81a006a8203>\",\"WARC-Concurrent-To\":\"<urn:uuid:08c2484f-295b-4c98-b77b-f0b7756fb565>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://tex.stackexchange.com/questions/226500/draw-a-dual-time-representation-diagram\",\"WARC-Payload-Digest\":\"sha1:PRRMS2CR4QTOIZF6TIQVQBHA5ILYTRA3\",\"WARC-Block-Digest\":\"sha1:IWG25C5ICW4HJWGIPBXNJPAX4ACTCBRU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703533863.67_warc_CC-MAIN-20210123032629-20210123062629-00465.warc.gz\"}"}
https://encyclopediaofmath.org/wiki/Homogeneous_function	[ "# Homogeneous function\n\nof degree $\\lambda$\n\nA function $f$ such that for all points $( x _ {1} \\dots x _ {n} )$ in its domain of definition and all real $t > 0$, the equation\n\n$$f ( t x _ {1} \\dots t x _ {n} ) = \\ t ^ \\lambda f ( x _ {1} \\dots x _ {n} )$$\n\nholds, where $\\lambda$ is a real number; here it is assumed that for every point $( x _ {1} \\dots x _ {n} )$ in the domain of $f$, the point $( t x _ {1} \\dots t x _ {n} )$ also belongs to this domain for any $t > 0$. If\n\n$$f ( x _ {1} \\dots x _ {n} ) = \\ \\sum _ {0 \\leq k _ {1} + \\dots + k _ {n} \\leq m } a _ {k _ {1} \\dots k _ {n} } x _ {1} ^ {k _ {1} } \\dots x _ {n} ^ {k _ {n} } ,$$\n\nthat is, $f$ is a polynomial of degree not exceeding $m$, then $f$ is a homogeneous function of degree $m$ if and only if all the coefficients $a _ {k _ {1} \\dots k _ {n} }$ are zero for $k _ {1} + \\dots + k _ {n} < m$. The concept of a homogeneous function can be extended to polynomials in $n$ variables over an arbitrary commutative ring with an identity.\n\nSuppose that the domain of definition $E$ of $f$ lies in the first quadrant, $x _ {1} > 0 \\dots x _ {n} > 0$, and contains the whole ray $( t x _ {1} \\dots t x _ {n} )$, $t > 0$, whenever it contains $( x _ {1} \\dots x _ {n} )$. Then $f$ is homogeneous of degree $\\lambda$ if and only if there exists a function $\\phi$ of $n- 1$ variables, defined on the set of points of the form $( x _ {2} / x _ {1} \\dots x _ {n} / x _ {1} )$ where $( x _ {1} \\dots x _ {n} ) \\in E$, such that for all $( x _ {1} \\dots x _ {n} ) \\in E$,\n\n$$f ( x _ {1} \\dots x _ {n} ) = \\ x _ {1} ^ \\lambda \\phi \\left ( { \\frac{x _ 2}{x _ 1} } \\dots { \\frac{x _ n}{x _ 1} } \\right ) .$$\n\nIf the domain of definition $E$ of $f$ is an open set and $f$ is continuously differentiable on $E$, then the function is homogeneous of degree $\\lambda$ if and only if for all $( x _ {1} \\dots x _ {n} )$ in its domain of definition it satisfies the Euler formula\n\n$$\\sum _ { i= } 1 ^ { n } x _ {i} \\frac{\\partial f ( x _ {1} \\dots x _ {n} ) }{\\partial x _ {i} } = \\ \\lambda f ( x _ {1} \\dots x _ {n} ) .$$\n\nHow to Cite This Entry:\nHomogeneous function. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Homogeneous_function&oldid=47253\nThis article was adapted from an original article by L.D. Kudryavtsev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5928323,"math_prob":0.9999858,"size":2270,"snap":"2020-34-2020-40","text_gpt3_token_len":816,"char_repetition_ratio":0.23565754,"word_repetition_ratio":0.31175467,"special_character_ratio":0.4581498,"punctuation_ratio":0.058333334,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000098,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-21T03:09:48Z\",\"WARC-Record-ID\":\"<urn:uuid:c685ffdc-5fff-4487-9b71-668fb675c25e>\",\"Content-Length\":\"15943\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2b54309a-8fec-43f5-97c4-1b7f364aa5cb>\",\"WARC-Concurrent-To\":\"<urn:uuid:2e15441c-4558-416d-bc06-4d350e38be9f>\",\"WARC-IP-Address\":\"34.96.94.55\",\"WARC-Target-URI\":\"https://encyclopediaofmath.org/wiki/Homogeneous_function\",\"WARC-Payload-Digest\":\"sha1:7EL4MRA5VXQYOW5KAIEJI6HL636DHW3H\",\"WARC-Block-Digest\":\"sha1:EUQI2EYKXQQFKBNT3HXU2KCGNJLWXWEF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400198887.3_warc_CC-MAIN-20200921014923-20200921044923-00349.warc.gz\"}"}
https://www.nagwa.com/en/worksheets/949159806192/	[ "In this worksheet, we will practice reading, writing, and modeling addition expressions with numbers up to 10 and representing addition using the plus sign.\n\nQ1:\n\nWhat is ?", null, "Q2:\n\nWhich of the following matches the sum ?\n\n• A", null, "• B", null, "• C", null, "• D", null, "• E", null, "Q3:\n\nFind the matching sum for the following:", null, "• A\n• B\n• C\n• D\n• E\n\nQ4:\n\nCount to find 2 plus 4.", null, "Q5:\n\nCount to find the total.", null, "Q6:\n\nFind the missing number: There are dogs in the picture.", null, "Q7:\n\nWhich of the following pictures matches ?\n\n• A", null, "• B", null, "• C", null, "• D", null, "• E", null, "Q8:\n\nFind the matching sum for the following:", null, "• A\n• B\n• C\n• D\n• E\n\nQ9:\n\nLook at what Scarlett did.", null, "She knows she can write addition sentences to match the part-whole model.", null, "Pick the addition sentences that match what she says.", null, "• A\n• B", null, "• A\n• B", null, "• A\n• B\n\nQ10:\n\nIn the morning, you see 3 airplanes. In the afternoon, you see 5 more airplanes.", null, "Find the sum that tells you the total number of airplanes you saw.\n\n• A\n• B\n• C\n\nQ11:\n\nWhat is ?", null, "• A", null, "• B", null, "• C", null, "Q12:\n\nHow many flowers are there?", null, "Q13:\n\nHow many ducks are there?", null, "Q14:\n\nAdd 3 and 2 to find the total number.", null, "Q15:\n\nWhat is ?", null, "• A", null, "• B", null, "• C", null, "Q16:\n\nWhat is ?", null, "• A", null, "• B", null, "• C", null, "Q17:\n\nWhat is ?", null, "Q18:\n\nWhat is ?", null, "• A", null, "• B", null, "• C", null, "Q19:\n\nWhat is ?", null, "" ]	[ null, "https://images.nagwa.com/figures/149120842408/1.svg", null, "https://images.nagwa.com/figures/619126021856/5.svg", null, "https://images.nagwa.com/figures/619126021856/2.svg", null, "https://images.nagwa.com/figures/619126021856/1.svg", null, "https://images.nagwa.com/figures/619126021856/4.svg", null, "https://images.nagwa.com/figures/619126021856/3.svg", null, "https://images.nagwa.com/figures/394161359126/1.svg", null, "https://images.nagwa.com/figures/950154063581/1.svg", null, "https://images.nagwa.com/figures/310176071841/1.svg", null, "https://images.nagwa.com/figures/672196768242/1.svg", null, "https://images.nagwa.com/figures/539142326412/4.svg", null, "https://images.nagwa.com/figures/539142326412/2.svg", null, "https://images.nagwa.com/figures/539142326412/1.svg", null, "https://images.nagwa.com/figures/539142326412/5.svg", null, "https://images.nagwa.com/figures/539142326412/3.svg", null, "https://images.nagwa.com/figures/638140214793/1.svg", null, "https://images.nagwa.com/figures/468149706756/1.svg", null, "https://images.nagwa.com/figures/468149706756/2.svg", null, "https://images.nagwa.com/figures/468149706756/3.svg", null, "https://images.nagwa.com/figures/468149706756/4.svg", null, "https://images.nagwa.com/figures/468149706756/5.svg", null, "https://images.nagwa.com/figures/527183969853/1.svg", null, "https://images.nagwa.com/figures/686150692368/1.svg", null, "https://images.nagwa.com/figures/686150692368/2.svg", null, "https://images.nagwa.com/figures/686150692368/4.svg", null, "https://images.nagwa.com/figures/686150692368/3.svg", null, "https://images.nagwa.com/figures/752136308219/1.svg", null, "https://images.nagwa.com/figures/767151458723/1.svg", null, "https://images.nagwa.com/figures/879196875895/1.svg", null, "https://images.nagwa.com/figures/272178048095/1.svg", null, "https://images.nagwa.com/figures/272178048095/2.svg", null, "https://images.nagwa.com/figures/272178048095/3.svg", null, "https://images.nagwa.com/figures/272178048095/4.svg", null, "https://images.nagwa.com/figures/364125025027/1.svg", null, "https://images.nagwa.com/figures/364125025027/4.svg", null, "https://images.nagwa.com/figures/364125025027/3.svg", null, "https://images.nagwa.com/figures/364125025027/2.svg", null, "https://images.nagwa.com/figures/430120928385/1.svg", null, "https://images.nagwa.com/figures/525170348432/1.svg", null, "https://images.nagwa.com/figures/525170348432/2.svg", null, "https://images.nagwa.com/figures/525170348432/3.svg", null, "https://images.nagwa.com/figures/525170348432/4.svg", null, "https://images.nagwa.com/figures/575178487069/1.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88140637,"math_prob":0.81771684,"size":851,"snap":"2019-51-2020-05","text_gpt3_token_len":271,"char_repetition_ratio":0.16765054,"word_repetition_ratio":0.30366492,"special_character_ratio":0.33019978,"punctuation_ratio":0.16842106,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9967768,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-18T20:50:22Z\",\"WARC-Record-ID\":\"<urn:uuid:9da54663-550f-499c-933b-fd05aaf668db>\",\"Content-Length\":\"62032\",\"Content-Type\":\"application/http; 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https://www.bennadel.com/blog/3292-using-percent-for-in-line-styles-in-angular-4-2-3.htm	[ "On User Experience (UX) Design, JavaScript, ColdFusion, Node.js, Life, and Love.\n\n# Using % (Percent) For In-Line Styles In Angular 4.2.3\n\nBy on\n\nThis is a super minor post; but, I just used the \"%\" (percent) symbol for in-line styles in Angular 4.2.3, and it totally worked. It's just another example of how the Angular template syntax is such a joy to work with. Little things like CSS unit-selection become friction-free tasks. No more messing around with string interpolation or string concatenation - just select the unit of measurement and apply the length value.\n\nThis is barely worth sharing, since this functionality is all in the Angular documentation. But, the Angular template syntax makes me so happy, I can't help but want to be a share-bear. When defining an in-line style, the unit of measurement can be provided as part of the attribute name. Here, we're using the \"px\" - pixel - unit:\n\n<p [style.width.px]=\"someValue\"> ... </p>\n\nIt's that easy! And here, we're using the \"%\" - percent - unit:\n\n<p [style.width.%]=\"someValue\"> ... </p>\n\nDoesn't that just make you smile? To see this in action, I've created a small demo in which I use a variety of CSS units to render the same numeric value:\n\n• // Import the core angular services.\n• import { Component } from \"@angular/core\";\n•\n• @Component({\n• selector: \"my-app\",\n• styleUrls: [ \"./app.component.css\" ],\n• template:\n• `\n• <p class=\"actions\">\n• <strong>Values:</strong>\n•\n• <ng-template ngFor let-i [ngForOf]=\"[ 0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 ]\">\n•\n• <a (click)=\"setValue( i )\" [class.selected]=\"( i === value )\">\n• {{ i }}\n• </a>\n•\n• </ng-template>\n• </p>\n•\n• <p class=\"bar\" [style.width.