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http://stackoverflow.com/questions/1726126/is-the-board-game-go-np-complete?answertab=active | 1,404,967,384,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776401658.57/warc/CC-MAIN-20140707234001-00054-ip-10-180-212-248.ec2.internal.warc.gz | 124,962,958 | 16,468 | # Is the board game “Go” NP complete?
There are plenty of Chess AI's around, and evidently some are good enough to beat some of the world's greatest players.
I've heard that many attempts have been made to write successful AI's for the board game Go, but so far nothing has been conceived beyond average amateur level.
Could it be that the task of mathematically calculating the optimal move at any given time in Go is an NP-complete problem?
-
Dunno why you got downvoted. This is a legit question. +1 – Mark Nov 12 '09 at 23:30
Well, current Monte Carlo and similar algorithms have pushed the frontier to average amateur level. Names to look for include Zenith, Many Faces of Go, Fuego, Leela, among others. – Svante Nov 12 '09 at 23:34
Chess and Go are both EXPTIME complete. IIRC, Go has more possible moves, so I think it a higher multiple of that complexity class than chess. Wikipedia has a good article on the complexity of Go.
-
You might want to mention that both results are for generalized versions of the games. The games with constant-sized boards can, trivially, both be solved in constant time. (Though the constant is too large for us to deal with right now, and possibly ever.) – ShreevatsaR Nov 12 '09 at 23:37
In terms of understanding what is meant when an expert says "Chess is EXPTIME-complete" ShreevatsaR's point is very important. – PeterAllenWebb Jan 7 '10 at 21:30
Neither Chess or Go AIs completely evaluate all possibilities before deciding on a move.
Chess AIs use various heuristics to narrow down the search space, and to quantify how 'good' a given position on the board happens to be. This can be done recursively by evaluating possible board positions 14-15 moves ahead and choosing a path that leads to a good position.
There's a bit of 'magic' in how a board position is quantified so the that at the top level, the AI can simply go Move A > Move B therefore lets do Move A. But since there's a limited number of pieces and they all have quantifiable value a 'good enough' algorithm can be implemented.
But it turns out to be a lot harder for a program to evaluate two possible board positions in Go and make that A > B calculation. Without that critical piece its a little hard to make the rest of the AI work.
-
Even if Go is merely in `P` it could still be something horrendous like `O(n^m)` where `n` is the number of spaces and `m` is some (large) fixed number. Even being in `P` doesn't make something reasonable to compute. | 592 | 2,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2014-23 | latest | en | 0.945776 |
http://www.slideserve.com/ivory/solutions | 1,506,369,275,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818693240.90/warc/CC-MAIN-20170925182814-20170925202814-00430.warc.gz | 559,643,283 | 13,889 | 1 / 4
# Solutions - PowerPoint PPT Presentation
Solutions. Solving Puzzles. Puzzle 1: Word Search. Sample Algorithm: Pick a word from the list of words you need to find. Go through each of the letters in the grid one by one and see if it’s the same letter as the first letter of the the word you’re trying to find.
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Solving Puzzles
Puzzle 1: Word Search
• Sample Algorithm:
• Pick a word from the list of words you need to find.
• Go through each of the letters in the grid one by one and see if it’s the same letter as the first letter of the the word you’re trying to find.
• If it is, check its 8 neighboring letters (up, down, left, right, and 4 diagonals) to see if it’s the second letter in the word you’re trying to find. Then, check for the 3rd, 4th… letters until you’ve found all of the letters that make the word. Go to step 3.
• If it’s not, go back to step 2 and keep going through the grid of letters.
• Cross out the word you’ve just found and move onto the next word. Repeat steps 1-3 until you’ve found and crossed out all of the words on your list.
• Algorithms may vary, but the idea is that they should have a series of “foolproof” steps that anyone would be able to follow to solve the puzzle.
Puzzle 2: Number Maze
Steps 1-4
Steps 5-8
Sample algorithm:
Start at the goal and look at all the tiles in its horizontal and vertical paths.
For each tile, see if it contains the correct number of steps to get to the goal.
For each tile that works, check its horizontal and vertical paths for tiles that contain the right number of steps to get to it.
Repeat until one of the tiles you get is the start tile.
You now have a reverse sequence of tiles, from goal to start. Reverse the path to get the sequence of tiles from start to goal.
Puzzle 3: How many handshakes?
Each dot represents a person, and all the dots must be connected to one another with no repeats:
Instructions may vary, but here’s one set of instructions that works:
Have everyone stand in a circle
The first person should go around the circle and shake every other person’s hand.
The next person should go around the circle and shake every other person’s hand except for the first person.
The next person should go around the circle and shake every other person’s hand except for the first two people.
The next person should go around the circle and shake every other person’s hand except for the first three people.
Repeat, with each successive person going around the circle to shake everyone else’s hand until you get to the second-to-last person, who will shake one hand and complete the handshakes required.
Puzzle 3: How many handshakes?
Extra Credit: If there were 20 people at the party, it would take 190 handshakes
Extra Credit: If there were N people at the party, there would be handshakes
[(N-1)] + [(N-1) – 1] + [(N-1) – 2] + …. + [(N-1) – (N-2)] + [(N-1) – (N-1)]
= (N-1) + (N-2) + (N-3) + …. + 2 + 1 + 0
= (N-1) + (N-2+1) + (N-3+2) +…. = (N-1) + (N-1) + (N-1) + …. = | 947 | 3,638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2017-39 | latest | en | 0.90682 |
https://www.onlinemathlearning.com/naming-angles.html | 1,579,798,351,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250611127.53/warc/CC-MAIN-20200123160903-20200123185903-00208.warc.gz | 1,012,450,771 | 10,547 | # Naming Angles
Related Topics:
Angle Games
Examples, solutions, videos, worksheets, stories, and songs to help Grade 6 students learn how to name angles.
The following diagram shows three ways that can be used to name angles. Scroll down the page for more examples and solutions on naming angles.
When an angle is presented with three labeled points, the angle would be named using the angle symbol ∠ and the three points written with the vertex in the middle.
The following angle could be correctly named ∠LMN or ∠NML, because in each case the vertex point is in the middle of the name.
Writing the points in alphabetical order is not a requirement and will not necessarily be correct in all cases, depending on the letter name of the vertex. The following angle could be named ∠BAC or ∠CAB. Writing the points in alphabetical order (∠ABC) in this case would be incorrect, as the vertex would not be in the middle.
Naming Angles
How to name an angle?
Naming Angles
Naming angles using the three letter system
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. | 298 | 1,419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-05 | latest | en | 0.896841 |
http://slideplayer.com/slide/7802961/ | 1,579,836,940,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250614880.58/warc/CC-MAIN-20200124011048-20200124040048-00439.warc.gz | 149,269,870 | 16,495 | # Units of Measurement SI – Le Systeme International d’Unites AKA: the Metrics System.
## Presentation on theme: "Units of Measurement SI – Le Systeme International d’Unites AKA: the Metrics System."— Presentation transcript:
Units of Measurement SI – Le Systeme International d’Unites AKA: the Metrics System
SI Base Units Adopted: 1960 Standard international measurement Powers of Ten
Mass vs. Weight Mass measures the quantity of matter Standard unit = g (gram) Instrument = Balance Weight measures the gravitational pull on matter Standard unit = N Newton is force Instrument = Spring Scale
Length Distance measurement Standard unit = m (meter) Instrument = Ruler
Volume Amount of space occupied by an object For non-liquid volumes, standard unit = m 3 For liquid volumes, standard unit = L (liter) 1 mL = 1 cm 3 Instrument = graduated cylinders or ruler
Temperature Standard unit = Celsius Using Celsius, water freezes at 0 degrees, and boils at 100 degrees F = 9/5C + 32 C = 5/9 (F – 32)
Conversions and Scale Larger ---------------------- Smaller (k) kilo (h) hecto (da) deca (d) deci (c) centi (m) milli Length kmhmdammdmcmmm Mass kghgdaggdgcgmg Liquid Volume kLhLdaLLdLcLmL
KILO 1000 Units HECTO 100 Units DEKA 10 Units DECI 0.1 Unit CENTI 0.01 Unit MILLI 0.001 Unit Meters Liters Grams Ladder Method How do you use the “ladder” method? 1 st – Determine your starting point. 2 nd – Count the “jumps” to your ending point. 3 rd – Move the decimal the same number of jumps in the same direction. 4 km = _________ m 1 2 3 How many jumps does it take? Starting Point Ending Point 4. 1 __. 2 3 = 4000 m
Try these conversions using the ladder method. 1000 mg = _______ g 1 L = _______ mL160 cm = _______ mm 14 km = _______ m109 g = _______ kg 250 m = _______ km Conversion Practice Compare using, or =. 56 cm 6 m 7 g 698 mg
Compare using, or =. 25) 63 cm 6 m 27) 5 g 508 mg 29) 1,500 mL 1.5 L 26) 536 cm 53.6 dm 28) 43 mg 5 g 30) 3.6 m 36 cm
Download ppt "Units of Measurement SI – Le Systeme International d’Unites AKA: the Metrics System."
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https://www.teacherspayteachers.com/Product/Solving-Trig-Equations-interactive-discovery-exercise-2248999 | 1,529,345,091,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860684.14/warc/CC-MAIN-20180618164208-20180618184208-00403.warc.gz | 913,298,616 | 18,497 | # Solving Trig Equations - interactive discovery exercise
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Solving trigonometric equations can be difficult for students to understand without good visual aids. Graphical solutions can provide highly effective support in this regard. Using either the pdf answer key sheets or a Geogebra file conceived and created by Mathew's House, students independently and effectively develop a more complete understanding of the presented topic while you facilitate!
Most students just accept concepts as a given in math without any internalizing or synthesis. But, in many cases, the theorems and equations are just an extension of what they already know or have previously studied. So, they are quite capable of discovering them on their own, given some guided instruction.
Using this unique Geogebra file, the user can explore math concepts with the click of a mouse. Geogebra is a software platform in the same sense as Power Point is. The user creates a file (this Geogebra file was created by Mathew's House) to demonstrate almost any idea in math. Then the student uses the interactive file to learn the desired concept.
Geogebra, a free program, is quickly and easily downloaded from the internet to both computers and tablets (as an app) and soon to phones as well. Students can work individually in a computer lab or you can use a single computer and projector and work through the exercise as a group. Alternatively, you could use the flipped classroom model and have them complete the exercise at home allowing class time to further develop the ideas presented.
In this exercise, when you move the appropriate sliders, the trigonometric functions will be dynamically transformed on screen. This allows the student to control the pace of learning and to clearly visualize the solutions within the given domain. This allows them to better understand the basic concepts behind this topic.
The graphics in Geogebra are visually appealing and each of the parameters are color coded to help students identify the different values. All very helpful to fully understand and to apply the concepts learned.
Two pdf pages of guided exercises as well as a Geogebra file and ten pdf answer key pages provide your students with a complete resource to fully and independently explore this topic.
Please check out the preview to get a more compete description and visual concept of what is included.
For Geogebra exercises in other topics click the title below:
High School Trigonometry Bundle ( 40% savings! )
Trig Translations
Pythagorean Theorem
Slope
Trig Ratios - Sine, Cosine & Tangent
Arc Length & Sector Area
Area of Any Triangle ( free ! )
The Unit Circle
Cosine Law
Sine Law
Ambiguous Case of the Sine Law
The Derivative - from First Principles
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2 pdf worksheets, 10 pdf answer key pages and 1 Geogebra file
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\$3.00 | 639 | 3,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-26 | latest | en | 0.935007 |
https://rdrr.io/cran/proportion/man/PlotcovpBA.html | 1,638,331,208,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359082.78/warc/CC-MAIN-20211201022332-20211201052332-00031.warc.gz | 558,830,577 | 8,285 | # PlotcovpBA: Graphs of Coverage Probability of the Bayesian method In proportion: Inference on Single Binomial Proportion and Bayesian Computations
## Description
Graphs of Coverage Probability of the Bayesian method
## Usage
`1` ```PlotcovpBA(n, alp, a, b, t1, t2, a1, a2) ```
## Arguments
`n` - Number of trials `alp` - Alpha value (significance level required) `a` - Beta parameters for hypo "p" `b` - Beta parameters for hypo "p" `t1` - Lower tolerance limit to check the spread of coverage Probability `t2` - Upper tolerance limit to check the spread of coverage Probability `a1` - Beta Prior Parameters for Bayesian estimation `a2` - Beta Prior Parameters for Bayesian estimation
## Details
The graphs of Coverage Probability of Bayesian method
Other Basic coverage probability methods: `PlotcovpAS`, `PlotcovpAll`, `PlotcovpEX`, `PlotcovpLR`, `PlotcovpLT`, `PlotcovpSC`, `PlotcovpTW`, `PlotcovpWD`, `covpAS`, `covpAll`, `covpBA`, `covpEX`, `covpLR`, `covpLT`, `covpSC`, `covpTW`, `covpWD`
```1 2 3 4 5``` ```## Not run: n= 10; alp=0.05; a=1;b=1; t1=0.93;t2=0.97;a1=1;a2=1 PlotcovpBA(n,alp,a,b,t1,t2,a1,a2) ## End(Not run) ``` | 369 | 1,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-49 | latest | en | 0.568151 |
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# Help with Rate & Work questions
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Help with Rate & Work questions [#permalink]
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15 Jan 2010, 11:18
Does anyone have any suggestions on learning R&W. I've read MGMAT Word Translation and can ans the basic questions, but get stumped on questions beyond that. I come up with ans that are not even part of the ans choices Any suggestions on how I can improve?
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Re: Help with Rate & Work questions [#permalink]
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15 Jan 2010, 12:29
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gi wrote:
Does anyone have any suggestions on learning R&W. I've read MGMAT Word Translation and can ans the basic questions, but get stumped on questions beyond that. I come up with ans that are not even part of the ans choices Any suggestions on how I can improve?
Well, coming to our point. Rate and Work problems, the one and only thing is translating the words of the problems into equations or numbers or proportions. They are just as good as any other algebraic problems, like age calculations, percentages and interest calculations etc. If a man can complete a task in 7 days, his rate is 1/7, of course, I can complete some tasks at my work-place much faster and some tasks in the bedroom very very late!!!!! If my partner is equally adept, or perhaps much faster in the bedroom at say a rate of 1/5, our combined rate would be 1/7 + 1/5, it's reciprocal will be the number of days we both will take to finish it off. That, my friend, is the funda of the entire maze of rate and work problems.
All the very best and practise well. Like they say, have fun and multiply.
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Re: Help with Rate & Work questions [#permalink]
### Show Tags
16 Jan 2010, 10:57
BarneyStinson wrote:
gi wrote:
Does anyone have any suggestions on learning R&W. I've read MGMAT Word Translation and can ans the basic questions, but get stumped on questions beyond that. I come up with ans that are not even part of the ans choices Any suggestions on how I can improve?
Well, coming to our point. Rate and Work problems, the one and only thing is translating the words of the problems into equations or numbers or proportions. They are just as good as any other algebraic problems, like age calculations, percentages and interest calculations etc. If a man can complete a task in 7 days, his rate is 1/7, of course, I can complete some tasks at my work-place much faster and some tasks in the bedroom very very late!!!!! If my partner is equally adept, or perhaps much faster in the bedroom at say a rate of 1/5, our combined rate would be 1/7 + 1/5, it's reciprocal will be the number of days we both will take to finish it off. That, my friend, is the funda of the entire maze of rate and work problems.
All the very best and practise well. Like they say, have fun and multiply.
_________________
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Re: Help with Rate & Work questions [#permalink] 16 Jan 2010, 10:57
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## Języki programowania obiektowego
### Roman Simiński
roman.siminski@us.edu.pl
www.programowanie.siminskionline.pl
(2)
### 100 [km] — x [litry]
(3)
#include <iostream> using namespace std; int main()
{
float dystans, paliwo;
cout << endl << "Obliczam ile Twoj pojazd spala paliwa na 100 km" << endl; cout << "Dystans: " << flush;
cin >> dystans;
cout << "Paliwo: " << flush; cin >> paliwo;
cout << "Spalanie " << ( paliwo*100 ) / dystans << " l na 100 km" << endl;
cout << "Nacisnij Enter by zakonczyc program..." << endl; cin.ignore(); cin.get(); return EXIT_SUCCESS; } #include <iostream> using namespace std; int main() {
float dystans, paliwo;
cout << endl << "Obliczam ile Twoj pojazd spala paliwa na 100 km" << endl; cout << "Dystans: " << flush;
cin >> dystans;
cout << "Paliwo: " << flush; cin >> paliwo;
cout << "Spalanie " << ( paliwo*100 ) / dystans << " l na 100 km" << endl;
cout << "Nacisnij Enter by zakonczyc program..." << endl; cin.ignore();
cin.get();
return EXIT_SUCCESS; }
Potencjalne dzielenie przez zero!
(4)
#include <iostream> using namespace std; int main()
{
float dystans, paliwo;
cout << endl << "Obliczam ile Twoj pojazd spala paliwa na 100 km" << endl; cout << "Dystans: " << flush;
cin >> dystans;
cout << "Paliwo: " << flush; cin >> paliwo;
if( dystans != 0 )
cout << "Spalanie " << (paliwo*100) / dystans << " l na 100 km" << endl; else
cout << "Nie dokonam obliczen dla zerowego dystansu." << endl;
cout << "Nacisnij Enter by zakonczyc program..." << endl; cin.ignore(); cin.get(); return EXIT_SUCCESS; #include <iostream> using namespace std; int main() {
float dystans, paliwo;
cout << endl << "Obliczam ile Twoj pojazd spala paliwa na 100 km" << endl; cout << "Dystans: " << flush;
cin >> dystans;
cout << "Paliwo: " << flush; cin >> paliwo;
if( dystans != 0 )
cout << "Spalanie " << (paliwo*100) / dystans << " l na 100 km" << endl; else
cout << "Nie dokonam obliczen dla zerowego dystansu." << endl;
cout << "Nacisnij Enter by zakonczyc program..." << endl; cin.ignore();
cin.get();
return EXIT_SUCCESS;
Instrukcja warunkowa
(5)
### Instrukcja warunkowa
if( W ) I if( W ) I if( W ) I if( W ) I
I
I
W
W tak
(6)
### Instrukcja alternatywy
if( W ) I1 else I2 if( W ) I1 else I2 if( W ) I1 else I2 if( W ) I1 else I2
### wykonaj instrukcję I2
I1 I1 W W tak nie I2 I2
(7)
(8)
### Instrukcja alternatywy a instrukcje warunkowe
float a, b, c, delta; . . . delta = b * b – 4 * a * c; float a, b, c, delta; . . . delta = b * b – 4 * a * c; if( delta < 0 )
cout << "Brak pierwiastków rzeczywistych" << endl; if( delta == 0 )
cout << "Jeden pierwiastek rzeczywisty" << endl; if( delta > 0 )
if( delta < 0 )
cout << "Brak pierwiastków rzeczywistych" << endl;
if( delta == 0 )
cout << "Jeden pierwiastek rzeczywisty" << endl;
if( delta > 0 )
if( delta < 0 )
cout << "Brak pierwiastków rzeczywistych" << endl; else
if( delta == 0 )
cout << "Jeden pierwiastek rzeczywisty" << endl; else
cout << "Dwa pierwiastki rzeczywiste" << endl;
if( delta < 0 )
cout << "Brak pierwiastków rzeczywistych" << endl;
else
if( delta == 0 )
cout << "Jeden pierwiastek rzeczywisty" << endl;
else
cout << "Dwa pierwiastki rzeczywiste" << endl;
(9)
### Uwaga na zagnieżdżone instrukcje warunkowe!
. . .
double kwota;
cout << "Podaj przychod: "; cin >> kwota; if( kwota >= 0 ) if( kwota > 0 ) cout << "Dochod"; else cout << "Strata"; cout << "Koniec” . . . . . . double kwota;
cout << "Podaj przychod: "; cin >> kwota; if( kwota >= 0 ) if( kwota > 0 ) cout << "Dochod"; else cout << "Strata"; cout << "Koniec” . . .
Podaj przychod: -100
(10)
### Uwaga na zagnieżdżone instrukcje warunkowe!
. . .
double kwota;
cout << "Podaj przychod: "; cin >> kwota; if( kwota >= 0 ) if( kwota > 0 ) cout << "Dochod"; else cout << "Strata"; cout << "Koniec” . . . . . . double kwota;
cout << "Podaj przychod: "; cin >> kwota; if( kwota >= 0 ) if( kwota > 0 ) cout << "Dochod"; else cout << "Strata"; cout << "Koniec” . . .
Podaj przychod: -100 Koniec
(11)
### jak sugerują wcięcia:
if( kwota >= 0 ) if( kwota > 0 ) cout << "Dochod"; else cout << "Strata"; if( kwota >= 0 ) if( kwota > 0 ) cout << "Dochod"; else cout << "Strata";
### A jest tak:
if( kwota >= 0 ) if( kwota > 0 ) cout << "Dochod"; else cout << "Strata"; if( kwota >= 0 ) if( kwota > 0 ) cout << "Dochod"; else cout << "Strata";
### Trzeba użyć instrukcji złożonej lub „sparować” if z else:
if( kwota >= 0 ) { if( kwota > 0 ) cout << "Dochod"; } else cout << "Strata"; if( kwota >= 0 ) { if( kwota > 0 ) cout << "Dochod"; } else cout << "Strata"; if( kwota >= 0 ) if( kwota > 0 ) cout << "Dochod"; else cout << "Zero!"; else cout << "Strata"; if( kwota >= 0 ) if( kwota > 0 ) cout << "Dochod"; else cout << "Zero!"; else cout << "Strata";
(12)
### Wykonanie takiej instrukcji polega na wyznaczeniu wartości danego wyrażenia.
x = 0; x = 0; a + b; a + b; x = a + b; x = a + b; ; ;
Oczywiście bez większego sensu, lecz legalne
Oczywiście bez większego
sensu, lecz legalne użyteczna instrukcja pustaNiepozorna, lecz bardzo Niepozorna, lecz bardzo użyteczna instrukcja pusta
(13)
### W obrębie zagnieżdżonych bloków następuje przesłanianie nazw.
{ int i = 0, j = 1, k; k = i + j; { float k = 10.2; cout << "k = " << k; } cout << "k = " << k; } { int i = 0, j = 1, k; k = i + j; { float k = 10.2; cout << "k = " << k; } cout << "k = " << k; }
(14)
### Od instrukcji warunkowych do instrukcji switch
int main() {
cout << "\nJaki typ nadwozia lubisz?"; cout << "\n1. Sedan\n2. SUV\n3. Coupe"; cout << "\nWpisz 1, 2 lub 3: ";
cout << "\nChyba lubisz eleganckie limuzyny!"; if( nadwozie == 2 )
cout << "\nWidze, ze ciagnie Cie w teren!"; if( nadwozie == 3 )
cout << "\nTy to pewnie lubisz szybka jazde!"; cout << "\n\nEnter=Koniec"; cin.ignore(); cin.get(); return EXIT_SUCCESS; } int main() { int nadwozie;
cout << "\nJaki typ nadwozia lubisz?"; cout << "\n1. Sedan\n2. SUV\n3. Coupe"; cout << "\nWpisz 1, 2 lub 3: ";
cout << "\nChyba lubisz eleganckie limuzyny!";
cout << "\nWidze, ze ciagnie Cie w teren!"; if( nadwozie == 3 )
cout << "\nTy to pewnie lubisz szybka jazde!"; cout << "\n\nEnter=Koniec"; cin.ignore(); cin.get(); return EXIT_SUCCESS; }
(15)
### Od instrukcji warunkowych do instrukcji switch
int main() {
cout << "\nJaki typ nadwozia lubisz?"; cout << "\n1. Sedan\n2. SUV\n3. Coupe"; cout << "\nWpisz 1, 2 lub 3: ";
case 1 : cout << "\nChyba lubisz eleganckie limuzyny!"; break;
case 2 : cout << "\nWidze, ze ciagnie Cie w teren!"; break;
case 3 : cout << "\nTy to pewnie lubisz szybka jazde!"; break; } cout << "\n\nEnter=Koniec"; cin.ignore(); cin.get(); return EXIT_SUCCESS; } int main() { int nadwozie;
cout << "\nJaki typ nadwozia lubisz?"; cout << "\n1. Sedan\n2. SUV\n3. Coupe"; cout << "\nWpisz 1, 2 lub 3: ";
case 1 : cout << "\nChyba lubisz eleganckie limuzyny!"; break;
case 2 : cout << "\nWidze, ze ciagnie Cie w teren!"; break;
case 3 : cout << "\nTy to pewnie lubisz szybka jazde!"; break; } cout << "\n\nEnter=Koniec"; cin.ignore(); cin.get(); return EXIT_SUCCESS; }
Instrukcja switch: jedna zmienna i porównanie z wartościami znanymi na etapie kompilacji Instrukcja switch: jedna zmienna i porównanie
(16)
### Instrukcja przełączająca switch
switch( wyra enie )ż {
case wyra enie_stałe : instrukcjeż case wyra enie_stałe : instrukcjeż . . .
default : instrukcje }
switch( wyra enie )ż {
case wyra enie_stałe : instrukcjeż case wyra enie_stałe : instrukcjeż . . .
default : instrukcje }
(17)
### Gdyby brakowało instrukcji break
. . .
case 1 : cout << "\nChyba lubisz eleganckie limuzyny!"; case 2 : cout << "\nWidze, ze ciagnie Cie w teren!"; case 3 : cout << "\nTy to pewnie lubisz szybka jazde!"; }
. . .
. . .
case 1 : cout << "\nChyba lubisz eleganckie limuzyny!"; case 2 : cout << "\nWidze, ze ciagnie Cie w teren!"; case 3 : cout << "\nTy to pewnie lubisz szybka jazde!"; }
. . .
cout << "\nChyba lubisz eleganckie limuzyny!";
cout << "\nChyba lubisz eleganckie limuzyny!";
cout << "\nWidze, ze ciagnie Cie w teren!"
cout << "\nWidze, ze ciagnie Cie w teren!"
cout << "\nTy to pewnie lubisz szybka jazde!"
cout << "\nTy to pewnie lubisz szybka jazde!"
(18)
### Rola instrukcji break
case 1 : cout << "\nChyba lubisz eleganckie limuzyny!"; break;
case 2 : cout << "\nWidze, ze ciagnie Cie w teren!"; break;
case 3 : cout << "\nTy to pewnie lubisz szybka jazde!"; break;
}
case 1 : cout << "\nChyba lubisz eleganckie limuzyny!"; break;
case 2 : cout << "\nWidze, ze ciagnie Cie w teren!"; break;
case 3 : cout << "\nTy to pewnie lubisz szybka jazde!"; break;
}
cout << "\nChyba lubisz eleganckie limuzyny!";
cout << "\nChyba lubisz eleganckie limuzyny!";
cout << "\nWidze, ze ciagnie Cie w teren!"
cout << "\nWidze, ze ciagnie Cie w teren!"
cout << "\nTy to pewnie lubisz szybka jazde!"
cout << "\nTy to pewnie lubisz szybka jazde!"
(19)
### Instrukcja iteracyjna do-while
do instrukcja while( wyrazenie ); do instrukcja while( wyrazenie ); do { ciag instrukcji } while( wyrazenie ); do { ciag instrukcji } while( wyrazenie );
wyrażenie
wyrażenie
instrukcja
instrukcja
(20)
### Zastosowanie instrukcji do-while
#include <iostream> using namespace std; int main()
{
float dystans, paliwo;
cout << endl << "Obliczam ile Twoj pojazd spala paliwa na 100 km" << endl; cout << "Dystans: " << flush;
cin >> dystans;
cout << "Paliwo: " << flush; cin >> paliwo;
if( dystans != 0 )
cout << "Spalanie " << (paliwo*100) / dystans << " l na 100 km" << endl; else
cout << "Nie dokonam obliczen dla takiego dystansu." << endl;
cout << "Nacisnij Enter by zakonczyc program..." << endl; cin.ignore(); cin.get(); return EXIT_SUCCESS; #include <iostream> using namespace std; int main() {
float dystans, paliwo;
cout << endl << "Obliczam ile Twoj pojazd spala paliwa na 100 km" << endl; cout << "Dystans: " << flush;
cin >> dystans;
cout << "Paliwo: " << flush; cin >> paliwo;
if( dystans != 0 )
cout << "Spalanie " << (paliwo*100) / dystans << " l na 100 km" << endl; else
cout << "Nie dokonam obliczen dla takiego dystansu." << endl;
cout << "Nacisnij Enter by zakonczyc program..." << endl; cin.ignore();
cin.get();
return EXIT_SUCCESS;
W tym miejscu użytkownik może wprowadzić nieprawidłowe dane. Nie pozwólmy mu na to!
W tym miejscu użytkownik może wprowadzić nieprawidłowe dane. Nie pozwólmy mu na to!
(21)
### Zastosowanie instrukcji do-while — weryfikacja danych
do {
cout << "Dystans: " << flush; cin >> dystans;
}
while( dystans <= 0 ); do
{
cout << "Paliwo: " << flush; cin >> paliwo;
}
while( paliwo <= 0 );
do {
cout << "Dystans: " << flush; cin >> dystans;
}
while( dystans <= 0 );
do {
cout << "Paliwo: " << flush; cin >> paliwo;
}
while( paliwo <= 0 );
Wykonuj wczytywanie dystansu, dopóki jest on nieprawidłowy.
Wykonuj wczytywanie dystansu, dopóki jest on nieprawidłowy.
Wykonuj wczytywanie il. paliwa, dopóki jest ona nieprawidłowa.
Wykonuj wczytywanie il. paliwa, dopóki jest ona nieprawidłowa.
(22)
### Ta wersja jest wyraźnie lepsza.
do {
cout << "Dystans: " << flush; cin >> dystans;
if( dystans <= 0 )
cout << "Dystans musi byc wiekszy od zera!" << endl; }
while( dystans <= 0 ); do
{
cout << "Paliwo: " << flush; cin >> paliwo;
if( paliwo <= 0 )
cout << "Ilosc paliwa musi byc wieksza od zera!" << endl; }
while( paliwo <= 0 );
do {
cout << "Dystans: " << flush; cin >> dystans;
if( dystans <= 0 )
cout << "Dystans musi byc wiekszy od zera!" << endl; }
while( dystans <= 0 );
do {
cout << "Paliwo: " << flush; cin >> paliwo;
if( paliwo <= 0 )
cout << "Ilosc paliwa musi byc wieksza od zera!" << endl; }
while( paliwo <= 0 );
Wykonuj wczytywanie dystansu, dopóki jest on nieprawidłowy, poinformuj o tym użytkownika.
Wykonuj wczytywanie dystansu, dopóki jest on nieprawidłowy, poinformuj o tym użytkownika.
Wykonuj wczytywanie il. paliwa, dopóki jest ona nieprawidłowa, poinformuj o tym użytkownika.
Wykonuj wczytywanie il. paliwa, dopóki jest ona nieprawidłowa, poinformuj o tym użytkownika.
(23)
(24)
### Zastosowanie instrukcji — gra w „za dużo, za mało”
Algorytm ogólny
Algorytm ogólny
Wczytana < Wylosowana
Wczytana Wylosowana<> false true true true false false Wczytana > Wylosowana Start
Wylosuj liczbę i zapamiętaj
Wczytaj liczbę
Wyświetl, że za dużo
(25)
### Zastosowanie instrukcji — gra w „za dużo, za mało”
#include <iostream> #include <ctime>
using namespace std; int main()
{
int wczytana, wylosowana;
cout << endl << "Witaj w grze w \"Za duzo, za malo\"";
cout << endl << "Odgadnij wylosowana liczbe (1 .. 100)" << endl; srand( ( unsigned )time( NULL ) );
wylosowana = rand() % 100 + 1; . . . return EXIT_SUCCESS; } #include <iostream> #include <ctime> using namespace std; int main() {
int wczytana, wylosowana;
cout << endl << "Witaj w grze w \"Za duzo, za malo\"";
cout << endl << "Odgadnij wylosowana liczbe (1 .. 100)" << endl;
srand( ( unsigned )time( NULL ) );
wylosowana = rand() % 100 + 1;
. . .
return EXIT_SUCCESS; }
Losowanie zalążka generatora liczb pseudolosowych zależnego ob aktualnej wartości timera
Losowanie zalążka generatora liczb pseudolosowych zależnego ob aktualnej wartości timera
Plik nagłówkowy zwykle potrzebny dla funkcji time
Plik nagłówkowy zwykle potrzebny dla funkcji time
Po co ten srand, po co ten %?
Losowanie liczby pseudolosowej, operator % to modulo
(26)
srand( 1000 );
srand( 1000 );
### przykładowo uzależniając go od bieżącego czasu:
srand( ( unsigned )time( NULL ) );
srand( ( unsigned )time( NULL ) );
(27)
### Zastosowanie instrukcji — gra w „za dużo, za mało”
int main() { . . . wylosowana = rand() % 100 + 1; do { cout << '>' << flush; cin >> wczytana;
if( wczytana > wylosowana ) cout << "Za duzo" << endl; else
if( wczytana < wylosowana ) cout << "Za malo" << endl; else
cout << "Brawo, to ta liczba!" << endl << "Nacisnij Enter by zakonczyc"; }
while( wylosowana != wczytana ); cin.ignore(); cin.get(); return EXIT_SUCCESS; } int main() { . . . wylosowana = rand() % 100 + 1; do { cout << '>' << flush; cin >> wczytana;
if( wczytana > wylosowana ) cout << "Za duzo" << endl; else
if( wczytana < wylosowana ) cout << "Za malo" << endl; else
cout << "Brawo, to ta liczba!" << endl << "Nacisnij Enter by zakonczyc"; }
while( wylosowana != wczytana );
cin.ignore(); cin.get();
return EXIT_SUCCESS;
} Wykonuj, dopóki liczba nie została odgadniętaWykonuj, dopóki liczba nie została odgadnięta
Wczytaj liczbę typowaną przez gracza
Wczytaj liczbę typowaną przez gracza
Porównaj wczytaną liczbę z wylosowaną i wyprowadź odpowiedni komunikat
Porównaj wczytaną liczbę z wylosowaną i wyprowadź odpowiedni komunikat
(28)
(29)
### Instrukcja switch + do-while = proste menu
. . .
char klawisz; do
{
cout << "\nFormatowanie dysku, wybierz opcje:\n1. Format";
cout << "\n2. Szybki format\n3. Diagnostyka\n4. Koniec\n>" << flush; cin >> klawisz;
switch( klawisz ) {
case '1' : cout << "\nWybrales formatowanie\n"; break;
case '2' : cout << "\nWybrales szybki format\n"; break;
case '3' : cout << "\nWybrales diagnostyke\n"; break;
} }
while( klawisz != '4' );
cout << "\n\nWybrales Koniec, nacisnij Enter by potwierdzic..."; cin.ignore(); cin.get(); . . . . . . char klawisz; do {
cout << "\nFormatowanie dysku, wybierz opcje:\n1. Format";
cout << "\n2. Szybki format\n3. Diagnostyka\n4. Koniec\n>" << flush; cin >> klawisz;
switch( klawisz ) {
case '1' : cout << "\nWybrales formatowanie\n"; break;
case '2' : cout << "\nWybrales szybki format\n"; break;
case '3' : cout << "\nWybrales diagnostyke\n"; break;
} }
while( klawisz != '4' );
cout << "\n\nWybrales Koniec, nacisnij Enter by potwierdzic..."; cin.ignore();
cin.get(); . . .
(30)
### Typ zmiennej sterującej a typ wartości przypadku
char klawisz; . . .
cin >> klawisz; switch( klawisz ) {
case '1' : cout << "\nWybrales formatowanie\n"; break;
case '2' : cout << "\nWybrales szybki format\n"; break;
case '3' : cout << "\nWybrales diagnostyke\n"; break; } char klawisz; . . . cin >> klawisz; switch( klawisz ) {
case '1' : cout << "\nWybrales formatowanie\n"; break;
case '2' : cout << "\nWybrales szybki format\n"; break;
case '3' : cout << "\nWybrales diagnostyke\n"; break; } int nadwozie; . . . cin >> nadwozie; switch( nadwozie ) {
case 1 : cout << "\nChyba lubisz eleganckie limuzyny!"; break;
case 2 : cout << "\nWidze, ze ciagnie Cie w teren!"; break;
case 3 : cout << "\nTy to pewnie lubisz szybka jazde!";
case 1 : cout << "\nChyba lubisz eleganckie limuzyny!"; break;
case 2 : cout << "\nWidze, ze ciagnie Cie w teren!"; break;
case 3 : cout << "\nTy to pewnie lubisz szybka jazde!"; break;
(31)
### Instrukcja iteracyjna while
while( wyra enie )ż instrukcja
while( wyra enie )ż instrukcja
while( wyra enie )ż {
ci g instrukcjią }
while( wyra enie )ż {
ci g instrukcjią }
wyrażenie
wyrażenie
instrukcja
instrukcja
(32)
### Instrukcja iteracyjna while — sylwestrowe odliczanie ;-)
int licznik = 10;
while( licznik > 0 ) {
cout << endl << licznik << "..."; licznik--;
}
cout << endl << "Nowy Rok!!!" << endl;
int licznik = 10;
while( licznik > 0 ) {
cout << endl << licznik << "..."; licznik--;
}
cout << endl << "Nowy Rok!!!" << endl;
licznik-– odpowiada licznik = licznik - 1 licznik-– odpowiada licznik = licznik - 1 int licznik = 10; do {
cout << endl << licznik << "..."; licznik--;
}
while( licznik > 0 )
cout << endl << "Nowy Rok!!!" << endl;
int licznik = 10; do
{
cout << endl << licznik << "..."; licznik--;
}
while( licznik > 0 )
cout << endl << "Nowy Rok!!!" << endl;
(33)
(34)
### Instrukcja iteracyjna for — ogólny schemat
inicjalizacja; while( warunek ) { ciało_iteracji modyfikacja } inicjalizacja; while( warunek ) { ciało_iteracji modyfikacja }
for( inicjalizacja; warunek; modyfikacja ) ciało_iteracji
for( inicjalizacja; warunek; modyfikacja ) ciało_iteracji
### W części inicjalizacyjnej w C++ wolno deklarować zmienne, w C nie wolno.
for( int licznik = 10; licznik > 0; cout << endl << licznik-- << "..." ) ;
for( int licznik = 10; licznik > 0; cout << endl << licznik-- << "..." ) ;
(35)
### Instrukcje break i continue
switch( klawisz ) { case '1' : full_disk_format(); break; case '2' : quick_disk_format(); break; case '3' : check_disk(); break; } switch( klawisz ) { case '1' : full_disk_format(); break; case '2' : quick_disk_format(); break; case '3' : check_disk(); break; }
### i przejście do wykonania następnego jej przebiegu od jego początku.
Dla iteracji while i do-while instrukcja continue powoduje przeniesienie sterowania do fazy testowania warunku kontynuacji. W przypadku iteracji for sterowanie przenoszone jest do wyrażenia modyfikującego a potem do testowania warunku.
Dla iteracji while i do-while instrukcja continue powoduje przeniesienie sterowania do fazy testowania warunku kontynuacji. W przypadku iteracji for sterowanie przenoszone jest do wyrażenia modyfikującego a potem do testowania warunku.
cin >> klawisz; while( klawisz != 'k' ) { if( klawisz == 'q' ) break; else . . . } cin >> klawisz; while( klawisz != 'k' ) { if( klawisz == 'q' ) break; else . . . }
(36)
### Przykład zastosowania instrukcji break i continue w iteracji
. . .
for( ; ; ) // Iteracja niesko czona czyli p tla ń ę {
cout << "\nFormatowanie dysku, wybierz opcje:\n1. Format";
cout << "\n2. Szybki format\n3. Diagnostyka\n4. Koniec\n>" << flush; cin >> klawisz;
if( klawisz == ’4’ ) // Czy wybrano klawisz ko ca?ń break; // Tak, wychodzimy z iteracji
if( !( klawisz >= ’0’&& klawisz <= ’9’) ) // Czy klawisz jest cyfr ?ą continue; // Klawisz nie jest cyfr , nie ma co dalej sprawdzaćą switch( klawisz ) { case '1' : full_disk_format(); break; case '2' : quick_disk_format(); break; case '3' : check_disk(); break; }// switch } . . .
for( ; ; ) // Iteracja niesko czona czyli p tla ń ę
{
cout << "\nFormatowanie dysku, wybierz opcje:\n1. Format";
cout << "\n2. Szybki format\n3. Diagnostyka\n4. Koniec\n>" << flush; cin >> klawisz;
if( klawisz == ’4’ ) // Czy wybrano klawisz ko ca?ń break; // Tak, wychodzimy z iteracji
if( !( klawisz >= ’0’&& klawisz <= ’9’) ) // Czy klawisz jest cyfr ?ą
continue; // Klawisz nie jest cyfr , nie ma co dalej sprawdzaćą
switch( klawisz ) { case '1' : full_disk_format(); break; case '2' : quick_disk_format(); break; case '3' : check_disk(); break; }// switch }
(37)
(38)
### Suplement — wykorzystanie instrukcji, przykład 1
#include <iostream> using namespace std;
const int MAKS_LB_KURSOW = 24; int main()
{
float kurs, sumaryczny; int lb_kursow;
cout << "\nWyznaczam dobowy sredn kurs waluty EURO."; cout << "\nWprowadz 24 dodatnie liczby -- kursy EURO";
cout << "\nzanotowane na poczatku kazdej godziny.\n" << flush;
for( sumaryczny = 0, lb_kursow = 1; lb_kursow <= MAKS_LB_KURSOW; lb_kursow++ ) {
cout << '>' << flush; cin >> kurs;
sumaryczny = sumaryczny + kurs; }//for
cout << "\nKurs sredni: " << sumaryczny / MAX_LB_KURSOW; cout << endl << "Nacisnij Enter by zakonczyc...";
cin.ignore(); cin.get();
#include <iostream> using namespace std;
const int MAKS_LB_KURSOW = 24;
int main() {
float kurs, sumaryczny; int lb_kursow;
cout << "\nWyznaczam dobowy sredn kurs waluty EURO."; cout << "\nWprowadz 24 dodatnie liczby -- kursy EURO";
cout << "\nzanotowane na poczatku kazdej godziny.\n" << flush;
for( sumaryczny = 0, lb_kursow = 1; lb_kursow <= MAKS_LB_KURSOW; lb_kursow++ ) {
cout << '>' << flush; cin >> kurs;
sumaryczny = sumaryczny + kurs; }//for
cout << "\nKurs sredni: " << sumaryczny / MAX_LB_KURSOW; cout << endl << "Nacisnij Enter by zakonczyc...";
cin.ignore(); cin.get();
Umieszczenie const przed zmienna sprawia, że jej wartość w trakcie wykonania programu nie może się zmieniać
Umieszczenie const przed zmienna sprawia, że jej wartość w trakcie wykonania programu nie może się zmieniać
Java — istnieje specyfikacja final pozwalające na uzyskanie podobnego efektu, to nie to samo co const z języka C++. W języku C# występuje słowo const interpretowane podobnie jak w języku C++.
Java — istnieje specyfikacja final pozwalające na uzyskanie podobnego efektu, to nie to samo co const z języka C++. W języku C# występuje słowo const interpretowane podobnie jak w języku C++.
(39)
(40)
### Suplement — wykorzystanie instrukcji, przykład 2
#include <iostream>
using namespace std;
const int MAKS_LB_KURSOW = 24; int main()
{
float kurs, sumaryczny, maksymalny, minimalny; int lb_kursow;
cout << "\nWyznaczam dobowy, sredni, minimalny i maksymalny kurs"; cout << "\nwaluty EURO. Wprowadz 24 dodatnie liczby -- kursy EURO"; cout << "\nzanotowane na poczatku kazdej godziny.\n>" << flush;
// Wczytanie pierwszego kursu cin >> kurs;
// Pierwszy i jedyny na razie kurs to kurs minimalny, maksymalny i sumaryczny maksymalny = minimalny = sumaryczny = kurs;
#include <iostream> using namespace std;
const int MAKS_LB_KURSOW = 24; int main()
{
float kurs, sumaryczny, maksymalny, minimalny; int lb_kursow;
cout << "\nWyznaczam dobowy, sredni, minimalny i maksymalny kurs"; cout << "\nwaluty EURO. Wprowadz 24 dodatnie liczby -- kursy EURO"; cout << "\nzanotowane na poczatku kazdej godziny.\n>" << flush;
// Wczytanie pierwszego kursu
cin >> kurs;
// Pierwszy i jedyny na razie kurs to kurs minimalny, maksymalny i sumaryczny
maksymalny = minimalny = sumaryczny = kurs;
1/2 1/2
(41)
### Suplement — wykorzystanie instrukcji, przykład 2
for( lb_kursow = 2; lb_kursow <= MAKS_LB_KURSOW; lb_kursow++ ) {
cout << '>' << flush; cin >> kurs;
sumaryczny = sumaryczny + kurs; if( kurs < minimalny )
minimalny = kurs; if( kurs > maksymalny ) maksymalny = kurs; }//for
cout << "\nKurs najwyzszy: " << maksymalny; cout << "\nKurs najnizszy: " << minimalny;
cout << "\nKurs sredni: " << sumaryczny / MAKS_LB_KURSOW; cout << endl << "Nacisnij Enter by zakonczyc...";
cin.ignore(); cin.get();
return EXIT_SUCCESS; }
for( lb_kursow = 2; lb_kursow <= MAKS_LB_KURSOW; lb_kursow++ ) {
cout << '>' << flush; cin >> kurs;
sumaryczny = sumaryczny + kurs;
if( kurs < minimalny ) minimalny = kurs; if( kurs > maksymalny ) maksymalny = kurs;
}//for
cout << "\nKurs najwyzszy: " << maksymalny; cout << "\nKurs najnizszy: " << minimalny;
cout << "\nKurs sredni: " << sumaryczny / MAKS_LB_KURSOW; cout << endl << "Nacisnij Enter by zakonczyc...";
cin.ignore(); cin.get();
return EXIT_SUCCESS;
} 2/2
(42)
(43)
(44)
### Suplement — wykorzystanie instrukcji, przykład 3
#include <cstdlib>
#include <iostream> using namespace std; int main()
{
float czas, sumaryczny, najlepszy, najgorszy; int licznik;
cout << "\nWprowadz kolejne czasy, a ja wyznacze nalepszy, najgorszy"; cout << "\noraz sredni. Podaj zerowy czas aby zakonczyc wprowadzanie."; cout << "\nNie rob jaj, nie wpisuj ujemnych czasow.\n>" << flush;
cin >> czas; #include <cstdlib> #include <iostream> using namespace std; int main() {
float czas, sumaryczny, najlepszy, najgorszy; int licznik;
cout << "\nWprowadz kolejne czasy, a ja wyznacze nalepszy, najgorszy"; cout << "\noraz sredni. Podaj zerowy czas aby zakonczyc wprowadzanie."; cout << "\nNie rob jaj, nie wpisuj ujemnych czasow.\n>" << flush;
cin >> czas;
1/3
(45)
### Suplement — wykorzystanie instrukcji, przykład 3
if( czas == 0 )
cout << "Nic do roboty"; else
{
sumaryczny = najgorszy = najlepszy = czas; licznik = 1; cout << ">" << flush; cin >> czas; while( czas != 0 ) {
sumaryczny = sumaryczny + czas; // lepiej: sumaryczny += czas; if( czas > najgorszy )
najgorszy = czas; if( czas < najlepszy ) najlepszy = czas;
licznik = licznik + 1; // lepiej: licznik++;
cout << ">" << flush; cin >> czas;
}//while
cout << "\nCzas najlepszy: " << najlepszy; cout << "\nCzas najgorszy: " << najgorszy;
cout << "\nCzas sredni: " << sumaryczny / licznik; }//if-else
if( czas == 0 )
cout << "Nic do roboty"; else
{
sumaryczny = najgorszy = najlepszy = czas; licznik = 1; cout << ">" << flush; cin >> czas; while( czas != 0 ) {
sumaryczny = sumaryczny + czas; // lepiej: sumaryczny += czas;
if( czas > najgorszy ) najgorszy = czas; if( czas < najlepszy ) najlepszy = czas;
licznik = licznik + 1; // lepiej: licznik++;
cout << ">" << flush; cin >> czas;
}//while
cout << "\nCzas najlepszy: " << najlepszy; cout << "\nCzas najgorszy: " << najgorszy;
cout << "\nCzas sredni: " << sumaryczny / licznik;
(46)
### Suplement — wykorzystanie instrukcji, przykład 3
cout << endl << "Nacisnij Enter by zakonczyc...";
cin.ignore(); cin.get();
return EXIT_SUCCESS; }
cout << endl << "Nacisnij Enter by zakonczyc..."; cin.ignore(); cin.get(); return EXIT_SUCCESS; } 2/3 2/3
(47)
### Teraz, albo:
Cytaty
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http://stackoverflow.com/questions/4084050/can-you-use-arithmetic-operators-to-flip-between-0-and-1/4084061 | 1,394,474,877,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010925635/warc/CC-MAIN-20140305091525-00028-ip-10-183-142-35.ec2.internal.warc.gz | 172,611,400 | 16,760 | # Can You Use Arithmetic Operators to Flip Between 0 and 1
Is there a way without using logic and bitwise operators, just arithmetic operators, to flip between integers with the value 0 and 1?
ie. "variable ?= variable" will make variable a 1 if it 0 or 0 if it is a one
-
``````x = 1 - x
``````
Will switch between 0 and 1.
-
Assuming that it is initialized as a 0 or 1:
``````x = 1 - x
``````
-
Edit: I misread the question, thought the OP could use `any` operator
A Few more...(ignore these)
``````x ^= 1 //bitwise operator
x = !x //logical operator
x = (x <= 0) // kinda the same as x != 1
``````
Without using an operator?
``````int arr[] = {1,0}
x = arr[x]
``````
-
^ is a bitwise operator and ! is a logical operator. – MAK Nov 3 '10 at 20:04
+1 for the second method – Fabio F. Nov 3 '10 at 20:24
@MAK, i misread the question :( `edited` – st0le Nov 4 '10 at 4:26
While your second idea works, x = 1 - x is a much better solution. Having an array is overkill/wasteful. – Khalos Nov 4 '10 at 4:33
Yet another way:
``````x = (x + 1) % 2
``````
-
+1 because this will always give you 0 or 1 even if you accidentally give an integer outside that range. – Nick Forge Nov 4 '10 at 4:30
``````int flip(int i){
return 1 - i;
};
``````
-
Just for a bit of variety:
``````x = 1 / (x + 1);
x = (x == 0);
x = (x != 1);
``````
Not sure whether you consider `==` and `!=` to be arithmetic operators. Probably not, and obviously although they work in C, more strongly typed languages wouldn't convert the result to integer.
