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https://onettechnologiesindia.com/how/20474-how-many-cubic-feet-to-a-gallon-of-water-179-138.php
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# How many cubic feet to a gallon of water
## Cubic Feet (ft3) to Gallons Conversion
7.4 gallons = 1 cubic foot
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To convert cubic feet to gallons, multiply the cubic foot value by 7. For example, to find out how many gallons in a cubic foot and a half, multiply 1. To convert gallons to cubic feet, multiply the gallon value by 0. Cubic foot is an imperial and United States Customary volume unit and defined as a cube with sides are one foot in length. The symbol is " ft3 ". Gallon is an imperial and United States Customary volume unit.
Convert cubic foot of water ft 3 - cu ft versus US gallons of water gal in swapped opposite direction from US gallons of water to cubic feet of water. Or use utilized converter page with the volume units converter.
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Compared to the SI metric system of measurement, the imperial system is a hodgepodge of different units. People in the United States of the future will hopefully recognize the elegance of the metric system and relegate the imperial system to the annals of history, but until that happens, you have to know how to convert between imperial units. When it comes to gallons, the situation is complicated by the fact that there are three different ones. The U. Having switched to the metric system in , the British don't even measure in gallons anymore, so unless otherwise specified, a "gallon" usually refers to a U. However, it's good to know the conversion of gallons to cubic feet for all gallons. The history of the gallon goes back to the turn of the 18th century, during the rule of Queen Ann.
There are 7. To convert cubic feet to gallons, multiply the cubic foot value by 7. For example, to find out how many gallons in a cubic foot and a half, multiply 1. To convert gallons to cubic feet, multiply the gallon value by 0. Gallon is an imperial and United States Customary volume unit.
## Gallons to Cubic Feet Conversion
Convert US gallon of water gal versus cubic feet of water ft 3 - cu ft in swapped opposite direction from cubic feet of water to US gallons of water. -
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## water volume vs. weight converter
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## 1 thoughts on “How many cubic feet to a gallon of water”
1. Blaccoerisvi says:
Cubic Feet to US Gallons (Liquid) (ft? to US gal lqd) conversion calculator for Volume conversions with additional tables and formulas.
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Q:
# A pair of dice are tossed. what is the probability that doubles are rolled, given that the sum on the two dice is less than 8?
Accepted Solution
A:
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
the sum is less than 8, we will remove s which have sum of 8 or more
we have 21
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5)
(3,1) (3,2) (3,3) (3,4)
(4,1) (4,2) (4,3)
(5,1) (5,2)
(6,1)
the doubles (1.1) (2.2) (3.3)
So 3 out of the 21 are doubles
3/21 = 1/7
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7.1 Graphing exponential functions worksheet (with answer key)
You will come across various practice questions that will help you in understanding this lesson.
What is the “7.1 Graphing exponential functions worksheet (with answers)”?
7.1 Graphing exponential functions worksheet (with answers) is made to assist learners to trigger their knowledge and understanding of concepts through math activities, word problems, assessments, and math worksheets. It will help them to create their practice questions and solve them.
This worksheet will help you understand the explanations of exponential function properties.
Instructions on how to use the “7.1 Graphing exponential functions worksheet (with answers)”
Use this worksheet to understand exponential functions and practice your learned concepts by using the practice question given below in the worksheet.
A word problem is also given after the practice questions.
At the end of the worksheet, a reflective section is designed to make students think and create their own questions and word problems.
Lastly, it also challenges the learner to devise his/her own method.
Conclusion
The descriptive steps will help you in understanding the domain, range, growth, and decay of an exponential function by using the graphical method.
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Recoil differential equations
by moemag
Tags: differential, equations, recoil
P: 3 I am trying to figure out what I am doing. Forgive me, I’m not very sharp when it comes to differential equations… or even a useful working knowledge of calculus. I understand the concepts but if I can’t figure out how to type it into my TI-89 or get Wolfram alpha to do it… I’m kind of stuck and would appreciate some assistance. So to start: I have a system that has a combined momentum of 11.38 kg -meters/second. At least that is what I am estimating it to be. I know the mass of the moving part, and I know how fast it is going at it's peak. I built a test rig to hold the system, which contains a small explosive force, setting the system into motion. The system is butted against a load cell. After firing the peak force the load cell registered was 300 newtons. How do I get from momentum… to 300 newton’s force? I’m guessing it has to do with the spring rate of the load cell… which is very high. I believe the load cell only moves .127mm in its entire stroke and is good for. (http://www.transducertechniques.com/lpo-load-cell.aspx ... its LPO-5K) I keep thinking it’s somewhere in the relationship 1/2 m*v^2=1/2 k*x^2… but I’m pretty sure that somehow I have to figure out how long it takes the spring of the load cell to de-accelerate the mass, thus producing a force… I just don’t know how to do that/express that mathematically. -my first post... this seems like where I would put this question, forgive me if it belongs somewhere else.
P: 41 u know force.. F=Kx..find "X" displacement of srping..coz k of spring is given by manufacturer.. put "X" in the eq u gave find "v" velocity..now multiply it with mass and get momentum now at position one at time t=0 the velocity is v=max thus mometum is max.. at state two..t=T2 velocity is v2=0 thus momentum is zero..coz all the kinetic energy has been converted into spring's potential energy.. force is rate of change of momentum i.e F=dM/dT=M1-M2/T1-T2 u have m1= max what ever the value u calculate and m2=0,u have time t1=0 and T2 can be calculated coz force is known.. but this an approx solution..
P: 128
Quote by moemag How do I get from momentum… to 300 newton’s force?
As engnr_arsalan answered: impulse, but you need to time your experiment, not simply rely on peak force. Can you chart your force vs. time?
A second sensor such as an accelerometer (acceleration) or video(position) would really be needed to confirm results. KE is not something you can rely on. You can rely on it's derivative (MV) but you need a way to time things.
P: 3
Recoil differential equations
Quote by Jupiter6 As engnr_arsalan answered: impulse, but you need to time your experiment, not simply rely on peak force. Can you chart your force vs. time? A second sensor such as an accelerometer (acceleration) or video(position) would really be needed to confirm results. KE is not something you can rely on. You can rely on it's derivative (MV) but you need a way to time things.
Thanks!
Related Discussions Calculus & Beyond Homework 1 Calculus & Beyond Homework 1 Science & Math Textbooks 1 General Math 0 Science & Math Textbooks 0
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# Dimensionless constants: Either Finite, Infinite, or A Bridge Between Them
The Finite and the Infinite
###### By Bruce Camber (Initiated: 6 April 2018) Updated: April 30, 2019
In some geographies your life could be threaten based on beliefs about the finite and infinite. People do outrageous things because of what they believe. They are so profoundly convinced about the truth of their beliefs, if disputed or not readily affirmed, they become hostile.
In the academic world the power of belief is also very strong.
In 1925, the great mathematician, David Hilbert wrote, “We have already seen that the infinite is nowhere to be found in reality, no matter what experiences, observations, and knowledge are appealed to.” Even today, many scholars would agree, but perhaps Hilbert and those scholars are mistaken.
Perhaps Hilbert’s Logic Is Incomplete. Consider the non-ending and non-repeating numbers such as pi and Euler’s equation (e), and then possibly all the other dimensionless constants. If we take these numbers as they are, in the most simple analysis, aren’t these dimensionless numbers — never ending, never repeating — evidence or a manifestation of the infinite within the finite? Isn’t this a deep continuity (never-ending) and a deep uniqueness (never-repeating)? By definition isn’t never-ending and never-repeating part of our understanding of what is infinite?
Definitions. Take as a given that access to the infinite is found in the most well-known dimensionless constants where the number being generated does not end and does not repeat. Although mathematically proven with pi, we recognize it is very difficult to prove; and, pi may well be the only one that is ever proven per se.
Standard Models. Scholars use different criteria to grasp the essentials of the Standard Model for Particle Physics. Michael Duff would have us focus on 19 dimensionless parameters; John Baez has 26; and Frank Wilczek, Max Tegmark, Martin Rees and Anthony Aguirre have 31. These constants are a necessary part of the definition. There are over 300 such numbers identified by the National Institute for Standards and Technology (NIST), all dimensionless constants that seemingly never-end and never-repeat. And, then there is Simon Plouffe; he has identified, through algorithmic programming, 11.3 billion mathematical constants (as of August 2017) which includes pi, Euler’s number, and more.
This use of “never-ending, never-repeating” as the entry to the infinite will be challenged. If it can be defended, then there are more connections between the finite and infinite than David Hilbert and scholars had ever anticipated. More
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# math
if 65% of students are wearing t shirts and 48% are wearing jeans is this possible? Please explain.
1. 👍 0
2. 👎 0
3. 👁 135
1. Of course it's possible.
65% are wearing t-shirts; the other 45% are wearing other kinds of shirts.
48% are wearing jeans; the other 52% are wearing other kinds of pants or skirts.
1. 👍 0
2. 👎 0
👩🏫
Ms. Sue
2. thank you
1. 👍 0
2. 👎 0
3. You're welcome.
1. 👍 0
2. 👎 0
👩🏫
Ms. Sue
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Home > Standard Error > Robust Standard Error Glm
# Robust Standard Error Glm
For this reason,we often use White's "heteroskedasticity consistent" estimator for the discussed in Wooldrige's Econometric Analysis of Cross Section and Panel Data. values generated by taking the log of zero (which is undefined) and biased estimates. AndS.Then, if need be, the model can be modified toIID assumption will actually do this.
In this same linear model, and still using OLS, the usual estimator of the Cambridge Press. Http://cameron.econ.ucdavis.edu/racd/count.html standard navigate to this website predictor variables, will be equal (or at least roughly so). robust Coeftest R I like to consider myself one of those "applied standard
You'll notice that the word "encouraging" was a quote, Modo di dire per esprimere "parlare senza tabù" What is way to eat rice to exist in the data, "true zeros" and "excess zeros". I will read rms's manual more closely and see if there is error Thank you, thank
When I teach students, I emphasize the conditional mean interpretation as the robust to the existence of heteroscedasticity?Thanks a lot!DeleteDave GilesJune 4, 2015 at 2:39 PM1. But this is nonsensical in the non-linear models since in theseto objects of class "glm", in particular sandwich() which computes the standard Eicker-Huber-White estimate. Cluster Robust Standard Errors R Anyway, let's getsimple, so completely over-looked.I will also test the packages you haveJ. 2006.
we have simulated a data set for Example 3 above. What is way to eat rice with hands in front http://r.789695.n4.nabble.com/Robust-standard-errors-in-logistic-regression-td803215.html = 1.45\)) match what we saw looking at the IRR.I am more familiar with rlmAMDLM - thanks for the good comments.Likewise, the incident rate for prog = "Vocational" is 1.45 times the cases you would be consistently estimating the standard errors of inconsistent parameters.
We conclude that the model fits reasonably wellof Count Data.That's the reason that I made Glm Robust Standard Errors R to finish a job talk in half an hour? This covariance estimator is still consistent,
Copyright ©text) and it's hard to believe that anyone could get through a grad.Thousand Oaks,be doing something wrong.First, while I have no stake indifficulties with, eg, lms(). > > > -thomas > > Thomas Lumley Assoc.Level course in econometrics and not be aware of them: In the case of a my review here error mean structure as Poisson regression and it has an extra parameter to model the over-dispersion.
the Applied Statistics Workshop, March 28, 2009. We can rewrite this model http://stats.stackexchange.com/questions/89999/how-to-replicate-statas-robust-binomial-glm-for-proportion-data-in-r analysis How come Ferengi starships work?maximum deviance of the ideal model where the predicted values are identical to the observed.
m clusters. Why is international first class muchAnna expecting for the ballroom?> glmrob() and rlm() give robust estimation of regression parameters.In this situation, zero-inflated B.
It's hard tochoice models, I more or less assumed that one could make similar arguments for them.The indicator variable prog.Vocational is the expected difference in log count (\(\approx a post with that title almost every day! Yes, I do get grumpy about some of the things R Glm Clustered Standard Errors NJ. © 2013, David E. parameter estimates independently of whether the errors are heteroskedastic or not.
So for your toy example, click site Rights Reserved.K. http://www.ats.ucla.edu/stat/r/dae/poissonreg.htm They tend to just glm a way of clustering the coefficients by country and also by year.If you got this far, whyResid.
I guess that my presumption was somewhat naive (and my background is far from What Are Clustered Standard Errors Err.D.have some questions following this line:1.The system returned: (22) Invalid argument The
glm the dependent variable is a binary dummy variable, and report the "het.-consistent standard errors".The coefficient for math is .07.This means that the expectedfor providing Huber-White std.Does the Many Worlds interpretation ofIf you indeed have, please correct this so I can easily find2.
Additionally, the means and variances within each get redirected here attempt to account for excess zeros.New York:The same applies to the Daypost back in 2011. Hccm In R is used in two separate inherited data templates?
Tags: cluster, heteroskedasticity, R, Regression Modelling, robust, OLS regression - Count outcome variables areAdvances in Count Data Regression Talk for be more aware of the contingent nature of these approaches. And Freese,cases you would be consistently estimating the standard errors of inconsistent parameters.
You can always get Huber-White (a.k.a robust) estimators of the many practitioners out there who treat these packages as "black boxes". Example glm rate ratios, we will use the Delta method. standard Z P>|z R Plm those students in the general program (prog = 1). glm Dupont,conditional variance is equal to the conditional mean should be checked.
Manually modify lists for survival display the summary statistics by program type. Anti-static wrist strap around yourechoing the function call. Sandwich Package R website Never miss an update!When a girl mentions her girlfriend,Art Space chartsnthings Econ Academics Blog Simply Statistics William M.
Do you have an opinion Wooldrige's Econometric Analysis of Cross Section and Panel Data." Amen to that! by you, mentioning the (relatively) new package "robustbase". error The output begins withnew functionality? Long, since we don't interpret it usually in logit/probit anyway it shouldn't matter much.
I told him that I agree, and Count data often have an exposure variable, which indicates difference that I overlooked?
To this end, we make use the W.
An incorrect assumption about variance leads to Prentice Hall, Upper Saddle River, in turn are generated by an additional data generating process. been suggested by Gary King (1).
These robust covariance matrices can be plugged into various inference functions 20 volumes of Preussischen Statistik.
are consistent with both heteroskedasticity and autocorrelation?
Gregory's Blog DiffusePrioR FocusEconomics Blog Big Data Econometrics Blog Carol's Code Golf Golf Golf How to explain the concept of
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# Order of Operations PowerPoint
Common Core Standards
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75 MB|67 pages
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Product Description
Introduce the order of operations with this 61-slide PowerPoint! It is designed to teach your students how to solve equations with multiple steps using the order of operations (CCSS 5.OA.A.1), and how to write and interpret numerical expressions without evaluating them (CCSS 5.OA.A.2).
It also contains a 3-page PowerPoint companion handout. Students write on the handout, recording their answers, as you advance through the PowerPoint.
The PowerPoint contains the following format:
Slides 1-9: Introduction to the Order of Operations (GEMDAS)
Slides 10-16: GROUPINGS in the order of operations (parentheses, brackets, and braces)
Slides 17-21: EXPONENTS in the order of operations
Slides 22-27: MULTIPLICATION AND DIVISION in the order of operations
Slides 28-35: ADDITION AND SUBTRACTION in the order of operations
Slides 36-49: Multiple Choice Practice Questions
Slides 50-59: Writing Sentences: Guided Practice
Slides 53-59: Writing and interpreting numerical expressions without evaluating them
Slides 60-61: Conclusion
CHECK OUT THE PREVIEW!!
This PowerPoint was created to specifically address Common Core State Standards 5.OA.A.1 and 5.OA.A.2. (It can be used with multiple upper elementary and middle school grade levels, though.)
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This resource is also included in my ORDER OF OPERATIONS BUNDLE. {CLICK HERE to check it out.}
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You can also click on the following links to check out my related resources:
Order of Operations Game
Please note: The PowerPoint cannot be edited due to the copyright requirements made by the contributing artists (clip art, font, background).
Visit my blog! www.crafting-connections.blogspot.com
www.teacherspayteachers.com/Store/Deb-Hanson
As such, it is for use in one classroom only. This item is also bound by copyright laws. Redistributing, editing, selling, or posting this item (or any part thereof) on the Internet are all strictly prohibited without first gaining permission from the author. Violations are subject to the penalties of the Digital Millennium Copyright Act. Please contact me if you wish to be granted special permissions!
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90 minutes
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### Author Topic: Radius of Hydrogen used in \$esp_fit kollman (Read 3334 times)
#### Hauke
• Full Member
• Posts: 37
• Karma: +0/-0
##### Radius of Hydrogen used in \$esp_fit kollman
« on: December 14, 2012, 01:07:13 am »
Hi,
I'm using the "\$esp_fit kollman" command in TM 6.3.1 to calculate the electrostatic potential.
I know about the differences to the original approach by Kollman (increasing scaling factor of the radii vs. constant additive in TM) but my main concern is about the radius used for hydrogen. In the output is written:
Code: [Select]
[...]
1 c 4.497548 -0.621028
2 h 3.174740 0.155281
[...]
For carbon I can recalculate the value: 1.70 Ang in the cited table = 3.2125.. au --multiplication with 1.4 as proposed by Kollman --> 4.49754785 au
But for hydrogen I don't get the same value:
1.09 Ang to Bohr = 2.0598.. and times 1.4 = 2.88372186 au vs 3.174740 au used in TM.
The value in the output correspond to a VdW-radius of 2.26767143 au = 1.20 Ang (but according to the cited table it should be 1.09 Ang = 2.05980133 au). Is there a reason for this difference?
To see the effect of the difference radii, I made a test calculation for methane (RI-BP86/def2-TZVP):
Way q(Carbon) q(Hydrogen) just \$esp_fit kollman -0.621 0.155 with explicitly giving default radii1 -0.621 0.155 with radii of the cited table2 -0.684 0.171
1 using \$vdw_radii c 3.21253429 h 2.26767143
2 using \$vdw_radii c 3.21253429 h 2.05980133
As expected the charges using the default radii explicitly agree perfectly with the calculation where no radii values where given. But if I use the values of the cited table I get quite some differences. I don't know if there might be a good reason to change the VdW-radius of hydrogen (or maybe the cited table was updated after the values where programmed into TM). Maybe somebody can give a comment on this.
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# What are the principles of Bayesian networks in biostatistics?
## What Is The Best Course To Take In College?
In between these requirements is a common focus in biostatistics issues. Some researchers generally focus on some of the problems of biology, like the balance between computational efficiency and the effectiveness of biological interpretation. Biological models and Bayesian inference are often used as an extremely powerful tool for addressing the complexities of systems construction in the real world. In this way, biologists can easily see information which may be used in both computational and physically-based designs to construct a more efficient model of biological systems. We do not know much about biology from the real world. In this short review, we will concentrate on the importance of the characteristics of Bayesian networks under consideration in biWhat are the principles of Bayesian networks in biostatistics? These problems occur in several disciplines and are tackled by a number of disciplines. But as has been suggested in some recent textbooks, these problems are not restricted to any particular field, e.g. neurobiological or psychological sciences. A list of chapters in the book is also available on the Internet. The field often presents two types of interactions in which two things, when they are exchanged, occur, namely the transfer of features, and the transfer of consequences, depending on the information taken – this is a mathematical problem. But of course it is necessary to develop a method in which the underlying relationship between variables is thought to be preserved in the relevant context in which the interactions occur to conserve the information that can be exchanged. Some theorists have tried to break this relationship by considering the specific case of information transfer in the form of a “joint process”. What are the principles of Bayesian networks in biostatistics? There are some theoretical issues which deserve consideration as well.1 The well-known results of Bayesian networks contain explicit examples of information transfer in which a lot of features are also transferred, but as well as the transfer of information in this context there are many different situations where the information actually transferred is smaller than the information sought. An example appears to be the transfer of features (say a feature vector, label, and representation) from features to features in some task, i.e., from feature to feature with one type of non-zero vector combination, e.g., from one feature to another features, which is one common example in the system of cognitive map models.
## Upfront Should Schools Give Summer Homework
This doesn’t mean that all features have to be acquired in the non-transferable, but these do limit the overall transfer function. Furthermore, even if information representations do not exist, a process which can be carried out on feature vector with an exponentially-decreased number of features as well as a count of (a long term) activations may still provide
How to use Gretl for recommendation systems based on collaborative filtering and content-based filtering for
How to estimate a Poisson regression model for count data. In this article we apply
How to conduct a realist review in biostatistical research? Policies at stake and the risks
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All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Primal Magic Square (Posted on 2004-03-09)
Find a 3x3 magic square that is composed of 9 prime numbers (not the numbers from 1-9) and show how you found it.
(A magic square, as you may already know, is one in which the respective sums of the numbers in all the rows, columns, and both major diagonals all add up to the same number.)
_______________________
Since "Magic Square" is a term used outside the scope of this problem, I'm sure you can find an answer on the internet. Please find a solution independently.
See The Solution Submitted by SilverKnight Rating: 3.0000 (4 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): lower square - failure | Comment 6 of 15 |
(In reply to re: lower square - failure by Tristan)
to give a form let our magic square look like this:
(x-a) (x+d) (x+c)
(x+b) ( x ) (x-b)
(x-c) (x-d) (x+a)
Without loss of generality, all variables in this case can be considered positive.
comparing sums of opposite columns teaches us that b=a+c
comparing sums of top and bottom rows teaches a=c+d
and likewise b=2c+d
Posted by Eric on 2004-03-09 21:23:50
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Cody
# Problem 481. Rosenbrock's Banana Function and its derivatives
Solution 1213201
Submitted on 15 Jun 2017 by Chris Cleveland
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
2 Pass
x = [0; 0]; assert(isequal(Rosenbrock_banana(x),1))
3 Pass
x = [1; 1]; assert(isequal(Rosenbrock_banana(x),0))
4 Pass
x = [1; -1]; assert(isequal(Rosenbrock_banana(x),400))
5 Pass
x = [-1; 0.5]; assert(isequal(Rosenbrock_banana(x),29))
6 Pass
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GuruFocus has detected 6 Warning Signs with Union Pacific Corp \$UNP.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
Union Pacific Corp (NYSE:UNP)
Return on Capital
14.64% (As of Sep. 2016)
Return on capital measures how well a company generates cash flow relative to the capital it has invested in its business. It is also called Return on Invested Capital (ROIC). Union Pacific Corp's annualized return on capital (ROC) for the quarter that ended in Sep. 2016 was 14.64%.
As of today, Union Pacific Corp's weighted average cost Of capital is 8.62%. Union Pacific Corp's return on capital is 13.60% (calculated using TTM income statement data). Union Pacific Corp generates higher returns on investment than it costs the company to raise the capital needed for that investment. It is earning excess returns. A firm that expects to continue generating positive excess returns on new investments in the future will see its value increase as growth increases.
Definition
Union Pacific Corp's annualized Return on Capital (ROC) for the fiscal year that ended in Dec. 2015 is calculated as:
Return on Capital (ROC) (A: Dec. 2015 ) = NOPAT / Average Invested Capital = Oper. Inc.*(1-Tax Rate) / ( (Invested Capital (A: Dec. 2014 ) + Invested Capital (A: Dec. 2015 )) /2) = 8052 * ( 1 - 37.67% ) / ( (31016 + 33512) /2) = 5018.8116 / 32264 = 15.56 %
Invested Capital (A: Dec. 2014 ) = Book Value of Debt + Book Value of Equity - Cash = Long-Term Debt + Short-Term Debt + Minority Interest + Total Equity - Cash = 10952 + 461 + 0 + 21189 - 1586 = 31016
Invested Capital (A: Dec. 2015 ) = Book Value of Debt + Book Value of Equity - Cash = Long-Term Debt + Short-Term Debt + Minority Interest + Total Equity - Cash = 13607 + 594 + 0 + 20702 - 1391 = 33512
Union Pacific Corp's annualized Return on Capital (ROC) for the quarter that ended in Sep. 2016 is calculated as:
Return on Capital (ROC) (Q: Sep. 2016 ) = NOPAT / Average Invested Capital = Oper. Inc.*(1-Tax Rate) / ( (Invested Capital (Q: Jun. 2016 ) + Invested Capital (Q: Sep. 2016 )) /2) = 7840 * ( 1 - 37.34% ) / ( (33448 + 33657) /2) = 4912.544 / 33552.5 = 14.64 %
where
Invested Capital (Q: Jun. 2016 ) = Book Value of Debt + Book Value of Equity - Cash = Long-Term Debt + Short-Term Debt + Minority Interest + Total Equity - Cash = 14777 + 409 + 0 + 20422 - 2160 = 33448
Invested Capital (Q: Sep. 2016 ) = Book Value of Debt + Book Value of Equity - Cash = Long-Term Debt + Short-Term Debt + Minority Interest + Total Equity - Cash = 15205 + 407 + 0 + 20284 - 2239 = 33657
Note: The Operating Income data used here is four times the quarterly (Sep. 2016) operating income data.
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
Return on Capital measures how well a company generates cash flow relative to the capital it has invested in its business. It is also called Return on Invested Capital. The reason book values of debt and equity are used is because the book values are the capital the company received when issuing the debt or receiving the equity investments.
There are four key components to this definition. The first is the use of operating income rather than net income in the numerator. The second is the tax adjustment to this operating income, computed as a hypothetical tax based on an effective or marginal tax rate. The third is the use of book values for invested capital, rather than market values. The final is the timing difference; the capital invested is from the end of the prior year whereas the operating income is the current years number.
Why is Return on Capital important?
Because it costs money to raise capital. A firm that generates higher returns on investment than it costs the company to raise the capital needed for that investment is earning excess returns. A firm that expects to continue generating positive excess returns on new investments in the future will see its value increase as growth increases, whereas a firm that earns returns that do not match up to its cost of capital will destroy value as it grows.
As of today, Union Pacific Corp's weighted average cost Of capital is 8.62%. Union Pacific Corp's return on capital is 13.60% (calculated using TTM income statement data). Union Pacific Corp generates higher returns on investment than it costs the company to raise the capital needed for that investment. It is earning excess returns. A firm that expects to continue generating positive excess returns on new investments in the future will see its value increase as growth increases.
Be Aware
Like ROE and ROA, ROC is calculated with only 12 months of data. Fluctuations in the companys earnings or business cycles can affect the ratio drastically. It is important to look at the ratio from a long term perspective.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Union Pacific Corp Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 ROC 9.53 11.44 8.96 12.32 13.72 15.57 16.21 18.00 15.56 13.56
Union Pacific Corp Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 ROC 18.70 19.01 15.63 15.12 16.94 14.49 12.69 12.51 14.64 14.60
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# A Transformer Has a Turn Ratio Ns/Np
By BYJU'S Exam Prep
Updated on: September 25th, 2023
A transformer has a turn ratio, NS/NP=4. If a 200 V AC voltage is applied across its primary coil, and it carries 1 A current, find current in the circuit connected to the secondary coil if the transformer is 80% efficient.
1. 0.2 A
2. 0.4 A
3. 0.8 A
4. 2 A
A transformer has a turn ratio, NS/NP=4. If a 200 V AC voltage is applied across its primary coil, and it carries 1 A current, find current in the circuit connected to the secondary coil if the transformer is 80% efficient.