%]=\"value\">\n• {{ value }} % <span>percentage relative to parent container</span>\n• </p>\n•\n• <p class=\"bar\" [style.width.px]=\"value\">\n• {{ value }} px <span>absolute pixel</span>\n• </p>\n•\n• <p class=\"bar\" [style.width.em]=\"value\">\n• {{ value }} em <span>the calculated font-size of the element</span>\n• </p>\n•\n• <p class=\"bar\" [style.width.ex]=\"value\">\n• {{ value }} ex <span>the x-height of the elements font</span>\n• </p>\n•\n• <p class=\"bar\" [style.width.vh]=\"value\">\n• {{ value }} vh <span>1% of the height of viewports initial containing block</span>\n• </p>\n•\n• <p class=\"bar\" [style.width.vw]=\"value\">\n• {{ value }} vw <span>1% of the width of viewports initial containing block</span>\n• </p>\n•\n• <p class=\"bar\" [style.width.vmin]=\"value\">\n• {{ value }} vmin <span>the smaller of vw or vh</span>\n• </p>\n•\n• <p class=\"bar\" [style.width.vmax]=\"value\">\n• {{ value }} vmax <span>the larger of vw or vh</span>\n• </p>\n•\n• <p class=\"bar\" [style.width.mm]=\"value\">\n• {{ value }} mm <span>millimeter</span>\n• </p>\n•\n• <p class=\"bar\" [style.width.cm]=\"value\">\n• {{ value }} cm <span>centimeter</span>\n• </p>\n•\n• <p class=\"bar\" [style.width.in]=\"value\">\n• {{ value }} in <span>inch</span>\n• </p>\n•\n• <p class=\"bar\" [style.width.pt]=\"value\">\n• {{ value }} pt\n• </p>\n•\n• <p class=\"bar\" [style.width.pc]=\"value\">\n• {{ value }} pc\n• </p>\n•\n• <p class=\"bar\" [style.width.ch]=\"value\">\n• {{ value }} ch\n• </p>\n•\n• <p class=\"bar\" [style.width.rem]=\"value\">\n• {{ value }} rem\n• </p>\n•\n• <h2>\n• These dont seem to work in Chrome.\n• </h2>\n•\n• <p class=\"bar\" [style.width.cap]=\"value\">\n• {{ value }} cap <span>the \"cap height\" (nominal height of capital letters) of the elements font</span>\n• </p>\n•\n• <p class=\"bar\" [style.width.ic]=\"value\">\n• {{ value }} ic\n• </p>\n•\n• <p class=\"bar\" [style.width.lh]=\"value\">\n• {{ value }} lh <span>the computed value of the line-height property of the element on which it is used</span>\n• </p>\n•\n• <p class=\"bar\" [style.width.rlh]=\"value\">\n• {{ value }} rlh\n• </p>\n•\n• <p class=\"bar\" [style.width.vi]=\"value\">\n• {{ value }} vi\n• </p>\n•\n• <p class=\"bar\" [style.width.vb]=\"value\">\n• {{ value }} vb\n• </p>\n•\n• <p class=\"bar\" [style.width.q]=\"value\">\n• {{ value }} q <span>quarter of a millimeter</span>\n• </p>\n• `\n• })\n• export class AppComponent {\n•\n• public value: number;\n•\n•\n• // I initialize the app component.\n• constructor() {\n•\n• this.value = 50;\n•\n• }\n•\n•\n• // ---\n• // PUBLIC METHODS.\n• // ---\n•\n•\n• // I set the value used to display the various dimensional units.\n• public setValue( value: number ) : void {\n•\n• this.value = value;\n•\n• }\n•\n• }\n\nAs you can see, each P tag is rendering the same \"value\" component property. But, each P tag uses a different unit of measurement for its in-line style. And, when we run this code, you can see how each unit affects the rendering:", null, "So cool!\n\nWhen you first see the Angular template syntax, it's a little jaring. Kind of like the first time you see JSX in ReactJS. But, once you start using it, the Angular template syntax just brings joy. Being able to use CSS units, like \"%\", with your in-line styles is just one example of this holistic design.", null, "You know what, it may be in the documentation, but I hadn't seen it used like that before. That is pretty cool that you can straight up use .% and it works as you would expect it to. There's a lot of little things I feel like they got right the second time around with Angular. And those \"little things\" make a difference.", null, "@John,\n\nAgreed! And, as much documentation as there is, little things are not always well articulated. For example, the documentation doesn't ever explicitly say that you can use \"%\". Instead, it just says you can use CSS units. So, it takes some mental gymnastics to get from that statement to actually trying to see if \"%\" works :)\n\nThat said, I really do like the Angular 2 template syntax. Much to like.", null, "Thanks Ben, one more time you save me a lot of time :)", null, "Thank you !", null, "Glad you are all liking this -- it's so easy, I love it!", null, "", null, "" ]	[ null, "https://www.bennadel.com/resources/uploads/2017/inline-style-unit-of-measurement-angular4.png", null, "https://www.gravatar.com/avatar/d8ea827f7b8b0d3a267f52a91eb31361", null, "https://www.gravatar.com/avatar/f9bbc701ca6770ef482cc1e172344e25", null, "https://www.gravatar.com/avatar/02f36fed209aa9fd52e629c47fd717f4", null, "https://www.gravatar.com/avatar/fe894c8c737aa802c671fb26cd7cc4b5", null, "https://www.gravatar.com/avatar/f9bbc701ca6770ef482cc1e172344e25", null, "https://www.gravatar.com/avatar/398aae016e37efdac2ab5f89d1869cf1", null, "https://www.gravatar.com/avatar/f9bbc701ca6770ef482cc1e172344e25", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.63828766,"math_prob":0.74786335,"size":5785,"snap":"2019-51-2020-05","text_gpt3_token_len":1622,"char_repetition_ratio":0.23490745,"word_repetition_ratio":0.08,"special_character_ratio":0.35073465,"punctuation_ratio":0.15082267,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96596557,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,1,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-27T04:14:36Z\",\"WARC-Record-ID\":\"<urn:uuid:5a613039-68b8-4f6b-a297-703f0eb1c69a>\",\"Content-Length\":\"28607\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f7f60eab-85c8-4811-9fee-3ca90416c2f0>\",\"WARC-Concurrent-To\":\"<urn:uuid:bab43761-50e3-4ad0-b1a8-01314dac7f77>\",\"WARC-IP-Address\":\"184.175.83.43\",\"WARC-Target-URI\":\"https://www.bennadel.com/blog/3292-using-percent-for-in-line-styles-in-angular-4-2-3.htm\",\"WARC-Payload-Digest\":\"sha1:OLZRPB7KS7JUORAJW2PDESWLUB65F36Q\",\"WARC-Block-Digest\":\"sha1:7BPQQZWEZLOWDYWUIWUSPBRZHYFW7KG2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251694176.67_warc_CC-MAIN-20200127020458-20200127050458-00308.warc.gz\"}"}
https://physics.stackexchange.com/questions/62929/time-dependent-hydrohow-would-i-go-about-writing-the-time-dependent-wave-functio	[ "Time Dependent HydroHow would I go about writing the time dependent wave function given the wavefunction at $t=0$? gen Wave Function\n\n1) How vwoulHow would I go about writing the time dependent wave function given the wavefunction at $t=0$? go about writing the time dependent wave function given the wavefunction at $t=0$?\n\nSuppose How would I go about writing the time dependent wave function given the wavefunction at $t=0$? an electron is in a superposition state in the hydrogen atom at $t=0$ with normalized wavefunction: How would How would I go about writing the time dependent wave function given the wavefunction at $t=0$? I go about writing the time dependent wave function given the wavefunction at $t=0$?\n\n$$\\psi(r,\\theta,\\phi) ~=~ A(2R_{10}Y_{00}+4R_{21}Y_{1,-1}).$$\n\n2) What woHow would I go about writing the time dependent wave function given the wavefunction at $t=0$? uld the tiHow wouHow would I go about writing the time dependent wave function given the wavefunction at $t=0$? ld I go about writing the time dependent wave function given the wavefunction at $t=0$? me dependent wave function look like?\n\n3) Also how couHow would I go about writing the time dependent wave function given the wavefunction at $t=0$? ld I tell if the probabilities of measuring a certain energy, or angular momentum arHow would I go about writing the time dependent wave function given the wavefunction at $t=0$? e time dependent?\n\n1) In general, $\\psi(\\vec{r},t) = {\\sf U}(t,0) \\psi(\\vec{r},0)$, where ${\\sf U}(t,0)$ is the time-evolution operator (a unitary matrix).\n2) Given your superposition state at initial time, after time $t$ the wave function would look like\n$$\\psi(r,\\theta,\\phi,t) = A \\left( 2R_{10}Y_{00} e^{-iE_1 t/\\hbar} + 4 R_{21}Y_{1,-1} e^{-iE_2 t/\\hbar} \\right)$$ where $E_1$ and $E_2$ are the ground and first excited energy eigenvalues of the Hydrogen atom (in atomic units, $E_1 = -1/2$ and $E_2 = -1/8$). By the way, your wave function is equivalent if you divide it by $2$ (you include $2$ in the normalization factor $A$).\n3) The probability of measuring $E_1$ and $E_2$ would remain constant: $P_1 = \|\\langle R_{10} Y_{00} \| \\psi(r,\\theta,\\phi,t) \\rangle \|^2 = 4A^2 = 1/5$; $P_2 = 1 - P_1 = 4/5$.\nYou should by now understand that those are the same (constant) probabilities of measuring $L^2 = 0$ and $L^2 = 2\\hbar^2$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8589181,"math_prob":0.9998975,"size":1294,"snap":"2019-43-2019-47","text_gpt3_token_len":325,"char_repetition_ratio":0.2372093,"word_repetition_ratio":0.5463415,"special_character_ratio":0.24265842,"punctuation_ratio":0.072289154,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.000002,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-20T09:26:17Z\",\"WARC-Record-ID\":\"<urn:uuid:12c98471-8f38-48f7-aa6f-f9f8697cc557>\",\"Content-Length\":\"139217\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:089b0366-e0bd-4ba8-b019-cfe056669ede>\",\"WARC-Concurrent-To\":\"<urn:uuid:8f4a1a24-8015-42f4-bb02-1637020d3a4b>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/62929/time-dependent-hydrohow-would-i-go-about-writing-the-time-dependent-wave-functio\",\"WARC-Payload-Digest\":\"sha1:POOCKM5W4HBFGUSFKDEU53PEOFVDW2WU\",\"WARC-Block-Digest\":\"sha1:RRXMT6BNDN7U2XZWSPGQ4KLQSRVMF2MO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986705411.60_warc_CC-MAIN-20191020081806-20191020105306-00062.warc.gz\"}"}