-
``````x = "\1"[x]
And if you like that, you'll love `x = x["\1"];` (C and C++). – Steve Jessop Nov 8 '10 at 22:03 | 550 | 1,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2014-10 | latest | en | 0.88998 |
https://scholar.archive.org/work/lctwhkv67baq7btbhexjzmeega | 1,642,384,844,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300253.51/warc/CC-MAIN-20220117000754-20220117030754-00684.warc.gz | 591,850,059 | 5,648 | ### Asymptotics of some nonlinear eigenvalue problems modelling a MEMS Capacitor. Part II: multiple solutions and singular asymptotics
A. E. LINDSAY, M. J. WARD
2010 European journal of applied mathematics
Some nonlinear eigenvalue problems related to the modelling of the steady-state deflection of an elastic membrane associated with a MEMS capacitor under a constant applied voltage are analyzed using formal asymptotic methods. These problems consist of certain singular perturbations of the basic membrane nonlinear eigenvalue problem ∆u = λ/(1 + u) 2 in Ω with u = 0 on ∂Ω, where Ω is the unit ball in R 2 . It is well-known that the radially symmetric solution branch to this basic membrane problem
more » ... ic membrane problem has an infinite fold-point structure with λ → 4/9 as ε ≡ 1 − ||u||∞ → 0 + . One focus of this paper is to develop a novel singular perturbation method to analytically determine the limiting asymptotic behaviour of this infinite fold-point structure in terms of two constants that must be computed numerically. This theory is then extended to certain generalizations of the basic membrane problem in the N -dimensional unit ball. The second main focus of this paper is to analyze the effect of two distinct perturbations of the basic membrane problem in the unit disk resulting from including either a bending energy term of the form −δ∆ 2 u to the operator, or inserting a concentric inner undeflected disk of radius δ. For each of these perturbed problems, it is shown numerically that the infinite fold-point structure for the basic membrane problem is destroyed when δ > 0, and that there is a maximal solution branch for which λ → 0 as ε ≡ 1 − ||u||∞ → 0 + . For δ > 0, a novel singular perturbation analysis is used in the limit ε → 0 + to construct the limiting asymptotic behaviour of the maximal solution branch for the biharmonic problem in the unit slab and the unit disk, and for the annulus problem in the unit disk. The asymptotic results for the bifurcation curves are shown to compare very favourably with full numerical results. | 468 | 2,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-05 | latest | en | 0.89264 |
https://ask.sagemath.org/answers/11923/revisions/ | 1,723,278,566,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00278.warc.gz | 91,895,934 | 6,084 | # Revision history [back]
It looks like Sage is (incorrectly? naively?) applying the chain rule to conjugate(q(x,t)), since conjugate is a pure symbolic thing in Sage a priori. In fact, the multiple variables are irrelevant:
sage: r = function('r',x)
sage: g = r*r.conjugate()
sage: g.derivative(x)
r(x)*D[0](conjugate)(r(x))*D[0](r)(x) + conjugate(r(x))*D[0](r)(x)
sage: r.conjugate().derivative(x)
D[0](conjugate)(r(x))*D[0](r)(x)
sage: r.derivative().conjugate()
conjugate(D[0](r)(x))
Ginac (and hence Pynac) does not appear to support doing this - see this. However, presumably one could implement a trivial conjugation to the fderivative or tderivative ... the problem is that I'm not sure whether we should automatically assume that this makes sense. See, for instance, the convoluted discussion at The Maple equivalent of ask.sagemath. On the other hand, Sympy allows this. If there is a consensus, a ticket should be opened, though!
It looks like Sage is (incorrectly? naively?) applying the chain rule to conjugate(q(x,t)), since conjugate is a pure symbolic thing in Sage a priori. In fact, the multiple variables are irrelevant:
sage: r = function('r',x)
sage: g = r*r.conjugate()
sage: g.derivative(x)
r(x)*D[0](conjugate)(r(x))*D[0](r)(x) + conjugate(r(x))*D[0](r)(x)
sage: r.conjugate().derivative(x)
D[0](conjugate)(r(x))*D[0](r)(x)
sage: r.derivative().conjugate()
conjugate(D[0](r)(x))
Ginac (and hence Pynac) does not appear to support doing this - see this. However, presumably one could implement a trivial conjugation to the fderivative or tderivative ... the problem is that I'm not sure whether we should automatically assume that this makes sense. See, for instance, the convoluted discussion at The Maple equivalent of ask.sagemath. On the other hand, Sympy allows this. If there is a consensus, a ticket should be opened, though! | 535 | 1,865 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-33 | latest | en | 0.760354 |
https://oeis.org/A115729/internal | 1,597,027,892,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738603.37/warc/CC-MAIN-20200810012015-20200810042015-00243.warc.gz | 426,300,583 | 2,960 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A115729 Number of subpartitions of partitions in Mathematica order. 12
%I
%S 1,2,3,3,4,5,4,5,7,6,7,5,6,9,9,10,9,9,6,7,11,12,10,13,14,10,13,12,11,
%T 7,8,13,15,14,16,19,16,16,17,19,14,16,15,13,8,9,15,18,18,15,19,24,23,
%U 22,19,21,26,22,23,15,21,24,18,19,18,15,9,10,17,21,22,20,22
%N Number of subpartitions of partitions in Mathematica order.
%C subpart([n^k]) = C(n+k,k); subpart([n,n-1,n-2,...,1]) = C_n = A000108(n).
%F For a partition P = [p_1,...,p_n] with the p_i in decreasing order, define b(i,j) to be the number of subpartitions of [p_1,...,p_i] with the i-th part = j (b(i,0) is subpartitions with less than i parts). Then b(1,j)=1 for j<=p_1, b(i+1,j) = Sum_{k=j}^{p_i} b(i,k) for 0<=k<=p_{i+1}; and the total number of subpartitions is sum_{k=1}^{p_n} b(n,k).
%e Partition 5 in Mathematica order is [2,1]; it has 5
%e subpartitions: [], [1], [2], [1^2] and [2,1] itself.
%o (PARI) /* Expects input as vector in decreasing order - e.g. [3,2,1,1] */ subpart2(p)=local(i,j,v,n,k);n=matsize(p)[2];if(n==0,1,v=vector(p[1]+1,i, 1);for(i=1,n,k=p[i];for(j=1,k,v[k+1-j]+=v[k+2-j]));v[1])
%Y Cf. A115728, A080577, A000108, A007318.
%K nonn
%O 0,2
%A _Franklin T. Adams-Watters_, Mar 11 2006
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Last modified August 9 22:42 EDT 2020. Contains 336335 sequences. (Running on oeis4.) | 686 | 1,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-34 | latest | en | 0.655957 |
https://pythonexamples.org/python-less-than/ | 1,618,375,057,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038076819.36/warc/CC-MAIN-20210414034544-20210414064544-00359.warc.gz | 560,509,406 | 11,312 | # Python Less Than (<) Operator
## Python Less Than Operator
Python Less Than operator is used to compare if an operand is less than other operand.
The syntax of less than comparison operator is
``operand_1 < operand_2``
Run
Less than operator returns a boolean value. True if operand_1 is less than operand_2 in value. Otherwise, it returns False. If the operands are sequences like strings, lists, tuple, etc., corresponding elements are compared to compute the result.
### Example 1: Less Than Operator
In this example, we will compare two integers, x and y, and check if x is less than y.
Python Program
``````x = 5
y = 12
result = x < y
print(result) #True
x = 8
y = 7
result = x < y
print(result) #False``````
Run
Output
``````True
False``````
For x = 5, and y = 12, x < y returns True.
For x = 8, and y = 7, x < y returns False.
### Example 2: Less Than Operator with String Operands
You can compare if a Python String is less than other string. Python considers lexicographic order of alphabets, or you could say the ASCII value. In that case, the alphabet ‘a’ is less than alphabet ‘b’, and ‘b’ is less than ‘c’, and so on. The same explanation holds for other possible characters in a string.
While comparing, the first character of each string is compared. If they are equal, next characters are compared, else the result is returned.
In this example, we will compare two strings, x and y, and check if one string is less than other.
Python Program
``````x = 'apple'
y = 'banana'
z = 'cherry'
k = 'Apple'
print(x < y) #True
print(y < z) #True
print(x < z) #True
print(x < k) #False``````
Run
Output
``````True
True
True
False``````
`'a'` is less than `'b'` and therefore `'apple'` is less than `'banana'`. So, x<y returned True. Similary for `y<z` ad `x<z`.
`'a'` is greater than `'A'`. Therefore `'apple' < 'Apple'` returned `False`.
### Example 3: Less Than Operator with Lists
Just like strings, Python Lists can be compared too. And the comparison happens in the same way.
In this example, we will compare some lists.
Python Program
``````x = [41, 54, 21]
y = [98, 8]
z = [41, 54, 4, 6]
print(x < y) #True
print(y < z) #False
print(x < z) #False``````
Run
Output
``````True
False
False``````
`[41, 54, 21]` less than `[98, 8]` first compares the elements `41` and `98`. The result is straight away `True` and the operator returns `True`.
`[98, 8]` less than `[41, 54, 4, 6]` first compares the elements `98` and `41`. The result is straight away `False` and the operator returns `False`.
`[41, 54, 21]` less than `[41, 54, 4, 6]` first compares the elements `41` and `41`. No result. Then `54` and `54` are compared. Still no result. Then `21` and `4` are compared. This returns False for `21 < 4` .
### Summary
In this tutorial of Python Examples, we learned how to compare two values or sequences like strings, lists, etc., using less than comparison operator. | 854 | 2,913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-17 | latest | en | 0.849055 |
https://www.calculatoratoz.com/en/coefficient-of-friction-between-the-cylinder-and-the-surface-of-inclined-plane-if-cylinder-is-rolling-without-slipping-down--calculator/Calc-508 | 1,611,332,787,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703530835.37/warc/CC-MAIN-20210122144404-20210122174404-00390.warc.gz | 696,929,877 | 24,694 | Anshika Arya
National Institute Of Technology (NIT), Hamirpur
Anshika Arya has created this Calculator and 400+ more calculators!
## < 3 Other formulas that you can solve using the same Inputs
Maximum Height attained
Maximum Height=((Initial Velocity*sin(Angle of Inclination))^2)/(2*Acceleration Due To Gravity*cos(Angle of plane)) GO
Time of Flight for Inclined Projectile
Time=(2*Initial Velocity*sin(Angle of Inclination))/(Acceleration Due To Gravity*cos(Angle of plane)) GO
Force of Friction between the cylinder and the surface of inclined plane if cylinder is rolling without slipping down a ramp
Force=(Mass*Acceleration Due To Gravity*sin(Angle of Inclination))/3 GO
## < 2 Other formulas that calculate the same Output
Coefficient of Friction (using forces)
Coefficient of Friction=(Centripetal Force*tan(Theta)+Tangential Force)/(Centripetal Force-Tangential Force*tan(Theta)) GO
Coefficient of Friction
Coefficient of Friction=Limiting Force/Normal reaction GO
### Coefficient of Friction between the cylinder and the surface of inclined plane if cylinder is rolling without slipping down Formula
Coefficient of Friction=(tan(Angle of Inclination))/3
More formulas
Moment of Inertia of a rod about an axis through its center of mass and perpendicular to rod GO
Moment of inertia of a circular ring about an axis through its center and perpendicular to its plane GO
Moment of inertia of a circular disc about an axis through its center and perpendicular to its plane GO
Moment of Inertia of a right circular solid cylinder about its symmetry axis GO
Moment of Inertia of a right circular hollow cylinder about its axis GO
Moment of Inertia of a solid sphere about its diameter GO
Moment of Inertia of a spherical shell about its diameter GO
Force of Friction between the cylinder and the surface of inclined plane if cylinder is rolling without slipping down a ramp GO
Moment of inertia of bob of pendulum, about an axis through the point of suspension GO
## How do you calculate the coefficient of friction?
The coefficient of friction (fr) is a number that is the ratio of the resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together. It is represented by the equation: fr = Fr/N.
## What is the largest coefficient of friction?
They being said, aluminum on aluminum when both surfaces are clean, smooth, and dry has a coefficient of static friction of between 1.05 and 1.35. That's higher than unity and extremely rare. It's about as high as it gets.
## How to Calculate Coefficient of Friction between the cylinder and the surface of inclined plane if cylinder is rolling without slipping down ?
Coefficient of Friction between the cylinder and the surface of inclined plane if cylinder is rolling without slipping down calculator uses Coefficient of Friction=(tan(Angle of Inclination))/3 to calculate the Coefficient of Friction, The Coefficient of Friction between the cylinder and the surface of inclined plane if cylinder is rolling without slipping down is the ratio defining the force that resists the motion of one body in relation to another body in contact with it. This ratio is dependent on material properties and most materials have a value between 0 and 1. Coefficient of Friction and is denoted by μ symbol.
How to calculate Coefficient of Friction between the cylinder and the surface of inclined plane if cylinder is rolling without slipping down using this online calculator? To use this online calculator for Coefficient of Friction between the cylinder and the surface of inclined plane if cylinder is rolling without slipping down , enter Angle of Inclination (θ) and hit the calculate button. Here is how the Coefficient of Friction between the cylinder and the surface of inclined plane if cylinder is rolling without slipping down calculation can be explained with given input values -> 83.39953 = (tan(89.771))/3.
### FAQ
What is Coefficient of Friction between the cylinder and the surface of inclined plane if cylinder is rolling without slipping down ?
The Coefficient of Friction between the cylinder and the surface of inclined plane if cylinder is rolling without slipping down is the ratio defining the force that resists the motion of one body in relation to another body in contact with it. This ratio is dependent on material properties and most materials have a value between 0 and 1 and is represented as μ=(tan(θ))/3 or Coefficient of Friction=(tan(Angle of Inclination))/3. Angle of Inclination is formed by the inclination of one line to another; measured in degrees or radians.
How to calculate Coefficient of Friction between the cylinder and the surface of inclined plane if cylinder is rolling without slipping down ?
The Coefficient of Friction between the cylinder and the surface of inclined plane if cylinder is rolling without slipping down is the ratio defining the force that resists the motion of one body in relation to another body in contact with it. This ratio is dependent on material properties and most materials have a value between 0 and 1 is calculated using Coefficient of Friction=(tan(Angle of Inclination))/3. To calculate Coefficient of Friction between the cylinder and the surface of inclined plane if cylinder is rolling without slipping down , you need Angle of Inclination (θ). With our tool, you need to enter the respective value for Angle of Inclination and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Coefficient of Friction?
In this formula, Coefficient of Friction uses Angle of Inclination. We can use 2 other way(s) to calculate the same, which is/are as follows -
• Coefficient of Friction=Limiting Force/Normal reaction
• Coefficient of Friction=(Centripetal Force*tan(Theta)+Tangential Force)/(Centripetal Force-Tangential Force*tan(Theta))
Let Others Know | 1,260 | 5,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-04 | latest | en | 0.805777 |
http://www.evi.com/q/how_many_kb_in_500mb | 1,386,774,431,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164037762/warc/CC-MAIN-20131204133357-00001-ip-10-33-133-15.ec2.internal.warc.gz | 331,351,752 | 18,908 | # How many kb in 500mb?
• the capacity 524,288 Kilobytes
• Assuming you meant
• the capacity 500 mebibytes and Kilobyte (derived from the SI prefix "kilo-", meaning 1,000), the unit of digital information storage equal to either 1,000 bytes (10) or 1,024 bytes (2), depending on context
Did you mean?
tk10npubl tk10ncanl
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• 500mb in kilobites | 431 | 1,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2013-48 | longest | en | 0.869003 |
http://math.stackexchange.com/questions/244049/0-lim-varepsilon-rightarrow-o-int-omega-frac-chi-u-varepsi | 1,469,766,749,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829970.64/warc/CC-MAIN-20160723071029-00264-ip-10-185-27-174.ec2.internal.warc.gz | 164,525,725 | 16,781 | # $0 = \lim_{\varepsilon \rightarrow o^+} \int_{\Omega} \frac{ \chi( u + \varepsilon \xi > 0 ) - \chi(\{ u>0 \} ) }{\varepsilon}?$
Let $\Omega$ be a domain, $u \in H^{1}(\Omega)$ and $\xi \in C^{\infty}_{0}(\{u > 0\})$ you can assume that $\{u > 0\}$ is an open set. I'd like to know if in this situation we can conclude that $$0 = \lim_{\varepsilon \rightarrow o^+} \int_{\Omega} \dfrac{ \chi( u + \varepsilon \xi >0 ) - \chi(\{ u>0 \} ) }{\varepsilon}?$$ Where $\chi(A)$ is the characteristc function of a set $A$. Or if can I assume another similar condition for that the limit above be true. This ask is motivated for instance by the question here. I've also read other things that motivate this suspicion.
-
user: There is no need to make your title larger than other titles to questions. It is perfectly legible as is. – amWhy Nov 25 '12 at 2:21 | 285 | 853 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2016-30 | latest | en | 0.810895 |
https://metsminorleagueblog.com/what-does-1037-mean-in-police-code/ | 1,719,074,680,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00565.warc.gz | 324,884,498 | 18,736 | # What does 1037 mean in Police Code?
## Introduction
In the world of law enforcement, communication is vital. Police officers use various codes and signals to convey information quickly and efficiently, allowing them to respond to incidents swiftly and maintain a sense of order. Among these codes is “1037,” a cryptic sequence that holds specific meaning within the police community. In this comprehensive blog post, we will unravel the enigma behind “1037” and explore its significance in police code.
## The Origins of Police Codes
Codes and signals have long been an integral part of police work. Originating from the need to communicate discreetly over radio frequencies, these codes emerged as a way to convey information quickly without divulging sensitive details to potential eavesdroppers. From the early days of police radio communication to the present, codes have evolved and expanded, becoming an essential tool for law enforcement officers worldwide.
Within this rich tapestry of codes, “1037” emerges as a curious numerical arrangement. Let us dive into the depths of police code to uncover its true meaning.
### The Symbolism of “10” Codes
In police parlance, “10” codes are widely used to communicate specific information concisely. These codes typically consist of a two-digit number, where the first digit indicates a general category, and the second digit denotes a particular situation or message. For example, “10-4” is a well-known code meaning “message received,” while “10-20” indicates a request for a location.
Understanding the symbolism of “10” codes is crucial to deciphering the meaning behind “1037.” By examining the context in which “1037” is used, we can discern its underlying message.
### a. Analyzing the First Digit: “10”
In the case of “1037,” the first digit is “10.” This suggests that the code falls under the category of general police information. Codes starting with “10” often encompass a wide range of messages, from routine administrative details to urgent alerts. Therefore, to unlock the secrets of “1037,” we must explore further.
### b. Decoding the Second Digit: “3”
The second digit of “1037” is “3.” This number, when used in conjunction with the “10” prefix, holds a specific significance within police code. It implies a message related to road and traffic matters. To proceed, we must dive deeper into the realm of traffic-related police codes.
### Traffic Enforcement Codes
In the realm of traffic enforcement, police codes play a crucial role in streamlining communication among officers and coordinating responses to various situations on the road. Let us explore the potential connections between “1037” and traffic-related codes, shedding light on its true meaning.
### a. Analyzing the First Digit: “1”
Considering the first digit of “1037” as “1,” we can infer that the code corresponds to a specific aspect of traffic enforcement. Codes starting with “1” often relate to traffic control, regulations, and violations, acting as a vital tool for officers to navigate the intricacies of road management.
### b. Decoding the Second Digit: “037”
The second digit, “037,” unveils a more nuanced layer of the code’s meaning. While not a standardized code on its own, “037” suggests a possible subcategory within the traffic enforcement context. To delve further, we must explore common traffic-related codes that align with this pattern.
### a. Speeding Offenses: “1037” and Speed Enforcement
When examining the possibilities within traffic enforcement, one potential interpretation of “1037” could be its association with speeding offenses. Speed enforcement is a critical component of maintaining road safety, and officers utilize specific codes to relay information about speed-related incidents. Codes like “1037-01” could indicate a speeding violation, while “1037-02” might signify a pursuit involving a speeding vehicle. The inclusion of “1037” in these codes reinforces its connection to traffic enforcement.
### b. Reckless Driving: Unraveling the Dangers of “1037”
Another plausible scenario tied to “1037” could involve instances of reckless driving. Codes related to reckless driving often highlight the gravity of the offense and aid officers in responding effectively. For example, “1037-03” might signify a driver displaying erratic behavior, endangering others on the road. By associating “1037” with reckless driving, law enforcement agencies can quickly communicate and address potential threats.
### a. Traffic Accidents: “1037” and Collision Response
Traffic accidents demand immediate attention from law enforcement, and codes play a pivotal role in coordinating accident responses. Within this framework, “1037” could signify an accident-related code, directing officers to a specific incident. For instance, “1037-04” might indicate a minor collision requiring assistance, while “1037-05” could denote a severe accident with potential injuries. By incorporating “1037” into accident-related codes, police departments ensure effective communication during critical moments.
### b. Road Hazards: Identifying “1037” in Hazard Situations
In the context of traffic incidents, “1037” might also be associated with codes relating to road hazards. These hazards encompass a wide range of situations, from fallen debris to malfunctioning traffic signals. Codes like “1037-06” could alert officers to hazardous road conditions, facilitating prompt intervention to ensure public safety. By employing “1037” in hazard-related codes, law enforcement agencies prioritize the effective management of potentially dangerous situations.
#### Conclusion
The enigmatic police code “1037” holds significant meaning within the realm of law enforcement and traffic control. By unraveling its intricacies and analyzing potential scenarios, we have gained insights into the possible interpretations of this code. From speeding violations and reckless driving to traffic accidents and road hazards, “1037” emerges as a versatile identifier within the extensive web of police codes.
Understanding the diverse applications of police codes enables law enforcement officers to communicate efficiently, respond promptly, and maintain public safety. As we continue to decode the mysteries behind police jargon, “1037” serves as a reminder of the importance of effective communication in the complex and ever-evolving field of law enforcement.
What does 1037 mean in Police Code?
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Limit theorem Chebyshev’s inequality \begin{align} {\sigma ^2} &= \int {{{\left( {x – \mu } \right)}^2}{f_X}\left( x \right)dx} \\ &\ge \int_{ – \infty }^{u – c} {{{\left( {x – \mu } \right)}^2}{f_X}\left( x \right)dx} + \int_{u + c}^\infty {{{\left( {x – \mu } \right)}^2}{f_X}\left( x \right)dx} \\ &\ge {c^2}\int_{ – \infty }^{u – c} {{f_X}\left( x \right)dx} {\rm{ + }}{c^2}\int_{u + c}^\infty {{f_X}\left( x \right)dx} \\ &={c^2}{\rm{P}}\left( {\left| {X – \mu } \right| \ge c} \right) \end{align} \[{\rm{P}}\left( {\left| {X –
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Construct a Perpendicular Bisector
Construct a Perpendicular Bisector. Geometry. Warm Up. Check your homework: work with your partner. Hold one partner’s work over another’s. Did your angles and bisectors coincide perfectly?. Vocabulary.
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Construct a Perpendicular Bisector
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Presentation Transcript
1. Warm Up • Check your homework: work with your partner. Hold one partner’s work over another’s. Did your angles and bisectors coincide perfectly?
2. Vocabulary • Perpendicular: Two lines are perpendicular if they intersect in one point, and any of the angles formed by the intersection of the lines is a 90˚ angle. Two segments or rays are perpendicular if the lines containing them are perpendicular lines. • Right Angle: An angle is called a right angle if its measure is 90˚.
3. A perpendicular bisector of a segment passes through the _____________________ of the segment and forms _____________________ with the segment. • Equidistant: A point is said to be equidistant from two different points and if . A point is said to be equidistant from a point and a line if the distance between and is equal to .
4. Perpendicular Bisector • We now investigate how to construct a perpendicular bisector of a line segment using a compass and straightedge. Using what you know about the construction of an angle bisector, experiment with your construction tools and the following line segment to establish the steps that determine this construction.
5. Steps • Label the endpoints of the segment A and B. • Draw circle A: center A, radius AB. • Draw circle B: center B, radius BA. • Label the points of intersections as C and D. • Draw line CD.
6. Exercise 1 • Now that you are familiar with the construction of a perpendicular bisector, we must make one last observation. Using your compass or ruler, examine the following pairs of segments: • , • , • , Based on your findings, fill in the observation below. Observation: • Any point on the perpendicular bisector of a line segment is _____________________ from the endpoints of the line segment.
7. Exercise 2 • Divide the following segment into 4 segments of equal length. Explain your steps.
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http://bluebulbprojects.com/CountOfThings/results.php?coeff=x&c=300000&sort=cntD&p=1 | 1,590,564,723,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347392141.7/warc/CC-MAIN-20200527044512-20200527074512-00412.warc.gz | 18,905,415 | 4,357 | Bluebulb Projects' The Measure of Things presents:
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It's about 200,000 times the number of Escalators in the State of Wyoming. In other words, the count of Escalators in the State of Wyoming is 0.00000500 times 300,000. (2013 figures) It's about 60,000 times the number of Living U.S. Presidents. In other words, the count of Living U.S. Presidents is 0.0000200 times 300,000. (a.k.a. Presidents of the United States; a.k.a. Presidents of the United States of America; a.k.a. POTUS), (2016 figures) (including current) It's about 35,000 times the number of Symphonies Composed by Beethoven. In other words, 300,000 is 33,333.33333330 times the count of Symphonies Composed by Beethoven, and the count of Symphonies Composed by Beethoven is 0.0000300 times that amount. (a.k.a. Ludwig van Beethoven) Over the course of his life, Beethoven composed nine symphonies over the course of a twenty-four year period. After his death in early 1827, an estimated 20,000 Austrian citizens attended his public funeral. It's about 20,000 times the number of Living Kings and Queens. In other words, the count of Living Kings and Queens is 0.0000500 times 300,000. (2015 figures) (distinct Monarchs)As of 2015, the world has fifteen Kings and Queens reigning over 45 countries. Of all these, Queen Elizabeth the II is the only reigning Queen; she has monarchical authority over the sixteen of the Commonwealth nations. It's about 10,000 times the number of Countries in the European Union. In other words, 300,000 is 11,000 times the count of Countries in the European Union, and the count of Countries in the European Union is 0.0000910 times that amount. (a.k.a. EU) (2016 figures) It's about 6,500 times the number of Teeth in a Great White Shark's Mouth. In other words, the count of Teeth in a Great White Shark's Mouth is 0.00016 times 300,000. (Carcharodon carcharias) (average; including developing rows) It's about 5,500 times the number of Countries in Africa. In other words, the count of Countries in Africa is 0.00018 times 300,000. (2016 figures) (sovereign states recognized by the UN only) | 607 | 2,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-24 | latest | en | 0.915678 |
https://aprovence.com/2023/08/27/ | 1,726,446,670,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.26/warc/CC-MAIN-20240915220324-20240916010324-00792.warc.gz | 88,030,681 | 16,613 | # Day: August 27, 2023
#### The Odds of Winning at Blackjack
Blackjack is a card game where players and the dealer each get two cards and have to decide whether they should ask for more (hit) or stick with their current hand (stand). It is important to understand the odds of a blackjack hand to make the best decision. The odds are calculated using the probability of obtaining a particular card and is expressed as a percentage. This number changes depending on the rules of the game, the house edge, and the number of decks in play. It is also possible to calculate the odds by dividing the probability of pulling an Ace by the probability of drawing a ten-valued card, such as 10, J, Q, or K (there are four of each in a standard 52-card pack).
The chances of a blackjack hand depend on both the dealer’s upcard and the player’s cards. In general, the dealer is likely to bust if their cards total 16 or more. However, the player’s cards are more likely to total 17 or higher than the dealer’s. In this situation, the player should hit to increase their chance of winning.
It is also important to note that the odds of blackjack are dependent on previous trials. This is different from roulette and dice games where the probabilities of a certain outcome are independent of each other. This is because the cards that are played change the composition of the deck and therefore influence the probability of specific hands in future trials.
Another factor that affects the odds of blackjack is the player’s strategy and knowledge of counting cards. Counting cards can increase the player’s odds of winning by indicating which cards are most likely to appear and which ones should be avoided. This technique is called expectation value or EV and it is an essential element of a sound blackjack strategy.
Besides understanding the odds, it is also important to understand how the game is played and the rules. This will help you avoid making mistakes that can lead to big losses. In addition to this, it is vital to have a good bankroll management strategy, as losing streaks can quickly deplete your account.
The key to winning in blackjack is to have a combination of skill, strategy, and luck. While some people believe that blind luck is enough to win, it is actually a lot more difficult to beat the casino than most of us think. In order to improve your chances of winning, you should learn about the various strategies that are available, and always remember that they can only take you so far. Fortunately, you can find many resources online that can help you master the game and make it profitable in the long run. This is particularly true if you use the right strategies in combination with proper money management skills. | 555 | 2,740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-38 | latest | en | 0.970396 |
https://www.physicsforums.com/threads/finding-angle-for-specific-moment.837613/ | 1,527,309,618,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867309.73/warc/CC-MAIN-20180526033945-20180526053945-00494.warc.gz | 782,149,633 | 19,158 | Homework Help: Finding angle for specific moment
1. Oct 14, 2015
werson tan
1. The problem statement, all variables and given/known data
i have question with part B
my working is
13.2 cos tetha (0.86x10^-3)+ 13.2 sin tetha (1.22x10^-3) = 1.95
i gt stucked here. how to proceed? i only have one equation , how to solve two unknown ?
2. Relevant equations
3. The attempt at a solution
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2. Oct 14, 2015
andrevdh
You should not assume the moment is 1.95 for part A. From the given information you should be able to see that theta is 30o, which enables you to calculate the moment of P about A for question a.
3. Oct 14, 2015
werson tan
i have done part A , i just have problem with part b , i did not assume the moment is 1.95 for part A. i gt -1.788N for part a
4. Oct 14, 2015
andrevdh
Another way is to use the formula τ = P r sin(θ) where θ is the angle between P and r. You need to make a drawing to figure our what the relationship between θ and α is then.
5. Oct 14, 2015
werson tan
(13.2cos (alpha) ) ( surd(0.086² +0.122² ) )= 1.95 , alpha = 8.23 degree , still wrong !
6. Oct 14, 2015
andrevdh
Did you take the square root to obtain r? The angle in the equation is not necessarily alpha. Make a drawing to discover the relationship.
7. Oct 30, 2015
werson tan
my working is 13.2 (1-(sin theta)^2) )^(0.5) (0.86x10^-3) + 13.2(sin theta)(122x10^-3) = 1.95 , so i gt the theta = 56.42 , but the ans given is 50.6 degree and 59.1 degree
8. Oct 30, 2015
andrevdh
I seem to get 47.5o for α
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9. Oct 30, 2015
haruspex
You do not have two unknowns. You have two functions, cos and sin, of one unknown.
Do you know the formula for sin(a+b)? Can you see how to apply it in reverse to an expression like $A\cos(\theta)+B\sin(\theta)$?
10. Oct 30, 2015
haruspex
Unfortunately the arc sin function gets very steep near 1, so it's hard to get enough precision. I get more like 46.6 or 63 degrees.
11. Oct 30, 2015
werson tan
can i do in this way ?
13.2 (1-(sin alpha)^2) )^(0.5) (0.86x10^-3) + 13.2(sin alpha)(122x10^-3) = 1.95 , so i gt the alpha = 56.42 , but the ans given is 50.6 degree and 59.1 degree
12. Oct 30, 2015
haruspex
All those numbers make it too hard to follow what you are doing. Use symbols and show your working to get that equation.
13. Oct 30, 2015
werson tan
sorry , pls ignore my previous working. i have changed the cos aplha = (1- (sin aplha )^2 )^0.5 , but i still coulnt get the ans , since the sin alpha value is greater than 1
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14. Oct 30, 2015
haruspex
$\sqrt{1-\sin^2(\alpha)}$ is not $1-\sin(\alpha)$. | 970 | 2,825 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2018-22 | latest | en | 0.851928 |
http://www.sermonindex.net/modules/newbb/viewtopic.php?topic_id=36915&forum=48 | 1,590,362,746,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347385193.5/warc/CC-MAIN-20200524210325-20200525000325-00178.warc.gz | 210,831,691 | 6,938 | Discussion Forum : News and Current Events : The Rapture Will Occur January 8th 2011!
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wayneman
Member
Joined: 2009/1/24
Posts: 454
Michigan
The Rapture Will Occur January 8th 2011! Brethren, I have been up all night and the Lord showed me UNIMPEACHABLE SCRIPTURAL PROOF that the Rapture will occur January 8th 2011. With the Lord a day is as a thousand years. Four days from Adam to Jesus, two days from Jesus to the Rapture, plus the millennium equals the Lords 7000-year week.The precise date of the cross was April 1, 33 A.D. when coordinated with our present modern calendar. There are 365.2422 days in a year. Thus all we have to do is multiply the number of years separating two events by the number 365.2422 to know the exact number of days between them. So from April 1, 33 A.D. to April 1, 2011 there are exactly 2011 33 = 1,978 years, each having 365.2422 days. This equals 722,579.07 days. Between April 1, 2011 and January 8th, 2011 inclusively (including the first day and the last day) are 79 days. Subtracting these 79 days from the number 722,579.07 gives us exactly 722,500.07 days. This number is enormously significant. The Bible uses numbers to symbolize spiritual points. Five is redemption; 10 is completion and 17 is The Rapture. 5 x 10 x 17=850. But the fact that the Bible says a day is a thousand years TWICE tells us that God wants us to square this figure. 5 x 10 x 17 x 5 x 10 x 17=722,500!!! That is, redemption times completion times The Rapture squared is 722,500, proving that The Rapture will occur January 8th 2011!Reader, you should be absolutely astonished at what you have just learned. And remember, I said the Lord showed me, so youre not allowed to argue with me._________________Wayne Kraus
2011/1/9 12:27Profile
Re: The Rapture Will Occur January 8th 2011! um, ahem, This is Sunday January 9th.Your clearly joking of course. There is a flaw though. Jesus was actually born around 4 B.C and crucified in A.D.30.
2011/1/9 12:34
learjet
Member
Joined: 2010/4/19
Posts: 447
Re: The Rapture Will Occur January 8th 2011! doh...:-)
2011/1/9 12:43Profile
Re: The Rapture Will Occur January 8th 2011! right here......Wayneman's posting is a TEXT EXAMPLE of what i mean by "weird" postings when i implored the forum to consider our ways when i wrote my "It's very sad what's been happening on this forum of late." post about a week ago.https://www.sermonindex.net/modules/newbb/viewtopic.php?topic_id=36878&forum=35&16what is wrong with you Wayne?Don't you fear God?you start playing games with His Word?, and in your post, i pray for your sake you didnt veer into the territory of committing the unpardonable sin.my advice? repent immediately, privately, then publicly, and ask the moderators to lock this thread.i dont know what impelled you to post as such; maybe you had 'i'm 5 years old' immature moment, or you're metally ill and/or demon possessed.....whatever the case, straighten up and fly right.as i type this there's 54 visitors here, nice witness bud.visitors, forgive our brother, he ate too many frosted flakes and the sugar made him a wee bit mental.
2011/1/9 13:04
amranger
Member
Joined: 2010/4/7
Posts: 71
Montana
Re: The Rapture Will Occur January 8th 2011! I think Wayneman is trying to make a point? He posted this on the 9th, maybe he's just trying to show how silly it is to predict the LORD's return when no man knows?_________________Andrew
2011/1/9 13:11Profile
learjet
Member
Joined: 2010/4/19
Posts: 447
Re: Quote:i dont know what impelled you to post as such; maybe you had 'i'm 5 years old' immature moment, or you're metally ill and/or demon possessed.....whatever the case, straighten up and fly right.Physician, heal thyself, you've murdered your brother in your heart. Yes, that's what anger does.He was referring to the uptick in 'The Rapture is going to occur on this day' articles that have been circulating in the press lately. They are all over, being reported about on every major news network.In Christ,Earl J.
2011/1/9 13:11Profile
wayneman
Member
Joined: 2009/1/24
Posts: 454
Michigan
Re: neil,Thanks for the ray of sunshine! Hopefully your outpouring of love has repaired the damage I've done to the forum. :-)_________________Wayne Kraus
2011/1/9 13:33Profile
Re: Wayne, no worries, i'll be making myself very very scarce around here.i love to laugh, love jokes, merriment,satire, but cmon man. was that post really THAT neccesary?anyway, God love you, i'm out.
2011/1/9 14:35
docseth1
Member
Joined: 2008/5/17
Posts: 173
Valdosta, Ga
Re: The Rapture Will Occur January 8th 2011! It's January the 9th, and I'm still here. Did I miss the rapture???_________________Cliff
2011/1/9 16:43Profile
wayneman
Member
Joined: 2009/1/24
Posts: 454
Michigan
Re: Quote:It's January the 9th, and I'm still here. Did I miss the rapture???Yes, but that is understandable since I only posted the warning this morning._________________Wayne Kraus
2011/1/9 17:04Profile | 1,465 | 4,998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2020-24 | latest | en | 0.940004 |
https://discuss.pytorch.org/t/how-can-i-compute-seq2seq-loss-using-mask/861 | 1,653,519,678,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662594414.79/warc/CC-MAIN-20220525213545-20220526003545-00655.warc.gz | 265,582,626 | 9,055 | # How can i compute seq2seq loss using mask?
I am working on image captioning task with PyTorch.
In seq2seq, padding is used to handle the variable-length sequence problems.
Additionally, mask is multiplied by the calculated loss (vector not scalar) so that the padding does not affect the loss.
In TensorFlow, i can do this as below.
``````# targets is an int64 tensor of shape (batch_size, padded_length) which contains word indices.
# masks is a tensor of shape (batch_size, padded_length) which contains 0 or 1 (0 if pad otherwise 1).
outputs = decoder(...) # unnormalized scores of shape (batch_size, padded_length, vocab_size)
outputs = tf.reshape(outputs, (-1, vocab_size))
targets = tf.reshape(targets, (-1))
losses = tf.nn.sparse_softmax_cross_entropy_loss(outputs, targets) # loss of shape (batch_size*padded_length)
``````
In PyTorch, `nn.CrossEntropyLoss()` returns a scalar not tensor so that i can not multiply loss by masks.
``````criterion = nn.CrossEntropyLoss()
outputs = decoder(features, inputs) # (batch_size, padded_length, vocab_size)
loss = criterion(outputs.view(-1, vocab_size), targets.view(-1)) # this gives a scalar not tensor
``````
How can i solve this problem?
3 Likes
A non-averaged cross-entropy loss is coming soon. Until then you can write your own using log_softmax and advanced indexing.
In addition, though I don’t think it helps you here, nn.LSTM now has support for variable-length sequences without including padding, meaning that sequence model results will not be affected by the influence of padding tokens even with bidirectional RNNs. There are utility functions provided for creating the packed array data structure (~TF’s TensorArray) needed for this.
1 Like
How can i use nn.LSTM with variable-length sequences and without padding?
In code below, torch.utils.data.DataLoader concatenates each single tensor to construct mini batch data. This makes me pad the each sequence to make the tensor of fixed size.
``````cap = CocoCaptions(root = './data/train2014resized',
annFile = './data/annotations/captions_train2014.json',
vocab = vocab,
transform=transform,
target_transform=transforms.ToTensor())
cap, batch_size=16, shuffle=True, num_workers=2)
# images: a tensor of shape (batch_size, 3, 256, 256).
1 Like
If `padded` is a Variable with padding in it and `lengths` is a tensor containing the length of each sequence, then this is how to run a (potentially bidirectional) LSTM over the sequences in a way that doesn’t include padding, then pad the result in order to use it in further computations.
``````import torch.nn.utils.rnn as rnn_utils
lstm = nn.LSTM(in_size, hidden_size, bidirectional, num_layers)
packed_out, packed_hidden = lstm(packed)
2 Likes
Like https://github.com/Element-Research/rnn has `MaskZero` module, won’t PyTorch also need a wrapper to deal with padded inputs and gradients between LSTM outputs and final layer? Is there any plan for that or is there other elegant way to deal with it?
@supakjk there’s no need for any wrapper. As long as the padded entries don’t contribute to your loss in any way, their gradient will always be 0.
1 Like
The cross-entropy loss for a particular vector of output scores and a target index is the value at that index of the negative log softmax of the vector of scores, so you can run negative log softmax on the whole score tensor, pick out the values you want using `gather` (advanced indexing was briefly semi-supported for this, and will be fully supported eventually), then sum/average the results
Could you let me know how to deal with the following example case?
Assume that I have an input, where maximum sequence length is 5, minibatch size is 4, and there are 3 possible labels. idxs has effective lengths of sequences in the input.
``````input = Variable(torch.randn(5,4,3))
idxs = [5,3,3,2]
target = Variable(torch.LongTensor(5,4))
# assume the target labels are assigned corresponding to the input.
``````
Then, In a sequence tagging task, I’d like to get the cross entropy errors for the whole time steps of the first sequence, 3 time steps of the second sequence, and so on considering the values of `idxs`.
How could I use advanced indexing for addressing this sequence tagging with variable lengthed input sequences?
(I thought `masked_select` might be used for my purpose but I wonder what would be the most elegant one at this moment before some other features are added.)
Thanks!
The `pack_padded_sequence` can be placed at any point in the DAG right? That is, given a sequence, apply some dense layer on it and then pack it and give it to the LSTM
@pranav it’s packed in an additional structure to hold the sequence lengths, but you might take out its `.data` attribute, compute a function on that, and rewrap it in a new `torch.nn.utils.rnn.PackedSequence` object.
2 Likes
Then the non-averaged cross-entropy loss hasn’t come yet, but this could be easily implemented by `torch.gather`.
For example, suppose
1. the outputs is the logit for each label, whose shape is `[batch_size, time_step, num_token]`(if not, you could always get them with `F.log_softmax`)
2. `targets` contains the targeted label for each step, whose shape is `[batch_size, time_step]`
Then the cross_entropy loss could calculated with
``````def nll_loss(outputs, targets):
``````
Then you can apply masks as is in `Tensorflow`.
Can you help me understand this torch.gather I have used the same tensors with similar shape but this line torch.gather(outputs, 1, targets.view(-1,1)) gives me error.
Thanks for sharing the code. I test the nll_loss function and get:
RuntimeError: Input tensor must have same dimensions as output tensor at /b/wheel/pytorch-src/torch/lib/TH/generic/THTensorMath.c:424
It seems that the input and index arguments of gather function should have same dimensions.
currently I use this way to compute masked loss for seq2seq model:
``````'''
target_sentences: [batch_size, max_length] int64
batch_target_len_array: [batch_size] int64, each element is the valid length of the corresponding target sentence.
'''
for step in range(max_length):
decoder_output, decoder_hidden = decoder(
decoder_input, decoder_hidden, encoder_outputs)
if np.sum(batch_target_len_array > step) == 0:
break
valid_output = torch.index_select(decoder_output, 0, index_vector)
valid_target = torch.index_select(target_sentences[:, step], 0, index_vector)
step_loss = criterion(valid_output, valid_target)
loss += step_loss
decoder_input = target_sentences[:, step] # Teacher forcing``````
4 Likes
In version 0.2.0 they added `ignore_index` in `CrossEntropyLoss`. I think this addition solves the problem. I use it like this:
``````loss_function = nn.CrossEntropyLoss(ignore_index=0)
``````
since i have zero padded targets.
I used to use this `masked_cross_entropy` workaround, but i tested them both and i observe the same behaviour (compared to `CrossEntropyLoss` without `ignore_index=0`).
10 Likes
I have tried, doesn’t work in my case. Could you post your code with a bit more details?
It is clear. I’ll try! The newest version 0.3 updated the cost function with a new Parameter “reduce”, and it’s easy to mask the loss
Can you explain how to mask the loss now?
With the parameter “reduce”, we can get the loss per batch element, but how can we use the mask on it?
For example, if I have a minibatch whose valid sequence lengths is [3,1,2], and with “reduce” we can get three loss values [L1, L2, L3]. But we need to mask the last two values in calculating L2 and last one value in calculating L3. How could it be achieved?
------------------------------edited----------------------------
Oh I see, we only need to use ignore_index=0 parameter in loss(), but don’t need the “reduce” parameter to achieve this. Is it right? | 1,875 | 7,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-21 | latest | en | 0.842718 |
https://ez.analog.com/mems/f/q-a/542507/gyro-data-angular-rate-integration | 1,620,411,476,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988802.93/warc/CC-MAIN-20210507181103-20210507211103-00412.warc.gz | 285,600,886 | 34,828 | # Gyro data(Angular rate) Integration
The IMU ADIS16475-2 is kept at stationary (@ Constant temperature) data logged at rate of 100Hz into CSV file. If we integrate gyro X,Y and Z independently at rate of 0.01sec.the gyroscope integrated values after 1 hr as follows. The integration was done after 10 sec from power ON by compensating each gyro axis OFFSET. This OFFSET is calculated for 10sec from POWER ON.
X --> -125.5 deg
Y--> 44 deg
Z-->90 deg.
If we integrate gyro X,Y and Z independently at rate of 100sec i.e integration time dt is taken as 100 and integrated for every 100sec gyro sample. The gyroscope integrated values after 1 hr as follows. The integration was done after 100 sec from power ON by compensating each gyro axis OFFSET. This OFFSET is calculated for 100sec from POWER ON.
X --> -112 deg
Y--> 52deg
Z-->89deg.
In ADIS16475-2 datasheet Bias instability is given as 2.5 deg/hr but in practical it is drifted too much.
Please suggest , is the above results are valid or not?
• Please give me the reply for above question
• Thank you, for your post. These devices have a drift behavior, in their bias, which we refer to as "turn-on drift." Since we have released this product, we have developed stronger understanding for this behavior. In general, we can model this behavior, using a first-order, exponential decay, where the initial bias error is typically within +/-0.75dps and the time constant is in the region of 5-20 minutes. Here is an example of the model, when the initial bias error is 0.6dps and the time constant is 10 minutes.
Bias Error = 0.6 x exp (-time/10minutes)
I hope that this helps!
• Thank you for answer
1.You said that the bias is exponential decay, but Gyro-Z in ADIS 16475-2 found to be exponential rise, i.e at turn ON its value is 0.06dps after 1hr it's value reached to 0.09dps.Why?
2.How to calculate the maximum drift angle/hour which is calculated by integrating gyro output(dps) at sample rate of 0.01sec in stationary conditions from ADIS IMU data sheet?
3.How much deviation can we expect in maximum drift angle/hour from stationary to dynamic conditions (Linear and Angular accelerations)?
please respond to above queries.
• Please give me the reply for above queries
• I am sorry for the lack of response on this. We lost a key staff member and are working to re-organize our team to provide quicker responses, in the future. We appreciate your persistence. Please see my preliminary responses, below.
1.You said that the bias is exponential decay, but Gyro-Z in ADIS 16475-2 found to be exponential rise, i.e at turn ON its value is 0.06dps after 1hr it's value reached to 0.09dps.Why?
NevadaMark-> In this case, it sounds like the model might be following this formula:
0.06dps + 0.3dps x (1 - exp(-time/TC)), where TC is in the region of 15-25minutes. Does this match the plot that you are observing? If not, can you share the plot?
2.How to calculate the maximum drift angle/hour which is calculated by integrating gyro output(dps) at sample rate of 0.01sec in stationary conditions from ADIS IMU data sheet?
NevadaMark -> The datasheet does not provide maximum values, nor can we offer them in this forum. However, we can offer that these typical values are conservative, with intent of covering >70% of the parts being as good as the Typical or better.
3.How much deviation can we expect in maximum drift angle/hour from stationary to dynamic conditions (Linear and Angular accelerations)?
NevadaMark -> Same comment on "maximum" specs. Generally speaking, the metrics are setup to support a second order model estimation: kg x a + VRE x a^2. What type of acceleration are you trying to quantify the impact for? | 931 | 3,717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-21 | latest | en | 0.916937 |
https://www.doorsteptutor.com/Exams/NEET/Physics/Questions/Topic-Work-Energy-and-Power-3/Subtopic-Kinetic-Energy-1/Part-1.html | 1,527,476,740,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870771.86/warc/CC-MAIN-20180528024807-20180528044807-00063.warc.gz | 728,607,838 | 11,015 | # Work, Energy and Power-Kinetic Energy (NEET (NTA)-National Eligibility cum Entrance Test (Medical) Physics): Questions 1 - 3 of 5
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## Question number: 1
» Work, Energy and Power » Kinetic Energy
MCQ▾
### Question
A slow moving electron collides elastically with a hydrogen atom at rest. The initial and final motions are along the same straight line. What fraction of electron’s kinetic energy is transferred to the hydrogen atom? The mass of hydrogen atom is 1850 times the mass of electron.
### Choices
Choice (4) Response
a.
2.17 %
b.
21.7%
c.
0.217%
d.
0.0217 %
## Question number: 2
» Work, Energy and Power » Kinetic Energy
MCQ▾
### Question
A body of mass M splits into two parts αM and (1 - a) M by an internal explosion, which generates kinetic energy T After explosion if the two parts move in the same direction as before, their relative speed will be –
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 3
» Work, Energy and Power » Kinetic Energy
MCQ▾
### Question
Which of the following hold when two particles of masses m 1 and m 2 undergo elastic collision?
(1) When m, = m 2 and m 2 is stationary, there is maximum transfer of kinetic energy in head on collision
(2) When m i = va 2 and m 2 is stationary, there is maximum transfer of momentum in head on collision
(3) When m l > > m 2 and m 2 is stationary, after head on collision m 2 moves with twice the velocity of m 1
(4) When the collision is oblique and m 1 = m 2 with m 2 stationary, after the collision the particle move in opposite directions.
Select the correct answers and mark it according to the following codes
### Choices
Choice (4) Response
a.
2 and 4 are correct
b.
1 and 2 are correct
c.
1 and 3 are correct
d.