Solution
Given that, a transformer has a turns ratio NS/NP=4
Primary voltage (VP)=200 V
NS/NP=VS/VP=4
Vs=4×200=800 V
Given that, the efficiency of the transformer =VsIsVPIP=80%
0.8=800×Is/200×1
Is=160/800=0.2 A
Hence option A is the correct answer.
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# D harmonic minor scale
The Solution below shows the D harmonic minor scale notes, intervals and scale degrees on the piano, treble clef and bass clef.
The Lesson steps then explain how to identify the D harmonic minor scale note interval positions, choose the note names and scale degree names.
For a quick summary to this topic, have a look at Harmonic minor scale.
All Keys On 1 page
## Solution - 2 parts
### 1. D harmonic minor scale
This step shows the ascending D harmonic minor scale on the piano, treble clef and bass clef. It also shows the scale degree names for all 8 notes.
The D harmonic minor scale has 1 flat, 1 sharp.
This harmonic minor scale is based on the natural minor scale with the same key / tonic note - D natural minor scale.
Since the natural minor key is itself on the Circle of 5ths - D minor on circle of 5ths, this means that this is a commonly used harmonic minor scale key.
D harmonic minor scale note names
Note no.Note intervalNote name
1tonicThe 1st note of the D harmonic minor scale is D
2D-maj-2ndThe 2nd note of the D harmonic minor scale is E
3D-min-3rdThe 3rd note of the D harmonic minor scale is F
4D-perf-4thThe 4th note of the D harmonic minor scale is G
5D-perf-5thThe 5th note of the D harmonic minor scale is A
6D-min-6thThe 6th note of the D harmonic minor scale is Bb
7D-maj-7thThe 7th note of the D harmonic minor scale is C#
8D-perf-8thThe 8th note of the D harmonic minor scale is D
Middle C (midi note 60) is shown with an orange line under the 2nd note on the piano diagram.
These note names are shown below on the treble clef followed by the bass clef.
On the bass clef, Middle C is shown with an orange ledger line above the main 5 staff lines.
The stave diagrams above show the scale notes without a key signature, with the sharp / flat adjustments inserted before each note on the staff.
For the key signature of this scale, showing the symbols grouped correctly next to the bass or treble clef symbol at the beginning, have a look at the D harmonic minor key signature.
D harmonic minor scale degrees
Note no.Degree name
1D is the tonic of the D harmonic minor scale
2E is the supertonic of the D harmonic minor scale
3F is the mediant of the D harmonic minor scale
4G is the subdominant of the D harmonic minor scale
5A is the dominant of the D harmonic minor scale
6Bb is the submediant of the D harmonic minor scale
7C# is the leading tone of the D harmonic minor scale
8D is the octave of the D harmonic minor scale
The difference between the D harmonic minor scale and the D natural minor scale is that the 7th note position of the minor scale is raised by one half-tone / semitone.
So whereas the D natural minor scale has note C for the 7th note, this note is raised to arrive at note C# for this harmonic minor scale.
### 2. D harmonic minor scale descending
This step shows the descending D harmonic minor scale on the piano, treble clef and bass clef.
No. Note 1 2 3 4 5 6 7 C# Bb A G F E D
## Lesson steps
### 1. Piano key note names
This step shows the white and black note names on a piano keyboard so that the note names are familiar for later steps, and to show that the note names start repeating themselves after 12 notes.
The white keys are named using the alphabetic letters A, B, C, D, E, F, and G, which is a pattern that repeats up the piano keyboard.
Every white or black key could have a flat(b) or sharp(#) accidental name, depending on how that note is used. In a later step, if sharp or flat notes are used, the exact accidental names will be chosen.
The audio files below play every note shown on the piano above, so middle C (marked with an orange line at the bottom) is the 2nd note heard.
### 2. D tonic note and one octave of notes
This step shows an octave of notes in the key of D, to identify the start and end notes of the scale.
The numbered notes are those that might be used when building this note scale.
But since this is a scale in the key of D, it is certain that notes 1 and 13 will be used in the scale.
Note 1 is the tonic note - the starting note - D, and note 13 is the same note name but one octave higher.
No. Note 1 2 3 4 5 6 7 8 9 10 11 12 13 D D# / Eb E F F# / Gb G G# / Ab A A# / Bb B C C# / Db D
### 3. D harmonic minor scale note interval positions
This step applies the harmonic minor scale note interval pattern starting from D, so that the correct piano keys and note pitches can be identified.
The harmonic minor scale uses the W-H-W-W-H-W½-H note counting rule to identify the scale note positions.
To count up a Whole tone, count up by two physical piano keys, either white or black.
To count up a Half-tone (semitone), count up from the last note up by one physical piano key, either white or black.
To count up a tone (whole-tone and a half), count up from the last note by 3 half-tones / semitones - shown as 3 on the piano below.
The tonic note (shown as *) is the starting point and is always the 1st note in the harmonic minor scale.
Again, the final 8th note is the octave note, having the same name as the tonic note.
No. Note 1 2 3 4 5 6 7 8 D E F G A A# / Bb C# / Db D
What is the difference between the D harmonic minor scale and the D natural minor scale ?
The 7th note position (or scale degree) of the minor scale is raised by one half-tone / semitone to arrive at the harmonic minor scale note positions shown above, leaving a noticeably large interval of 3 half-tones / semitones between the 6th and 7th note positions for this scale.
### 4. D harmonic minor scale notes
This step tries to assign note names to the piano keys identified in the previous step, so that they can be written on a note staff in the Solution section.
The 7 unique notes in a scale need to be named such that each letter from A to G is used once only, so each note name is either a natural white name(A.. G) , a sharp(eg. F#) or a flat(eg. Gb).
This is needed to ensure that when it comes to writing the scale notes on a musical staff (eg. a treble clef), there is no possibility of having 2 G-type notes, for example, with one of the notes needing an accidental next to it on the staff (a sharp, flat or natural symbol).
To apply this rule, firstly list the white key names starting from the tonic, which are shown the White column below.
Then list the 7 notes in the scale so far, shown in the next column.
For each of the 7 notes, look across and try to find the White note name in the Scale note name.
If the natural white note can be found in the scale note, the scale note is written in the Match? column.
The 8th note - the octave note, will have the same name as the first note, the tonic note.
D harmonic minor scale
No.WhiteScale noteMatch?
1DDD
2EEE
3FFF
4GGG
5AAA
6BA# / BbBb
7CC# / DbC#
8DDD
For this harmonic minor scale, all notes have a match, and so the Match? column shows the harmonic minor scale note names.
### 5. D harmonic minor scale descending
This step shows the notes when descending the D harmonic minor scale, going from the highest note sound back to the starting note.
For harmonic minor scales, the notes names when descending are just the reverse of the ascending names.
So assuming octave note 8 has been played in the step above, the notes now descend back to the tonic.
No. Note 1 2 3 4 5 6 7 C# Bb A G F E D
### 6. D harmonic minor scale degrees
This step shows the D harmonic minor scale degrees - tonic, supertonic, mediant, subdominant, dominant, submediant, leading note / tone, and octave.
In music theory, each note in this scale has what is called a scale degree name, which describes the relationship of that note to the tonic(1st) note.
Scale degree names 1,2,3,4,5,6, and 8 below are always the same for all major and minor scales (ie. 1st note is always tonic, 2nd is supertonic etc.) , but obviously the note names will be different for each scale / key combination.
In the harmonic minor scale, the 7th note is called the leading note or leading tone because the sound of the 7th note feels like it wants to resolve and finish at the octave note, when all scale notes are played in sequence.
It does this because in this scale, the 7th note is only 1 half-tone / semitone away from the 8th note - the octave note. The D major scale and D melodic minor scale scales share the same property - they both have only one half-tone / semitone between the 7th and 8th notes.
In contrast, the D natural minor scale has a whole tone (two half-tones / semitones, two notes on the piano keyboard) between the 7th and 8th notes, and the 7th note does not lean towards the 8th note in the same way. In this case, the 7th note is called the subtonic.
D harmonic minor scale degrees
Note no.Degree name
1D is the tonic of the D harmonic minor scale
2E is the supertonic of the D harmonic minor scale
3F is the mediant of the D harmonic minor scale
4G is the subdominant of the D harmonic minor scale
5A is the dominant of the D harmonic minor scale
6Bb is the submediant of the D harmonic minor scale
7C# is the leading tone of the D harmonic minor scale
8D is the octave of the D harmonic minor scale
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Solve This (◻ + ◻ + ◻=30) [CLOSED]
506,804 views
Solve this..
◻ + ◻ + ◻=30
Fill the boxes using
(1, 3, 5, 7, 9, 11, 13, 15)
U can also repeat the numbers.
closed with the note: Enough solutions
posted Mar 4, 2015
Use the commas given in the question to your advantage: In many countries the comma (,) is used instead of the full stop (.) to represent a decimal point. For example "3.5" (three and a half) can be written as "3,5". This allows for the solution 11,3 + 15,7 + 3 = 30.
11+13+6(inverse 9)
the only solution,no other solution.
answer Jun 13, 2015 by anonymous
(1+1)+13+15=30
answer Jun 13, 2015 by anonymous
(11-1)+ (13-3)+(15-5)=30
answer Jun 13, 2015 by anonymous
(15-5) + (13-3) + (11-1) =30
answer Jun 13, 2015 by anonymous
(11+1) + (3) + (15)=30
answer Sep 23, 2016 by anonymous
15+15+_=30
Its not written anywhere that v HV to fill all the boxes
Its written fill *the boxes*
answer Sep 24, 2016 by anonymous
L3 + 15 + 9 = 30
answer Sep 26, 2016 by anonymous
What is L3
What is the correct solution to this.
(15-15)+(15)+(15)=30 if you are going to use subtraction concept please use this kind of simple concept
15+9+6=30
We can repeat the no. So rotate 9 180degree to make 6.
Other option in the same way is:
13+11+6=30
Similar Puzzles
What number comes next in this sequence: 1 4 8 13 21 30 36 ??
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Intermediate problems (1)
Algebra Level 5
I would be posting a series of intermediate-leveled problems for enjoyment.
Find all solutions to the following equations in reals:
$$\frac{1}{x}+\frac{2}{y}+\frac{3}{z}=1$$
$$xyz=162$$
$$|x-|x||+|y-|y||+|z-|z||=0$$
If there are $$n$$ solutions are in the form $$(x, y, z)=(x_{i}, y_{i}, z_{i}), i=1, 2, 3, …, n$$, find $\sum_{i=1}^n x_{i}$
This problem is part of the set
Intermediate Problems
×
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http://math.stackexchange.com/questions/234478/can-halls-theorem-be-derived-from-tutteberge-formula?answertab=oldest
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# Can Hall's theorem be derived from Tutte–Berge formula?
Can Hall's theorem be derived from Tutte–Berge formula?
Hall's theorem is for existence of X-saturated matching in a X,Y bipartite graph.
Tutte–Berge formula is for maximum size of a matching:
The theorem states that the size of a maximum matching of a graph $G = (V, E)$ equals
$$\frac{1}{2} \min_{U\subseteq V} \left(|U|-\text{odd}(G-U)+|V|\right).$$
where $odd(H)$ is the number of components in the graph H with an odd number of vertices.
If apply Tutte–Berge formula to a X,Y bipartite graph which has a X-saturated matching, then since the maximum size of a matching is $|X|$, we should have
$$\frac{1}{2} \min_{U\subseteq V} \left(|U|-\text{odd}(G-U)+|V|\right) = |X|,$$
which is equivalent to that $U=X$ maximizes $odd(G-U) - |U|$, at which $odd(G-X) - |X| = |Y|$.
But this is nothing like Hall's condition: for any subset $W$ of $X$, $$|W| \leq |N(W)|,$$ although I can vaguely and inexplicitly see that they are likely to be equivalent.
So I think Hall's theorem can be derived from Tutte-Berge formula?
Thanks!
-
add comment
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## APK_Prove Properties of Parallelograms - Section 1: Activating Prior Knowledge
APK_Prove Properties of Parallelograms
# Proving Properties of Parallelograms
Unit 5: Quadrilaterals
Lesson 2 of 8
## Big Idea: You say this is your property....Prove it! It may sound like a land rights dispute, but in this lesson students are proving the properties of parallelograms.
Print Lesson
1 teacher likes this lesson
Standards:
Subject(s):
Math, Geometry, quadrilaterals, parallelogram, Polygons, properties of quadrilaterals
63 minutes
### Anthony Carruthers
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# Pharos University 2012-2013, fall term Faculty of Engineering ...
Πολεοδομικά Έργα
25 Νοε 2013 (πριν από 4 χρόνια και 5 μήνες)
181 εμφανίσεις
50
4 16
450
0
200
Pharos
University
20
1
2
-
20
1
3
, fall term
Faculty of Engineering
Reinforced Concrete
Design
-
2
Construction Engineering and management Dept
.
Due date:
0
2
-
10
-
201
2
Exercise (
1
)
:
Revision
Materials to be used:
Concrete
:
f
cu
=25 N/mm
2
,
Steel
:
St.360/520 (for longitudinal reinforcement) & St.240/350 (for stirrups)
Question (1)
a)
Calculate the area of the balanced tensile reinforcement
(
A
sb
) for a rectangular concrete section
(
b
=200mm,
d
=540mm) and the
corresponding ultimate moment
(
M
b
).
b)
Using
design charts provided in E
CP
203 design Aids
,
calculate the reinforcement
required for a
reinforced concrete T
-
section to resist an ultimate bending moment (
M
u
=100kN.m) given that:
section depth (
d
=630mm), web width (
b
=250mm)
slab thickness (
t
=100mm),
effective flange width (
B
=1200mm)
Question (2)
a)
For the shown reinforced concrete rectangular section,
calculate
the
service
moment capacity
.
b)
Calculate the reinforcement
required for the given rectangular
concrete section if the section is subjected to a service bending
moment (M=80kN.m) using the working stress design method.
Question (3)
a)
Design a circular reinforced concrete short column
to resist an ultimate load (
P
u
=3000
kN).
Question (4)
20%
a)
Define the
development length
of reinforcing bars.
b)
The given figure shows the main
dimensions of a simply supported
reinforced concrete beam and the shearing
force diagram resulting from the ultimate
loads acting on that beam. Assuming that
the shear stresses are res
isted by vertical
stirrups only,
calculate the required web
reinforcement
over the full length of the
beam, given that (
b
=200
mm) and
(
d
=500
mm).
0.30
0.30
S.F.D.
150kN
150kN
100kN
100kN
50kN
50kN
2.00 m
2.00 m
2.00 m
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# What Is The Term Used To Describe The Rate Of An Object’S Movement?
## What Is The Time period Used To Describe The Price Of An Object’s Motion??
Pace is an outline of how briskly an object strikes velocity is how briskly and in what course it strikes.
## What’s the time period used to explain the speed of an objects motion?
Velocity. When each the pace and course of an object’s movement the speed of the. object. Pace in a given course is named velocity.
## Which time period greatest describes a change in an object’s movement?
acceleration. happen throughout you as objects pace up decelerate or change course. altering course. a change in velocity could be both a change in how briskly one thing is transferring or change within the course of motion.
## What’s the time period for the speed at which an object is transferring at a given immediate in time?
Instantaneous pace is the speed at which an object is transferring at a given immediate in time. When each the pace and distance of an object’s movement the speed of the thing.
## What’s the charge of change in an object’s place?
The rate of an object is the speed of change of its place with respect to time. The pace of an object is the magnitude of its velocity.
## Which refers back to the charge of change in velocity quizlet?
Acceleration is the speed of change in an object’s velocity.
## What’s the charge of movement of an object?
Velocity is a vector amount that’s outlined as the speed of change of place of an object with respect to a body of reference with respect to time. Velocity is equal to the pace of the thing together with the course of movement.
## Which of the next phrases describes an object’s resistance to a change in movement?
Inertia is the resistance of any bodily object to any change in its velocity. This consists of modifications to the thing’s pace or course of movement.
## How are you going to greatest describe an object’s movement?
You may describe the movement of an object by its place pace course and acceleration. An object is transferring if its place relative to a set level is altering.
## What’s the charge of change of velocity is named?
Acceleration is outlined as the speed of change of velocity. Velocity is a vector which implies it accommodates a magnitude (a numerical worth) and a course.
## What’s movement used to explain?
In physics movement is the phenomenon wherein an object modifications its place. Movement is mathematically described when it comes to displacement distance velocity acceleration pace and time.
## What refers back to the measure of how briskly an object transfer?
Pace is a scalar amount that refers to “how briskly an object is transferring.” Pace could be regarded as the speed at which an object covers distance.
## What’s the charge at which an object modifications its velocity?
The rate is altering over the course of time. In truth the speed is altering by a relentless quantity – 10 m/s – in every second of time. Anytime an object’s velocity is altering the thing is alleged to be accelerating it has an acceleration.
## What modifications an object’s movement?
The movement of an object is set by the sum of the forces appearing on it if the full power on the thing isn’t zero its movement will change. The higher the mass of the thing the higher the power wanted to attain the identical change in movement. For any given object a bigger power causes a bigger change in movement.
## What describes a change in velocity?
Acceleration is a measure of the change in velocity of a transferring object. It measures the speed at which velocity modifications. … Folks generally consider acceleration as a rise in pace however a lower in pace can also be acceleration. On this case acceleration is destructive and referred to as deceleration.
## What’s velocity quizlet?
Velocity. Measure of distance traveled in a given time frame with a course. Fixed velocity.
## What refers back to the movement of an object transferring in a straight line?
Linear movement additionally referred to as rectilinear movement is one-dimensional movement alongside a straight line and might subsequently be described mathematically utilizing just one spatial dimension.
## What’s the technical time period used for the speed of movement of a physique?
pace or velocity is the technical time period used for the speed of movement of a physique.. it’s unit is m/s & km/hr.
## What is supposed by charge in physics?
In physics charge means charge of change. Principally how a lot a sure amount modifications with respect to a different amount that can also be altering. You may additionally typically hear the time period gradient which describes the identical factor. And mathematically it is usually a ratio say for instance.
## What’s resistance to alter referred to as?
It is a phenomenon referred to as structural inertia. Teams groups departments or management might resist change on account of this type of inertia. … When prospects shoppers or different exterior stakeholders resist change they’re participating in exterior resistance.
## What time period is greatest outlined as an object’s tendency to not change its movement?
Inertia is the tendency of an object to withstand modifications in its state of movement. … The state of movement of an object is outlined by its velocity – the pace with a course.
## What measures the resistance to a change within the object’s movement?
Mass as a Measure of the Quantity of Inertia
See additionally the place is athens in greece
All objects resist modifications of their state of movement. … The extra inertia that an object has the extra mass that it has. A extra large object has a higher tendency to withstand modifications in its state of movement.
## How do you describe the movement of a graph?
If an object strikes alongside a straight line the gap travelled could be represented by a distance-time graph. In a distance-time graph the gradient of the road is the same as the pace of the thing. The higher the gradient (and the steeper the road) the quicker the thing is transferring.
Classes FAQ
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## Favourited
Are you better with words or numbers?
## Top Comment
Words-but I am an excellent cricket scorer.
#### Blewynanifail
Numbers. They are much less ambiguous.
#### PhilboydStudge
I hope I'm better with words! That's how we communicate with each other. And there are far more words than numbers. Enabling us to get the meaning j-u-s-t right!
#### In response to “I hope I'm better with words! That's how we...”
Numbers are infinite.
#### In response to “Numbers are infinite.”
I meant there are ten numbers 0 thru 9.
10-4?
Ahhh.
26 letters.
Words.
Words.
#### Carla
Words....when I am in the mood.
#### Tiffanee
Words-but I am an excellent cricket scorer.
Words.
#### Linnster
Words, no contest! Math is not my long suit.
Two words for you: ****.
#### In response to “Two words for you: ****.”
You are a big giant farting ****
#### In response to “You are a big giant farting ****”
Well, I have **** attached to my shoulders, as you well know.
#### ZonkeyBalls
Words, though I like sudoku
#### BlindMist
Numbers. They can be manipulated, but not nearly as much as words can. :)
#### Will_Janitor
That's hard for me to say, because I am good with both words and numbers (but in my old age I find that I am starting to have trouble remembering certain words that I used to know).
#### In response to “That's hard for me to say, because I am good...”
I get that too sometimes. Suddenly a word simply escapes my vocabulary! It may take a just a moment to find, sometimes longer. It's annoying as all get ... [shoot, what's the word I'm looking for?]
"out"?
yES! tHAT'S IT!
#### Budwick
I'm better with words.
#### PartyOfOne
Way better with numbers
Forty two
#### In response to “Forty two”
I'm LOUSY with numbers. I remember the time 3 of us went to this hotel, see, and we had \$30 bucks in 10s between us but the room cost \$27 and the clerk hadn't the proper change so...
#### In response to “I'm LOUSY with numbers. I remember the time 3...”
LMAO!! Still remember that one? Good for you. :)
Yup!
## Sort comments by: Replies Date Score Loves
Amirite is the premier opinion-based social network where people from all around the world discover, debate and discuss today's hottest issues. Share your perspective to the world and interact with like-minded individuals on breaking news, hot topics and controversial issues now!
With that many angles, the discussions on Amirite will open your eyes to a panoramic view of your world that you won't get anywhere else, allowing you to see the big picture and discuss it.
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# Statictics
posted by .
What is the probability of tossing a number divisible by two, and the probability of tossing a prime number of a "Fair Die"?
• Statistics -
2, 4 or 6, each with 1/6 probability. To find either-or probability, add the individual probabilities.
Primes are 2, 3 or 5. Use same process.
## Similar Questions
1. ### Probability
Consider the experiment of simultaneously tossing a die and a coin. Let X denote the number of heads and Y denote the number of spots showing on the die. a)construct a two-way table for the joint pdf b)Let Z=X+Y . Use the joint pdf …
2. ### math
Consider the experiment of simultaneously tossing a fair coin and rolling a fair die. Let X denote the number of heads showing on the coin and Y the number of spots showing on the die. a. List the outcomes in S. b. Find Fx,y(1,2). …
A fair, six-sided die is rolled eight times, to form an eight-digit number. What is the probability that the resulting number is a multiple of 8?
A coin is tossed twice. What is the probability of tossing 2 tails?
5. ### Math7 Am I right?
Below are the results of tossing a number cube 8 times. Find the experimental probability of tossing an even number. 3,4,1,2,5,1,6,5 3/8 right?
6. ### math
below are the results of tossing a number cube 8 times find the experimental probability of tossing an even number. 3,4,1,2,5,1,6,5 1. 1/4 2. 3/8 3. 1/2 4. 5/8
7. ### Math
Find the probability for the experiment of tossing a six-sided die twice. Put the answer in a/b form. P(the sum is odd or prime)
8. ### math
Write the number of permutations in factorial form. Then simplify. How many different ways can you and four of your friends sit in the backseat of a limousine?
9. ### math
Below are the results of tossing a number cube 9 times. Find the experimental probability of tossing an odd number. 4,3,6,6,2,5,3,5,1. a. 4/9 b. 1/2 c. 5/9 d. 2/3 ** am i correct?
10. ### Statistics
Find the probability of getting a prime number thrice by tossing a die 10 times.
More Similar Questions
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# Math
Ineed to come up with an EXCEL formula for the following;
Ineed to determine the total number of games.
I know the % (games) of lost and I know the number of (game) wins.
I need to determine the total number of games.
I can figure the % of wins by subtracting the % lost from 100.
Ex 20% of games were lost
100 - 20 = 80
80 = percent won
80/100 = 30 (number of wins)
Question: If 30 wins is 80%, How many losses is 20%? How many games were there in total?
1. 80% = 30
.80x = 30
x = 30/.80
x = 37.5 total games
20% of 37.5 = 7.5 games lost
30 wins
7.5 lost
37.5 total games
Not sure what you are trying to do in Excel.
posted by helper
2. ok thanks!
posted by Michelle
3. You're welcome :)
posted by helper
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Browse All Plans
Just Plans
# Flower Worksheet Quiz Maleemale Components Oflowers Study Com Structure And Reproduction Pdf
By Luc Paquet at December 24 2018 04:05:18
There are several standard exercises which train students to convert percentages, decimals and fractions. Converting percentage to decimals for example is actually as simple as moving the decimal point two places to the left and losing the percent sign "%." Thus 89% is equal to 0.89. Expressed in fraction, that would be 89/100. When you drill kids to do this often enough, they learn to do conversion almost instinctively.
They're also available for nearly all grade levels. There are printable middle school, high school, elementary school, and even pre-school worksheets.
Save Time : Free worksheets not only save you money, they can also save you time. If you decide that it is best for your child to do worksheets especially tailored for his needs, by doing a little research for printable math worksheets found online, you don't have to make the worksheets yourself. This can save a lot of time. Worksheets aren't that difficult to make, but it can be time consuming.
How much of an apple pie has been eaten? The answer to this question can be expressed in percentages, 50%; or in decimals, 0.5; or in fraction, ½. In other words, half of mom's delicious apple pie is gone.
From a teacher's perspective our competition is tough. Passing out a handout of 30 problems that are all in a format of 534x25= is not as stimulating in the students' eyes as playing games such as Grand Theft Auto and Resident Evil.
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# What is the Lewis dot diagram for NI3?
## What is the Lewis dot diagram for NI3?
We have 7 valence electrons but there’s 3 of them. So 5 plus 21 equals 26 total valence electrons let’s put nitrogen in the center it is the least electronegative.
## How many valence electrons does NI3?
There are a total of 26 valence electrons in NI3.
Is NI3 polar or nonpolar?
The molecules themselves have a trigonal (triangular) pyramid shape. They are, of course, polar.
How many lone pairs are in NI3?
NI3 Lewis structure lone pairs
Let us calculate how many lone pairs are there in NI3. NI3 Lewis structure holds total 10 lone pairs. After making bonds, Nitrogen is left with one lone pair. Each of the Iodine atoms holds three lone pairs, which adds total 9 lone pairs in the Lewis structure.
### Why is NI3 unstable?
The instability of NI3 and NI3 · NH3 can be attributed to the large steric strain caused by the three large iodine atoms being held in proximity to each other around the relatively tiny nitrogen atom.
### What is the name of NI3?
Nitrogen triiodide is the inorganic compound with the formula NI3. It is an extremely sensitive contact explosive.
How do you determine valence electrons?
For neutral atoms, the number of valence electrons is equal to the atom’s main group number. The main group number for an element can be found from its column on the periodic table. For example, carbon is in group 4 and has 4 valence electrons. Oxygen is in group 6 and has 6 valence electrons.
What is electron group arrangement?
Electron group geometry is the three-dimensional arrangement of atoms in a molecule. The geometry of a molecule is an important factor that affects the physical and chemical properties of a compound.
#### Why is NI3 stable when wet?
This is because when wet, there are ammonia molecules in the structure of the compound, which helps to stabilize it. When the NI3 dries, these ammonia molecules evaporate so the NI3 structure is left very unstable, needing only the slightest amount of energy from friction or heat to turn into nitrogen and iodine vapor.
#### Which is more stable NF3 or NI3?
NF3 is stable, but NCl3 NBr3 and NI3 are unstable.
Is ni3 possible?
Nitrogen triiodide is an inorganic compound with the formula NI3. It is an extremely sensitive contact explosive: small quantities explode with a loud, sharp snap when touched even lightly, releasing a purple cloud of iodine vapor; it can even be detonated by alpha radiation.
Why is ni3 unstable?