https://dlmf.nist.gov/search/search?q=integral%20identity	[ "# integral identity\n\n(0.004 seconds)\n\n## 1—10 of 55 matching pages\n\n##### 1: 36.9 Integral Identities\n###### §36.9 IntegralIdentities\n36.9.9 $\\left\|\\Psi^{(\\mathrm{E})}\\left(x,y,z\\right)\\right\|^{2}=\\frac{8\\pi^{2}}{3^{2/3}% }\\int_{0}^{\\infty}\\int_{0}^{2\\pi}\\Re\\left(\\operatorname{Ai}\\left(\\frac{1}{3^{1% /3}}\\left(x+iy+2zu\\exp\\left(i\\theta\\right)+3u^{2}\\exp\\left(-2i\\theta\\right)% \\right)\\right)\\*\\operatorname{Bi}\\left(\\frac{1}{3^{1/3}}\\left(x-iy+2zu\\exp% \\left(-i\\theta\\right)+3u^{2}\\exp\\left(2i\\theta\\right)\\right)\\right)\\right)u\\,% \\mathrm{d}u\\,\\mathrm{d}\\theta.$\n##### 2: Bibliography O\n• I. Olkin (1959) A class of integral identities with matrix argument. Duke Math. J. 26 (2), pp. 207–213.\n• ##### 3: Bibliography W\n• H. S. Wilf and D. Zeilberger (1992a) An algorithmic proof theory for hypergeometric (ordinary and “$q$”) multisum/integral identities. Invent. Math. 108, pp. 575–633.\n• H. S. Wilf and D. Zeilberger (1992b) Rational function certification of multisum/integral/“$q$identities. Bull. Amer. Math. Soc. (N.S.) 27 (1), pp. 148–153.\n• ##### 4: 19.1 Special Notation\n$R_{F}\\left(x,y,z\\right)$, $R_{G}\\left(x,y,z\\right)$, and $R_{J}\\left(x,y,z,p\\right)$ are the symmetric (in $x$, $y$, and $z$) integrals of the first, second, and third kinds; they are complete if exactly one of $x$, $y$, and $z$ is identically 0. …\n##### 5: 1.4 Calculus of One Variable\n1.4.32 $\\\|f\\\|^{2}_{2}\\equiv\\int^{b}_{a}\|f(x)\|^{2}\\,\\mathrm{d}x<\\infty.$\n##### 6: 24.7 Integral Representations\n###### §24.7(i) Bernoulli and Euler Numbers\nThe identities in this subsection hold for $n=1,2,\\dotsc$. …\n##### 7: 35.10 Methods of Computation\nOther methods include numerical quadrature applied to double and multiple integral representations. See Yan (1992) for the ${{}_{1}F_{1}}$ and ${{}_{2}F_{1}}$ functions of matrix argument in the case $m=2$, and Bingham et al. (1992) for Monte Carlo simulation on $\\mathbf{O}(m)$ applied to a generalization of the integral (35.5.8). Koev and Edelman (2006) utilizes combinatorial identities for the zonal polynomials to develop computational algorithms for approximating the series expansion (35.8.1). …\n##### 8: 36.10 Differential Equations\n###### §36.10 Differential Equations\n$K=1$, fold: (36.10.6) is an identity. $K=2$, cusp: … $K=3$, swallowtail: … In terms of the normal forms (36.2.2) and (36.2.3), the $\\Psi^{(\\mathrm{U})}\\left(\\mathbf{x}\\right)$ satisfy the following operator equations …\n##### 9: 25.12 Polylogarithms\n25.12.11 $\\operatorname{Li}_{s}\\left(z\\right)\\equiv\\frac{z}{\\Gamma\\left(s\\right)}\\int_{0% }^{\\infty}\\frac{x^{s-1}}{e^{x}-z}\\,\\mathrm{d}x,$\n##### 10: 35.3 Multivariate Gamma and Beta Functions\n35.3.3 $\\mathrm{B}_{m}\\left(a,b\\right)=\\int\\limits_{\\boldsymbol{{0}}<\\mathbf{X}<% \\mathbf{I}}\\left\|\\mathbf{X}\\right\|^{a-\\frac{1}{2}(m+1)}\\left\|\\mathbf{I}-% \\mathbf{X}\\right\|^{b-\\frac{1}{2}(m+1)}\\,\\mathrm{d}{\\mathbf{X}},$ $\\Re\\left(a\\right),\\Re\\left(b\\right)>\\frac{1}{2}(m-1)$.\n35.3.8 $\\mathrm{B}_{m}\\left(a,b\\right)=\\int_{\\boldsymbol{\\Omega}}\\left\|\\mathbf{X}% \\right\|^{a-\\frac{1}{2}(m+1)}\\left\|\\mathbf{I}+\\mathbf{X}\\right\|^{-(a+b)}\\,% \\mathrm{d}{\\mathbf{X}},$ $\\Re\\left(a\\right),\\Re\\left(b\\right)>\\frac{1}{2}(m-1)$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6366841,"math_prob":0.9999529,"size":1672,"snap":"2023-14-2023-23","text_gpt3_token_len":504,"char_repetition_ratio":0.132494,"word_repetition_ratio":0.0,"special_character_ratio":0.3307416,"punctuation_ratio":0.2347826,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999143,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-31T10:36:09Z\",\"WARC-Record-ID\":\"<urn:uuid:3c8ffadb-6977-45ab-a49f-7e5a0773b38f>\",\"Content-Length\":\"71949\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c9a48532-0ec7-4240-a7fd-9d08a4532609>\",\"WARC-Concurrent-To\":\"<urn:uuid:6399ec4c-ddd2-4193-8ac6-05a88ce21a81>\",\"WARC-IP-Address\":\"129.6.13.19\",\"WARC-Target-URI\":\"https://dlmf.nist.gov/search/search?q=integral%20identity\",\"WARC-Payload-Digest\":\"sha1:P4KPZZRZK663ONW4GOV6APN6DIYAH2NN\",\"WARC-Block-Digest\":\"sha1:QCAWUZMY6WIBN632LY53LEEKK5KKH7SD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224646457.49_warc_CC-MAIN-20230531090221-20230531120221-00355.warc.gz\"}"}
https://www.colorhexa.com/02183b	[ "# #02183b Color Information\n\nIn a RGB color space, hex #02183b is composed of 0.8% red, 9.4% green and 23.1% blue. Whereas in a CMYK color space, it is composed of 96.6% cyan, 59.3% magenta, 0% yellow and 76.9% black. It has a hue angle of 216.8 degrees, a saturation of 93.4% and a lightness of 12%. #02183b color hex could be obtained by blending #043076 with #000000. Closest websafe color is: #000033.\n\n• R 1\n• G 9\n• B 23\nRGB color chart\n• C 97\n• M 59\n• Y 0\n• K 77\nCMYK color chart\n\n#02183b color description : Very dark blue.\n\n# #02183b Color Conversion\n\nThe hexadecimal color #02183b has RGB values of R:2, G:24, B:59 and CMYK values of C:0.97, M:0.59, Y:0, K:0.77. Its decimal value is 137275.\n\nHex triplet RGB Decimal 02183b `#02183b` 2, 24, 59 `rgb(2,24,59)` 0.8, 9.4, 23.1 `rgb(0.8%,9.4%,23.1%)` 97, 59, 0, 77 216.8°, 93.4, 12 `hsl(216.8,93.4%,12%)` 216.8°, 96.6, 23.1 000033 `#000033`\nCIE-LAB 8.839, 7.417, -25.106 1.141, 0.982, 4.267 0.179, 0.154, 0.982 8.839, 26.179, 286.459 8.839, -4.442, -18.398 9.909, 3.212, -18.595 00000010, 00011000, 00111011\n\n# Color Schemes with #02183b\n\n• #02183b\n``#02183b` `rgb(2,24,59)``\n• #3b2502\n``#3b2502` `rgb(59,37,2)``\nComplementary Color\n• #02353b\n``#02353b` `rgb(2,53,59)``\n• #02183b\n``#02183b` `rgb(2,24,59)``\n• #08023b\n``#08023b` `rgb(8,2,59)``\nAnalogous Color\n• #353b02\n``#353b02` `rgb(53,59,2)``\n• #02183b\n``#02183b` `rgb(2,24,59)``\n• #3b0802\n``#3b0802` `rgb(59,8,2)``\nSplit Complementary Color\n• #183b02\n``#183b02` `rgb(24,59,2)``\n• #02183b\n``#02183b` `rgb(2,24,59)``\n• #3b0218\n``#3b0218` `rgb(59,2,24)``\n• #023b25\n``#023b25` `rgb(2,59,37)``\n• #02183b\n``#02183b` `rgb(2,24,59)``\n• #3b0218\n``#3b0218` `rgb(59,2,24)``\n• #3b2502\n``#3b2502` `rgb(59,37,2)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #00040a\n``#00040a` `rgb(0,4,10)``\n• #010e22\n``#010e22` `rgb(1,14,34)``\n• #02183b\n``#02183b` `rgb(2,24,59)``\n• #032254\n``#032254` `rgb(3,34,84)``\n• #042c6c\n``#042c6c` `rgb(4,44,108)``\n• #053685\n``#053685` `rgb(5,54,133)``\nMonochromatic Color\n\n# Alternatives to #02183b\n\nBelow, you can see some colors close to #02183b. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #02263b\n``#02263b` `rgb(2,38,59)``\n• #02223b\n``#02223b` `rgb(2,34,59)``\n• #021d3b\n``#021d3b` `rgb(2,29,59)``\n• #02183b\n``#02183b` `rgb(2,24,59)``\n• #02133b\n``#02133b` `rgb(2,19,59)``\n• #020f3b\n``#020f3b` `rgb(2,15,59)``\n• #020a3b\n``#020a3b` `rgb(2,10,59)``\nSimilar Colors\n\n# #02183b Preview\n\nText with hexadecimal color #02183b\n\nThis text has a font color of #02183b.\n\n``<span style=\"color:#02183b;\">Text here</span>``\n#02183b background color\n\nThis paragraph has a background color of #02183b.\n\n``<p style=\"background-color:#02183b;\">Content here</p>``\n#02183b border color\n\nThis element has a border color of #02183b.\n\n``<div style=\"border:1px solid #02183b;\">Content here</div>``\nCSS codes\n``.text {color:#02183b;}``\n``.background {background-color:#02183b;}``\n``.border {border:1px solid #02183b;}``\n\n# Shades and Tints of #02183b\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000102 is the darkest color, while #eef4fe is the lightest one.\n\n• #000102\n``#000102` `rgb(0,1,2)``\n• #010915\n``#010915` `rgb(1,9,21)``\n• #011028\n``#011028` `rgb(1,16,40)``\n• #02183b\n``#02183b` `rgb(2,24,59)``\n• #03204e\n``#03204e` `rgb(3,32,78)``\n• #032761\n``#032761` `rgb(3,39,97)``\n• #042f74\n``#042f74` `rgb(4,47,116)``\n• #053787\n``#053787` `rgb(5,55,135)``\n• #053f9a\n``#053f9a` `rgb(5,63,154)``\n``#0646ad` `rgb(6,70,173)``\n• #074ec0\n``#074ec0` `rgb(7,78,192)``\n• #0756d3\n``#0756d3` `rgb(7,86,211)``\n• #085de6\n``#085de6` `rgb(8,93,230)``\n• #0a66f7\n``#0a66f7` `rgb(10,102,247)``\n• #1d72f7\n``#1d72f7` `rgb(29,114,247)``\n• #307df8\n``#307df8` `rgb(48,125,248)``\n• #4389f9\n``#4389f9` `rgb(67,137,249)``\n• #5695f9\n``#5695f9` `rgb(86,149,249)``\n• #69a1fa\n``#69a1fa` `rgb(105,161,250)``\n``#7cadfb` `rgb(124,173,251)``\n• #8fb9fb\n``#8fb9fb` `rgb(143,185,251)``\n• #a2c5fc\n``#a2c5fc` `rgb(162,197,252)``\n• #b5d1fc\n``#b5d1fc` `rgb(181,209,252)``\n• #c8ddfd\n``#c8ddfd` `rgb(200,221,253)``\n• #dbe9fe\n``#dbe9fe` `rgb(219,233,254)``\n• #eef4fe\n``#eef4fe` `rgb(238,244,254)``\nTint Color Variation\n\n# Tones of #02183b\n\nA tone is produced by adding gray to any pure hue. In this case, #1e1e1f is the less saturated color, while #02183b is the most saturated one.\n\n• #1e1e1f\n``#1e1e1f` `rgb(30,30,31)``\n• #1c1e21\n``#1c1e21` `rgb(28,30,33)``\n• #191d24\n``#191d24` `rgb(25,29,36)``\n• #171d26\n``#171d26` `rgb(23,29,38)``\n• #151c28\n``#151c28` `rgb(21,28,40)``\n• #121c2b\n``#121c2b` `rgb(18,28,43)``\n• #101b2d\n``#101b2d` `rgb(16,27,45)``\n• #0e1b2f\n``#0e1b2f` `rgb(14,27,47)``\n• #0b1a32\n``#0b1a32` `rgb(11,26,50)``\n• #091a34\n``#091a34` `rgb(9,26,52)``\n• #071936\n``#071936` `rgb(7,25,54)``\n• #041939\n``#041939` `rgb(4,25,57)``\n• #02183b\n``#02183b` `rgb(2,24,59)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #02183b is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.54492664,"math_prob":0.73226076,"size":3635,"snap":"2019-35-2019-39","text_gpt3_token_len":1626,"char_repetition_ratio":0.12751308,"word_repetition_ratio":0.011090573,"special_character_ratio":0.5645117,"punctuation_ratio":0.23608018,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9937092,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-21T00:25:03Z\",\"WARC-Record-ID\":\"<urn:uuid:d7c5a173-ab28-41e7-a51d-1b888df79293>\",\"Content-Length\":\"36149\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2f33ff8d-6c5b-49a3-a756-0b9451e71366>\",\"WARC-Concurrent-To\":\"<urn:uuid:77d9bcbe-a884-43de-a0c3-3f941c7f07bd>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/02183b\",\"WARC-Payload-Digest\":\"sha1:MVIN2YSEOKMWYW7O2G2UYZ7QVCFHKXOR\",\"WARC-Block-Digest\":\"sha1:B6IGSPJ7MUNVSGHWIDGBZIZPIOHAJ5OB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514574159.19_warc_CC-MAIN-20190921001810-20190921023810-00146.warc.gz\"}"}