1,2 and 3 are correct
f Page | 533 | 2,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-22 | latest | en | 0.837116 |
https://gzipwtf.com/how-big-should-a-steam-shower-be/ | 1,686,381,813,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657144.94/warc/CC-MAIN-20230610062920-20230610092920-00069.warc.gz | 341,904,036 | 42,317 | # How big should a steam shower be?
## How big should a steam shower be?
Minimum Shower Size For a Steam System Many different sized showers will fit in a 5’x5′ bathroom. However, for a shower enclosure utilizing a steam system, the minimum size we recommend is 36” x 36” x 7′. That way you have enough space to stand or sit on a fold-up seat installed away from the steam outlet head.
How do you figure out the cubic feet of a steam shower?
Steam Generator Sizing
1. Calculate the cubic footage of the enclosure. Length X Width X Height (in feet) = Cubic feet.
2. Add the following adjustments to the cubic footage calculated in the step above to account for your specific design features. (
3. Compare the total calculated in #2 above with the chart below.
### Can a steam shower replace a regular shower?
Can a steam shower be used as a regular shower? Yes! A steam generator can be added to your standard shower, so you can choose your relaxation experience day by day – no need to install a separate unit or stall.
Do steam showers use a lot of water?
Steam showers are also easy on the water bill. In 20 minutes of use, a steam shower only consumes about 2 gallons of water compared to the 50 or more gallons that a conventional shower would use over the same duration.
## How do I choose a steam shower?
5 Steps to Choosing the Right Steam Shower for Your Home
1. Consider Your Enclosure. Thanks to modern day steam generator technology, steam is not hard to introduce into an existing shower enclosure — but it must be enclosed, insulated and sealed.
3. Calculate the Size You Need.
What size is a 2 person shower?
As far as size goes, two of the most common sizes are 3-feet by 5-feet and 4-feet by 6-feet; there are no minimum requirements. It is whatever the couple in question feels comfortable with as far as space goes.
ACF (Adjusted Cubic Feet) – The adjusted size (in cubic feet) of a steam shower enclosure determined by a calculation that considers many variables including construction materials, unique room features and actual cubic footage in order to determine the appropriate Kilowatt rating and model of steam shower generator.
Steam rooms create an environment that warms the mucous membrane and encourages deep breathing. As a result, using one can help break up congestion inside your sinuses and lungs. But steam rooms are relatively safe in comparison, as long as you don’t stay inside too long.
## Does a steam shower need a tile ceiling?
The best material for the interior of a steam shower is a non-porous tile. Ceramic and porcelain tile works best as it won’t absorb the steam. Using material like marble or granite will absorb the heat, making it dissipate and causing your steam generator to work that much harder to fill the stall with steam.
How small can a steam shower be?
Steam showers require more space than a standard shower does. The very smallest shower size for a steam shower would be three square feet, with a height of seven feet tall. This is the bare minimum for a typical person to be able to sit and stand in the shower.
### Is a steam shower worth the extra cost?
Steam showers are worth the money if you value the physical and mental health benefits they offer. When it comes to selling your home, you could make an extra 30% or more on what it cost you to build. How much does it cost to repair a steam room?
How much water does a steam shower use?
The steam shower itself requires: Steam rooms can also be a water-efficient way to bathe; on average a 20-minute steam bath requires about two gallons of water. Even a water-wise shower head can use up to 50 gallons of water in the same period of time.
## How to build a steam shower?
Have A Seat in Your Steam Shower. It would be best to have a seat in your steam shower.
• Consider the Temperature Limits When Choosing the Materials.
• Consider Installing A Shower Door.
• Steam Generator Installation.
• Call the Help from the Professionals.
• How much does a steam shower cost?
Steam Shower Prices. Custom steam showers can be as high as \$6,200 depending on the features of the shower itself and the plumbing company performing the installation. If you opt for a prefabricated steam shower, the cost lies between \$1,000-\$5,000.
Begin typing your search term above and press enter to search. Press ESC to cancel. | 936 | 4,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-23 | latest | en | 0.921276 |
https://oeis.org/A183644/internal | 1,721,235,372,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.19/warc/CC-MAIN-20240717151625-20240717181625-00014.warc.gz | 390,459,791 | 3,147 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A183644 Number of (n+1) X 2 0..4 arrays with every 2 X 2 subblock summing to 8. 1
%I #10 Mar 31 2018 14:34:29
%S 85,325,1333,5725,25405,115525,535333,2517805,11982925,57575125,
%T 278766133,1358125885,6649985245,32693825125,161264049733,
%U 797557733965,3952911584365,19625567713525,97573430562133,485654176126045
%N Number of (n+1) X 2 0..4 arrays with every 2 X 2 subblock summing to 8.
%C Column 1 of A183652.
%H R. H. Hardin, <a href="/A183644/b183644.txt">Table of n, a(n) for n = 1..200</a>
%F Empirical: a(n) = 15*a(n-1) - 85*a(n-2) + 225*a(n-3) - 274*a(n-4) + 120*a(n-5).
%F Conjectures from _Colin Barker_, Mar 31 2018: (Start)
%F G.f.: x*(85 - 950*x + 3683*x^2 - 5770*x^3 + 3000*x^4) / ((1 - x)*(1 - 2*x)*(1 - 3*x)*(1 - 4*x)*(1 - 5*x)).
%F a(n) = 2 + 2^(2+n) + 2^(3+2*n) + 2*3^(1+n) + 5^(1+n).
%F (End)
%e Some solutions for 3 X 2:
%e 4 2 3 1 2 3 3 0 1 4 1 4 0 4 1 3 3 1 1 3
%e 2 0 0 4 0 3 1 4 1 2 2 1 3 1 3 1 1 3 0 4
%e 2 4 1 3 1 4 3 0 2 3 2 3 0 4 1 3 4 0 1 3
%Y Cf. A183652.
%K nonn
%O 1,1
%A _R. H. Hardin_, Jan 06 2011
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Last modified July 17 12:34 EDT 2024. Contains 374377 sequences. (Running on oeis4.) | 699 | 1,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-30 | latest | en | 0.531234 |
https://training.incf.org/search?subjects%5B53%5D=53&%3Bf%5B0%5D=topics%3A43&f%5B0%5D=difficulty_level%3Abeginner&f%5B1%5D=topics%3A22&f%5B2%5D=topics%3A38&f%5B3%5D=topics%3A39&f%5B4%5D=topics%3A48&f%5B5%5D=topics%3A58&f%5B6%5D=topics%3A63&f%5B7%5D=topics%3A65&f%5B8%5D=topics%3A71&f%5B9%5D=topics%3A73&filter= | 1,591,189,684,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347434137.87/warc/CC-MAIN-20200603112831-20200603142831-00153.warc.gz | 567,114,088 | 16,408 | ## Difficulty level
Lecture title:
The probability of a hypothesis, given data.
Difficulty level: Beginner
Duration: 7:57
Speaker: : Barton Poulson
Lecture title:
Why math is useful in data science.
Difficulty level: Beginner
Duration: 1:35
Speaker: : Barton Poulson
Lecture title:
Why statistics are useful for data science.
Difficulty level: Beginner
Duration: 4:01
Speaker: : Barton Poulson
Lecture title:
Statistics is exploring data.
Difficulty level: Beginner
Duration: 2:23
Speaker: : Barton Poulson
Lecture title:
Graphical data exploration
Difficulty level: Beginner
Duration: 8:01
Speaker: : Barton Poulson
Lecture title:
Numerical data exploration
Difficulty level: Beginner
Duration: 5:05
Speaker: : Barton Poulson
Lecture title:
Simple description of statistical data.
Difficulty level: Beginner
Duration: 10:16
Speaker: : Barton Poulson
Lecture title:
Basics of hypothesis testing.
Difficulty level: Beginner
Duration: 06:04
Speaker: : Barton Poulson
Lecture title:
Enabling neuroscience research using high performance computing
Difficulty level: Beginner
Duration: 39:27
Speaker: : Subha Sivagnanam
Lecture title:
This lecture covers structured data, databases, federating neuroscience-relevant databases, ontologies.
Difficulty level: Beginner
Duration: 1:30:45
Speaker: : Maryann Martone
Lecture title:
This primer on optogenetics primer discusses how to manipulate neuronal populations with light at millisecond resolution and offers possible applications such as curing the blind and "playing the piano" with cortical neurons.
Difficulty level: Beginner
Duration: 59:06
Speaker: : Clay Reid
Lecture title:
This lecture covers describing and characterizing an input-output relationship.
Difficulty level: Beginner
Duration: 1:35:33
Part 1 of 2 of a tutorial on statistical models for neural data
Difficulty level: Beginner
Duration: 1:45:48
Speaker: : Jonathan Pillow
Part 2 of 2 of a tutorial on statistical models for neural data.
Difficulty level: Beginner
Duration: 1:50:31
Speaker: : Jonathan Pillow
Lecture title:
From the retina to the superior colliculus, the lateral geniculate nucleus into primary visual cortex and beyond, this lecture gives a tour of the mammalian visual system highlighting the Nobel-prize winning discoveries of Hubel & Wiesel.
Difficulty level: Beginner
Duration: 56:31
Speaker: : Clay Reid
Lecture title:
From Universal Turing Machines to McCulloch-Pitts and Hopfield associative memory networks, this lecture explains what is meant by computation.
Difficulty level: Beginner
Duration: 55:27
Speaker: : Christof Koch
Lecture title:
Ion channels and the movement of ions across the cell membrane.
Difficulty level: Beginner
Duration: 25:51
Speaker: : Carl Petersen
Lecture title:
How does the brain learn? This lecture discusses the roles of development and adult plasticity in shaping functional connectivity.
Difficulty level: Beginner
Duration: 1:08:45
Speaker: : Clay Reid
Introduction to neurons, synaptic transmission, and ion channels.
Difficulty level: Beginner
Duration: 46:07
2nd part of the lecture. Introduction to cell receptors and signalling cascades
Difficulty level: Beginner
Duration: 41:38 | 743 | 3,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-24 | latest | en | 0.68671 |
https://www.gradesaver.com/textbooks/math/other-math/CLONE-547b8018-14a8-4d02-afd6-6bc35a0864ed/chapter-6-percent-6-3-using-the-percent-proportion-and-identifying-the-components-in-a-percent-problem-6-3-exercises-page-406/16 | 1,725,763,142,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650958.30/warc/CC-MAIN-20240908020844-20240908050844-00227.warc.gz | 760,939,394 | 13,436 | ## Basic College Mathematics (10th Edition)
Published by Pearson
# Chapter 6 - Percent - 6.3 Using the Percent Proportion and Identifying the Components in a Percent Problem - 6.3 Exercises - Page 406: 16
43.7
#### Work Step by Step
$\frac{x}{115} = \frac{38}{100}$ $100x = 115\times38$ $\frac{100x}{100} = \frac{4370}{100}$ $x = 43.7$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 145 | 501 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-38 | latest | en | 0.784982 |
https://hackage.haskell.org/package/hgeometry-0.5.0.0/docs/Data-Geometry-Box-Internal.html | 1,575,852,600,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540515344.59/warc/CC-MAIN-20191208230118-20191209014118-00051.warc.gz | 403,954,696 | 4,402 | hgeometry-0.5.0.0: Geometric Algorithms, Data structures, and Data types.
Data.Geometry.Box.Internal
Synopsis
# d-dimensional boxes
data Box d p r Source
Constructors
Box Fields_minP :: Min (Point d r) :+ p _maxP :: Max (Point d r) :+ p
Instances
PointFunctor (Box d p) Source Coordinate r => IpeReadText (Rectangle () r) Source (Num r, Ord r) => IsIntersectableWith (Rectangle p r) (Rectangle p r) Source (Eq r, Eq p, Arity d) => Eq (Box d p r) Source (Ord r, Ord p, Arity d) => Ord (Box d p r) Source (Show r, Show p, Arity d) => Show (Box d p r) Source (Arity d, Ord r, Semigroup p) => Semigroup (Box d p r) Source (Num r, AlwaysTruePFT d) => IsTransformable (Box d p r) Source type IntersectionOf (Line 2 r) (Boundary (Rectangle p r)) = (:) * NoIntersection ((:) * (Point 2 r) ((:) * (Point 2 r, Point 2 r) ((:) * (LineSegment 2 () r) ([] *)))) Source type IntersectionOf (Line 2 r) (Rectangle p r) = (:) * NoIntersection ((:) * (Point 2 r) ((:) * (LineSegment 2 () r) ([] *))) Source type NumType (Box d p r) = r Source type Dimension (Box d p r) = d Source type IntersectionOf (Box d p r) (Box d q r) = (:) * NoIntersection ((:) * (Box d () r) ([] *)) Source
minP :: forall d p r. Lens' (Box d p r) ((:+) (Min (Point d r)) p) Source
maxP :: forall d p r. Lens' (Box d p r) ((:+) (Max (Point d r)) p) Source
fromCornerPoints :: (Point d r :+ p) -> (Point d r :+ p) -> Box d p r Source
Given the point with the lowest coordinates and the point with highest coordinates, create a box.
# Functions on d-dimensonal boxes
minPoint :: Box d p r -> Point d r :+ p Source
maxPoint :: Box d p r -> Point d r :+ p Source
inBox :: (Arity d, Ord r) => Point d r -> Box d p r -> Bool Source
Check if a point lies a box
````>>> ````origin `inBox` (boundingBoxList' [point3 1 2 3, point3 10 20 30] :: Box 3 () Int)
```False
`>>> ````origin `inBox` (boundingBoxList' [point3 (-1) (-2) (-3), point3 10 20 30] :: Box 3 () Int)
```True
```
extent :: Arity d => Box d p r -> Vector d (Range r) Source
Get a vector with the extent of the box in each dimension. Note that the resulting vector is 0 indexed whereas one would normally count dimensions starting at zero.
````>>> ````extent (boundingBoxList' [point3 1 2 3, point3 10 20 30] :: Box 3 () Int)
```Vector3 [Range {_lower = Closed 1, _upper = Closed 10},Range {_lower = Closed 2, _upper = Closed 20},Range {_lower = Closed 3, _upper = Closed 30}]
```
size :: (Arity d, Num r) => Box d p r -> Vector d r Source
Get the size of the box (in all dimensions). Note that the resulting vector is 0 indexed whereas one would normally count dimensions starting at zero.
````>>> ````size (boundingBoxList' [origin, point3 1 2 3] :: Box 3 () Int)
```Vector3 [1,2,3]
```
widthIn :: forall proxy p i d r. (Arity d, Num r, Index' (i - 1) d) => proxy i -> Box d p r -> r Source
Given a dimension, get the width of the box in that dimension. Dimensions are 1 indexed.
````>>> ````widthIn (C :: C 1) (boundingBoxList' [origin, point3 1 2 3] :: Box 3 () Int)
```1
`>>> ````widthIn (C :: C 3) (boundingBoxList' [origin, point3 1 2 3] :: Box 3 () Int)
```3
```
widthIn' :: (Arity d, KnownNat d, Num r) => Int -> Box d p r -> Maybe r Source
Same as `widthIn` but with a runtime int instead of a static dimension.
````>>> ````widthIn' 1 (boundingBoxList' [origin, point3 1 2 3] :: Box 3 () Int)
```Just 1
`>>> ````widthIn' 3 (boundingBoxList' [origin, point3 1 2 3] :: Box 3 () Int)
```Just 3
`>>> ````widthIn' 10 (boundingBoxList' [origin, point3 1 2 3] :: Box 3 () Int)
```Nothing
```
# Rectangles, aka 2-dimensional boxes
width :: Num r => Rectangle p r -> r Source
height :: Num r => Rectangle p r -> r Source
corners :: Num r => Rectangle p r -> (Point 2 r :+ p, Point 2 r :+ p, Point 2 r :+ p, Point 2 r :+ p) Source
Get the corners of a rectangle, the order is: (TopLeft, TopRight, BottomRight, BottomLeft). The extra values in the Top points are taken from the Top point, the extra values in the Bottom points are taken from the Bottom point
# Constructing bounding boxes
class IsBoxable g where Source
Methods
boundingBox :: (Monoid p, Semigroup p, Ord (NumType g)) => g -> Box (Dimension g) p (NumType g) Source
Instances
IsBoxable (Point d r) Source Arity d => IsBoxable (LineSegment d p r) Source Arity d => IsBoxable (PolyLine d p r) Source
boundingBoxList' :: (IsBoxable g, Monoid p, Ord (NumType g), IsAlwaysTrueBoundingBox g p) => [g] -> Box (Dimension g) p (NumType g) Source
Unsafe version of boundingBoxList, that does not check if the list is non-empty | 1,476 | 4,537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-51 | latest | en | 0.51617 |
http://vlab.amrita.edu/?sub=1&brch=194&sim=354&cnt=1 | 1,606,150,250,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141163411.0/warc/CC-MAIN-20201123153826-20201123183826-00369.warc.gz | 104,853,575 | 7,496 | . .
.
Newton's Law of Cooling
.
.
# Aim
1. The aim of the experiment is to verify Newton's Law of Cooling of different materials and different liquids.
2. To draw the cooling curve.
# Theory
Temperature difference in any situation results from energy flow into a system or energy flow from a system to surroundings. The former leads to heating, whereas latter leads to cooling of an object.
Newton’s Law of Cooling states that the rate of temperature of the body is proportional to the difference between the temperature of the body and that of the surrounding medium. This statement leads to the classic equation of exponential decline over time which can be applied to many phenomena in science and engineering, including the discharge of a capacitor and the decay in radioactivity.
Newton's Law of Cooling is useful for studying water heating because it can tell us how fast the hot water in pipes cools off. A practical application is that it can tell us how fast a water heater cools down if you turn off the breaker when you go on vacation.
Suppose that a body with initial temperature T1°C, is allowed to cool in air which is maintained at a constant temperature T2°C.
Let the temperature of the body be T°C at time t.
Then by Newton’s Law of Cooling,
(1)
Where k is a positive proportionality constant. Since the temperature of the body is higher than the temperature of the surroundings then T-T2 is positive. Also the temperature of the body is decreasing i.e. it is cooling down and rate of change of temperature is negative.
The constant ‘k’ depends upon the surface properties of the material being cooled.
Initial condition is given by T=T1 at t=0
Solving (1)
(2)
Applying initial conditions;
Substituting the value of C in equation (2) gives
This equation represents Newton’s law of cooling.
If k <0, lim t --> ∞, e-kt = 0 and T= T2 ,
Or we can say that the temperature of the body approaches that of its surroundings as time goes.
The graph drawn between the temperature of the body and time is known as cooling curve. The slope of the tangent to the curve at any point gives the rate of fall of temperature.
In general,
where,
T(t) = Temperature at time t,
TA = Ambient temperature (temp of surroundings),
TH = Temperature of hot object at time 0,
k = positive constant and
t = time.
## Example of Newton's Law of Cooling:
This kind of cooling data can be measured and plotted and the results can be used to compute the unknown parameter k. The parameter can sometimes also be derived mathematically.
### Applications
1. To predict how long it takes for a hot object to cool down at a certain temperature.
2. To find the temperature of a soda placed in a refrigerator by a certain amount of time.
3. It helps to indicate the time of death given the probable body temperature at the time of death and current body temperature.
Cite this Simulator:
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Worksheet
Budgeting and personal finance
1
A weekly budget is shown in the table:
a
Calculate the weekly expenses.
b
Calculate the weekly savings.
c
Calculate the percentage of income saved correct to two decimal places.
d
Calculate the percentage of income spent on mortgage payments correct to two decimal places.
e
Calculate the amount paid on the mortgage in one year.
2
James's monthly expenses include: rent, food, pay TV, travel, insurance, clothing, repairs.
a
Determine whether the following expenses are fixed:
i
Pay TV
ii
Travel
iii
Insurance
iv
Clothing
v
Repairs
vi
Rent
b
Determine whether the following expenses are variable:
i
Pay TV
ii
Travel
iii
Insurance
iv
Clothing
v
Repairs
vi
Rent
3
Amelia earns a weekly salary of \$723 and receives income from other sources of \$126. Her weekly expenses are: rent \$199, groceries \$51, petrol \$33, electricity \$33, telephone \$23, magazines \$13 and clothes \$62. Calculate: a Her weekly income b Her weekly expenses c Her weekly savings d Her annual savings 4 Julie earns \$610 per week. Her weekly work related expenses are \$185 on travel and \$86 on lunch. What percentage of her weekly income goes towards work related expenses? Round your answer to two decimal places.
5
Amerie pays rent of \$239 per week, and must budget for electricity and water costs of \$533 per quarter. How much should she put aside each week to cover these expenses?
6
Sally earns a weekly salary of \$1195 and receives income from other sources of \$181. Her weekly expenses are on the right:
Calculate:
a
The percentage of income saved, correct to two decimal places.
b
The percentage of income spent on magazines and clothes, correct to two decimal places.
c
The amount of rent paid annually.
d
The number of full weeks needed to accumulate enough savings to pay for a holiday that costs \$3800. • Rent \$122
• Groceries \$97 • Petrol \$60
• Electricity \$46 • Telephone \$28
• Magazines \$10 • Clothes \$79
7
Last month Neil saved \$235, which was 15\% of his income. Calculate his income last month. 8 Lauren earns a gross annual salary of \$34\,500. She wants to put aside equal monthly amounts towards starting her own business. If she must pay income tax of 24\% and other expenses adding up to \$19\,650 per annum, how much does she have left to set aside each month? 9 Luke’s weekly expenses vary slightly from week to week. His expenses in the first 3 weeks are \$225, \$218 and \$208. If he can only afford to have an average weekly expense of \$221 every four weeks, calculate the maximum expense he can have in the fourth week. 10 A company’s advertising cost is 2 times its technology cost, and production cost is 8 times the advertising cost. If advertising, technology and production cost \$142\,000 in total, calculate the cost of advertising.
11
Christa’s expenses currently represent 30\% of her \$920 weekly income. If her expenses increase by 5\%, what percentage of her income will expenses take up? 12 Todd's current income of \$327 is to increase by \$6.54, but his current expenses of \$114.45 are also set to increase by 4\%.
a
What percentage of his income is currently consumed by expenses?
b
Calcuate the percentage of income consumed by expenses after the increases. Give your answer to two decimal places if necessary.
c
Calculate the change in percentage of his income consumed by expenses.
13
Tina wants to take out a car loan, making regular monthly repayments. She expects the other costs to be:
• \$50 per week for petrol • \$110 per month for insurance
• \$40 per annum for the renewal of her driver's license • \$310 every six months for servicing
• \$750 per annum for her Green Slip (third party insurance) • \$600 per annum for registration
a
Calculate the equivalent annual cost of the following:
i
Petrol
ii
Insurance
iii
iv
Servicing
v
Green Slip
vi
Registration
b
What is the total annual cost of running the car?
c
If Tina has an annual income of \$59\,000, and can spend 19\% of it on buying and running the car, calculate the maximum monthly repayment she can afford to make. Overtime and extra income 14 The normal pay rate is \$13.40 per hour.
a
What is the time-and-a-half rate?
b
What is the double time rate?
15
For an hourly rate of \$30, calculate the income earned for working: a 35 hours at normal rates and 6 hours at time-and-a-half. b 28 hours at normal rates and 5 hours at double time. c 29 hours at normal rates, 9 hours at time-and-a-half and 7 hours at double time. d 8 hours at time-and-a-half and 7 hours at double time. 16 Calculate the equivalent number of hours worked at normal pay for working: a 33 hours at normal rates and 10 hours at time-and-a-half b 37 hours at normal rates and 5 hours at double time c 21 hours at normal rates, 10 hours at time-and-a-half and 8 hours at double time d 7 hours at time-and-a-half and 10 hours at double time 17 A call centre worker is paid \$20 per hour for a normal 37-hour week, and time-and-a-half for each hour worked thereafter. Calculate his income for a week in which he works:
a
27 hours
b
45 hours
18
For each week, Sarah earns a wage of \$18 per hour for the first 23 hours, time-and-a-half for the next 9 hours, and double time after that. Calculate her weekly income if she works: a 18 hours per week. b 27 hours per week. c 37 hours per week 19 Dave, a delivery driver, is paid \$10 per hour, plus a uniform allowance of \$15 per week, \$92 per week for depreciation on his car, and a travel allowance of 37 cents per kilometre. Calculate his weekly income, if he worked 28 hours and travelled 1284 km.
20
A job advertisement states that the hourly wage based on a 35-hour working week is \$38 and that there is a 17\% holiday loading on 4 weeks wages. How much holiday loading would the employee receive? 21 Calculate the value of a 7\% annual bonus, for an income of: a \$88\,000 per annum.
b
\$7000 per month. c \$3900 per fortnight.
d
\$1500 per week. 22 Carl earns \$910 per week. Calculate his normal hourly rate of pay if he works:
a
34 hours at normal rates and 10 hours at time-and-a-half.
b
35 hours at normal rates and 10 hours at double time.
c
34 hours at normal rates, 7 hours at time-and-a-half and 6 hours at double time.
23
Jack needs to earn \$381.15 this week to cover his expenses. His normal hourly rate is \$12.10 and he is rostered to work 24 normal hours from Monday to Friday.
a
How much will he earn from Monday through to Friday?
b
Calculate the amount he needs to earn on Saturday.
c
Hence, calculate how many hours Jack must work on Saturday, where he will earn time-and-a-half.
24
An employee is given a 13\% bonus on his annual income. This bonus amounted to \$19\,000. Calculate his annual income. 25 An advertising executive is offered a bonus in the form of a percentage of the increase in sales of the product. If sales increase by \$173\,950 and the executive receives a bonus of \$11\,654.65, what was the percentage rate of his bonus, correct to one decimal place? Tax and other deductions 26 For a gross annual income of \$52\,000, calculate the net income when:
a
Tax is \$11\,100 and superannuation is \$2500 per year.
b
Tax is \$14\,000, superannuation is \$3200 and insurance is \$2600 per year. c Tax is \$12\,600, rent is \$12\,500 and union fees are \$800 per year.
27
Bill earns \$843.50 per week and has the following deductions from his gross wage: tax \$253.05, superannuation, \$85, and union fees, \$25.61. Calculate Bill’s net pay.
28
Amanda earns \$2563.65 every two weeks and has the following deductions from her gross wage: tax of \$559.17, superannuation of \$79, and union fees of \$22.47.
a
Calculate Amanda's net pay.
b
Express her income tax as a percentage of her gross pay, correct to one decimal place.
29
Calculate the net weekly wage of a carpenter who must pay 16\% tax on an annual salary of \$38\,080. 30 If tax is stated as being 15.9 cents in the dollar, what percentage tax is this? 31 Brad has a gross weekly wage of \$982.32, and pays \$36.27 for health insurance and \$9.76 for union membership. His net weekly pay is \$781.57. a How much tax does Brad pay each week? b What percentage of Brad's gross wage is paid in tax? Write your answer correct to one decimal place. c What percentage of Brad's gross wage goes towards union fees? Write your answer correct to three decimal places. 32 A train conductor is paid \$21.66 per hour for a 40 hour week, with overtime paid at a time and a half rate.
a
Find her net pay for a week when she worked for 43 hours and had PAYG tax deductions of \$109.28. b What percentage of her gross pay was paid in tax? Round your answer to the nearest whole number. 33 Calculate the percentage of gross income going to taxes, correct to two decimal places, when: a Gross income is \$56\,000 and income tax is \$15\,000. b Gross income is \$49\,000, income tax is \$13\,000 and superannuation is \$5400.
34
Express net income as a percentage of gross income, correct to two decimal places, for each of the following cases:
a
Gross income is \$45\,000 and net income is \$35\,000.
b
Gross income is \$50\,000 and income tax is \$14\,000.
c
Gross income is \$42\,000, income tax is \$10\,000, superannuation is \$2100 and insurance is \$1100.
35
A self-employed tradesman must put aside 9\% of each fortnightly wage towards superannuation, and pay 10\% GST to the government. If he earns \$1067 in a fortnight, calculate his net income after these deductions. 36 A lifeguard earns a gross hourly wage of \$19 during normal working hours, and earns time-and-a-half on weekends. Every fortnight, he works 19 normal hours and 14 hours on weekends.
a
Find his gross fortnightly wage.
b
Find his net fortnightly income, given that he is taxed at a rate of 12\%.
37
Bob received his weekly pay slip. His hourly rate is \$23.21. Complete the third column of the following table: 38 Ben is a window cleaner and is paid a gross wage of \$28.55 per hour for 52 hours per fortnight. Each fortnight \$162.92 is deducted as PAYG tax instalments as well as \$43.28 for in home and contents insurance. Find Ben's net fortnightly pay.
39
Valentina, a retiree, receives annual superannuation payments of 64\% of her final year's salary of \$55\,000. Calculate: a Her annual superannuation income. b Her annual disposable superannuation income, if her income tax is 24\% of her gross income. 40 Michael receives a weekly gross salary of \$530. If he pays 19\% of this in income tax, 9\% in superannuation and \$7.70 in medical insurance, calculate his net weekly pay. 41 If 9\% tax is payable on income but not inheritance, how much must be deducted from an income and inheritance that totals \$7350, of which 46\% is inheritance?
42
A labourer is paid a gross annual salary of \$31\,300. Each week he contributes 8\% of his gross pay to a voluntary superannuation fund. His other weekly deductions are \$103.50 PAYG tax and \$9.50 for union fees. Calculate: a His total weekly deductions. b His net weekly income. 43 In the last financial year, Carl was paid a wage every month, and each week there was \$61 deducted as PAYG tax.
a
How much did Carl pay in total in PAYG tax last financial year?
b
At the end of the financial year, he calculates his total tax payable to be \$3049. Will he receive a refund or have tax owing? 44 Amelia’s employer deducts a percentage of her income in tax. The fortnightly amount deposited into Amelia's account is \$1964 and she earns an annual gross income of \$63\,830.00. a Calculate her net annual income. b Hence, calculate the rate at which she is being taxed. Round your answer to the nearest whole percentage. 45 Roxanne’s gross fortnightly pay is \$2845.18. Each fortnight her employer deducts \$592.04 in PAYG tax instalments and she contributes \$47.63 to a voluntary superannuation fund for her retirement. Find Roxanne's weekly net pay.
46
Employers are legally obliged to pay an amount worth 9.5\% of an employee's income into a superannuation fund. If Eileen's annual income is \$43\,000 and she chooses to contribute a further 5\% of her income into her superannuation fund, calculate: a The amount paid into Eileen's superannuation fund in one year. b The amount extra paid into Eileen's superannuation fund the following year, when she receives a 9\% pay rise. 47 Yuri has a gross salary of \$69\,941.69 with an annual income tax deduction of \$15\,759.98, a superannuation contribution of 9.5\% of the gross salary and a health insurance contribution of \$92 per week, for 52 weeks. Calculate his net annual income.
48
Neil is self-employed and must put aside his own superannuation and tax each month. He estimates his income in the first half of the year to be \$57\,600 and for it to grow by 2.8\% in the second half of the year. He is to be taxed at a rate of 30\% and must put aside 9.5\% of his gross annual income into a superannuation fund. a Calculate his expected gross income for the year. b Calculate how much he must deduct from his gross income each month to pay his tax and superannuation. Income tax 49 The following table shows the tax rates for 2018 to 2019: Resident Tax Rate (2018-2019) Table Calculate the income tax payable for the following taxable incomes: a \$8326
b
\$36\,810 c \$53\,395
d
\$146\,664 e \$194\,388
f
\$158\,810 50 Last financial year, Roxanne's taxable income was \$157\,872. Based on the Resident Tax Rate (2018-2019) table, calculate:
a
The income tax payable on Roxanne's income.
b
The Medicare levy payable, given that it is 2\% of taxable income.
c
The total amount Roxanne needs to pay in income tax and Medicare levy.
51
Ivan is a teacher who earns an annual salary of \$40\,800. He earns an additional \$4520 per year from private tutoring. He is allowed various tax deductions (for resource books etc.) amounting to \$786, and he has paid tax instalments of \$4880 during the year. Find:
a
His total gross income.
b
His taxable income.
c
His tax payable using the Resident Tax Rate (2018-2019) table.
d
The tax amount he still owes.
52
Hermione earned a taxable income of \$91\,354 in the last financial year. Throughout the year, her employer deducted a total of \$22\,386.15 in PAYG tax instalments.
a
Calculate how much Hermione must pay in income tax for the year using the Resident Tax Rate (2018-2019) table.
b
Hermione also owes a Medicare levy of \$1827.08, in addition to her income tax. Determine whether she will receive a refund or have tax owing. c Find the tax amount she still owes. 53 Sharon is a project manager with a taxable income of \$97\,192. Throughout the year, her employer has deducted \\$553 per week in tax instalments.
a
Calculate Sharon's tax payable using the Resident Tax Rate (2018-2019) table.
b
Calculate the Medicare Levy, which is 2\% of taxable income.
c
Since Sharon's employer has deducted more than is needed from her income, Sharon is entitled to a refund for the year. Find the amount of this refund.
### Outcomes
#### ACMEM001
solve practical problems requiring basic number operations
#### ACMEM002
apply arithmetic operations according to their correct order
#### ACMEM008
evaluate decimal fractions to the required number of decimal places | 3,852 | 15,377 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2022-05 | longest | en | 0.90526 |
https://eagerclub.com/15-of-42/ | 1,726,148,031,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00810.warc.gz | 205,690,447 | 15,506 | # What is 15 of 42? Find 15 Percent of 42 (15% of 42)
In this article, we will find out what is 15 of 42 (15% of 42) is. Without wasting time let’s calculate 15 percent of 42.
Contents
15 of 42 is 6.3
## How to calculate 15 of 42?
15% of 42?
= (15/100) x 42
= 630/100
= 6.3
Thus, 15% of 42 is 6.3.
## Formula
(Value/100) x Total Value
Example
What is 15 percent of 42?
15 percent of 42 is
= (15/100) x 42
= 630/100
= 6.3
Answer: 15 percent of 42 is 6.3.
Now you can easily find out the answer to 15 of 42.
On Eagerclub you will get to know more interesting topics like these.
## FAQ
### How Can You Find 15% Of 42?
15 percent of 42 is
(15/100) x 42 = 6.3
### How To Find 15 Of 42?
(15/100) x 42 = 6.3
### What Is 15 As A Percentage Of 42?
The total answers count 42 – it’s 100%, so we to get a 1% value, divide 42 by 100 to get 0.42. Next, calculate the percentage of 15: divide 15 by 1% value (0.42), and you get 35.71% – it’s your percentage grade.
### What Percent Is 15 Out Of 42?
Solution: 15/42 as a percent is 35.714%
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What is the percentage of 15 in 42 | 470 | 1,270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-38 | latest | en | 0.85132 |
https://slideplayer.com/slide/6840612/ | 1,581,928,018,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141749.3/warc/CC-MAIN-20200217055517-20200217085517-00478.warc.gz | 572,732,271 | 23,660 | # Circuit Analysis. Circuit Analysis using Series/Parallel Equivalents 1.Begin by locating a combination of resistances that are in series or parallel.
## Presentation on theme: "Circuit Analysis. Circuit Analysis using Series/Parallel Equivalents 1.Begin by locating a combination of resistances that are in series or parallel."— Presentation transcript:
Circuit Analysis
Circuit Analysis using Series/Parallel Equivalents 1.Begin by locating a combination of resistances that are in series or parallel. Often the place to start is farthest from the source. 2.Redraw the circuit with the equivalent resistance for the combination found in step 1. 3.Repeat steps 1 and 2 until the circuit is reduced as far as possible. Often (but not always) we end up with a single source and a single resistance. 4.Solve for the currents and voltages in the final equivalent circuit.
Working Backward
Find current flowing each resistor
Voltage Division
Application of the Voltage- Division Principle
Current Division
Application of the Current-Division Principle
Voltage division Voltage division and current division
Current division
Although they are very important concepts, series/parallel equivalents and the current/voltage division principles are not sufficient to solve all circuits.
Mesh Current Analysis Sources
Definition of a loop Definition of a mesh
Choosing the Mesh Currents When several mesh currents flow through one element, we consider the current in that element to be the algebraic sum of the mesh currents.
Writing Equations to Solve for Mesh Currents If a network contains only resistors and independent voltage sources, we can write the required equations by following each current around its mesh and applying KVL.
For mesh 1, we have For mesh 2, we obtain For mesh 3, we have
Determine the two mesh currents, i 1 and i 2, in the circuit below. For the left-hand mesh, -42 + 6 i 1 + 3 ( i 1 - i 2 ) = 0 For the right-hand mesh, 3 ( i 2 - i 1 ) + 4 i 2 - 10 = 0 Solving, we find that i 1 = 6 A and i 2 = 4 A. (The current flowing downward through the 3- resistor is therefore i 1 - i 2 = 2 A. )
Mesh Currents in Circuits Containing Current Sources *A common mistake is to assume the voltages across current sources are zero. Therefore, loop equation cannot be set up at mesh one due to the voltage across the current source is unknown Anyway, the problem is still solvable.
As the current source common to two mesh, combine meshes 1 and 2 into a supermesh. In other words, we write a KVL equation around the periphery of meshes 1 and 2 combined. Mesh 3: It is the supermesh. Three linear equations and three unknown
Find the three mesh currents in the circuit below. Creating a “supermesh” from meshes 1 and 3: -7 + 1 ( i 1 - i 2 ) + 3 ( i 3 - i 2 ) + 1 i 3 = 0 [1] Around mesh 2: 1 ( i 2 - i 1 ) + 2 i 2 + 3 ( i 2 - i 3 ) = 0[2] Rearranging, i 1 - 4 i 2 + 4 i 3 = 7[1] -i 1 + 6 i 2 - 3 i 3 = 0[2] i 1 - i 3 = 7[3] Solving, i 1 = 9 A, i 2 = 2.5 A, and i 3 = 2 A. Finally, we relate the currents in meshes 1 and 3: i 1 - i 3 = 7[3]
supermesh of mesh1 and mesh2 current source branch current
Three equations and three unknown.
Mesh-Current Analysis 1. If necessary, redraw the network without crossing conductors or elements. Then define the mesh currents flowing around each of the open areas defined by the network. For consistency, we usually select a clockwise direction for each of the mesh currents, but this is not a requirement. 2. Write network equations, stopping after the number of equations is equal to the number of mesh currents. First, use KVL to write voltage equations for meshes that do not contain current sources. Next, if any current sources are present, write expressions for their currents in terms of the mesh currents. Finally, if a current source is common to two meshes, write a KVL equation for the supermesh. 3. If the circuit contains dependent sources, find expressions for the controlling variables in terms of the mesh currents. Substitute into the network equations, and obtain equations having only the mesh currents as unknowns. 4. Put the equations into standard form. Solve for the mesh currents by use of determinants or other means. 5. Use the values found for the mesh currents to calculate any other currents or voltages of interest.
Superposition Superposition Theorem – the response of a circuit to more than one source can be determined by analyzing the circuit’s response to each source (alone) and then combining the results Insert Figure 7.2
Superposition Insert Figure 7.3
Superposition Analyze Separately, then Combine Results
Use superposition to find the current i x. Current source is zero – open circuit as I = 0 and solve i Xv Voltage source is zero – short circuit as V= 0 and solve i Xv
Use superposition to find the current i x. The controlled voltage source is included in all cases as it is controlled by the current i x.
Voltage and Current Sources Insert Figure 7.7
Voltage and Current Sources Insert Figure 7.8
Voltage and Current Sources Insert Figure 7.9
Source Transformation Under what condition, the voltage and current of the load is the same when operating at the two practical sources? For voltage source, For current source We have,
Voltage and Current Sources Equivalent Voltage and Current Sources – for every voltage source, there exists an equivalent current source, and vice versa
Thevenin’s Theorem Thevenin’s Theorem – any resistive circuit or network, no matter how complex, can be represented as a voltage source in series with a source resistance
Thevenin’s Theorem Thevenin Voltage (V TH ) – the voltage present at the output terminals of the circuit when the load is removed Insert Figure 7.18
Thevenin’s Theorem Thevenin Resistance (R TH ) – the resistance measured across the output terminals with the load removed
Thévenin Equivalent Circuits
Finding the Thévenin Resistance Directly When zeroing a voltage source, it becomes a short circuit. When zeroing a current source, it becomes an open circuit. We can find the Thévenin resistance by zeroing the sources in the original network and then computing the resistance between the terminals.
Computation of Thévenin resistance
Equivalence of open-circuit and Thévenin voltage
A circuit and its Thévenin equivalent
Superposition As the voltage source does not contribute any output voltage, Only the current source has the effect.
Determine the Thévenin and Norton Equivalents of Network A in (a). Source transformation
Find the Thévenin equivalent of the circuit shown in (a). As i = -1, therefore, the controlled voltage source is -1.5V. Use nodal analysis at node v, v Thus, R th =v/I = 0.6/1 = 0.6 ohms
Applications of Thevenin’s Theorem Load Voltage Ranges – Thevenin’s theorem is most commonly used to predict the change in load voltage that will result from a change in load resistance
Applications of Thevenin’s Theorem Maximum Power Transfer –Maximum power transfer from a circuit to a variable load occurs when the load resistance equals the source resistance –For a series-parallel circuit, maximum power occurs when R L = R TH
Applications of Thevenin’s Theorem Multiload Circuits Insert Figure 7.30
Norton’s Theorem Norton’s Theorem – any resistive circuit or network, no matter how complex, can be represented as a current source in parallel with a source resistance
Norton’s Theorem Norton Current (I N ) – the current through the shorted load terminals Insert Figure 7.35
Computation of Norton current
Norton’s Theorem Norton Resistance (R N ) – the resistance measured across the open load terminals (measured and calculated exactly like R TH )
Norton’s Theorem Norton-to-Thevenin and Thevenin-to-Norton Conversions Insert Figure 7.39
Step-by-step Thévenin/Norton- Equivalent-Circuit Analysis 1. Perform two of these: a. Determine the open-circuit voltage V t = v oc. b. Determine the short-circuit current I n = i sc. c. Zero the sources and find the Thévenin resistance R t looking back into the terminals.
2. Use the equation V t = R t I n to compute the remaining value. 3. The Thévenin equivalent consists of a voltage source V t in series with R t. 4. The Norton equivalent consists of a current source I n in parallel with R t.
Maximum Power Transfer The load resistance that absorbs the maximum power from a two-terminal circuit is equal to the Thévenin resistance.
Graphical representation of maximum power transfer Power transfer between source and load
Download ppt "Circuit Analysis. Circuit Analysis using Series/Parallel Equivalents 1.Begin by locating a combination of resistances that are in series or parallel."
Similar presentations | 2,060 | 8,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2020-10 | latest | en | 0.87416 |
https://convertilo.com/imperial-tablespoons-to-liters | 1,713,184,319,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816977.38/warc/CC-MAIN-20240415111434-20240415141434-00625.warc.gz | 170,908,336 | 5,457 | Imperial Tablespoons to Liters - uk tbsp to L
How to convert from Imperial Tablespoons to Liters
The conversion factor between Imperial Tablespoons and Liters is 0.01420653125. To convert an amount in Imperial Tablespoons in Liters, multiply the the desired amount in Imperial Tablespoons by 0.01420653125:
Amount(Imperial Tablespoons) × 0.01420653125(Conversion Factor) = Result(Liters)
Definition of units
Let's see how both units in this conversion are defined, in this case Imperial Tablespoons and Liters:
Imperial Tablespoon (uk tbsp)
An imperial tablespoon (abbreviation tbsp, "Tb." or T.) is exactly 15 ml (0.51 US fl oz). A tablespoon is a large spoon used for serving or eating. In many English-speaking regions, the term now refers to a large spoon used for serving, however, in some regions, including parts of Canada, it is the largest type of spoon used for eating. By extension, the term is used as a measure of volume in cooking.
Liter (L)
The liter (also written "litre"; SI symbol L or l) is a non-SI metric system unit of volume. It is equal to 1 cubic decimeter (dm3), 1,000 cubic centimeters (cm3) or 1/1,000 cubic meter. The mass of one liter liquid water is almost exactly one kilogram. A liter is defined as a special name for a cubic decimeter or 10 centimeters × 10 centimeters × 10 centimeters, thus, 1 L ≡ 1 dm3 ≡ 1000 cm3.
Imperial Tablespoons to Liters conversion table
Below is the conversion table you can use to convert from Imperial Tablespoons to Liters
Imperial Tablespoons (uk tbsp) Liters (L)
1 Imperial Tablespoons 0.0142 Liters
2 Imperial Tablespoons 0.0284 Liters
3 Imperial Tablespoons 0.0426 Liters
4 Imperial Tablespoons 0.0568 Liters
5 Imperial Tablespoons 0.071 Liters
6 Imperial Tablespoons 0.0852 Liters
7 Imperial Tablespoons 0.0994 Liters
8 Imperial Tablespoons 0.114 Liters
9 Imperial Tablespoons 0.128 Liters
10 Imperial Tablespoons 0.142 Liters | 530 | 1,909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-18 | latest | en | 0.803499 |
http://code.activestate.com/recipes/425044/ | 1,519,523,624,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816083.98/warc/CC-MAIN-20180225011315-20180225031315-00107.warc.gz | 66,385,600 | 7,458 | Welcome, guest | Sign In | My Account | Store | Cart
NOTE: Recipes have moved! Please visit GitHub.com/activestate/code for the current versions.
Split up a sequence in same-size(if possible) parts.
Python, 6 lines
```1 2 3 4 5 6``` ```# Guyon Morée # http://gumuz.looze.net/ def split_seq(seq,size): """ Split up seq in pieces of size """ return [seq[i:i+size] for i in range(0, len(seq), size)] ```
It's very straight forward:
``````>>> lst = [1,2,3,4,5,6,7,8,9,10]
>>> split_seq(lst,2)
[[1, 2], [3, 4], [5, 6], [7, 8], [9,10]]
``````
It takes a list, re-groups them in another list, 2 items per group. The source list doesn't have to be evenly dividable though:
``````>>> lst = [1,2,3,4,5,6,7,8,9,10]
>>> split_seq(lst,3)
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10]]
``````
Of course, a string is a sequence too. That's what I used it for last time. I had a string of bits, which I needed to divide in pieces of 8, bytes:
``````>>> bits = '101011101001011010010110'
>>> split_seq(bits,8)
['10101110', '10010110', '10010110']
``````
Nick Matsakis 12 years, 8 months ago
One issue with this recipe is that you wind up with a list at the end that could be much shorter than what you have. Consider this alternative:
``````def split_seq(seq, size):
newseq = []
splitsize = 1.0/size*len(seq)
for i in range(size):
newseq.append(seq[int(round(i*splitsize)):int(round((i+1)*splitsize))])
return newseq
``````
This produces the following results, which may be preferred for some applications:
``````>>> split_seq([1,2,3,4,5,6,7,8,9,10], 3)
[[1, 2, 3], [4, 5, 6, 7], [8, 9, 10]]
``````
Guyon Morée (author) 12 years, 8 months ago
very useful as well. Thanx Nick, that's a cool one as well. Do you want me to add your version to the recipe or are you going to add this as a recipe as well?
Cheers, Guyon Morée http://gumuz.looze.net/
Nick Matsakis 12 years, 8 months ago
I just submitted it as a separate recipe, though I'm not sure I like it so much. I wonder if it could be done using only integer math.
Gene tani 12 years, 8 months ago
Created by Guyon Morée on Tue, 7 Jun 2005 (PSF)
### Required Modules
• (none specified) | 710 | 2,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-09 | latest | en | 0.879124 |
https://google-developers.appspot.com/apps-script/reference/optimization/linear-optimization-engine | 1,579,585,217,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250601615.66/warc/CC-MAIN-20200121044233-20200121073233-00181.warc.gz | 460,815,144 | 18,904 | # Class LinearOptimizationEngine
LinearOptimizationEngine
The engine used to model and solve a linear program. The example below solves the following linear program:
Two variables, `x` and `y`:
`0 ≤ x ≤ 10`
`0 ≤ y ≤ 5`
Constraints:
`0 ≤ 2 * x + 5 * y ≤ 10`
`0 ≤ 10 * x + 3 * y ≤ 20`
Objective:
Maximize `x + y`
```var engine = LinearOptimizationService.createEngine();
// Add two variables, 0 <= x <= 10 and 0 <= y <= 5
// Create the constraint: 0 <= 2 * x + 5 * y <= 10
constraint.setCoefficient('x', 2);
constraint.setCoefficient('y', 5);
// Create the constraint: 0 <= 10 * x + 3 * y <= 20
constraint.setCoefficient('x', 10);
constraint.setCoefficient('y', 3);
// Set the objective to be x + y
engine.setObjectiveCoefficient('x', 1);
engine.setObjectiveCoefficient('y', 1);
// Engine should maximize the objective
engine.setMaximization();
// Solve the linear program
var solution = engine.solve();
if (!solution.isValid()) {
Logger.log('No solution ' + solution.getStatus());
} else {
Logger.log('Value of x: ' + solution.getVariableValue('x'));
Logger.log('Value of y: ' + solution.getVariableValue('y'));
}```
### Methods
MethodReturn typeBrief description
`addConstraint(lowerBound, upperBound)``LinearOptimizationConstraint`Adds a new linear constraint in the model.
`addConstraints(lowerBounds, upperBounds, variableNames, coefficients)``LinearOptimizationEngine`Adds constraints in batch to the model.
`addVariable(name, lowerBound, upperBound)``LinearOptimizationEngine`Adds a new continuous variable to the model.
`addVariable(name, lowerBound, upperBound, type)``LinearOptimizationEngine`Adds a new variable to the model.
`addVariable(name, lowerBound, upperBound, type, objectiveCoefficient)``LinearOptimizationEngine`Adds a new variable to the model.
`addVariables(names, lowerBounds, upperBounds, types, objectiveCoefficients)``LinearOptimizationEngine`Adds variables in batch to the model.
`setMaximization()``LinearOptimizationEngine`Sets the optimization direction to maximizing the linear objective function.
`setMinimization()``LinearOptimizationEngine`Sets the optimization direction to minimizing the linear objective function.
`setObjectiveCoefficient(variableName, coefficient)``LinearOptimizationEngine`Sets the coefficient of a variable in the linear objective function.
`solve()``LinearOptimizationSolution`Solves the current linear program with the default deadline of 30 seconds.
`solve(seconds)``LinearOptimizationSolution`Solves the current linear program.
## Detailed documentation
### `addConstraint(lowerBound, upperBound)`
Adds a new linear constraint in the model. The upper and lower bound of the constraint are defined at creation time. Coefficients for the variables are defined via calls to `LinearOptimizationConstraint.setCoefficient(variableName, coefficient)`.