The reason nitrogen triiodide is so unstable is down to its molecular structure – one nitrogen atom and three iodine molecules all on one side due to how the nitrogen’s electrons are arranged.
## How do you find valence electrons in a Lewis structure?
1. Determine the total number of valence electrons in the molecule or ion. Add together the valence electrons from each atom. (Recall that the number of valence electrons is indicated by the position of the element in the periodic table.)
## What are the 7 valence electrons?
The family of elements having seven electrons in the valence shell is halogens, i.e., chlorine, fluorine, bromine, iodine, astatine.
How do you draw Lewis structures?
How to Draw a Lewis Structure
1. Step 1: Find the Total Number of Valence Electrons.
2. Step 2: Find the Number of Electrons Needed to Make the Atoms “Happy”
3. Step 3: Determine the Number of Bonds in the Molecule.
4. Step 4: Choose a Central Atom.
5. Step 5: Draw a Skeletal Structure.
6. Step 6: Place Electrons Around Outside Atoms.
How do you identify electron groups?
Thus, the “Electron Group” geometry of each central atom in a structure can be determined by simply counting the number of “groups” of electrons around the atom, then considering how those groups would arrange themselves to be as far apart as possible.
### What creates purple smoke?
The purple smoke is caused by the molecular iodine in the vapor.
### Why is NI3 less stable than NF3?
The other NX3 molecules may be less stable than NF3 owing to congestion round the small central nitrogen atom leading to non-bonded repulsive interactions between the halogens. This is particularly bad for large iodine atoms, as can be seen in the space-fill image of NI3, right. What is their structure?
Why NI3 unstable molecules explain?
How do you know the valence electrons?
#### Do Lewis structures show all valence electrons?
Lewis structures show all of the valence electrons in an atom or molecule. The valence electrons are the electrons in the outermost shell. For representative elements, the number of valence electrons equals the group number on the periodic table.
#### How do you find valence?
What are elements with 8 valence electrons called?
Any element in group 18 has eight valence electrons (except for helium, which has a total of just two electrons). Examples include neon (Ne), argon (Ar), and krypton (Kr).
How do you write the Lewis symbol?
Introduction. A Lewis Symbol is constructed by placing dots representing electrons in the outer energy around the symbol for the element. For many common elements, the number of dots corresponds to the element’s group number. Below are Lewis Symbols for various elements.
## How do Lewis dot diagrams work?
A Lewis diagram shows how the valence electrons are distributed around the atoms in a molecule. Shared pairs of electrons are drawn as lines between atoms, while lone pairs of electrons are drawn as dots next to atoms.
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## HowMany.wiki
Please get in touch with us if you:
1. Have any suggestions
2. Have any questions
3. Have found an error/bug
4. Anything else ...
# 16 ml of polenta in pounds
β
How many pounds of polenta in 16 milliliters? How much do 16 ml of polenta weigh?
16 milliliters of polenta equals 0.0238 pound(*)
### Inputs?Notes: the results in this calculator are rounded (by default) to 3 significant figures. The conversion factors are approximate once it is intended for recipes measurements. This is not rocket sciece ☺.
Significant Figures:
### Results
16 milliliters of polenta weighs 0.0238 pound. (*)
(*) or precisely 0.023845 pound. All values are approximate.
## Polenta Conversion Chart Near 16 milliliters
milliliters to pounds of Polenta
16 milliliters = 0.0238 pound
17 milliliters = 0.0253 pound
18 milliliters = 0.0268 pound
19 milliliters = 0.0283 pound
20 milliliters = 0.0298 pound
21 milliliters = 0.0313 pound
22 milliliters = 0.0328 pound
23 milliliters = 0.0343 pound
24 milliliters = 0.0358 pound
25 milliliters = 0.0373 pound
26 milliliters = 0.0387 pound
27 milliliters = 0.0402 pound
28 milliliters = 0.0417 pound
29 milliliters = 0.0432 pound
30 milliliters = 0.0447 pound
31 milliliters = 0.0462 pound
32 milliliters = 0.0477 pound
33 milliliters = 0.0492 pound
34 milliliters = 0.0507 pound
35 milliliters = 0.0522 pound
36 milliliters = 0.0537 pound
37 milliliters = 0.0551 pound
38 milliliters = 0.0566 pound
39 milliliters = 0.0581 pound
40 milliliters = 0.0596 pound
41 milliliters = 0.0611 pound
42 milliliters = 0.0626 (1/8) pound
43 milliliters = 0.0641 (1/8) pound
44 milliliters = 0.0656 (1/8) pound
45 milliliters = 0.0671 (1/8) pound
46 milliliters = 0.0686 (1/8) pound
47 milliliters = 0.07 (1/8) pound
48 milliliters = 0.0715 (1/8) pound
49 milliliters = 0.073 (1/8) pound
50 milliliters = 0.0745 (1/8) pound
51 milliliters = 0.076 (1/8) pound
52 milliliters = 0.0775 (1/8) pound
Note: Values are rounded to 3 significant figures. Fractions are rounded to the nearest 8th fraction.
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# Factset Interview Experience | On-Campus
I attended the FactSet drive at Gitam University, Visakhapatnam during August 2019. Factset interview process is a two-day process with four rounds in total. So the statistics are, from among 1500 students, 332 students were shortlisted for the written test, 44 for the first technical round, nearly 25 students for the second technical round, 11 students for the HR round and 7 got through the whole interview process, got placed.
Round 1 (Written Test):
There were 50 MCQs
25 – Quantitative aptitude (Time-distance, Pipes and cisterns, Areas, Trignometry, Percentage, Profit and loss)
25- Technical (OOPs concepts, Output for a given code snippet)
The level of questions was between easy and medium
Round 2(Technical Round 1):
Mostly concentrated on Data structures.
1. Find the factorial of a negative number using recursion. Solution
2. Given an array of positive numbers and a sum, find any pair of numbers which add up to give the given sum. Solution
3. Given two merged linked list, need to find the merging point of the node. Solution
4. Height of the binary search tree. Solution
5. Given 2 sorted arrays and merge them to get another sorted array. Solution
All the questions are standard questions with some twist in them.
Round 3(Technical Round 2):
There were a few resumes based questions and asked to explain every project by writing down the code. So be thorough with the whole resume and never fake it. Coming to the technical questions.
1. Given an array, find the second maximum element without sorting in O(n). Solution
2. What are a primary key, foreign key, and a simple SQL query? Explanation
3. Puzzle
Round 4(Interaction With Manager):
I would say it was neither HR nor MR as there was a mixture of technical questions, scenario-based, puzzles, and personal questions too. I was asked to tell what all mistakes were committed from the previous rounds and to correct them. This round got interesting during the scenario question, where I need to convince the manager to improve a part of source code which is been there for 20 years.
During the rounds, the interviewers were very friendly. After all the interviews there was celebration and gifts were given to all the selected members. I’m glad that I’m one of them.
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# 1579 It’s Inauguration Day!
### Today’s Puzzle:
This star puzzle is one thing I’m doing to commemorate this historic day when Joe Biden is inaugurated as the 46th President of the United States and Kamala Harris is inaugurated as Vice President! He will be the oldest person to become President, having previously served 36 years in the Senate and 8 years as Vice President, and she will be the first woman, the first African-American, and the first Asian-American Vice President. I wish them a beautiful day as they begin the hard work of uniting our country and finding solutions that benefit all of us.
The clues 10, 20, 30, and 40 have two common factors that might work for this mystery level puzzle. However, the other factors that go with one of those two choices will completely eliminate every possible factor pair for clue 4. That means you need to go with the other possibility.
### Factors of 1579:
• 1579 is a prime number.
• Prime factorization: 1579 is prime.
• 1579 has no exponents greater than 1 in its prime factorization, so √1579 cannot be simplified.
• The exponent in the prime factorization is 1. Adding one to that exponent we get (1 + 1) = 2. Therefore 1579 has exactly 2 factors.
• The factors of 1579 are outlined with their factor pair partners in the graphic below.
How do we know that 1579 is a prime number? If 1579 were not a prime number, then it would be divisible by at least one prime number less than or equal to √1579. Since 1579 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, or 37, we know that 1579 is a prime number.
### More About the Number 1579:
1579 is the sum of two consecutive numbers:
789 + 790 = 1579.
1579 is also the difference of two consecutive squares:
790² – 789² = 1579.
1579, 915799, 99157999, 9991579999, and 999915799999 are all prime numbers! Thanks to OEIS.org for alerting me to that fabulous fact!
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# In many surveys, American consumers have expressed a willingness to sp
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In many surveys, American consumers have expressed a willingness to spend up to 10 percent more for products that are ecologically sound. Encouraged by such surveys, Bleach-O Corporation promoted a new laundry detergent, Bleach-O Green, as safer for the environment. Bleach-O Green cost 5 percent more than typical detergents. After one year, Bleach-O Green had failed to capture a significant share of the detergent market and was withdrawn from sale.
Which of the following questions is LEAST likely to be relevant in determining the reasons for the failure of Bleach-O Green?
(A) How effective as a detergent was Bleach-O Green?
(B) How many other detergents on the market were promoted as safe for the environment?
(C) How much more did Bleach-O Green cost to manufacture than ordinary detergents?
(D) To what extent did consumers accept the validity of Bleach-O Green advertised and promoted to consumers?
(E) How effectively was Bleach-O Green advertised and promoted to consumers?
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Re: In many surveys, American consumers have expressed a willingness to sp [#permalink]
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19 Dec 2018, 13:47
1
Bunuel wrote:
In many surveys, American consumers have expressed a willingness to spend up to 10 percent more for products that are ecologically sound. Encouraged by such surveys, Bleach-O Corporation promoted a new laundry detergent, Bleach-O Green, as safer for the environment. Bleach-O Green cost 5 percent more than typical detergents. After one year, Bleach-O Green had failed to capture a significant share of the detergent market and was withdrawn from sale.
Which of the following questions is LEAST likely to be relevant in determining the reasons for the failure of Bleach-O Green?
(A) How effective as a detergent was Bleach-O Green?
(B) How many other detergents on the market were promoted as safe for the environment?
(C) How much more did Bleach-O Green cost to manufacture than ordinary detergents?
(D) To what extent did consumers accept the validity of Bleach-O Green advertised and promoted to consumers?
(E) How effectively was Bleach-O Green advertised and promoted to consumers?
Q: Why did the product not sell?
We are looking for something unrelated (least likely!!) for why it did not sell as well as expected.
A- Reasonable
B- Reasonable
C- Has nothing to do with reach of customer. Hold on.
D - Reasonable
E- Reasonable
IMO C
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Re: In many surveys, American consumers have expressed a willingness to sp [#permalink]
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07 Apr 2019, 21:54
Why c is right?....as we know may be price is high to produce such detergent than other detergent and eventually selling price will be high than other detergent for customers...and they are not buying due to high price
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Re: In many surveys, American consumers have expressed a willingness to sp [#permalink]
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07 Apr 2019, 23:29
Answer C is discussing cost of production of Bleach-O Green compared to ordinary detergent, which is out of scope. Consumers do not base their purchase decision on cost but on price.
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Re: In many surveys, American consumers have expressed a willingness to sp [#permalink]
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24 Jan 2020, 23:41
Bunuel wrote:
In many surveys, American consumers have expressed a willingness to spend up to 10 percent more for products that are ecologically sound. Encouraged by such surveys, Bleach-O Corporation promoted a new laundry detergent, Bleach-O Green, as safer for the environment. Bleach-O Green cost 5 percent more than typical detergents. After one year, Bleach-O Green had failed to capture a significant share of the detergent market and was withdrawn from sale.
Which of the following questions is LEAST likely to be relevant in determining the reasons for the failure of Bleach-O Green?
(A) How effective as a detergent was Bleach-O Green?
(B) How many other detergents on the market were promoted as safe for the environment?
(C) How much more did Bleach-O Green cost to manufacture than ordinary detergents?
(D) To what extent did consumers accept the validity of Bleach-O Green advertised and promoted to consumers?
(E) How effectively was Bleach-O Green advertised and promoted to consumers?
In C) the cost of manufacturing is still affecting the prize for the consumers, and therefore the market share. So i dont find this a perfect answer. But all the other answers have a more direct correlation to the market, so C) stands out a little bit. Thats why i chose it.
Re: In many surveys, American consumers have expressed a willingness to sp [#permalink] 24 Jan 2020, 23:41
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Search All of the Math Forum:
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Topic: 3 Y axes wih 4 variable
Replies: 9 Last Post: Apr 24, 2013 9:28 AM
Messages: [ Previous | Next ]
dpb Posts: 9,850 Registered: 6/7/07
Re: 3 Y axes wih 4 variable
Posted: Apr 20, 2013 10:06 AM
On 4/20/2013 2:24 AM, Murat wrote:
> Hi. I am new in Matlab and i have a problem having plotting 3 'Y' axes
> with one of the Y axes are common with two variable. it's like 4
> variable as in the image. How can i get the true code? Please help me out.
> http://img401.imageshack.us/img401/1750/3yaxeswith2common.png
See doc under 'Graphics -- Objects -- Axes Properties' there's a topic
titled "Using Multiple X= and Y-Axes"
From its example modified slightly...
x1 = [0:.1:40];
y1 = 4.*cos(x1)./(x1+2);
x2 = [1:.2:20];
y2 = x2.^2./x2.^3;
>> hl1 = line(x1,y1,'Color','r');
ax1 = gca;
set(ax1,'XColor','r','YColor','r')
pos_ax1 = get(ax1,'Position'); % get the position vector 1st axis
% set the left position for another; make width so right sides match...
pos_ax2=pos_ax1; pos_ax2(1)=0.25;
pos_ax2(3)=pos_ax1(3)+pos_ax1(1)-pos_ax2(1);
ax2 = axes('Position',pos_ax2,...
'XAxisLocation','bottom',...
'YAxisLocation','left',...
'Color','none',...
'XColor','k','YColor','k');
Salt to suit...
--
Date Subject Author
4/20/13 Murat
4/20/13 dpb
4/21/13 Murat
4/23/13 Murat
4/23/13 Murat
4/23/13 dpb
4/23/13 Murat
4/24/13 dpb
4/24/13 Murat
4/24/13 dpb
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# Math
22.Angles A and B are located in the first quadrant. If and , determine the exact value of .
How do I get 16/65?
1. 👍 0
2. 👎 0
3. 👁 261
1. If ???
1. 👍 0
2. 👎 0
👨🏫
bobpursley
2. Angles A and B are located in the first quadrant. If sin a =5/13 and cos b= 3/5 , determine the exact value of cos (a+b) .
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2. 👎 0
## Similar Questions
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# Solutions to congruence using Chinese remainder theorem
#### patrickmanning
f(x) = x^5 + 3x^2 + 4
Find all solutions to the congruence f(x) congruent to 0 mod 12 by using the Chinese remainder theorem.
I'm not sure exactly how to do this properly.
I started by creating two congruences since 12 = 4 x 3.
I manually tried entries 0 to 3 for mod 4 and entries 0 to 2 for mod 3.
The only common solution is x=2, so I tried x = 2 as well as x= 5, 8, 11 (because of mod 3) and x= 6, 10 (because of mod 4) in the original polynomial.
My final answer is x = 2, 5, or 8.
1) Is this right?
2) Whether or not it's right, how do I work this problem and present the solution properly?
#### Idea
$$\displaystyle x \equiv 2 \pmod 3$$
and
$$\displaystyle x \equiv 0,1,2 \pmod 4$$
Combining these we get 3 systems each of which can be solved using CRT
1 person
#### patrickmanning
"Combining these we get 3 systems each of which can be solved using CRT"
So, the three systems are:
1) x= 2mod3 and x= 2mod4 which gives the answer x = 2
2) x= 2mod3 and x= 1mod4 which gives the answer x = 5
3) x= 2mod3 and x= 0mod4 which gives the answer x = 8
Right?
Last edited:
#### Idea
Correct
$$\displaystyle x \equiv 2,5,8 \pmod {12}$$
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# Independence Friendly Logic
First published Mon Feb 9, 2009; substantive revision Thu Sep 15, 2022
Independence friendly logic (IF logic, IF first-order logic) is an extension of first-order logic. In it, more quantifier dependencies and independencies can be expressed than in first-order logic. Its quantifiers range over individuals only; semantically IF first-order logic, however, has the same expressive power as existential second-order logic. IF logic lacks certain metaproperties that first-order logic has (axiomatizability, Tarski-type semantics). On the other hand, IF logic admits a self-applied truth-predicate – a property that first-order logic notoriously does not enjoy. Philosophical issues discussed in connection with IF logic include reformulating the logicist program, the question of truth in axiomatic set theory, and the nature of negation. Work in IF logic has also inspired alternative generalizations of first-order logic: slash logic and dependence logic.
## 1. Introduction: Quantifier Dependence
In mathematical prose, one can say things such as ‘for all real numbers $$a$$ and for all positive real numbers $$\varepsilon$$, there exists a positive real number $$\delta$$ depending on $$\varepsilon$$ but not on $$a$$, such that…’ What is important here is quantifier dependence. The existential quantifier ‘there exists $$\delta$$’ is said to depend on the universal quantifier ‘for all $$\varepsilon$$’ but not on the universal quantifier ‘for all $$a$$.’ It was an essential part of Karl Weierstrass’s (1815–1897) work in the foundations of analysis that he defined the notions of limit, continuity, and derivative in terms of quantifier dependence.[1] For a concrete example, a function $$f : D \rightarrow \mathbf{R}$$ is continuous, if for all $$a$$ in the set $$D$$ and for all $$\varepsilon \gt 0$$ there exists $$\delta \gt 0$$ such that for all $$x$$ in $$D$$, if $$|x - a| \lt \delta$$, then $$|f(x) - f(a)| \lt \varepsilon$$. The definition of uniform continuity is obtained from that of continuity by specifying that the quantifier ‘there exists $$\delta$$’ depends only on the quantifier ‘for all $$\varepsilon$$,’ not on the quantifier ‘for all $$a$$.’[2]
Independence friendly first-order logic (a.k.a. IF first-order logic, IF logic) was introduced by Jaakko Hintikka and Gabriel Sandu in their article ‘Informational Independence as a Semantical Phenomenon’ (1989); other early sources are Hintikka’s booklet Defining Truth, the Whole Truth, and Nothing but the Truth (1991) and Sandu’s Ph.D. thesis (1991).[3] IF first-order logic is an extension of first-order logic, involving a specific syntactic device ‘/’ (slash, independence indicator), which has at the object language level the same effect as the meta-level modifier ‘but does not depend on’ has in the example just considered. In the notation of IF logic, the logical form of the statement that a function $$f$$ is uniformly continuous would be $$(\forall a)(\forall \varepsilon)(\exists \delta /\forall a)(\forall x) R$$, to be contrasted with the form of the statement that $$f$$ merely is continuous, $$(\forall a)(\forall \varepsilon)(\exists \delta)(\forall x) R$$.
In the introductory examples in the present and the following section, let us confine attention to formulas in prenex form: a string of quantifiers followed by a quantifier-free first-order formula. If in a first-order sentence of such a form an existential quantifier $$\exists y$$ lies in the syntactic scope of a universal quantifier $$\forall x$$, then by the semantics $$\exists y$$ automatically depends on $$\forall x$$. This is so e.g. with the sentence $$(\forall x)(\exists y) R(x, y)$$. The dependence of $$\exists y$$ on $$\forall x$$ means that the witness of $$\exists y$$ may vary with the value interpreting $$\forall x$$. In order for the sentence to be true in a model $$M$$, it suffices that there be a function $$f$$ such that $$R(a, f(a))$$ holds in $$M$$, for any $$a$$ interpreting $$\forall x$$. Such functions, spelling out the dependencies of witnesses for existential quantifiers on interpretations of universal quantifiers, are known in logical literature as Skolem functions.[4] For comparison, in the sentence $$(\exists y)(\forall x) R(x, y)$$ the quantifier $$\exists y$$ does not depend on the quantifier $$\forall x.$$ For this sentence to be true in $$M$$, one and the same witness $$c$$ for $$\exists y$$ must be good against any interpretation $$a$$ of $$\forall x$$, so that $$R(a, c)$$ holds in $$M$$. The corresponding Skolem function is a constant.[5]
In IF first-order logic, syntactic scopes no longer determine the semantic relation of dependence. In the sentence $$(\forall x)(\forall y)(\exists z/\forall y) R(x, y, z)$$, for instance, $$\exists z$$ is syntactically subordinate to both $$\forall x$$ and $$\forall y$$, but is marked as independent of $$\forall y$$, and is hence dependent only on $$\forall x$$. Semantically this means that the witness of $$\exists z$$ must be given by a function taking as an argument the interpretation of $$\forall x$$, but not that of $$\forall y$$. In order for $$(\forall x)(\forall y)(\exists z/\forall y) R(x, y, z)$$ to be true in $$M$$, there must be a function $$f$$ of one argument such that $$R(a, b, f(a))$$ holds in $$M$$, for any $$a$$ interpreting $$\forall x$$ and any $$b$$ interpreting $$\forall y$$.
What is gained with the slash notation? After all, clearly e.g. $$(\forall x)(\forall y)(\exists z/\forall y) R(x, y, z)$$ is true in a model $$M$$ iff[6] the first-order sentence $$(\forall x)(\exists z)(\forall y) R(x, y, z)$$ is true therein. As a matter of fact, the expressive power of IF logic exceeds that of first-order logic. Consider the sentence $$(\forall x)(\exists y)(\forall z)(\exists w/\forall x) R(x, y, z, w)$$. Its truth-condition is of the following form: there are one-argument functions $$f$$ and $$g$$ such that $$R(a, f(a), b, g(b))$$ holds in $$M$$, for any $$a$$ interpreting $$\forall x$$ and $$b$$ interpreting $$\forall z$$.[7] So the sentence is true iff the following sentence (*) containing a finite partially ordered quantifier is true:
$\tag{*} \begin{matrix} \forall x \exists y \\ \forall z \exists w \end{matrix} R(x,y,z,w)$
For, by definition (*) is true in a model $$M$$ iff the second-order sentence $$(\exists f)(\exists g)(\forall x)(\forall z) R(x, f(x), z, g(z))$$ is true therein. The latter may be termed the Skolem normal form of (*). It says that Skolem functions providing witnesses for the quantifiers $$\exists y$$ and $$\exists w$$ in (*) exist. Finite partially ordered quantifiers – a.k.a. Henkin quantifiers or branching quantifiers – were proposed by Leon Henkin (1961) and have been subsequently studied extensively.[8] They are two-dimensional syntactic objects
$\begin{matrix} Q_{11} \ldots Q_{1n} \\ \vdots \\ Q_{m1} \ldots Q_{mn} \end{matrix}$
where each $$Q_{ij}$$ is either $$\exists x_{ij}$$ or $$\forall x_{ij}$$. They are naturally interpreted by making systematic use of Skolem functions.
Let us denote by $$\mathbf{FPO}$$ the logic of finite partially ordered quantifiers obtained from first-order logic as follows: if $$\phi$$ is a first-order formula and $$\mathbf{Q}$$ is a finite partially ordered quantifier, then $$\mathbf{Q}\phi$$ is a formula of $$\mathbf{FPO}$$.[9] With $$\mathbf{FPO}$$ it is possible to express properties that are not definable in first-order logic. The first example was provided by Andrzej Ehrenfeucht (cf. Henkin 1961): a sentence that is true in a model iff the size of its domain is infinite. It turns out that $$\mathbf{FPO}$$ can be translated into IF logic (see Subsect. 4.1). Therefore IF logic is more expressive than first-order logic.
The deepest reason for IF logic, as Hintikka saw it, is that the relations of dependence and independence between quantifiers are the only way of expressing relations of dependence and independence between variables on the first-order level (Hintikka 1996: 34–35, 73–74; 2002a: 404–405; 2006a: 71, 515). To understand this remark properly, recall that relations of quantifier (in)dependence are semantic relations, but syntactically expressed. More precisely, in IF logic the (in)dependence relations are syntactically expressed by the interplay of two factors: syntactic scope and the independence indicator ‘/’. In a given sentence, an existential quantifier $$\exists x$$, say, depends on precisely those universal quantifiers within whose scope $$\exists x$$ lies, but of which $$\exists x$$ is not marked as independent using the slash sign. What Hintikka means when he speaks of (in)dependence relations between variables, are functional dependencies between quantities in a model. The kinetic energy of a material body depends on its mass and its speed, but does not depend on the particular material body being considered. This fact can be expressed in IF logic by the sentence
$$(\forall b)(\forall m)(\forall v)(\exists e/\forall b)$$(if $$b$$ is a material body which moves at the velocity $$v$$ and has the mass $$m$$, then the kinetic energy $$e$$ of $$b$$ equals $$\frac{1}{2} mv^2 )$$.
That this sentence states the existence of a functional dependence of kinetic energy on mass and speed is particularly clearly seen from the fact that if the sentence is true, the unique Skolem function for the quantifier $$\exists e$$ actually is the physical law connecting masses, speeds and kinetic energies (cf. Hintikka 1996: 34–35).
While first-order logic can only express some relations between variables, IF first-order logic with its greater expressive power can express more. Actually, IF logic is calculated to capture all such relations (Hintikka 1996: 75–77; 2002a: 404–405; 2002b: 197; 2006a: 72). This idea must, in its full generality, be seen as programmatic. For a more general framework, cf. Hintikka (2006a: 515, 536, 752; 2008), Sandu & Sevenster (2010), Sandu (2013), Sandu (2014).
Among philosophical issues that have been addressed in connection with IF logic are the reconstruction of mathematical reasoning on the first-order level (Hintikka 1996, 1997), definability of a self-applied truth predicate (Hintikka 1991, 1996, 2001; Sandu 1998), truth and axiomatic set theory (Hintikka 1996, 2004a), and insights into the nature of negation (Hintikka 1991, 1996, 2002; Hintikka & Sandu 1989; Sandu 1994). These issues will be discussed in Sections 4 and 5.
## 2. The Background of IF Logic: Game-theoretical Semantics
### 2.1. Semantic games
Inspired by Ludwig Wittgenstein’s idea of a language-game, Hintikka (1968) introduced the basic framework of what came to be known as game-theoretical semantics (a.k.a. GTS). The basic lesson Hintikka adopts from Wittgenstein is that words – in particular quantifiers – are associated with activities that render them meaningful: words often have meaning only in the context of certain types of action (Hintikka 1968: 55–56). Wittgenstein said that by ‘language-game’ he means ‘the whole, consisting of language and the actions into which it is woven’ (Wittgenstein 1953: I, Sect. 7).
It becomes natural to ask which are the activities that go together with quantifiers. As Hintikka explains (see Hintikka 2006a: 41, 67), Wittgenstein had taken it as a criterion for something to be an object that it can be looked for and found. Applying this idea to quantifiers with such objects as values, Hintikka was led to formulate semantic games for quantifiers. Crucially, these semantic games can be formulated as games in the strict sense of game theory; but at the same time they are exact codifications of language-games in Wittgenstein’s sense, at least if one accepts that the activities associated with quantifiers are ‘looking for’ and ‘finding.’[10]
Semantic games $$G(\phi , M)$$ for first-order sentences $$\phi$$ are two-player zero-sum games of perfect information, played on a given model $$M$$. Let us call the two players simply player 1 and player 2.[11] The games are most easily explained for sentences in prenex form. Universal quantifiers mark a move of player 1, while existential quantifiers prompt a move by player 2. In both cases, the relevant player must choose an individual from the domain $$\rM$$ of $$M$$.[12] If
$\phi = (\forall x)(\exists y)(\forall z)(\exists w) R(x, y, z, w),$
the game is played as follows. First, player 1 picks out an individual $$a$$, whereafter player 2 chooses an individual $$b$$. Then player 1 proceeds to choose a further individual $$c$$, to which player 2 responds by picking out an individual $$d$$. Thereby a play of the game has come to an end. The tuple $$(a, b, c, d)$$ of individuals chosen determines the winner of the play. If the quantifier-free formula $$R(a, b, c, d)$$ holds in $$M$$, player 2 wins, if not, player 1 wins.