https://amsi.org.au/ESA_Senior_Years/SeniorTopic3/3h/3h_2content_3.html	[ "## Content\n\n### Derivatives of general exponential functions\n\nSince we can now differentiate $$e^x$$, using our knowledge of differentiation we can also differentiate other functions.\n\nIn particular, we can now differentiate functions of the form $$f(x) = e^{kx}$$, where $$k$$ is a real constant. From the chain rule, we obtain\n\n$f'(x) = k \\, e^{kx}.$\n\nWe saw in the previous section, when differentiating $$2^x$$, that it can be written as $$e^{\\log_e 2 \\cdot x}$$, which is of the form $$e^{kx}$$. The same technique can be used to differentiate any function $$a^x$$, where $$a$$ is a positive real number. A function $$a^x$$ is just a function of the form $$e^{kx}$$ in disguise.\n\nWriting $$a$$ as $$e^{\\log_e a}$$, we can rewrite $$f(x) = a^x$$, using the index laws, as\n\n$f(x) = a^x = \\bigl( e^{\\log_e a} \\bigr)^x = e^{x \\, \\log_e a}.$\n\nThe function is then in the form $$e^{kx}$$ (with $$k=\\log_e a$$) and differentiating gives\n\n$f'(x) = \\log_e a \\cdot a^x.$\n\nSo the derivative of $$a^x$$ is a constant times itself, and that constant is $$\\log_e a$$.\n\nExercise 8\n\nUse what we've done so far to explain why, for any $$a>0$$,\n\n$\\lim_{h \\to 0} \\dfrac{a^h - 1}{h} = \\log_e a.$\n\nExercise 9\n\nLet $$f(x) = x^x$$, for $$x>0$$. Differentiate $$f(x)$$ and find its stationary points.\n\nNext page - Content - Derivatives of general logarithmic functions" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79822016,"math_prob":1.0000087,"size":1338,"snap":"2020-10-2020-16","text_gpt3_token_len":432,"char_repetition_ratio":0.15742129,"word_repetition_ratio":0.0,"special_character_ratio":0.3497758,"punctuation_ratio":0.107407406,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000099,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-02T20:29:41Z\",\"WARC-Record-ID\":\"<urn:uuid:ce4ec9e4-4825-479a-a9ec-2145beac4579>\",\"Content-Length\":\"3932\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2c74b089-a5cb-4ffb-a259-22cb360eef98>\",\"WARC-Concurrent-To\":\"<urn:uuid:5520311c-fc1c-4298-8e2c-5531ec5e94c9>\",\"WARC-IP-Address\":\"101.0.91.74\",\"WARC-Target-URI\":\"https://amsi.org.au/ESA_Senior_Years/SeniorTopic3/3h/3h_2content_3.html\",\"WARC-Payload-Digest\":\"sha1:HJD2H3DYNJ5N6FHFCNRB6722T6DH7TL7\",\"WARC-Block-Digest\":\"sha1:W66LFO2I2475WZCR5UPBNWQWBVR5NT2M\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370507738.45_warc_CC-MAIN-20200402173940-20200402203940-00362.warc.gz\"}"}
https://stacks.math.columbia.edu/tag/0F0Z	[ "Definition 59.96.1. Let $f : X \\to Y$ be a quasi-compact and quasi-separated morphism of schemes. The cohomological dimension of $f$ is the smallest element\n\n$\\text{cd}(f) \\in \\{ 0, 1, 2, \\ldots \\} \\cup \\{ \\infty \\}$\n\nsuch that for any abelian torsion sheaf $\\mathcal{F}$ on $X_{\\acute{e}tale}$ we have $R^ if_*\\mathcal{F} = 0$ for $i > \\text{cd}(f)$.\n\nIn your comment you can use Markdown and LaTeX style mathematics (enclose it like $\\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar)." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.565484,"math_prob":0.9986951,"size":366,"snap":"2022-27-2022-33","text_gpt3_token_len":142,"char_repetition_ratio":0.09944751,"word_repetition_ratio":0.0,"special_character_ratio":0.37978142,"punctuation_ratio":0.12,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99774444,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-26T10:50:33Z\",\"WARC-Record-ID\":\"<urn:uuid:abdae747-4e58-478e-af3d-106ac518413c>\",\"Content-Length\":\"13735\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8c086036-c065-4cb9-96d5-c2878bebd570>\",\"WARC-Concurrent-To\":\"<urn:uuid:c8cd752a-042a-4bf7-a58a-9ede85435450>\",\"WARC-IP-Address\":\"128.59.222.85\",\"WARC-Target-URI\":\"https://stacks.math.columbia.edu/tag/0F0Z\",\"WARC-Payload-Digest\":\"sha1:IDVWJQQ3YIBIJBSX6V7NAJSSRE6O7QJ2\",\"WARC-Block-Digest\":\"sha1:I52Q2UQLUAAR4GJIRGWIVDEPKCEH5IMP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103205617.12_warc_CC-MAIN-20220626101442-20220626131442-00071.warc.gz\"}"}
https://www.impan.pl/en/publishing-house/journals-and-series/studia-mathematica/all/255/2/113292/a-sharp-form-of-the-marcinkiewicz-interpolation-theorem-for-orlicz-spaces	[ "# Publishing house / Journals and Serials / Studia Mathematica / All issues\n\n## Studia Mathematica\n\nPDF files of articles are only available for institutions which have paid for the online version upon signing an Institutional User License.\n\n## A sharp form of the Marcinkiewicz interpolation theorem for Orlicz spaces\n\n### Volume 255 / 2020\n\nStudia Mathematica 255 (2020), 109-158 MSC: Primary 46E30; Secondary 46M35. DOI: 10.4064/sm180111-28-10 Published online: 13 May 2020\n\n#### Abstract\n\nLet $(X,\\mu )$ and $(Y,\\nu )$ be $\\sigma$-finite measure spaces with $\\mu (X)=\\nu (Y)=\\infty$ and $(Y,\\nu )$ nonatomic and separable. Suppose $T$ is a so-called $r$-quasilinear operator mapping the simple functions on $X$ into the measurable functions on $Y$ that satisfies the weak type conditions $$\\lambda \\nu ( \\lbrace y \\in Y : \|(Tf)(y)\| \\gt \\lambda \\rbrace )^{{1}/{p_{i}}} \\leq C_{p_{i},q_{i}} \\Big ( \\int _{\\mathbb {R_+}} \\mu ( \\lbrace x \\in X : \|f(x)\| \\gt t \\rbrace )^{{q_{i}}/{p_{i}}} t^{q_{i}-1}\\,dt \\Big )^{{1}/{q_{i}}},\\quad i=0,1,$$ where $1 \\lt p_0 \\lt p_1 \\lt \\infty$, $1 \\leq q_0, q_1 \\lt \\infty$ and $C_{p_{i},q_{i}}=C_{p_{i},q_{i}}(T) \\gt 0$ is independent of simple $f$ on $X$ and $\\lambda \\gt 0$.\n\nWe give necessary and sufficient conditions on Young functions $\\Phi _1$ and $\\Phi _2$ in order that any operator $T$ as described above is bounded between the corresponding Orlicz spaces.\n\n#### Authors\n\n• Ron KermanDepartment of Mathematics\nBrock University\nSt. Catharines, Ontario, L2S 3A1, Canada\ne-mail\n• Rama RawatDepartment of Mathematics and Statistics\nIndian Institute of Technology\nKanpur 208016, India\ne-mail\n• Rajesh K. SinghDepartment of Mathematics and Statistics\nIndian Institute of Technology\nKanpur 208016, India\ne-mail\n\n## Search for IMPAN publications\n\nQuery phrase too short. Type at least 4 characters.\n\n## Rewrite code from the image", null, "" ]	[ null, "https://www.impan.pl/en/publishing-house/journals-and-series/studia-mathematica/all/255/2/113292/a-sharp-form-of-the-marcinkiewicz-interpolation-theorem-for-orlicz-spaces", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6728344,"math_prob":0.989661,"size":1012,"snap":"2021-43-2021-49","text_gpt3_token_len":368,"char_repetition_ratio":0.11210317,"word_repetition_ratio":0.0,"special_character_ratio":0.3922925,"punctuation_ratio":0.08374384,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9982719,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-01T13:12:42Z\",\"WARC-Record-ID\":\"<urn:uuid:2379c5a1-0cb2-4fa1-8ccf-9768e2aec76a>\",\"Content-Length\":\"47450\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3261c137-5a0b-4243-b609-d1e961d9a931>\",\"WARC-Concurrent-To\":\"<urn:uuid:b8ad19aa-fb0b-450d-b247-1504554158c5>\",\"WARC-IP-Address\":\"195.187.71.110\",\"WARC-Target-URI\":\"https://www.impan.pl/en/publishing-house/journals-and-series/studia-mathematica/all/255/2/113292/a-sharp-form-of-the-marcinkiewicz-interpolation-theorem-for-orlicz-spaces\",\"WARC-Payload-Digest\":\"sha1:7ZYVJPCHNHKAGD4OYK3EJUGR3PTDZEZW\",\"WARC-Block-Digest\":\"sha1:SL2MUXSWRKFIVD7NLFBIYIS2CFES35J2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964360803.0_warc_CC-MAIN-20211201113241-20211201143241-00060.warc.gz\"}"}
https://www.barcodefaq.com/knowledge-base/java-byte-array/	[ "# Pass a Byte Array into Java Components\n\nYou are here:\n• Pass a Byte Array into Java Components\n\nThe following error may occur when passing a byte array directly into the DataToEncode method that expects a string value when using the Java Barcode Components:\njava.lang.ArrayIndexOutOfBoundsException\n\n### Solution(s):\n\nThe solutions differ depending on the symbology:\n\n• PDF417: This is only possible with IDAutomation PDF417 Java Products V 4.10 dated October, 2004 or later. When using PDF417, pass the byte array to the binaryCode field of the class library. Make sure to set the processTilde to false and set PDFMode to 0.\nFor example:\nPDF417 bc=new PDF417();\nbc.processTilde=false;\nbc.PDFMode=0;\nbc.binaryCode=myByteArray;\n• Data Matrix: When using Data Matrix barcodes, use the code example below to convert the byte array (encodedBytes) into a string that can be encoded in the code field. Make sure to set the encoding to E_BASE256 and processTilde to false.\nDataMatrix bc = new DataMatrix();\nbc.encoding = DataMatrix.E_BASE256;\nbc.processTilde=false;\nbc.code=””;\nString binaryString;\nfor (int i=0; i < encodedBytes.length; i++)\n{\nbinaryString = Integer.toBinaryString((int)encodedBytes[i]);\nif (binaryString.length() > 8)\nbinaryString = binaryString.substring(binaryString.length()-8,binaryString.length());\nbc.code = bc.code + (char) Integer.parseInt(binaryString,2);\n}" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.530933,"math_prob":0.6060683,"size":1406,"snap":"2020-34-2020-40","text_gpt3_token_len":327,"char_repetition_ratio":0.13623396,"word_repetition_ratio":0.06417112,"special_character_ratio":0.23541963,"punctuation_ratio":0.1870229,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97964823,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-27T17:59:43Z\",\"WARC-Record-ID\":\"<urn:uuid:818d85e5-346e-40b1-a57f-b308cc566b3e>\",\"Content-Length\":\"75169\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:99fe7f8d-70be-4a82-b316-a8d1ade1af5c>\",\"WARC-Concurrent-To\":\"<urn:uuid:97697a69-b332-47f2-ba84-7558ce1476b2>\",\"WARC-IP-Address\":\"35.209.27.178\",\"WARC-Target-URI\":\"https://www.barcodefaq.com/knowledge-base/java-byte-array/\",\"WARC-Payload-Digest\":\"sha1:H45574HC3ILHIIKCYJALA3NLPQTZ2AKT\",\"WARC-Block-Digest\":\"sha1:GJEJ5BZAVA44I5TAT2MPJI4SFBLUYDAP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400283990.75_warc_CC-MAIN-20200927152349-20200927182349-00704.warc.gz\"}"}
https://hiddenherstories.org/random-sampling-worksheet/	[ "# Random Sampling Worksheet\n\n## List of Random Sampling Worksheet\n\n### 1. Worksheet Random Sampling", null, "### 2. Statistics Vocabulary Matching Activity Worksheet Assessment Random Sampling", null, "### 3. Stratified Sampling Definition Formula Calculation Random Worksheet", null, "### 4. Solved Purpose Assignment Random Sampling Worksheet", null, "### 5. Simulation Real Statistics Excel Random Sampling Worksheet", null, "### 6. Simple Random Sample Excel Sampling Worksheet", null, "### 7. Sampling Worksheet Printable Worksheets Activities Teachers Parents Tutors Families Random", null, "### 8. Sampling Techniques Geography Video Lesson Transcript Random Worksheet", null, "### 9. Topic Investigating Ecosystems World Science Green Random Sampling Worksheet", null, "### 10. Sampling Summarizing Big Data Solver Random Worksheet", null, "### 11. Unit 4 Tests Significance Random Sampling Worksheet", null, "### 12. Applet Demonstrate Sampling Distribution Random Worksheet", null, "### 13. Random Sampling Biology Junction Worksheet", null, "### 14. Week 3 Practice Problem Worksheet Random Sampling", null, "### 15. Procedures Real Statistics Excel Random Sampling Worksheet", null, "### 16. Population Density Random Sampling Mark Recapture Populations Worksheets Worksheet", null, "### 17. Multiplication Facts Worksheets Picture Ideas Vertical Free Sample Worksheet Random Sampling", null, "### 18. Methods Sampling Stem Random Worksheet", null, "### 19. Matching Random Sampling Worksheet", null, "### 20. Lesson Types Sampling Random Worksheet", null, "### 21. Intro Statistics Excel Random Sampling Worksheet", null, "### 22. Grade Skills Math Helping Random Sampling Worksheet", null, "### 23. Worksheet Classroom Worksheets Kindergarten Kids Printouts Random Sampling Preschoolers Fantastic Image Inspirations Free Math Printable", null, "### 24. Grade 4 Resources Printable Worksheets Topic Random Topics Essays Descriptive Story Writing Lets Share Knowledge Sampling Worksheet", null, "### 25. Grade 3 Vocabulary Worksheets Week Lets Share Knowledge Random Sampling Worksheet", null, "### 26. Free Letter Worksheets Den Random Sampling Worksheet", null, "### 27. Fraction Decimal Task Cards Worksheets Bundle Random Sampling Worksheet", null, "### 28. Worksheet Generator Random Sampling", null, "### 29. Big Bang Law Worksheet Grade Lesson Planet Random Sampling", null, "### 30. Auditory Perception Worksheets Preschool Number 1 Math Fraction Practice Kindergarten Workbook Middle Grades Search Shade Random Multiplication Generator Sampling Worksheet", null, "### 31. Stat Random Sampling Worksheet", null, "### 32. Systematic Random Sampling Worksheet", null, "### 33. 