```var engine = LinearOptimizationService.createEngine();
// Create a linear constraint with the bounds 0 and 10
// Create a variable so we can add it to the constraint
// Set the coefficient of the variable in the constraint. The constraint is now:
// 0 <= 2 * x <= 5
constraint.setCoefficient('x', 2);```
#### Parameters
NameTypeDescription
`lowerBound``Number`lower bound of the constraint
`upperBound``Number`upper bound of the constraint
#### Return
`LinearOptimizationConstraint` — the constraint created
### `addConstraints(lowerBounds, upperBounds, variableNames, coefficients)`
Adds constraints in batch to the model.
```var engine = LinearOptimizationService.createEngine();
// Add a boolean variable 'x' (integer >= 0 and <= 1) and a real (continuous >= 0 and <= 100)
variable 'y'.
engine.addVariables(['x', 'y'], [0, 0], [1, 100],
[LinearOptimizationService.VariableType.INTEGER,
LinearOptimizationService.VariableType.CONTINUOUS]);
// 0 <= x + y <= 3
// 1 <= 10 * x - y <= 5
engine.addConstraints([0.0, 1.0], [3.0, 5.0], [['x', 'y'], ['x', 'y']], [[1, 1], [10, -1]]);```
#### Parameters
NameTypeDescription
`lowerBounds``Number[]`lower bounds of the constraints
`upperBounds``Number[]`upper bounds of the constraints
`variableNames``String[][]`the names of variables for which the coefficients are being set
`coefficients``Number[][]`coefficients being set
#### Return
`LinearOptimizationEngine` — a linear optimization engine
### `addVariable(name, lowerBound, upperBound)`
Adds a new continuous variable to the model. The variable is referenced by its name. The type is set to `VariableType.CONTINUOUS`.
```var engine = LinearOptimizationService.createEngine();
// Add a boolean variable (integer >= 0 and <= 1)
// Add a real (continuous) variable. Notice the lack of type specification.
#### Parameters
NameTypeDescription
`name``String`unique name of the variable
`lowerBound``Number`lower bound of the variable
`upperBound``Number`upper bound of the variable
#### Return
`LinearOptimizationEngine` — a linear optimization engine
### `addVariable(name, lowerBound, upperBound, type)`
Adds a new variable to the model. The variable is referenced by its name.
```var engine = LinearOptimizationService.createEngine();
// Add a boolean variable (integer >= 0 and <= 1)
// Add a real (continuous) variable
#### Parameters
NameTypeDescription
`name``String`unique name of the variable
`lowerBound``Number`lower bound of the variable
`upperBound``Number`upper bound of the variable
`type``VariableType`type of the variable, can be one of `VariableType`
#### Return
`LinearOptimizationEngine` — a linear optimization engine
### `addVariable(name, lowerBound, upperBound, type, objectiveCoefficient)`
Adds a new variable to the model. The variable is referenced by its name.
```var engine = LinearOptimizationService.createEngine();
// Add a boolean variable (integer >= 0 and <= 1)
// The objective is now 2 * x.
// Add a real (continuous) variable
// The objective is now 2 * x - 5 * y.```
#### Parameters
NameTypeDescription
`name``String`unique name of the variable
`lowerBound``Number`lower bound of the variable
`upperBound``Number`upper bound of the variable
`type``VariableType`type of the variable, can be one of `VariableType`
`objectiveCoefficient``Number`objective coefficient of the variable
#### Return
`LinearOptimizationEngine` — a linear optimization engine
### `addVariables(names, lowerBounds, upperBounds, types, objectiveCoefficients)`
Adds variables in batch to the model. The variables are referenced by their names.
```var engine = LinearOptimizationService.createEngine();
// Add a boolean variable 'x' (integer >= 0 and <= 1) and a real (continuous >=0 and <= 100)
// variable 'y'.
engine.addVariables(['x', 'y'], [0, 0], [1, 100],
[LinearOptimizationService.VariableType.INTEGER,
LinearOptimizationService.VariableType.CONTINUOUS]);```
#### Parameters
NameTypeDescription
`names``String[]`unique names of the variables
`lowerBounds``Number[]`lower bounds of the variables
`upperBounds``Number[]`upper bounds of the variables
`types``VariableType[]`types of the variables, can be one of `VariableType`
`objectiveCoefficients``Number[]`objective coefficients of the variables
#### Return
`LinearOptimizationEngine` — a linear optimization engine
### `setMaximization()`
Sets the optimization direction to maximizing the linear objective function.
```var engine = LinearOptimizationService.createEngine();
// Add a real (continuous) variable. Notice the lack of type specification.
// Set the coefficient of 'y' in the objective.
// The objective is now 5 * y
engine.setObjectiveCoefficient('y', 5);
// We want to maximize.
engine.setMaximization();```
#### Return
`LinearOptimizationEngine` — a linear optimization engine
### `setMinimization()`
Sets the optimization direction to minimizing the linear objective function.
```var engine = LinearOptimizationService.createEngine();
// Add a real (continuous) variable. Notice the lack of type specification.
// Set the coefficient of 'y' in the objective.
// The objective is now 5 * y
engine.setObjectiveCoefficient('y', 5);
// We want to minimize
engine.setMinimization();```
#### Return
`LinearOptimizationEngine` — a linear optimization engine
### `setObjectiveCoefficient(variableName, coefficient)`
Sets the coefficient of a variable in the linear objective function.
```var engine = LinearOptimizationService.createEngine();
// Add a real (continuous) variable. Notice the lack of type specification.
// Set the coefficient of 'y' in the objective.
// The objective is now 5 * y
engine.setObjectiveCoefficient('y', 5);```
#### Parameters
NameTypeDescription
`variableName``String`name of variable for which the coefficient is being set
`coefficient``Number`coefficient of the variable in the objective function
#### Return
`LinearOptimizationEngine` — a linear optimization engine
### `solve()`
Solves the current linear program with the default deadline of 30 seconds. Returns the solution found.
```var engine = LinearOptimizationService.createEngine();
// ...
// Solve the linear program
var solution = engine.solve();
if (!solution.isValid()) {
throw 'No solution ' + solution.getStatus();
}
Logger.log('Value of x: ' + solution.getVariableValue('x'));```
#### Return
`LinearOptimizationSolution` — solution of the optimization
### `solve(seconds)`
Solves the current linear program. Returns the solution found. and if it is an optimal solution.
```var engine = LinearOptimizationService.createEngine();
// ...
// Solve the linear program
var solution = engine.solve(300);
if (!solution.isValid()) {
throw 'No solution ' + solution.getStatus();
}
Logger.log('Value of x: ' + solution.getVariableValue('x'));```
#### Parameters
NameTypeDescription
`seconds``Number`deadline for solving the problem, in seconds; the maximum deadline is 300 seconds
#### Return
`LinearOptimizationSolution` — solution of the optimization | 2,316 | 9,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-05 | latest | en | 0.50612 |
https://gmatclub.com/forum/a-study-comparing-a-group-of-chronically-depressed-101899.html | 1,487,508,709,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501169769.33/warc/CC-MAIN-20170219104609-00618-ip-10-171-10-108.ec2.internal.warc.gz | 722,878,860 | 59,429 | A study comparing a group of chronically depressed : GMAT Critical Reasoning (CR)
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# A study comparing a group of chronically depressed
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A study comparing a group of chronically depressed [#permalink]
### Show Tags
28 Sep 2010, 21:11
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A study comparing a group of chronically depressed individuals with an otherwise matched group of individuals free from depression found significantly more disorders of the immune system among the depressed group. According to the researchers, these results strongly support the hypothesis that mental states influence the body's vulnerability to the infection.
Q: Which of the following, if true, cast the most serious doubt on the researcher's intepretation?
A: The researchers' view does little more than echo a familiar theme in folklore an literature
B Chronically depressed individuals are no less careful than others to avoid exposure to infections
C: Discorders of the immune system cause many of those inviduals who have them to become chronically depressed
D: Inviduals who have previously been free from depression can become depressed quite suddenly
E A high frequency of infections can stem from an unusally high level of exposure rather than from any disorder of the immune system
OA to come...
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D Day is April 23rd, 2010
Be humble, be focused, and be calm!
If you have any questions
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### Show Tags
28 Sep 2010, 21:26
A- irrelevant
B- Strengthens the argument
C- Cause effect reversal. Right answer.
D- Not relevant
E- Its not about how infections stem. Its about how mental states affect body's vulnerability.
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### Show Tags
28 Sep 2010, 21:31
C for me too.
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### Show Tags
28 Sep 2010, 21:38
C to me:
Conclusion sates: "mental states influence the body's vulnerability to the infection"
Option C states that infection is causing the depressed mental state....
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28 Sep 2010, 21:52
OA plz...
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29 Sep 2010, 03:27
Used POE and only C makes sense.
Plus, researchers assumed that poor mental health will cause vunerability to infection. However, if it was the other way round, then the conclusion is invalid.
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29 Sep 2010, 03:35
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29 Sep 2010, 03:54
Classic case of denying a reverse cause and effect.
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29 Sep 2010, 05:05
It has to be C
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Its your duty to post OA afterwards; some one must be waiting for that...
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29 Sep 2010, 09:04
IMO C
OA plz
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29 Sep 2010, 09:27
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Thank you everyone.
Oa is c
So for weaken type of CR questions, it's ok for the right answer to be totally different from the premise?
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29 Sep 2010, 10:02
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A correct answer choice to the weaken question will do one of the following:
1. Breaks down causality (As in this case, since it has reversed the cause-effect relationship by saying that they are depressed BECAUSE they have lesser immune capabilities, as opposed to being the other way around). This could mean reversal of cause-effect or proving that a relationship between cause and effect is non existent.
3. Break down an assumption
In all these cases the key thing is to realize that the ANSWER choices are taken to be true. So yes, the stimulus can be "wrong" and very different from the answer choice so to speak.
Hope this helps.
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### Show Tags
29 Sep 2010, 20:08
MBAwannabe10 wrote:
A study comparing a group of chronically depressed individuals with an otherwise matched group of individuals free from depression found significantly more disorders of the immune system among the depressed group. According to the researchers, these results strongly support the hypothesis that mental states influence the body's vulnerability to the infection.
Q: Which of the following, if true, cast the most serious doubt on the researcher's intepretation?
A: The researchers' view does little more than echo a familiar theme in folklore an literature
B Chronically depressed individuals are no less careful than others to avoid exposure to infections
C: Discorders of the immune system cause many of those inviduals who have them to become chronically depressed
D: Inviduals who have previously been free from depression can become depressed quite suddenly
E A high frequency of infections can stem from an unusally high level of exposure rather than from any disorder of the immune system
Conclusion:
mental states influence the body's vulnerability to the infection.
Premise:
group of chronically depressed individuals with an otherwise matched group of individuals free from depression found significantly more disorders of the immune system among the depressed group
So to weaken the argument we have to show that mental state is not the only cause of the body's vulnerability to the infection
Option C defies the argument by stating that its the disorders of the immune system those cause the individual to become chronically depressed.
So C is totally opposite to the Argument
C is the ans
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01 Oct 2010, 22:00
this question is very easy in terms of comparing other options..so once can eliminate all options except the right one
so i guess answer should be C.
Re: Infection and Depression [#permalink] 01 Oct 2010, 22:00
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18 A recent study compared property crime rates 13 01 Nov 2013, 04:52
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Display posts from previous: Sort by | 2,180 | 8,220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2017-09 | longest | en | 0.901127 |
https://personalstatementhelp.online/creative-writing/2-step-word-problems-for-third-grade | 1,679,799,455,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945381.91/warc/CC-MAIN-20230326013652-20230326043652-00084.warc.gz | 494,810,200 | 16,070 | ## Math Word Problems for Third Graders
Help kids master basic math and boost problem-solving skills.
• Math Tutorials
• Pre Algebra & Algebra
• Exponential Decay
Word problems allow students the opportunity to apply their math skills in authentic situations. All too often, children who are able to solve numeric problems find themselves at a loss when faced with a word problem. Some of the best problems to work with are those in which the unknown factor is located in either the beginning or the middle of the problem. For example, instead of saying, "I have 29 balloons and the wind blew eight of them away," and then asking "How many do I have left?" try something like this instead: "I had a lot of balloons but the wind blew eight of them away. Now I only have 21 balloons left. How many did I have to begin with?" Or, "I had 29 balloons, but the wind blew some away, and I only have 21 now. How many balloons did the wind blow away?"
## Word Problem Examples
As teachers and parents, we're often very good at creating or using word problems in which the unknown value is located at the end of the question. Unfortunately, this type of problem can prove too challenging for young children. By changing the position of the unknown you can create problems that are easier for beginning math students to solve.
Another type of problem that's great for young learners is a two-step problem, which requires them to solve for one unknown before solving for another. Once young students have mastered basic word problems, they can practice two-step (and three-step) problems to work on more challenging concepts. These problems help students learn how to process and relate complex sets of information. Here are some examples:
Students will often need to re-read a question to make sure they have all of the information they need. They should also be encouraged to read the question again to be certain that they actually understand what the question is asking them to solve for.
## Worksheet #1
Deb Russell
This worksheet features several basic word problems for young math students.
## Worksheet #2
Deb Russell
This worksheet features a set of intermediate word problems for young students who have already mastered basic skills. To solve these problems, students will need to have an understanding of how to count money.
## Worksheet #3
Deb Russell
This worksheet features several multi-step problems for advanced students.
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## 2 Step Addition and Subtraction Word Problems - Digital Distance Learning
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## Multiple-Step Word Problems
These multiple-step word problems require students to use reasoning and critical thinking skills to determine how each problem can be solved.
## Basic (Grades 2 - 3)
Logged in members can use the Super Teacher Worksheets filing cabinet to save their favorite worksheets.
## Intermediate (Grades 3 - 4)
These word problems are sorted by type: addition, subtraction, multiplication, division, fractions and more.
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S.T.W. has thousands of worksheets. Visit the full math index to find them all, sorted by topic.
## Solving Two Step Word Problems 3rd Grade Worksheets
Solving two-step word problems 3rd grade worksheets provide a basic understanding and practice of how to solve a word problem by applying two arithmetic operations. A two-step word problem requires applying two equations to resolve. Such problems challenge students to use their reasoning and critical thinking abilities to interpret and convert statements into equations. The answer keys included in these worksheets implement step-by-step easy explanation of this process for better learning and self-assessment.
## Benefits of 3rd Grade Solving Two-Step Word Problems Worksheets
By solving two-step word problems 3rd grade worksheets, kids can easily form the basis for solving complex multi-level algebraic equations. These 3rd grade math worksheets are worded in a simple style to help children grasp concepts quickly and potentially apply them in life. The word problems available in these worksheets familiarize kids with mathematical language and terms like fewer, altogether, difference, more, share, multiply, subtract, equal, reduced, etc. Solving two-step word problems grade 3 worksheets establish higher-level mathematical thinking in kids required for advanced math concepts.
## Printable PDFs for Grade 3 Solving Two-Step Word Problems Worksheets
Explore more topics at Cuemath's Math Worksheets .
## 3rd Grade Math Word Problems Worksheets
Math word problems help deepen a student's understanding of mathematical concepts by relating mathematics to everyday life.
These worksheets are best attempted after a student has studied the underlying skill; for example, our 'addition in columns" word problem worksheets should not be attempted until students are comfortable with addition in columns.
In many of our word problems we intentionally include superfluous data , so that students need to read and think about the questions carefully, rather than simply applying a computation pattern to solve the problems.
Mixed add and subtract word problems
## Subtraction word problems
Simple subtraction word problems
Subtraction in columns word problems
## Multiplication word problems
Simple multiplication word problems
Multiples of 10
Multiplying in columns
More multiplication word problems
Mixed multiply & divide word problems
## Division word problems
Simple division word problems
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## Fraction word problems
Identifying and comparing fractions word problems
Adding and subtracting fractions word problems
## Mixed 3rd grade word problems
The following worksheets contain a mix of grade 3 addition, subtraction, multiplication and division word problems . Mixing math word problems tests the understanding mathematical concepts, as it forces students to analyze the situation rather than mechanically apply a solution.
Mixed word problems - mental math
Mixed word problems - column math
Mixed word problems - simpler form (shorter texts, no superfluous data)
## Measurement word problems for grade 3
These word problems combine the 4 operations with real world units of length, time, volume and mass. There is no conversion of units.
Length word problems
Time word problems
Mass & weight word problems
Volume & capacity word problems.
## Word problems with variables
These grade 3 word problems introduce students to using variables ("x, y, etc.") to represent unknowns. The problems are relatively simple, but emphasize the use of variables and the writing of equations.
Word problems with variables (variable is chosen for the student)
Writing variables to solve word problems (student chooses the variable)
Sample Grade 3 Math Word Problems Worksheet
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## 2-step word problems
#### IMAGES
1. 3rd Grade Two Step Word Problems with Multiplication and Division Task Cards
2. 2-Step Word Problems
3. 2-step word problems--FREE--Christmas math for 2nd grade
4. Two Step Word Problems 3rd Grade Math Videos
5. 3rd Grade Math Two-Step Word Problem Test: 3.OA.8 and 3.MD.3 by Ms Samantha
6. Following Optimism in 2nd Grade: Two Step Word Problems
#### VIDEO
1. Math Grade 2 2.OA.1 Two-Step Word Problems
2. 2 step word problems with equations
3. Eureka math grade 5 module 4 lesson 19 homework
4. Day 2 Lesson Plan 2
5. Money 2 Step Word Problems
6. 2 step word problems
1. What Is the Average Words Per Minute One Can Read, by Grade Level?
According to Jan Hasbrouck and Gerald Tindal, an average student in the middle of the school year can read 23 words per minute at grade 1, which increases gradually up to 151 words per minute at grade 8.
2. Third Grade Christmas Word Problems
Word problems and problem-solving math questions help students to put the computations into authentic practice. Try these 3rd grade Christmas word problems. Word problems and problem-solving questions help students to put the computations i...
3. 3rd Grade Math Word Problems
Word problems are an essential part of grade 3 common core standards. They help students master basic math and problem-solving skills. Word problems allow students the opportunity to apply their math skills in authentic situations. All too ...
4. 3rd Grade 2-step Word Problems Teaching Resources
One of the most important skills for 3rd grade is the ability to solve 2-step word problems using all 4 operations.
5. Search 3rd Grade Two-Step Word Problem Educational Resources
Browse 3rd Grade Two-Step Word Problem Educational Resources. Award winning educational materials designed to help kids succeed. Start for free now!
6. Multiple-Step Word Problems
Students will need to add and subtract 2-digit and 3-digit numbers to find the answers to these multi-step word problems. 2nd and 3rd Grades.
7. Solving Two Step Word Problems 3rd Grade Worksheets
A two-step word problem requires applying two equations to resolve. Such problems challenge students to use their reasoning and critical thinking abilities to
8. Solving 2-step Word Problems
Word problems can make us nervous, but not if we break them down into smaller steps. This video shows how to use our knowledge of #addition
Visit http://www.educationgalaxy.com to learn more.Education Galaxy provides online assessment, instruction, and practice for elementary
10. Two-Step Problems Using the Four Operations
across grades as it builds on problem solving with.
11. Two-step mixed operation word problems
Learn with an example. review. To run a mile, Jamal must run 4 laps around the track. His goal is to run 3 miles. Jamal has run 9 laps so far. How many more
12. 3rd Grade Math Word Problems Worksheets
Printable third grade word problem worksheets including addition, subtraction, multiplication, division and fraction word problems.
13. 2-step word problems (practice) | 2,884 | 13,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2023-14 | latest | en | 0.971938 |
https://socratic.org/questions/how-do-you-determine-the-number-of-ways-a-computer-can-randomly-generate-an-prim | 1,582,941,880,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148163.71/warc/CC-MAIN-20200228231614-20200229021614-00489.warc.gz | 537,548,123 | 5,928 | # How do you determine the number of ways a computer can randomly generate an prime integer from 1 through 12?
No. of ways to arrange them randomly are = 5!
The prime numbers from $1$ to $12$ are $2 , 3 , 5 , 7 , 11.$
No. of ways to arrange them randomly are = n!
No. of ways to arrange them randomly are = 5! | 91 | 310 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2020-10 | longest | en | 0.92359 |
https://vustudents.ning.com/forum/topics/cs502-assignment-no-3-due-date-is-july-27-2018?groupUrl=cs502fundamentalsofalgorithms&commentId=3783342%3AComment%3A6103126&groupId=3783342%3AGroup%3A59371 | 1,620,580,568,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989006.71/warc/CC-MAIN-20210509153220-20210509183220-00361.warc.gz | 639,456,885 | 17,689 | www.vustudents.ning.com
# CS502 ASSIGNMENT NO. 3 - Due Date is July 27, 2018
Assignment # 3
Dated:Jul 20, 18
Dear Students, It is to notify that assignment no. 03 of CS502 has been uploaded. You are required to download assignment file from VULMS and upload your solutions within due date (July 27, 2018). Please read the instructions carefully before submitting assignment. Assignments submitted after due date will not be entertained. For any query regarding the assignment, contact at cs502@vu.edu.pk Please do not post queries related to assignment on MDB. Regards, CS502 Team
Views: 6214
Attachments:
### Replies to This Discussion
solution will also be uploaded soon
solution plz cs502
Q.2
Part A : Graph in the Adjacency List format
222.jpg
Part B is the opposite of A
Q.1: Greedy Algorithm for activity selection
111.jpg
Q.2
Part A : Graph in the Adjacency List format
222.jpg
Part B is the opposite of part A
Mery Solution ky bary main kia khial hy , tik hy ya nehi.
Part A is idea of solution.
Q2.
Part: B:- Solution
3333.jpg
CS502 assignment No 3 Solution Spring 2018 Last date 27 07 2018
yar app ny tu mery wala copy kia hy.
Q1 tu tik hy lakan ya jo Q2 Part A hy ya srif idea hy.
Q2 Part B bhi app ny sahi copy nehi kia hy,
ur is mian bhi two edges kam hy.
this solution confirm ok
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30 minutes ago | 590 | 1,913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-21 | latest | en | 0.777907 |
https://www.examrace.com/SPSC/Uttar-Pradesh-PSC/UPPSC-MCQs/Statistics-Questions/Continuous-Distributions-Part-12.html | 1,620,620,470,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989030.87/warc/CC-MAIN-20210510033850-20210510063850-00082.warc.gz | 786,260,842 | 5,187 | # Statistics MCQs – Continuous Distributions Part 12
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221. A statistical analysis of long-distance telephone calls indicates that the length of these calls is normally distributed with a mean of 240 seconds and a standard deviation of 40 seconds. What is the length of a particular call (in seconds) if only 5 % of calls are shorter?
a. 146.95
b. 157.85
c. 174.21
d. 333.05
e. 305.79
222. A statistical analysis of long-distance telephone calls indicates that the length of these calls is normally distributed with a mean of 240 seconds and a standard deviation of 40 seconds. What is the length of a particular call (in seconds) if only 1 % of calls are longer?
a. 146.95
b. 157.85
c. 174.21
d. 333.05
e. 305.79
223. A statistical analysis of long-distance telephone calls indicates that the length of these calls is normally distributed with a mean of 240 seconds and a standard deviation of 40 seconds. What is the length of a particular call (in seconds) if only 5 % of calls are longer?
a. 146.95
b. 157.85
c. 174.21
d. 333.05
e. 305.79
224. If X ~N (μ, 25) and p (X > 12) = 0.3446. What is the value of μ?
a. 10.00
b. 5.90
c. 1.80
d. 8.05
e. 4.65
225. If X ~N (μ, 25) and p (X > 12) = 0.1112. What is the value of μ?
a. 10.00
b. 5.90
c. 1.80
d. 8.05
e. 4.65
226. If X ~N (μ, 25) and p (X > 12) = 0.0207. What is the value of μ?
a. 10.00
b. 5.90
c. 1.80
d. 8.05
e. 4.65
227. If X ~N (μ, 25) and p (X > 12) = 0.2148. What is the value of μ?
a. 10.00
b. 5.90
c. 1.80
d. 8.05
e. 4.65
228. If X ~N (μ, 25) and p (X > 12) = 0.0708. What is the value of μ?
a. 10.00
b. 5.90
c. 1.80
d. 8.05
e. 4.65 | 664 | 1,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2021-21 | latest | en | 0.715665 |
https://community.intel.com:443/t5/Intel-Embree-Ray-Tracing-Kernels/Duplicated-leaf-node-in-custom-BVH8/m-p/1443375#M944 | 1,675,355,266,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500028.12/warc/CC-MAIN-20230202133541-20230202163541-00703.warc.gz | 204,213,005 | 37,160 | Intel® Embree Ray Tracing Kernels
Discussion forum on the open source ray tracing kernels for fast photo-realistic rendering on Intel® CPUs.
## Duplicated leaf node in custom BVH8
Beginner
222 Views
I'm following bvh_builder tutorial to use rtcNewBVH to build my own custom BVH8. However when I calculated leaf counts under each node I found that given the input of bounding box, the leaf nodes of BVH tree might get duplicated, for example if input bounding boxes are 13, final leaf counts is 25. Here is my BVH8 creation code. I can provide whole code if you want to reproduce it.
Normally input counts should be equal to leaf counts, so what could be wrong?
``````struct Node
{
public:
unsigned int id;
bool isLeaf;
virtual float SAH() = 0;
};
struct InnerNode : public Node
{
embree::BBox3fa bounds[8];
Node* children[8];
InnerNode()
{
isLeaf = false;
for (size_t i = 0; i < 8; ++i)
{
bounds[i] = embree::empty;
children[i] = nullptr;
}
}
float SAH()
{
float sum_area = 0.0f;
embree::BBox3fa merged_bounds(embree::empty);
for (size_t i = 0; i < 8; i++)
{
if (children == nullptr) break;
if (children[i] == nullptr) continue;
sum_area += area(bounds[i]) * children[i]->SAH();
merged_bounds.extend(bounds[i]);
}
return 1.0f + sum_area / area(merged_bounds);
//return 1.0f + (area(bounds[0]) * children[0]->sah() + area(bounds[1]) * children[1]->sah()) / area(merge(bounds[0], bounds[1]));
}
static void* Create(RTCThreadLocalAllocator alloc, unsigned int numChildren, void* userPtr)
{
//assert(numChildren == 8);
void* ptr = rtcThreadLocalAlloc(alloc, sizeof(InnerNode), 16);
return (void*) new (ptr) InnerNode;
}
static void SetChildren(void* nodePtr, void** childPtr, unsigned int numChildren, void* userPtr)
{
//assert(numChildren == 8);
for (size_t i = 0; i < numChildren; i++)
((InnerNode*)nodePtr)->children[i] = (Node*)childPtr[i];
}
static void SetBounds(void* nodePtr, const RTCBounds** bounds, unsigned int numChildren, void* userPtr)
{
//assert(numChildren == 8);
for (size_t i = 0; i < numChildren; i++)
((InnerNode*)nodePtr)->bounds[i] = *(const embree::BBox3fa*)bounds[i];
}
};
struct LeafNode : public Node
{
embree::BBox3fa bounds;
unsigned int indexToInstanceData;
LeafNode(const embree::BBox3fa& bounds, unsigned int indexToInstanceData)
:
bounds(bounds),
indexToInstanceData(indexToInstanceData)
{
isLeaf = true;
}
float SAH()
{
return 1.0f;
}
static void* Create(RTCThreadLocalAllocator alloc, const RTCBuildPrimitive* prims, size_t numPrims, void* userPtr)
{
assert(numPrims == 1);
void* ptr = rtcThreadLocalAlloc(alloc, sizeof(LeafNode), 16);
return (void*) new (ptr) LeafNode(*(embree::BBox3fa*)prims, prims->geomID);
}
};
bool memoryMonitor(void* userPtr, ssize_t bytes, bool post)
{
return true;
}
bool buildProgress(void* userPtr, double f)
{
return true;
}
void splitPrimitive(const RTCBuildPrimitive* prim, unsigned int dim, float pos, RTCBounds* lprim, RTCBounds* rprim, void* userPtr)
{
assert(dim < 3);
assert(prim->geomID == 0);
*(embree::BBox3fa*)lprim = *(embree::BBox3fa*)prim;
*(embree::BBox3fa*)rprim = *(embree::BBox3fa*)prim;
(&lprim->upper_x)[dim] = pos;
(&rprim->lower_x)[dim] = pos;
}
void BVH8::Build(embree::avector<RTCBuildPrimitive>& prims_i, const size_t& extraSpace)
{
rtcSetDeviceMemoryMonitorFunction(m_device, memoryMonitor, nullptr);
m_bvh = rtcNewBVH(m_device);
embree::avector<RTCBuildPrimitive> prims;
prims.reserve(prims_i.size() + extraSpace);
prims.resize(prims_i.size());
/* settings for BVH build */
RTCBuildArguments arguments = rtcDefaultBuildArguments();
arguments.byteSize = sizeof(arguments);
arguments.buildFlags = RTC_BUILD_FLAG_NONE;
arguments.buildQuality = RTC_BUILD_QUALITY_HIGH;
arguments.maxBranchingFactor = 8;
arguments.maxDepth = 1024;
arguments.sahBlockSize = 8;
arguments.minLeafSize = 1;
arguments.maxLeafSize = 1;
arguments.traversalCost = 1.0f;
arguments.intersectionCost = 1.0f;
arguments.bvh = m_bvh;
arguments.primitives = prims.data();
arguments.primitiveCount = prims.size();
arguments.primitiveArrayCapacity = prims.capacity();
arguments.createNode = InnerNode::Create;
arguments.setNodeChildren = InnerNode::SetChildren;
arguments.setNodeBounds = InnerNode::SetBounds;
arguments.createLeaf = LeafNode::Create;
arguments.splitPrimitive = splitPrimitive;
arguments.buildProgress = nullptr;
arguments.userPtr = nullptr;
/* we recreate the prims array here, as the builders modify this array */
for (size_t j = 0; j < prims.size(); j++)
prims[j] = prims_i[j];
std::cout << "Building BVH over " << prims.size() << " primitives, " << std::flush;
double t0 = embree::getSeconds();
m_root = (Node*)rtcBuildBVH(&arguments);
double t1 = embree::getSeconds();
const float sah = m_root ? m_root->SAH() : 0.0f;
std::cout << 1000.0f * (t1 - t0) << "ms, " << 1E-6 * double(prims.size()) / (t1 - t0) << " Mprims/s, sah = " << sah << " [DONE]" << std::endl;
}``````
1 Solution
Moderator
204 Views
This is expected behaviour in RTC_BUILD_QUALITY_HIGH mode, which will perform spatial splits. Use RTC_BUILD_QUALITY_MEDIUM if you do not want the BVH builder to perform spatial splits.
Moderator
205 Views
This is expected behaviour in RTC_BUILD_QUALITY_HIGH mode, which will perform spatial splits. Use RTC_BUILD_QUALITY_MEDIUM if you do not want the BVH builder to perform spatial splits. | 1,495 | 5,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-06 | latest | en | 0.547728 |
http://mathhelpforum.com/calculus/38031-complex-analysis-question.html | 1,524,182,432,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937074.8/warc/CC-MAIN-20180419223925-20180420003925-00403.warc.gz | 193,139,899 | 10,666 | 1. ## complex analysis question
p is the radius of convergence of Σa_n*z^n (n≥0) and let r be a real number less than p. prove that the series converges normally on the disc abs (z)<r and diverges for abs (z)>p. *converging normally means that the sup-norm converges.
2. What would this mean? $\displaystyle \sum_{n=0}^{\infty} || a_n z^n ||$ converges? If so then how do we interpert $\displaystyle ||a_n z^n||$?
I can show $\displaystyle \sum_{n=0}^{\infty} |a_nz^n|$ would converge, for $\displaystyle |z|<r$. But I am not sure if this is what you want.
3. i'm sorry for not being clear, but this would mean that llull=sup abs (u(z)) where z is in E(subset of complex C) converges. here llull would be called the sup-norm of u.
4. Originally Posted by squarerootof2
i'm sorry for not being clear, but this would mean that llull=sup abs (u(z)) where z is in E(subset of complex C) converges. here llull would be called the sup-norm of u.
The sup-norm, $\displaystyle ||a_nz^n||$, over $\displaystyle E = \{ z\in \mathbb{C}: |z| < r\}$ would be $\displaystyle \sup\{ |a_n z^n| : z\in E \} = \sup \{ |a_n||z^n| : z\in E\} = |a_n|r^n$.
Thus, we need to show $\displaystyle \sum_{n=0}^{\infty} |a_n| r^n$ converges.
It does converge by using the theorem that if $\displaystyle \sum_{n=0}^{\infty} a_n z^n$ is convergent on $\displaystyle |z|<\rho$ then it is absolutely convergent on $\displaystyle |z| \leq r < \rho$.
5. is this theorem we're supposed to use the abel's lemma?
6. Originally Posted by squarerootof2
is this theorem we're supposed to use the abel's lemma?
I am not sure what Abel's lemma you are using here.
If you never seen that before, how about you try proving this.
It happens to be not so difficult to show.
7. oh and how would we deal with divergence of the series? i was actually getting more stuck on that part... thanks.
8. Originally Posted by squarerootof2
oh and how would we deal with divergence of the series? i was actually getting more stuck on that part... thanks.
Same idea. If $\displaystyle E = \{ z\in \mathbb{C} : |z| < r\}$ then $\displaystyle ||a_n z^n||_E = \sup \{ |a_n z^n|: z\in E \} = |a_n|r^n$.
Suppose that $\displaystyle \sum_{n=0}^{\infty} |a_n|r^n$ would converge then by the comparision test it would mean $\displaystyle \sum_{n=0}^{\infty}a_n r^n$ would converge. But that is clearly impossible since $\displaystyle r>\rho$, so it diverges. | 750 | 2,402 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-17 | latest | en | 0.901422 |
https://clickanswer.us/question/a-mixture-of-helium-gas-and-argon-gas-occupies-80-0-l-at-398-k-and-3-50-atm-if-the-mass-of-helium-gas-is-equal-to-the-mass-of-argon-gas-in-the-mixture-how-many-moles-of-helium-does-the-mixture-conta/ | 1,685,384,219,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644907.31/warc/CC-MAIN-20230529173312-20230529203312-00509.warc.gz | 209,991,170 | 19,820 | # A mixture of helium gas and argon gas occupies 80.0 L at 398 K and 3.50 atm. If the mass of helium gas is equal to the mass of argon gas in the mixture, how many moles of helium does the mixture contain
The search for the correct answer in this moment is easy. clickanswer.us is a reliable source for questions and answers. we offer a concise answer key and complete with the discussion. We provide a range of answer keys that span from elementary, junior high and high school. We offer subjects like biology, math, physics, economics, history and more. below are the questions and answer keys that have been summarized from different sources found online.
## Question:
A mixture of helium gas and argon gas occupies 80.0 L at 398 K and 3.50 atm. If the mass of helium gas is equal to the mass of argon gas in the mixture, how many moles of helium does the mixture contain
7.79 moles
Explanation:
Let the mass of helium gas = Mass of argon gas = x g
Moles of helium = moles
Moles of argon = moles
Total moles =
Given that:
Temperature = 398 K
V = 80.0 L
Pressure = 3.50 atm
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L atm/ K mol
Applying the equation as:
x=31.16 g
Moles of helium = 31.16 / 4 = 7.79 moles
Use the answer key below as a reference when studying at home and at school. We appreciate your visit and I hope that it is useful for all of us. | 392 | 1,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-23 | latest | en | 0.930775 |
https://ourpastimes.com/how-to-read-munsell-color-chart-numbers-12404203.html | 1,653,250,107,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662546071.13/warc/CC-MAIN-20220522190453-20220522220453-00492.warc.gz | 498,011,086 | 26,614 | # How to Read Munsell Color Chart Numbers
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Munsell's color system is a model used for identifying every existing color in the spectrum. While various color models exist, Munsell's is the most recognized. The system was developed by Albert H. Munsell, who conducted numerous scientific experiments to measure the way in which humans perceive color. He concluded that color is comprised of three components: hue, value and chroma. Munsell created a chart comprised of numbers and letters to represent the levels of hue, value and chroma in a color. Before attempting to read Munsell's color chart numbers, it is important to understand the three dimensions of color.
Identify the color chart number that you would like to read. An example of a Munsell color chart number would be "5P 4/10." The color chart number refers to the three elements of color: hue, value and chroma.
Identify the hue. Hue is measured around a horizontal circle, comprised of five main hues: red (R), purple (P), blue (B), green (G) and yellow (Y). Additional hues are positioned halfway between the principle hues: RP, PB, BG, GY and YR. The hues are further divided by 10, resulting in a total of 100 hues around the hue circle.
In the example in Step 1, the "P" refers to the hue name, purple. The "5" directly in front of the "P" refers to the position of the hue around the circumference of the horizontal hue circle. Hues closer to PB would have a different number before the "P."
Identify the value. Value is the lightness and darkness of a color, which is measured in a vertical line through the center of the hue circle. At the bottom of the value measure is black, with a value of 0. At the top is white, with a value of 10.
In the sample color chart number, the number directly after the "P" indicates the value number, in this case "4." The value of a color is measured vertically from the center of the hue circle. A value of 4 would indicate that this color is darker than the median, since the value scale positions black at 0 and white at 10.
Identify the chroma. Chroma is measured through the radius of the of the hue circle. Chroma refers to the purity of color. A lower chroma number indicates that the color is less saturated or pure.
The last number in the color chart number refers to the chroma. The chroma measures the purity of a color. A low chroma number appears more washed out and pastel-like, and a high chroma number appears saturated. In the case of the sample color chart number, a chroma of "10" would be more pure and saturated than a chroma of "1." | 606 | 2,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-21 | latest | en | 0.925248 |
https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_II_Problems/Problem_5&diff=cur&oldid=153739 | 1,631,929,067,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056120.36/warc/CC-MAIN-20210918002951-20210918032951-00251.warc.gz | 171,971,856 | 18,647 | # Difference between revisions of "2021 AIME II Problems/Problem 5"
## Problem
For positive real numbers $s$, let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$. The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$. Find $a^2+b^2$.
## Solution 1
We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given $4$ and $10$ as the sides, so we know that the 3rd side is between $6$ and $14$, exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less than the 3rd side. So the triangles' sides are between $6$ and $\sqrt{84}$ exclusive, and the larger bound is between $\sqrt{116}$ and $14$, exclusive. The area of these triangles are from $0$ (straight line) to $2\sqrt{84}$ on the first "small bound" and the larger bound is between $0$ and $20$. $0 < s < 2\sqrt{84}$ is our first equation, and $0 < s < 20$ is our 2nd equation. Therefore, the area is between $\sqrt{336}$ and $\sqrt{400}$, so our final answer is $\boxed{736}$.
~ARCTICTURN
## Solution 2 (Inequalities and Casework)
If $a,b,$ and $c$ are the side-lengths of an obtuse triangle with $a\leq b\leq c,$ then both of the following must be satisfied:
• Triangle Inequality Theorem: $a+b>c$
• Pythagorean Inequality Theorem: $a^2+b^2
For one such obtuse triangle, let $4,10,$ and $x$ be its side-lengths and $K$ be its area. We apply casework to its longest side:
Case (1): The longest side has length $\boldsymbol{10,}$ so $\boldsymbol{0
By the Triangle Inequality Theorem, we have $4+x>10,$ from which $x>6.$
By the Pythagorean Inequality Theorem, we have $4^2+x^2<10^2,$ from which $x<\sqrt{84}.$
Taking the intersection produces $6 for this case.
At $x=6,$ the obtuse triangle degenerates into a straight line with area $K=0;$ at $x=\sqrt{84},$ the obtuse triangle degenerates into a right triangle with area $K=\frac12\cdot4\cdot\sqrt{84}=2\sqrt{84}.$ Together, we obtain $0 or $K\in\left(0,2\sqrt{84}\right).$
Case (2): The longest side has length $\boldsymbol{x,}$ so $\boldsymbol{x\geq10.}$
By the Triangle Inequality Theorem, we have $4+10>x,$ from which $x<14.$
By the Pythagorean Inequality Theorem, we have $4^2+10^2 from which $x>\sqrt{116}.$
Taking the intersection produces $\sqrt{116} for this case.
At $x=14,$ the obtuse triangle degenerates into a straight line with area $K=0;$ at $x=\sqrt{116},$ the obtuse triangle degenerates into a right triangle with area $K=\frac12\cdot4\cdot10=20.$ Together, we obtain $0 or $K\in\left(0,20\right).$
It is possible for noncongruent obtuse triangles to have the same area. Therefore, all such positive real numbers $s$ are in exactly one of $\left(0,2\sqrt{84}\right)$ or $\left(0,20\right).$ Taking the exclusive disjunction, the set of all such $s$ is $$[a,b)=\left(0,2\sqrt{84}\right)\oplus\left(0,20\right)=\left[2\sqrt{84},20\right),$$ from which $a^2+b^2=\boxed{736}.$
~MRENTHUSIASM
## Solution 3
We have the diagram below.
$[asy] draw((0,0)--(1,2*sqrt(3))); draw((1,2*sqrt(3))--(10,0)); draw((10,0)--(0,0)); label("A",(0,0),SW); label("B",(1,2*sqrt(3)),N); label("C",(10,0),SE); label("\theta",(0,0),NE); label("\alpha",(1,2*sqrt(3)),SSE); label("4",(0,0)--(1,2*sqrt(3)),WNW); label("10",(0,0)--(10,0),S); [/asy]$
We proceed by taking cases on the angles that can be obtuse, and finding the ranges for $s$ that they yield .
If angle $\theta$ is obtuse, then we have that $s \in (0,20)$. This is because $s=20$ is attained at $\theta = 90^{\circ}$, and the area of the triangle is strictly decreasing as $\theta$ increases beyond $90^{\circ}$. This can be observed from $$s=\frac{1}{2}(4)(10)\sin\theta$$by noting that $\sin\theta$ is decreasing in $\theta \in (90^{\circ},180^{\circ})$.
Then, we note that if $\alpha$ is obtuse, we have $s \in (0,4\sqrt{21})$. This is because we get $x=\sqrt{10^2-4^2}=\sqrt{84}=2\sqrt{21}$ when $\alpha=90^{\circ}$, yileding $s=4\sqrt{21}$. Then, $s$ is decreasing as $\alpha$ increases by the same argument as before.
$\angle{ACB}$ cannot be obtuse since $AC>AB$.
Now we have the intervals $s \in (0,20)$ and $s \in (0,4\sqrt{21})$ for the cases where $\theta$ and $\alpha$ are obtuse, respectively. We are looking for the $s$ that are in exactly one of these intervals, and because $4\sqrt{21}<20$, the desired range is $$s\in [4\sqrt{21},20)$$giving $$a^2+b^2=\boxed{736}\Box$$
## Solution 4
Note: Archimedes15 Solution which I added an answer here are two cases. Either the $4$ and $10$ are around an obtuse angle or the $4$ and $10$ are around an acute triangle. If they are around the obtuse angle, the area of that triangle is $<20$ as we have $\frac{1}{2} \cdot 40 \cdot \sin{\alpha}$ and $\sin$ is at most $1$. Note that for the other case, the side lengths around the obtuse angle must be $4$ and $x$ where we have $16+x^2 < 100 \rightarrow x < 2\sqrt{21}$. Using the same logic as the other case, the area is at most $4\sqrt{21}$. Square and add $4\sqrt{21}$ and $20$ to get the right answer $$a^2+b^2= \boxed{736}\Box$$
## Solution 5 (Diagrams)
For $\triangle ABC,$ we fix $AB=10$ and $BC=4.$ Without the loss of generality, we consider $C$ on only one side of $\overline{AB}.$
As shown below, all locations for $C$ at which $\triangle ABC$ is an obtuse triangle are indicated in red, excluding the endpoints.
• The region in which $\angle B$ is obtuse is determined by construction.
• The region in which $\angle C$ is obtuse is determined by the corollaries of the Inscribed Angle Theorem.
For any fixed value of $s,$ the height from $C$ is fixed. We need obtuse $\triangle ABC$ to be unique, so there can only be one possible location for $C.$ As shown below, all possible locations for $C$ are on minor arc $\widehat{C_1C_2},$ including $C_1$ but excluding $C_2.$
Let the brackets denote areas:
• If $C=C_1,$ then $[ABC]$ will be minimized (attainable). By the same base and height and the Inscribed Angle Theorem, we have \begin{align*} [ABC]&=[ABD] \\ &=\frac12\cdot BD\cdot DA \\ &=\frac12\cdot BD\cdot \sqrt{AB^2-BD^2} \\ &=\frac12\cdot 4\cdot \sqrt{10^2-4^2} \\ &=2\sqrt{84}. \end{align*}
• If $C=C_2,$ then $[ABC]$ will be maximized (unattainable). For this right triangle, we have \begin{align*} [ABC]&=\frac12\cdot AB\cdot BC \\ &=\frac12\cdot 10\cdot 4 \\ &=20. \end{align*}
Finally, we get $$s\in[a,b)=\left[2\sqrt{84},20\right),$$ from which $a^2+b^2=\boxed{736}.$
~MRENTHUSIASM (credit given to Snowfan) | 2,148 | 6,565 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 135, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2021-39 | latest | en | 0.882071 |
https://hirevideoeditor.net/water-flows-at-0-20-l-s-through-a-12-m-long-garden-hose-2-5-cmin-diameter-that-is/ | 1,660,380,917,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571911.5/warc/CC-MAIN-20220813081639-20220813111639-00470.warc.gz | 299,645,012 | 11,576 | # Water flows at 0.20 L/s through a 12-m-long garden hose 2.5 cm in diameter that is…
Water flows at 0.20 L/s through a 12-m-long garden hose 2.5 cm
in diameter that is lying flat on the ground. The temperature of
the water is 20 ∘C.
What is the gauge pressure of the water where it enters the
hose? | 91 | 300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-33 | latest | en | 0.942259 |
https://www.gradesaver.com/textbooks/math/calculus/thomas-calculus-13th-edition/chapter-16-integrals-and-vector-fields-section-16-5-surfaces-and-area-exercises-16-5-page-989/7 | 1,679,890,073,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946637.95/warc/CC-MAIN-20230327025922-20230327055922-00542.warc.gz | 882,614,539 | 13,301 | ## Thomas' Calculus 13th Edition
$r(u,v)=\lt \sqrt 3 \sin u \cos v,\sqrt 3 \sin u \sin v, \sqrt 3 \cos u \gt$ and $\dfrac{\pi}{3} \le u \le \dfrac{2\pi}{3};\\0 \le v \le 2 \pi$
Use spherical coordinates as: $x= l \sin \phi \cos \theta; y= l \sin \phi \sin \theta; z= l \cos \phi$ ; $0 \le \phi \le \pi; \\0 \le \theta \le 2 \pi$. We know that $r(r, \theta)=xi+yj+zk$ or, $r^2=x^2+y^2+z^2$ We have: $x^2+y^2+z^2 =3$ $x= \sqrt 3 \sin u \cos v, y= \sqrt 3 \sin u \sin v, z= \sqrt 3 \cos u$ So, $r(u,v)=\lt \sqrt 3 \sin u \cos v,\sqrt 3 \sin u \sin v, \sqrt 3 \cos u \gt$ and $\dfrac{\pi}{3} \le u \le \dfrac{2\pi}{3};\\0 \le v \le 2 \pi$ | 308 | 635 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2023-14 | latest | en | 0.216122 |
https://www.jiskha.com/questions/1866916/translate-this-phrase-into-an-algebraic-expression-one-more-than-the-product-of-8-and-a | 1,657,002,570,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104514861.81/warc/CC-MAIN-20220705053147-20220705083147-00553.warc.gz | 880,936,085 | 4,426 | # Expression
Translate this phrase into an algebraic expression. One more than the product of 8 and a number
1. 👍
2. 👎
3. 👁
4. ℹ️
5. 🚩
1. the product of 8 and a number ... 8x
now, what's one more than that?
1. 👍
2. 👎
3. ℹ️
4. 🚩
👤
oobleck
2. The product of a number is just multiplication.
So, we have 8 times an unknown value, which we can call x.
Now, just add 1 to that.
= 8x + 1
1. 👍
2. 👎
3. ℹ️
4. 🚩
3. Two more than the product of a number and 6 is 8
1. 👍
2. 👎
3. ℹ️
4. 🚩
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https://forallx.openlogicproject.org/html/Ch34.html | 1,718,907,527,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861989.79/warc/CC-MAIN-20240620172726-20240620202726-00465.warc.gz | 225,526,348 | 7,748 | # Chapter 34 Reasoning about interpretations
We can show that a sentence is not a validity just by providing one carefully specified interpretation: an interpretation in which the sentence is false. To show that something is a validity, on the other hand, it would not be enough to construct ten, one hundred, or even a thousand interpretations in which the sentence is true. A sentence is only a validity if it is true in every interpretation, and there are infinitely many interpretations. We need to reason about all of them, and we cannot do this by dealing with them one by one!
Sometimes, we can reason about all interpretations fairly easily. For example, we can offer a relatively simple argument that ‘$R(a,a)\vee\neg R(a,a)$’ is a validity:
Any relevant interpretation will give ‘$R(a,a)$’ a truth value. If ‘$R(a,a)$’ is true in an interpretation, then ‘$R(a,a)\vee\neg R(a,a)$’ is true in that interpretation. If ‘$R(a,a)$’ is false in an interpretation, then $\neg R(a,a)$ is true, and so ‘$R(a,a)\vee\neg R(a,a)$’ is true in that interpretation. These are the only alternatives. So ‘$R(a,a)\vee\neg R(a,a)$’ is true in every interpretation. Therefore, it is a validity.
This argument is valid, of course, and its conclusion is true. However, it is not an argument in FOL. Rather, it is an argument in English about FOL: it is an argument in the metalanguage.
Note another feature of the argument. Since the sentence in question contained no quantifiers, we did not need to think about how to interpret ‘$a$’ and ‘$R$’; the point was just that, however we interpreted them, ‘$R(a,a)$’ would have some truth value or other. (We could ultimately have given the same argument concerning TFL sentences.)