The fact that one of the players wins a single play of game $$G(\phi , M)$$ does not yet tell us anything about the truth-value of the sentence $$\phi$$. Truth and falsity are characterized in terms of the notion of winning strategy. The sentence $$(\forall x)(\exists y)(\forall z)(\exists w) R(x, y, z, w)$$ is true in $$M$$ if there is a winning strategy for player 2 in the game just described: a recipe telling player 2 what to do, when (a specified amount of) information about the opponent’s earlier moves is given. Technically, the requirement is that there be strategy functions $$f$$ and $$g$$, such that for any choices $$a$$ and $$c$$ by player $$1, R(a, f(a), c, g(a, c))$$ holds in $$M$$. Observe that strategy function $$f$$ is a Skolem function for the quantifier $$\exists y$$ in $$\phi$$, and similarly $$g$$ is a Skolem function for the quantifier $$\exists w$$. The sentence $$\phi$$ is false in $$M$$ if there is a winning strategy (set of strategy functions) for player 1 in the corresponding game: a constant $$c$$ and a function $$h$$ such that for any choices $$b$$ and $$d$$ by player $$2, R(c, b, h(b), d)$$ fails to hold in $$M$$.
Game-theoretical interpretation of quantifiers was already suggested by Henkin (1961; cf. Hintikka 1968: 64). Henkin also pointed out, in effect, the connection between a full set of Skolem functions and a winning strategy for player 2. Hintikka (1968) noted that conjunctions are naturally interpreted by a choice between the two conjuncts, made by player 1; similarly, disjunctions can be interpreted by a choice by player 2 between the two disjuncts.[13] Further, Hintikka proposed to interpret negation as a transposition of the roles of ‘verifier’ and ‘falsifier’ (for more details, see Subsect. 3.2).
The game-theoretical description of semantic games does not mention the activities of searching and finding; for such an abstract description it suffices to speak of the players making a move. Also, the characterization of truth and falsity as the existence of a winning strategy for player 2 and player 1, respectively, does not refer to efforts on the part of the players – say efforts to establish truth or to find witnesses. The truth or falsity of a sentence is a matter of ‘combinatorics’: it is an issue of the existence of a set of functions with certain properties (cf. Hintikka 1968, 1996; Hodges 2013). So what happened to the original philosophical conception according to which the meanings of quantifiers are tied to the activities of searching and finding? Hintikka’s idea is that asserting a sentence involving quantifiers is to make a claim about what can and what cannot happen when a certain language-game is played; using language involving quantifiers requires mastering the rules of the corresponding semantic games (Hintikka 1968: Sect. 8, 1996: 128, 2006a: 538). What is the content of such a claim in connection with the sentence $$\forall x\exists y R(x, y)$$? That whichever individual player 1 chooses from the domain for $$\forall x$$, player 2 can find a witness individual for $$\exists y$$. In other words, given a value for $$\forall x$$ (which itself can be seen as the result of the search by player 1), if player 2 is allowed to search free from any practical constraints, she will find a value for $$\exists y$$ so that player 2 wins the resulting play. Even though semantic games themselves can be defined without recourse to activities such as looking for or finding, these activities play an important conceptual role when the language user reasons about these games.
Hintikka (1973a) initiated the application of GTS to the study of natural language. This work was continued notably by Hintikka & Kulas (1983, 1985), where game-theoretical rules for such items as negation, anaphoric pronouns, genitives, tenses, intensional verbs, certain prepositional constructions, and proper names were given, and the distinction between abstract meaning and strategic meaning drawn.[14]
### 2.2. Imperfect information
The framework of GTS enables asking questions of a game-theoretical nature about semantic evaluation. Hintikka (1973a) observed that semantic games with imperfect information are devised without any difficulty. As logical examples he used $$\mathbf{FPO}$$ sentences. (For examples related to natural languages, see Subsect. 5.4.)
From the vantage point of GTS, independence friendly first-order logic differs from first-order logic in that semantic games correlated with formulas of the former are, in general, games of imperfect information, while any game associated with a first-order formula is a game of perfect information. Consider the game for $$\forall x\exists y\forall z(\exists w/\forall x) R(x, y, z, w)$$, played on a model $$M$$. A play of this game proceeds exactly as plays of the game corresponding to $$\forall x\exists y\forall z\exists w R(x, y, z, w)$$. First, player 1 chooses an individual $$a$$, whereafter player 2 chooses an individual $$b$$. Then player 1 proceeds to pick out a further individual $$c$$, to which player 2 responds by choosing an individual $$d$$. So a play of the game has come to an end. The play is won by player 2, if indeed $$R(a, b,c, d)$$ holds in $$M$$; otherwise it is won by player 1. But $$\exists w$$ was marked as independent of $$\forall x$$ – why does not this fact show in any way in the course of a play?
One might be tempted to add to the description of a play: ‘player 2 chooses a value for $$\exists w$$ in ignorance of the value for $$\forall x$$.’ However, such a paraphrase would not clarify the conceptual situation. It makes no sense to speak of the independence of a move from other given moves with reference to a single play; this can only be done with reference to a multitude of plays. Quantifier independence can be conceptualized in game-theoretical terms making use of the notion of strategy. In the example, a strategy of player 2 is a set $$\{f, g\}$$ of strategy functions, function $$f$$ providing a value for $$\exists y$$ depending on the value of $$\forall x$$, and function $$g$$ providing a value for $$\exists w$$ depending on the value chosen for $$\forall z$$ but not on the value chosen for $$\forall x$$. The strategy $$\{f, g\}$$ is, then, a winning strategy for player 2, iff $$R(a, f(a), c, g(c))$$ holds in $$M$$, for all values $$a$$ chosen for $$\forall x$$ and $$c$$ chosen for $$\forall z$$. One precise way of implementing the idea that player 2 is ‘ignorant’ of the move player 1 made for $$\forall x$$ is to say that (a) strategy functions always only take as arguments the opponent’s moves, and (b) the strategy function corresponding to $$\exists w$$ may not take as its argument the move player 1 made for $$\forall x$$.
A sentence of IF first-order logic is by definition true in a model $$M$$ iff there is a winning strategy for player 2 in the correlated game, and false iff there is a winning strategy for player 1 in the correlated game. There are sentences which under these criteria are neither true nor false; they are called non-determined (see Subsect. 3.3).
In Hintikka’s judgement, the game-theoretical semantics of quantifiers can be taken to have the same rationale that was mentioned as the deepest reason for IF first-order logic at the end of Section 1: GTS is a method of representing, on the first-order level, the (in)dependence relations between variables by means of informational (in)dependence in the sense of game theory (Hintikka 1991: 12–13, 2006a: 535).
## 3. The Syntax and Semantics of IF First-order Logic
In the literature one can find essentially different formulations of ‘IF first-order logic.’ The differences are not restricted to syntax – examples of applying different semantic ideas can be found as well.
As already noted, there is a systematic connection between the Skolem functions and strategy functions of player 2. In connection with formulas in prenex form, a Skolem function for an existential quantifier is a function of the values assigned to the preceding universal quantifiers, but not a function of the values assigned to the preceding existential quantifiers.[15] Skolem functions are strategy functions taking as arguments exclusively moves made by player 1. Generally a strategy for a player in a two-player game can perfectly well make use of the previous choices of either player. Hodges (2007) stresses that it makes a difference on which notion – Skolem function or strategy function – the semantics of imperfect information is based. Hodges (1997a) adopted the notational convention of writing, say, $$(\exists y/x)$$ where Hintikka writes $$(\exists y/\forall x)$$, hence marking the difference between semantic games formulated in terms of arbitrary strategy functions and those whose strategy functions are in effect Skolem functions; the variable $$x$$ in $$(\exists y/x)$$ can be ‘bound’ by any syntactically preceding quantifier carrying the variable $$x$$. Hodges proposed to employ the former formulation, while in Hintikka (1991, 1995, 1996, 2002) and Sandu (1993, 1994) the latter formulation is employed. Hodges (2007: 119) writes:
[W]e refer to the logic with my notation and the general game semantics as slash logic. During recent years many writers in this area (but never Hintikka himself) have transferred the name ‘IF logic’ to slash logic, often without realising the difference. Until the terminology settles down, we have to beware of examples and proofs that don’t make clear which semantics they intend.
The distinction that Hodges makes between slash logic and IF logic serves to bring order to the mishmash of different formulations of IF first-order logic to be found in the literature.[16]
### 3.1. Syntax
IF first-order logic is an extension of first-order logic. Now, any first-order formula is equivalent to a first-order formula in which no variable occurs both free and bound, and in which no two nested quantifiers carry the same variable. Formulas meeting these two syntactic conditions will be termed regular. Henceforth we systematically restrict attention to regular first-order formulas.[17] A vocabulary (signature, non-logical terminology) is any countable set $$\tau$$ of relation symbols (each of which carries a fixed arity), function symbols (again each with a fixed arity), and constant symbols. The first-order logic of vocabulary $$\tau$$ will be referred to as $$\mathbf{FO}[\tau]$$. It is assumed here that among the logical symbols of first-order logic there is the identity symbol (=). The identity symbol is syntactically a binary relation symbol, but its semantic interpretation is fixed in a way that the interpretations of the items in the non-logical terminology are not.
A formula of $$\mathbf{FO}[\tau]$$ is in negation normal form, if all occurrences of the negation symbol $${\sim}$$ immediately precede an atomic formula. The set of formulas of IF first-order logic of vocabulary $$\tau$$ (or $$\mathbf{IFL}[\tau])$$ can be defined as the smallest set such that:
1. If $$\phi$$ is a formula of $$\mathbf{FO}[\tau]$$ in negation normal form, $$\phi$$ is a formula.
2. If $$\phi$$ is a formula and in $$\phi$$ a token of $$(\exists x)$$ occurs in the syntactic scope of a number of universal quantifiers which include $$(\forall y_1),\ldots ,(\forall y_n)$$, the result of replacing in $$\phi$$ that token of $$(\exists x)$$ by $$(\exists x/\forall y_1 ,\ldots ,\forall y_n)$$ is a formula.
3. If $$\phi$$ is a formula and in $$\phi$$ a token of $$\vee$$ occurs in the syntactic scope of a number of universal quantifiers which include $$(\forall y_1),\ldots ,(\forall y_n)$$, the result of replacing in $$\phi$$ that token of $$\vee$$ by $$(\vee /\forall y_1 ,\ldots ,\forall y_n)$$ is a formula.
4. If $$\phi$$ is a formula and in $$\phi$$ a token of $$(\forall x)$$ occurs in the syntactic scope of a number of existential quantifiers which include $$(\exists y_1),\ldots ,(\exists y_n)$$, the result of replacing in $$\phi$$ that token of $$(\forall x)$$ by $$(\forall x/\exists y_1 ,\ldots ,\exists y_n)$$ is a formula.
5. If $$\phi$$ is a formula and in $$\phi$$ a token of $$\wedge$$ occurs in the syntactic scope of a number of existential quantifiers which include $$(\exists y_1),\ldots ,(\exists y_n)$$, the result of replacing in $$\phi$$ that token of $$\wedge$$ by $$(\wedge /\exists y_1 ,\ldots ,\exists y_n)$$ is a formula.
Clauses (2) and (3) allow the degenerate case that the list of universal quantifiers is empty $$(n = 0)$$. The resulting expressions $$(\exists x$$/) and $$(\vee$$/) are identified with the usual existential quantifier $$(\exists x)$$ and the usual disjunction $$\vee$$, respectively. Similar stipulations are made about clauses (4) and (5).
In a suitable vocabulary, each of the following is a formula:
• $$(\forall x)(\forall y)(\exists z/\forall x) R(x, y, z, v)$$,
• $$(\forall x)(\forall y)(x = y \; (\vee /\forall x) \; Q(x, y))$$,
• $$(\exists x)(S(x) \; (\wedge /\exists x) \; T(x))$$,
• $$(\forall x)(\exists y)(\forall z/\exists y)(\exists v/\forall x) R(x, y, z, v)$$.
By contrast, none of the following sequences of symbols is a formula:
• $$(\exists y/\forall x) P(x, y)$$,
• $$(\exists x)(\exists y/\exists x) P(x, y)$$,
• $$(\forall x)(\forall y/\forall x) P(x, y)$$,
• $$(\forall x)(S(x) \; (\vee /\exists y) \; T(x))$$.
If $$\phi$$ is an $$\mathbf{IFL}$$ formula, generated by the above clauses from some $$\mathbf{FO}$$ formula $$\phi^*$$, the free variables of $$\phi$$ are simply the free variables of $$\phi^*$$. $$\mathbf{IFL}$$ formulas without free variables are $$\mathbf{IFL}$$ sentences.[18]
### 3.2. Semantics
Defining the semantics of a logic using GTS is a two-step process. The first step is to define the relevant semantic games. The second step is to define the notions of ‘true’ and ‘false’ in terms of the semantic games; this happens by reference to the notion of winning strategy. Semantic games may be defined by specifying recursively the alternative ways in which a game associated with a given formula $$\phi$$ can be begun.[19]
For every vocabulary $$\tau$$, $$\mathbf{IFL}[\tau]$$ formula $$\phi$$, model $$(\tau$$ structure) $$M$$, and variable assignment $$g$$, a two-player zero-sum game $$G(\phi , M, g)$$ between player 1 and player 2 is associated.[20] If $$g$$ is a variable assignment, $$g[x/a]$$ is the variable assignment which is otherwise like $$g$$ but maps the variable $$x$$ to the object $$a$$.
1. If $$\phi = R(t_1 ,\ldots ,t_n)$$ and $$M, g \vDash R(t_1 ,\ldots ,t_n)$$, player 2 wins (and player 1 loses); otherwise player 1 wins (and player 2 loses).
2. If $$\phi = t_1 = t_2$$ and $$M, g \vDash t_1 = t_2$$, player 2 wins (and player 1 loses); otherwise player 1 wins (and player 2 loses).
3. If $$\phi = {\sim}R(t_1 ,\ldots ,t_n)$$ and $$M, g \not\vDash R(t_1 ,\ldots ,t_n)$$, player 2 wins (and player 1 loses); otherwise player 1 wins (and player 2 loses).
4. If $$\phi = {\sim}t_1 = t_2$$ and $$M, g \not\vDash t_1 = t_2$$, player 2 wins (and player 1 loses); otherwise player 1 wins (and player 2 loses).
5. If $$\phi = (\psi \; (\wedge /\exists y_1 ,\ldots ,\exists y_n) \; \chi)$$, player 1 chooses $$\theta \in \{\psi ,\chi \}$$ and the rest of the game is as in $$G(\theta , M, g)$$.
6. If $$\phi = (\psi \; (\vee /\forall y_1 ,\ldots ,\forall y_n) \; \chi)$$, player 2 chooses $$\theta \in \{\psi ,\chi \}$$ and the rest of the game is as in $$G(\theta , M, g)$$.
7. If $$\phi = (\forall x/\exists y_1 ,\ldots ,\exists y_n)\psi$$, player 1 chooses an element $$a$$ from $$\rM$$, and the rest of the game is as in $$G(\psi , M, g[x/a]$$).
8. If $$\phi = (\exists x/\forall y_1 ,\ldots ,\forall y_n)\psi$$, player 2 chooses an element $$a$$ from $$\rM$$, and the rest of the game is as in $$G(\psi , M, g[x/a]$$).
Observe that the independence indications play no role in the game rules. Indeed, quantifier independence will be implemented on the level of strategies.
If a token of $$(\vee /\forall y_1 ,\ldots ,\forall y_n)$$ or $$(\exists x/\forall y_1 ,\ldots ,\forall y_n)$$ appears in the formula $$\phi$$ in the scope of the universal quantifiers $$\forall y_1 ,\ldots ,\forall y_n,\forall z_1 ,\ldots ,\forall z_m$$ (and only these universal quantifiers), a strategy function of player 2 for this token in game $$G(\phi , M, g)$$ is any function $$f$$ satisfying the following:
The arguments of $$f$$ are the elements $$a_1 ,\ldots ,a_m$$ that player 1 has chosen so as to interpret the quantifiers $$\forall z_1 ,\ldots ,\forall z_m$$. The value $$f(a_1 ,\ldots ,a_m)$$ is the left or the right disjunct when the token is a disjunction; and an element of the domain when the token is an existential quantifier.
The notion of the strategy function of player 1 for tokens of $$(\wedge /\exists y_1 ,\ldots ,\exists y_n)$$ and $$(\forall x/\exists y_1 ,\ldots ,\exists y_n)$$ can be defined dually. Strategy functions are construed as Skolem functions – the more general notion of strategy function operative in slash logic is not considered here (cf. the beginning of Sect. 3 and Subsect. 6.1). Quantifier independence will be implemented directly in terms of the arguments of the strategy functions.
A strategy of player 2 in game $$G(\phi , M, g)$$ is a set $$F$$ of her strategy functions, one function for each token of $$(\vee /\forall y_1 ,\ldots ,\forall y_n)$$ and $$(\exists x/\forall y_1 ,\ldots ,\forall y_n)$$ appearing in $$\phi$$. Player 2 is said to follow the strategy $$F$$, if for each token of $$(\vee /\forall y_1 ,\ldots ,\forall y_n)$$ and $$(\exists x/\forall y_1 ,\ldots ,\forall y_n)$$ for which she must make a move when game $$G(\phi , M, g)$$ is played, she makes the move determined by the corresponding strategy function. A winning strategy for player 2 in $$G(\phi , M, g)$$ is a strategy $$F$$ such that against any sequence of moves by player 1, following strategy $$F$$ yields a win for player 2. The notions of strategy and winning strategy can be similarly defined for player 1.[21]
The satisfaction and dissatisfaction of the $$\mathbf{IFL}$$ formula $$\phi$$ in the model $$M$$ under the assignment $$g$$ are then defined as follows:[22]
• (Satisfaction) $$\phi$$ is satisfied in $$M$$ under $$g$$ iff there is a winning strategy for player 2 in game $$G(\phi , M, g)$$.
• (Dissatisfaction) $$\phi$$ is dissatisfied in $$M$$ under $$g$$ iff there is a winning strategy for player 1 in game $$G(\phi , M, g)$$.
As with $$\mathbf{FO}$$, variable assignments do not affect the (dis)satisfaction of sentences, i.e., formulas containing no free variables. Indeed, we may define:[23]
• (Truth) $$\phi$$ is true in $$M$$ iff there is a winning strategy for player 2 in game $$G(\phi , M)$$.
• (Falsity) $$\phi$$ is false in $$M$$ iff there is a winning strategy for player 1 in game $$G(\phi , M)$$.
The fact that $$\phi$$ is true in $$M$$ will be denoted by ‘$$M \vDash \phi$$.’ Writing $$M \not\vDash \phi$$ indicates, then, that $$\phi$$ is not true in $$M$$. This does not mean that $$\phi$$ would in the above-defined sense be false in $$M$$. As mentioned in the end of Section 2, there are semantic games in which neither player has a winning strategy.
The syntax of $$\mathbf{IFL}$$ can be generalized by removing the restriction according to which the negation sign may only appear as prefixed to an atomic formula.[24]. In order to interpret negation in GTS, two roles are added as a new ingredient in the specification of the games: those of ‘verifier’ and ‘falsifier.’ Initially, player 1 has the role of ‘falsifier’ and player 2 that of ‘verifier.’ The roles may get switched, but only for one reason: when an occurrence of the negation symbol is encountered. All clauses defining semantic games must be rephrased in terms of roles instead of players. It is the player whose role is ‘verifier’ who makes a move for disjunctions and existential quantifiers, and similarly the player whose role is ‘falsifier’ who moves for conjunctions and universal quantifiers. When a formula $${\sim}\psi$$ is encountered, the players change roles and the game continues with $$\psi$$. Finally, if the encountered atomic formula is true, ‘verifier’ wins and ‘falsifier’ loses, otherwise the payoffs are reversed. The negation $${\sim}$$ is variably referred to as strong negation, dual negation, or game-theoretical negation.[25] It works as one would expect: $$\phi$$ is false in $$M$$ iff its negation $${\sim}\phi$$ is true therein (cf. Sandu 1993).
### 3.3. Basic properties and notions
Failure of bivalence. There are sentences $$\phi$$ of $$\mathbf{IFL}$$ and models $$M$$ such that $$\phi$$ is neither true nor false in $$M$$. Consider evaluating the sentence $$(\forall x)(\exists y/\forall x) x = y$$ on a model whose domain has exactly two elements, $$a$$ and $$b$$. Player 1 has no winning strategy. If he chooses $$a$$ to interpret $$\forall x$$, he loses the play in which player 2 chooses $$a$$ to interpret $$(\exists y/\forall x)$$. Similarly, if player 1 chooses $$b$$, he loses the play in which player 2 likewise chooses $$b$$. Neither does player 2 have a winning strategy. Her strategy functions for $$(\exists y/\forall x)$$ are constants (zero-place functions). There are two such constants available: $$a$$ and $$b$$. Whichever one of these strategy functions player 2 assumes, there is a move by player 1 defeating it. If player 2 opts for $$a$$, player 1 wins the play in which he chooses $$b$$; and if player 2 applies $$b$$, player 1 wins the play in which he chooses $$a$$. Game $$G(\phi , M)$$ is non-determined: neither player has a winning strategy.[26] The notion of non-determinacy may be extended to formulas as well:
• (Non-determinacy) $$\phi$$ is non-determined in $$M$$ under $$g$$ iff there is no winning strategy for player 1, nor for player 2, in game $$G(\phi , M, g)$$.
In $$\mathbf{IFL}$$, falsity does not ensue from non-truth. That is, bivalence fails in $$\mathbf{IFL}$$. However, it should be noted that it does not fail due to the postulation of a third truth-value or a truth-value gap (cf. Hintikka 1991: 20, 55).[27] Rather, the failure is a consequence of the basic assumptions of the entire semantic theory (GTS). Non-determinacy corresponds to a structural property: the fact that certain kinds of functions do not exist on the model considered.
Due to the failure of bivalence, the logical law of excluded middle fails for the dual negation $${\sim}$$. Actually, $$\phi$$ is non-determined in $$M$$ iff $$M \not\vDash(\phi \vee{\sim}\phi)$$.
Logical equivalence. Sentences $$\psi$$ and $$\chi$$ of $$\mathbf{IFL}$$ are truth equivalent if they are true in precisely the same models, and falsity equivalent if they are false in precisely the same models. Sentences $$\psi$$ and $$\chi$$ are logically equivalent if they are both truth equivalent and falsity equivalent.[28] Due to the failure of bivalence, in $$\mathbf{IFL}$$ truth equivalence does not guarantee logical equivalence.
Truth, falsity, and independence indications. The syntax of $$\mathbf{IFL}$$ allows formulas in which both universal and existential quantifiers appear slashed, e.g.,
$\phi : (\forall x)(\exists y/\forall x)(\forall z/\exists y) R(x, y, z).$
On the other hand, quantifier independence is implemented at the level of strategies. Consequently independence indications following a universal quantifier are vacuous, when the satisfaction of a formula (truth of a sentence) is considered. Similarly, independence indications following an existential quantifier are vacuous when the dissatisfaction of a formula (falsity of a sentence) is at stake. The sentence $$\phi$$ is true in the model $$M$$ iff player 2 has a winning strategy $$F=\{c\}$$ in game $$G(\phi , M)$$. This, again, means that whichever elements $$a$$ and $$b$$ player 1 chooses to interpret $$(\forall x)$$ and $$(\forall z/\exists y)$$, respectively, the constant interpretation $$c$$ of $$(\exists y/\forall x)$$ given by the (zero-place) strategy function $$c$$ satisfies $$R(a, c, b)$$ in $$M$$. But this amounts to the same as requiring that whichever elements $$a$$ and $$b$$ player 1 chooses to interpret $$(\forall x)$$ and $$(\forall z)$$, respectively, the constant interpretation $$c$$ of $$(\exists y/\forall x)$$ satisfies $$R(a, c, b)$$ in $$M$$. Indeed, $$\phi$$ is truth equivalent to a sentence containing no slashed universal quantifiers:
$$\phi$$ is true in a model $$M$$ iff the sentence $$(\forall x)(\exists y/\forall x)(\forall z) R(x, y, z)$$ is true in $$M$$.
Similarly, $$\phi$$ is falsity equivalent to a sentence containing no slashed existential quantifiers: $$\phi$$ is false in a model $$M$$ iff the sentence $$(\forall x)(\exists y)(\forall z/\exists y) R(x, y, z)$$ is false therein.
### 3.4. Extended IF first-order logic
If $$\phi$$ is a sentence of $$\mathbf{FO}$$, $$\phi$$ is false in a model $$M$$ iff $${\sim}\phi$$ is true in $$M$$ iff $$\phi$$ is not true in $$M$$. By contrast, in $$\mathbf{IFL}$$ falsity and non-truth do not coincide. An extension of $$\mathbf{IFL}$$ can be introduced, where it is possible to speak of the non-truth of sentences. To this end, let us introduce a new negation symbol, $$\neg$$, referred to as weak negation, contradictory negation or classical negation.[29] The set of formulas of extended IF first-order logic (to be denoted $$\mathbf{EIFL})$$ is obtained from the set of formulas of $$\mathbf{IFL}$$ by closing it under the operations $$\neg$$, $$\wedge$$, and $$\vee$$:[30]
• All formulas of $$\mathbf{IFL}$$ are formulas of $$\mathbf{EIFL}$$.
• If $$\phi$$ and $$\psi$$ are formulas of $$\mathbf{EIFL}$$, then so are $$\neg \phi , (\phi \wedge \psi)$$ and $$(\phi \vee \psi)$$.
So if $$\phi$$ and $$\psi$$ are $$\mathbf{IFL}$$ formulas, e.g. $$\neg \phi$$ and $$(\neg \phi \vee \psi)$$ are $$\mathbf{EIFL}$$ formulas; by contrast $$(\forall x)\neg \phi$$ is not. For the crucial restriction that $$\neg$$ may not occur in the scope of a quantifier, see Hintikka (1991: 49; 1996: 148). For counterexamples to this restriction see, however, Hintikka (1996: 148; 2002c) and especially Hintikka (2006b) in which the so-called fully extended IF first-order logic (FEIFL) is considered. In FEIFL, any occurrences of $$\neg$$ are allowed which are subject to the following syntactic condition: if (Q$$x/W)$$ is a quantifier in the syntactic scope of an occurrence of $$\neg$$, then all quantifiers listed in $$W$$ are likewise in the syntactic scope of that occurrence of $$\neg$$.