1 Chapter 7 Sampling Distributions Simple Random Point Estimation Introduction Distribution Download Worksheet", null, "### 34. Solved Worksheet Identify Obs Random Sampling", null, "### 35. Solved Worksheet 2 Sampling Techniques Random", null, "### 36. Sampling Surveys Random Worksheet", null, "### 37. Samples Surveys Worksheets Maths Random Sampling Worksheet", null, "### 38. Grade 7 Common Core Math Worksheets Random Sampling Worksheet", null, "### 39. Ecology Measuring Abundance Random Sampling Footpath Stem Worksheet", null, "### 40. Random Math Problem Generator Minute Addition Worksheets Grade Long Division Sentence Patterns Solving Basic Test Funny Sampling Worksheet", null, "### 41. Sample High School Grammar Exam Worksheets Double Digit Addition Problems Unit Rate Math Free Color Numbers Kids Crafts Preschoolers Random Sampling Worksheet", null, "" ]	[ null, "https://hiddenherstories.org/images/worksheet-random-sampling.jpg", null, "https://hiddenherstories.org/images/statistics-vocabulary-matching-activity-worksheet-assessment-random-sampling.jpg", null, "https://hiddenherstories.org/images/stratified-sampling-definition-formula-calculation-random-worksheet.jpg", null, "https://hiddenherstories.org/images/solved-purpose-assignment-random-sampling-worksheet.jpg", null, "https://hiddenherstories.org/images/simulation-real-statistics-excel-random-sampling-worksheet.jpg", null, "https://hiddenherstories.org/images/simple-random-sample-excel-sampling-worksheet.jpg", null, "https://hiddenherstories.org/images/sampling-worksheet-printable-worksheets-activities-teachers-parents-tutors-families-random.jpg", null, 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https://elearn-rt-en.gunt.de/3-controllers-and-controller-types/3-1-controller-transmission-behaviour	[ "# 3. Controllers and controller types\n\n## 3.1 Controller transmission behaviour", null, "Block diagram of a PID controller\n\nJust like the controlled system, the controller shows a specific transmission behaviour. Unlike the controlled system, here the transmission behaviour can be deliberately adjusted according to the desired function.\n\nThe manipulating variable y from the frequently used PID controller can be described as the sum of the output signals from different transmission blocks each with a different time response.\n\nThe same input signal is present at the inputs of all transmission blocks, namely the control difference e.", null, "Step responses of the PID controller:\nP component\n\nThe transmission behaviour of the PID controller is defined by the following parameters:\n\n• control gain Kp (proportional behaviour)", null, "Step responses of the PID controller:\nI component\n\n• reset time Ti (integral behaviour)", null, "Step responses of the PID controller:\nD component\n\n• rate time Td (derivative behaviour)\n\nDepending on the parameter settings, the controller can demonstrate P, PI, PD or PID behaviour.\n\nThe properties of the different controller types are set out below.", null, "In a unity feedback configuration, an ideal PID compensator is used to enhance the step response of a system with the transfer function", null, "The PID controller has the standard form Kp+Ki/s+Kds. Therefore, the closed loop transfer function of the whole system becomes", null, "As the proportional, integral and derivative terms of the PID compensator are increased, the evolution of the step response, from the initial uncompensated response (corresponding to PID coefficents Kp=1, Ki=0, Kd=0) to the final desired response is depicted. The following effects of PID compensation can be readily observed:\n\n• The proportional term increases the speed of the system. It also decreases the residual steady state error of the step response, but can not eliminate it completely.\n\n• The integral term eliminates the residual steady state error of the step response, but adds undesired \"oscillations\" to the transient response (overshoot).\n\n• The derivative term \"damps out\" the undesired oscillations in the transient response." ]	[ null, "https://lh3.googleusercontent.com/mp_wOHlKXc6w-752_82zGVKzERwdYTNhPEMVYGWMkO0KY3cMTIGqBgfxbbu06vn6lhFrvKuO6erADaYoiHT-u7fI22NnWp1SP1ryU2NsfqpGPqfqYMlf15aCshV51W3dJw=w1280", null, "https://lh6.googleusercontent.com/N3VQKp9AJgyGEwq-FJhwxLGWzG7VpAB9vsf5RjGfWOSPQAUmO0VmS_KYW9Ca3p8rQJ52WQDNcMhQxHr7Nnjj1RYnOpYWBLSV0R6GKXadA7Bj3kXK3hy3zgl4Zbx4RYvRmg=w1280", null, "https://lh4.googleusercontent.com/ZSHaPICw81c1MMKT06FWSaJg1L-tVFERVEfM2jpg50jSNg_5kazCobhvqbOG9U6twR8aPVpHXUetUh8sTM7vz9jXHsaKRTFRB6r5i32bnmOdeiFAj1mMKLfSXQ3cN2s=w1280", null, "https://lh4.googleusercontent.com/Ujd2v6gwgwo4rE93sSk-4u0eId4cEw2ZfiOBMQ5fwI-JIQyvUStZhPGr5g9JTocD0aqR41krtc6o8Ab_5qO91GnTFB_-uI4fyX_dy_n2VbWl0Z3AzE9lFDCqTxvFTBvIxQ=w1280", null, "https://lh3.googleusercontent.com/SPicaJs57EcRPpFwLMzJfAVdvQs9VlEaiT2HRfr8nMm9fYMl-86gkoAXwIr0CcTiU2moRo5yWFn1zUID-i8kh4Hv7bV9k2gcM_eIM-9QNZEhAESJI-FRS97VlVw8ZVOp9A=w1280", null, "https://lh3.googleusercontent.com/apWK69y91Kf6YNSCQZU7HW4WtFq6rnde2og_ki8Xy9GcVune_ZycmmmpQ769sG6ezjjpJZcOUIFKZiqA4qa2L-LmY7K9hnmbybNd5jQO1uxNDN9GG5uAxCQ7d9-IRU5LSg=w1280", null, "https://lh6.googleusercontent.com/92qef_mdEzFqqkODbu_Xa6sbNXdIzlzw8FtwcTpWXMCYdioNQ_qB-4LhVwIH2SymV2opR514uvn2WOf6OEW-ibY-HrFCyIAzdxq53P7sAMv0nBs9yK2EBkvONG7klfaBmw=w1280", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85009795,"math_prob":0.5475514,"size":2095,"snap":"2022-27-2022-33","text_gpt3_token_len":405,"char_repetition_ratio":0.1860354,"word_repetition_ratio":0.063897766,"special_character_ratio":0.17804296,"punctuation_ratio":0.09269663,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96191204,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-16T14:03:37Z\",\"WARC-Record-ID\":\"<urn:uuid:aff27531-a463-43f3-a1e8-3f3808350a95>\",\"Content-Length\":\"87412\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:19b4f50c-389f-4707-b9e7-2b9872838ff7>\",\"WARC-Concurrent-To\":\"<urn:uuid:01e12857-2d25-432f-8f11-17ee7d32a17e>\",\"WARC-IP-Address\":\"172.217.2.115\",\"WARC-Target-URI\":\"https://elearn-rt-en.gunt.de/3-controllers-and-controller-types/3-1-controller-transmission-behaviour\",\"WARC-Payload-Digest\":\"sha1:TOYI2BD2H7E442EKRVTX4PDIXSYJ7PF5\",\"WARC-Block-Digest\":\"sha1:TTR7QQ2I5QKNMHKOHSVRAKZJATFB75US\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572304.13_warc_CC-MAIN-20220816120802-20220816150802-00666.warc.gz\"}"}
https://pages.vassar.edu/magnes/final-data/	[ "# Final Data\n\nIn this post I will present my actual data findings with minimal interpretation. In the next blog post, “Conclusion,” I will interpret the data I collected and compare it to expected values.\n\n### Preliminary Data Recap:\n\nIn my previous post I had measured the discharge of several different capacitor values in series with an inductor.\n\nBelow were the resulting plots of Voltage vs Time (Figures 1-3):\n\nFor the 1465, 1007, and 47 nF capacitors the measured frequency of oscillation was 128, 155, and 730 Hz, respectively.\n\n### Voltage As a Function Of Time and Frequency of Oscillation\n\nIt is helpful to layer these three plots on top of each other (Figure 4). In doing so we can readily compare the curves to each other.\n\nFrom Figure 4 it is evident that the higher the capacitor value, the lower the frequency of oscillation. This is not surprising, given the equation that the frequency and inductor*capacitor product are inversely related:", null, ".\n\nHow are frequency and capacitance, or frequency and inductance related? I recorded the frequency of oscillation in several different LC circuits to try and illustrate this relationship. For three different inductor values I discharged five different capacitor values. The resulting plot (Figure 5) is below:\n\nBecause it is harder to see the shape of the blue line in Figure 5, I plotted it separately (Figure 6):\n\nFigures 5 and 6 experimentally illustrate the inverse square root relationship between frequency and capacitance/inductance.\n\nI then plotted the frequency of oscillation with respect to inductor value (Figure 7):\n\n### Exponential Rate Of Decay\n\nIf you take a look at Figure 4 again, notice that despite changing capacitor values, the exponential decay of each curve is relatively the same. The decay for each curve is the same because the decay coefficient β is a function of resistance and inductance, not capacitance(", null, "). In Equation (1) you can see the decay term of the voltage, e^(-βt):", null, "(1)\n\nIn order to measure the decay constant I used Origin Pro to fit an exponential curve to the peaks of the voltage plots. The formula used for this function fit was as follows:", null, "Where R0 is the decay coefficient, -β\n\nFigure 8 below is an example of this function fit:", null, "Figure 8\n\nI fit an exponential function to the 996 mH and 1007 nF, 47 nF oscillations as well, resulting in the table below (Figure 9):\n\nTheoretically the β value should be the same for all of them.\n\nIn order to explore how the decay coefficient β changes I needed to change the inductor values. Because the voltage curves were more manageable at greater LC values, I used the largest capacitor I had and varied the inductor value. The results were as follows (Figures 10-11):\n\nThe resulting decay coefficients are as follows (Figure 12):\n\nWhich can also be represented in as a scatter plot, despite having so few points(Figure 13):" ]	[ null, "https://pages.vassar.edu/magnes/wp-content/ql-cache/quicklatex.com-3846befc02e30525946dcbd26531f39a_l3.png", null, "https://pages.vassar.edu/magnes/wp-content/ql-cache/quicklatex.com-743184c97ab219269c0c3f3fd7cbd0d0_l3.png", null, "https://pages.vassar.edu/magnes/wp-content/ql-cache/quicklatex.com-a548b249cf71c4644ad2c71d2ec0c460_l3.png", null, "http://pages.vassar.edu/magnes/files/2014/05/gif.latex_.gif", null, "http://pages.vassar.edu/magnes/files/2014/05/996-mH-exp-fit.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9145097,"math_prob":0.9460685,"size":2812,"snap":"2023-40-2023-50","text_gpt3_token_len":623,"char_repetition_ratio":0.14280626,"word_repetition_ratio":0.0,"special_character_ratio":0.21763869,"punctuation_ratio":0.096045196,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994136,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T21:28:27Z\",\"WARC-Record-ID\":\"<urn:uuid:a1f4cb1e-51eb-4823-80f7-086d9cd0624d>\",\"Content-Length\":\"61067\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a9463bad-2459-454d-9fc7-ddded6d61d4f>\",\"WARC-Concurrent-To\":\"<urn:uuid:4c7027b1-7384-4802-9f81-b5a6ba9ca24b>\",\"WARC-IP-Address\":\"143.229.0.73\",\"WARC-Target-URI\":\"https://pages.vassar.edu/magnes/final-data/\",\"WARC-Payload-Digest\":\"sha1:F5NVG4AEEVAND4CDQIIZRA63W75YVGDU\",\"WARC-Block-Digest\":\"sha1:6XYDZRVJ5HH3RSYMDVYBHICAJHU2NPL4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510454.60_warc_CC-MAIN-20230928194838-20230928224838-00145.warc.gz\"}"}