Let’s have another example. The sentence ‘$\forall x(R(x,x)\vee\neg R(x,x))$’ should obviously be a validity. However, saying precisely why is quite tricky. We cannot say that ‘$R(x,x)\vee\neg R(x,x)$’ is true in every interpretation, since ‘$R(x,x)\vee\neg R(x,x)$’ is not even a sentence of FOL (remember that ‘$x$’ is a variable, not a name). Instead, we should say something like this:
Consider some arbitrary interpretation. ‘$\forall x(R(x,x)\vee\neg R(x,x))$’ is true in our interpretation iff ‘$R(x,x)\vee\neg R(x,x)$’ is satisfied by every object of its domain. Consider some arbitrary member of the domain, which, for convenience, we will call Fred. Either Fred satisfies ‘$R(x,x)$’ or it does not. If Fred satisfies ‘$R(x,x)$’, then Fred also satisfies ‘$R(x,x)\vee\neg R(x,x)$’. If Fred does not satisfy ‘$R(x,x)$’, it does satisfy ‘$\neg R(x,x)$’ and so also ‘$R(x,x)\vee\neg R(x,x)$’.11 1 We use here the fact that the truth conditions for connectives also apply to satisfaction: $a$ satisfies $\mathscr{A}(\mathscr{x})\lor\mathscr{B}(\mathscr{x})$ iff $a$ satisfies $\mathscr{A}(\mathscr{x})$ or $\mathscr{B}(\mathscr{x})$, etc. So either way, Fred satisfies ‘$R(x,x)\vee\neg R(x,x)$’. Since there was nothing special about Fred—we might have chosen any object—we see that every object in the domain satisfies ‘$R(x,x)\vee\neg R(x,x)$’. So ‘$\forall x(R(x,x)\vee\neg R(x,x))$’ is true in our interpretation. But we chose our interpretation arbitrarily, so ‘$\forall x(R(x,x)\vee\neg R(x,x))$’ is true in every interpretation. It is therefore a validity.
This is quite long-winded, but, as things stand, there is no alternative. In order to show that a sentence is a validity, we must reason about all interpretations.
## 34.2 Other cases
Similar points hold of other cases too. Thus, we must reason about all interpretations if we want to show:
• that a sentence is a contradiction (this requires that it is false in every interpretation);
• that two sentences are logically equivalent (this requires that they have the same truth value in every interpretation);
• that some sentences are jointly unsatisfiable (this requires that there is no interpretation in which all of those sentences are true together, i.e., that, in every interpretation, at least one of those sentences is false);
• that an argument is valid (this requires that the conclusion is true in every interpretation where the premises are true);
• that some sentences entail another sentence.
The problem is that, with the tools available to you so far, reasoning about all interpretations is a serious challenge! For a final example, here is a perfectly obvious entailment:
$\forall x(H(x)\wedge J(x))\vDash\forall x\,H(x)$
After all, if everything is both $H$ and $J$, then everything is $H$. But we can only establish the entailment by considering what must be true in every interpretation in which the premise $\forall x(H(x)\wedge J(x))$ is true. To show this, we would have to reason as follows:
Consider an arbitrary interpretation in which ‘$\forall x(H(x)\wedge J(x))$’ is true. It follows that ‘$H(x)\wedge J(x)$’ is satisfied by every object in this interpretation. ‘$H(x)$’ will, then, also be satisfied by every object.22 2 Here again we make use of the fact that any object that satisfies $\mathscr{A}(\mathscr{x})\land\mathscr{B}(\mathscr{x})$ must satisfy both $\mathscr{A}(\mathscr{x})$ and $\mathscr{B}(\mathscr{x})$. So it must be that ‘$\forall x\,H(x)$’ is true in the interpretation. We’ve assumed nothing about the interpretation except that it was one in which ‘$\forall x(H(x)\wedge J(x))$’ is true. So any interpretation in which ‘$\forall x(H(x)\wedge J(x))$’ is true is one in which ‘$\forall x\,H(x)$’ is true.
Even for a simple entailment like this one, the reasoning is somewhat complicated. For more complicated entailments, the reasoning can be extremely torturous.
The following table summarizes whether a single interpretation or counter-interpretation suffices, or whether we must reason about all interpretations.
Yes No validity? all interpretations one counter-interpretation contradiction? all interpretations one counter-interpretation equivalent? all interpretations one counter-interpretation satisfiable? one interpretation all interpretations valid? all interpretations one counter-interpretation entailment? all interpretations one counter-interpretation
You might want to compare this table with the table at the end of chapter 15. The key difference resides in the fact that TFL concerns truth tables, whereas FOL concerns interpretations. This difference is deeply important, since each truth-table only ever has finitely many lines, so that a complete truth table is a relatively tractable object. By contrast, there are infinitely many interpretations for any given sentence(s), so that reasoning about all interpretations can be a deeply tricky business. | 1,635 | 6,628 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 47, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-26 | latest | en | 0.920723 |
https://nrich.maths.org/public/leg.php?code=130&cl=2&cldcmpid=12208 | 1,568,717,158,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573070.36/warc/CC-MAIN-20190917101137-20190917123137-00317.warc.gz | 580,700,279 | 8,086 | # Search by Topic
#### Resources tagged with Compound transformations similar to Regular Rings 2:
Filter by: Content type:
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### There are 36 results
Broad Topics > Transformations and constructions > Compound transformations
### Midpoint Triangle
##### Age 7 to 11 Challenge Level:
Can you cut up a square in the way shown and make the pieces into a triangle?
##### Age 7 to 11 Challenge Level:
How can the same pieces of the tangram make this bowl before and after it was chipped? Use the interactivity to try and work out what is going on!
### Twice as Big?
##### Age 7 to 11 Challenge Level:
Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too.
### Square to L
##### Age 7 to 11 Challenge Level:
Find a way to cut a 4 by 4 square into only two pieces, then rejoin the two pieces to make an L shape 6 units high.
### Square Tangram
##### Age 7 to 11 Challenge Level:
This was a problem for our birthday website. Can you use four of these pieces to form a square? How about making a square with all five pieces?
### 2001 Spatial Oddity
##### Age 11 to 14 Challenge Level:
With one cut a piece of card 16 cm by 9 cm can be made into two pieces which can be rearranged to form a square 12 cm by 12 cm. Explain how this can be done.
### Counting Triangles
##### Age 11 to 14 Challenge Level:
Triangles are formed by joining the vertices of a skeletal cube. How many different types of triangle are there? How many triangles altogether?
### Triangular Tantaliser
##### Age 11 to 14 Challenge Level:
Draw all the possible distinct triangles on a 4 x 4 dotty grid. Convince me that you have all possible triangles.
### Chess
##### Age 11 to 14 Challenge Level:
What would be the smallest number of moves needed to move a Knight from a chess set from one corner to the opposite corner of a 99 by 99 square board?
### Bow Tie
##### Age 11 to 14 Challenge Level:
Show how this pentagonal tile can be used to tile the plane and describe the transformations which map this pentagon to its images in the tiling.
### Decoding Transformations
##### Age 11 to 14 Challenge Level:
See the effects of some combined transformations on a shape. Can you describe what the individual transformations do?
### Screwed-up
##### Age 11 to 14 Challenge Level:
A cylindrical helix is just a spiral on a cylinder, like an ordinary spring or the thread on a bolt. If I turn a left-handed helix over (top to bottom) does it become a right handed helix?
### Cut and Make
##### Age 7 to 11 Challenge Level:
Cut a square of paper into three pieces as shown. Now,can you use the 3 pieces to make a large triangle, a parallelogram and the square again?
### Transforming the Letters
##### Age 7 to 11 Challenge Level:
What happens to these capital letters when they are rotated through one half turn, or flipped sideways and from top to bottom?
### Simplifying Transformations
##### Age 11 to 14 Challenge Level:
How many different transformations can you find made up from combinations of R, S and their inverses? Can you be sure that you have found them all?
### Flight of the Flibbins
##### Age 11 to 14 Challenge Level:
Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . .
### Squares, Squares and More Squares
##### Age 11 to 14 Challenge Level:
Can you dissect a square into: 4, 7, 10, 13... other squares? 6, 9, 12, 15... other squares? 8, 11, 14... other squares?
### Sorting Symmetries
##### Age 7 to 11 Challenge Level:
Find out how we can describe the "symmetries" of this triangle and investigate some combinations of rotating and flipping it.
### Matching Frieze Patterns
##### Age 11 to 14 Challenge Level:
Sort the frieze patterns into seven pairs according to the way in which the motif is repeated.
### Cutting Corners
##### Age 7 to 11 Challenge Level:
Can you make the most extraordinary, the most amazing, the most unusual patterns/designs from these triangles which are made in a special way?
### Combining Transformations
##### Age 11 to 14 Challenge Level:
Does changing the order of transformations always/sometimes/never produce the same transformation?
### Friezes Using Logo
##### Age 11 to 14 Challenge Level:
Experimenting with variables and friezes.
### Mirror, Mirror...
##### Age 11 to 14 Challenge Level:
Explore the effect of reflecting in two parallel mirror lines.
### ...on the Wall
##### Age 11 to 14 Challenge Level:
Explore the effect of reflecting in two intersecting mirror lines.
### Who Is the Fairest of Them All ?
##### Age 11 to 14 Challenge Level:
Explore the effect of combining enlargements.
### Trees and Friezes
##### Age 11 to 14 Challenge Level:
This problem is based on the idea of building patterns using transformations.
### Making Rectangles, Making Squares
##### Age 11 to 14 Challenge Level:
How many differently shaped rectangles can you build using these equilateral and isosceles triangles? Can you make a square?
### Transformation Game
##### Age 11 to 14 Challenge Level:
Why not challenge a friend to play this transformation game?
### The Frieze Tree
##### Age 11 to 16
Patterns that repeat in a line are strangely interesting. How many types are there and how do you tell one type from another?
### Mathematical Patchwork
##### Age 7 to 14
Jenny Murray describes the mathematical processes behind making patchwork in this article for students.
### Going Places with Mathematicians
##### Age 7 to 14
This article looks at the importance in mathematics of representing places and spaces mathematics. Many famous mathematicians have spent time working on problems that involve moving and mapping. . . .
### Maurits Cornelius Escher
##### Age 7 to 14
Have you ever noticed how mathematical ideas are often used in patterns that we see all around us? This article describes the life of Escher who was a passionate believer that maths and art can be. . . .
### Transformations Tables
##### Age 7 to 11 Challenge Level:
These grids are filled according to some rules - can you complete them?
### Frieze Patterns in Cast Iron
##### Age 11 to 16
A gallery of beautiful photos of cast ironwork friezes in Australia with a mathematical discussion of the classification of frieze patterns.
### Grouping Transformations
##### Age 11 to 18
An introduction to groups using transformations, following on from the October 2006 Stage 3 problems.
### Paint Rollers for Frieze Patterns.
##### Age 11 to 16
Proofs that there are only seven frieze patterns involve complicated group theory. The symmetries of a cylinder provide an easier approach. | 1,564 | 6,819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2019-39 | latest | en | 0.894782 |
http://openstudy.com/updates/50f9abc3e4b027eb5d9aa376 | 1,448,715,997,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398452560.13/warc/CC-MAIN-20151124205412-00317-ip-10-71-132-137.ec2.internal.warc.gz | 177,098,661 | 9,828 | ninovich 2 years ago The probability that Juan hits the target is 0.7. If Juan got three tries to hit the target, what is the probability that he will hit the target only in the first two tries? Find probability of hitting each time is same.
1. cwrw238
these probabilities (0.7) are independent so for a series of attempts they are multiplued P(hitting first and missing the third) = 0.7 * 0.7 * 0.3 note probability of missing = 1 - 0.7 = 0.3
2. ninovich
oh, thanks :')
3. cwrw238
yw | 143 | 490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2015-48 | longest | en | 0.900169 |
http://www.physics.umd.edu/courses/Phys131/Redish/schedule.htm | 1,586,024,541,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370524604.46/warc/CC-MAIN-20200404165658-20200404195658-00297.warc.gz | 285,319,108 | 4,705 | # Physics 131 Schedule Fall 2011
## (Note: This schedule is TENTATIVE and subject to frequent change)
Class Class number Reading Assignments (online with commentary in MP due 9 PM prev eve.) Assignments (Due at beginning of lecture) Content Lab Recitation Week 1 No lab-- Survey (FMCE) 8/31 1 R110831 Why is this class different from all other classes? 9/2 2 R110902 Mechanism, Measurement, and Math: Dimensions and units Week 2 No lab-- Scaling and estimation 9/7 3 R110907 HW 1 Coordinates, Graphs, Vectors 9/9 4 R110909 Rate of change and velocity -- instantaneous and average Week 3: Lab 1 -- Measurement Cheetah and antelope 9/12 5 R110912 Graphs: Velocity and consistency Quiz 1 9/14 6 R110914 HW 2 Acceleration 9/16 7 None Physical content of Newton's laws Week 4 Lab 2-- Grandfather clock:1 Woodpecker 9/19 8 R110919 What's a force? Quiz 2 9/21 9 R110921 HW 3 Newton 0, 1, and 2 9/23 10 R110923 The implications game; Newton's 3rd Law Week 5 Lab 3 -- Grandfather clock:2 Water coat forces 9/26 11 R110926 Forces: Springs Quiz 3 9/28 12 R110928 HW 4 Forces: Tension and normal forces 9/30 13 R110930 Forces: Friction, Viscosity, and Drag Week 6 Lab 4 -- Let it roll 10/3 14 R111003 Forces: Gravity Quiz 4 10/5 15 None HW 5 Review 10/7 16 MIDTERM EXAM 1 Week 7 Lab 4 -- Let it roll part 2 10/10 17 None Go over exam 1 10/12 18 R111012 HW 6 Electric charge 10/14 19 R111014 Forces: Electricity 10/14 Make up exam 1 Week 8 Lab 5 -- Endangered creatures 10/17 20 R111017 Electrical examples Quiz 5 10/19 21 R111019 HW 7 Momentum and momentum conservation 10/21 22 R111021 Properties of solids Week 9 Lab 6 -- No free launch:1 10/24 23 R111024 Basics of fluids: Pressure Quiz 6 10/26 24 R111026 HW 8 Buoyancy 10/27 25 R111028 Fluid flow Week 10 Lab 7 --No free launch:2 10/31 26 R111031 Work and Energy Quiz 7 11/2 27 R111102 HW 9 Potential energy 11/4 28 R111104 Various kinds of PE Week 11 Lab 8 -- Roller coaster:1 11/7 29 R111107 Mechanical energy conservation Quiz 8 11/9 30 HW 10 Review 11/11 31 MIDTERM EXAM 2 Week 12 Lab 9 -- Roller coaster:2 11/14 32 Go over exam 2 11/16 33 R111116 Molecular forces and energies 11/18 34 R111118 Chemical energy:bonds Week 13 No lab-- 11/21 35 R111121 Heat and temperature Quiz 9 11/21 Makeup exam 2 11/23 36 None HW 11 Guest lecture: RNA Polymerase, the Ribosome and the Work-Energy theorem (Prof. La Porta) Week 14 Lab 10 -- Gravity:1 11/28 37 R111128 Kinetic theory of gas: Molecular interpretation of pressure and temperature 11/30 38 R111130 HW 12 Energy -- 1st law 12/2 39 R111202 Random motion and diffusion Week 15 Lab 11 -- Gravity:2 12/5 40 R111205 Fick's law Quiz 10 12/7 41 R111207 HW 13 Motivating the 2nd law 12/9 42 R111209 Entropy 12/12 43 R111212 HW 14 Transforming organization: The second law 12/21 FINAL EXAM (8:00-12:00)
*
UP="Understanding Physics" by Cummings et al.
K = "College Physics" by Knight, Jones, & Field
T = "Introductory Physics: Building Understanding" by Touger
## RETURNS
University of Maryland Physics Department Physics 131 Home | 988 | 3,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-16 | longest | en | 0.73673 |
https://www.gradesaver.com/textbooks/math/algebra/linear-algebra-and-its-applications-5th-edition/chapter-1-linear-equations-in-linear-algebra-1-2-exercises-page-22/8 | 1,534,473,038,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221211403.34/warc/CC-MAIN-20180817010303-20180817030303-00131.warc.gz | 868,125,615 | 12,616 | ## Linear Algebra and Its Applications (5th Edition)
$\begin{cases}x_1=-9 \\ x_2=4 \\ x_3 \text{ is free} \end{cases}$
To find the general solution, we use row operations on the augmented matrix $\left[ \begin{array}{cccc} 1 & 4 & 0 & 7\\ 2 & 7 & 0 & 10 \end{array} \right]$ to obtain an equivalent matrix in reduced row echelon form. Then we solve the corresponding system of equations. First, we want to eliminate the 2 in the second row since it is below the leading term of the first row. To do this, we use row-replacement. So we replace row 2 with -2*(row 1)+(row 2), which gives $$\left[ \begin{array}{cccc} 1 & 4 & 0 & 7\\ 0 & -1 & 0 & -4 \end{array} \right].$$ Next, we need to eliminate the 4 above the leading one of the second row. We will again use row-replacement. We replace row 1 with 4*(row 2) + (row 1) to get $$\left[ \begin{array}{cccc} 1 & 0 & 0 & -9\\ 0 & -1 & 0 & -4 \end{array} \right].$$ Thirdly, we multiply row 2 by -1 to obtain the equivalent matrix $$\left[ \begin{array}{cccc} 1 & 0 & 0 & -9\\ 0 & 1 & 0 & 4 \end{array} \right].$$ Now we solve the corresponding system of equations, which is $$x_1=-9\\ x_2=4.$$ From the reduced row echelon form of the matrix, we see that $x_1$ and $x_2$ are the leading variables (since they correspond to the pivot positions of the matrix) and $x_3$ is free. Our system is already solved for the leading variables, and we know $x_3$ is free. So our general solution is $$\begin{cases}x_1=-9 \\ x_2=4 \\ x_3 \text{ is free} \end{cases}.$$ | 500 | 1,504 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2018-34 | longest | en | 0.802081 |
https://static.javadoc.io/org.specs2/specs2-common_sjs0.6_2.11/4.3.3/org/specs2/data/HopcroftKarp$.html | 1,571,136,830,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986658566.9/warc/CC-MAIN-20191015104838-20191015132338-00053.warc.gz | 731,422,581 | 3,818 | # HopcroftKarp
### Related Doc: package data
#### object HopcroftKarp
Hopcroft-Karp (https://en.wikipedia.org/wiki/Hopcroft%E2%80%93Karp_algorithm) algorithm for finding the maximum matching in a bipartite graph
This is used for "contain" matchers to find the best matches between actual and expected elements in 2 lists
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@throws( ... ) | 554 | 2,015 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-43 | latest | en | 0.396632 |
https://complementarytraining.net/analyzing-time-series-of-individual-data-2-using-harmfultrivialbeneficial-chances/ | 1,721,893,743,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518579.47/warc/CC-MAIN-20240725053529-20240725083529-00344.warc.gz | 152,681,008 | 22,992 | Analyzing Time-Series of Individual Data – Part2: Using Harmful/Trivial/Beneficial Chances - Complementary Training
Analyzing Time-Series of Individual Data – Part2: Using Harmful/Trivial/Beneficial Chances
# Analyzing Time-Series of Individual Data
## Part 2: Using Harmful/Trivial/Beneficial Chances
Just a quick update – after a comment by John Fitzpatrick (@JFitz138) on the use of Will Hopkins’s approach of calculating chances I decided to give it a shot.
For a Typical Error (used together with SWC to calculate chances) I used SD of the Rolling Average.
To calculate chances I have coded simple VBA function that use NORM.DIST function of Excel.
To calculate chances, one must assume that data points in Rolling average are normally distributed (and that might not be the case!) around their mean with SD. Here is my sketch:
My wonderful drawing skills
Even simpler approach than this might involve pure counting of days (data points) above, within and below baseline and SWC, especially for non-normally distributed data set (someone correct me if I said something stupid). Using Box and Whisker plot and interquartile ranges might also seem possible solution.
Anyway, here is the short video of the workbook and below you can find updated download link. | 279 | 1,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-30 | latest | en | 0.860127 |
https://www.ecologycenter.us/population-dynamics-2/flux-and-mixed-boundary-conditions.html | 1,563,347,902,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525094.53/warc/CC-MAIN-20190717061451-20190717083451-00061.warc.gz | 663,912,915 | 8,395 | ## Flux and Mixed Boundary Conditions
We now show a common trick that can be used to construct methods of the same accuracy as above when boundary conditions other than [3] are used; namely, while [3] specifies the value of the unknown at the boundaries (recall that it was set to 0 without loss of generality), in some problems it may be the flux of u, or a linear combination of « and its flux, that are to be specified. Since the flux of u is proportional to 8xu, it is the calculation of 8xu at the boundary with accuracy O(h2) that requires special treatment. Thus, consider the condition at, say, the left boundary, which replaces the first condition in [3]:
where q(t) and b(t) are known for any t. When q(t) = 0, this is called a Neumann boundary condition, and when q(t) = 0, a mixed, or Robin, boundary condition. One discretizes [24] by introducing a fictitious node at x =—h :
where U—1 — u(—h, tn); the accuracy of this approximation is O(h2) (cf. [14]). This one equation introduces two new unknowns, U0 and U— j. Therefore, one more equation involving these quantities must be supplied. Such an equation is given by the first equation in [5] with m — 0 for explicit methods and by [9] with m — 0 for implicit ones. For explicit methods, this concludes the treatment of [24]. For implicit methods, one more step is needed. Recall that to render the solution of the matrix equation [11] time-efficient, the matrix on its LHS must be tridiagonal. If [25] and [9] with m — 0 are both included separately into the counterpart of [11] arising for boundary conditions [24], the resulting matrix is not tridiagonal. To circumvent this problem, one should, instead of including the aforementioned two equations separately, solve [25] for U— 1 and substitute the result into [9] with m — 0. Then the matrix in question becomes tridiagonal and the analog of [11] can be solved time-efficiently.
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Get My Free Ebook | 562 | 2,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2019-30 | longest | en | 0.932956 |
https://www.physicsforums.com/threads/double-slit-interference-problem.592950/ | 1,534,547,441,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213158.51/warc/CC-MAIN-20180817221817-20180818001817-00012.warc.gz | 965,423,822 | 13,790 | # Homework Help: Double-Slit Interference Problem
1. Apr 2, 2012
### KendrickLamar
1. The problem statement, all variables and given/known data
A physics instructor wants to produce a double-slit interference pattern large enough for her class to see. For the size of the room, she decides that the distance between successive bright fringes on the screen should be at least 2.81 cm. If the slits have a separation d = 0.0410 mm, what is the minimum distance from the slits to the screen when 590 nm light is used?
2. Relevant equations
L = x .d / m* λ
3. The attempt at a solution
L = x .d / m* λ
[(2.81)(.0410x10^-1)] / [(1)(590x10^-7) = 195.27 cm
i dont understand why this is wrong, we have this online quiz (which accepts answers within 3% of the correct answer) and some of my friends did it the same way, used the same exact formula plugged my numbers in for me too and they got the same answer as me, but for their quiz got it correct so i feel they just coincidentally got it right, and that it was w/in 3% for them but theres something im doing wrong some slight thing thats turning out to make my answer incorrect.
Can anyone help me please ASAP? | 302 | 1,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-34 | latest | en | 0.935078 |
https://dsp.stackexchange.com/questions/17011/frequency-spectrum-analysis | 1,719,244,216,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865401.1/warc/CC-MAIN-20240624151022-20240624181022-00802.warc.gz | 188,357,936 | 39,566 | # Frequency Spectrum Analysis
How do you find the phase and inverse Fourier transform of this frequency spectrum?
• Is this a homework assignment? Commented Jun 23, 2014 at 19:15
• no it isn't.... Commented Jun 23, 2014 at 19:18
• The spectrum is usually complex. Sometimes the definition implies "magnitude", which is real, but in that case it would be positive. Perhaps you could clear this up a little from the context of the book? In this meant to be a Fourier representation of a signal which happens to be purely real? Commented Jun 23, 2014 at 22:45
1. The magnitude is given by ${\left| X \left( f \right) \right|}$ hence it is 6.
• The magnitude is $|X(f)|$, not $|X(f)|^2$ (which is the squared magnitude). So its value is 6, not 36. Commented Jun 24, 2014 at 7:43 | 219 | 777 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-26 | latest | en | 0.935723 |
https://www.divorce-attorney-fort-worth.net/3-5-as-a-percent/ | 1,591,437,756,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348513230.90/warc/CC-MAIN-20200606093706-20200606123706-00212.warc.gz | 683,668,793 | 9,904 | # 3.5 As A Percent
· In this lesson, students learn to write percents as fractions and fractions as percents. For example, to write 6% as a fraction, remember that percent means "out of 100", so 6% can be written as.
Primary care spending represents a small percentage — less than 3% — of total fee-for-service Medicare spending, though it varies substantially across populations and states.
“After dropping for three month's in a row, Arkansas' unemployment rate was unchanged at 3.5 percent in October. The stable jobless rate.
· What is 3/8 in percentage form? Guest Apr 28, 2014. 0 users composing answers.. Best Answer #3 +6 . Converting 3/8 (or any fraction) into a percentage is a two-step process. First, calculate the "decimal equivalent" of 3/8. Second, multiply the decimal equivalent by 100 to convert to the percentage value.
where annual sales dropped from \$5.3 million in 2016 to \$3.96 million last year and are projected to drop again to \$3.5.
Strengthening Knowledge and Understanding of Dietary Supplements. Health Information Health Information. Making decisions; supplement fact sheets; frequently asked Questions
· In this short tutorial, you will find many helpful details about Excel percent format and learn how to format existing values as per cents, how to show percentage in empty cell and change numbers to percentages as you type. In Microsoft Excel, displaying values as.
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Three-fifths, otherwise written as 3/5, can also be written in decimal form as 0.6. Decimal form can be determined by dividing the numerator of a fraction by the denominator using a calculator. While using a calculator is the easiest way to turn a fraction into a decimal number, certain numbers can be converted by manipulation and inspection.
Fha Loans Near Me Best Mortgage Quote How to Find the Best Mortgage Rates in 2019 – The Simple Dollar – A note about mortgage points: One way to get the best mortgage rates is to pay "points," or upfront interest paid to the bank that secures a lower long-term interest rate on your home loan. One point generally costs 1% of the total loan amount, so paying 1 point on a \$200,000 mortgage would add \$2,000 in upfront costs.Fha Credit Score Limit Mortgage Q&A: FHA sets new jumbo loan limits – The new limits. FHA lender. These apply to FHA jumbo loans in excess of \$417,000. m Only 3 percent down payment required for a purchase price up to \$752,000. m For refinances, 85 percent loans are.3 Months, 3 Housing trends: fast buyers, Higher Rates, Tapping Equity – On top of specialized state programs, the Federal Housing Administration and the Department of veterans affairs offer FHA loans and VA loans. expects rates to remain near the current level for the.. How Housing Discrimination Created the Idea of Whiteness – As Nightingale told me, Chicago was a "lab" for segregation.
· how do you convert 12.55 to percent? i get 1,255% is that correct? or should it be 125.5%? If the 12.55 is a decimal, then .1255 is the percent. Per Cent means "per 100". the math problem asked me to find the percent increase of . asked by Dena on April 25, 2007; math. 14/25 please convert to a decimal and percent
Equipment spending was revised up to a 3.5 percent rise from a 0.4 percent gain, while investment in structures showed a 1.7 percent drop compared with a previously reported decline of 7.9 percent. Net exports subtracted 1.91 percentage point from growth, while inventories added provided a 2.27 point.
This version: https://www.w3.org/TR/2019/CR-css-values-3-20190606/ Latest published version: https://www.w3.org/TR/css-values-3/ Editor’s Draft: | 1,057 | 4,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-24 | latest | en | 0.913298 |
https://journal.ugm.ac.id/juliet/article/view/81744/0 | 1,719,035,238,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862252.86/warc/CC-MAIN-20240622045932-20240622075932-00360.warc.gz | 287,551,551 | 8,367 | ### Linierisasi Model Sistem Ball and Beam dan Metode Kendali LQG Berbasis MATLAB-Simulink
https://doi.org/10.22146/juliet.v4i1.81744
Hanifah Suwardi(1), Anisa Septyaning Choir(2), Salima Nurrahma(3), Imroatul Hudati(4*)
(1) Departemen Teknik Elektro dan Informatika, Universitas Gadjah Mada
(2) Departemen Teknik Elektro dan Informatika, Universitas Gadjah Mada
(3) Departemen Teknik Elektro dan Informatika, Universitas Gadjah Mada
(4) Departemen Teknik Elektro dan Informatika, Universitas Gadjah Mada
(*) Corresponding Author
#### Abstract
Abstract – There are some unstable control systems, it can be an important problem in control, like systems in aerospace and flight. This can be studied simply in the ball and beam system. The ball and beam system uses an LQG (Linear Quadratic Gaussian) controller to determine the effect of stability and optimal system to reach the set point and the results of the slope angle on the beam at the set point. There is a combination of LQR with the element matrix parameters Q and R which can be searched using a trial and error technique with the condition that the Q matrix is a positive semidefinite while the R matrix is a positive definite matrix. As well as the use of gain Kalman filter as an estimator. Two experiments are given on the Q and R matrix elements which will be used in three distance/position displacement tests. The design of LQG on ball and beam systems is carried out in the MATLAB-Simulink software. Then the results of the angle of inclination of the beam (beam) obtained 14° to make the ball reach a set point of 10 cm, 17° for a set point of 20 cm, and 28° for a set point of 30 cm.
Keywords – ball and beam, angle, LQG, LQR, linearization
IntisariSistem kontrol yang tidak stabil menjadi sebuah masalah yang penting dalam pengontrolan, seperti sistem dalam aerospace dan penerbangan. Hal ini dapat dipelajari secara sederhana pada sistem ball and beam. Pada sistem ball and beam digunakan pengontrol LQG (linear quadratic gaussian) untuk mengetahui pengaruh stabilitas maupun optimalnya sistem mencapai set point serta hasil kemiringan sudut pada balok (beam) pada titik set point. Terdapat kombinasi LQR dengan parameter matriks elemen Q dan R yang dapat dicari menggunakan teknik trial and error dengan syarat matriks Q ialah semidefinite positive sedangkan matriks R ialah matrik definit positif Serta penggunaan gain filter Kalman sebagai estimator. Dilakukan dua eksperimen pada elemen matriks Q dan R yang akan digunakan pada tiga pengujian perpindahan jarak/posisi. Rancangan LQG pada sistem ball dan beam dilakukan di software MATLAB-Simulink. Diperoleh hasil sudut kemiringan beam sebesar 14° untuk membuat ball mencapai di posisi set point 10 cm, 17° untuk set point 20 cm, dan 28° untuk set point 30 cm.
Kata kunci ball and beam, sudut, LQG, LQR, linierisasi
PDF
#### References
[1] N. O. Dewi and P. W. Rusimamto, “Perancangan Sistem Pengendalian Ball and Beam Menggunakan Perangkat Lunak LabVIEW Berbasis Fuzzy Logic Controller,” Jurnal Teknik Elektro, vol. 8, no. 2, 2019.
[2] D. Martinez and F. Ruiz, “Nonlinear model predictive control for a Ball&Beam,” 2012 IEEE 4th Colombian Workshop on Circuits and Systems, CWCAS 2012 - Conference Proceedings, no. 2, 2012, doi: 10.1109/CWCAS.2012.6404073.
[3] T. Trapp and W. Selby, “Ball and Beam Balance Final Report,” 2009.
[4] R. Soni and Sathans, “Optimal control of a ball and beam system through LQR and LQG,” Proceedings of the 2nd International Conference on Inventive Systems and Control, ICISC 2018, no. Icisc, pp. 179–184, 2018, doi: 10.1109/ICISC.2018.8399060.
[5] J. Dofela, “DESAIN KENDALI OPTIMAL METODE LINEAR QUADRATIC GAUSSIAN (LQG) UNTUK PENGENDALIAN SISTEM SUSPENSI KENDARAAN SEPEREMPAT.” Universitas Islam Negri Sultan Syarif Kasim Riau, 2019.
[6] A. Srivastava and B. Pratap, “Nonlinear robust observers for ball and beam system: A comparative analysis,” 2015 2nd International Conference on Recent Advances in Engineering and Computational Sciences, RAECS 2015, no. December, pp. 22–27, 2016, doi: 10.1109/RAECS.2015.7453281.
[7] “Ball & Beam : System Modelling,” Control Tutorials for MATLAB & SIMULINK.
[8] T. TAUFIKURRAHMAN, “DESAIN KONTROLER LINEAR QUADRATIC GAUSSIAN (LQG) PADA SISTEM PENDULUM TERBALIK.” University of Muhammadiyah Malang, 2017.
[9] F. L. Q. Regulator, “Cerita singkat tentang Linear Quadratic Regulator”.
[10] F. L. Lewis, Applied Optimal Control & Estimation (Digital Design & Implementation). Prentice Hall, 1992.
[11] K. Ryu and Y. Oh, “Balance control of ball-beam system using redundant manipulator,” 2011 IEEE International Conference on Mechatronics, ICM 2011 - Proceedings, pp. 403–408, 2011, doi: 10.1109/ICMECH.2011.5971319.
[12] T. TAUFIKURRAHMAN, “DESAIN KONTROLER LINEAR QUADRATIC GAUSSIAN (LQG) PADA SISTEM PENDULUM TERBALIK.” University of Muhammadiyah Malang, 2017.
[13] J. Dofela, “DESAIN KENDALI OPTIMAL METODE LINEAR QUADRATIC GAUSSIAN (LQG) UNTUK PENGENDALIAN SISTEM SUSPENSI KENDARAAN SEPEREMPAT.” Universitas Islam Negri Sultan Syarif Kasim Riau, 2019.
[14] G. Welch and G. Bishop, “An introduction to the Kalman filter,” 1995.
[15] D. Kinoshita and K. Yoshida, “Stabilizing Control for a ball and beam system considering the restricted beam angle and ball speed,” 2019 58th Annual Conference of the Society of Instrument and Control Engineers of Japan, SICE 2019, pp. 170–176, 2019, doi: 10.23919/SICE.2019.8859847.
[16] K. T. Lee, S. Y. Oh, and H. L. Choi, “Bounded control of a ball and beam system in the absence of feedback,” International Conference on Control, Automation and Systems, vol. 2017-October, no. Iccas, pp. 1167–1169, 2017, doi: 10.23919/ICCAS.2017.8204389.
DOI: https://doi.org/10.22146/juliet.v4i1.81744
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SINTA 4 accredited based on Decree of the Minister of Research, Technology and Higher Education, Republic of Indonesia Number 225/E/KPT/2022, Vol. 2 No. 1 (2020) - Vol. 6 No. 1 (2025)
e-ISSN: 2746-2536 | 1,839 | 6,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.527797 |
https://www.moneylife.in/article/how-bid-and-offer-rates-work/63878.html | 1,718,582,435,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861674.39/warc/CC-MAIN-20240616233956-20240617023956-00427.warc.gz | 802,539,385 | 20,539 | How ‘Bid & Offer’ Rates Work
Amitabha Banerjee 17 May 2021
In everyday life, you face only one-way prices, either to buy or to sell. Be it buying groceries, or selling old newspapers, you get just one price.
You can haggle on the price, but you cannot sell groceries to the kirana shop or buy old newspapers. Yes, you might find both buy and sell rates in a jewellery shop, which sells gold at Rs4,900 and buys old gold at Rs4,300.
But, in the world of finance, you will get both buying and selling prices at the same time. These are called ‘bid-offer’ (B&O) prices.
Some examples:
ACC shares - Rs1,890 – Rs1,910
1 month interest rate – 2.28% – 2.34%
Option premium – Rs4.45 – Rs4.80
Which is which?
A simple rule: Whichever you want to do, buy, or sell, you will get the worse of the two rates.
If you are buying ACC shares, the better rate for you is Rs1,890. Sorry, you will have to pay Rs1,910.
If you are selling, you will get the worse (for you) rate – Rs1,890.
If you are borrowing, you will pay 2.34%, if lending you will earn 2.28%.
The difference between the two rates is called the spread.
The gold prices in first para have a spread of Rs600, a huge 13%. In the international market, spreads are tiny.
A pound/dollar rate of 1.3815-16 means that the bid-offer rates are Pound 1 = Dollar 1.3815 and 1.3816.
Here the spread is just 0.0001, commonly called one pip.
One more thing to remember.
The left-side price (never mind what it is called) is ALWAYS lower, and the right-side price is higher.
If the RHS bid seems lower than the LHS bid, that is not so. For example, the ACC quote of Rs1,895-05 does not means 1,895-1,805. It means 1,895-1,905. Similarly, 1,890-00 means 1,890-1,900.
Now you can face any bid-offer price confidently.
If you are the curious and questioning type, you are probably asking yourself “what is in it for the guy who quotes these bid-offer prices?”
The person who quotes makes money on the spread. Here is how.
Consider a guy who quotes ACC all day long. He is a market-maker (MM). Others come to him to buy or sell because he always has a firm price for them to transact. Every item has a standard ‘market lot’, say, 100 shares of ACC.
Take a day when the market opens. ACC has been trading at fairly steady prices and had closed at 1,850 the previous evening.
MM will start with an opening quote of 1,845-55, meaning he buys ACC at 1,845 and sells at 1,855, 100 shares per transaction.
The first customer comes along, asks the price, and sells 100 shares to MM at 1,845. Now MM is plus 100 ACC shares, or ‘long 100’, at 1,845.
MM changes his quote to 1,843-53. Please note that he shifts his quote a little bit, to the side where he has been ‘hit’.
If the next customer buys ACC at 1,853, MM squares off with Rs800 profit. But if this customer also sells ACC, this time at 1,843, MM becomes 200 ‘long’, and his next quote is 1,841-51.
The other MMs dealing in ACC may still be quoting 1,845-55, or something close to that.
Hence, compared to the market price for you to buy ACC shares, viz., 1,855, our MM friend is quoting 1,851, whereas the market will buy ACC at 1,845 while our MM will pay only 1,841.
Thus, our MM’s quote is the best for anyone buying ACC, and the worst for anyone selling.
If the next customer in the market is a seller of ACC, he will not come to MM, but if he is a buyer, he will. MM’s quote has become one-sided, intentionally, because MM wants someone to buy ACC from him.
Sure enough, the next customer for our MM is a buyer of ACC, who buys 100 shares at the best rate going – 1,851.
Now our MM is only 100 ‘long’.
Look at his ‘position’ after these three transactions.
Our MM:
- Bought 100 shares at 1,845
- Bought 100 shares at 1,843
- Sold 100 shares at 1,851.
Net position – he owns 100 shares at a price of 1,837 (1,845 + 1,843 – 1,851). Already he is in profit, because if other MMs are quoting 1,845-55, he can sell his 100 shares at 1,845 and earn Rs800.
Better still, he can continue to offer an attractive price to anyone buying ACC, say 1,842-52, which is still better for a buyer of ACC than the other market quotes (1,852 versus 1,855).
If a customer now buys 100 shares from our MM at 1,852, he ‘squares’ his position, i.e., brings his holding of shares to zero, and makes a profit of Rs1,500.
The MMs buy and sell all day long, making a small profit every time. Multiply this by hundreds of transactions, involving lakhs of shares, and yes, MMs do make money.
Today, a lot of this is done by machines through what are called algorithms.
In a steady market the spread is tighter, maybe Rs5 for ACC. If the market is volatile, the spread will widen, perhaps to Rs20.
Mind you, our MM does not have any particular ‘view’ about ACC shares.
In fact, he does not care whether ACC goes up or down, as long as people buy and sell through him.
If the price is volatile, so much the better, because more people will buy or sell, his spread is bigger, and he earns even more.
If our MM does have a ‘view’ or an overnight ‘position’, he will slant his starting quote accordingly.
If he is ‘long’, or expects ACC to fall, he will make his quote attractive to buyers, e.g., 1,840-50 when the other MMs are at 1,845-55.
Of course, MMs can be hit by a sudden catastrophe.
When Ronald Reagan was shot, the US dollar dropped sharply within a minute. But such calamities can happen in any business at any time.
Now that you know all about the B&O stuff, do not let the numbers faze you.
(Deserting engineering after a year in a factory, Amitabha Banerjee did an MBA in the US and returned to India. Choosing work-to-live over live-to-work, he joined banking and worked for various banks in India and the Middle East. Post retirement, he returned to his hometown Kolkata and is now spending his golden years travelling the world (until Covid, that is), playing bridge, befriending Netflix & Prime Video and writing in his wife’s travel blog.)
Amitabha Banerjee
3 years ago
Dear Mr Garg, you are right. However, the purpose of my article was to explain the basic concept of Bid-Offer rates to a person who is not familiar with this subject. If I tried to describe online trading systems, combined quotes etc the article would become too complicated and far exceed the limited number of words it is meant to have. Hence I had to limit myself to the basic concepts, not the sophisticated online application. I trust you understand. Regards
Kamal Garg
3 years ago
When you open your terminal, the exchanges give/show the 'best' prices in descending order for buy and reverse in case of sell order. And the moment you place the order, 'price' first and 'time of order' later, the orders are matched and the transaction is completed.
Not able to understand when you say, some one is buying from 'MM-1' because the buyers' and sellers' names are not visible on the computer screen/electronic trading platforms. Of course, in the back end, all these orders are having unique codes and strings attached so as to determine who sold how much, exchange liability, delivery liability and failed delivery liability.
Ramesh Popat
3 years ago
buyer beware always, and loser too.
in shares its both buyer and seller!
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1. Mar 29, 2005
### vitaly
I'm having difficulty with this question. All help is appreciated.
*The cross section of television antenna dish is a parabola and the receiver is located at the focus.
A. If the receiver is located 5 feet above the vertex, assume the vertex is the origin, find an equation for the cross section of the dish.
Okay, I know the vertex is 0,0. The focus is 0, 5. The equation is x^2=4ay.
I don't know where to go from there, or what equation is needed to find the cross section.
2. Mar 29, 2005
### vitaly
Actually, I figured it out. x^2 = 4ay, and a must equal 5 because the focus is (0,5).
That means teh equation is x^2 = 4(5)y or x^2 = 20y.
What I can't figure out is part B:
If the dish is 10 feet wide, how deep is it?
I have never had a question like this before. How do you know how "deep" a dish is?
3. Mar 29, 2005
### Kamataat
So the equation of the parabola is $y=x^2/20$. If it's 10 feet wide and centered at the origin, then it's cross section is between -5 and 5 on the x-axis. So, to find the depth, you need to calculate "y" for x=5... that is, if I understand the question correctly.
- Kamataat
4. Mar 29, 2005
### vitaly
Thank you for the help. I think that's right. Solving for y, it would be 1.25, which is the answer. I just didn't know how to come to it and show my work. Thanks again. | 412 | 1,373 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-09 | longest | en | 0.97706 |
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# For members of the seventeenth century Ashanti nation in Africa, anima
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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15 Jul 2008, 11:45
goalsnr wrote:
For members of the seventeenth-century Ashanti nation in Africa, animal-hide shields with wooden frames were essential items of military equipment, a method to protect warriors against enemy arrows and spears.
A. a method to protect
B. as a method protecting
C. protecting
D. as a protection of
E. to protect
IMO C is the correct answer here
the issue with usage of E emanates from the fact that it will commit a modifier error
here is how
"Military equipments" to protect
i think this is an OG problem
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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15 Jul 2008, 15:57
OA is C.
here is my explanation:
This SC is in the form of DC,IC,DC
"a method to protect warriors against enemy arrows and spears "- is a sentence that can standby itself. We convert the independent clause
to a subordinate clause by adding "participle" -"protecting"
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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02 Apr 2009, 07:47
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306. For members of the seventeenth-century Ashanti nation in Africa, animal-hide shields with wooden frames were essential items of military equipment, a method to protect warriors against enemy arrows and spears.
(A) a method to protect
(B) as a method protecting
(C) protecting
(D) as a protection of
(E) to protect
please explain .. how do you know/conclude the answer has to be a regular clause or modifier here? or explain a way how you come to the conclusion to get the answer?
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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02 Apr 2009, 09:42
ugimba wrote:
306. For members of the seventeenth-century Ashanti nation in Africa, animal-hide shields with wooden frames were essential items of military equipment, a method to protect warriors against enemy arrows and spears.
(A) a method to protect
(B) as a method protecting
(C) protecting
(D) as a protection of
(E) to protect
please explain .. how do you know/conclude the answer has to be a regular clause or modifier here? or explain a way how you come to the conclusion to get the answer?
C,imo. "protecting" refers to "animal hide-shields"
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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01 Mar 2011, 08:32
bmwhype2 wrote:
For members of the seventeenth century Ashanti nation in Africa, animal-hide shields with wooden frames were essential items of military equipment, protecting warriors against enemy arrows and spears.
Dependent clause, independent clause, participial phrase.
participial phrases are extremely important on the GMAT test. Once I knew what they were, I realized that many questions contain this structure.
Hi, could anyone please explain what participle phrases/clauses mean? I am not at all clear on that concept. It'll be great if anyone could help on this. Thanks.
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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24 May 2011, 08:31
prasforgmat wrote:
Is it universal rule that infinitives (to) cannot come just after comma?
No. There are certainly possibilities where this can be done.
"For lizards, organisms that inhabit the earth, to [X] is to [Y]."
"For lizards, organisms that inhabit the earth, to [die] is to [leave your remains for the rest of the ecosystem]."
In this example "to" is right after a comma.
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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04 May 2012, 02:50
Ron said,
abstract noun can refer/modify the preceding clause. So, in A, I can understand that "a method " modify preceding clause.
A is not incorrect.
pls, help explain.
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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19 Sep 2012, 14:09
In Original text "a method to protect warriors against enemy arrows and spears" is a noun modifier modifying military equipment, which is certainly not a method.The original intention of this modifier is to modify "animal-hide Shields"
(A) As explained above
(B)same problem
(C)correct.comma with -Ing form correctly modifying the subject of the previous sentence that is "animal-hide Shields" here.Please note that without the comma it would have been wrong.
(D)same problem
(E)with the comma before it does not convey the intended meaning.
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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27 Dec 2012, 08:24
well the bigger issue with E is that use of 'to+verb' ie to protect changes the meaning of the sentence altogether.
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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09 Jan 2013, 22:56
this problem is in og10 and so maybe old to study
I find a problem
C is correct, I agree.
but
in A, "a method..." can modify the whole preceding clause and is correct. A is also correct
a similar example is
I learn gmat, a good thing many persons know.
am I correct, pls help. i am confused.
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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10 Jan 2013, 00:43
thangvietnam wrote:
this problem is in og10 and so maybe old to study
I find a problem
C is correct, I agree.
but
in A, "a method..." can modify the whole preceding clause and is correct. A is also correct
a similar example is
I learn gmat, a good thing many persons know.
am I correct, pls help. i am confused.
Unless the phrase is an Absolute Phrase, it cannot modify the preceding clause.
The preceding clause can be entirely modified by:
1) Absolute Phrase
2) ", verb-ing"
Apart from this, I don't think that the preceding clause can me modified.
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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22 Aug 2013, 23:41
Sachin9 wrote:
Hi deepak,
So participial phrases are basically -ING verb phrases that are separated with a comma (,) from the rest of the sentence.
So in this case, "protecting warriors against enemy arrows and spears" is the participial phrase.
What is this phrase describing? Structurally, it is positioned to describe "animal-hide shields."
If you're not sure, you can always flip it and read it like this:
"Protecting warriors against enemy arrows, animal-hide shields with wooden frames were essential items of military equipment."
Ask yourself if this makes sense.
Do "shields" "protect warriors against enemy arrows"? Sure! So you know this makes sense.
You can read more of a detailed explanation of this question here: http://www.gmatpill.com/practice-questi ... orrection/
amazing.. Why is D wrong? [/quote][/quote]
Because comma is fixed before underlined portion, infinitive will not work here.
Pls note: if the clause before comma is non essential then it depends on the main clause.
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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26 Aug 2013, 02:43
For members of the seventeenth-century Ashanti nation in Africa, animal-hide shields with wooden frames were essential items of military equipment, a method to protect warriors against enemy arrows and spears.
A. a method to protect (military equipment cannot be a method)
B. as a method protecting ( Method doesn't do protecting instead the animal shields do the protecting)
C. protecting ( This one correctly modifiers the noun in the preceding clause "animal-shields")
D. as a protection of ( Although grammatically correct choice, wordiness is the issue here)
E. to protect ( "to protect " is the purpose and shouldn't follow a comma)
Although I cannot argue against the OG's explanation, I have a query related to answer choice (E):
-ING modifier has three forms:
(1). Modifies NOUN
(2). Modifies verb and verb's subject
(3). Result of the main clause
Here the -ING modifier functions clearly as (1). as it modifies "Shields"
Meaning: Shields protecting warriors against enemy arrows and spears with wooden frames were essential items of military equipment
But the option (E) explanation by OG is
Although the infinitive to protect would work if it were not preceded by a comma, it cannot act as a nonrestrictive adjectival
phrase modifying items
We aren't modifying ITEMS here , we are modifying "SHIELDS" here. So why this explanation?
Furthermore, the explanation for the correctness of option (C) is
Correct. In this sentence, protecting properly
introduces a modifying phrase revealing the
purpose of the items.
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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06 Apr 2014, 12:33
"Animal-hide shields were essential items of military equipment, protecting warriors against enemy arrows and spears"
Even above sentence is correct, this type of sentence (simplified) is rarely or never read by me before in English. I request you to provide me few similar examples of this type.
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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31 Jul 2016, 09:38
TGC wrote:
For members of the seventeenth-century Ashanti nation in Africa, animal-hide shields with wooden frames were essential items of military equipment, a method to protect warriors against enemy arrows and spears.
A. a method to protect
B. as a method protecting
C. protecting
D. as a protection of
E. to protect
Not convinced with the old discussions on this, plz throw some light and concepts
'a method to protect' must modify the previous noun- 'military equipment'. But, military equipment is not a 'method'. A and B are out.
'to protect' must not be followed by ','. E is out.
'as a protection of' what?? Not right.
C is the right option.
Infact if we flip the sentence, we see that 'protecting' correctly modifies following sentence.
Protecting warriors against enemy arrows and spears, animal-hide shields with wooden frames were essential items of military equipment.
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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31 Jul 2016, 23:58
A - Wrong --> "a method" is wrong here.
animal-hide shields with wooden
frames were essential items of military equipment --> it's not a method.
E is wrong because comma before "to" is wrong.
I believe if sentence is written like this:
animal-hide shields with wooden
frames were essential items of military equipment to protect warriors against enemy arrows
and spears. --> then it's right.