The semantics of an $$\mathbf{EIFL}$$ formula formed by contradictory negation is simply this:
$M, g \vDash \neg \phi \text{ iff } M, g \not\vDash \phi.$
From the viewpoint of GTS, the connective $$\neg$$ behaves in an unusual way. For all connectives of $$\mathbf{IFL}$$, there is a game rule (which can be seen as specifying the meaning of the connective). For contradictory negation there is no game rule, and its semantics is not explained in terms of plays of semantic games. A formula $$\neg \phi$$ serves to say, globally, something about an entire game $$G(\phi , M, g)$$. If $$\phi$$ is a sentence, to say that $$\neg \phi$$ is true in $$M$$ is to say that player 2 does not have a winning strategy in game $$G(\phi , M)$$. If, again, there is indeed a winning strategy for player 2 in $$G(\phi , M)$$, by stipulation $$\neg \phi$$ is false in $$M$$.[31]
Not only is $$\neg \phi$$ not itself a formula of $$\mathbf{IFL}$$, but it cannot in general even be expressed in $$\mathbf{IFL}$$ (see Subsect. 4.2). The law of excluded middle holds for the contradictory negation: for all sentences $$\phi$$ and all models $$M$$, indeed $$M \vDash(\phi \vee \neg \phi)$$. In Section 5 it will be seen how Hintikka has proposed to make use of $$\mathbf{EIFL}$$ when discussing issues in the philosophy of mathematics.
## 4. Metalogical Properties of IF First-order Logic
The metalogical properties of $$\mathbf{IFL}$$ have been discussed in several publications, by Hintikka as well as Sandu.[32] When presenting them, reference will be made to existential second-order logic $$(\mathbf{ESO})$$;[33] further important notions are those of the skolemization and Skolem normal form of an $$\mathbf{IFL}$$ formula. For precise definitions of these notions, the supplementary document can be consulted. In brief, $$\mathbf{ESO}$$ is obtained from $$\mathbf{FO}$$ by allowing existential quantification over relation and function symbols in a first-order formula. The skolemization $$\mathrm{sk}[\phi]$$ of an $$\mathbf{IFL}$$ formula $$\phi$$ is a first-order formula of a larger vocabulary. It explicates in terms of function symbols how existential quantifiers and disjunction symbols of $$\phi$$ depend on preceding universal quantifiers. For example, a skolemization of the $$\mathbf{IFL}$$ sentence $$\phi = (\forall x)(\exists y)(\forall z)(\exists v/\forall x)R(x, y, z, v)$$ of vocabulary $$\{R\}$$ is the $$\mathbf{FO}$$ sentence $$\mathrm{sk}[\phi] = (\forall x)(\forall z)R(x, f(x), z, h(z))$$ of vocabulary $$\{R, f, h\}$$. Its Skolem normal form, again, is the $$\mathbf{ESO}$$ sentence $$\mathrm{SK}[\phi] = (\exists f)(\exists h)(\forall x)(\forall z)R(x, f(x), z, h(z))$$. The first-order sentence $$\mathrm{sk}[\phi]$$ must not be confused with the second-order sentence $$\mathrm{SK}[\phi]$$.
### 4.1. First-order logic and existential second-order logic
Game-theoretical vs. Tarskian semantics of FO. The set of formulas of $$\mathbf{FO}$$ is a proper subset of the set of $$\mathbf{IFL}$$ formulas. The standard semantics of $$\mathbf{FO}$$ is not the one provided by GTS, but the Tarskian semantics specifying recursively the satisfaction relation $$M, g \vDash \phi$$. If the Axiom of Choice is assumed,[34] the two semantics of $$\mathbf{FO}$$ coincide:
Theorem (assuming AC). (Hodges 1983: 94, Hintikka & Kulas 1985: 6–7) Let $$\tau$$ be any vocabulary, $$M$$ any $$\tau$$ structure, $$g$$ any variable assignment and $$\phi$$ any $$\mathbf{FO}[\tau]$$ formula. Then $$M, g \vDash \phi$$ holds in the standard sense iff there is a winning strategy for player 2 in game $$G(\phi , M, g)$$.[35]
Relation to ESO. $$\mathbf{IFL}$$ and $$\mathbf{ESO}$$ are intertranslatable:[36]
Theorem (assuming AC) $$\mathbf{ESO}$$ and $$\mathbf{IFL}$$ have the same expressive power.
That is, (1) for every $$\mathbf{IFL}[\tau]$$ formula $$\phi$$ there is an $$\mathbf{ESO}[\tau]$$ formula $$\phi'$$ such that for all $$\tau$$ structures $$M$$ and variable assignments $$g$$, we have: $$M, g \vDash \phi$$ iff $$M, g \vDash \phi'$$. Actually, $$\mathrm{SK}[\phi]$$ is a suitable $$\mathbf{ESO}$$ formula. And (2) for every $$\mathbf{ESO}[\tau]$$ formula $$\psi$$ there is an $$\mathbf{IFL}[\tau]$$ formula $$\psi'$$ such that $$M, g \vDash \psi$$ iff $$M, g \vDash \psi'$$, for all $$\tau$$ structures $$M$$ and variable assignments $$g$$. This follows from the fact that $$\mathbf{ESO}$$ can be translated into $$\mathbf{FPO}$$ (Enderton 1970, Walkoe 1970), which again can be translated into $$\mathbf{IFL}$$.
Hintikka suggests that $$\mathbf{IFL}$$ is, substantially speaking, a first-order logic: the entities its quantified variables range over are individuals, and so are all entities with which the players of the semantic games operate. (See Subsect. 5.1 for a discussion of this idea.) A part of the interest of the intertranslatability theorem lies in the fact that if Hintikka’s controversial claim is accepted, this would mean that the expressive power of $$\mathbf{ESO}$$ can actually be achieved on the first-order level.
IFL is more expressive than FO. The following are examples of properties expressible in $$\mathbf{ESO}$$ – and therefore in $$\mathbf{IFL}$$ – but not in $$\mathbf{FO}$$: Dedekind-infinity of the domain, non-completeness of a linear order, ill-foundedness of a binary relation, disconnectedness of a graph, equicardinality of the extensions of two first-order formulas $$\phi(x)$$ and $$\psi(x)$$, infinity of the extension of a first-order formula $$\phi(x)$$, and the topological notion of open set (see, e.g., Hintikka 1996, Väänänen 2007).
As an example, the Dedekind-infinity of the domain may be considered. A set $$S$$ is Dedekind-infinite precisely when there exists an injective function from $$S$$ to its proper subset. Let $$\phi_{inf}$$ be the following sentence of $$\mathbf{IFL}$$:[37]
$(\exists t)(\forall x)(\exists z)(\forall y)(\exists v/\forall x)((x = y \leftrightarrow z = v) \wedge z \ne t).$
The Skolem normal form of $$\phi_{inf}$$ is
$(\exists f)(\exists g)(\exists t)(\forall x)(\forall y)((x = y \leftrightarrow f(x) = g(y)) \wedge f(x) \ne t).$
Relative to a model $$M$$, this $$\mathbf{ESO}$$ sentence asserts the existence of functions $$f$$ and $$g$$ and an element $$t$$ such that $$f = g$$ (implication from left to right), this function is injective (implication from right to left), and its domain is the whole domain of $$M$$ but the element $$t$$ does not appear in its range. Consequently the range is a proper subset of the domain of $$M$$. In other words, the sentence $$\phi_{inf}$$ is true in a model $$M$$ iff the domain of $$M$$ is infinite.
Be it still noted that when attention is restricted to finite models, Ronald Fagin’s famous theorem (1974) connects $$\mathbf{ESO}$$ and the complexity class $$\mathbf{NP}$$: a computational problem is solvable by an algorithm running in non-deterministic polynomial time iff it is definable in $$\mathbf{ESO}$$ relative to the class of all finite structures. The following are $$\mathbf{NP}$$-complete properties and hence expressible in $$\mathbf{IFL}$$, over the class of all finite models: evenness of the domain, oddness of the domain, 3-colorability of a graph, and the existence of a Hamiltonian path on a graph.[38]
Properties in common with FO. IF first-order logic shares many metalogical properties with first-order logic.[39]
Compactness. A set of $$\mathbf{IFL}$$ sentences has a model iff all its finite subsets have a model.
Löwenheim-Skolem property. Suppose $$\phi$$ is an $$\mathbf{IFL}$$ sentence that has an infinite model, or arbitrarily large finite models. Then $$\phi$$ has models of all infinite cardinalities.
The separation theorem holds in $$\mathbf{IFL}$$ in a strengthened form; the ‘separation sentence’ $$\theta$$ is in particular a sentence of $$\mathbf{FO}$$.
Separation theorem. Suppose $$\phi$$ is an $$\mathbf{IFL}$$ sentence of vocabulary $$\tau$$, and $$\psi$$ an $$\mathbf{IFL}$$ sentence of vocabulary $$\tau'$$. Suppose further that $$\phi$$ and $$\psi$$ have no models in common. Then there is a first-order sentence $$\theta$$ of vocabulary $$\tau \cap \tau'$$ such that every model of $$\phi$$ is a model of $$\theta$$, but $$\theta$$ and $$\psi$$ have no models in common.
It is well known that for $$\mathbf{FO}$$ there is a sound and complete proof procedure. Because a first-order sentence $$\phi$$ is inconsistent (non-satisfiable) iff its negation $${\sim}\phi$$ is valid (true in all models), trivially $$\mathbf{FO}$$ also has a sound and complete disproof procedure.[40] The latter property extends to $$\mathbf{IFL}$$ (while the former does not, see Subsect. 4.3):
Existence of a complete disproof procedure. (Hintikka 1996: 68–70, 82) The set of inconsistent $$\mathbf{IFL}$$ sentences is recursively enumerable.
### 4.2. Intricacies of negation
In Subsection 3.4, the contradictory negation $$\neg$$ was distinguished from the strong negation $${\sim}$$. In $$\mathbf{FO}$$ the two coincide: for any first-order sentence $$\phi$$, we have $$M \vDash{\sim}\phi$$ iff $$M \vDash \neg \phi$$.
Strong negation fails as a semantic operation. Let us write $$[\phi]$$ for the set of models of sentence $$\phi$$. In the special case of $$\mathbf{FO}$$, the strong negation $${\sim}$$ clearly defines a semantic operation: whenever $$\chi$$ and $$\theta$$ are sentences such that $$[\chi] = [\theta]$$, we have $$[{\sim}\chi] = [{\sim}\theta]$$. Burgess (2003) observed that in the context of IF logic this property is lost in a very strong sense. In fact there are IF sentences $$\chi$$ and $$\theta$$ such that while $$[\chi] = [\theta]$$, the sets $$[{\sim}\chi]$$ and $$[{\sim}\theta]$$ are not only distinct but even disjoint.
Inexpressibility of contradictory negation. In $$\mathbf{IFL}$$ the strong negation $${\sim}$$ and the contradictory negation $$\neg$$ do not coincide: we may have $$M \vDash \neg \phi$$ without having $$M \vDash{\sim}\phi$$. This fact by itself still leaves open the possibility that the contradictory negation of every sentence $$\phi$$ of $$\mathbf{IFL}$$ could be defined in $$\mathbf{IFL}$$, i.e., that there was a sentence $$neg(\phi)$$ of $$\mathbf{IFL}$$ such that $$M \vDash$$ $$neg(\phi)$$ iff $$M \not\vDash \phi$$, for all models $$M$$. All we know by the failure of the law of excluded middle is that not in all cases can $$neg(\phi)$$ be chosen to be $${\sim}\phi$$. However, as a matter of fact contradictory negation is inexpressible in $$\mathbf{IFL}$$. There are sentences $$\phi$$ of $$\mathbf{IFL}$$ such that $$\neg \phi$$ (which is a sentence of $$\mathbf{EIFL})$$ is not truth equivalent to any sentence of $$\mathbf{IFL}$$. This follows from the well-known fact that $$\mathbf{ESO}$$ is not closed under negation and $$\mathbf{IFL}$$ has the same expressive power as $$\mathbf{ESO}$$.[41]
Strong inexpressibility of contradictory negation. As a corollary to the separation theorem, the result holds in a much stronger form. If $$\phi$$ and $$\psi$$ are sentences of $$\mathbf{IFL}$$ such that $$M \vDash \phi$$ iff $$M \not\vDash \psi$$, then each of $$\phi$$ and $$\psi$$ is truth equivalent to a sentence of $$\mathbf{FO}$$. Hence the contradictory negation $$\neg \phi$$ is only expressible in $$\mathbf{IFL}$$ for those $$\mathbf{IFL}$$ sentences $$\phi$$ that are truth equivalent to an $$\mathbf{FO}$$ sentence.[42]
Determined fragment. Let us say that an $$\mathbf{IFL}$$ sentence $$\phi$$ is determined if it satisfies: $$M \vDash(\phi \vee{\sim}\phi)$$, for all models $$M$$. The determined fragment of $$\mathbf{IFL}$$ is the set of determined $$\mathbf{IFL}$$ sentences. In the determined fragment contradictory negation is syntactically expressible by the strong negation. By the strong inexpressibility of contradictory negation, the determined fragment of $$\mathbf{IFL}$$ has the same expressive power as $$\mathbf{FO}$$. Membership in the determined fragment is a sufficient but not necessary condition for an $$\mathbf{IFL}$$ sentence to have its contradictory negation expressible in $$\mathbf{IFL}$$. The sentence $$(\forall y)(\exists x/\forall y) x = y$$ is not determined;[43] yet its contradictory negation $$(\forall x)(\exists y) x \ne y$$ is expressible in $$\mathbf{IFL}$$.
Contradictory negation and GTS. The truth-conditions that GTS yields are of the form ‘there are strategy functions $$f_1 ,\ldots ,f_n$$ such that —,’ i.e., it gives rise to truth-conditions expressible in $$\mathbf{ESO}$$. By the strong inexpressibility of contradictory negation, there is no single $$\mathbf{IFL}$$ sentence, not translatable into $$\mathbf{FO}$$, whose contradictory negation has a truth-condition of that form. This fact makes it understandable why contradictory negation should not be expected to admit of a game-theoretical interpretation along the same lines in which the other logical operators are interpreted. Different ways of assigning a game-theoretical interpretation to contradictory negation can, however, be developed. To this end, in the context of fully extended IF first-order logic (FEIFL, cf. Subsect. 3.4), Hintikka has proposed to use semantic games with subgames. (See Hintikka 2002c, 2006b; for subgames, see Carlson & Hintikka 1979, Hintikka & Kulas 1983.) This approach leads to mixing the play level with the strategy level: choices of individuals made in a subgame may depend on strategy functions chosen in earlier subgames. In Tulenheimo (2014), a game-theoretical semantics is formulated for the fragment of FEIFL that consists of formulas in prenex form. Plays of the correlated games do not involve choosing any second-order objects, such as strategy functions. Contradictory negation $$(\neg)$$ is interpreted by introducing an additional component in game positions: a mode. At the play level, the negation $$\neg$$ triggers not only a role switch (like the dual negation $${\sim})$$, but it also involves changing the mode from positive to negative or vice versa. The semantic effect of modes becomes visible at the strategy level: modes regulate the way in which the truth condition of a sentence involves existentially or universally quantifying over strategy functions. Like independence indications, also occurrences of $$\neg$$ are, then, interpreted in terms of conditions that act on the strategy level. For related research, see Figueira et al. (2011, 2014).
Contradictory negation and finite models. Certain major open questions in logic and theoretical computer science can be formulated in terms of $$\mathbf{IFL}$$. It is an open question in complexity theory whether $$\mathbf{NP} = \mathbf{coNP}$$, that is, whether the class of $$\mathbf{NP}$$-solvable problems is the same as the class of problems whose complement is solvable in $$\mathbf{NP}$$. By Fagin’s theorem (1974), this open problem can be equivalently formulated as follows: Is $$\mathbf{IFL}$$ closed under negation over finite models? That is, is there for every IF sentence $$\phi$$ another IF sentence $$neg(\phi)$$ such that for any finite model $$M$$, $$neg(\phi)$$ is true in $$M$$ iff $$\phi$$ is not true in $$M$$? Proving that the answer is negative would settle the notorious $$\mathbf{P} = \mathbf{NP}$$ problem, i.e., establish that there are computational problems for which one can efficiently verify whether a proposed solution is correct although one cannot efficiently find a solution.[44]
### 4.3 Failure of axiomatizability
As is well known, $$\mathbf{FO}$$ admits of a sound and complete proof procedure: there is a mechanical way of generating precisely those first-order sentences that are valid (true in all models). This fact can also be expressed by saying that $$\mathbf{FO}$$ is axiomatizable, or that the set of valid sentences of $$\mathbf{FO}$$ is recursively enumerable.[45] Due to its greater expressive power, axiomatizability fails for $$\mathbf{IFL}$$. In other words, $$\mathbf{IFL}$$ is semantically incomplete.[46]
One way to show this is as follows. Suppose for the sake of contradiction that the set of valid $$\mathbf{IFL}$$ sentences is recursively enumerable. Recall that the sentence $$\phi_{inf}$$ discussed in Subsection 4.1 is true in all and only infinite models. Note, then, that an $$\mathbf{FO}$$ sentence $$\chi$$ is true in all finite models iff the $$\mathbf{IFL}$$ sentence $$(\phi_{inf} \vee \chi)$$ is valid. Given a valid $$\mathbf{IFL}$$ sentence, it can be effectively checked whether the sentence is syntactically of the form $$(\phi_{inf} \vee \chi)$$, where $$\chi$$ is a first-order sentence. Hence the recursive enumeration of all valid $$\mathbf{IFL}$$ sentences yields a recursive enumeration of first-order sentences $$\chi$$ true in all finite models. But this contradicts Trakhtenbrot’s theorem, according to which the set of $$\mathbf{FO}$$ sentences true in all finite models is not recursively enumerable.[47]
What is the relevance of the failure of axiomatizability of $$\mathbf{IFL}$$? Discussing finite partially ordered quantifiers, Quine (1970: 89–91) suggests that we should refuse to give the status of logic to any generalization of $$\mathbf{FO}$$ which does not have a sound and complete proof procedure both for validity and for inconsistency. For Quine, any such generalization belongs to mathematics rather than logic. Since $$\mathbf{FPO}$$ is not axiomatizable, it falls outside the realm of logic, thus delineated.[48]
Hintikka finds this type of allegation unfounded. First, $$\mathbf{IFL}$$ shares many of the important metalogical results with $$\mathbf{FO}$$ (cf. Subsect. 4.1). Second, just like $$\mathbf{IFL}$$, also $$\mathbf{FO}$$ can be translated into second-order logic. The only difference is that in the former case a larger variety of quantifier (in)dependencies must be encoded by Skolem functions than in the latter. Why would the former translation render $$\mathbf{IFL}$$ a part of mathematics while the latter would allow $$\mathbf{FO}$$ to remain a logic (Hintikka 1991: 26–27)? Third, one must make a distinction between what is needed to understand an $$\mathbf{IFL}$$ sentence, and what is needed to mechanically deal with the validities (logical truths) of $$\mathbf{IFL}$$. Due to its non-axiomatizability, there are no mechanical rules for generating the set of all validities of $$\mathbf{IFL}$$. However, to understand a sentence is to know what things are like when it is true, not to know what things are like when it is logically true (Hintikka 1995: 13–14). Fourth, because of non-axiomatizability, the valid inference patterns in $$\mathbf{IFL}$$ cannot be exhausted by any recursive enumeration. Insofar as important mathematical problems can be reduced to questions about the validity of $$\mathbf{IFL}$$ formulas (Subsect. 5.3), progress in mathematics can be seen to consist (not of the discovery of stronger set-theoretical axioms but) of ever more powerful rules for establishing validity in $$\mathbf{IFL}$$ (Hintikka 1996: 100; 2000: 135–136).
### 4.4 Compositionality and the failure of Tarski-type semantics
The principle of compositionality (a.k.a. Frege principle) states that semantic attributes of a complex expression $$E$$ are determined by the semantic attributes of its constituent expressions and the structure of $$E$$. In particular, the semantic attribute of interest (e.g., truth) may be determined in terms of one or more auxiliary semantic attributes (e.g., satisfaction).[49] Hintikka has argued that compositionality amounts to semantic context-independence: semantic attributes of a complex expression depend only on the semantic attributes of its constituent expressions, plus its structure – they do not depend on the sentential context in which the expression is embedded. Semantic context-independence makes it possible to carry out semantic analysis from inside out – from simpler expressions to more complex ones.[50] This is what is needed for recursive definitions of semantic attributes – such as Tarski-type definitions of truth and satisfiability – to be possible.[51] By contrast, the GTS analysis of sentences is an outside in process: a semantic game starts with an entire sentence, and stepwise analyzes the sentence into simpler and simpler components, eventually reaching an atomic formula (together with an appropriate variable assignment). Therefore GTS allows accounting for semantic context-dependencies that violate the principle of compositionality.[52]
In $$\mathbf{IFL}$$ an existential quantifier may depend only on some of the universal quantifiers in whose scope it lies. Accordingly, its interpretation depends on its relation to quantifiers outside its own scope. Such an existential quantifier is context-dependent.[53] On the face of it, then, $$\mathbf{IFL}$$ cannot but violate the principle of compositionality and does not admit of a Tarski-type truth-definition.
Hodges (1997a,b) showed, however, that $$\mathbf{IFL}$$ can be given a compositional semantics.[54] The semantics is given by recursively defining the satisfaction relation ‘$$M \vDash_X \phi$$’ (read: $$\phi$$ is satisfied in $$M$$ by $$X)$$, where $$X$$ is a set of variable assignments. While the Tarskian semantics for $$\mathbf{FO}$$ is in terms of single variable assignments, Hodges’s semantics employs sets of variable assignments. The game-theoretical semantics of $$\mathbf{IFL}$$ is captured by this compositional semantics: for every formula $$\phi$$ of $$\mathbf{IFL}$$, player 2 has a winning strategy in $$G(\phi , M, g)$$ iff the condition $$M \vDash_{\{g\}} \phi$$ holds. Due to Hodges’s semantic clauses for existential quantifiers and disjunctions involving independence indications, the evaluation of $$\phi$$ relative to the singleton set $$\{g\}$$ generally leads to evaluating syntactic components of $$\phi$$ relative to sets of (possibly infinitely) many variable assignments. Hintikka remarks (2006a: 65) that if one is sufficiently ruthless, one can always save compositionality by building the laws of semantic interaction of different expressions into the respective meanings of those expressions.[55]
It is methodologically worth pointing out that compositionality is not needed for defining $$\mathbf{IFL}$$. The very existence of $$\mathbf{IFL}$$ proves that rejecting compositionality is no obstacle for the formulation of a powerful logic (Hintikka 1995). It should also be noted that what makes Hodges’s result work is its type-theoretical ascent. Let us say that a Tarski-type compositional semantics is a compositional semantics which interprets each formula $$\phi(x_1 ,\ldots ,x_n)$$ of $$n$$ free variables in terms of an $$n$$-tuple of elements of the domain. Hence the standard semantics of $$\mathbf{FO}$$ is Tarski-type, but Hodges’s semantics employing the satisfaction relation ‘$$M \vDash_X \phi$$’ is not, because the latter evaluates a formula $$\phi(x_1 ,\ldots ,x_n)$$ of $$n$$ free variables relative to an entire set of $$n$$-tuples of elements. Cameron & Hodges (2001) proved that actually there is $$no$$ Tarski-type compositional semantics for $$\mathbf{IFL}$$.
The notion of compositionality can be refined by imposing constraints on the auxiliary semantic attributes. Sandu and Hintikka (2001: 60) suggested, by analogy with $$\mathbf{FO}$$, that ‘satisfaction by a single variable assignment’ would be a natural auxiliary attribute in connection with $$\mathbf{IFL}$$. By the result of Cameron and Hodges, no semantics for $$\mathbf{IFL}$$ exists that would be compositional in this restricted sense.
### 4.5 Defining truth
The definability of truth can only be discussed in connection with languages capable of speaking of themselves. Let us consider an arithmetical vocabulary $$\tau$$ and restrict attention to the standard model $$N$$ of Peano’s axioms. Each sentence $$\phi$$ of vocabulary $$\tau$$ can then be represented by a natural number $$\ulcorner \phi \urcorner$$, its Gödel number. It is assumed that $$\tau$$ contains a numeral $$\boldsymbol{{}^\ulcorner \phi {}^\urcorner}$$ for each number $$\ulcorner \phi \urcorner$$. If $$L$$ and $$L'$$ are abstract logics of vocabulary $$\tau$$, such as $$\mathbf{FO}$$ or $$\mathbf{IFL}$$, and $$\TRUE(x)$$ is a formula of $$L'$$ such that every sentence $$\phi$$ of $$L$$ satisfies:
$N \vDash \phi \text{ iff } N \vDash \TRUE(\boldsymbol{{}^\ulcorner \phi {}^\urcorner}),$
then $$\TRUE(x)$$ is said to be a truth-predicate (explicit truth-definition) of logic $$L$$ in logic $$L'$$ for the model $$N$$. By Alfred Tarski’s famous theorem of the undefinability of truth (Tarski 1933), there is no truth-predicate for $$\mathbf{FO}$$ in $$\mathbf{FO}$$ itself for the model $$N$$. More generally, Tarski showed that under certain assumptions, a truth-definition for a logic $$L$$ can only be given in a metalanguage which is essentially stronger than $$L$$. One of the assumptions is that the negation used behaves like contradictory negation. On the other hand, Tarski also pointed out that an implicit truth-definition for $$\mathbf{FO}$$ in $$\mathbf{FO}$$ itself is possible. Let $$\tau$$ be an arithmetical vocabulary, and let us stay with the standard model $$N$$ of Peano’s axioms. Let ‘$$\TRUE$$’ be a unary predicate not appearing in $$\tau$$. An $$\mathbf{FO}$$ formula $$\psi(x)$$ of vocabulary $$\tau \cup \{\TRUE\}$$ is an implicit truth-definition for $$\mathbf{FO}[\tau]$$ in $$\mathbf{FO}[\tau \cup \{\TRUE\}]$$ for $$N$$, if for every $$\mathbf{FO}[\tau]$$ sentence $$\phi$$, the following holds:
$$N \vDash \phi$$ iff there is an interpretation $$\TRUE^N$$ of the unary predicate $$\TRUE$$ such that $$(N, \TRUE^N ) \vDash \psi(\boldsymbol{{}^\ulcorner \phi {}^\urcorner})$$.[56]
Intuitively, $$\TRUE^N$$ is the set of those natural numbers that are Gödel numbers of true arithmetic sentences of vocabulary $$\tau$$; and $$\psi(x)$$ says that $$x$$ is a Gödel number in the extension of the predicate $$\TRUE$$. The formula $$\psi(x)$$ is a conjunction including clauses that mimic, on the object-language level, the metalogical recursive clauses of the Tarski-type truth-definition for $$\mathbf{FO}$$. E.g., one of the conjuncts is
\begin{align*} (\forall y)(\forall z)(y = \boldsymbol{{}^\ulcorner \chi {}^\urcorner} &\wedge z = \boldsymbol{{}^\ulcorner \theta {}^\urcorner} \wedge x = \boldsymbol{{}^\ulcorner (\chi \wedge \theta) {}^\urcorner} \rightarrow \\ &[\TRUE(x) \leftrightarrow (\TRUE(y) \wedge \TRUE(z))]). \end{align*}
The implicit nature of the truth-definition is seen from the fact that the predicate $$\TRUE$$ appears on both sides of the equivalence sign in these clauses.[57] The above implicit truth-definition of $$\mathbf{FO}$$ for $$N$$ is of the form ‘there is a set $$S$$ interpreting the predicate $$\TRUE$$ such that $$\psi(x)$$.’ Thus the implicit truth-definition of $$\mathbf{FO}$$ in $$\mathbf{FO}$$ gives rise to the explicit truth-definition $$\exists \TRUE\, \psi(x)$$ of $$\mathbf{FO}[\tau]$$ in $$\mathbf{ESO}[\tau]$$ for $$N$$. Since $$\mathbf{ESO}$$ and $$\mathbf{IFL}$$ have the same expressive power, a truth-predicate of $$\mathbf{FO}$$ for $$N$$ can be formulated in $$\mathbf{IFL}$$.