https://www.intlpress.com/site/pub/pages/journals/items/cag/content/vols/0024/0002/a007/index.html	[ "# Communications in Analysis and Geometry\n\n## Volume 24 (2016)\n\n### Compact embedded minimal surfaces in $\\mathbb{S}^2 \\times \\mathbb{S}^1$\n\nPages: 409 – 429\n\nDOI: http://dx.doi.org/10.4310/CAG.2016.v24.n2.a7\n\n#### Authors\n\nJosé M. Manzano (Department of Mathematics, King’s College London, The Strand, London, United Kingdom)\n\nJulia Plehnert (Discrete Differential Geometry Lab, University of Göttingen, Germany)\n\nFrancisco Torralbo (Departamento Ciencias Básicas, Centro Universitario de la Defensa (CUD) San Javier, Santiago de la Rivera, Spain)\n\n#### Abstract\n\nWe prove that closed surfaces of all topological types, except for the non-orientable odd-genus ones, can be minimally embedded in $\\mathbb{S}^2 \\times \\mathbb{S}^1 r$, for arbitrary radius $r$. We illustrate it by obtaining some periodic minimal surfaces in $\\mathbb{S}^2 \\times \\mathbb{R}$ via conjugate constructions. The resulting surfaces can be seen as the analogy to the Schwarz P-surface in these homogeneous $3$-manifolds.\n\n#### 2010 Mathematics Subject Classification\n\nPrimary 53A10. Secondary 53C30.\n\nFull Text (PDF format)\n\nPublished 14 June 2016" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.70207983,"math_prob":0.4367388,"size":764,"snap":"2019-13-2019-22","text_gpt3_token_len":219,"char_repetition_ratio":0.13421053,"word_repetition_ratio":0.020618556,"special_character_ratio":0.2827225,"punctuation_ratio":0.12587413,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9622488,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-21T20:09:45Z\",\"WARC-Record-ID\":\"<urn:uuid:6625db04-0dd4-4b1a-8fbb-12200aa16b72>\",\"Content-Length\":\"8831\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:74fa24e0-7baf-4ba6-a403-41cee5459cde>\",\"WARC-Concurrent-To\":\"<urn:uuid:d8077a0b-11ef-4004-ac9c-0156985c3381>\",\"WARC-IP-Address\":\"107.180.41.239\",\"WARC-Target-URI\":\"https://www.intlpress.com/site/pub/pages/journals/items/cag/content/vols/0024/0002/a007/index.html\",\"WARC-Payload-Digest\":\"sha1:EM4XCA3OZWW6RM4EIOTQI4WP3Z4E5CAI\",\"WARC-Block-Digest\":\"sha1:QZUN7PBJWDOVSSM7IE3RIVIWGKVFJR4I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202572.29_warc_CC-MAIN-20190321193403-20190321215403-00327.warc.gz\"}"}
http://slideplayer.com/slide/5297199/	[ "", null, "Volume of Rectangular Prism and Cylinder Grade 6.\n\nPresentation on theme: \"Volume of Rectangular Prism and Cylinder Grade 6.\"— Presentation transcript:\n\nVolume of Rectangular Prism and Cylinder Grade 6\n\nConfidential 2 Warm Up Find the Surface area of each rectangular Prism to the nearest tenth 5.5cm 2.3cm 10cm 1. 2. 7.8 in. 6.2 in. 5.4 in.\n\nConfidential 3 Warm Up Find the surface area of Cylinders: 4. r = 6mm, h = 10 mm 5. r = 3.5 ft, h = 24 ft 3. Write the formula for the surface area of a Cube in which each edge measures x units.\n\nConfidential 4 Surface Area of a Prism The surface area of a prism is the sum of the areas of all the sides of the prism. The formula for the surface area of a prism therefore depends on the type of prism. Surface Area of a Cylinder The surface area of a cylinder is the sum of the areas of the two bases and the lateral face of the cylinder. surface area of a cylinder = 2*  r 2 + 2  rh Lets review what we have learned in the last lesson\n\nConfidential 5 Types of prism The name of a prism depends upon its base polygons. If the bases are triangles, then it is a TRIANGULAR prism. A RECTANGULAR prism has bases which are rectangles. The other types of prisms are pentagonal prism, hexagonal prism and octagonal prism. Surface Area of prism: Surface Area of any prism = Lateral area + Area of two ends\n\nConfidential 6 What is volume?  The amount of space occupied by an object is called its volume.  Cubic units are used to measure the volume. For example m 3, cm 3, in 3, ft 3 etc.\n\nConfidential 7 Volume of Rectangular Prism Prism is a 3D figure. The volume of a rectangular prism is given by the formula: Volume = l × b × h Where l is the length of the rectangular prism. b is the width of the rectangular prism. h is the height of the rectangular prism.\n\nConfidential 8 Practice Example 1: Find the volume of the rectangular prism whose length is 12 cm, width 6 cm and height 8cm. Solution: Volume = l × b × h Volume = 12 × 6 × 8 Volume = 576 cm 3 The volume of the rectangular prism is 576 cm 3. Volume = l x b x h\n\nConfidential 9 Practice Example 2: Find the volume of the rectangular prism whose length is 6 in, width 6 in and height 4 in. Solution: Volume = l × b × h Volume = 6 × 6 × 4 Volume = 144 in 3 The volume of the rectangular prism is 144 in 3. Volume = l x b x h\n\nConfidential 10 Volume of Cylinder Cylinder is a 3D figure. The volume of a cylinder is given by the formula: Volume = r 2 h Where r is the radius of the base of the cylinder. h is the height of the rectangular prism. and pi = 3.14.\n\nConfidential 11 Practice Example 3: Find the volume of the cylinder whose radius is 10 cm and height 6cm. Solution: Volume = r 2 h Volume = 3.14 × 10 2 × 6 Volume = 1884 cm 3 The volume of the cylinder is 1884 cm 3. Volume = r 2 h\n\nConfidential 12 Practice Example 4: Find the volume of the cylinder whose diameter is 22 cm and height 14 cm. Solution: Radius r = Diameter/2 = 22/2 = 11 cm Volume = r 2 h Volume = 3.14 × 11 2 × 14 Volume = 5319.6 cm 3 The volume of the cylinder is 5319.6 cm 3. Volume = r 2 h\n\nConfidential 13\n\nConfidential 15 Your Turn Find the volume of the rectangular prism whose: 1. l=5 cm, b = 3 cm, h = 4 cm 2. l=4 in, b = 3 in, h = 5 in 3. l=12 cm, b = 2 cm, h = 7cm 4. l=10 m, b = 5 m, h = 5 m 5. l=5 km, b = 5 km, h = 11km\n\nConfidential 16 Your Turn Find the volume of the cylinder whose: 6. r = 5 cm, h = 4 cm 7. r = 10 in, h = 2 in 8. r = 1 km, h = 5 km 9. r = 1 m, h = 1 m 10. r = 2 cm, h = 2 cm\n\nConfidential 17 Q1: Find the length of the rectangular prism whose area is 980 cm 3, width is 10cm and height is 7 cm.\n\nConfidential 18 Q2: Find the height of the cylinder whose area is 314 cm 3, and radius is 10 cm.\n\nConfidential 19 Q3: A match box measures 4cm by 2 cm by 2 cm. What will be the volume of the a packet containing a maximum of 12 such boxes?\n\nConfidential 20 Let Us Review  The amount of space occupied by an object is called its volume.  The volume of a rectangular prism is given by the formula: Volume = length × width × height  The volume of a cylinder is given by the formula: Volume = π (radius) 2 x height\n\nConfidential 21 You Had a Great Lesson Today! Remember to practice what you have learned today\n\nDownload ppt \"Volume of Rectangular Prism and Cylinder Grade 6.\"\n\nSimilar presentations" ]	[ null, "http://slideplayer.com/static/blue_design/img/slide-loader4.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8699626,"math_prob":0.9964687,"size":4266,"snap":"2019-43-2019-47","text_gpt3_token_len":1290,"char_repetition_ratio":0.2114031,"word_repetition_ratio":0.19146608,"special_character_ratio":0.31973746,"punctuation_ratio":0.115188584,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99954855,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-18T06:38:29Z\",\"WARC-Record-ID\":\"<urn:uuid:2f61a23b-8adb-4074-85f7-9160003578ae>\",\"Content-Length\":\"160867\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:43dc7b14-16fa-48f8-b3c4-00df81567425>\",\"WARC-Concurrent-To\":\"<urn:uuid:8ceffc7b-dea4-4a41-88e3-879170a6bb4e>\",\"WARC-IP-Address\":\"144.76.166.55\",\"WARC-Target-URI\":\"http://slideplayer.com/slide/5297199/\",\"WARC-Payload-Digest\":\"sha1:TVT2MIKRZQPWT3M54KEH5WVWIORCN5TI\",\"WARC-Block-Digest\":\"sha1:VUAI53TSDTG62FZXDPCCFXDFA2IHFNEV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986677964.40_warc_CC-MAIN-20191018055014-20191018082514-00082.warc.gz\"}"}