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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12 Mar 2017, 02:07
As per the meaning of the sentence, shields were important for them as these shields protect then from something.
Clearly, C maintains the meaning and this the correct answer.
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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12 Mar 2017, 04:38
abhimahna wrote:
As per the meaning of the sentence, shields were important for them as these shields protect then from something.
Clearly, C maintains the meaning and this the correct answer.
I'm still not getting the answer. I studied "to verb vs for verb-ing notes" also but this question is still confusing
Here if you want to go with "to verb" then you should have a purpose in the passage. Here in the main question the soldiers hide their weapons to protect from enemies.
They hide weapons is the context. Purpose is to protect them from enemies.
But the answer is "for protecting". According to you if i replace "to protect" in the actual sentence the meaning will change. Can you help me in this?
abhimahna egmat
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink]
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12 Mar 2017, 05:33
sai897 wrote:
I'm still not getting the answer. I studied "to verb vs for verb-ing notes" also but this question is still confusing
Here if you want to go with "to verb" then you should have a purpose in the passage. Here in the main question the soldiers hide their weapons to protect from enemies.
They hide weapons is the context. Purpose is to protect them from enemies.
But the answer is "for protecting". According to you if i replace "to protect" in the actual sentence the meaning will change. Can you help me in this?
abhimahna egmat
I understand your point but here as per the question the meaning is certain tools are important items and then we are told why are they important/how are they useful. They are important because they protect warriors against enemies.
Also, answer is not 'for protecting' but 'protecting'. Meaning tools are essential as they have a consequence of protecting.
Take these examples:
I want to give GMAT to get an admit in Harvard. Here the purpose/intend is given.
I woke up late in the morning, missing my GMAT exam. Here the consequence is given.
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Re: For members of the seventeenth century Ashanti nation in Africa, anima [#permalink] 12 Mar 2017, 05:33
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From Gallons to Liters: Unveiling the Mystery of Fluid Measurements
Fluid measurements have always been an integral part of our daily lives. Whether it’s cooking, filling up our vehicles, or simply quenching our thirst, understanding fluid measurements is crucial. One of the most common conversions that can sometimes perplex individuals is the conversion between gallons and liters. In this article, we will unravel the mystery behind these fluid measurements and provide a clear understanding of how to convert between the two.
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On the other hand, liters are the standard unit of measurement for volume in most countries worldwide, including the majority of Europe. One liter is equivalent to 0.264 US gallons or 0.220 imperial gallons. This means that if you have a quantity in liters and want to convert it to gallons, you would divide the number of liters by the appropriate conversion factor.
To convert gallons to liters, you would multiply the number of gallons by the corresponding conversion factor. For example, if you have 2.5 US gallons and want to find out how many liters it is, you would multiply 2.5 by 3.785, resulting in 9.463 liters. Similarly, if you have 4 imperial gallons, you would multiply 4 by 4.546, giving you 18.184 liters.
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Contents
### Today’s Puzzle:
I debated with myself about whether or not I should mention the famous poem about 1492 in this post. Then I looked at the puzzle I had already created for today and noticed it would not be too much of a stretch to say the twelve clues look a little bit like three ships. So whether or not I should, I decided to go ahead and mention the Niña, the Pinta, and the larger Santa Maria in the title. The ships themselves are innocent of any barbaric acts and might just be the three most famous ships in world history.
Start at the top of this level 3 puzzle and work your way down cell by cell using logic until you have written all the factors from 1 to 12 in both the first column and the top row.
Here’s the same puzzle without color:
### Factors of 1492:
• 1492 is a composite number.
• Prime factorization: 1492 = 2 × 2 × 373, which can be written 1492 = 2² × 373
• 1492 has at least one exponent greater than 1 in its prime factorization so √1492 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1492 = (√4)(√373) = 2√373
• The exponents in the prime factorization are 2 and 1. Adding one to each exponent and multiplying we get (2 + 1)(1 + 1) = 3 × 2 = 6. Therefore 1492 has exactly 6 factors.
• The factors of 1492 are outlined with their factor pair partners in the graphic below.
### Other Facts About the Number 1492:
1492 is the sum of two squares:
36² + 14² = 1492
1492 is the hypotenuse of a Pythagorean triple:
1008-1100-1492, which is 4 times (252-275-373)
and can also be calculated from 2(36)(14), 36² – 14², 36² + 14².
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 487 | 1,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2022-33 | latest | en | 0.908977 |
https://www.jiskha.com/questions/574694/variable-to-represent-one-quantity-and-second-a-theater-sold-220-more-adult-tickets-than | 1,582,308,894,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145534.11/warc/CC-MAIN-20200221172509-20200221202509-00107.warc.gz | 772,788,111 | 5,254 | # algebra
variable to represent one quantity and second a theater sold 220 more adult tickets than childrens tickets
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3. 👁 78
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2. 12d^2+34d+20=0 solve equation
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posted by ange
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More Similar Questions | 820 | 3,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2020-10 | latest | en | 0.955653 |
http://blog.podsnap.com/hollywood-typecasting.html | 1,542,363,043,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743007.0/warc/CC-MAIN-20181116091028-20181116113028-00310.warc.gz | 44,705,314 | 13,275 | I am broadly sympathetic to view that scalable systems must be built with statically typed langages, for reasons outlined in this wonderful screed, and, until recently, that has made it difficult for me to recommend clojure for institutional use.
With the introduction of core.typed, that has changed. The author has says that `core.typed` is now production-ready, and I agree. It's not perfect, but it will find bugs in your code without breaking it or causing performance problems. It's also pretty cool, and in many ways more expressive than type declarations in "normal" statically typed languages.
That said, the documentation is not, uhm, exhaustive. I decided to teach myself what I could using this toy project, which started life as a coding challenge for a job interview. (It didn't work out, though probably not for reasons connected with the challenge. Thanks for asking.) The task is to compute movie degrees of separation using data scraped from IMDB: two actors are considered to be directly connected if they appear in the same film together; if they're connected via a third actor with whom each of the first two appeared in a film, then it's a 2nd degree connection, etc. (One of the reasons this is not a particularly good challenge question is that IMDB limits your scraping rate so much that it is literally impossible to complete the calculation in the allotted time for non-trivial examples.)
Many people would say that clojure is not an obvious choice for this task, since it would seem at first that Dijkstra is a naturally stateful algorithm, but persistent data structures are wonderful things. We can store the graph as a nested map of links, every "update" to which is effectively a brand new graph.
As you've gathered, this post is about three totally unrelated things, in descending order of emphasis:
• Typed clojure
• Graph algorithms with functional data structures
• Scraping IMDB
So hang on tight.
Combining the first two topics, I now have the luxury of explaining the graph structure by showing you type signatures:
```(t/def-alias Node "Node representing an actor or film"
(HMap :mandatory {:id String
:title String}
:optional {:distance t/AnyInteger
:path (t/Seq String)}
:complete? true))
(t/def-alias Graph "Graph of tokens to nodes"
(Map String Node))
```
This is almost self-explanatory. Every node has a string id, along with a title and a set of ids of nodes to which it is linked. As the Dijkstra algorithm proceeds, we add fields containing the updated estimtated distance from the starting node and the path by which we got to it. So far, so obvious.
What's not immediately obvious is that this declaration contains more type information than we're used to seeing in standard OO languages. Remember that this isn't a structure or class - just an ordinary hash map. The field names are being represented here as types; that is, a type can specify not just a particular sort of entity but the values that the entity is allowed to take! Outide of map structures, you may not find yourself taking advantage of this ability at first, but when `core.typed` infers type information within your code, it will be as conservative as possible, down to the value level if it has that information. For example, if I had `(if (> x 0.0) (Math/log x) "Bleh")`, the type might be inferred as `(U double "Bleh")`. A more general implication, about which I'll say more a bit later, is that there are often multiple approaches to typing within a given piece of code.
Note that the enforcement of this typing is strictly at our discretion. To typecheck a namespace, run `(check-ns)`; for a more automated experience, `lein typed check` (using a plugin you'll find in my `project.clj`); or examine individual expressions in the REPL using `cf`. Actual compilation can proceed irrespective of whether we did any of these things, or whether they succeeded, and the compiled code won't differ either.
Let's take a look at some more type declarations in the project. One possibly unnecessary complication I introduced was a Mongo caching layer over IMDB, so it wouldn't be necessary to scrape the same thing more than once. There's a very nice Mongo interface available, but it doesn't (yet) come typed out of the box. One of the great things about the separation of definition and type declaration is that I can provide remedial typing of external libraries:
```(t/ann ^:no-check monger.core/connect!
(Fn [(HMap :optional {:port t/AnyInteger :host String} :complete? true) -> (t/Option com.mongodb.MongoClient)]
[-> (t/Option com.mongodb.MongoClient)]))
(t/ann ^:no-check monger.core/get-db [String -> com.mongodb.DBApiLayer])
(t/ann ^:no-check monger.core/set-db! [com.mongodb.DBApiLayer -> com.mongodb.gridfs.GridFS])
```
As you can see, this adds type information to functions in another namespace.
• The `:no-check` metadata tells `core.typed` not to bother checking within the definitions of these functions (which would certainly fail).
• `Option` is shorthand for `(U Something nil)` The monadic allusion may be overplayed here, but explictly noting that `nil` is a possible outcome allows a type error if my code doesn't handle it.
• As you can see, qualified Java class names can be types.
I also needed external libraries for web queries and parsing. My `httpkit` annotation
```(t/ann ^:no-check org.httpkit.client/request
['{:url String :method (Value :get)}, [Any -> Any] ->
(t/Atom1 '{:status Number :body String})])
```
does not even attempt completeness. As long as I err on the overly restrictive side, this does nobody any harm. Worst case, I'll realize I need to use the function more generally and have to expand my annotation. (In truth, `[Any->Any]` is too broad for the callback function, so I've sort of broken my own rule. I think it should be `(Value identity)`, but `core.typed` doesn't permit function constants at this point.)
I used `enlive` to convert the raw content stream from the imdb.com into a nested map. The declaration language is rich enough to describe this object recursively
```(t/def-alias Resource "Parsed DOM resource"
(Rec [x] (t/Map Keyword (U String x (t/Seq x))))
```
where `x` stands for the full declaration of `Resource`, and it's easy enough to declare the functions I use:
```(t/ann ^:no-check net.cgrand.enlive-html/html-resource [java.io.Reader -> Resource])
(t/ann ^:no-check net.cgrand.enlive-html/select [Resource (t/Vec Keyword) -> (t/Seq Resource)])
```
The actual parsing, however, is a bit ugly. Either we rigorously validate all the assumptions we make about the layout of the document, or we punt on the type checking. Extraction of links to actors or films starts a bit like this:
``` (-> resource
(html/select [:div.title])
(as-> doc (map #(-> % :content second :attrs :href) doc)))
```
That is, we expect to find a list of `div`s of class `title`, the second item in the content of each of which is supposed to be the link we're interested in. Expressing this with higher order functions and threading is relatively compact and readable, but the type inference turns into a complete mess. If we truly cared about Blythe Danner's relation to Meatloaf, it would be worth doing a host of extra validations, including but hardly limited to type checking, but there's no guarantee that IMDB would be using the same format by the time we were done. (We might hope that they don't: IMDB pages are a total rat's nest. I make life a little easier by scraping the mobile pages, but there are still nuances such as different URLs for actors and actresses, despite no explicit data for gender.) In any case, I wrap my link extraction routine in a big fat `no-check`.
```(t/ann ^:no-check resource->links [String Resource -> (t/Set String)])
```
which nicely isolates the most typologically problematic portion of the code while allowing me to make progress on the rest.
The final external library I'll use is `data.priority-map`. At each iteration of the Dijkstra algorithm, we visit the node that currently has the smallest estimated distance, and it's nice to be able to find that node in something less than O(n). I could use `java.util.PriorityQueue`, but This lovely package gives me a more idiomatic clojure solution. The object created by `(priority-map)` behaves like a map, but you can use `peek` and `pop` to retrieve and discard the entry with the smallest value.
Once again, of course, the package doesn't come pre-typed, but `core.typed` lets me correct for that:
```(t/def-alias Queue (t/Map String t/AnyInteger))
(t/ann ^:no-check clojure.data.priority-map/priority-map [ -> Queue])
(t/ann ^:no-check pmap-assoc [Queue String (U nil t/AnyInteger) -> Queue])
(def pmap-assoc assoc)
(t/ann ^:no-check pmap-peek [Queue -> (Vector* String t/AnyInteger)])
(def pmap-peek peek)
(t/ann ^:no-check pmap-pop [Queue -> Queue])
(def pmap-pop pop)
```
The full cache layering can be summarized with these declarations:
```(t/def-alias Graph "Graph of tokens to nodes" (t/Map String Node))
(t/ann fetch-node-from-imdb [String -> Node])
(t/ann fetch-node-from-mongo-fallback-to-imdb [String -> Node])
(t/ann fetch-node-from-graph-fallback-to-mongo [Graph String -> (Vector* Graph Node)])
```
And now for the core of the algorithm. Since I have complete control over my own code, there will be none of this `^:no-check` nonsense; every single line is type-checked. As the following function illustrates, `core.typed` is doing a lot of work on my behalf:
```(t/ann update-queue [Queue Graph String (t/Seqable String) -> Queue])
(defn update-queue [queue graph id links]
(reduce (t/fn> :- Queue [q :- Queue id :- String]
(pmap-assoc q id (:distance (get graph id)))) queue links))
```
To verify that this code does indeed produce a `Queue`, type must be properly inferred through 4 layers of function calls, one of them higher-order. I find that impressive.
Had I fat-fingered the last line as
``` ... (pmap-assoc id q (:distance (get graph id)))) queue links)
```
I would have learned that immediately
``` Type Error (imdb.core:219:18) Type mismatch:
Expected: Queue
Actual: String
in: (imdb.core/pmap-assoc id412107 q (:distance (clojure.lang.RT/get graph id412107)))
Type Error (imdb.core:219:18) Type mismatch:
Expected: String
Actual: Queue
in: (imdb.core/pmap-assoc id412107 q (:distance (clojure.lang.RT/get graph id412107)))
```
rather than having to deduce it from some downstream catastrophe.
The rest of the algorithm contains few surprises. As promised, it's purely functional, with multiple iterations of the graph emerging from an recursion and reduction.
```(defn update-distance
"Update :distance field in graph node id to be one greater than di if appropriate
and return the updated graph."
[entry graph id]
(let [di (distance entry)
[graph node] (fetch-node-from-graph graph id)
path (or (:path entry) '())]
(if (<= (distance node) di) graph
(-> graph
(set-node id :distance (inc di))
(set-node id :path (conj path id))))))
(t/ann visit-node [Graph Queue String -> (Vector* Graph Queue)])
(defn visit-node
"Visits node id, and update the :distance fields of its neighbors.
queue is a priority queue of id=>distance, containing entries only for unvisited
nodes. We will use this to determine what to visit next, Dijkstra style."
[graph queue id]
(let [[graph entry] (fetch-node-from-graph graph id)
graph (reduce (partial update-distance entry) graph links)
queue (update-queue queue graph id links)
graph (set-node graph id :visited true)]
[graph queue]))
(t/ann find-distance [String String -> (t/Option Any)])
(defn find-distance
"E.g. (find-distance \"/name/nm0000257/\" \"/name/nm0000295/\")
Use Dijkstra algorithm to find shortest path from id to target."
[id target]
(let [ [graph node] (fetch-node-from-graph {} id)
graph (set-node graph id :distance 0)
queue (priority-map)]
(t/loop> [graph :- Graph graph
queue :- Queue queue
id :- String id]
(if (= id target) (get graph id)
(let [[graph queue] (visit-node graph queue id)
closest (first (pmap-peek queue))]
(if (empty? queue)
"Couldn't find a path!"
(recur graph (pmap-pop queue) closest)))))))
```
I mentioned above that the `core.typed`'s flexibility gives us some unexpected choices. To investigate this in the REPL, we'll use a couple of extra tools. First, `(t/ann-form #{"a"} (t/Set String))` specifies inline that `#{a}` is to be treated as a set of String, rather than the more conservatively inferred `(t/Set (Value "a"))`, and `(t/cf someform)` checks a form and displays its inferred type.
Here's how `union` might be annotated
```(t/ann ^:no-check clojure.set/union (All [a] (Fn [(t/Set a) * -> (t/Set a)])))
```
and here's how that would be interpreted:
```imdb.core> (t/cf (clojure.set/union (t/ann-form #{1} (t/Set Integer)) (t/ann-form #{"one"} (t/Set String))))
(t/Set (U Integer String))
```
That certainly isn't what I expected, given apparently analogous declarations in other languages. For example, scala's sets have this `union` method
```def union(that: GenSet[A]): Set[A]
```
which of course bombs on heterogenous sets:
```scala> HashSet(1).union(HashSet(2))
res0: scala.collection.immutable.HashSet[Int] = Set(1, 2)
scala> HashSet(1).union(HashSet("one"))
<console>:9: error: type mismatch;
found : String("one")
required: Int
```
To get the familiar behavior from clojure, we'd have to specify an equality bound on the type parameter:
```(t/ann clojure.set/union (All [x [x1 :< x :> x]]
(Fn [(t/Set x) (t/Set x1) * -> (t/Set x1)])))
```
Now:
```imdb.core> (t/cf (clojure.set/union (t/ann-form #{1} (t/Set Integer))
(t/ann-form #{"one"} (t/Set String))))
AssertionError Assert failed: 1: Inferred type String is not between bounds Integer and Integer
(and (subtype? inferred upper-bound) (subtype? lower-bound inferred)) clojure.core.typed.cs-gen/subst-gen/fn--10414 (cs_gen.clj:1333)
```
Neither of these alternate declarations is obviously more correct, and both will catch our coding errors, though in slightly different ways. Suppose we had the following solecism:
```(t/cf (let [s (clojure.set/union #{1} #{"a"})]
(map inc s)))
```
With the bounded annotation, the call to `union` is simply illegal:
```AssertionError Assert failed: 1: Inferred type (Value "a") is not between bounds (Value 1) and (Value 1)
(and (subtype? inferred upper-bound) (subtype? lower-bound inferred)) clojure.core.typed.cs-gen/subst-gen/fn--10414 (cs_gen.clj:1333)
```
With the first, more general declaration, the call to `union` is fine, but `inc` will have problems:
```Type Error (imdb.core:2:6) Polymorphic function clojure.core/map could not be applied to arguments:
Domains:
(Fn [a b ... b -> c]) (t/NonEmptySeqable a) (t/NonEmptySeqable a) ... b
(Fn [a b ... b -> c]) (U nil (clojure.lang.Seqable a)) (U nil (clojure.lang.Seqable b)) ... b
Arguments:
(Fn [t/AnyInteger -> t/AnyInteger] [Number -> Number]) (t/Set (U (Value 1) (Value "a")))
Ranges:
(t/NonEmptyLazySeq c)
(clojure.lang.LazySeq c)
with expected type:
Any
in: (clojure.core/map clojure.core/inc s)
```
In both cases, our error is caught, but at different stages of analysis.
This flexibility is immensely powerful, and it encourage more nuanced thinking about static types. In many situations, finding ourselves with a heterogeous set would be a sign that something had gone wrong, but not always. For instance, with our more general annotation, we could do this:
```imdb.core> (t/cf (let [s (clojure.set/union #{1} #{1.0})]
(map inc s)))
(clojure.lang.LazySeq Number)
```
I think I've demonstrated that `core.typed` is cool. I know that it has improved my own code. I hope that it will ultimately remove one of the greatest barriers to broad adoption of clojure. | 3,900 | 15,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-47 | longest | en | 0.967248 |
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# Georgias New Grail pic 3 mid/eve straight only
Topic closed. 41 replies. Last post 10 years ago by emilyg.
Page 3 of 3
mississippi
United States
Member #34478
March 3, 2006
5903 Posts
Offline
Posted: April 11, 2007, 6:06 am - IP Logged
You are welcome Em..and LuckyStar..I hope they help out...
"Attention all Mathematicians: Check your degree at the door because when it comes to whole numbers you are the Amateur"
mississippi
United States
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March 3, 2006
5903 Posts
Offline
Posted: April 14, 2007, 4:44 pm - IP Logged
11=531
12=683
13=373
14=573
15=731
16=530
Georgia if I checked this right..you have First digit 2 and first digit 0, groups left to give you a straight..wish I could provide all of you with a day..but I cant..so..just keep an eye out is all I can say..
Georgia midday April 14 th..239..here is your first digit 2 straight..Georgia I believe the only group you have left is first digit 0 group..good luck...
"Attention all Mathematicians: Check your degree at the door because when it comes to whole numbers you are the Amateur"
mississippi
United States
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March 3, 2006
5903 Posts
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Posted: April 16, 2007, 1:10 pm - IP Logged
Georgia midday April 14 th..239..here is your first digit 2 straight..Georgia I believe the only group you have left is first digit 0 group..good luck...
georgia midday 013 that completes your 0-9 STRAIGHT sequence run Georgia..looks like about 1 MONTH time frame for the states that have 2 draws a day..like I said..when you know what is coming and its RIGHT ORDER OF ARRANGEMENT..it gives you a HUGE ADVANTAGE....
"Attention all Mathematicians: Check your degree at the door because when it comes to whole numbers you are the Amateur"
Georgia
United States
Member #38040
April 23, 2006
1643 Posts
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Posted: April 17, 2007, 2:43 am - IP Logged
Thank you, Lotterybraker!!!
Peace. And, Good Luck!!!
mississippi
United States
Member #34478
March 3, 2006
5903 Posts
Offline
Posted: April 27, 2007, 12:55 pm - IP Logged
11=531
12=683
13=373
14=573
15=731
16=530
Georgia if I checked this right..you have First digit 2 and first digit 0, groups left to give you a straight..wish I could provide all of you with a day..but I cant..so..just keep an eye out is all I can say..
17=958
18=525
19=239
20=446
21=013
22=679
23=537
24=252
25=317
26=357
27=714
28=706
29=706
30=704
31=365
32=983
33=948
34=467
35=857
I hope they helped out some...good luck everyone..
"Attention all Mathematicians: Check your degree at the door because when it comes to whole numbers you are the Amateur"
United States
Member #14
November 9, 2001
31351 Posts
Offline
Posted: April 27, 2007, 1:08 pm - IP Logged
17=958
18=525
19=239
20=446
21=013
22=679
23=537
24=252
25=317
26=357
27=714
28=706
29=706
30=704
31=365
32=983
33=948
34=467
35=857
I hope they helped out some...good luck everyone..
Thanks LB.
love to nibble those micey feet.
mississippi
United States
Member #34478
March 3, 2006
5903 Posts
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Posted: April 27, 2007, 1:13 pm - IP Logged
you are most welcome Em...
"Attention all Mathematicians: Check your degree at the door because when it comes to whole numbers you are the Amateur"
Georgia
United States
Member #38040
April 23, 2006
1643 Posts
Offline
Posted: May 1, 2007, 7:47 am - IP Logged
Ok my friends in Georgia I finished up your new Cycle Thursday nite..so lets see what she will give us this time..if she gets say 5 or 6 straights next 5 or 6 days I will say she is running fine..so lets see how she does..I am looking for at least 5 STRAIGHTS for each First digit group..lets Rock n Roll..lol
1.020,021,030,031,002,012,003,013
2.022,023,032,033
3.024,042,025,052,034,043,035,053
4.026,062,027,072,036,063,037,073
5.028,082,029,092,038,083,039,093
6.160,106,161,116,170,107,117,171
7.162,126,163,136,172,127,173,137
8.164,146,165,156,174,147,175,157
9.166,167,176,177
10.168,186,169,196,178,187,179,197
11.220,202,221,212,230,203,213,231
12.222,223,232,233
13.224,242,225,252,234,243,235,253
14.226,262,227,272,236,263,237,273
15.228,282,229,292,238,283,239,293
16.360,306,361,316,370,307,317,371
17.362,326,363,336,372,327,373,337
18.364,346,365,356,374,347,357,375
19.366,367,376,377
20.368,386,369,396,378,387,379,397
21.460,406,416,461,470,407,417,471
22.462,426,463,436,472,427,473,437
23.464,446,465,456,474,447,475,457
24.466,467,476,477
25.468,486,469,496,478,487,479,497
26.520,502,503,530,512,521,513,531
27.522,523,532,533
28.524,542,525,552,534,543,535,553
29.526,562,527,572,536,563,537,573
30.528,582,529,592538,583,539,593
31.680,608,609,690,618,681,619,691
32.682,628,683,638,692,629,693,639
33.684,648,685,658,694,649,695,659
34.686,668,687,678,696,669,697,679
35.688,689,698,699
36.700,701,710,711
37.702,720,703,730,712,721,713,731
38.704,740,705,750,714,741,715,751
39.706,760,707,770,716,761,717,771
40.708,780,709,790,718,781,719,791
41.840,804,850,805,841,814,815,851
42.842,824,843,834,852,825,853,835
43.844,845,854,855
44.846,864,847,874,856,865,857,875
45.848,884,849,894,858,885,859,895
46.980,908,990,909,981,918,991,919
47.982,928,983,938,992,929,993,939
48.984,948,985,958,994,949,995,959
49.986,968,987,978,996,969,997,979
50.988,989,998,999
36 = 718
37 = 197
38 = 168
39 = 386
State specific. Straight. Six "hits" in a row!!! Has there been more than two "skips" between "hits"?
Amazing!!!
Peace. And, Good Luck!!!
mississippi
United States
Member #34478
March 3, 2006
5903 Posts
Offline
Posted: May 1, 2007, 8:29 am - IP Logged
To be Honest Lucky Star I have no idea..I havent even been keeping up with this list..just watching it from time to time to see if it was still running is all..she has done very well and has to be getting close to finishing this run up...
"Attention all Mathematicians: Check your degree at the door because when it comes to whole numbers you are the Amateur"
Stone Mountain, GA
United States
Member #3189
January 1, 2004
28242 Posts
Offline
Posted: May 1, 2007, 1:21 pm - IP Logged
Ok my friends in Georgia I finished up your new Cycle Thursday nite..so lets see what she will give us this time..if she gets say 5 or 6 straights next 5 or 6 days I will say she is running fine..so lets see how she does..I am looking for at least 5 STRAIGHTS for each First digit group..lets Rock n Roll..lol
1.020,021,030,031,002,012,003,013
2.022,023,032,033
3.024,042,025,052,034,043,035,053
4.026,062,027,072,036,063,037,073
5.028,082,029,092,038,083,039,093
6.160,106,161,116,170,107,117,171
7.162,126,163,136,172,127,173,137
8.164,146,165,156,174,147,175,157
9.166,167,176,177
10.168,186,169,196,178,187,179,197
11.220,202,221,212,230,203,213,231
12.222,223,232,233
13.224,242,225,252,234,243,235,253
14.226,262,227,272,236,263,237,273
15.228,282,229,292,238,283,239,293
16.360,306,361,316,370,307,317,371
17.362,326,363,336,372,327,373,337
18.364,346,365,356,374,347,357,375
19.366,367,376,377
20.368,386,369,396,378,387,379,397
21.460,406,416,461,470,407,417,471
22.462,426,463,436,472,427,473,437
23.464,446,465,456,474,447,475,457
24.466,467,476,477
25.468,486,469,496,478,487,479,497
26.520,502,503,530,512,521,513,531
27.522,523,532,533
28.524,542,525,552,534,543,535,553
29.526,562,527,572,536,563,537,573
30.528,582,529,592538,583,539,593
31.680,608,609,690,618,681,619,691
32.682,628,683,638,692,629,693,639
33.684,648,685,658,694,649,695,659
34.686,668,687,678,696,669,697,679
35.688,689,698,699
36.700,701,710,711
37.702,720,703,730,712,721,713,731
38.704,740,705,750,714,741,715,751
39.706,760,707,770,716,761,717,771
40.708,780,709,790,718,781,719,791
41.840,804,850,805,841,814,815,851
42.842,824,843,834,852,825,853,835
43.844,845,854,855
44.846,864,847,874,856,865,857,875
45.848,884,849,894,858,885,859,895
46.980,908,990,909,981,918,991,919
47.982,928,983,938,992,929,993,939
48.984,948,985,958,994,949,995,959
49.986,968,987,978,996,969,997,979
50.988,989,998,999
Still hittin'
Georgia Midday 3 Tue, May 1, 2007 1-3-6
Now, will that 127 finally fall!!!!!!
mississippi
United States
Member #34478
March 3, 2006
5903 Posts
Offline
Posted: May 1, 2007, 1:25 pm - IP Logged
Wish I could help you out with that one Toney..maybe she will..or maybe she will wait til the next time she comes up to actually show..heh..I read you caught a double/double in Georgia under Em's Thread..very nice pick up...wtg!!!
"Attention all Mathematicians: Check your degree at the door because when it comes to whole numbers you are the Amateur"
United States
Member #14
November 9, 2001
31351 Posts
Offline
Posted: May 1, 2007, 1:38 pm - IP Logged
Wish I could help you out with that one Toney..maybe she will..or maybe she will wait til the next time she comes up to actually show..heh..I read you caught a double/double in Georgia under Em's Thread..very nice pick up...wtg!!!
Congrats!
love to nibble those micey feet.
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https://www.math.wisc.edu/wiki/index.php?title=AMS_Student_Chapter_Seminar&diff=17195&oldid=17189 | 1,591,489,183,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348521325.84/warc/CC-MAIN-20200606222233-20200607012233-00070.warc.gz | 801,705,462 | 9,690 | Difference between revisions of "AMS Student Chapter Seminar"
The AMS Student Chapter Seminar is an informal, graduate student seminar on a wide range of mathematical topics. Pastries (usually donuts) will be provided.
Everyone is welcome to give a talk. To sign up, please contact one of the organizers with a title and abstract. Talks are 30 minutes long and should avoid assuming significant mathematical background beyond first-year graduate courses.
The schedule of talks from past semesters can be found here.
Spring 2019
February 6, Xiao Shen (in VV B139)
Title: Limit Shape in last passage percolation
Abstract: Imagine the following situation, attached to each point on the integer lattice Z^2 there is an arbitrary amount of donuts. Fix x and y in Z^2, if you get to eat all the donuts along an up-right path between these two points, what would be the maximum amount of donuts you can get? This model is often called last passage percolation, and I will discuss a classical result about its scaling limit: what happens if we zoom out and let the distance between x and y tend to infinity.
February 13, Michel Alexis (in VV B139)
Title: An instructive yet useless theorem about random Fourier Series
Abstract: Consider a Fourier series with random, symmetric, independent coefficients. With what probability is this the Fourier series of a continuous function? An $L^{p}$ function? A surprising result is the Billard theorem, which says such a series results almost surely from an $L^{\infty}$ function if and only if it results almost surely from a continuous function. Although the theorem in of itself is kind of useless in of itself, its proof is instructive in that we will see how, via the principle of reduction, one can usually just pretend all symmetric random variables are just coin flips (Bernoulli trials with outcomes $\pm 1$).
February 20, Geoff Bentsen
Title: An Analyst Wanders into a Topology Conference
Abstract: Fourier Restriction is a big open problem in Harmonic Analysis; given a "small" subset $E$ of $R^d$, can we restrict the Fourier transform of an $L^p$ function to $E$ and retain any information about our original function? This problem has a nice (somewhat) complete solution for smooth manifolds of co-dimension one. I will explore how to start generalizing this result to smooth manifolds of higher co-dimension, and how a topology paper from the 60s about the hairy ball problem came in handy along the way.
February 27, James Hanson
Title: What is...a Topometric Space?
Abstract: Continuous first-order logic is a generalization of first-order logic that is well suited for the study of structures with a natural metric, such as Banach spaces and probability algebras. Topometric spaces are a simple generalization of topological and metric spaces that arise in the study of continuous first-order logic. I will discuss certain topological issues that show up in topometric spaces coming from continuous logic, as well as some partial solutions and open problems. No knowledge of logic will be required for or gained from attending the talk.
March 6, Working Group to establish an Association of Mathematics Graduate Students
Title: Introducing GRAMS (Graduate Representative Association of Mathematics Students)
Abstract: Over the past couple months, a handful of us have been working to create the UW Graduate Representative Association of Mathematics Students (GRAMS). This group, about 5-8 students, is intended to be a liaison between the graduate students and faculty, especially in relation to departmental policies and decisions that affect graduate students. We will discuss what we believe GRAMS ought to look like and the steps needed to implement such a vision, then open up the floor to a Q&A. Check out our website for more information.
March 13, Connor Simpson
Title: Counting faces of polytopes with algebra
Abstract: A natural question is: with a fixed dimension and number of vertices, what is the largest number of d-dimensional faces that a polytope can have? We will outline a proof of the answer to this question.
March 26 (Prospective Student Visit Day), Multiple Speakers
Eva Elduque, 11-11:25
Title: Will it fold flat?
Abstract: TBD
Soumya Sankar, 11:30-11:55
Title: An algebro-geometric perspective on integration
Abstract: TBD
Chun Gan, 12:00-12:25
Title: Extension theorems in complex analysis
Abstract: TBD
Jenny Yeon, 2:00-2:25
Title: Application of Slope Field
Abstract: TBD
Rajula Srivastava, 2:30-2:55
Title: The World of Wavelets
Abstract: TBD
Shengyuan Huang, 3:00-3:25
Title: Group objects in various categories
Abstract: TBD
Ivan Ongay Valverde, 3:30-3:55
Title: Games and Topology
Abstract: TBD
Sun Woo Park, 4:00-4:25
Title: Reconstruction of character tables of Sn
Abstract: TBD
Max Bacharach, 4:30-4:55
Title: Clothes, Lice, and Coalescence
Abstract: TBD
Title: TBD
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Abstract: TBD | 1,246 | 5,296 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-24 | latest | en | 0.925706 |
https://fixthefool.com/sainte-anne-des-plaines/applications-of-linear-algebra-in-engineering.php | 1,653,261,274,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662550298.31/warc/CC-MAIN-20220522220714-20220523010714-00410.warc.gz | 300,202,303 | 9,410 | suggest me a real world application of linear algebra to My Linear Algebra teacher has assigned an individual Linear Algebra in Computer Engineering. I looked online for different application notes regarding
## Applications Of Linear Algebra In Engineering
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My Linear Algebra teacher has assigned an individual Linear Algebra in Computer Engineering. I looked online for different application notes regarding Engineering: Linear algebra is the simplest way to look at functions of many variables, which usually arise in engineering by the discretization of a concept stated in terms of a continuum, e.g. the law governing the relation between stresses and strains in a structure. Linear Algebra is used quite heavily in Structural Engineering.
Designed for advanced engineering, physical science, and applied mathematics students, this innovative textbook is an introduction to both the theory and the practical application of linear algebra and functional analysis. The book is self-contained, beginning with elementary … 2010-11-05 · I'm currently an electrical engineering major taking Calc 2 and Physics I right now. Linear Algebra is not required at my school but I kinda want to take it. I...
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Linear algebra is also used in most sciences and engineering areas, Bretscher, Otto (2004), Linear Algebra with Applications (3rd ed.), Prentice Hall, Applications of Linear Algebra in Engineering are extremely important while creating engineering models (i.e. jet engines) There are primarily two types of instabilities that occur in the flow through the …
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## Applications of Methods of Numerical Linear Algebra in
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The very first class I had was Linear Algebra. eigenvalues and all their applications. Ken is a professional engineer, engineering mentor, Linear Algebra for Engineering or Science MATH1104B, Fall 2014 School of Mathematics and Statistics, Linear Algebra and its applications, by David C. Lay.
Applications of Linear Algebra in Engineering are extremely important while creating engineering models (i.e. jet engines) There are primarily two types of instabilities that occur in the flow through the … An Interesting Application of Linear Algebra scientists and engineers This is one of the greatest benefits to be had from the application of Linear Algebra
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Math 2220: Linear Algebra and Engineering Applications Spring 2009 . To use linear algebra to solve this system we will first write down the augmented Linear Algebra for Engineering or Science MATH1104B, Fall 2014 School of Mathematics and Statistics, Linear Algebra and its applications, by David C. Lay.
Linear algebra is also used in most sciences and engineering areas, Bretscher, Otto (2004), Linear Algebra with Applications (3rd ed.), Prentice Hall, My Linear Algebra teacher has assigned an individual Linear Algebra in Computer Engineering. I looked online for different application notes regarding
There is a large gap between the engineering course in tensor algebra on the one hand and the treatment of linear transformations within classical linear algebra on Methods of numerical linear algebra are concerned with the theory and practical aspects of computing solutions of mathematical problems in engineering such as image
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Designed for advanced engineering, physical science, and applied mathematics students, this innovative textbook is an introduction to both the theory and the practical application of linear algebra and functional analysis. The book is self-contained, beginning with elementary … Linear Algebra for Engineering or Science MATH1104A, Fall 2014 School of Mathematics and Statistics, Linear Algebra and its applications, by David C. Lay.
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### Tensor Algebra and Tensor Analysis for Engineers With
Free Applications Of Linear Algebra In Electrical. 2010-11-05В В· I'm currently an electrical engineering major taking Calc 2 and Physics I right now. Linear Algebra is not required at my school but I kinda want to take it. I... Linear Algebra is used quite heavily in Structural Engineering. This is for a very simple reason. The analysis of a structure in equilibrium involves writing down many equations in many unknowns. Often these equations are linear, even when material deformation (i.e. bending) is considered..
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Linear Algebra for Engineering or Science MATH1104B, Fall 2014 School of Mathematics and Statistics, Linear Algebra and its applications, by David C. Lay. Applications of Linear Algebra in Engineering are extremely important while creating engineering models (i.e. jet engines) There are primarily two types of instabilities that occur in the flow through the …
LINEAR ALGEBRA AND ITS APPLICATIONS theory or linear algebra to other branches of mathematics and to other sciences. Dept. of Mathematical Engineering, Electrical and Computer Engineering majors study a variety of areas including signal analysis and circuits. Signal analysis rests, to a significant extent, on application of Fourier Series (Lathi), for example. Fourier Series, in turn, is a product of the concept of orthogonal basis in linear algebra.
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Linear Algebra for Engineering or Science MATH1104A, Fall 2014 School of Mathematics and Statistics, Linear Algebra and its applications, by David C. Lay. Linear algebra is also used in most sciences and engineering areas, Bretscher, Otto (2004), Linear Algebra with Applications (3rd ed.), Prentice Hall,
2011-04-15В В· im doing a bachelor's in electrical engineering and i already have the linear algebra credit, but i totally forgot everything. should i retake it just in The book contains all the material necessary for a first year graduate or advanced undergraduate course on numerical linear algebra with numerous applications to engineering and science.
Engineering: Linear algebra is the simplest way to look at functions of many variables, which usually arise in engineering by the discretization of a concept stated in terms of a continuum, e.g. the law governing the relation between stresses and strains in a structure. Linear Algebra is used quite heavily in Structural Engineering. LINEAR ALGEBRA AND ITS APPLICATIONS theory or linear algebra to other branches of mathematics and to other sciences. Dept. of Mathematical Engineering,
engineers. In linear algebra one studies sets of linear equations and their transformation properties. It is possible to consider the analysis of rotations in space, selected curve fitting techniques, differential equation solutions, as well as many other problems in science … My Linear Algebra teacher has assigned an individual Linear Algebra in Computer Engineering. I looked online for different application notes regarding
2011-04-15В В· im doing a bachelor's in electrical engineering and i already have the linear algebra credit, but i totally forgot everything. should i retake it just in Linear Algebra is used quite heavily in Structural Engineering. This is for a very simple reason. The analysis of a structure in equilibrium involves writing down many equations in many unknowns. Often these equations are linear, even when material deformation (i.e. bending) is considered.
Engineering: Application Areas. Linear Algebra. Browse Category : Linear Algebra. Collision detection between toolholder and workpiece on ball nut grinding Title: Free Applications Of Linear Algebra In Electrical Engineering (PDF, ePub, Mobi) Author: Bella Books Subject: Applications Of Linear Algebra In Electrical
LINEAR ALGEBRA AND ITS APPLICATIONS theory or linear algebra to other branches of mathematics and to other sciences. Dept. of Mathematical Engineering, Numerical and Computational Challenges in Science and Engineering Workshop on Numerical Linear Algebra in Scientific and Engineering Applications
Code Review Stack Exchange is a question and answer site for peer programmer code reviews. Java implementation of spell-checking algorithm. Dictionary.java. Java source code for dictionary application Waverley Beach Source code definition, program instructions that must be translated by a compiler, interpreter, or assembler into object code before execution. See more. | 3,841 | 21,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-21 | longest | en | 0.891585 |
https://www.esaral.com/q/tick-the-correct-answer-35078 | 1,725,904,598,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651133.92/warc/CC-MAIN-20240909170505-20240909200505-00190.warc.gz | 739,112,237 | 11,597 | # Tick (✓) the correct answer:
Question:
If (a + b) = 12 and ab = 14, then (a2 + b2) = ?
(a) 172
(b) 116
(c) 165
(d) 126
Solution:
(b) 116
$(a+b)=12$
$\Rightarrow$ Squaring both the sides:
$\Rightarrow(a+b)^{2}=(12)^{2}$
$\Rightarrow\left(a^{2}+b^{2}+2 a b\right)=144$
$\Rightarrow\left(a^{2}+b^{2}\right)=144-2 a b$
$\Rightarrow\left(a^{2}+b^{2}\right)=144-2(14)$
$\Rightarrow\left(a^{2}+b^{2}\right)=144-28$
$\Rightarrow\left(a^{2}+b^{2}\right)=116$ | 200 | 467 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-38 | latest | en | 0.173687 |
https://classroom.thenational.academy/lessons/combined-events-and-sample-spaces-part-1-69gp6d?step=3&activity=worksheet | 1,708,886,537,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474641.34/warc/CC-MAIN-20240225171204-20240225201204-00801.warc.gz | 175,772,515 | 26,523 | Combined events and sample spaces (Part 1)
In this lesson, we will represent the probabilities of combined events in a sample space.
Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.True or false: I am more likely to flip a coin and get "Heads" than I am to roll a six-sided dice and get a 4.
1/5
Q2.True or false: I am more likely to flip a coin and get "Tails" than I am to roll a six-sided dice and get an even number.
2/5
Q3.True or false: I am more likely to roll a 3 than a 5 on a six sided dice
3/5
Q4.True or false: I am more likely to roll an even number than a prime number on an eight sided dice.
4/5
Q5.True or false: I am more likely to roll a 4 on a six sided dice than on an 8 sided dice.
5/5
Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.True or false: I am more likely to flip a coin and get "Heads" than I am to roll a six-sided dice and get a 4.
1/5
Q2.True or false: I am more likely to flip a coin and get "Tails" than I am to roll a six-sided dice and get an even number.
2/5
Q3.True or false: I am more likely to roll a 3 than a 5 on a six sided dice
3/5
Q4.True or false: I am more likely to roll an even number than a prime number on an eight sided dice.
4/5
Q5.True or false: I am more likely to roll a 4 on a six sided dice than on an 8 sided dice.
5/5
Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
Combined Events and Sample Spaces
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
Q1.What is the probability of getting a 5?
1/5
Q2.What is the probability of getting a 1?
2/5
Q3.What is the probability of getting greater than 4?
3/5
Q4.What is the probability of getting a multiple of 3?
4/5
Q5.What is the probability of getting a prime number?
5/5
Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
Combined Events and Sample Spaces
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
Q1.What is the probability of getting a 5?
1/5
Q2.What is the probability of getting a 1?
2/5
Q3.What is the probability of getting greater than 4?
3/5
Q4.What is the probability of getting a multiple of 3?
4/5
Q5.What is the probability of getting a prime number?
5/5
Lesson summary: Combined events and sample spaces (Part 1)
Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Climb stairs
On the spot:
Dance | 924 | 3,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-10 | latest | en | 0.915567 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/1584/4/ | 1,719,168,649,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862488.55/warc/CC-MAIN-20240623162925-20240623192925-00305.warc.gz | 751,628,527 | 95,316 | # Properties
Label 1584.4 Level 1584 Weight 4 Dimension 90437 Nonzero newspaces 32 Sturm bound 552960 Trace bound 25
## Defining parameters
Level: $$N$$ = $$1584 = 2^{4} \cdot 3^{2} \cdot 11$$ Weight: $$k$$ = $$4$$ Nonzero newspaces: $$32$$ Sturm bound: $$552960$$ Trace bound: $$25$$
## Dimensions
The following table gives the dimensions of various subspaces of $$M_{4}(\Gamma_1(1584))$$.
Total New Old
Modular forms 209600 91165 118435
Cusp forms 205120 90437 114683
Eisenstein series 4480 728 3752
## Trace form
$$90437 q - 48 q^{2} - 48 q^{3} - 68 q^{4} - 61 q^{5} - 64 q^{6} - 63 q^{7} + 36 q^{8} + 24 q^{9} + O(q^{10})$$ $$90437 q - 48 q^{2} - 48 q^{3} - 68 q^{4} - 61 q^{5} - 64 q^{6} - 63 q^{7} + 36 q^{8} + 24 q^{9} - 12 q^{10} + 30 q^{11} - 144 q^{12} + 39 q^{13} - 300 q^{14} - 6 q^{15} + 444 q^{16} + 97 q^{17} + 216 q^{18} - 379 q^{19} - 892 q^{20} - 678 q^{21} - 476 q^{22} - 728 q^{23} - 1536 q^{24} - 1365 q^{25} - 1700 q^{26} + 708 q^{27} - 996 q^{28} + 499 q^{29} + 936 q^{30} - 267 q^{31} + 1932 q^{32} - 117 q^{33} + 1800 q^{34} + 201 q^{35} - 1592 q^{36} + 1715 q^{37} - 2404 q^{38} + 402 q^{39} + 1468 q^{40} + 1699 q^{41} + 1296 q^{42} - 1704 q^{43} + 2156 q^{44} + 346 q^{45} + 3476 q^{46} - 6741 q^{47} + 4824 q^{48} - 2569 q^{49} + 7152 q^{50} - 3968 q^{51} + 3876 q^{52} + 237 q^{53} + 376 q^{54} - 799 q^{55} - 5880 q^{56} - 2316 q^{57} - 4580 q^{58} + 8645 q^{59} - 15928 q^{60} + 2247 q^{61} - 9516 q^{62} + 5034 q^{63} - 3332 q^{64} + 3678 q^{65} + 4848 q^{66} - 5776 q^{67} + 6820 q^{68} + 5626 q^{69} - 136 q^{70} - 1763 q^{71} + 15840 q^{72} - 1929 q^{73} + 17632 q^{74} + 1000 q^{75} + 4708 q^{76} + 7088 q^{77} + 888 q^{78} + 1523 q^{79} - 20256 q^{80} + 808 q^{81} - 18444 q^{82} - 353 q^{83} - 19168 q^{84} - 2375 q^{85} - 31260 q^{86} + 438 q^{87} - 23088 q^{88} - 8438 q^{89} - 2368 q^{90} + 6105 q^{91} + 14220 q^{92} - 1822 q^{93} + 13372 q^{94} + 1149 q^{95} + 12856 q^{96} - 4865 q^{97} + 33948 q^{98} + 1515 q^{99} + O(q^{100})$$
## Decomposition of $$S_{4}^{\mathrm{new}}(\Gamma_1(1584))$$
We only show spaces with even parity, since no modular forms exist when this condition is not satisfied. Within each space $$S_k^{\mathrm{new}}(N, \chi)$$ we list available newforms together with their dimension.