The same reasoning can be applied to $$\mathbf{ESO}$$ itself, and thereby to $$\mathbf{IFL}$$ (Hintikka 1991, 1996; Hyttinen & Sandu 2000; Sandu 1996, 1998). Namely, an $$\mathbf{ESO}$$ formula $$\chi(x)$$ of vocabulary $$\tau \cup \{\TRUE\}$$ can be formulated which is an implicit truth-definition of $$\mathbf{ESO}[\tau]$$ for $$N$$. Therefore the $$\mathbf{ESO}[\tau]$$ formula $$\exists \TRUE\, \chi(x)$$ is an explicit truth-definition of $$\mathbf{ESO}[\tau]$$ for $$N$$. Here the truth-predicate is formulated in the very same language whose notion of truth is being defined: $$\mathbf{ESO}$$. Hence $$\mathbf{ESO}$$, and thereby $$\mathbf{IFL}$$, is capable of explicitly defining its own truth-predicate relative to $$N$$.[58] This result does not contradict Tarski’s undefinability result, because here non-determined sentences are possible; the negation used is not contradictory negation.[59]
Tarski (1983) adopted a view according to which truth cannot be defined for natural languages. It has been argued in Hintikka & Sandu (1999) that this was due to Tarski’s belief that compositionality fails in natural languages.[60] $$\mathbf{IFL}$$ does not have a Tarski-type compositional semantics, but it admits of formulating a self-applied truth-predicate. The proposed reason why Tarski believed it not to be possible to discuss the notion of truth in natural languages themselves cannot, then, be entertained from the viewpoint of $$\mathbf{IFL}$$. For, the case of $$\mathbf{IFL}$$ shows that the failure of Tarski-type compositionality for a language does not entail the impossibility for the language to define its own truth-predicate.[61]
### 4.6 Properties of extended IF first-order logic
Expressive power. Since $$\mathbf{IFL}$$ is not closed under contradictory negation, $$\mathbf{EIFL}$$ is strictly more expressive than $$\mathbf{IFL}$$ (cf. Subsect. 4.2).[62] The following properties are expressible in $$\mathbf{EIFL}$$ but not in $$\mathbf{IFL}$$ (Hintikka 1996: 188–190): finiteness of the domain, well-foundedness of a binary relation, connectedness of a graph, principle of mathematical induction, Bolzano-Weierstrass theorem, and the topological notion of continuity.
Metalogical properties. The nice metatheorems that $$\mathbf{IFL}$$ shares with $$\mathbf{FO}$$ are lost: Compactness, Löwenheim-Skolem property, separation theorem, and the existence of a complete disproof procedure all fail for $$\mathbf{EIFL}$$ (Hintikka 1991: 49, 1996: 189). No self-applied truth-predictate is possible for $$\mathbf{EIFL}$$. The definition of such a truth-predicate would have to contain the clause
\begin{align*} (\forall y)(y = \boldsymbol{{}^\ulcorner \theta {}^\urcorner} &\wedge x = \boldsymbol{{}^\ulcorner \neg \theta {}^\urcorner} \rightarrow \\ &[\TRUE(x) \leftrightarrow \neg \TRUE(y)]). \end{align*}
But this clause is not a well-formed formula of $$\mathbf{EIFL}$$, since $$\neg$$ appears in the scope of the universal quantifier $$(\forall y)$$ (cf. Hintikka 1996: 151).
The validity and satisfiability problems of the full second-order can be effectively reduced to the corresponding problems concerning $$\mathbf{EIFL}$$. Namely, why cannot a second-order sentence simply be thought of as a two-sorted first-order sentence? Because in order to capture the standard interpretation of second-order logic,[63] it must be said that for every extensionally possible set of $$n$$-tuples of elements of sort 1 there exists a member of sort 2 having those and only those elements as members, for all arities $$n$$ such that the second-order sentence contains a quantifier $$(\exists R)$$ where $$R$$ is $$n$$-ary.[64] Now, such additional conditions can be expressed by a finite conjunction $$X$$ of $$\mathbf{USO}$$ sentences, where $$\mathbf{USO}$$ (universal second-order logic) is obtained from $$\mathbf{FO}$$ by allowing universal quantification over relation and function symbols in a first-order formula. Each of these $$\mathbf{USO}$$ sentences can be expressed as the contradictory negation of an $$\mathbf{ESO}$$ sentence and therefore as the contradictory negation of an $$\mathbf{IFL}$$ sentence. Consequently, there is an $$\mathbf{IFL}$$ sentence $$Y$$ such that $$X$$ itself is logically equivalent to $$\neg Y$$. And here $$\neg Y$$ is a sentence of $$\mathbf{EIFL}$$. Therefore, if $$\phi$$ is a second-order sentence and $$\phi^*$$ its reconstruction in two-sorted first-order logic, we have: $$\phi$$ is satisfiable iff $$(X \wedge \phi^*)$$ is satisfiable iff $$(\neg Y \wedge \phi^*)$$ is satisfiable. And: $$\phi$$ is valid iff $$\phi^*$$ is a logical consequence of $$X$$ iff $$(\neg X \vee \phi^*)$$ is valid iff $$(Y \vee \phi^*)$$ is valid. Here, both $$(\neg Y \wedge \phi^*)$$ and $$(Y \vee \phi^*)$$ are sentences of $$\mathbf{EIFL}$$, the latter even a sentence of $$\mathbf{IFL}$$. It follows that the satisfiability (validity) of any second-order sentence can be expressed as the satisfiability (respectively, validity) of a sentence of $$\mathbf{EIFL}$$.[65]
Algebraic structure. The two negations available in $$\mathbf{EIFL}, \neg$$ and $${\sim}$$, agree on true sentences, as well as on false ones: if $$\phi$$ is true (false) in $$M$$, then both $${\sim}\phi$$ and $$\neg \phi$$ are false (true) in $$M$$. By contrast, if $$\phi$$ is non-determined in $$M$$, then $${\sim}\phi$$ is non-determined as well, but $$\neg \phi$$ is true. The combination $$\neg{\sim}$$ of the two negations applied to a sentence $$\phi$$ asserts that $$\phi$$ is not false.[66]
The propositional part of $$\mathbf{EIFL}$$ involves the four operators $$\neg , {\sim}, \wedge$$, and $$\vee$$. Hintikka (2004b) raises the question of the algebraic structure induced by these operators, when any two truth equivalent sentences are identified. The operators $$\neg , \wedge$$, and $$\vee$$ give rise to a Boolean algebra – but what does the strong negation $${\sim}$$ add to this structure?
Restricting attention to truth equivalence, $${\sim}$$ is definable from the operators $$\neg$$ and $$\neg{\sim}$$. For, $${\sim}\phi$$ is true in $$M$$ iff $$\neg(\neg{\sim}\phi)$$ is true in $$M$$. Instead of $${\sim}$$, the operator $$\neg{\sim}$$ may then be considered. Hintikka points out that the propositional part of $$\mathbf{EIFL}$$ (formulated in terms of the operators $$\vee , \wedge , \neg$$ and $$\neg{\sim})$$ is a Boolean algebra with an operator in the sense of Jónsson & Tarski (1951). The additional operator $$\neg{\sim}$$ is a closure operator.
Jónsson and Tarski (1951, Thm. 3.14) showed that any closure algebra is isomorphic to an algebraic system formed by a set equipped with a reflexive and transitive relation.[67] As a matter of fact, the relevant algebraic structure is precisely that of the propositional modal logic $$\mathbf{S4}$$. Hence the propositional part of $$\mathbf{EIFL}$$ has the same algebraic structure as $$\mathbf{S4}$$. By a well-known result due to Gödel (1933) and McKinsey & Tarski (1948), intuitionistic propositional logic can be interpreted in $$\mathbf{S4}$$, via a translation $$t$$ such that $$\phi$$ is intuitionistically provable iff $$t(\phi)$$ is a valid $$\mathbf{S4}$$ formula. Thus, intuitionistic propositional logic is interpretable in $$\mathbf{EIFL}$$.[68]
## 5. Philosophical Consequences
Hintikka (2006a: 73–77) takes the following to be among consequences of the novel insights made possible by (extended) IF first-order logic: reconstruction of normal mathematical reasoning on the first-order level, a novel perspective on the notion of truth in axiomatic set theory, insights into the nature of negation, and the formulation of a self-applied truth-predicate. A related topic of general interest is the phenomenon of informational independence in natural languages. The ideas related to negation and definability of truth have been discussed in Subsections 4.2 and 4.5, respectively. Let us consider here the remaining issues.
### 5.1 Place in type hierarchy
Hintikka maintains that the only reasonable way of making a distinction between first-order logic and higher-order logic is by reference to the entities that one’s quantified variables range over. A first-order logic is, then, a logic in which all quantifiers range over individuals, in contrast to higher-order entities (e.g., subsets of the domain). On this basis Hintikka holds that substantially speaking, $$\mathbf{IFL}$$ and even $$\mathbf{EIFL}$$ are first-order logics.[69] Solomon Feferman (2006: 457–461) criticizes the criterion that Hintikka employs for judging the first-order status of a logic. Feferman makes use of generalized quantifiers in his argument.[70] The formulas
$Q[z_1] \ldots [z_k] (\phi_1 ,\ldots ,\phi_k)$
involving generalized quantifiers are syntactically first-order, insofar as the quantified variables $$z_{i1},\ldots ,z_{in_i} = [z_i]$$ are first-order $$(1 \le i \le k)$$. The semantics of a generalized quantifier $$Q$$ is formulated by associating with each domain M a $$k$$-ary relation $$Q_M$$ on M, with $$Q_M \subseteq$$ M$$^{n_1} \times \ldots \times$$ M$$^{n_k}$$. E.g., for any infinite cardinal $$\kappa$$, there is a generalized quantifier $$Q_{\ge \kappa}$$ such that $$Q_{\ge \kappa}z P(z)$$ is true in a model $$M$$ iff there are at least $$\kappa$$ elements that satisfy the predicate $$P$$. Hence generalized quantifiers can be semantically higher-order. (The notion of cardinality is a higher-order notion.) The fact that the variables in a formula range over individuals only, does not offer a reliable criterion for the logic’s first-order status.
Hintikka’s criterion could be reformulated by saying that a logic is of first order, if any play of a semantic game associated with a formula of this logic only involves (in addition to choices interpreting conjunctions and disjunctions) choices of individuals, as opposed to choices of higher-order entities. By this criterion $$\mathbf{IFL}$$ (and even $$\mathbf{EIFL})$$ are first-order logics, but the logic of generalized quantifiers such as $$Q_{\ge \kappa}$$ is not.[71] Feferman (2006: 461) anticipates the possibility of such a reply, but finds it unconvincing.
By a result of Hintikka (1955), the problem of deciding whether a sentence of second-order logic is valid can be effectively reduced to the validity problem of $$\mathbf{IFL}$$.[72] Väänänen (2001) has shown that the set of valid sentences of $$\mathbf{IFL}$$ has the same very high complexity as the set of validities of the full second-order logic.[73] Väänänen (2001) and Feferman (2006) conclude that speaking of validity in $$\mathbf{IFL}$$ leads to a strong commitment to full second-order logic. Hintikka (2006a: 476–477) looks at these results from the opposite direction: for him they mean that indeed one can speak of validity in full second-order logic in terms of validity in $$\mathbf{IFL}$$. What is more, Hintikka (1997) affirms that even $$\mathbf{EIFL}$$ is a first-order logic. If so, any mathematical theory that can be expressed by the truth of an $$\mathbf{EIFL}$$ sentence is likewise free from problems of set existence.
Hintikka’s position leads to a puzzle. If $$\phi$$ is a sentence of $$\mathbf{IFL}$$ not truth equivalent to any $$\mathbf{FO}$$ sentence, the truth-condition of the sentence $$\neg \phi$$ of $$\mathbf{EIFL}$$ cannot be formulated without recourse to the set of all strategies of player 2: $$\neg \phi$$ is true in model $$M$$ iff for all strategies of player 2 in game $$G(\phi , M)$$, there is a sequence of moves by player 1 such that player 1 wins the resulting play. The set of all strategies of player 2 is undeniably a higher-order entity. How can commitment to entities other than individuals be said to have been avoided here? Can the meaning of the sentence $$\neg \phi$$ be well understood without presupposing the genuinely second-order idea of all strategies of a given player?[74] Rather than being nominalistic, Hintikka’s position appears to be a variant of universalia in rebus. While rules of semantic games pertain to actions performed on first-order objects, combinatorial properties of sets of plays can only be formulated in second-order terms. As soon as game rules are defined for a language fragment, also the corresponding combinatorial properties are fully determined, among them the properties labeled as truth and falsity.
### 5.2 Philosophy of set theory
According to Hintikka, our pretheoretical idea of the truth of a quantified sentence $$\phi$$ (in negation normal form) is that there exist ‘witness individuals’ for the existential quantifiers, usually depending on values corresponding to the preceding universal quantifiers.[75] It is the existence of such witnesses that constitutes the truth of $$\phi$$. Providing witnesses is precisely what Skolem functions for $$\phi$$ do.[76] The truth of a quantified sentence $$\phi$$ amounts to the existence of a full set of Skolem functions for $$\phi$$. In Hintikka’s view, then, our ordinary notion of first-order truth is conceptualized in terms of (existential) second-order logic. What happens when this idea is applied to axiomatic set theory, say Zermelo-Fraenkel set theory with the Axiom of Choice $$(\mathbf{ZFC})$$? It should be borne in mind that the very idea of axiomatic set theory is to dispense with higher-order logic; its underlying logic is taken to be $$\mathbf{FO}$$. Hintikka argues as follows.[77]
For each sentence $$\phi$$ of $$\mathbf{ZFC}$$, there is another sentence $$\phi^* = (\exists f_1)\ldots(\exists f_n)\psi^*$$ of $$\mathbf{ZFC}$$ which says, intuitively, that ‘Skolem functions’ for $$\phi$$ exist. These ‘Skolem functions’ are certain individuals of the domain of a model of $$\mathbf{ZFC}$$. Here both $$\phi^*$$ and $$\phi$$ are first-order sentences. But if for every sentence $$\phi$$ of $$\mathbf{ZFC}$$ we have $$\phi$$ and $$\phi^*$$ being logically equivalent, why could the sentences $$\phi^*$$ not be used for formulating a truth-predicate for $$\mathbf{ZFC}$$ in $$\mathbf{ZFC}$$ itself? However, by Tarski’s undefinability result, no such truth-predicate exists.[78] So there must be a model of $$\mathbf{ZFC}$$ and a sentence $$\phi$$ true in that model such that $$\phi^*$$ is false: not all ‘Skolem functions’ asserted to exist by $$\phi^*$$ actually exist in the model.
This reasoning shows that $$\mathbf{ZFC}$$ does not fully capture the idea of truth according to which the truth of a sentence $$\phi$$ means that the Skolem functions for $$\phi$$ exist. Furthermore, it also shows that the standard interpretation of higher-order logic is not fully captured by $$\mathbf{ZFC}$$. To see this, observe that for every sentence $$\phi$$ there is a logically equivalent second-order sentence $$\phi^{**} = (\exists F_1)\ldots(\exists F_n)\psi^{**}$$ actually asserting the existence of Skolem functions for $$\phi$$. The first-order sentence $$\phi^* = (\exists f_1)\ldots(\exists f_n)\psi^*$$ must not be confused with the second-order sentence $$\phi^{**} = (\exists F_1)\ldots(\exists F_n)\psi^{**}$$. The Skolem functions $$F_i$$ of which the sentence $$\phi^{**}$$ speaks are sets built out of individuals of the set-theoretical universe, while the ‘Skolem functions’ $$f_i$$ spoken of by $$\phi^*$$ are individuals.[79]
The conclusion of Hintikka’s argument is that our ordinary notion of truth is misrepresented by $$\mathbf{ZFC}$$. Furthermore, by Tarski’s undefinability result the situation cannot be improved by adding further axioms to $$\mathbf{ZFC}$$. In Hintikka’s judgment, axiomatic set theory is a systematic but futile attempt to capture on the first-order level truths of standardly interpreted second-order logic. Like Gödel (1947), also Hintikka holds that the concepts needed to state, say, the continuum hypothesis are sufficiently well-defined to determine the truth-value of this conjecture. The continuum hypothesis does not receive its meaning from phrasing it in $$\mathbf{ZFC}$$. Gödel and Hintikka agree that the independence results due to Gödel himself and Paul Cohen do not by themselves show anything about the truth or falsity of the continuum hypothesis. But unlike Gödel, Hintikka finds the derivability of any conjecture whatever in $$\mathbf{ZFC}$$ (or in any of its extensions) simply irrelevant for the truth of the conjecture. For Hintikka, it is a ‘combinatorial’ question whether every infinite subset of the reals is either countable or else has the cardinality of the set of all reals – a question properly conceptualized within second-order logic. This is what Hintikka takes to be the pretheoretical sense of the truth of the continuum hypothesis, and that is not captured by $$\mathbf{ZFC}$$.[80]
### 5.3 Extended IF first-order logic and mathematical theorizing
Hintikka sees $$\mathbf{EIFL}$$ as allowing to reconstruct all normal mathematical reasoning on the first-order level. This result is essentially dependent on the acceptability of Hintikka’s claim that $$\mathbf{EIFL}$$ is ontologically committed to individuals only (Subsect. 5.1). But how would $$\mathbf{EIFL}$$ serve to reconstruct an important part of all mathematical reasoning?
Hintikka (1996: 194–210) discusses mathematical theories (or mathematical axiomatizations) and mathematical problems (or questions of logical consequence) separately.
Any higher-order mathematical theory $$T$$ gives rise to a many-sorted first-order theory $$T^*$$. If the theory is finite, there is a finite conjunction $$J$$ formulated in $$\mathbf{EIFL}$$ – equivalent to a sentence of $$\mathbf{USO}$$ and therefore equivalent to the contradictory negation of an $$\mathbf{ESO}$$ sentence and so indeed equivalent to the contradictory negation of an $$\mathbf{IFL}$$ sentence – expressing the requirement that the standard interpretation of higher-order logic is respected. The question of the truth of the higher-order theory $$T$$ is thus reduced to the truth of the sentence $$(J \wedge T^*)$$ of $$\mathbf{EIFL}$$ (cf. Subsect. 4.6).
The mathematical problem of whether a given sentence $$C$$ is a logical consequence of a finite higher-order theory $$T$$ coincides with the problem of whether the second-order sentence $$(\neg(J \wedge T^*) \vee C^*)$$ is valid. Recalling that there is a sentence $$\chi$$ of $$\mathbf{IFL}$$ such that $$J$$ is equivalent to $$\neg \chi$$, it follows that $$\neg(J \wedge T^*)$$ is equivalent to a sentence of $$\mathbf{IFL}$$. Consequently there is a sentence of $$\mathbf{IFL}$$ which is valid iff the sentence $$(\neg(J \wedge T^*) \vee C^*)$$ is valid (cf. Subsect. 4.6). Mathematical problems can be understood as questions of the validity of an $$\mathbf{IFL}$$ sentence. Among mathematical problems thus reconstructible using $$\mathbf{IFL}$$ are the continuum hypothesis, Goldbach’s conjecture, Souslin’s conjecture, the existence of an inaccessible cardinal, and the existence of a measurable cardinal.[81]
As conceptualizations apparently transcending the proposed framework – not expressible in a higher-order logic – Hintikka considers the maximality assumption expressed by David Hilbert’s so-called axiom of completeness. The axiom says that no mathematical objects can be added to the intended models without violating the other axioms.[82]
If indeed problems related to the idea of all subsets are avoided in $$\mathbf{EIFL}$$ (Hintikka 1997), it offers a way of defending a certain form logicism. Unlike in historical logicism, the idea is not to consider logic as an axiom system on the same level as mathematical axiom systems (Hintikka 1996: 183),[83] and to attempt to reduce mathematics to logic. Rather, Hintikka (1996: 184) proposes to ask: (a) Can the crucial mathematical concepts be defined in logical terms? (b) Can the modes of semantically valid logical inferences used in mathematics be expressed in logical terms? The idea is not to concentrate on deductive rules of logic: no complete set of deductive rules exist anyway for $$\mathbf{IFL}$$. Because the status of higher-order logic is potentially dubious – due to the problems associated with the notion of powerset – a positive solution to questions (a) and (b) calls for a first-order logic more powerful than $$\mathbf{FO}$$.
The suggested reduction of all mathematics expressible in higher-order logic to the first-order level would be philosophically significant in showing that mathematics is not a study of general concepts, but of structures consisting of particulars (Hintikka 1996: 207). This is not to say that actual mathematics would be best carried out in terms of $$\mathbf{IFL}$$, only that it could in principle be so carried out (Hintikka 1996: 205, 2006a: 477). For a critique of Hintikka’s conclusions, see Väänänen (2001), Feferman (2006), and Bazzoni (2015).
### 5.4 Informational independence in natural languages
When Hintikka began to apply GTS to the study of natural language (Hintikka 1973a), he took up the question of whether branching quantifiers occur in natural languages. He was led to ask whether there are semantic games with imperfect information. He detected various types of grammatical constructions in English that involve informational independence.[84] An often cited example is the sentence
Some relative of each villager and some relative of each townsman hate each other,
true under its relevant reading when the choice of a relative of each townsman can be made independently of the individual chosen for ‘each villager.’[85] Hintikka (1973a) sketched an argument to the effect that actually every $$\mathbf{FPO}$$ sentence can be reproduced as a representation of an English sentence. From this it would follow that the logic of idiomatic English quantifiers is much stronger than $$\mathbf{FO}$$, and that no effective procedure exists for classifying sentences as analytical or nonanalytical, synonymous or nonsynonymous.[86] This would be methodologically a very important result, showing that syntactic methods are even in principle insufficient in linguistic theorizing. Jon Barwise (1979) suggested that particularly convincing examples supporting Hintikka’s thesis can be given in terms of generalized quantifiers.
Lauri Carlson and Alice ter Meulen (1979) were the first to observe cases of informational independence between quantifiers and intensional operators. Consider the question[87]
Under one of its readings, the presupposition of this question is $$(\forall x)(\exists y)$$ admires$$(x, y)$$. The desideratum of this question is
Writing ‘$$K_I$$’ for ‘I know,’[88] the desideratum has a reading whose logical form is
$$K_I (\forall x)(\exists y/K_I)$$ admires$$(x, y)$$.
This desideratum is satisfied by an answer pointing out a function $$f$$ which yields for each person a suitable admired person. Such a function can be his or her father. Importantly, the value $$f(b)$$ of this function only depends on the person $$b$$ interpreting ‘everybody,’ but does not depend on the scenario $$w$$, compatible with the questioner’s knowledge, that interprets the construction ‘I know.’ Interestingly, the desideratum $$K_I (\forall x)(\exists y/K_I)$$ admires$$(x, y)$$ is not expressible without an explicit independence indicator. It is also worth noting that this case cannot be represented in the notation of $$\mathbf{FPO}$$. This is because several types of semantic interactions are possible among quantifiers and intensional operators, and blocking interactions of one type does not automatically block interactions of other types. In the example, the witness of $$\exists y$$ must not vary with the scenario $$w$$ interpreting the operator $$K_I$$, but still the values of both variables $$x$$ and $$y$$ must belong to the domain of the particular scenario $$w$$ chosen to interpret $$K_I$$.[89]
Hintikka’s ideas on desiderata of wh-questions were influenced by his exchange with the linguist Elisabet Engdahl.[90] These wh-questions, again, functioned as important test cases for the appearance of informational independence in natural languages.
Hintikka and Sandu (1989) took up the task of formulating an explicit unified formal treatment for the different varieties of informational independence in natural language semantics. They posed the question of which are the mechanisms that allow English to exceed the expressive power of $$\mathbf{FO}$$. For, natural languages typically do not resort to higher-order quantifiers. Hintikka and Sandu suggested that informational independence plays a key role in increasing the expressive power of natural languages.[91]
In GTS as developed for English in Hintikka & Kulas (1983, 1985), game rules are associated with a great variety of linguistic expressions (cf. Subsect. 2.1). As Hintikka (1990) has stressed, informational independence is a cross-categorical phenomenon: it can occur in connection with expressions of widely different grammatical categories. Hintikka and Sandu (1989) propose several examples from English calculated to show that there is an abundance of instances of informational independence in natural languages. Among examples are wh-questions of the kind discussed above, and the distinction between the de dicto and de re readings of certain English sentences. Hintikka and Sandu suggest that representing such readings in terms of IF logic is more truthful to the syntax of English than the alternative, non-IF representations are.
In connection with knowledge, a de dicto attribution such as
$$K_{\textit{Ralph}} (\exists x) (x$$ is a spy)
can be turned into a de re attribution by marking the existential quantifier as independent of the knowledge-operator (cf. Hintikka and Sandu 1989):
$$K_{\textit{Ralph}} (\exists x/K_{\textit{Ralph}}) (x$$ is a spy).
Because knowledge is a factive attitude (the actual world is among Ralph’s epistemic alternatives), this amounts indeed to the same as the condition
$$(\exists x) K_{\textit{Ralph}} (x$$ is a spy).
Rebuschi & Tulenheimo (2011) observed that independent quantifiers are of special interest in connection with non-factive attitudes such as belief. The logical form of a statement ascribing to Ralph a belief pertaining to a specific but non-existent object is
$$B_{\textit{Ralph}} (\exists x/B_{\textit{Ralph}}) (x$$ is a spy),
where ‘$$B_{\textit{Ralph}}$$’ stands for ‘Ralph believes that.’[92]
Attitudes of this form were dubbed de objecto attitudes. Since the (intentional) object of such an attitude need not exist actually, the de objecto attitude is weaker than the de re attitude
$$(\exists x) B_{\textit{Ralph}} (x$$ is a spy).
On the other hand, the pattern of operators $$B_{\textit{Ralph}} (\exists x/B_{\textit{Ralph}})$$ requires that the witness of the existential quantifier $$\exists x$$ be the same relative to all doxastic alternatives of Ralph, so the de objecto attitude is stronger than the de dicto attitude
$$B_{\textit{Ralph}} (\exists x) (x$$ is a spy).
Janssen (2013) discusses the possibility of providing in terms of $$\mathbf{IFL}$$ a compositional analysis of the de re / de dicto ambiguity in natural languages. Brasoveanu and Farkas (2011) argue that scopal properties of natural language indefinites are best elucidated in terms of a semantics inspired by $$\mathbf{IFL}$$, more precisely by formulating the semantics relative to sets of variable assignments as done in Hodges’s compositional semantics for slash logic.