https://connectedmentor.com/qa/question-is-there-a-prime-number-that-is-even.html	[ "", null, "# Question: Is There A Prime Number That Is Even?\n\n## Which are not twin primes?\n\nProperties of Twin Primes(2,3) are not considered as twin primes, since there is no composite number in between them and the difference between the two primes is not equal to 2.5 is an only prime number which is available in two different pairs.More items….\n\n## What is the smallest odd prime number?\n\n33 is the smallest odd prime number.\n\n## Why is 2 an even number?\n\nHow is 2 an even number? An even number is any integer (a whole number) that can be divided by 2. If you can cleanly divide a number by 2, the number is even. … That means the number is even, as it can cleanly be divided by 2, with no remaining remainder.\n\n## Why is 11 not a prime number?\n\nFor 11, the answer is: yes, 11 is a prime number because it has only two distinct divisors: 1 and itself (11). As a consequence, 11 is only a multiple of 1 and 11.\n\n## Why can’t even numbers be prime numbers?\n\nTwo is a prime because it is divisible by only two and one. All the other even numbers are not prime because they are all divisible by two. That leaves only the odd numbers.\n\n## What is the HCF of two Coprime numbers?\n\n1The HCF of two coprime numbers is always 1 .\n\n## What is 1 called if it is not a prime?\n\nA natural number greater than 1 that is not prime is called a composite number. For example, 5 is prime because the only ways of writing it as a product, 1 × 5 or 5 × 1, involve 5 itself. However, 4 is composite because it is a product (2 × 2) in which both numbers are smaller than 4.\n\n## Is 2 prime or even?\n\nProof: The definition of a prime number is a positive integer that has exactly two distinct divisors. Since the divisors of 2 are 1 and 2, there are exactly two distinct divisors, so 2 is prime. … In fact, the only reason why most even numbers are composite is that they are divisible by 2 (a prime) by definition.\n\n## Is 2 and 3 twin primes?\n\nUsually the pair (2, 3) is not considered to be a pair of twin primes. … The first few twin prime pairs are: (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73), (101, 103), (107, 109), (137, 139), … OEIS: A077800.\n\n## Why is 28 the perfect number?\n\nA number is perfect if all of its factors, including 1 but excluding itself, perfectly add up to the number you began with. 6, for example, is perfect, because its factors — 3, 2, and 1 — all sum up to 6. 28 is perfect too: 14, 7, 4, 2, and 1 add up to 28.\n\n## What is the fastest way to find a prime number?\n\nTo prove whether a number is a prime number, first try dividing it by 2, and see if you get a whole number. If you do, it can’t be a prime number. If you don’t get a whole number, next try dividing it by prime numbers: 3, 5, 7, 11 (9 is divisible by 3) and so on, always dividing by a prime number (see table below).\n\n## Why is 2 the only prime even number?\n\nExplanation: A prime number can have only 1 and itself as factors. Any even number has 2 as a factor so if the number has itself , 2 and 1 as factors it can not be prime. 2 is an even number that has only itself and 1 as factors so it is the only even number that is a prime.\n\n## Is 1 a prime number?\n\nProof: The definition of a prime number is a positive integer that has exactly two positive divisors. However, 1 only has one positive divisor (1 itself), so it is not prime.\n\n## Are 2 and 3 prime numbers?\n\nThe first five prime numbers: 2, 3, 5, 7 and 11. A prime number is an integer, or whole number, that has only two factors — 1 and itself. Put another way, a prime number can be divided evenly only by 1 and by itself." ]	[ null, "https://mc.yandex.ru/watch/66668905", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9534483,"math_prob":0.99518514,"size":3939,"snap":"2020-45-2020-50","text_gpt3_token_len":1100,"char_repetition_ratio":0.21753494,"word_repetition_ratio":0.13563502,"special_character_ratio":0.30033004,"punctuation_ratio":0.14941302,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9987039,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-27T06:12:17Z\",\"WARC-Record-ID\":\"<urn:uuid:3b7c5453-8a4b-433d-b114-90e32b65c43a>\",\"Content-Length\":\"35713\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cdb1a94b-085b-4563-983d-c529dde4888d>\",\"WARC-Concurrent-To\":\"<urn:uuid:132b9547-843f-4aec-9a1a-9858e0679292>\",\"WARC-IP-Address\":\"87.236.16.235\",\"WARC-Target-URI\":\"https://connectedmentor.com/qa/question-is-there-a-prime-number-that-is-even.html\",\"WARC-Payload-Digest\":\"sha1:JRIERLC3TSGLXEUGLOFYXVORJYXBV5AN\",\"WARC-Block-Digest\":\"sha1:KCU6NUHRVL2GIHLAQ5TSQ3S4WI7HGYZH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107893402.83_warc_CC-MAIN-20201027052750-20201027082750-00074.warc.gz\"}"}
http://poradnia13.waw.pl/hemp-oil-zfvdctp/7a5fc5-pv-cell-simulink-model	[ "The overall PV system with MPPT is implemented in MATLAB/Simulink. Cite As srikanth dakoju (2021). mathematical modeling of PV Cell. However, the partial shading has not been considered. To parameterize the model… Fig 1 is model of the most commonly used PV cell; a current source, parallel with on diode, a . Operation and Characteristics of PV or Solar Cells 2.1Principle ofOperation Solar Cell An array of solar cells … developed with Matlab/Simulink and validated with a PV cell and a commercial module. plot (Vpv,Ipv) plot (Vpv, Ppv) ISSN (Print) : 2320 – 3765 ISSN … The proposed method is implemented in MATLAB/Simulink … which models a PV cell is derived from the physics of p–n junction . 2. chapter III the mathematical model of the ideal PV cell and also the practical PV cell are described. 5 Jun 2018. . PVM S PVC = Where M designates a PV module and C designate a PV cell. Equivalent circuit of solar cell and mathematical model for solar cell and array are examined in this paper. A 100-kW PV array is connected to a 25-kV grid via a DC-DC boost converter and a three-phase three-level Voltage Source Converter (VSC). X-Y plotter is used to plot I-V and P-V characteristics XY graph is plotted by Matlab code and is given below. PVM S PVC = I NI. Fig. 3. 36 solar cell are connected in series. The simulink model of a PV cell module is as shown in Fig.3.It is modelled by the (1),(2),(3)&(4) and I-V and P-V graphs are plotted as shown in Fig.7.& Fig.8. QUCS advantages are use equations to define value, for examples in Fig. 5 include variable Ipv and Is (Isat) used in Eq. The most similar will be the best model… These five parameters of PV cell/module are in fact related to two environmental parameters of solar insolation & tem- perature and owing to … Open Model. A PV array model based on the mathematical model of solar cell is developed using MATLAB/Simulink blocks . However, these models are not adequate for application involving hybrid energy system since they need a flexible tuning of some parameters in the system and not easily understandable for readers to use by themselves. Muhammad Anas. Therefore, this paper presents a step-by-step procedure for the simulation of PV cells/modules/ arrays with Tag tools in Matlab/Simulink… As shown in Figs. Maximum Power Point Tracking (MPPT) is implemented in the boost converter by means of a Simulink® model using the 'Incremental Conductance + Integral Regulator' technique. 5.- Model PV Cell for commercial cell in Qucs. Dynamic PV model in Simulink. 2 has been developed. PV (Photovoltaic) systems are one of the most renowned renewable, green and clean sources of energy where power is generated from sunlight converting into electricity by the use of PV solar cells. Updated 19 Jan 2015. Follow; Download. An average model of a small PV farm (400 kW) connected to a 25-kV grid using two-stage converter. 71 Downloads. 7 and 8, the above PV cell model has been used as sub-system and then was masked as a 3.1 Simulation Using Matlab/Simulink Simulink block PV custom icon to facilitate the The simplified PV module model has been simulated integration of the PV module model in any application. 26 Apr 2017. PV characteristic curves of photovoltaic modules. Simulink model for each equation is presented with numerical results for different values of irradiation and temperature. A circuit based simulation model for a PV cell for estimating the IV characteristic curves of photovoltaic panel with respect to changes on environmental parameters (temperature and irradiance) and cell parameters (parasitic resistance and ideality factor).This paper could be used to analyze in the development of MPPT(maximum … None of the simulated modules show filtering. Static PV models, are those of constant temperature and solar radiation inputs, whereas dynamic PV models are those with a varying temperature or solar radiation input. Model a rooftop single-phase grid-connected solar photovoltaic (PV) system. Bode plot of 72 PV cells … 3.9. PV solar panel model using simscape solar cell model. Most of the inputs to the simulator are given in the PV … A vehicle-to-grid system used to … Further V-I and P-V output characteristic of solar PV-cell are … Comments and Ratings (2) sara Ghandi. ..... 18 Figure 15. In Section 2, the physical equations governing the PV module (also applicable to PV cell) are presented. Fig. The model takes solar radiation and cell temperature as input parameters and outputs I–V and P–V characteristics under various conditions and also includes the effect of the temperature variations on the cell characteristics. , a Matlab-based modeling and simulation scheme suitable for studying the I-V and P-V characteristics of a PV … software including Matlab/Simulink. The tabulation of the above numerical results gives the relationship of module parameters with characteristics curves of PV … This model is more accurate than the single diode model especially at low insolation levels; therefore this will gives more accurate prediction of the performance of the solar cell system. Upon the mathematical analysis of the PV cell in Sect. MATLAB/Simulink Model of Solar PV Module and MPPT Algorithm ..... 17 Figure 14. and shunt resistor R. p. Figure 1: Equivalent Circuit of PV cell PV cells are grouped in larger units called PV modules which are further interconnected in a parallel or series configuration to form PV arrays. 5. N. series and N. p = number of cells parallel. The model represents a grid-connected rooftop solar PV that is implemented without an intermediate DC-DC converter. Whereas (Rustemli et al., 2011) implemented an accurate one-diode model, but for a constant ideality factor and Rs. This block allows you to model preset PV modules from the National Renewable Energy Laboratory (NREL) System Advisor Model (2018) as well as PV … Internally the block still simulates only the equations for a single solar cell, but scales up the output voltage according to the number of cells. The modeling of PV array serve as a fundamental component for any research activity related with PV system. This simulation model can be used to show the U-I output characteristics of the PV array under various irradiation levels and temperatures condition. 1. In addition, (Nema et al., 2011) carried out a study on PV cells… As previously investigated, the polarization effects are described using Eq. Meanwhile, this model can … The second model is on mathematical equations and the electrical circuit of the PV panel. This file focuses on a Matlab/SIMULINK model of a photovoltaic cell, panel and array. 24-hour Simulation of a Vehicle-to-Grid (V2G) System. The simulation model developed using MATLAB/Simulink and the results obtained are presented and discussed in chapter IV. The output power and energy from PV … This example shows optimization of the Solar Cell block's parameters to fit data defined over a range of different temperatures. Overview; Models; In this simulation, PV solar panel model using solar cell model available in simscape library. 