Label $$\chi$$ Newforms Dimension $$\chi$$ degree
1584.4.a $$\chi_{1584}(1, \cdot)$$ 1584.4.a.a 1 1
1584.4.a.b 1
1584.4.a.c 1
1584.4.a.d 1
1584.4.a.e 1
1584.4.a.f 1
1584.4.a.g 1
1584.4.a.h 1
1584.4.a.i 1
1584.4.a.j 1
1584.4.a.k 1
1584.4.a.l 1
1584.4.a.m 1
1584.4.a.n 1
1584.4.a.o 1
1584.4.a.p 1
1584.4.a.q 1
1584.4.a.r 1
1584.4.a.s 1
1584.4.a.t 1
1584.4.a.u 1
1584.4.a.v 1
1584.4.a.w 2
1584.4.a.x 2
1584.4.a.y 2
1584.4.a.z 2
1584.4.a.ba 2
1584.4.a.bb 2
1584.4.a.bc 2
1584.4.a.bd 2
1584.4.a.be 2
1584.4.a.bf 2
1584.4.a.bg 2
1584.4.a.bh 2
1584.4.a.bi 2
1584.4.a.bj 2
1584.4.a.bk 2
1584.4.a.bl 3
1584.4.a.bm 3
1584.4.a.bn 3
1584.4.a.bo 3
1584.4.a.bp 3
1584.4.a.bq 4
1584.4.a.br 4
1584.4.b $$\chi_{1584}(593, \cdot)$$ 1584.4.b.a 2 1
1584.4.b.b 2
1584.4.b.c 6
1584.4.b.d 6
1584.4.b.e 6
1584.4.b.f 6
1584.4.b.g 8
1584.4.b.h 18
1584.4.b.i 18
1584.4.d $$\chi_{1584}(287, \cdot)$$ 1584.4.d.a 10 1
1584.4.d.b 10
1584.4.d.c 20
1584.4.d.d 20
1584.4.f $$\chi_{1584}(793, \cdot)$$ None 0 1
1584.4.h $$\chi_{1584}(1495, \cdot)$$ None 0 1
1584.4.k $$\chi_{1584}(1079, \cdot)$$ None 0 1
1584.4.m $$\chi_{1584}(1385, \cdot)$$ None 0 1
1584.4.o $$\chi_{1584}(703, \cdot)$$ 1584.4.o.a 2 1
1584.4.o.b 2
1584.4.o.c 2
1584.4.o.d 4
1584.4.o.e 4
1584.4.o.f 8
1584.4.o.g 8
1584.4.o.h 12
1584.4.o.i 24
1584.4.o.j 24
1584.4.q $$\chi_{1584}(529, \cdot)$$ n/a 360 2
1584.4.r $$\chi_{1584}(307, \cdot)$$ n/a 716 2
1584.4.u $$\chi_{1584}(397, \cdot)$$ n/a 600 2
1584.4.v $$\chi_{1584}(683, \cdot)$$ n/a 480 2
1584.4.y $$\chi_{1584}(197, \cdot)$$ n/a 576 2
1584.4.z $$\chi_{1584}(289, \cdot)$$ n/a 356 4
1584.4.bc $$\chi_{1584}(175, \cdot)$$ n/a 432 2
1584.4.be $$\chi_{1584}(329, \cdot)$$ None 0 2
1584.4.bg $$\chi_{1584}(23, \cdot)$$ None 0 2
1584.4.bh $$\chi_{1584}(439, \cdot)$$ None 0 2
1584.4.bj $$\chi_{1584}(265, \cdot)$$ None 0 2
1584.4.bl $$\chi_{1584}(815, \cdot)$$ n/a 360 2
1584.4.bn $$\chi_{1584}(65, \cdot)$$ n/a 428 2
1584.4.bq $$\chi_{1584}(127, \cdot)$$ n/a 360 4
1584.4.bs $$\chi_{1584}(233, \cdot)$$ None 0 4
1584.4.bu $$\chi_{1584}(71, \cdot)$$ None 0 4
1584.4.bx $$\chi_{1584}(343, \cdot)$$ None 0 4
1584.4.bz $$\chi_{1584}(361, \cdot)$$ None 0 4
1584.4.cb $$\chi_{1584}(575, \cdot)$$ n/a 288 4
1584.4.cd $$\chi_{1584}(17, \cdot)$$ n/a 288 4
1584.4.cf $$\chi_{1584}(155, \cdot)$$ n/a 2880 4
1584.4.cg $$\chi_{1584}(461, \cdot)$$ n/a 3440 4
1584.4.cj $$\chi_{1584}(43, \cdot)$$ n/a 3440 4
1584.4.ck $$\chi_{1584}(133, \cdot)$$ n/a 2880 4
1584.4.cm $$\chi_{1584}(49, \cdot)$$ n/a 1712 8
1584.4.cn $$\chi_{1584}(413, \cdot)$$ n/a 2304 8
1584.4.cq $$\chi_{1584}(179, \cdot)$$ n/a 2304 8
1584.4.cr $$\chi_{1584}(37, \cdot)$$ n/a 2864 8
1584.4.cu $$\chi_{1584}(19, \cdot)$$ n/a 2864 8
1584.4.cw $$\chi_{1584}(497, \cdot)$$ n/a 1712 8
1584.4.cy $$\chi_{1584}(47, \cdot)$$ n/a 1728 8
1584.4.da $$\chi_{1584}(25, \cdot)$$ None 0 8
1584.4.dc $$\chi_{1584}(7, \cdot)$$ None 0 8
1584.4.dd $$\chi_{1584}(119, \cdot)$$ None 0 8
1584.4.df $$\chi_{1584}(41, \cdot)$$ None 0 8
1584.4.dh $$\chi_{1584}(79, \cdot)$$ n/a 1728 8
1584.4.dl $$\chi_{1584}(157, \cdot)$$ n/a 13760 16
1584.4.dm $$\chi_{1584}(139, \cdot)$$ n/a 13760 16
1584.4.dp $$\chi_{1584}(29, \cdot)$$ n/a 13760 16
1584.4.dq $$\chi_{1584}(59, \cdot)$$ n/a 13760 16
"n/a" means that newforms for that character have not been added to the database yet
## Decomposition of $$S_{4}^{\mathrm{old}}(\Gamma_1(1584))$$ into lower level spaces
$$S_{4}^{\mathrm{old}}(\Gamma_1(1584)) \cong$$ $$S_{4}^{\mathrm{new}}(\Gamma_1(1))$$$$^{\oplus 30}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(2))$$$$^{\oplus 24}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(3))$$$$^{\oplus 20}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(4))$$$$^{\oplus 18}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(6))$$$$^{\oplus 16}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(8))$$$$^{\oplus 12}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(9))$$$$^{\oplus 10}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(11))$$$$^{\oplus 15}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(12))$$$$^{\oplus 12}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(16))$$$$^{\oplus 6}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(18))$$$$^{\oplus 8}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(22))$$$$^{\oplus 12}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(24))$$$$^{\oplus 8}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(33))$$$$^{\oplus 10}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(36))$$$$^{\oplus 6}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(44))$$$$^{\oplus 9}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(48))$$$$^{\oplus 4}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(66))$$$$^{\oplus 8}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(72))$$$$^{\oplus 4}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(88))$$$$^{\oplus 6}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(99))$$$$^{\oplus 5}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(132))$$$$^{\oplus 6}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(144))$$$$^{\oplus 2}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(176))$$$$^{\oplus 3}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(198))$$$$^{\oplus 4}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(264))$$$$^{\oplus 4}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(396))$$$$^{\oplus 3}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(528))$$$$^{\oplus 2}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(792))$$$$^{\oplus 2}$$$$\oplus$$$$S_{4}^{\mathrm{new}}(\Gamma_1(1584))$$$$^{\oplus 1}$$ | 3,869 | 7,577 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-26 | latest | en | 0.145152 |
https://crypto.stackexchange.com/questions/27062/shuffleless-prng-function-with-non-repeating-values | 1,568,701,851,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573053.13/warc/CC-MAIN-20190917061226-20190917083226-00169.warc.gz | 439,300,192 | 32,241 | # Shuffleless PRNG function with non-repeating values?
I need a simple PRNG function of type: Integer = PRNG(n, maxval) as I would like to count from 0 to maxval, not in a linear manner, but in a pseudorandom manner where I still use every value, but only once. For example, "0 to maxval=9" might result in "3,8,4,5,1,9,0,2,6,7", and if I asked the function for "Integer = PRNG(3)" in this case it would return the third value which is 4. (For my actual need maxval is always 64bit so 0xFFFFFFFFFFFFFFFF)
Im hoping this can be accomplished without shuffling a pre-filled array? My googling suggests perhaps a "feedback register" of sorts could be what I'm after but ive got no idea which. Thankyou
ps. are there any CPRNG secure versions of this? I dont need that level of security - PRNG is fine, but just wondering!
• "without shuffles/arrays" seems like a rather arbitrary requirement. $\;$ – user991 Jul 22 '15 at 1:22
• So there's no way of doing this without creating a 64bit array with values 0 to FFFFFFFFFFFFFFFF and shuffling that? seems expensive and inefficient! – David Jul 22 '15 at 1:47
• There is a much cheaper and more efficient method; it just only does the $\hspace{1.76 in}$ currently-relevant parts of each shuffle. $\;$ – user991 Jul 22 '15 at 5:49
• Don't get me wrong, but we're not Google. It's not clear to me what your request for a PRNG recommendation has to do with cryptography... maybe you could edit your question and explain the exact scenario, what you've tried, what you've researched, and what exactly you need help with. For more infos on "how to ask", please take a look at our help center. – e-sushi Jul 22 '15 at 6:00
• @David No, it is entirely possible to do for 64 bit values in constant space - see my answer. – Thomas M. DuBuisson Jul 22 '15 at 16:20 | 493 | 1,800 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-39 | longest | en | 0.920888 |
http://www.onlinemathlearning.com/4-digit-dividend.html | 1,493,221,094,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121453.27/warc/CC-MAIN-20170423031201-00206-ip-10-145-167-34.ec2.internal.warc.gz | 635,694,161 | 9,297 | # Four-Digit Dividend
Videos to help Grade 4 students learn how to represent numerically four-digit dividend division with divisors of 2, 3, 4, and 5, decomposing a remainder up to three digits.
Common Core Standards: 4.OA.3, 4.NBT.6, 4.NBT.1
New York State Common Core Math Module 3, Grade 4, Lesson 29
NYS Math Module 3 Grade 4 Lesson 29 Concept Development
Problem 1
Solve 4,325 ÷ 3 using the standard algorithm. Multiply to check the answer.
Problem 2
Ellie bought two packs of beads. Altogether she has 1,254 beads. If the number of beads in each bag is the same, how many beads are in three packs?
Lesson 29 Application Problem
Janet uses 4 feet of ribbon to decorate each pillow. The ribbon comes in 225-foot rolls. How many pillows will she be able to decorate with one roll of ribbon? Will there be any ribbon left over?
Lesson 29 Homework
1. Divide, then check using multiplication.
2. A truck has four crates of apples. Each crate has an equal number of apples. Altogether, the truck is carrying 1,728 apples. How many apples are in three crates?
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. | 357 | 1,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2017-17 | longest | en | 0.890903 |
https://mathspace.co/textbooks/syllabuses/Syllabus-875/topics/Topic-19465/subtopics/Subtopic-260342/?textbookIntroActiveTab=guide&activeTab=interactive | 1,643,271,296,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305242.48/warc/CC-MAIN-20220127072916-20220127102916-00283.warc.gz | 429,780,651 | 38,743 | # 2.04 Practical problems with rational numbers
## Interactive practice questions
Consider the following sentence:
$-6$6 is added to the product of $2$2 and $-4$4.
a
Without simplifying the result, translate this sentence into a mathematical expression.
b
Evaluate the expression.
Easy
Approx 2 minutes
Consider the following sentence:
The sum of $-8$8 and the quotient of $-40$40 and $4$4
Consider the following phrase:
The quotient of $-3$3 and the sum of $7$7 and $6$6 .
Determine whether the following statements are true or false:
### Outcomes
#### NY-7.NS.3
Solve real-world and mathematical problems involving the four operations with rational numbers. Notes: Computations with rational numbers extend the rules for manipulating fractions to complex fractions limited to (a/b)/(c/d) where a, b, c, and d are integers and b, c, and d ≠ 0.
#### NY-7.EE.3
Solve multi-step real-world and mathematical problems posed with positive and negative rational numbers in any form (whole numbers, fractions, and decimals), using tools strategically. Apply properties of operations to calculate with numbers in any form; convert between forms as appropriate. Assess the reasonableness of answers using mental computation and estimation strategies. | 287 | 1,259 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2022-05 | latest | en | 0.870612 |
http://rpg.stackexchange.com/questions/26289/drawing-a-weapon-and-attacking-the-same-turn | 1,466,857,047,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783393146.70/warc/CC-MAIN-20160624154953-00018-ip-10-164-35-72.ec2.internal.warc.gz | 256,124,046 | 20,834 | # Drawing a weapon and attacking the same turn
I'm playing Pathfinder for the first time and was wondering exactly what I can do with one turn in combat. My character has a reach weapon and so I an trying to figure out what my options are once a opponent has come within 5 ft. It looks like I can drop my current weapon (free action), draw another(movement action) and attack all in one turn. Is that correct?
Also on a side note, would I also get an attack of opportunity when a opponent leaves the area I threaten (10 ft from me) and comes within 5 ft? I assume because my reach weapon cannot be used at this range I do not threaten within 5 ft, and thus the enemy would be leaving a square I threaten?
Thanks!
-
My character has a reach weapon and so I an trying to figure out what my options are once a opponent has come within 5 ft. It looks like I can drop my current weapon (free action), draw another(movement action) and attack all in one turn. Is that correct?
Yes, that's accurate.
Would I also get an attack of opportunity when a opponent leaves the area I threaten (10 ft from me) and comes within 5 ft? I assume because my reach weapon cannot be used at this range I do not threaten within 5 ft, and thus the enemy would be leaving a square I threaten?
If they only take a 5-foot step, they don't provoke your opportunity attack since a 5-foot step never does that.
If they're not taking a 5-foot step: based on a question I just asked about it because it's really not clear, they do provoke an opportunity attack and you get to attack them whilst they're still at a distance at which you can attack.
-
The withdraw action would also allow somebody to leave the threatened square safely, but it wouldn't make much sense to approach somebody that way. It's more for running away. – Bradd Szonye Jun 12 '13 at 8:46
Withdraw only gets them out of their initial square for free. Someone starting adjacent to you (if you are wielding a reach weapon) would provoke an attack of opportunity on leaving the second square since you still threaten that square. – Jacob Proffitt Jun 12 '13 at 15:13
@JacobProffitt I was talking about somebody starting in the threatened ring, not adjacent to the opponent. – Bradd Szonye Jun 12 '13 at 18:27
@BraddSzonye Yeah, I just wanted to clarify given the author's original example of having a reach weapon. – Jacob Proffitt Jun 13 '13 at 15:09
Yes, you can combine any given set of standard, move, and free actions in that manner. Additionally, if you have a BAB of +1 or higher, you can draw a weapon as a free action when moving. So if you want to, you can drop a weapon, move + draw, and attack. Or you can get Quick Draw, drop, do some other move-equivalent action, draw, and attack.
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+1 for pointing out the move+draw option. – Bradd Szonye Jun 12 '13 at 8:35
Yes, you can drop your weapon (free), ready a new one (move), and attack (standard), all in the same turn. However, if you're switching between a reach weapon and a one-handed weapon, you can actually do better than that:
Two-Handed Weapons: What kind of action is it to remove your hand from a two-handed weapon or re-grab it with both hands? Both are free actions. For example, a wizard wielding a quarterstaff can let go of the weapon with one hand as a free action, cast a spell as a standard action, and grasp the weapon again with that hand as a free action; this means the wizard is still able to make attacks of opportunity with the weapon (which requires using two hands).
Therefore, you can remove your main hand from the reach weapon (free), ready a one-handed weapon (move), and attack (standard). When you want to use the reach weapon again, you can sheathe the backup weapon (move), put both hands back on the reach weapon (free), and attack (standard). You can switch between weapons and attack in the same turn, without needing to drop either.
Yes, opponents provoke an attack of opportunity if they pass through your threatened space:
Provoking an Attack of Opportunity: Two kinds of actions can provoke attacks of opportunity: moving out of a threatened square and performing certain actions within a threatened square.
Moving: Moving out of a threatened square usually provokes attacks of opportunity from threatening opponents. There are two common methods of avoiding such an attack—the 5-foot step and the withdraw action.
Note that this applies any time they leave a threatened square, regardless of whether they move toward you, away from you, or side to side. It's especially dangerous to move near somebody with reach from size or a spiked chain, since you can't easily escape with a 5-foot step or withdraw action.
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Scratch "many GMs" re holding in one hand – that's the official rule. – KRyan Jun 12 '13 at 13:34
@KRyan Thanks! That's the rule I used in D&D3, but I wasn't sure whether it was still correct for Pathfinder. – Bradd Szonye Jun 12 '13 at 18:01
This hasn't been answered so far, so i'll mention it.
You can also take a 5' step (free action) away from your opponent, and attack them with your reach weapon. This is the usual course of action for reach-weapon wielders - create some space, and strike at the foe.
You also intimated that you think the opponent must leave your entire threatened area before you get an attack of opportunity. In actuality, a foe leaving any threatened square provokes an Attack of Opportunity (although, keep in mind, you can only provoke one AoO from any individual character/creature per turn from movement - i.e. you can provoke from other things (like spellcasting), but leaving multiple threatened squares only ever provokes one attack per turn per attacker).
- | 1,302 | 5,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2016-26 | latest | en | 0.973987 |
https://aprove.informatik.rwth-aachen.de/eval/JAR06/JAR_INN/TRS/AG01/%233.52.trs.wst2005.html.lzma | 1,726,760,397,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652031.71/warc/CC-MAIN-20240919125821-20240919155821-00081.warc.gz | 78,457,595 | 1,903 | Term Rewriting System R:
[x, y, z]
f(0, 1, x) -> f(s(x), x, x)
f(x, y, s(z)) -> s(f(0, 1, z))
Innermost Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
F(0, 1, x) -> F(s(x), x, x)
F(x, y, s(z)) -> F(0, 1, z)
Furthermore, R contains one SCC.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Usable Rules (Innermost)`
Dependency Pairs:
F(x, y, s(z)) -> F(0, 1, z)
F(0, 1, x) -> F(s(x), x, x)
Rules:
f(0, 1, x) -> f(s(x), x, x)
f(x, y, s(z)) -> s(f(0, 1, z))
Strategy:
innermost
As we are in the innermost case, we can delete all 2 non-usable-rules.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳UsableRules`
` →DP Problem 2`
` ↳Size-Change Principle`
Dependency Pairs:
F(x, y, s(z)) -> F(0, 1, z)
F(0, 1, x) -> F(s(x), x, x)
Rule:
none
Strategy:
innermost
We number the DPs as follows:
1. F(x, y, s(z)) -> F(0, 1, z)
2. F(0, 1, x) -> F(s(x), x, x)
and get the following Size-Change Graph(s):
{1} , {1}
3>3
{2} , {2}
3=2
3=3
which lead(s) to this/these maximal multigraph(s):
{1} , {1}
3>3
{2} , {1}
3>3
{1} , {2}
3>2
3>3
DP: empty set
Oriented Rules: none
We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial
with Argument Filtering System:
s(x1) -> s(x1)
We obtain no new DP problems.
Innermost Termination of R successfully shown.
Duration:
0:00 minutes | 554 | 1,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-38 | latest | en | 0.684376 |
http://statkat.org/stattest.php?t=9&t2=12&t3=20 | 1,606,313,437,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141182794.28/warc/CC-MAIN-20201125125427-20201125155427-00401.warc.gz | 85,469,133 | 9,222 | # Two sample t test - equal variances not assumed - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
Two sample $t$ test - equal variances not assumed
Two way ANOVA
Logistic regression
Independent/grouping variableIndependent/grouping variablesIndependent variables
One categorical with 2 independent groupsTwo categorical, the first with $I$ independent groups and the second with $J$ independent groups ($I \geqslant 2$, $J \geqslant 2$)One or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variables
Dependent variableDependent variableDependent variable
One quantitative of interval or ratio levelOne quantitative of interval or ratio levelOne categorical with 2 independent groups
Null hypothesisNull hypothesisNull hypothesis
H0: $\mu_1 = \mu_2$
Here $\mu_1$ is the population mean for group 1, and $\mu_2$ is the population mean for group 2.
ANOVA $F$ tests:
• H0 for main and interaction effects together (model): no main effects and interaction effect
• H0 for independent variable A: no main effect for A
• H0 for independent variable B: no main effect for B
• H0 for the interaction term: no interaction effect between A and B
Like in one way ANOVA, we can also perform $t$ tests for specific contrasts and multiple comparisons. This is more advanced stuff.
Model chi-squared test for the complete regression model:
• H0: $\beta_1 = \beta_2 = \ldots = \beta_K = 0$
Wald test for individual regression coefficient $\beta_k$:
• H0: $\beta_k = 0$
or in terms of odds ratio:
• H0: $e^{\beta_k} = 1$
Likelihood ratio chi-squared test for individual regression coefficient $\beta_k$:
• H0: $\beta_k = 0$
or in terms of odds ratio:
• H0: $e^{\beta_k} = 1$
in the regression equation $\ln \big(\frac{\pi_{y = 1}}{1 - \pi_{y = 1}} \big) = \beta_0 + \beta_1 \times x_1 + \beta_2 \times x_2 + \ldots + \beta_K \times x_K$. Here $x_i$ represents independent variable $i$, $\beta_i$ is the regression weight for independent variable $x_i$, and $\pi_{y = 1}$ represents the true probability that the dependent variable $y = 1$ (or equivalently, the proportion of $y = 1$ in the population) given the scores on the independent variables.
Alternative hypothesisAlternative hypothesisAlternative hypothesis
H1 two sided: $\mu_1 \neq \mu_2$
H1 right sided: $\mu_1 > \mu_2$
H1 left sided: $\mu_1 < \mu_2$
ANOVA $F$ tests:
• H1 for main and interaction effects together (model): there is a main effect for A, and/or for B, and/or an interaction effect
• H1 for independent variable A: there is a main effect for A
• H1 for independent variable B: there is a main effect for B
• H1 for the interaction term: there is an interaction effect between A and B
Model chi-squared test for the complete regression model:
• H1: not all population regression coefficients are 0
Wald test for individual regression coefficient $\beta_k$:
• H1: $\beta_k \neq 0$
or in terms of odds ratio:
• H1: $e^{\beta_k} \neq 1$
If defined as Wald $= \dfrac{b_k}{SE_{b_k}}$ (see 'Test statistic'), also one sided alternatives can be tested:
• H1 right sided: $\beta_k > 0$
• H1 left sided: $\beta_k < 0$
Likelihood ratio chi-squared test for individual regression coefficient $\beta_k$:
• H1: $\beta_k \neq 0$
or in terms of odds ratio:
• H1: $e^{\beta_k} \neq 1$
AssumptionsAssumptionsAssumptions
• Within each population, the scores on the dependent variable are normally distributed
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
• Within each of the $I \times J$ populations, the scores on the dependent variable are normally distributed
• The standard deviation of the scores on the dependent variable is the same in each of the $I \times J$ populations
• For each of the $I \times J$ groups, the sample is an independent and simple random sample from the population defined by that group. That is, within and between groups, observations are independent of one another
• Equal sample sizes for each group make the interpretation of the ANOVA output easier (unequal sample sizes result in overlap in the sum of squares; this is advanced stuff)
• In the population, the relationship between the independent variables and the log odds $\ln (\frac{\pi_{y=1}}{1 - \pi_{y=1}})$ is linear
• The residuals are independent of one another
• Variables are measured without error
Also pay attention to:
• Multicollinearity
• Outliers
Test statisticTest statisticTest statistic
$t = \dfrac{(\bar{y}_1 - \bar{y}_2) - 0}{\sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}} = \dfrac{\bar{y}_1 - \bar{y}_2}{\sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}}$
Here $\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $s^2_1$ is the sample variance in group 1, $s^2_2$ is the sample variance in group 2, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2. The 0 represents the difference in population means according to the null hypothesis.
The denominator $\sqrt{\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}}$ is the standard error of the sampling distribution of $\bar{y}_1 - \bar{y}_2$. The $t$ value indicates how many standard errors $\bar{y}_1 - \bar{y}_2$ is removed from 0.
Note: we could just as well compute $\bar{y}_2 - \bar{y}_1$ in the numerator, but then the left sided alternative becomes $\mu_2 < \mu_1$, and the right sided alternative becomes $\mu_2 > \mu_1$.
For main and interaction effects together (model):
• $F = \dfrac{\mbox{mean square model}}{\mbox{mean square error}}$
For independent variable A:
• $F = \dfrac{\mbox{mean square A}}{\mbox{mean square error}}$
For independent variable B:
• $F = \dfrac{\mbox{mean square B}}{\mbox{mean square error}}$
For the interaction term:
• $F = \dfrac{\mbox{mean square interaction}}{\mbox{mean square error}}$
Note: mean square error is also known as mean square residual or mean square within.
Model chi-squared test for the complete regression model:
• $X^2 = D_{null} - D_K = \mbox{null deviance} - \mbox{model deviance}$
$D_{null}$, the null deviance, is conceptually similar to the total variance of the dependent variable in OLS regression analysis. $D_K$, the model deviance, is conceptually similar to the residual variance in OLS regression analysis.
Wald test for individual $\beta_k$:
The wald statistic can be defined in two ways:
• Wald $= \dfrac{b_k^2}{SE^2_{b_k}}$
• Wald $= \dfrac{b_k}{SE_{b_k}}$
SPSS uses the first definition.
Likelihood ratio chi-squared test for individual $\beta_k$:
• $X^2 = D_{K-1} - D_K$
$D_{K-1}$ is the model deviance, where independent variable $k$ is excluded from the model. $D_{K}$ is the model deviance, where independent variable $k$ is included in the model.
n.a.Pooled standard deviationn.a.
-\begin{aligned} s_p &= \sqrt{\dfrac{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2}{N - (I \times J)}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned} -
Sampling distribution of $t$ if H0 were trueSampling distribution of $F$ if H0 were trueSampling distribution of $X^2$ and of the Wald statistic if H0 were true
Approximately the $t$ distribution with $k$ degrees of freedom, with $k$ equal to
$k = \dfrac{\Bigg(\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}\Bigg)^2}{\dfrac{1}{n_1 - 1} \Bigg(\dfrac{s^2_1}{n_1}\Bigg)^2 + \dfrac{1}{n_2 - 1} \Bigg(\dfrac{s^2_2}{n_2}\Bigg)^2}$
or
$k$ = the smaller of $n_1$ - 1 and $n_2$ - 1
First definition of $k$ is used by computer programs, second definition is often used for hand calculations.
For main and interaction effects together (model):
• $F$ distribution with $(I - 1) + (J - 1) + (I - 1) \times (J - 1)$ (df model, numerator) and $N - (I \times J)$ (df error, denominator) degrees of freedom
For independent variable A:
• $F$ distribution with $I - 1$ (df A, numerator) and $N - (I \times J)$ (df error, denominator) degrees of freedom
For independent variable B:
• $F$ distribution with $J - 1$ (df B, numerator) and $N - (I \times J)$ (df error, denominator) degrees of freedom
For the interaction term:
• $F$ distribution with $(I - 1) \times (J - 1)$ (df interaction, numerator) and $N - (I \times J)$ (df error, denominator) degrees of freedom
Here $N$ is the total sample size.
Sampling distribution of $X^2$, as computed in the model chi-squared test for the complete model:
• chi-squared distribution with $K$ (number of independent variables) degrees of freedom
Sampling distribution of the Wald statistic:
• If defined as Wald $= \dfrac{b_k^2}{SE^2_{b_k}}$: approximately the chi-squared distribution with 1 degree of freedom
• If defined as Wald $= \dfrac{b_k}{SE_{b_k}}$: approximately the standard normal distribution
Sampling distribution of $X^2$, as computed in the likelihood ratio chi-squared test for individual $\beta_k$:
• chi-squared distribution with 1 degree of freedom
Significant?Significant?Significant?
Two sided:
Right sided:
Left sided:
• Check if $F$ observed in sample is equal to or larger than critical value $F^*$ or
• Find $p$ value corresponding to observed $F$ and check if it is equal to or smaller than $\alpha$
For the model chi-squared test for the complete regression model and likelihood ratio chi-squared test for individual $\beta_k$:
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
For the Wald test:
• If defined as Wald $= \dfrac{b_k^2}{SE^2_{b_k}}$: same procedure as for the chi-squared tests. Wald can be interpret as $X^2$
• If defined as Wald $= \dfrac{b_k}{SE_{b_k}}$: same procedure as for any $z$ test. Wald can be interpreted as $z$.
Approximate $C\%$ confidence interval for $\mu_1 - \mu_2$n.a.Wald-type approximate $C\%$ confidence interval for $\beta_k$
$(\bar{y}_1 - \bar{y}_2) \pm t^* \times \sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}$
where the critical value $t^*$ is the value under the $t_{k}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20).
The confidence interval for $\mu_1 - \mu_2$ can also be used as significance test.
-$b_k \pm z^* \times SE_{b_k}$
where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval).
n.a.Effect sizeGoodness of fit measure $R^2_L$
-
• Proportion variance explained $R^2$:
Proportion variance of the dependent variable $y$ explained by the independent variables and the interaction effect together:
\begin{align} R^2 &= \dfrac{\mbox{sum of squares model}}{\mbox{sum of squares total}} \end{align} $R^2$ is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population.
• Proportion variance explained $\eta^2$:
Proportion variance of the dependent variable $y$ explained by an independent variable or interaction effect:
\begin{align} \eta^2_A &= \dfrac{\mbox{sum of squares A}}{\mbox{sum of squares total}}\\ \\ \eta^2_B &= \dfrac{\mbox{sum of squares B}}{\mbox{sum of squares total}}\\ \\ \eta^2_{int} &= \dfrac{\mbox{sum of squares int}}{\mbox{sum of squares total}} \end{align} $\eta^2$ is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population.
• Proportion variance explained $\omega^2$:
Corrects for the positive bias in $\eta^2$ and is equal to:
\begin{align} \omega^2_A &= \dfrac{\mbox{sum of squares A} - \mbox{degrees of freedom A} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\ \\ \omega^2_B &= \dfrac{\mbox{sum of squares B} - \mbox{degrees of freedom B} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\ \\ \omega^2_{int} &= \dfrac{\mbox{sum of squares int} - \mbox{degrees of freedom int} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\ \end{align} $\omega^2$ is a better estimate of the explained variance in the population than $\eta^2$. Only for balanced designs (equal sample sizes).
• Proportion variance explained $\eta^2_{partial}$: \begin{align} \eta^2_{partial\,A} &= \frac{\mbox{sum of squares A}}{\mbox{sum of squares A} + \mbox{sum of squares error}}\\ \\ \eta^2_{partial\,B} &= \frac{\mbox{sum of squares B}}{\mbox{sum of squares B} + \mbox{sum of squares error}}\\ \\ \eta^2_{partial\,int} &= \frac{\mbox{sum of squares int}}{\mbox{sum of squares int} + \mbox{sum of squares error}} \end{align}
$R^2_L = \dfrac{D_{null} - D_K}{D_{null}}$
There are several other goodness of fit measures in logistic regression. In logistic regression, there is no single agreed upon measure of goodness of fit.
Visual representationn.a.n.a.
--
n.a.ANOVA tablen.a.
- -
n.a.Equivalent ton.a.
-OLS regression with two categorical independent variables and the interaction term, transformed into $(I - 1)$ + $(J - 1)$ + $(I - 1) \times (J - 1)$ code variables.-
Example contextExample contextExample context
Is the average mental health score different between men and women?Is the average mental health score different between people from a low, moderate, and high economic class? And is the average mental health score different between men and women? And is there an interaction effect between economic class and gender?Can body mass index, stress level, and gender predict whether people get diagnosed with diabetes?
SPSSSPSSSPSS
Analyze > Compare Means > Independent-Samples T Test...
• Put your dependent (quantitative) variable in the box below Test Variable(s) and your independent (grouping) variable in the box below Grouping Variable
• Click on the Define Groups... button. If you can't click on it, first click on the grouping variable so its background turns yellow
• Fill in the value you have used to indicate your first group in the box next to Group 1, and the value you have used to indicate your second group in the box next to Group 2
• Continue and click OK
Analyze > General Linear Model > Univariate...
• Put your dependent (quantitative) variable in the box below Dependent Variable and your two independent (grouping) variables in the box below Fixed Factor(s)
Analyze > Regression > Binary Logistic...
• Put your dependent variable in the box below Dependent and your independent (predictor) variables in the box below Covariate(s)
JamoviJamoviJamovi
T-Tests > Independent Samples T-Test
• Put your dependent (quantitative) variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable
• Under Tests, select Welch's
• Under Hypothesis, select your alternative hypothesis
ANOVA > ANOVA
• Put your dependent (quantitative) variable in the box below Dependent Variable and your two independent (grouping) variables in the box below Fixed Factors
Regression > 2 Outcomes - Binomial
• Put your dependent variable in the box below Dependent Variable and your independent variables of interval/ratio level in the box below Covariates
• If you also have code (dummy) variables as independent variables, you can put these in the box below Covariates as well
• Instead of transforming your categorical independent variable(s) into code variables, you can also put the untransformed categorical independent variables in the box below Factors. Jamovi will then make the code variables for you 'behind the scenes'
Practice questionsPractice questionsPractice questions | 4,432 | 15,895 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2020-50 | latest | en | 0.752409 |
http://www.ionizationx.com/index.php?PHPSESSID=2sehfecbcohag5famrf1umsch4&topic=2606.0 | 1,643,395,122,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306335.77/warc/CC-MAIN-20220128182552-20220128212552-00393.warc.gz | 98,639,232 | 7,289 | ### Author Topic: What else? (Read 4752 times)
0 Members and 1 Guest are viewing this topic.
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##### What else?
« on: January 24, 2013, 06:28:36 am »
E= 1J = 1W*1s Joules
L= E/A^2 = (1V*1s)/1A Henrie
C= E/V^2 = (1A*1s)/1V Faraday
t/2= Pi*sqrt(L*C) Second
A= V/R Ampere
XL= sqrt(L/C) Ohm
F= 1/t Hertz
V= Q/C Volt
http://en.wikipedia.org/wiki/Henry_(unit)
http://en.wikipedia.org/wiki/Farad
http://en.wikipedia.org/wiki/Joule
it= dQ/dt = C dV/dt
Vt= L di/dt
http://en.wikipedia.org/wiki/Inductor
http://en.wikipedia.org/wiki/Capacitor
τ= L/R = C*R
C= dQ/dV
« Last Edit: January 25, 2013, 05:03:29 am by sebosfato »
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• let the voltage do the work
##### Re: What else?
« Reply #1 on: January 24, 2013, 23:17:06 pm »
it's not tau it's τ
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##### 1amp aplied for 1 second into 1 faraday capacitor equal 1 volt...
« Reply #2 on: January 25, 2013, 05:07:29 am »
The definition says the voltage in a capacitor equals the charge applied to the capacitor.
1amp aplied for 1 second into 1 faraday capacitor equal 1 volt...
what is restriction of amps than?
A 1henrie inductor will have 1amp flowing thru it if 1 volt is applied for 1 second...
What is the doubled pulse than?
In a resonant circuit the energy becomes all magnetic at one time and all eletcric at the other...
A diode can prevent a capacitor from discharge....
than?
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##### Re: 1amp aplied for 1 second into 1 faraday capacitor equal 1 volt...
« Reply #3 on: January 26, 2013, 00:43:02 am »
The definition says the voltage in a capacitor equals the charge applied to the capacitor.
1amp aplied for 1 second into 1 faraday capacitor equal 1 volt...
what is restriction of amps than?
A 1henrie inductor will have 1amp flowing thru it if 1 volt is applied for 1 second...
What is the doubled pulse than?
In a resonant circuit the energy becomes all magnetic at one time and all eletcric at the other...
A diode can prevent a capacitor from discharge....
than?
A diode can prevent a perfect capacitor fom discharge. Since there is no perfect capacitor, then?
TS
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##### Re: What else?
« Reply #4 on: January 26, 2013, 17:55:55 pm »
to Ts,make a real water capacitor can solve that.
thanks
geenee
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##### Re: What else?
« Reply #5 on: January 26, 2013, 19:31:13 pm »
to Ts,make a real water capacitor can solve that.
thanks
geenee
Huh?
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##### Re: 1amp aplied for 1 second into 1 faraday capacitor equal 1 volt...
« Reply #6 on: January 27, 2013, 06:43:31 am »
The definition says the voltage in a capacitor equals the charge applied to the capacitor.
1amp aplied for 1 second into 1 faraday capacitor equal 1 volt...
what is restriction of amps than?
A 1henrie inductor will have 1amp flowing thru it if 1 volt is applied for 1 second...
What is the doubled pulse than?
In a resonant circuit the energy becomes all magnetic at one time and all eletcric at the other...
A diode can prevent a capacitor from discharge....
than?
Volts and Amps are different units that are not directly interchangeable. You cannot say 1A = 1V in any context. It's like saying 1 lb/sq inch air pressure equals 1 cubic inch of air volume. It doesn't make sense because the two units are not interchangeable this way.
TS
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##### Re: What else?
« Reply #7 on: January 28, 2013, 05:30:16 am »
of course, but i'm not saying that.
i'm only saying that for a capacitor of 1 faraday to get 1 volt it must flow 1 coulomb of charge, if its done in one second 1 ampere will be the average current...
thats the definition...
same for henrie
what is one henrie?
is the inductance that will let 1 ampere of current to flow if 1 volt is applied for 1 second..
the voltage across a capacitor will equal the charge stored in this capacitor
the current flowing in a coil similarly equal the voltage charge aplied.. | 1,258 | 4,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2022-05 | latest | en | 0.736934 |
https://photolens.tech/calculator-using-variable-names/ | 1,653,429,020,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662577259.70/warc/CC-MAIN-20220524203438-20220524233438-00431.warc.gz | 523,534,065 | 22,611 | # Calculator using variable names
I’ve recently been assigned an assignment to create a word calculator, and given my knowledge in python is still quite lacking, I want to ask if anybody has any better ideas for any possible solutions.
The question is here:
Jimmy has invented a new kind of calculator that works with words
rather than numbers.
• Input is read from stdin and consists of up to 1000 commands, one
per line.
• Each command is a definition, a calculation or `clear`.
• All tokens within a command are separated by single spaces.
• A definition has the format `def x y` where x is a variable name and y is an integer in the range [-1000, 1000].
• Existing definitions are replaced by new ones i.e. if x has been defined previously, defining x again erases its old definition.
• Variable names consist of 1-30 lowercase characters.
• No two variables are ever defined to have the same value at the
same time.
• The `clear` command erases all existing variable definitions.
• A calculation command starts with the word `calc`, and is followed
by one or more variable names separated by addition or subtraction
operators.
• The end of a calculation command is an equals sign.
The goal is to write a program for Jimmy’s calculator. Some rules are:
• The program should produce no output for definitions, but for calculations it should output the value of the calculation.
• Where there is no word for the result, or some word in a calculation has not been defined, then the output should be
`unknown`. (The word `unknown` is never used as a variable name.)
• Your solution may only import content from the `sys` module.
• Your solution may not use the `eval()` function.
Here is a sample input below:
``````calc foo + bar =
def bar 7
def programming 10
calc foo + bar =
def is 4
def fun 8
calc programming - is + fun =
def fun 1
calc programming - is + fun =
clear
``````
And the corresponding output:
``````foo + bar = unknown
foo + bar = programming
programming - is + fun = unknown
programming - is + fun = bar
``````
My current solution is this:
``````import sys
d_define = {}
numsInDict = []
operators = ["+", "-", "="]
# IN : word to checked it is in the dictionary
# OUT : returns unknown or value for word
def lookup(word):
if word in d_define:
return d_define[word]
else:
return "unknown"
# IN : answer to check if there is a word assigned to it a dictionary
# OUT : returns unknown or word assigned to answer
for k, v in d_define.items():
return k
return "unknown"
# IN : All values to calc (includes operators)
# OUT : print unknown or word if in dict
def calc(args):
equation = 0
lastOperator = "+"
for word in args:
if word not in operators:
res = lookup(word)
if res == "unknown":
return "unknown"
else:
#print(res)
if lastOperator == "+":
equation += res
else:
equation -= res
else:
lastOperator = word
if equation in numsInDict:
return res
else:
return "unknown"
# IN : word to be added and its value
# OUT : updated dictionary
def define(word, num):
num = int(num)
if num in numsInDict or word in d_define:
# print(f'NEEDS REPLACE')
# print(f'same value -> {num in numsInDict}')
# print(f'same word -> {word in d_define}')
# print(f'same word -> {d_define}')
# print(f'same word -> {numsInDict}')
topop = ""
for k, v in d_define.items():
if k == word:
d_define[word] = num # Update Word with new value
elif v == num:
topop = k # Saves value to pop later
if topop != "":
d_define.pop(topop)
d_define[word] = num
else:
d_define[word] = num
numsInDict.append(num)
#print(f'{word} - {d_define[word]}')
for line in lines:
#print(f'-------------------------------------- LOOP START ------------------------------------')
line = line.rstrip().split()
#print(f'Line Split - {line}')
if len(line) == 3 and line[0] == "def":
define(line[1], line[2])
elif len(line) > 1 and line[len(line) - 1] == "=":
result = calc(line[1:])
print(f'{" ".join(line[1:]) + " " + result}')
elif len(line) == 1 and line[0] == "clear":
d_define = {}
wordsInDict = []
numsInDict = []
#print(f'Cleared d_define - {d_define} {wordsInDict}')
#print(d_define)
#print(f'--------------------------------------- LOOP END -------------------------------------')
``````
It does feel quite clunky but it gets the job done. I am just wondering if anybody has any better ways in which it could be improved upon.
I think your solution does work, but it’s very long-winded. Your `IN: ` and `OUT: ` comments should be moved to docstrings `""" """` in the first line of the function body.
`getAnswer` should be `get_answer` by PEP8.
Your `getAnswer` doesn’t need to loop if you maintain an inverse dictionary of values to names.
Consider using the built-in `operator.add` and `.sub`. This violates the letter (though perhaps not the spirit) of Your solution may only import content from the sys module; if that’s actually a problem just define the `add` and `sub` functions yourself.
Consider writing unit tests to validate your output. This can be done by operating on the `stdin` and `stdout` streams passed as parameters, or yielding lines.
## Suggested
``````from operator import add, sub
from typing import Iterable, Iterator
def rewritten(in_stream: Iterable[str]) -> Iterator[str]:
OPS = {'+': add, '-': sub}
state, inverse_state = {}, {}
for line in in_stream:
command, *args = line.split()
if command == 'def':
name, val = args
val = int(val)
state[name] = val
inverse_state[val] = name
elif command == 'clear':
state.clear()
inverse_state.clear()
elif command == 'calc':
result, *values = (state.get(name) for name in args[::2])
if result is not None:
for op, value in zip(args[1:-1:2], values):
if value is None:
result = None
break
result = OPS[op](result, value)
prefix = line.rstrip().split(maxsplit=1)[1]
result = inverse_state.get(result, 'unknown')
yield f'{prefix} {result}'
def test() -> None:
sample_in = (
'''def foo 3
calc foo + bar =
def bar 7
def programming 10
calc foo + bar =
def is 4
def fun 8
calc programming - is + fun =
def fun 1
calc programming - is + fun =
clear'''
).splitlines()
actual = '\n'.join(rewritten(sample_in))
assert actual == (
'''foo + bar = unknown
foo + bar = programming
programming - is + fun = unknown
programming - is + fun = bar'''
)
if __name__ == '__main__':
test()
``````
## Even more lookups
As @Stef suggests, it is possible to add a lookup for the command. It’s awkward, because only one of the three functions actually produces a result. One way to express this is an unconditional outer `yield from`, and all functions as generators, only one being non-empty. I don’t particularly recommend this; it’s just a demonstration:
``````def rewritten(in_stream: Iterable[str]) -> Iterator[str]:
def clear() -> Iterator[str]:
state.clear()
inverse_state.clear()
return; yield
def define(name: str, val: str) -> Iterator[str]:
val = int(val)
state[name] = val
inverse_state[val] = name
return; yield
def calc(*args: str) -> Iterator[str]:
result, *values = (state.get(name) for name in args[::2])
if result is not None:
for op, value in zip(args[1:-1:2], values):
if value is None:
result = None
break
result = OPS[op](result, value)
prefix = line.rstrip().split(maxsplit=1)[1]
result = inverse_state.get(result, 'unknown')
yield f'{prefix} {result}'
OPS = {'+': add, '-': sub}
COMMANDS = {'clear': clear, 'def': define, 'calc': calc}
state, inverse_state = {}, {}
for line in in_stream:
parts = line.split()
command, *args = parts
yield from COMMANDS[command](*args)
``````
Or just use `Optional`s:
``````def rewritten(in_stream: Iterable[str]) -> Iterator[str]:
def clear() -> Optional[str]:
state.clear()
inverse_state.clear()
def define(name: str, val: str) -> Optional[str]:
val = int(val)
state[name] = val
inverse_state[val] = name
def calc(*args: str) -> Optional[str]:
result, *values = (state.get(name) for name in args[::2])
if result is not None:
for op, value in zip(args[1:-1:2], values):
if value is None:
result = None
break
result = OPS[op](result, value)
prefix = line.rstrip().split(maxsplit=1)[1]
result = inverse_state.get(result, 'unknown')
return f'{prefix} {result}'
OPS = {'+': add, '-': sub}
COMMANDS = {'clear': clear, 'def': define, 'calc': calc}
state, inverse_state = {}, {}
for line in in_stream:
parts = line.split()
command, *args = parts
result = COMMANDS[command](*args)
if result is not None:
yield result
`````` | 2,151 | 8,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-21 | longest | en | 0.898792 |
http://forums.worden.com/keeploggedin.aspx?g=posts&t=69566 | 1,555,908,936,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578534596.13/warc/CC-MAIN-20190422035654-20190422061654-00463.warc.gz | 74,123,806 | 16,350 | Minimize Sorting Time for an Indicator Rate this Topic:
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danielbender
Posted : Wednesday, May 23, 2018 12:30:43 PM
Platinum Customer
Joined: 1/14/2006
Posts: 436
Bruce
How do I optimize the following indicator to only look at the last 500 bars, to minimize the time needed to sort?
Thank you... Dan
++++++++++
'|******************************************************************
'|*** StockFinder RealCode Indicator - Version 5.1 www.worden.com
'|*** Copy and paste this header and code into StockFinder *********
'|*** Indicator:UpBar Count
'|*** Example: plot = price.close - price.close(1)
'|******************************************************************
'# Component = UserInput.Integer = 1
Static Count As Single
Static Marker As Single
Static Lo As Single
Static Entry As Single
Static Pop As Single
If isfirstBar Then
Count = Single.NaN
Marker = 0
Pop = 0
Else If (Marker = 0 AndAlso _
(SMI.value > SMI.Value(1) AndAlso SMI.Value(1) < 0 OrElse _
Price.Close - Price.Open > 3 * (Price.Close(1) - Price.Open(1))))Then
Marker = 1
Count = 1
Lo = Price.Low
Else If (Marker = 1 AndAlso _
(Price.High > Price.High(1) AndAlso Price.Low > Price.Low(1) OrElse _
Price.Open > Price.Open(1) AndAlso Price.Close > Price.Close(1) OrElse _
Price.Close >= 1.0 * Price.XAVGC(8))) Then
Count += 1
If Count = 3 Then
Pop = Price.Close - Lo
Entry = Price.Close
End If
Else
Count = 0
Marker = 0
Pop = 0
End If
'If Count >= 2 Then
' Plot = Count
'Else
' Plot = 0
'End If
If Component = 1 Then
Plot = Count
Else If Component = 2 Then
If Count > 3
Plot = Price.Close - Entry
Else
Plot = 0
End If
End If
Bruce_L
Posted : Wednesday, May 23, 2018 12:56:21 PM
Worden Trainer
Joined: 10/7/2004
Posts: 65,138
You won't be able to do anything to the code as the SMI indicator should have the '#Cumulative property since it is based on exponential moving averages. This means it will try to use all of the available data.
There are two ways to limit the amount of data used.
You can go to Settings | Data Manager and adjust the number of bars for the entire program.
You can edit the Block Diagram of an indicator to add a Length Limit Block to the Length Limit input of any Prices or Volume bars in a Block Diagram.
Unfortunately while you may be able to edit the Block Diagram for the RealCode Indicator (or possibly not), it technically isn't editable and such changes will not survive a program restart (you might have better luck editing the Block Diagram for the SMI indicator itself).
-Bruce
Personal Criteria Formulas
TC2000 Support Articles
danielbender
Posted : Wednesday, May 23, 2018 5:23:34 PM
Platinum Customer
Joined: 1/14/2006
Posts: 436
Bruce, If I did not have SMI indicator in the indicator, could I limit the amount of data backtested?
Bruce_L
Posted : Thursday, May 24, 2018 9:36:28 AM
Worden Trainer
Joined: 10/7/2004
Posts: 65,138
Without a '#Cumulative indicator (or tag in the RealCode), a WatchList Column / sort will automatically try to minimize the amount of data used (usually drops to about 50 bars).
You can override the number of bars to use by adding something similar to the following between the Inherits and Public Overrides lines in the Class tab of the RealCode Editor (what to use instead of 20 is up to you):
Sub New
NumOfBars = 20
End Sub
The RealCode Programmers Reference covers this and more advanced techniques such as creating your own manual loops instead of using the autoloop in more detail.
-Bruce
Personal Criteria Formulas
TC2000 Support Articles
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Forum Jump Customer Training & Support - Ask a Trainer - TC2000 version 12/18 - Ask a Trainer - TC2000 version 7 - Ask a Trainer - StockFinder 5.0 - PCFs, EasyScan and Custom Indicators General Discussions - Stock and Market Talk - TC2000 version 12 or 18 - TC2000 version 7 - StockFinder 5.0 - RealCode for StockFinder 5.0 Tutorial Videos - TC2000 version 12 tutorial videos - TC2000 version 7 tutorial videos You cannot post new topics in this forum. You cannot reply to topics in this forum. You cannot delete your posts in this forum. You cannot edit your posts in this forum. You cannot create polls in this forum. You cannot vote in polls in this forum. | 1,119 | 4,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-18 | latest | en | 0.695668 |
https://blog.taiwolskit.com/leetcode-solution-medium-725-split-linked-list-in-parts | 1,726,313,927,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00004.warc.gz | 119,397,800 | 43,341 | # 725. Split Linked List in Parts
### 題目敘述
Given the `head` of a singly linked list and an integer `k`, split the linked list into `k` consecutive linked list parts.
The length of each part should be as equal as possible: no two parts should have a size differing by more than one. This may lead to some parts being null.
The parts should be in the order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal to parts occurring later.
Return an array of the `k` parts.
Example 1:
``````Input: head = [1,2,3], k = 5
Output: [[1],[2],[3],[],[]]
Explanation:
The first element output[0] has output[0].val = 1, output[0].next = null.
The last element output[4] is null, but its string representation as a ListNode is [].
``````
Example 2:
``````Input: head = [1,2,3,4,5,6,7,8,9,10], k = 3
Output: [[1,2,3,4],[5,6,7],[8,9,10]]
Explanation:
The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts.
``````
Constraints:
• The number of nodes in the list is in the range `[0, 1000]`.
• `0 <= Node.val <= 1000`
• `1 <= k <= 50`
Hint 1:
If there are N nodes in the list, and k parts, then every part has N/k elements, except the first N%k parts have an extra one.
### 解法解析
Time complexity：`O(N + k)`. Space complexity：`O(k)`
#### 程式範例
##### Python
``````# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def splitListToParts(self, head: Optional[ListNode], k: int) -> List[Optional[ListNode]]:
for N in range(1001):
if not cur:
break
cur = cur.next
width, remainder = divmod(N, k)
ans = []
for i in range(k):
tmp = cur
for _ in range(width + (i < remainder) - 1):
if cur:
cur = cur.next
if cur:
cur.next, cur = None, cur.next
ans.append(tmp)
return ans
``````
##### JavaScript
``````/**
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {number} k
* @return {ListNode[]}
*/
var splitListToParts = function (head, k) {
let len = 0;
while (cur) {
++len;
cur = cur.next;
}
let remainder = 0;
let width = Math.floor(len / k);
if (width === 0) width = 1;
else remainder = len % k;
let result = [];
for (let i = 0; i < k; i++) {
const tmp = cur;
let j = 0;
while (j < width + (i < remainder) - 1) {
if (cur) {
cur = cur.next;
}
j++;
}
if (cur) {
const temp = cur.next;
cur.next = null;
cur = temp;
}
result.push(tmp);
}
return result;
};
``````
##### Go
``````/**
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func splitListToParts(head *ListNode, k int) []*ListNode {
len := 0
for cur := head; cur != nil; cur = cur.Next {
len++
}
width, remainder := len/k, len%k
ans := make([]*ListNode, k)
for i := 0; i < k && cur != nil; i++ {
ans[i] = cur
var rc int
if remainder > 0 {
rc = 1
}
for j := 1; j < width+rc; j++ {
cur = cur.Next
}
remainder--
t := cur.Next
cur.Next = nil
cur = t
}
return ans
}
``````
##### Swift
``````/**
* public class ListNode {
* public var val: Int
* public var next: ListNode?