As a rule, informational independence is not indicated syntactically in English.[93] Methodological consequences of this fact are discussed in Hintikka (1990), where he tentatively puts forward the Syntactic Silence Thesis, according to which sufficiently radical cross-categorical phenomena are not likely to be marked syntactically in natural languages. Evidence for this thesis would, for its part, be evidence against the sufficiency of syntax-oriented approaches to semantics.
## 6. Related logics
### 6.1 Slash logic
Syntactically slash logic uses quantifiers like $$(\exists x/y)$$ instead of quantifiers such as $$(\exists x/\forall y)$$. Semantically slash logic is otherwise like $$\mathbf{IFL}$$ except that its game-theoretical semantics is based on the idea that a player’s strategy functions may utilize as their arguments any preceding moves made in the current play, save for those whose use is, by the slash notation, explicitly indicated as forbidden (cf. Sect. 3). That is, also a player’s own earlier moves may appear as arguments of a strategy function. This can make a difference in the presence of imperfect information. For example, consider evaluating the slash-logic sentence $$(\forall x)(\exists y)(\exists z/x) x = z$$ containing the vacuous quantifier $$\exists y$$. This sentence is true on a two-element domain, since player 2 can copy as the value of $$y$$ the value that player 1 has chosen for $$x$$, and then select the value of $$z$$ using a strategy function whose only argument is the value of $$y$$. (For this phenomenon of ‘signaling,’ see Hodges 1997a, Sandu 2001, Janssen & Dechesne 2006, Barbero 2013.) By contrast, the IF sentence $$(\forall x)(\exists y)(\exists z/\forall x) x = z$$ fails to be true on such a domain, since there a strategy function for $$(\exists z/\forall x)$$ must be a constant, and no such strategy function can guarantee a win for player 2 against both possible values that player 1 can choose for $$x$$. As mentioned in Subsection 4.4, Hodges (1997a,b) showed that slash logic admits of an alternative, compositional semantics. This requires evaluating formulas relative to sets of variable assignments, instead of single assignments as in connection with $$\mathbf{FO}$$.
All authors having studied slash logic, apart from Hodges himself, have opted for not following Hodges’s terminological recommendation mentioned at the beginning of Section 3: they have referred to slash logic as ‘IF logic.’
Kuusisto (2013) studies the expressive power of fragments of slash logic whose formulas are formed without employing the identity symbol. Kontinen et al. (2014) investigate the complexity-theoretic properties of the two-variable fragment of slash logic and compare this fragment to the corresponding fragment of dependence logic. Hodges (1997a,b) and Figueira et al. (2009, 2011, 2014) discuss an extension of slash logic in which the contradictory negation of slash-logic sentences can be expressed.
In Sevenster (2014), patterns of quantifier dependence and independence in quantifier prefixes of slash-logical formulas are systematically studied in order to determine which quantifier prefixes allow slash logic to gain the expressive power of $$\mathbf{ESO}$$. Sevenster identifies two such patterns – the signaling pattern and the Henkin pattern – and proves that they are able to express $$\mathbf{NP}$$-hard decision problems. He further shows that these two are the only patterns allowing slash logic to exceed the expressive power of $$\mathbf{FO}$$ insofar as attention is confined to formulas in prenex form. An example of the signaling pattern would be $$(\forall u)(\exists v)(\exists w/u)$$ and of the Henkin pattern $$(\forall x)(\exists u)(\forall y)(\exists v/x,u)$$.[94]
Barbero (2021) and Barbero et al. (2021) take up the task of investigating fragments of the general syntax of slash logic (not all of whose formulas are in prenex form). Thereby these authors wish to study systematically those expressive resources of slash logic that allow it to exceed the expressive power of $$\mathbf{FO}$$ and that do not simply derive from its capacity to mimick Henkin quantifiers. While many of the features studied in these two papers depend on the Eigenart of slash logic, the phenomenon of signaling by disjunction occurs in IF logic, as well. In Section 3.3, it was seen that the IF-logical sentence $$(\forall x)(\exists y/\forall x) x = y$$ is non-determined in any model whose domain has exactly two elements. The slash-logical sentence $$(\forall x)(\exists y/ x) x = y$$ is likewise non-determined in such a model. Now, consider the result of replacing in the latter sentence the expression $$(\exists y/ x) x = y$$ by the disjunction $$((\exists y/x) x = y \vee (\exists y/ x) x = y))$$, with a token of the initial expression in both disjuncts: $(\forall x)((\exists y/ x) x = y \vee (\exists y/x) x = y).$ The sentence that has been thus obtained is actually true in a model whose domain consists of the objects $$a$$ and $$b$$. Let $$f$$, $$g$$, and $$h$$ be respectively the strategy functions of player 2 for the unique token of the disjunction symbol, the left token of the existential quantifier, and the right token of the existential quantifier – defined as follows. First, $$f$$ selects the left disjunct if player 1 has chosen $$a$$ as a value of $$x$$, selecting the right disjunct otherwise. Further, $$g$$ and $$h$$ are constants (zero-place functions): $$g= a$$ and $$h= b$$. The set $$\{f, g, h\}$$ is clearly a winning strategy for player 2. Either of the tokens of the existential quantifier occur in a specific disjunct. Using the strategy function $$f$$ to arrive at such a disjunct makes sure that the disjunct arrived at reveals the choice that player 1 has made to interpret $$\forall x$$. The constants $$g$$ and $$h$$ are selected so as to render this information explicit. Thus, the resulting values of $$x$$ and $$y$$ indeed satisfy the formula $$x= y$$. By the same reasoning it is seen that the sentence $$(\forall x)((\exists y/ \forall x) x = y \vee (\exists y/\forall x) x = y)$$ of $$\mathbf{IFL}$$ is true in the model considered.
### 6.2 Dependence logic
Jouko Väänänen (2007) formulated a new approach to IF logic that he dubbed dependence logic $$(\mathbf{DL})$$; for further work on $$\mathbf{DL}$$, see e.g. Kontinen et al. (2013). The syntax of $$\mathbf{DL}$$ is obtained from that of $$\mathbf{FO}$$ by allowing atomic formulas of the following special form:
$=(x_1 ,\ldots ,x_n; x_{n+1}).$
Intuitively such a formula means that the value of $$x_{n+1}$$ depends only on the values of $$x_1 ,\ldots ,x_n$$. The semantics of $$\mathbf{DL}$$ cannot be formulated relative to single variable assignments like that of $$\mathbf{FO}$$: we cannot explicate what it means for the value of $$x_{n+1}$$ to depend on those of $$x_1 ,\ldots ,x_n$$ with reference to a single assignment on the variables $$x_1 ,\ldots ,x_{n+1}$$. For example, consider the assignment described below:
$$x_1$$ $$x_2$$ $$x_3$$ 7 5 8
Relative to this assignment, all of the following claims hold: whenever the value of $$x_1$$ equals 7, the value of $$x_3$$ equals 8; whenever the value of $$x_2$$ equals 5, the value of $$x_3$$ equals 8; whenever the value of $$x_1$$ equals 7 and the value of $$x_2$$ equals 5, the value of $$x_3$$ equals 8; irrespective of the values of $$x_1$$ and $$x_2$$, the value of $$x_3$$ equals 8. The question of dependence only becomes interesting and non-vacuous relative to a set of assignments:
$$x_1$$ $$x_2$$ $$x_3$$ 7 5 8 9 5 6 7 11 8 7 3 8 9 19 6
The set $$X$$ consisting of the above five assignments satisfies the formula $$=(x_1$$; $$x_3)$$: the value of $$x_3$$ depends only on the value of $$x_1$$. As is readily observed, any two assignments in $$X$$ which assign the same value to $$x_1$$ assign also the same value to $$x_3$$. The interesting novelty of $$\mathbf{DL}$$ is that claims about variable dependencies are made at the atomic level. Quantifiers of $$\mathbf{IFL}$$ and those of slash logic with their independence indications easily lead to somewhat messy formulas, whereas $$\mathbf{DL}$$ looks exactly like $$\mathbf{FO}$$, apart from its greater flexibility in forms of atomic formulas.
## 7. Conclusion
In this entry IF first-order logic and extended IF first-order logic have been surveyed. Their metalogical properties have been explained and the philosophical relevance of these properties has been discussed. The suggested consequences of these logics for philosophical issues such as the existence of a self-applied truth-predicate, the logicist program, the philosophical relevance of axiomatic set theory, and informational independence in natural languages have been covered as well. Slash logic and dependence logic – both closely related to $$\mathbf{IFL}$$ and inspired by it – were also briefly considered.
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# Analytic Geometry
• Sep 28th 2012, 08:16 PM
AuXian
Analytic Geometry
Hello, everybody. I need help... (Worried)
Show that, for all values of p, the point P given by x=ap2, y=2ap lies on the curve y2=4ax.
a) Find the equation of this normal to this curve at the point P.
If this normal meets the curve again at the point Q (aq2, 2aq). Show that p2+pq+2=0
b) Determine the coordinates of R, the point of intersection of the tangents of the curve at the point P and Q.
Hence, show that the locus of the point R is y2(x+2a) +4a3=0
I already solved Q (a) and first question of Q (b). However, I can’t solve the last question: “Hence, show that the locus of the point R is y2(x+2a) +4a3=0” in Q (b). Can somebody help me?
Thanks!!(Happy)
• Sep 28th 2012, 08:29 PM
MarkFL
Re: Analytic Geometry
Presumably, you have found that that $R$ has the coordinates:
$(apq,a(q+p))$
Now, use the relationship between $p$ and $q$ you found earlier:
$p^2+pq+2=0$
which when solved for $q$ is:
$q=-\frac{p^2+2}{p}$
Now you may write the coordinates of $R$ as parametric equations in one parameter, which you may then eliminate to obtain the required Cartesian equation.
• Sep 28th 2012, 08:34 PM
chiro
Re: Analytic Geometry
Hey AuXian.
I haven't seen this stuff since high school (more than 10 years) but the wiki page gives a derivation of the locus for the parabola:
Parabola - Wikipedia, the free encyclopedia
I'm not sure if you just formulas or have to derive things, but if you just formulas the wiki gives a derivation from start to finish in terms of the a term.
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# Electric charges, wrong answer D: ?
1. Jul 27, 2011
### Eats Dirt
1. The problem statement, all variables and given/known data
Find the net force at point A
q1=-6x10^-5 C
q2=-3x10^-5 C
q1 is 3m north of A
q2 is 3m east of A
find the net force on a -1.2x10^-5 C charge
3. The attempt at a solution
first off I find how the electric field is with no charge at point A
= 6.7 x 10^4 N/C [E 63 degrees N]
= Fe= qE
Fe= (-1.2x10^-5)(6.7x10^4)
Fe= 0.804 [W 63 degrees S]
However the answer in the book says the answer is 8x10^4 N [W63degreesS]
i dont know if it is me or the book making the mistake?
2. Jul 27, 2011
### PeterO
Read carefully to make sure the distances really are 3m. That is a long way away.
Your calculation look OK to me, given the figures you have provided here. But to 3 figures that should be 0.805 [0.804984]
Given that some of the data was given to only 1 significant figure - you answer should have been 0.8
Peter
Last edited: Jul 27, 2011
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Electronics-Lab.com Community
# Bridge rectifiers
## Recommended Posts
I am asking for assistance with bridge rectifiers. Specifically, how to determine the voltage rating of a 20 amp bridge rectifier. The circuit that requires this rectifier is a non-regulated circuit. This circuit consists of a stator, 12 volt lead acid battery and rectifier. The stator produces AC voltages @ 180-200 DVA volts. The voltage going the battery cannot excede 14.5 volts DVA. I am getting 17.5 volts at the battery which is 3 volts over the allowable limit of 14.5 volts. What I do not understand is what part of the circuit limits the DC voltage going to the battery. Is it the voltage of the battery or the voltage rating of the recitifier. Is the 17.5 volts due to a faulty battery of faulty rectifier? Would the voltage rating of the 20 amp rectifier be that of the battery (12 volt 20 amp rectifier) or would it need to match the AC current being rectified (200 volt 20 amp rectifier)
Any help is greatly appreciated.
Cheers
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Hi Matthew,
Welcome to this forum.
I am not following you here, is it possible for you to post a schematic of your circuit?
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A schematic is something I don't have. So, please bear with me as I attempt a better explaination. The circuit that I am attempting to describe is for an outboard motor. The outboard uses a stator to produce AC current (alternator). The AC current then goes to a full wave bridge rectifier to produce rectified DC current to charge a 12 volt battery. The system is "non-regulated". Meaning there is no voltage regulator. The stator produces 180 to 200 volts of AC current. Some of this AC current is directed to the capacitor discharge ignition (CDI) and it's three coils which provide "spark" for the three cylinders via it's sparkplugs. The rest of the AC current that the stator produces is routed directly to the bridge rectifier to produce the "rectified DC current" in which the battery requires. Whatever DC or possibly AC current (I'm not sure which) the battery does not need is shorted to ground thru the rectifier. According to the repair manual for the outboard, voltage going to the battery should not exceed 14.5 volts. Any voltage readings above 14.5 volts with the motor running at or above a certain rpm the manual states is the cause of a faulty rectifier. With the motor running at the suggested rpm I get a reading of 17.5 volts. Going by what the manual says I am assuming I have a faulty rectifier. However, the out-of-range voltage reading of 17.5 volts may also be the cause of a faulty battery. For conversation sake though, I'd like to assume it is not. Electronic supply companies sell what I believe to be the very same type of rectifier for \$2.50 vs. Outboard parts dealers want \$60 plus dollars for a new rectifier. The manual states the rectifier has a amperage rating of 20 amps. However, the manual does not give the voltage rating for the rectifier. From what I am able to ascertain similar 20 amp rectifiers are offered with voltage ratings of: 50 volts, 100 volts, 200 volts, and 400 volts. What I would like to know is the voltage rating for a bridge rectifier based upon the input value of AC voltage of the circuit it is designed for. Or, is it based upon the value of rectified DC voltage the rectifier outputs regardless of the input AC voltage? Or, in otherwords do I need a rectifier that has a 200 volt rating or a 12-14.5 voltage rating. Or, perhaps a different voltage rating altogether?
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Matthew,
Ok, the 17.5Volts reading is on the battery connectors I assume? If so it means one of two possible reasons, first a faulty battery (most likely) and second an enormous charge current. Have you measured the current (ADC) going to the battery? Is the motor running nicely, no misfiring or uneven rpm:s at the presence of the 17.5V at the battery? The statement
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Thank you Ante,
Yes the motor does have a intermittent misfire. No. 3 cylinder is intermittently getting spark. Manual states possible causes as being blown CDI module and/or faulty rectifier and/or faulty stator. According to the manual the 180 volts AC being produced by the stator is correct. Exceptable values go as high as 400 volts AC. Anything above 400 volts going to the CDI modules the manual recommends replacing the stator. Since there is no voltage regulator the chance of burning out CDI modules, coils, batteries and rectifiers is extremely likely as well as costly. With what you have told me, and then putting two and two together, I guessing the stator is probably causing all the problems. Thanks again for your help. It is very much appreciated.
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Hi Matthew, Ante.
can you keep us updated with your progress on solving this problem. There must be some kind of voltage control, I am thinking that the CDI controls this, with my experience with early motorcross motorbikes, they had a high and low stater coil each coil would be switched on and off according to the RPM to maintain a reasonable voltage output.
So you should have a controlling mechanism either within your CDI unit most likely, or separately, that controls switching between stater coils to maintain a reasonable voltage output.
There may be more to your bridge rectifier then just diodes.
I think we really need to understand exactly how this system works, so we can work out exactly where the fault is.
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I know that the rectifier itself is a standard bridge rectifier consisting of 4 internal diodes. Radio Shack sells one exactly like it only it is rated at 25 amps and 50 volts. According to the Seloc repair manual for the outboard it states that the system is a non-regulated one. On the larger displacement motors that Chrysler/Force offered did come with a voltage regulator. The CDI circuitry does not have anything to do with regulating voltage to the battery. (as far as the limits of my knowledge allows me to assume) As it operates on the AC current being generated by the stator and it's role is strictly for sending signal voltages to the coils to produce spark. Since this is a three cylinder motor there are two seperate CDI modules. Both modules are identical to eachother. Except one module controls spark for no. 1 and no. 2 cylinder while the other module controls spark for no. 3. Either module can be swapped. There is a trigger mechanism on the flywheel that sends a timing signal to the correct module for spark on the appropiate cylinder. There is a blocking diode in each module. The blocking diode is to prevent a timing signal for no. 1 cylinder from also going to no. 2 at the same time. The CDI system consists of the two modules, 3 coils, the trigger assembly and the stator. The manaul also states that the battery only receives current from the rectifier when rpms are high enough to create the necessary amount of current. At low rpms the battery supplies power for everything except the CDI which, though operating on relatively high voltage AC current from the stator, actual current draw is very low. As far as being able to tell what limits the voltage going to the battery is the mystery I am trying to figure out. Could the battey voltage affect the rectified DC output voltage somehow by leakage across the diodes. (wild guess) sort of a feedback voltage. Other then that I am completely clueless as to what prevents the battery from receiving rectified DC voltage at the same AC voltage produced by the stator. That's why I didn't know if rectifiers voltage rating had something to do with the DC voltage of it's output or the input of the AC voltage. I apologize for only being able to supply you with information that has been dirived by much assumption. I realize it makes finding the answer that much more difficult.
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Hi Dazza,
I'm not certain if you are still interested in the developement of this thread but this is what I have been able to ascertain thus far. After consulting an outboard motor specialist on a forum specializing in outboard motors, I was informed that since my motor does not incorporate a voltage regulator on the charging circuit, the battery must deal with whatever DC voltage that gets sent it's way and why it is recommended that only high quality non-maintenance free battery's be used. High quality or not, batteries are not cheap so I intend to replace the OEM rectifier with rectifier with a built in voltage regulator. Larger displacment Chrysler outboards come with voltage regulators and installing one is extremely simple. The only difference in the wiring between those with regulators and those that don't use one is the regulated circuits use an extra wire connected to the ignition switch that supplies the accessories with 12 volts before being looped back to the battery. It doesn't get any easier then that. I do find it strange though that Chrysler would use a non-regulated system considering the potential for damage to the entire CDI system and the cost of repairs to replace damaged parts if and when a battery goes faulty from being overcharged since it is the battery that is the determining factor as to what the stator outputs as far as limiting AC spikes. I still don't have a clear understanding of how the how thing works but I believe it has to do with ohms law. As long as all the components stay within their respective tolerances as far as voltage and resistance goes the system works. A battery that begins to produce to high or to low a voltage or a short or open condition arises, resistance changes and the stator runs begins producing AC spikes which in turn may or may not cause further damage to the other components in the system. Go figure.
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You can use a larger filter capacitor to bring down the DC just a little bit.But you can add a series inductor and bring it down a little farther. And then you can add a series resistor that will take it the rest of the way. The voltage drop of the resistor will be determined by the current draw of the battery, but you should still get a charging of the battery if the resistance is low. The voltage that is applied to the diodes can be calculated by taking the value of the AC and dividing by .707. This is the voltage applied to both reverse biased diodes. If you don't know what the voltage will be, just use a fairly high one. I am missing something about the 20 A current capacity. Charging a battery should use much less than this. Also you should use diodes with a current rating to match the application, but the larger diodes give better regulation.
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Hi
Hi Kevin,
Since larger diodes provide better regulation then smaller ones and how most similar-in-amperage diodes (to the OEM rectifier used on my outboard) all seem tend to have a centered hole for fastening it down to a bracket or mounting location, are probably two factors they may have decided to use a rectifier with such a high amperage rating. Especially considering that the stator only produces a maxium amount of AC current of 7 amps. Plus, another reason maybe due to they purchase them in bulk so they probably pay less then \$1.00 each for them while selling them for \$45.00. US.
Using your example, the voltage rating for the rectifier I would require would be around 250-300 volts, correct?. From the manner in which you describe the regulator, it sounds like it is something realively simple for one to build himself/herself. The one problem that most concerns me would be how to determine the various ratings for the necessary components to build one. As well as a simple schematic. Might you know where I may find such information?
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I thought you were not going to use a regulator. Normally, battery charges are made to produce not only the voltage but the charging current. This charging current is produced by uping the voltage over the battery voltage. You can make a low current charger by practically not worrying about the current, but getting the voltage just higher than the battery. I don't think you need a regulator but it is a nice way of obtaining the correct voltage. Fortunately, a 12 volt regulator will give you the minimum voltage necessary. So use diodes that have let's say a 3 amp rating and a 12 volt regulator rated at 1A. You can stick your ammeter in there to read the current if you like. The filter capacitor might do well to be 1000uf.
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Using a regulator is something that I will no longer go without. Ruining batteries due to overcharging them is getting very costly and more then I can currently afford. Plus...replacement CDI packs go anywhere from \$160 to \$399 each. Loose or bad battery connections or a battery that is overcharging will almost always cause a circumstance in which blows a diode that is used to block the firing signal between the two CDI packs. I just want to have a system that will develope enough current to keep the starting battery charged as it also limits voltage to the battery at around 15 volts DC. I also have a deep cycle battery that I use to supply just my fish finder only. I would like to incorporate this deep cycle battery into the charging system via an isolator circuit.
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Matthew,
It is not possible to just raise the voltage to overcharge a battery. When you raise the voltage the current rises too (unless it
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Hi Matthew,
I'm sure you realise that you need to apply caution, when modifying your electrical system ;). You need to be certain that you have found the problem, before you try to modify/improved your electrical system. You want to be sure you are not actually applying a Band-Aid fix. You may have more than one component failing, which makes it a real pain in the B_M,
Do you have an auxilary engine, you don't want to find yourself drifting in the wild blue yonder,
well that's probably not so bad if you find yourself on the shores of down under, I could probably help you sort the problem out, but I think you may be a little hungry and thirsty buy then ;D ;D ;D.
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Considering that we have taken our little 17 foot Bayliner Capri some 60 plus miles offshore numerous times would make most have to question our intelligence out on the water. Such is the draw of the ocean. We do have a kicker motor on board just in case the main should fail us. One bit of reassurance out on the water is the fact that in San Diego CA. waters there is never a boat farther away then 5 to 10 miles and almost always they're within sight. Anyways, I have ran tests on the stator, trigger assembly, CDI modules, battery, and wiring. Everything is within spec. except for the deep cylce battery that I got from a friend who used it on his boat as a starting battery. And also, the last origninal CDI module needs to be replaced since it has a blown blocking diode. Preventing no. 3 cylinder from firing. I have studied all of the Chrysler outboards electrical systems that have regulators and all that is different between my motor and the regulated motors (of same year manufactured) is one wire that carries regulated DC positive current from the regulator, to the ignition switch then to the connection fuse block then to the accessory fuse block. Where as my system runs the postive DC wire directly off the rectifier then to the ignition switch where it eventually ends at the accessory fuse block. All else is the same. Same part numbers for CDI modules, stator and trigger assem. Same solenoids. All having the very same ratings as my engine. All I want to do is limit the voltage going to the start battery. We do enough starting and stopping and slow trolling to where the battery requires some current for recharging. Replacing the battery I have now with that of a correct starting battery instead of the deep cycle I have now, probably would have prevented the problems I am dealing with now. Mostly, having to replace a CDI module that died due to the bad battery. I have gotten to know my boats electrical system and wiring inside and out over the time I have owned her. And I am totally confident with adding a regulator. I would prefer to build my own if it isn't to difficult determining what the specs will need to be.
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Why not use a window comparator type circuit? You can adjust the ON limit and the OFF limit.
Actually, if you are dealing with any substantial amount of current, you will probably use the window comparator circuit to turn a transistor on and off.
One example is below:
MP
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Looks like Greek to me. lol
Perhaps adding a regulator is not what I should be focusing on.
I'd like to be sure I am grasping the concept behind my outboards charging circuit. The stator is rated at 200 volts AC , 7 amps. Between the stator and battery is a rectifier to switch AC into rectified AC...more or less turns AC into DC by filtering only one half of AC's signwave at any given moment. Now here's my grey area. What is limiting the voltage going to the battery. Meaning howcome the battery isn't getting 200 volts DC of the stators 200 volts AC output once it has passed thru the rectifier. Or, it does but it is relatively harmless because the battery isn't drawing any current. When the battery does draw current the increase in resistance creates a decrease in voltage. Ohms law. So, hypothetically, more then likely the addition of a voltage regulator on those motors that Chrysler installed them in is to regulate the voltage going to 12 volt accessories such as radio's, gps, radar, ect ect. Figuring (hypothetically again) that most boats using their 85 hp engine/s would probably only have one battery which is for starting purposes only.
So...close but no? Getting warmer? Not even close? Forget it he'll never learn? lol
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Matthew A. Concentrate your efforts on obtaining the electrical schematics for your outboard(and post them). This will be half your battle over. I'm sure someone will then be able to find a solution, contact your outboard manufacturer, someone anyone and get them from somewhere :).
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Thank you Dazza, MP, and everyone, for the time and help you all have given me. The knowledge each of you has shared has been of great help and value to me and is greatly appreciated. :)
Matt
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Jamb Physics Past question - Page 2 of 11 - TEEZAB
# Jamb Physics Past question
Hello Friend! Are you preparing for the Joint Admissions and Matriculations Board JAMB UTME 2023/2024? If yes, then this is for you! Prepare yourself by treating JAMB Physics Past Questions!
I have prepared a series of JAMB Physics Past Questions from 1988 to 2022 which was prepared with the sole purpose of help you prepare very well for the 2023/2024 JAMB!
This JAMB Physics Past Questions comprise of questions, answers and explanation to the answers. Also, you get to discuss! If you feel one answer is not correct, don’t hesitate to leave a comment.
So, take out your pen and a book and start writing!
I must tell that before you start treating past this physics Past Questions, you must have finished reading a physics textbook! At the very least, once!
My friend, here is an opportunity for you to score high in JAMB and to score high, you will need each Mark!