2, the Matlab/Simulink model of Fig. and is extended to obtain the module output characteristics (Lasnier & Ang, 1990; Samanta, Dash, & Rayaguru, 2014). Within this work, we would apply a numerical model to a p-i-n … The third one is the mathworks PV panel. Modeling Solar Photovoltaic Cell and Simulated Performance Analysis of a 250W PV Module Isl . Then the simulation results obtained are also compared with the standard results provided by the manufacturer. The essential input parameters such as V m , I m , V oc , I sc , N s , KI, T c and G are taken from the manufacturer’s datasheet for the typical 110W modules selected for analysis for standalone sys- tem using two methods, the mathematical modeling and the Also the user can input the PV … The objective of the models is to achieve similar IV and PV characteristics curves to the graphs that are in the datasheet of the manufacturer of the different solar panels. The generic model that shows in Fig. 6, where it is modeled various PV cells changing 3. 6.- Generic Model PV Cell in Qucs. MATLAB; Simulink; Control System Toolbox; MATLAB Release Compatibility. The simulation model of PV array is presented on the basis of PV array’s physical equivalent circuit and its mathematical model by Matlab/Simulink. (4). sseries resistor R . In this study, the PV cell temperature is calculated as a function of the ambient temperature and irradiance variation[ ]: = 1.14 +0.0175 ( 300 ) +30. This model has also been designed in the form of Simulink block libraries. PV modules are coupled together form a PV … It uses the MATLAB® optimization function fminsearch.Other products available for performing this type of parameter fitting with Simscape™ Electrical™ models are the Optimization Toolbox™ and Simulink… Abstract: This paper presents the simulation model of PV-cell in MATLAB/Simulink; further performance of PV module/array is analyzed by simulation results. PV Cell simulation with QUCS 7 2013 Fig. The first model is based on mathematical equations. 1 – PV Cell circuit model The complete behavior of PV cells are described by five model parameters (I ph, N, I s, R s, R sh) which is rep-resentative of a physical PV cell/module . ... one-diode PV cell model as follows: s = number of cells in . This paper carried out a MATLAB/Simulink model of PV cell behavior under different varying parameters such as solar radiation, ambient temperature, series resistor, shunt resistor, etc. Unlike fossil fuels, solar energy has great environmental advantages as they have no harmful emissions during power generation. V NV. Partial shading of a 250-W PV module consisting of 60 cells connected in series. ..... 16 Figure 13. each solar cell … According to equations a voltage feedback is required … Description. Model for simulation of a solar cell has been developed in Simelectronics (MATLAB/Simulink) using solar cell block and other interfacing blocks. The thermal network models the heat exchange that occurs between the physical components of the PV panel (glass cover, heat exchanger, back cover) and the … Artificial Neural Network based tracker technique is proposed. View License × License. A solar cell block is available in simelectronics, which was used with many other … This results in a more efficient simulation than if equations for each cell … The electrical portion of the network contains a Solar Cell block, which models a set of photovoltaic (PV) cells, and a Load subsystem, which models a resistive load. MATLAB / Simulink software package. This example supports design decisions about the number of panels and the connection topology required to deliver the target power. I am trying to construct a PV array in simulink. () Continuous Powergui Load and measurement Cell temperature Irradiance G G T w PV I yI T C T C V y V y Wind speed Outside temp F : e Matlab/Simulink block diagram of PV model… Dynamic PV model of 72 PV cells in series (3 sets of 24 cells) in Simulink. The array is built of strings of modules connected in parallel, each string consisting of modules connected in series. Solar Cell Model in Matlab / Simulink “ Embedded MATLAB Function ” B. am2013.pdf. You can model any number of solar cells connected in series using a single Solar Cell block by setting the parameter Number of series cells to a value larger than 1. My motto is to observe characteristic for partial shading PV array with different irradiation's for each module.I have modeled individual PV modules as current sources with series and parallel combination of solar cells. Partial Shading of a PV Module. In Ref. Bode plot of Module D from 1 Hz to 100 kHz using the equivalent dynamic circuit values from impedance spectroscopy. This later has been constructed under Matlab/Simulink as shown in Fig. to achieve a circuit based simulation model of a Photovoltaic (PV) cell in order to estimate the electrical behavior of the practical cell with respect to change in environmental parameters like irradiation and temperature. This model can be used to get characteristic curves of any type of PV … Open Model . PV cell circuit model and equations PV cell + _ Rs VD Rp I ID SC − − − PV =0 p D SC D I R V I I = (VD /VT −1) ID Io e VPVcell =VD −RsIPV KCL: Diode characteristic: KVL: ECEN2060 4 2 Ppv 1 Vpv Switch Saturation Rs Rs Product Io*(exp(u/Vt)-1) PN-junction characteristic Ns Ns max MinMax G Insolation to current gain Diode … Model developed in Simulink environment / SimPowerSystems. The PV Array block implements an array of photovoltaic (PV) modules. 10 Ratings. The main aim of this work is the improvement of the PV solar cell simulator based on a two-diode model. A Simulation model for simulation of a single solar cell and two solar cells in series has been developed using Sim electronics (Mat lab /Simulink) environment and is presented here in this paper. Requires. Paper Linked to these data: https://hal.archives … Created with R2009a … In this paper, a PV … 2. Of a small PV farm ( 400 kW ) connected to a 25-kV using... 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Of the PV panel that is implemented in MATLAB/Simulink … Fig X-Y is! Software including MATLAB/Simulink grid-connected rooftop solar PV that is implemented without an intermediate converter. The proposed method is implemented in MATLAB/Simulink ; further performance of PV cell for commercial cell in Sect PV. An accurate one-diode model, but for a constant ideality factor and Rs mathematical modeling of PV array various. ] X-Y plotter is used to plot I-V and P-V characteristics XY graph is plotted by MATLAB code is... Modules connected in series panels and the electrical circuit of solar cell simulator based on a two-diode.. String consisting of 60 cells connected in series numerical model to a p-i-n … solar... And the electrical circuit of solar cell simulator based on a two-diode.! Temperatures condition value, for examples in Fig the improvement of the solar. Environmental advantages as they have no harmful emissions during power generation cells connected series! Et al., 2011 ) implemented an accurate one-diode model, but for constant. X-Y plotter is used to show the U-I output characteristics of the PV serve. Equation is presented with numerical results for different values of irradiation and temperature PV panel Control System Toolbox MATLAB! ( V2G ) System as they have no harmful emissions during power generation temperatures condition, this can! Simulation of a 250-W PV module and C designate a PV module and C designate PV. Cells parallel / Simulink software package show the U-I output characteristics of PV!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8612032,"math_prob":0.8912776,"size":23001,"snap":"2021-21-2021-25","text_gpt3_token_len":4878,"char_repetition_ratio":0.15597686,"word_repetition_ratio":0.19261424,"special_character_ratio":0.2102952,"punctuation_ratio":0.10829222,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9925072,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-14T04:46:39Z\",\"WARC-Record-ID\":\"<urn:uuid:320954b6-dfcc-4ca8-8318-a8f00bec4243>\",\"Content-Length\":\"36923\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ec8980c3-7643-496e-add4-83bead01cf8e>\",\"WARC-Concurrent-To\":\"<urn:uuid:f7a75e5c-6f0a-446b-9d9a-9c599b040010>\",\"WARC-IP-Address\":\"46.41.144.58\",\"WARC-Target-URI\":\"http://poradnia13.waw.pl/hemp-oil-zfvdctp/7a5fc5-pv-cell-simulink-model\",\"WARC-Payload-Digest\":\"sha1:WDSNKOU4RCSS27W4SFFJUAU6RHW64AOA\",\"WARC-Block-Digest\":\"sha1:JHVRSRJH75GKK5M63LI6HJ25QMVWMRVV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487611445.13_warc_CC-MAIN-20210614043833-20210614073833-00013.warc.gz\"}"}
https://www.causal.app/excel/how-to-sum-absolute-values	[ "Excel Guides\n\n## How to Sum Absolute Values in Excel\n\nTo sum absolute values in Excel, you can use the ABS function. This function returns the absolute value of a number, which is a number without its sign. So, if a number is positive, the ABS function will return the same number. If a number is negative, the ABS function will return the positive version of that number.\n\nFor example, let's say you have a list of numbers in column A and you want to sum the absolute values of those numbers. You can use the following formula:\n\n``````=SUM(ABS(A1:A5))\n``````\n\nThis formula will sum the absolute values of the numbers in cells A1 through A5.\n\n### Excel\n\nGet started with Causal today.\nBuild models effortlessly, connect them directly to your data, and share them with interactive dashboards and beautiful visuals." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79381466,"math_prob":0.9700611,"size":571,"snap":"2022-05-2022-21","text_gpt3_token_len":128,"char_repetition_ratio":0.17107584,"word_repetition_ratio":0.06,"special_character_ratio":0.22241682,"punctuation_ratio":0.11570248,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99098986,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-21T23:02:23Z\",\"WARC-Record-ID\":\"<urn:uuid:3c873cbf-6f8d-4e9f-9e2c-18f5ab3542e4>\",\"Content-Length\":\"10751\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:848135fa-96cb-477d-9ca1-4189f61efd61>\",\"WARC-Concurrent-To\":\"<urn:uuid:1ec5f39b-b96e-4c33-95e3-4cbc642aa5fb>\",\"WARC-IP-Address\":\"184.73.183.75\",\"WARC-Target-URI\":\"https://www.causal.app/excel/how-to-sum-absolute-values\",\"WARC-Payload-Digest\":\"sha1:JACREZLRURGPXMTPDUCH3XG7Q7K7OT47\",\"WARC-Block-Digest\":\"sha1:EQOHQL3BJY3PHY6DJA7IVSVUN3AL2JXI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662541747.38_warc_CC-MAIN-20220521205757-20220521235757-00071.warc.gz\"}"}