* public init() { self.val = 0; self.next = nil; }
* public init(_ val: Int) { self.val = val; self.next = nil; }
* public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
* }
*/
class Solution {
func splitListToParts(_ head: ListNode?, _ k: Int) -> [ListNode?] {
var ret = [ListNode?]()
var N = 0
while node != nil {
node = node!.next
N += 1
}
var r = N % k, q = N / k
for i in 0..<k {
ret.append(cur)
var size = q + (i < r ? 1 : 0)
var j = 1
while j < size {
cur = cur?.next
j += 1
}
let next = cur?.next
cur?.next = nil
cur = next
}
return ret
}
}
``````
##### Kotlin
``````/**
* Example: var li = ListNode(5) var v = li.`val` Definition for singly-linked list. class
* ListNode(var `val`: Int) {
*
``````
• var next: ListNode? = null
• ```
• } */ class Solution { fun splitListToParts(head: ListNode?, k: Int): Array {
`````` val listSize = size(head)
val splitSize = listSize / k
val plusOne = listSize % k
val result = Array<ListNode?>(k) { null }
for (i in 1..k) {
result[i - 1] = itr
itr = run(itr, splitSize - 1 + if (plusOne >= i) 1 else 0)
val next = itr?.next
itr?.next = null
itr = next
}
return result
``````
}
private fun size(head: ListNode?): Int {
`````` var count = 0
while (itr != null) {
itr = itr.next
count++
}
return count
``````
}
private fun run(head: ListNode?, num: Int): ListNode? {
`````` var itr = head
repeat(num) { itr = itr?.next }
return itr
``````
} } ``` | 1,473 | 4,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-38 | latest | en | 0.824082 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1-common-core-15th-edition/chapter-7-exponents-and-exponential-functions-7-2-multiplying-powers-with-the-same-base-practice-and-problem-solving-exercises-page-431/56 | 1,685,952,842,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224651325.38/warc/CC-MAIN-20230605053432-20230605083432-00159.warc.gz | 853,849,598 | 14,130 | ## Algebra 1: Common Core (15th Edition)
$A$
$...\times 10^{-1}$ means "move the decimal point one place to the left" $2.06\rightarrow\qquad .206\rightarrow\qquad 0.206\qquad ...\qquad A$ | 64 | 188 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-23 | latest | en | 0.582945 |
https://us.metamath.org/nfegif/sbcimg.html | 1,716,835,229,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.25/warc/CC-MAIN-20240527175049-20240527205049-00750.warc.gz | 496,871,139 | 4,110 | New Foundations Explorer < Previous Next > Nearby theorems Mirrors > Home > NFE Home > Th. List > sbcimg Unicode version
Theorem sbcimg 3087
Description: Distribution of class substitution over implication. (Contributed by NM, 16-Jan-2004.)
Assertion
Ref Expression
sbcimg
Proof of Theorem sbcimg
Dummy variable is distinct from all other variables.
StepHypRef Expression
1 dfsbcq2 3049 . 2
2 dfsbcq2 3049 . . 3
3 dfsbcq2 3049 . . 3
42, 3imbi12d 311 . 2
5 sbim 2065 . 2
61, 4, 5vtoclbg 2915 1
Colors of variables: wff setvar class Syntax hints: wi 4 wb 176 wceq 1642 wsb 1648 wcel 1710 wsbc 3046 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1546 ax-5 1557 ax-17 1616 ax-9 1654 ax-8 1675 ax-6 1729 ax-7 1734 ax-11 1746 ax-12 1925 ax-ext 2334 This theorem depends on definitions: df-bi 177 df-or 359 df-an 360 df-tru 1319 df-ex 1542 df-nf 1545 df-sb 1649 df-clab 2340 df-cleq 2346 df-clel 2349 df-nfc 2478 df-v 2861 df-sbc 3047 This theorem is referenced by: sbceqal 3097 sbcimdv 3107 sbc19.21g 3110 sbcss 3660 iota4an 4358
Copyright terms: Public domain W3C validator | 517 | 1,156 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-22 | latest | en | 0.199217 |
https://learner.org/courses/learningmath/measurement/session7/solutions_a.html | 1,571,206,188,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986664662.15/warc/CC-MAIN-20191016041344-20191016064844-00382.warc.gz | 552,786,185 | 8,891 | Teacher resources and professional development across the curriculum
Teacher professional development and classroom resources across the curriculum
Solutions for Session 7, Part A
See solutions for Problems: A1 | A2 | A3 | A4 | A5 | A6 | A7 | A8 | A9 | A10
Problem A1
Here is the completed table:
Design 1
Design 2
Design 3
Diameter of the Circle 2 cm 4 cm 6 cm Perimeter of the Hexagon 6 cm 12 cm 18 cm Perimeter of the Square 8 cm 16 cm 24 cm Approximate Circumference of the Circle 6.3 cm 12.6 cm 18.9 cm
Problem A2 The measurements stay in scale. In all three, the diagonal of the hexagon is twice the length of the hexagon's side. Also, as we move from one design to the next, the length of each side of the inscribed hexagon increases by 1; the length of each side of the inscribed hexagon is equal to the radius of the circle (as shown by the inscribed equilateral triangles): The length of each side of the square is the same as the diameter of the circle inscribed within; the ratio of the length of the diameter of the circle to the length of one side of the hexagon is 2/1 for all three designs.
Problem A3
a. The perimeter of the hexagon is three times the diameter of the circle, and the perimeter of the square is four times the diameter of the circle. b. The circumference of the circle is between these two values, and closer to the hexagon's perimeter. c. The circumference appears to be between 3.1 and 3.2 times larger than the diameter. If a circle had a diameter of 7 cm, you might predict its circumference to be somewhere near 22 cm. As we explore this further, we will see that the relationship between the circumference and diameter is a constant value.
Problem A4
a. Answers will vary. b. The measured ratio of C/d should be approximately the same for all circular objects. It seems that the relationship between diameter and circumference is linear, and there is some number k so that C = k • d for every circle's circumference and diameter. For now, we can say that this number is just slightly larger than 3.
Problem A5
a. Answers will vary. The points should roughly form a straight line. If you were to place a line of best fit onto your scatter plot, the line would be y = 3.14 • x. Notice that the ratio C/d is about 3.14 no matter what the size of the circle. This is called a constant ratio since the value is constant, regardless of the circle. Constant change is represented by a straight-line graph and is sometimes referred to as a linear relationship. If the ratio between circumference and diameter differed for every circle, the graph would not be a straight line. b. This suggests that the diameter and circumference are in direct variation; that is, the circumference is a direct multiple of the diameter. Note that since you measured the circumference and diameter, there are likely to be measurement errors which will affect the graphed data.
Problem A6 Answers will vary, but should be close to 3.14 (an approximation for ). Finding the mean minimizes any measurement errors in the calculations of Problem A5.
Problem A7 We've seen that = C/d, where C is the circumference and d is the diameter of a circle. We can multiply both sides of the equation by d to get a new equation as C = d. Because the diameter of a circle is always twice its radius, we can write the new equation as C = • 2r, which is what we wanted.
Problem A8 These forms make calculations involving easier by using an approximation. In cases where there may already be measurement error, it doesn't make sense to use an overly accurate version of . Fractions like 22/7 or decimals like 3.14 do the job nicely in different situations. In practical terms, it is impossible to buy, for example, a length of fencing that measures 4. In applications, we often want an approximation that we can measure and work with. In mathematics problems, however, it is almost always preferable to use the symbol .
Problem A9 One or the other may be rational, but not both. If they were both rational, their ratio (which is ) would also have to be rational, which it is not. A circle may have a diameter of exactly 12 cm with an irrational circumference, or a circumference of exactly 100 m with an irrational diameter. They can, however, both be irrational.
Problem A10 This answer provides the only way to write the answer exactly. Additionally, it is easier to perform arithmetic with 4 than with a decimal approximation of it. An approximation may be substituted later if needed. | 1,022 | 4,512 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2019-43 | latest | en | 0.921023 |
http://drkwongolf.info/biom/grf.html | 1,695,632,315,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233508959.20/warc/CC-MAIN-20230925083430-20230925113430-00433.warc.gz | 14,726,779 | 3,256 | Home > Foundation > Ground Reaction Force/Moment Ground Reaction Force and Moment Ground reaction force is the reaction supplied by the ground: i.e., the reaction to the force the golfer exerts to the ground. The foot interacts with the ground through the entire contact area (Figure 1A). The blue arrows in Figure 1A are the forces the foot exerts to the ground while the black arrows are their reactions. This multi-force system (formed by individual reaction forces) can be reduced to a simpler but mechanically equivalent system with a net force (ground reaction force; GRF; F in Figure 1B) and a net moment (ground reaction moment; GRM; T in Figure 1B) only. GRF is the sum of all reaction forces acting on the foot. GRM, on the other hand, is the sum of the moments produced by the individual reaction forces about the center of pressure (COP). COP is GRF's point of action. Figure 1. Ground reaction force (F) and moment (T), and center of pressure (COP) COP is not just an arbitrary point within the contact area but a unique point that should suffice specific requirements. As shown in Figure 2A, each reaction force () generates a moment about the COP (). GRM is the sum of all these. COP in fact is the point of balance of these moments so the horizontal components of the net moment become zero (Figure 2B). As a result, GRM has only one non-zero component, the vertical component (). Figure 2. Center of pressure (COP). Ground reaction moment (A) is sum of the moments produced by individual reaction forces about the COP. The horizontal components of the GRM are all zero, which makes the COP the balance point of the horizontal moments (B). A typical pressure mat system (such as Swing Catalyst®) has an array of sensors and measures the vertical forces acting on the individual sensors. The location of the COP can be computed from these vertical forces (Figure 3). This is beacuse the vertical forces acting on the ground () can still generate horizontal moments about a given point on the ground within the contact area and the COP can be located by balancing the horizontal moments. In this case, however, computation of the vertical moment () is not possible as the vertical forces acting on the ground cannot generate moment about the vertical axis. A pressure mat provides the COP location and the vertical component of the GRF as well as the pressure map. Figure 3. COP revisited. A pressure mat (or plate) provides the COP location as it can be computed from the vertical components of the reaction forces acting under the foot. The GRFs acting on individual feet can be combined into single force, combined GRF (Figure 4). The combined GRF acts at the combined COP. The combined COP must lie on the straight line connecting the individual COPs (lead side and trail side). The location of the combined COP is essentially determined by the magnitudes of the vertical GRFs acting on the individual feet. For example, if 80% and 20% of the combined vertical force are acting on the trail foot and the lead foot, respectively, the combined COP divides the line connecting the individual COPs 20:80. This way the horizontal moments generated by the individual GRFs about the combined COP become balanced. Figure 4. Combined COP. The location of the cpmbined COP reflects the relative magnitudes of the vertical GRFs acting on the feet. Remember here that the location of the combined COP only reflects the relative magnitudes of the vertical forces acting on the individual feet and has nothing to do with the position of the center of mass (COM) of the golfer's body. Therefore, 'COP shift' and 'weight shift' shouldn't be used synonymously. A pressure mat reports the combined and individual COP locations and the combined and individual vertical forces. A 'force plate' reports the combined GRF/GRM (all three components) and the combined COP location. Use of two force plates, one foot on each plate (which is the standard in typical biomechanical research), enables the calculation of various moments golfers generate through the golfer-ground interaction. See the Golfer-Ground Interaction Moments page for this. Top | 877 | 4,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-40 | latest | en | 0.911124 |
https://www.bartleby.com/essay/W3-Quiz-Solutions-F36J2PCTC | 1,556,275,186,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578765115.93/warc/CC-MAIN-20190426093516-20190426115516-00147.warc.gz | 628,767,373 | 12,223 | # W3 Quiz Solutions Essay
668 Words Feb 12th, 2015 3 Pages
BUSN 5200 Quiz for week 3
Note: In the questions below, the correct answer is identified with an asterisk
1. (See Exhibit 2-2 on page 58 in your BUSN 5200 Custom text Version 2) Assume Main Street Store’s Net Sales in 2010 were \$1,000,000 and it’s Net Income in 2010 was \$17,000. Thus, between 2010 and 2011 Main Street Store’s net sales increased 20%. During the same period what percentage did net income increase?
a. 5.6% From Exhibit 2-2 on page 58:
b. 17.0% 2010 net income: \$17,000 2011 net income: \$18,000 c. 5.9% * Change: \$ 1,000 Percent change = \$1,000 / \$17,000 = .0588, or 5.9% d. 94.4%
2. (See Exhibit 2-2 on page 58 in your BUSN 5200 Custom text Version 2) In 2011 Main Street Store’s
b. Paying off debt no pay down of debt in 2011
c. Increasing inventory * \$(170,000) See Exhibit 2-4 on page 62
d. Paying dividends * \$(5,000)
7. (See Exhibit 2-4 on page 62 in your BUSN 5200 Custom text Version 2) What happened to Main Street Store’s Cash Account in 2011?
a. It increased * From Exhibit 2-4 on page 62: Increase in cash during 2011 = \$34,000 b. It decreased
c. It stayed the same
d. You can’t tell without further information
8. Which of the following is true about McDonalds Corporation?
a. Between 2010 and 2011 total revenue increased by a higher percentage than net income. * 2010 2011 Change Percent Change McDonalds Total Revenue \$24,074.6 \$27,006.0 \$2,931.4 .1218, or 12.18% McDonalds Net Income \$4,946.3 \$5,503.1 \$556.8 .1126, or 11.26% http://www.sec.gov/Archives/edgar/data/63908/000119312511046701/d10k.htm#toc37660_19 Scroll down to page 28
b. Between 2010 and 2011 total revenue increased by a lower percentage than net income.
c. McDonalds’ sales decreased between 2010 and 2011.
d. McDonalds’ net income decreased between 2010 and 2011.
9. McDonalds’ net worth (in millions) at the end of 2011 was:
a. \$16.6
b. \$0
c. \$14,390.2 * See http://www.sec.gov/Archives/edgar/data/63908/000119312511046701/d10k.htm#toc37660_19
Scroll down to page 29
d. 32,989.9
10. Which of the following is | 702 | 2,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2019-18 | longest | en | 0.842003 |
https://www.physicsforums.com/threads/expression-for-the-force-on-a-particle.219620/ | 1,713,013,343,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816734.69/warc/CC-MAIN-20240413114018-20240413144018-00100.warc.gz | 919,551,207 | 16,309 | # Expression for the force on a particle
• Nightrider55
In summary, the problem is asking for the force exerted on a particle at a given time, given its momentum at that time. This can be found using Newton's second law, which states that force equals the rate of change of momentum. Therefore, the expression for Fx(t) is 12t kg m/s^2. This may seem like a simple problem, but it can be more difficult if not familiar with calculus and the general form of Newton's law.
Nightrider55
## Homework Statement
A particle of mass m is at rest at t=0. Its momentum for t>0 is given by Px=6t^2 kg m/s, where t is in s. Find an expression for Fx(t), the force exerted on the particle as a function of time.
Px=MVx
## The Attempt at a Solution
The question seems really simple but momentum confuses me. I know that momentum is related to the area under the Fx(t)curve between Ti and Tf by Pfx-Pix or the change in momentum, but I don't know where to go from there. Something to get me headed in the right direction would be greatly appreciated!
Hint: Newton's second law states that:
$$\vec{F} = \frac{d \vec{p}}{dt}$$
so then it would be F=12t. That seems a little too easy. When they say Px would I also have to find Py in order to find the mag of P?
Nightrider55 said:
so then it would be F=12t. That seems a little too easy.
Maybe it is. What level of physics are you taking? Is it calculus-based? It's clear that you already know how to differentiate. If your prof expected you to be familar with calculus and to be aware that F = dp/dt was the true (most general) form of Newton's law, then yes, he has assigned a trivial problem. However, if you prof did not expect you to be familiar with calculus or F = dp/dt, then maybe he thought he had given you a stumper.
By the way, WERE you aware that F = dp/dt before I told you? If not, can you see that F = ma follows from this relation provided the mass of the particle is constant?
Nightrider55 said:
When they say Px would I also have to find Py in order to find the mag of P?
No. Why would you? They don't ask you for P or Py. They only ask for Px.
cepheid said:
Maybe it is. What level of physics are you taking? Is it calculus-based? It's clear that you already know how to differentiate. If your prof expected you to be familar with calculus and to be aware that F = dp/dt was the true (most general) form of Newton's law, then yes, he has assigned a trivial problem. However, if you prof did not expect you to be familiar with calculus or F = dp/dt, then maybe he thought he had given you a stumper.
By the way, WERE you aware that F = dp/dt before I told you? If not, can you see that F = ma follows from this relation provided the mass of the particle is constant?
No. Why would you? They don't ask you for P or Py. They only ask for Px.
I am in calculus physics and I talked to my teacher and he didn't realize that he assigned such an easy problem :rofl:
I was aware that F = dp/dt before you told me but I didn't use it because I thought I was missing something because it seemed too simple of a problem considering it was among the more difficult problems at the end of the chapter.
## 1. What is the expression for the force on a particle?
The expression for the force on a particle is F = ma, where F is the force, m is the mass of the particle, and a is the acceleration.
## 2. How is the expression for the force on a particle derived?
The expression for the force on a particle is derived from Newton's Second Law of Motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
## 3. Can the expression for the force on a particle be used for all types of forces?
Yes, the expression for the force on a particle can be used for all types of forces, including gravitational, electromagnetic, and frictional forces.
## 4. What are the units of the expression for the force on a particle?
The units of the expression for the force on a particle are Newtons (N), which is equivalent to kg*m/s^2.
## 5. How is the expression for the force on a particle used in real-world applications?
The expression for the force on a particle is used in a variety of real-world applications, such as calculating the forces acting on a rocket during launch, determining the impact of a collision between two objects, and analyzing the motion of objects in a gravitational field.
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6K | 1,229 | 4,984 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-18 | latest | en | 0.967785 |
https://www.nag.co.uk/numeric/cl/nagdoc_latest/examples/source/f08zece.c.html | 1,511,496,203,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807084.8/warc/CC-MAIN-20171124031941-20171124051941-00080.warc.gz | 814,597,950 | 3,572 | ```/* nag_dggqrf (f08zec) Example Program.
*
* Copyright 2017 Numerical Algorithms Group.
*
* Mark 26.1, 2017.
*/
#include <stdio.h>
#include <nag.h>
#include <nag_stdlib.h>
#include <nagf07.h>
#include <nagf08.h>
#include <nagf16.h>
int main(void)
{
/* Scalars */
double alpha, beta, rnorm;
const double zero = 0.0;
Integer i, j, m, n, nm, p, pda, pdb, pdd, pnm, zrow;
Integer exit_status = 0;
/* Arrays */
double *a = 0, *b = 0, *d = 0, *taua = 0, *taub = 0, *y = 0;
/* Nag Types */
NagError fail;
Nag_OrderType order;
#ifdef NAG_COLUMN_MAJOR
#define A(I, J) a[(J-1)*pda + I - 1]
#define B(I, J) b[(J-1)*pdb + I - 1]
order = Nag_ColMajor;
#else
#define A(I, J) a[(I-1)*pda + J - 1]
#define B(I, J) b[(I-1)*pdb + J - 1]
order = Nag_RowMajor;
#endif
INIT_FAIL(fail);
printf("nag_dggqrf (f08zec) Example Program Results\n\n");
/* Skip heading in data file */
scanf("%*[^\n]");
scanf("%" NAG_IFMT "%" NAG_IFMT "%" NAG_IFMT "%*[^\n]", &n, &m, &p);
if (n < 0 || m < 0 || p < 0) {
printf("Invalid n, m or p\n");
exit_status = 1;
goto END;
}
#ifdef NAG_COLUMN_MAJOR
pda = n;
pdb = n;
pdd = n;
#else
pda = m;
pdb = p;
pdd = 1;
#endif
/* Allocate memory */
if (!(a = NAG_ALLOC(n * m, double)) ||
!(b = NAG_ALLOC(n * p, double)) ||
!(d = NAG_ALLOC(MAX(n, m), double)) ||
!(taua = NAG_ALLOC(MIN(m, n), double)) ||
!(taub = NAG_ALLOC(MIN(n, p), double)) || !(y = NAG_ALLOC(p, double)))
{
printf("Allocation failure\n");
exit_status = -1;
goto END;
}
/* Read A, B and d from data file */
for (i = 1; i <= n; ++i)
for (j = 1; j <= m; ++j)
scanf("%lf", &A(i, j));
scanf("%*[^\n]");
for (i = 1; i <= n; ++i)
for (j = 1; j <= p; ++j)
scanf("%lf", &B(i, j));
scanf("%*[^\n]");
for (i = 0; i < n; ++i)
scanf("%lf", &d[i]);
scanf("%*[^\n]");
/* Compute the generalized QR factorization of (A,B) as
* A = Q*(R), B = Q*(T11 T12)*Z
* (0) ( 0 T22)
* using nag_dggqrf (f08zec).
*/
nag_dggqrf(order, n, m, p, a, pda, taua, b, pdb, taub, &fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_dggqrf (f08zec).\n%s\n", fail.message);
exit_status = 1;
goto END;
}
/* Solve weighted least squares problem for case n > m */
if (n <= m)
goto END;
nm = n - m;
pnm = p - nm;
/* Multiply Q^T through d = Ax + By to get
* (c1) = Q^T * d = (R) * x + (T11 T12) * Z * (y1)
* (c2) (0) ( 0 T22) (y2)
* Compute C using nag_dormqr (f08agc).
*/
nag_dormqr(order, Nag_LeftSide, Nag_Trans, n, 1, m, a, pda, taua, d, pdd,
&fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_dormqr (f08agc).\n%s\n", fail.message);
exit_status = 1;
goto END;
}
/* Let Z*(y1) = (w1) and solving for w2 we have to solve the triangular sytem
* (y2) = (w2)
* T22 * w2 = c2
* This is done by putting c2 in y2 and backsolving to get w2 in y2.
*
* Copy c2 (at d[m]) into y2 using nag_dge_copy (f16qfc).
*/
nag_dge_copy(Nag_ColMajor, Nag_NoTrans, nm, 1, &d[m], n - m, &y[pnm], nm,
&fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_dge_copy (f16qfc).\n%s\n", fail.message);
exit_status = 1;
goto END;
}
/* Solve T22*w2 = c2 using nag_dtrtrs (f07tec).
* T22 is stored in a submatrix of matrix B of dimension n-m by n-m
* with first element at B(m+1,p-(n-m)+1). y2 is stored from y[p-(n-m)].
*/
nag_dtrtrs(order, Nag_Upper, Nag_NoTrans, Nag_NonUnitDiag, nm, 1,
&B(m + 1, pnm + 1), pdb, &y[pnm], nm, &fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_dtrtrs (f07tec).\n%s\n", fail.message);
exit_status = 1;
goto END;
}
/* set w1 = 0 for minimum norm y. */
nag_dload(m + p - n, zero, y, 1, &fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_dload.\n%s\n", fail.message);
exit_status = 1;
goto END;
}
/* Compute estimate of the square root of the residual sum of squares
* norm(y) = norm(w2) with y1 = 0 using nag_dge_norm (f16rac).
*/
nag_dge_norm(Nag_ColMajor, Nag_FrobeniusNorm, n - m, 1, &y[pnm],
nm, &rnorm, &fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_dge_norm (f16rac).\n%s\n", fail.message);
exit_status = 1;
goto END;
}
/* The top half of the system remains:
* (c1) = Q^T * d = (R) * x + (T11 T12) * ( 0)
* (w2)
* => c1 = R * x + T12 * w2
* => R * x = c1 - T12 * w2;
*
* first form d = c1 - T12*w2 where c1 is stored in d
* using nag_dgemv (f16pac).
*/
alpha = -1.0;
beta = 1.0;
nag_dgemv(order, Nag_NoTrans, m, nm, alpha, &B(1, pnm + 1), pdb, &y[pnm], 1,
beta, d, 1, &fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_dgemv (f16pac).\n%s\n", fail.message);
exit_status = 1;
goto END;
}
/* Next, solve R * x = d for x (in d) where R is stored in leading submatrix
* of A in a. This gives the least squares solution x in d.
* Using nag_dtrtrs (f07tec).
*/
nag_dtrtrs(order, Nag_Upper, Nag_NoTrans, Nag_NonUnitDiag, m, 1, a, pda, d,
pdd, &fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_dtrtrs (f07tec).\n%s\n", fail.message);
exit_status = 1;
goto END;
}
/* Compute the minimum norm residual vector y = (Z^T)*w
* using nag_dormrq (f08ckc).
*/
zrow = MAX(1, n - p + 1);
nag_dormrq(order, Nag_LeftSide, Nag_Trans, p, 1, MIN(n, p), &B(zrow, 1),
pdb, taub, y, pdd, &fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_dormrq (f08ckc).\n%s\n", fail.message);
exit_status = 1;
goto END;
}
/* Print least squares solution x */
printf("Generalized least squares solution\n");
for (i = 0; i < m; ++i)
printf(" %11.4f%s", d[i], i % 7 == 6 ? "\n" : "");
/* Print residual vector y */
printf("\n");
printf("\nResidual vector\n");
for (i = 0; i < p; ++i)
printf(" %10.2e%s", y[i], i % 7 == 6 ? "\n" : "");
/* Print estimate of the square root of the residual sum of squares. */
printf("\n\nSquare root of the residual sum of squares\n");
printf("%11.2e\n", rnorm);
END:
NAG_FREE(a);
NAG_FREE(b);
NAG_FREE(d);
NAG_FREE(taua);
NAG_FREE(taub);
NAG_FREE(y);
return exit_status;
}
``` | 2,199 | 5,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-47 | latest | en | 0.414302 |
https://publicapp.in/category/discrete-mathematics/ | 1,696,077,969,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510676.40/warc/CC-MAIN-20230930113949-20230930143949-00604.warc.gz | 514,155,996 | 14,681 | # Category: Discrete Mathematics
## Basic Counting Principles
Sum Rule Principle Suppose that an event E can occur in m ways and a second event F can occur in n ways and suppose both events cannot occur simultaneously. Then E or F can occur in m+n ways. Example –...
## Logical Equivalent
Two logical statements p and q are said to be logical equivalents. It is denoted by p≡q when if p⇔q is a tautology. Note: Do not write p = q; instead write p ≡ q. Note: It doesn’t mean that p and... | 122 | 491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-40 | latest | en | 0.921164 |
https://engineerexcel.com/highest-coefficient-of-friction/ | 1,708,507,208,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473401.5/warc/CC-MAIN-20240221070402-20240221100402-00236.warc.gz | 253,874,870 | 153,734 | Highest Coefficient of Friction: Applications and Analysis
Friction is a force that resists relative motion between surfaces in contact. In the study of statics and dynamics, friction between contact surfaces is quantified using a dimensionless parameter called the coefficient of friction.
In this article, the focus is on exploring materials with the highest coefficients of friction, their applications across various industries, and conducting an in-depth analysis of the factors influencing this parameter.
Defining The Highest Coefficient Of Friction
When two bodies come into direct contact and touch each other’s surfaces, their interaction can be described by two types of contact forces: normal force and frictional force.
The normal force is the force exerted perpendicular to the surfaces in contact. On the other hand, the frictional force is the resistance encountered when two surfaces move or attempt to move past each other; hence, it always acts in the direction opposite to this relative motion or attempted motion.
These contact forces are illustrated in the diagram below.
The magnitude of the force of friction depends on the normal force. The higher the normal force, the more friction there will be. In general, it was found that this relationship is linear, and the proportional constant that relates the two contact forces is called the coefficient of friction.
The higher the coefficient of friction, the higher the friction force will be, given a normal force.
Types Of Friction
The value of the coefficient of friction depends on the type of friction present in the system. In general, friction is divided into two types: kinetic and static.
Kinetic friction refers to friction experienced by surfaces that are already in relative motion with each other. In this case, the proportional constant between the friction force and the normal force is called the coefficient of kinetic friction, as shown in the following formula:
Where:
• fk = kinetic friction force [N]
• FN = normal force [N]
• μk = coefficient of kinetic friction [unitless]
On the other hand, static friction refers to friction that acts to prevent relative motion between two surfaces when an external force is applied. As the applied force is increased, the static friction force also increases to match it, until a point where the applied force is just enough to overcome the maximum static friction. At this point, the object will start to move, and static friction transitions to kinetic friction.
In this case, the proportional constant that relates the maximum static friction force to the normal force is called the coefficient of static friction, as shown in the following formula:
Where:
• fs = static friction force [N]
• μs = coefficient of static friction [unitless]
In general, the coefficient of static friction is greater than the coefficient of kinetic friction. This means that it is usually easier to maintain the movement of an object than to initiate its motion from a state of rest.
However, there are some exceptions to this rule. For instance, consider the aluminum-on-aluminum dry interface, where the kinetic coefficient of friction is 1.4, and the static coefficient of friction is 1.05, as shown in the table below.
It is important to note that the coefficient of friction is a relational property between two materials in contact, not an intrinsic property of a single material. Therefore, it can only be defined and measured in the context of two interacting surfaces.
Can Friction Coefficient Be Greater Than 1
Contact materials commonly have a coefficient of friction below 1. Nevertheless, it is worth noting that a coefficient of friction greater than 1 is also possible.
The coefficient of friction simply represents the ratio of friction force to the normal force. Hence, when the friction force exceeds the normal force, the coefficient of friction can be greater than 1. In such cases, the contact materials are generally considered slip-resistant.
Maximum Value Of The Coefficient Of Friction
The coefficient of friction is influenced by the microscopic properties of the materials in contact as well as other factors such as the local temperature, relative velocity, pressure, and surface structure. At the microscopic level, friction occurs due to electrostatic interactions between the electrical fields of the materials. This means that stronger electrostatic interactions lead to a higher coefficient of friction.
In theory, it is possible to bring atoms extremely close together without reaching a singularity, allowing for electrostatic interactions to approach infinity. Therefore, there is theoretically no limit to the coefficient of friction. It can potentially be infinite, constrained only by the physical limitations of atomic interactions.
Contact Materials With The Highest Coefficients Of Friction
Although there is theoretically no limit to the magnitude of the coefficient of friction, in reality, the highest observed coefficient of friction occurs between rubber and other solid materials. This coefficient can range from 1 to more than 4.
For example, consider the graphs below showing the coefficient of friction of rubber-on-glass and rubber-on-steel interfaces at various sliding speeds.
Some other contact materials with high coefficients of friction observed include silver on silver at 1.4, aluminum on aluminum at 1.05-1.35, platinum on platinum at 1.2, cast iron on cast iron at 1.1, copper on cast iron at 1.05, copper on copper at 1.0, iron on iron at 1.0, and steel on lead at 0.95.
Applications Of High Coefficients Of Friction
High coefficients of friction are desirable in applications where preventing slippage is important. Common applications include braking systems, tires, belt drives, climbing gear, and sports equipment.
Brakes depend on friction to transform the kinetic energy of the rotating wheel into heat in order to decelerate the vehicle. Therefore, a high coefficient of friction between the brake pads and the brake rotors is essential for optimal braking performance.
Similarly, tires require a sufficiently high coefficient of friction with the road surface to ensure secure grip and prevent slippage, thereby enhancing road safety. In the case of belt drives, a high coefficient of friction is important to prevent belt slippage and transmit power between rotating shafts more efficiently.
In sports equipment like golf clubs, baseball bats, and tennis rackets, a high coefficient of friction is needed to prevent them from slipping out of the player’s grasp. Additionally, climbing gear, such as ropes, relies on friction to protect climbers from dangerous slips.
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1. Problem Definition
My colleague and I are planning to implement phone survey for data collection for our study’s sample. Prior the real survey, we have to test the phone questionnaire whether the survey can be finished in 20 minutes. The questionnaire consists of 30 questions, so we picked random respondents to examine how long it takes to finish the survey.
Problem Statement: does the duration of the survey within the expected duration?
2. The Feasible Alternatives
Based on the problem, I identify three possible results:
1. The result shows the actual duration of the survey is the same with the expected duration.
2. The result shows the actual duration is less than expected duration.
3. The result shows the actual duration is more than the expected duration.
3. Tools and Technique
To answer the problem, several steps to solve the problem:
1. Collect sample data;
2. Determine number of class and frequency;
3. Draw the histogram from the frequency table; and
4. Interpret the histogram.
4. Selection of the Acceptable Criteria.
Table 1 shows the frequency table which we collected from respondents.
Table 1: Frequency Table
Then, we decided the number of classes that contribute to our analysis, next construct the frequency table.
Table 2: Number of Classes and Frequency
Then, we draw the histogram.
Figure 1: The Histogram
Figure 1 shows the frequency within 10-15 minutes is the highest. We are confidence the survey’s respondent can finish the interview less than 20 minutes with current questionnaire.
5. Post-Evaluation of the Result
Based on the calculation above, the result shows the chance of interview can be finished in 10-15 minutes is higher. We have to scrutinize the current questionnaire and make sure it collects the information we needed for the report.
6. References
Brassard, M. & Ritter, D. (2010). The Memory Jogger 2: Tools for Continuous Improvement and Effective Planning, pp. 91-100.
Henning, J. 2009. Do phone surveys have a future? Retrieved from: http://www.research-live.com/features/do-phone-surveys-have-a-future?/4000692.article
Vicente, P., Reis, E., & Santos, M. Using Mobile Phones for Survey Research: A Comparative Analysis between Data Collected via Mobile Phones and Fixed Phones. Retrieved from: http://homepages.wmich.edu/~wmartz/assets/mobile-phones-survey-research.pdf
# W13.0_Ary_HV Transmission Line Losses Reduction – Economic Analysis using PW Method
Problem Recognition and Definition
Power loss in high voltage transmission line has a linear relationship to the loads and has exponential relationship to the current flows in the lines. Several methods available in compensating transmission losses and one the most popular method is utilizing Static Var compensation (SVC) equipment installed at the load side. It is important to analyze the econimical benefit of SVC implementation to solve transmission line loss problem.
Possible Alternatives
As mentioned above, SVC is the most practical used in electric transmission & distribution system while another system has been introduced (new technology) by using AC-DC-AC converter equipped with SVC system.
In this economic analysis both alternatives will be analyzed based on cost savings given and opportunities available from the features of these equipment.
Analysis and Criteria
The attached graph showing increasing trend of MVAr supply from hydro power generation through with distance about 30 kilometers long from the center of loads.
Figure 13-1. MVar generated trend
The average MVAr delivery is 80 MVAr and in this study, reduction to half of this value is technically accepted. Loss from 40 MVAr transmission line is worth to \$499,000 per year.
Below are two alternatives with technical capability of both equipment and opportunities provided if the equipment installed.
SVC installation is designed to reduce MVAr delivery from power generation while the AC-DC-AC converter (DC Link) does the same with additional capability to transfer active power (MW) from hydro bus to thermal power station bus system. It provides extended opportunity to shutdown diesel generating plant.
Cost comparison of two generating stations is figured below and will be used as a basis to calculate cost savings:
With opportunities (savings) from the two alternatives are figured below:
And summarized data for this analysis is shown on below table
Table 13-1. Basic data of two alternatives
Equivalent worth method will be used to select the best economical option selected for implementation. And alternatives with the highest PW value at defined MARR will be proposed to management.
MARR 25% will be used in this study with 10 years useful life of equipment.
Assessment Result
Below figure shows calculation result among two alternatives. Alternative 1 (SVC) provided negative PW value and alternative 2 (DC Link) provide positive (higher) PW value despite higher investment and yearly operating cost compare to the 1st alternative.
Table 13-2. Study result using PW method
Post Evaluation
Extended economical analysis using another method can be done in separate page to provide strong justification. More technical study and alternative provided to solve the problem might be done and obtained as part of the next study.
References:
1. Sullivan, W.G. Wicks, E.M. Koelling, C.P (2012). Engineering Economy, Fifteenth Ed. (chapter 6, page 233-239).New Jersey: Prentice Hall.
2. Deng, Yongan. Reactive Power Compensation of Transmission Lines. Concordia University: retrieved from: http://users.encs.concordia.ca/~lalopes/Courses/IGEE401-F07/TL_Compensation.pdf
# W14_IQB_Theories of Motivation
I was in a Leadership workshop when I found out that there are theories of motivation which I never heard before until I found them in the pre exam pass by my mentor. There are many theories regarding how to motivate others but the most prominent theories are:
• X and Y theory
• 2 Factor theory
• Equity Theory
• Expectancy theory
• Goal Setting theory
X and Y Theory
This theory was proposed in early 60’s by Douglas McGregor, an American social psychologist, from his book “The Human Side of Enterprise”. McGregor suggest that there are 2 style in managing people, using “authoritarian style” or known as X theory which the management will use power to direct his/her subordinate to achieve the organization’s goals. He/she will give a detail direction what and how to do things. It will result in an un-ambitious, lack of responsibility subordinate. All responsibility will be under the manager. No sense of belongings to the result of the work from the subordinate.
The other style is “participative management” or known as Y theory, which the management allows his/her subordinates to express their ideas, self directed and self control. The management will only give a general objectives and timeline/duration. The subordinates will be responsible to achieve the objectives.
Those 2 (X and Y) theories are well-matched to certain conditions/level of organization. When we face the situation where the subordinates is the lowest level of organization with inadequate knowledge and capability, then use the X theory, give them detail direction and closely monitor the progress. But for the medium level of organization where the knowledge and capability were adequate, then use the Y theory, give them the objectives and general direction and let them solve it.
2 Factors Theory
Herzberg shares his theory about the cause of employee’s satisfaction and dissatisfaction. In his theory, he explains that there are 2 factors which influence employee’s motivation. They are motivation and hygiene. Motivation factor is the factor that the absence of this factor will not result in dissatisfaction of the employee but when this factor provided by the company, it will heighten the employee’s motivation. This factor included status, recognition, bonuses, promotion etc.
And not in the opposite of the motivation factor but different factor that gives effect to employee’s motivation is hygiene factor. Hygiene factor is factor that the absence of this factor will result in dissatisfaction of the employee but when this factor provided by the company there will be no change to their motivation since this is the basic needs. This factor includes wages, salary, security of works, quality of inter personnel relations etc.
Equity Theory
This theory was presented by John Stacey Adam in 1963, it said that people will be highly motivated when he receive fairly and equally treatment in every aspects of his life. There are 3 elements in equity theory, they are: 1. Input, are every aspect that given by the worker such as effort, skills, commitment, and hard work etc, 2. Outcomes, are every aspects that the worker thought that those are his rewards such as payment, salary, wages, and promotion etc, 3. Comparison person, is individual whose worker compare his ratio to.
The equity theory is dependant to comparison between an individual ratio of reward/investment (input/outcomes) to ratio by other (Comparison Person) in similar condition. He/she will naturally compare his/her ratio to other ratio which any inequity due to comparison will result in disappointment worker.
Expectancy Theory
This theory was brought by Victor H. Vroom in 1964, it said that an individual tend to motivated when he knows that in the end there is something he can expected to become. This theory is based on 3 aspects, they are: 1. Expectancy, is perception that the current effort will lead to good performance, 2. Instrumentality, is perception that the performance will lead to reward, 3. Valence, is the expected value will be received.
Conclusion
From the above theories, we can conclude that every theory has similarities to each other. There are many ways to encourage the worker but all theory emphasize that reward id the easiest way to motivate the worker.
Bibliography
# W18_RW_Mutually Exclusive Project Selection with Benefit Cost Ratio
Disclaimer: All value in this blog is not real, it used to simulate to determine the economic life calculation.
Problem Statement:
My company has 5 alternatives investment projects with certain cost and benefit calculated previously, the data shown in the table below:
My boss ask me to compare those alternatives by using Benefit Cost Ratio (B/C) to determine which alternatives that has B/C>1.1
Acceptable Criteria
1. Alternative which has B/C>1.1
Tools and Techniques:
From Sullivan’s Engineering Economy 15th Edition, chapter 10, section 10-9, Comparison of Mutually Exclusive Project.
Calculation:
Step 1. Rank the Cost of Alternatives from the smallest to the largest cost.
Step 2. Calculate the B/C ration from the smallest cost
Alternative E has 1.13>1.1 (acceptable) now it became a baseline to compare with other ranked alternatives.
Alternative D: (benefit D – benefit E) / (Cost D – Cost E) = 1.28 >1.1 (acceptable) now it became a new baseline
Alternative B: (benefit B – benefit D) / (Cost B – Cost D) = 1.25 >1.1 (acceptable) now it became a new baseline
Alternative A: (benefit A – benefit B) / (Cost A – Cost B) = – 4.5 <1.1 (not acceptable)
Alternative C: (benefit C – benefit B) / (Cost C – Cost B) = 0.33 <1.1 (not acceptable)
The conclusion:
From the above alternative comparison which mutually exclusive alternative using the incremental analysis we know that Alternative B has B/C=1.25>1.1, and this alternative B is recommended.
References:
Sullivan, W.G., Wicks, E.M., Koelling, C.P. (2009). Engineering Economy, Fifteenth edition. Pearson International Edition, 2012, Chapter 10, pages 436.
# W17_RW_When Should I Replace my Car
Disclaimer: All value in this blog is not real, it used to simulate to determine the economic life calculation.
Problem Statement:
I just bought 3 years old used car, this car was bought with \$20,000 it was in good condition, from the manual book I see there is a potential to replacing some major spare part beside the routine preventive maintenance. The detail data shown in table below, assume I have 15% personal interest rate per year. When should I replace this car?
Feasible Alternative:
1. Continue using that Car until it cannot be used
Tools and Techniques:
From Sullivan’s Engineering Economy 15th Edition, chapter 9, section 9-6, Determining the economic life
Selection Criteria:
Equivalent Uniform Annual Cost (EUAC) > total margin cost is the best choice.
Calculation:
Deterimine the potential loss in Market value and cost of capital to get the total marginal cost.
Marginal Cost = Major Maintenance + Preventive Maintenance + Loss in Market Value + Cost of Capital
And now calculate the EUAC with 15% MARR.
The conclusion:
Using the car until it scrap is not a right choice at it will drain your pocket for increasing maintenance cost and lowering the market value. Sell the car at the 2nd year and purchase the new car is a good choice, in term of economic value it is the optimum years where we can get the reasonable market value compare with the maintenance cost, where the total Marginal cost is less then the EUAC (15%)
References:
Sullivan, W.G., Wicks, E.M., Koelling, C.P. (2009). Engineering Economy, Fifteenth edition. Pearson International Edition, 2012, Chapter 9, section 9-6 pp 391. | 2,859 | 13,368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2019-51 | latest | en | 0.831024 |
https://sciencedocbox.com/Physics/117231857-Ece-342-electronic-circuits-lecture-34-cmos-logic.html | 1,660,395,414,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571950.76/warc/CC-MAIN-20220813111851-20220813141851-00154.warc.gz | 460,449,174 | 23,399 | # ECE 342 Electronic Circuits. Lecture 34 CMOS Logic
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1 ECE 34 Electronic Circuits Lecture 34 CMOS Logic Jose E. Schutt-Aine Electrical & Computer Engineering University of Illinois 1
2 De Morgan s Law Digital Logic - Generalization ABC... ABC... ABC... ABC... Distributive Law AB AC BC BD A( B C) B( C D) General Procedure 1. Design PDN to satisfy logic function. Construct PUN to be complementary of PDN in every way 3. Optimize using distributive rule
3 CMOS Logic Gate Circuits Two Networks Pull-down network (PDN) with NMOS Pull-up network (PUN) with PMOS PUN conducts when inputs are low and consists of PMOS transistors PDN consists of NMOS transistors and is active when inputs are high PDN and PUN utilize devices In parallel to form OR functions In series to form AND functions 3
4 Pull-Down Networks Y AB Y AB 4
5 Pull-Up Networks Y AB Y AB 5
6 Basic Logic Function Basic Function INVERTER NOR NAND Symbol # Devices PUN 1 PMOS PMOS-Series PMOS-Parallel # Devices PDN 1 NMOS NMOS-Parallel NMOS-Series Truth Table 6
7 Pull-Down and Pull-Up Functions Pull-up network (PUN) Pull-down network (PDN) Key features When PDN switch is on, PUN switch is off and vice versa Conditions for being on and off are complementary 7
8 Pull-Down and Pull-Up PDN-parallel NMOS PUN-series PMOS Truth Tables YDP AB YUS AB 8
9 Pull-Down and Pull-Up When Y DP in PDN-parallel is low, this means that either A or B (or both) is high. When either A or B (or both) is high, either transistor (or both) in PUNseries are offy US =low When Y DP in PDN-parallel is high, both A and B are low. Both transistors in PUN-Series are on creating a path to VDD. Y US =highy US =Y DP. PDN-Parallel and PUN-series are complementary 9
10 Pull-Down and Pull-Up PDN-Series NMOS PUN-Parallel PMOS YDS AB Truth Tables Y A B UP 10
11 Pull-Down and Pull-Up If Y DS is low, both A and B must be high in which case both transistors in PUN-Parallel are off providing no path to V DD Y UP =lowy UP =Y DS. If Y DS is high, then either A or B (or both) are off (low) in which case either Q PA or Q PB in PUN-Parallel will be on and present a path to V DD ; thus Y UP =high Y UP =Y DS PDN-Series and PUN-Parallel are complementary 11
12 Two-Input NOR Gate Y AB AB 1
13 Two-Input NOR Gate Actual Ideal 13
14 Two-Input NAND Gate Y AB AB 14
15 Y ABCD Using De Morgan s Law Example Y ABCD ABCD( AB) ( C D) Pull-down network Pull-up network 15
16 Example 1 Evaluate Logic Function Y ( A B) C from pull down from pull up Y AB C AB C AB C ( A B) C 16
17 pull down Example Implement the function Y AB C pull up Y ABC ABC ( AB) C 17
18 Exclusive-OR (XOR) Function Y AB AB Y ( AB)( AB) XOR pull down A B Y pull up 18
19 Transistor Sizing Objectives PDN provides discharge current of at least that of an NMOS PUN provides charging current of at least that of a PMOS Worst case gate delay equal to that of basic inverter Find combination that results in lowest output current For transistors in parallel aspect ratios add For transistors in series, inverses of aspect ratios add Series :... W / L W / L W / L W / L eq 1 M Parallel : W / L W / L W / L... W / L eq 1 n W L p W L / p / n 19 M
20 Transistor Sizing NOR NAND 0
21 Transistor Sizing Example 1 Two approaches to realizing the OR function of six input variables. Assuming that the transistors in both circuits are properly sized to provide each gate with a current-driving capability equal to that of the basic matched inverter, find the number of transistors and the total area of each circuit. Assume the basic inverter to have a (W/L) n ratio of 1. m/0.8 m and a (W/L) p ratio of 3.6 m/0.8 m
22 Transistor Sizing Example 1 For design (a), there are (6)+=14 transistors: All 7 NMOS use (W/L) n = n 1 PMOS uses (W/L) p = p 6 PMOS use (W/L) p = 6p Total Area = 7(1.)0.8+ 1(3.6) (6)(3.6)0.8 = m For design (b), there are (3) + 1()=16 transistors: 6 NMOS use (W/L) n = n 6 PMOS use (W/L) p = 3p PMOS use (W/L) p = p NMOS use (W/L) n = n Total Area = 70(1.)0.8 = 67. m, or 59% of (a)
23 Transistor Sizing Example Transistors in two-input NOR gate are properly sized so that the current-driving capability in each direction is equal to that of a matched inverter. For V t = 1 V and V DD = 5 V, find the gate threshold in the cases for which (a) input terminal A is connected to ground and (b) the two input terminals are tied together. Neglect the body effect in Q PB
24 Transistor Sizing Example Corresponding to a matched inverter characterized by n and p where k p =k n = k, the two-input NOR uses transistors n and p where k p = k n a) For A grounded, V thb occurs near V DD /, with Q PB and Q NB in saturation and Q PA in triode. Let V th = v, and the voltage across Q PA be x Thus and and id k p 51 xx / i D kp xv i 1 1 D kn v 4
25 Transistor Sizing Example For k k, i k 4 xx / k 8xx p n D n n and and id kn 4 xv i 1 1 D kn v (1) () (3) v xv From ) 3): Thus, 1.707v x x v or, 0.93v 3.93 x x v 5
26 Transistor Sizing Example Now x 0, in which case, v / (ok) or v 3.93 / (Clearly too large) Thus, x v (4) v xx Now from 1) 3) : 1 8 With 4), v v v or v v v v5.83v or 6.83v v or 6.83v 6.81v
27 1/ whence v / (6.83) (1) Check: Transistor Sizing Example ( ) / V.5 V probably ok since one PMOS is full on Thus V.65V th b) For A and B joined, the PMOS can be approximated as a single device with twice the length, for which the width is twice that in a matched inverter. Thus, for the equivalent PMOS device, (W/L) peq = p and k p =k. For each of the two NMOS, (W/L) n = n and k n = k. Thus, at V th = v with all devices in saturation: k k id v1 5v1 7
28 Transistor Sizing Example v v and v v 1 4, 1 4 Thus 1.414v v,.414v whence V v.4v th See this is reduced from the single-input value (of.65v)! Note that this fact can be used to control the relative threshold of multiple gates connected to a single fan-out node, in order to guarantee operation sequence for slowly changing signals 8
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Designing ombinational Logic ircuits dapted from hapter 6 of Digital Integrated ircuits Design Perspective Jan M. Rabaey et al. opyright 2003 Prentice Hall/Pearson 1 ombinational vs. Sequential Logic In | 8,935 | 32,752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2022-33 | latest | en | 0.828249 |
https://www.jiskha.com/questions/565631/in-triangle-ABC-side-AB-IS-2cm-shorter-than-side-AC-while-side-BC-is-1cm-if-the | 1,571,295,441,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672723.50/warc/CC-MAIN-20191017045957-20191017073457-00325.warc.gz | 963,670,048 | 5,081 | # geometric problem-math
in triangle ABC,side AB IS 2cm. shorter than side AC, while side BC is 1cm. if the perimeter is 62 find the lenght of three sides? i need the answer tomorrow pls... help me
1. 👍 0
2. 👎 0
3. 👁 67
1. P=sum three sides
P=BC+Ab+Ac
62=AC+AC-2+1 solve for AC first, then AC-2 (which is AB) Are you certain about BC?
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2. 👎 0
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