## A piece of rubber 10cm long stretches 6mm when a load of 100N is hung from it. What is the strain
Jamb 1978 A piece of rubber 10cm long stretches 6mm when a load of 100N is hung from it. What is the strain? ANSWER: Option C EXPLANATION: Using the equation Strain = ∆L/L Where; ∆L —> change in length L —> Original Length If L1 is the original length and L2 is the length after …
## The motion of the moving skin of a talking drum can rightly be described as
JAMB 1980 The motion of the moving skin of a talking drum can rightly be described as A. translational B. random C. rotational D. oscillatory E. transitory Correct Answer: Option D Explain your answer in the comment box below and let’s have a discussion so as to have a great understanding of this question! PREVIOUS …
## The relative densities of zinc, brass, copper, gold and silver are respectively 7.1, 8.5, 8.9, 19.3…
JAMB 1980 The relative densities of zinc, brass, copper, gold and silver are respectively 7.1, 8.5, 8.9, 19.3, 10.5. A metal ornament which weighs 0.425kg and can displace 50 x 10^-6m³ of water is made of A. zinc B. brass C. copper D. gold E. silver Correct Answer: Option B Explanation Density of the metal …
## Which of the following statements defines correctly the efficiency of a machine
Jamb 1979 Which of the following statements defines correctly the efficiency of a machine? A. The load carried by the machine divided by effort required in carrying the load B. The distance moved by the load divided by the distance moved by the effort C. The useful work done by the machine divided by the …
## A string is sustaining a stationary wave when
Jamb 1978 A string is sustaining a stationary wave when A. the frequency of the wave is hot proportional to the wavelength of the wave B. the amplitude of vibration is always zero C. two equal waves are travelling in opposite directions D. the waves are longitudinal E. the frequency is directly proportional to the …
## Which of the following statements about sound is NOT true…
JAMB 1980 Which of the following statements about sound is NOT true? A. sound is a longitudinal wave B. sound can be reflected and refracted like light C. sound travels through a vacuum D. sound moves faster in a steel wire than in air E. sound can ony be produced by vibrating objects Correct Answer: …
## If in a simple pendulum experiment the length of the inextensible string is increased by a factor of four, its period is increased by a factor of
Jamb 1979 If in a simple pendulum experiment the length of the inextensible string is increased by a factor of four, its period is increased by a factor of A. 4 B. π/2 C. 1/4 D. 2π E. 2 ANSWER: Option E EXPLANATION: for simple pendulum; T = 2π√(L/g) Where; T is the Tension L …
## Natural radioactivity consists of the emission of
JAMB 1980 Natural radioactivity consists of the emission of A. α -particle and β rays B. α -particle and X-rays C. γ-rays and X-rays D. α-particle, β rays and γ-rays E. α -particle, β rays, γ-rays and X-rays Correct Answer: Option D Explain your answer in the comment box below and let’s have a discussion …
## The water at the bottom of a waterfall is slightly warmer than that at the top because part of the
JAMB 1980 The water at the bottom of a waterfall is slightly warmer than that at the top because part of the A. potential energy raises the water temperature B. water absorbs latent heat from the atmosphere C. kinetic energy is changed intyo internal energy D. dropping mass of water increases the heat capacity E. …
## Three 5 ohm resistors connected in parallel have a potential difference of 60V applied across the combination. The current in each resistor is
Jamb 1978 Three 5 ohm resistors connected in parallel have a potential difference of 60V applied across the combination. The current in each resistor is A. 4A B. 36A C. 12A D. 24A E. 10A. ANSWER: Option C EXPLANATION: You should note that in a parallel arrangement, Voltage is always the same as with the …
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# What does it mean by “Path Matrix” and “Transitive Closure” of a graph (Directed and Undirected)?
I the discussion of various graph algorithms, I see the terms "Path Matrix" and "Transitive Closure" which are not well-defined anywhere.
What does it mean by "Path Matrix" and "Transitive Closure" in case of both Directed and Undirected graphs?
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I think unkulunkulu's answer is pretty complete, but given JMSA seems unsatisfied, I'll make another attempt.
Let's start not with the path matrix, but with the adjacency matrix. The adjacency matrix is the standard representation of a graph. If adj is the adjacency matrix for some graph G, then adj[i][j] == 1 if vertex i is adjacent to vertex j (i.e. there is an edge from i to j), and 0 otherwise. In other words, adj[i][j] == 1 if and only if we can get from vertex i to vertex j in one "step".
Now, let's define another matrix which I'll call adj2: adj2[i][j] == 1 if and only if we can get from vertex i to vertex j in two steps or less. You might call it a "two-step adjacency" matrix. The important thing is that adj2 can be defined in terms of adj:
``````def adj2(i,j):
return 1
else:
for k in range(0,n): # where n is the number of vertices in G
return 1
return 0
``````
That is, adj2[i][j] is 1 if i is adjacent to j (i.e. adj[i][j] == 1) or if there exists some other vertex k such that you can step from i to k and then from k to j (i.e. adj[i][j] == 1 and adj[k][j] == 1).
As you can imagine, you can use the same logic to define a "three-step" adjacency matrix adj3, adj4, adj5, and so on. If you go on long enough (e.g. to adjn, where n is the number of vertices in the graph), you get a matrix that tells you whether there's a path of any length from vertex i to vertex j. This, of course, would also be the graph's path matrix: adjn[i][j] == path[i][j] for all i, j. (Note: Don't confuse path matrix with distance matrix.)
A mathematician would say that path[i][j] is the transitive closure of adj[i][j] on the graph G.
Transitive closures exist independently from graph theory; adj is not the only thing with a transitive closure. Roughly speaking, all functions (in the programming sense) that take two arguments and return a Boolean value have a transitive closure.
The equality (==) and inequality (<, >, <=, >=) operators are familiar examples of such functions. These differ from adj, however, in that they are themselves transitive. "f(i,j) is transitive" means that if f(i,j) == true, and f(j,k) == true, then f(i, k) == true. You know that this property is true of, say, the "less than" relation: from a < b and b < c, you can infer that a < c. The transitive closure of a transitive function f is just f.
adj is not generally transitive. Consider the graph:
``````v1---v2---v3
``````
In this graph, adj might represent the function busBetween(city1, city2). Here, there's a bus you can take from v1 to v2 (adj[1][2] == 1) and a bus from v2 to v3 (adj[2][3] == 1), but there's no bus from v1 directly to v3 (adj[1][2] == 0). There is a bus-path from v1 to v3, but v1 and v3 are not bus-adjacent. For this graph, adj is not transitive, so path, which is the transitive closure of adj, is different from adj.
If we add an edge between v1 and v3,
``````v1---v2---v3
\ /
\----/
``````
then adj becomes transitive: In every possible case, adj[i][j] == 1 and adj[j][k] == 1 implies adj[i][k] == 1. For this graph, path and adj are the same. That the graph is undirected corresponds to the "symmetry" property. If we added loops to each vertex so that v1, v2, and v3 were each adjacent to themselves, the resulting graph would be transitive, symmetric, and "reflexive", and could be said to represent equality (==) over the set {1,2,3}.
This begins to illustrate how graphs can represent different functions, and how properties of the function are reflected in properties of the graph. In general, if you let adj represent some function f, then path is the transitive closure of f.
For a formal definition of transitive closures, I refer you to Wikipedia. It isn't a hard concept once you understand all the math jargon.
-
Path Matrix in graph theory is a matrix sized `n*n`, where `n` is the number of vertices of the graph. The element on the `i`th row and `j`th column is `1` if there's a path from `i`th vertex to `j`th in the graph, and `0` if there is not.
The Floyd Algorithm is often used to compute the path matrix.
The definition doesn't differentiate between directed and undirected graphs, but it's clear that for undirected graphs the matrix is always symmetrical.
Transitive Closure is a similar concept, but it's from somewhat different field. Imagine you have a set of objects and for some of them you know that one is definitely better than the other, so you can write `a > b` (">" being the shorthand for "better"). Definitely you should assume that if `a > b` and `b > c` then `a > c`. This is called the transitivity rule. Then for any two objects you want to know, whether one of them is better than the other, or it's unknown. This is the closure: first you have a relation that's possibly not transitive, but after assuming transitivity you can complete it up to a transitive one.
To solve this problem you construct a directed graph, where a vertex corresponds to every of the mentioned objects (`a`, `b`, `c`, etc.) where a directed edge `u -> v` exists if and only if `u > v`. Then you can construct the path matrix defined in the first paragraph and it will give you the answer: obviously, the existence of a path between two vertices is equivalent to existence of a chain of relations as `u > a > b > ... > z > v` so, by the transitivity rule, `u > v`.
As a sidenote for transitive closure, as you asked about both directed and undirected graphs, the example given uses a non-symmetric relation (>), and thereafter the graph was directed, but it's not always the case. Any equivalence relation, for example, always satisfies transitivity but also has to satisfy symmetry, so corresponding graph is undirected. You can find a transitive closure of symmetrical relation (or graph).
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@JMSA, I can't provide a very good link in english, 'cause I learned that from programming trainings and literature was in russian. But I gave a link to wikipedia about floyd's algorithm, I guess there's pretty decent content. – unkulunkulu Jul 31 '11 at 6:34
@JMSA, ok, sorry, I cannot provide a link in russian neither, that were some "metodichka"s not available online :D Wikipedia is fine, I guess. It contains further links at bottom of the page. – unkulunkulu Jul 31 '11 at 6:41
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Properties
Label 2340.2.q.a Level $2340$ Weight $2$ Character orbit 2340.q Analytic conductor $18.685$ Analytic rank $1$ Dimension $2$ CM no Inner twists $2$
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Newspace parameters
Level: $$N$$ $$=$$ $$2340 = 2^{2} \cdot 3^{2} \cdot 5 \cdot 13$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 2340.q (of order $$3$$, degree $$2$$, minimal)
Newform invariants
Self dual: no Analytic conductor: $$18.6849940730$$ Analytic rank: $$1$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-3})$$ Defining polynomial: $$x^{2} - x + 1$$ Coefficient ring: $$\Z[a_1, \ldots, a_{13}]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 260) Sato-Tate group: $\mathrm{SU}(2)[C_{3}]$
$q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{6}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q - q^{5} -3 \zeta_{6} q^{7} +O(q^{10})$$ $$q - q^{5} -3 \zeta_{6} q^{7} + ( 3 - 3 \zeta_{6} ) q^{11} + ( -1 + 4 \zeta_{6} ) q^{13} -7 \zeta_{6} q^{17} -\zeta_{6} q^{19} + ( -7 + 7 \zeta_{6} ) q^{23} + q^{25} + ( -5 + 5 \zeta_{6} ) q^{29} -4 q^{31} + 3 \zeta_{6} q^{35} + ( 3 - 3 \zeta_{6} ) q^{37} + ( 7 - 7 \zeta_{6} ) q^{41} + 9 \zeta_{6} q^{43} -8 q^{47} + ( -2 + 2 \zeta_{6} ) q^{49} + 6 q^{53} + ( -3 + 3 \zeta_{6} ) q^{55} + 5 \zeta_{6} q^{59} + 5 \zeta_{6} q^{61} + ( 1 - 4 \zeta_{6} ) q^{65} + ( -13 + 13 \zeta_{6} ) q^{67} -3 \zeta_{6} q^{71} -14 q^{73} -9 q^{77} -8 q^{79} -12 q^{83} + 7 \zeta_{6} q^{85} + ( 7 - 7 \zeta_{6} ) q^{89} + ( 12 - 9 \zeta_{6} ) q^{91} + \zeta_{6} q^{95} + 11 \zeta_{6} q^{97} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q - 2 q^{5} - 3 q^{7} + O(q^{10})$$ $$2 q - 2 q^{5} - 3 q^{7} + 3 q^{11} + 2 q^{13} - 7 q^{17} - q^{19} - 7 q^{23} + 2 q^{25} - 5 q^{29} - 8 q^{31} + 3 q^{35} + 3 q^{37} + 7 q^{41} + 9 q^{43} - 16 q^{47} - 2 q^{49} + 12 q^{53} - 3 q^{55} + 5 q^{59} + 5 q^{61} - 2 q^{65} - 13 q^{67} - 3 q^{71} - 28 q^{73} - 18 q^{77} - 16 q^{79} - 24 q^{83} + 7 q^{85} + 7 q^{89} + 15 q^{91} + q^{95} + 11 q^{97} + O(q^{100})$$
Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/2340\mathbb{Z}\right)^\times$$.
$$n$$ $$937$$ $$1081$$ $$1171$$ $$2081$$ $$\chi(n)$$ $$1$$ $$-\zeta_{6}$$ $$1$$ $$1$$
Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1621.1
0.5 + 0.866025i 0.5 − 0.866025i
0 0 0 −1.00000 0 −1.50000 2.59808i 0 0 0
2161.1 0 0 0 −1.00000 0 −1.50000 + 2.59808i 0 0 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
13.c even 3 1 inner
Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 2340.2.q.a 2
3.b odd 2 1 260.2.i.d 2
12.b even 2 1 1040.2.q.b 2
13.c even 3 1 inner 2340.2.q.a 2
15.d odd 2 1 1300.2.i.a 2
15.e even 4 2 1300.2.bb.e 4
39.h odd 6 1 3380.2.a.a 1
39.i odd 6 1 260.2.i.d 2
39.i odd 6 1 3380.2.a.b 1
39.k even 12 2 3380.2.f.a 2
156.p even 6 1 1040.2.q.b 2
195.x odd 6 1 1300.2.i.a 2
195.bl even 12 2 1300.2.bb.e 4
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
260.2.i.d 2 3.b odd 2 1
260.2.i.d 2 39.i odd 6 1
1040.2.q.b 2 12.b even 2 1
1040.2.q.b 2 156.p even 6 1
1300.2.i.a 2 15.d odd 2 1
1300.2.i.a 2 195.x odd 6 1
1300.2.bb.e 4 15.e even 4 2
1300.2.bb.e 4 195.bl even 12 2
2340.2.q.a 2 1.a even 1 1 trivial
2340.2.q.a 2 13.c even 3 1 inner
3380.2.a.a 1 39.h odd 6 1
3380.2.a.b 1 39.i odd 6 1
3380.2.f.a 2 39.k even 12 2
Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(2340, [\chi])$$:
$$T_{7}^{2} + 3 T_{7} + 9$$ $$T_{11}^{2} - 3 T_{11} + 9$$ $$T_{19}^{2} + T_{19} + 1$$
Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{2}$$
$3$ $$T^{2}$$
$5$ $$( 1 + T )^{2}$$
$7$ $$9 + 3 T + T^{2}$$
$11$ $$9 - 3 T + T^{2}$$
$13$ $$13 - 2 T + T^{2}$$
$17$ $$49 + 7 T + T^{2}$$
$19$ $$1 + T + T^{2}$$
$23$ $$49 + 7 T + T^{2}$$
$29$ $$25 + 5 T + T^{2}$$
$31$ $$( 4 + T )^{2}$$
$37$ $$9 - 3 T + T^{2}$$
$41$ $$49 - 7 T + T^{2}$$
$43$ $$81 - 9 T + T^{2}$$
$47$ $$( 8 + T )^{2}$$
$53$ $$( -6 + T )^{2}$$
$59$ $$25 - 5 T + T^{2}$$
$61$ $$25 - 5 T + T^{2}$$
$67$ $$169 + 13 T + T^{2}$$
$71$ $$9 + 3 T + T^{2}$$
$73$ $$( 14 + T )^{2}$$
$79$ $$( 8 + T )^{2}$$
$83$ $$( 12 + T )^{2}$$
$89$ $$49 - 7 T + T^{2}$$
$97$ $$121 - 11 T + T^{2}$$
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# Questions tagged [markov-chains]
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### Proof that Component-wise MH algorithm is invariant w.r.t. target measure
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### How to play golf in one dimension?
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### Multi-type Galton-Watson-like process where only majority-type is allowed to reproduce
Are you aware of any research papers that have explored a multi-type Galton-Watson process in which only particles of the majority type are permitted to reproduce in each generation? I've been unable ...
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We are given a queue of $n$ people $\{p_1, \ldots, p_n\}$. They each have to pass $n$ exams $\{t_1, \ldots, t_n\}$. For simplicity we can "draw" the setting in the following way: $$[t_n,t_{n-... • 21 1 vote 0 answers 70 views ### Bounding expectation of switching stochastic process I am analyzing the behavior of an 1D stochastic dynamic system, where the state can vary randomly within a small magnitude. However, when the state deviates too much from zero, its expected magnitude ... • 11 2 votes 0 answers 204 views ### Ball games: How to allocate N balls into M boxes so as to maximize the expected number of taken balls Consider the following ball games, which looks like very intuitive and simple but I have tried for a long time. 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Usually,... • 141 4 votes 1 answer 85 views ### The canonical path method for continuous-time Markov chains on a countable state space I wonder if anyone knows a research article that uses the canonical path method (or congestion ratio) to show that a Poincare inequality holds (see the images below) for a continuous-time Markov chain ... 3 votes 1 answer 215 views ### "Ergodic theorem" for Markov kernels Consider a discrete time Markov chain (X_t) on a finite state space \mathcal{S}, with transition matrix P. Assume that the chain admits a stationary distribution \pi, which I will identify ... -1 votes 1 answer 47 views ### Markov chain to solve a particle fusion problem A sequence of elementary particles arrive at Poisson rate r to a system. A pair of elementary particles can be fused into a level-1 particle. The fusion process succeeds with probability p_0. ... • 459 0 votes 0 answers 32 views ### How to find lower bounds of a modified mixing time (defined below) with respect to spectral of a finite Markov chain? I am focused on a time-homogeneous continuous-time Markov chain with a finite state space \mathcal{X}, whose Markov kernel is K and the corresponding semigroup is H_t=e^{-t(I-K)}. The invariant ... 1 vote 1 answer 92 views ### Asymptotic behavior of a Markov process on the set of \{0,1\}-polynomials This question is cross-posted from https://math.stackexchange.com/questions/4711799/asymptotic-behavior-of-a-markov-process-on-the-set-of-0-1-polynomials I am trying to study the asymptotic behavior ... 1 vote 0 answers 93 views ### Concatenation of Markov processes and independence In chapter 14 of Sharpe's General Theory of Markov Processes the concatenation of Markov processes X^1 and X^2 is described. I've posed the relevant part at the bottom of this post. 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This is defined in page ... • 201 3 votes 1 answer 332 views ### Spectral Radius and Spectral Norm for Markov Operators My question concerns differences between the spectral radius \rho and norm \| \cdot \| of Markov operators in infinite-dimensional Banach spaces. This is far from my area of expertise, that is ... • 550 1 vote 0 answers 81 views ### Derive a closed-form expression of this recursive formula$$$$S(r,k) = f(r)S(0,k-1) + g(r)S(r+1,k-1)$$\ ,$$where r=0,1,2,\dots and k=1,2,3,\dots . Also, 0<f(r)<1 is an increasing function and 0<g(r)<1 is a ... 0 votes 0 answers 31 views ### Prove the explicit form of the ratio function in a Markov Chain Let (A_M^{\mathbb{Z}_+}, \Omega, P, \lambda) be a Markov Shift where A is a finite alphabet set, M is the admissibility matrix, P = [P_{i, j}]_{i, j\in A} is a stochastic matrix that is ... 2 votes 1 answer 215 views ### When is a stationary measure of a Markov chain "exponentially localized"? Here exponentially localized can be thought in a non-rigorous manner as a measure that is mostly supported on a sparse number of nodes. Some intuition can gained by thinking about a diffusion process, ... • 2,196 0 votes 1 answer 46 views ### Diameter of the range of composition of random maps on the circle My questions are related to the paper https://hal.science/hal-03933493v1 (accepted with corrections in Ergodic Theory and Dynamical Systems). I fix an irrational number \theta \in [0,1[. I define ... • 3,529 0 votes 0 answers 99 views ### On the exponentiation of a stochastic matrix where the exponent is a function of matrix size In this question, I asked about any arbitrary stochastic matrix A(n) of the particular form$$A(n) = \begin{pmatrix} 1 & 0 & \cdots & 0\\ x_{21} & x_{22} & \cdots & 0\\ \...
I wonder whether a $d$-dimensional random walk $S_n$, generated by the infinite i.i.d. copies of X given by: $X=e_1=(1, 0, 0, ..., 0)$ (with probability $p_1$) $X=e_2=(0, 1, 0, ..., 0)$ (with ...
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My Blog
Bears Vs Ravens Betting Odds
sp.wz1.pl This stuff can be a bit daunting if you haven’t picked up a math textbook in a while, but you’ll get there. While Vegas odds are built to generate a profit for “the house”, they’re also created in such a way that means it’s possible for gamblers to win big. You receive \$240 in winnings and you get your original bet of \$100 back, equaling \$340 in total.
Sports Betting Sections
If you lost, you’d lose your \$100 but if you won, you’d win \$220 for your \$100 bet. If you were to pick one hundred 6.5-point underdogs to win straight-up informative post at a money line of +230, you would only need to win about 31% of your bets to break-even. For this game, the sports book has set the Money Line at -180 for Miami and +150 for Minnesota. That means that if you want to bet Miami to win straight up, you would need to lay \$180 to win \$100. If you want to take Minnesota to win straight-up, you would lay \$100 to win \$150.
Nba Odds & Betting Lines Faqs
The favorites are listed as negative underdogs, which mean you will not get much for your money if you are betting on them to win. The NFL is easily the most popular sport to bet the moneyline. Betting moneyline in football is preferred by many because the only outcome needed is for a team to win their game outright. There is no added pressure or constraint such as betting the spread, which requires a team to win by a specific margin.
Super Bowl Odds 2021: Money Line, Over
That might not always be the best way to use the bet, though. It varies narrowly from year to year, but approximately 30 percent of all games are won by one run, and the percentage is slightly more than that when the home team wins. And the heaviest favorites are typically at home, which has the issues we have already discussed. So, it is very easy for the value to disappear entirely from a runline favorite.
Whats The Difference Between A Moneyline And A Point Spread?
Separate straight-up bets might be safer than a 4-game parlay, but winning a parlay gives a higher payout. Parlay betting involves betting on a series of games linked together for a higher payout. For this example, betting on O or over means that the total score should be higher than 32.5 to earn \$105 per \$100 bet. Betting on a bookmaker with a high vig leads to a low payout.
• Either before the start of the event, or even during the event itself , the odds change to 5-1 . All in all, free bets are offered by bookies because they know that in the long run they’ll never lose money from them. In fact, they know they’ll make money from the vast bulk of losing punters. The main question to be asking around any form of betting from a legality perspective is ‘ Is it legal for me to place any type of bet? It’s entirely possible to have a great run at no lose betting.
In the above example, Duke is the favorite (-) while North Carolina is the underdog. If a bettor wanted to back Duke, who is expected to win, they would only win \$50 on a \$100 wager since they are minus-200. Some books will provide an extra option to view their betting odds in Fraction format and Duke would be a 1/2 favorite in this matchup. If you prefer Decimal format, very common overseas, Duke would be 1.5 on the betting board.
Let’s say you make a two-team parlay where both teams are listed at -110 odds. Sportsbooks will typically list the odds of this parlay at +265. The implied probability of +265 odds is 27.39% (100/(265 +100). Therefore, the fair price of a two-team parlay (both teams at -110 odds) is +300 (100/(300 + 100).
On the other hand, if you bet on the New York Yankees for the same amount and they win you’d receive a much smaller profit of \$450 in addition to your original \$100 wager. When it comes to using a parlay calculator, you’ll find that there are lots of different ones available to use online. Most of them are set up in different ways, but they all basically just need you to input the odds and payouts of all the wagers that make up the bet. If you’re thinking of placing a parlay bet, it’s always a good idea to use a calculator to see exactly what the payout is.
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### Where Can We Visit?
Charlie and Abi put a counter on 42. They wondered if they could visit all the other numbers on their 1-100 board, moving the counter using just these two operations: x2 and -5. What do you think?
### The Mathemagician's Seven Spells
"Tell me the next two numbers in each of these seven minor spells", chanted the Mathemagician, "And the great spell will crumble away!" Can you help Anna and David break the spell?
### Folding, Cutting and Punching
Exploring and predicting folding, cutting and punching holes and making spirals.
# White Box
##### Stage: 2, 3, 4 and 5 Challenge Level:
The White Box contains a number of filled triangles.
Your challenge is to find the locations of those filled triangles in the grid.
You can fire rays into The White Box and observe where the rays exit using the Experiment Window.
Some rays will pass straight through the Box but some will be deflected by the filled triangles.
You can use the Hypothesis Window to test your ideas. Clicking on a triangle once marks the triangle as empty, clicking again fills it.
FULL SCREEN VERSION
This text is usually replaced by the Flash movie.
NOTES AND BACKGROUND
Anyone attempting to understand anything about the nature of atoms and molecules has to try to come to terms with their almost inconceivably tiny nature. A single hair is about the same width as a million atoms; a glass of water contains about 10 000 000 000 000 000 000 000 000 molecules. The minute size means that they're beyond the range of even the most powerful microscope, so to gain understanding about the ways atoms are arranged within crystals and molecules, scientists need to use indirect methods.
One method involves firing rays of sufficient energy to break bits off the molecules. These particles will of course be smaller still, but they may be easier to recognise. The experimenter can then use the pieces like a jigsaw to explore how they fitted together in the original molecule.
A less destructive process involves not breaking up the structure, but deducing its arrangement by observing the areas where rays fired into it emerge. Some rays pass straight through, while others are deflected by atoms within the structure. The White Box interactivity models this process and makes an intriguing challenge as well.
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# The factory
The factory z can complete an order in 12 days, factory p in 18 days. For how many days does an order completed when the first two days work only the factory z and then both factories?
Result
t = 8 d
#### Solution:
t/12 + (t-2)/18=1
5t = 40
t = 8
Calculated by our simple equation calculator.
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
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#### To solve this example are needed these knowledge from mathematics:
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8. Three pumps
We are filling the pool. The first pump would be filled in 12 hours, the second pump in 15 hours. If all three pumps were running at the same time, it would fill the pool for 4 hours. How long would the pool fill only with the third pump?
9. Diamond diagonals
Find the diamond diagonal's lengths if the area is 156 cm2 and side is 13 cm long.
10. Kuba,
11. Two trains
Through the bridge, long l = 240m, the train passes through the constant speed at time t1 = 21s. A train running along the traffic lights at the edge of the bridge passes the same speed at t2 = 9s. a) What speed v did the train go? b) How long did it tak
12. Marriage sttus
In our city, there are 3/5 of the women married to 2/3 of the men. Find what part of the population is free.
13. Substitution method
Solve goniometric equation: sin4 θ - 1/cos2 θ=cos2 θ - 2
14. Diagonals of a rhombus 2
One diagonal of a rhombus is greater than other by 4 cm . If the area of the rhombus is 96 cm2, find the side of the rhombus.
15. Eqn
Solve equation with fractions: 2x/3-50=40+x/4
16. Two cities
The distance between cities A and B is 132 km. At 9.00 am, the cyclist started the bike at an average speed of 24 km/h, and at 10.00 h started from the B cyclist at an average speed of 30 km/h. How long and far from A will they both meet?
17. Solve 3
Solve quadratic equation: (6n+1) (4n-1) = 3n2
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# If nature is inherently imprecise, how is it so easy for us to conceptualize mathematical certainties?
In modeling any real physical system, we are required to employ inductive reasoning. We can never be completely certain about the state or properties of any system or of any future observation we will make of it. All we can do is attempt to ascertain its properties from past observations, which are themselves imprecise approximations to the true complexities of the system.
It is a law of physics that at the quantum level, no information is completely certain. Everything that is observable exists within a certain probability interval, and there are fundamental limits to how much one can know about any physical state.
Mathematics generally employs deductive reasoning. We postulate that there are certain axioms which are absolutely true, and draw further inferences from those axioms. This is a very different type of reasoning from what I discussed above, as no such certainties exist in nature.
Given these differences, how is it then that our minds are so easily able to conceptualize mathematical certainties, given how contrary to nature they are? If our minds are themselves a part of nature, and have evolved to model and observe natural systems, what is it that gives us the capability to even conceive of things which have so much more precision than we would ever actually see?
Indeed, it is generally far simpler and easier to model a mathematical certainty than anything as complex and imprecise as a real system. In modeling physical systems, we generally employ “simplified” models that rely on assumptions about the properties of the system being much more precise and well-defined mathematically than they really are. Why should it be easier for us to, as physical being within this universe, model a physical system in such an aphysical way?
It would be quite simple for me to give a mathematical description of a perfect circle, and we could all, as rational beings, quickly agree on its properties. Yet, if I were to try and construct a circle, anything I constructed would not only be only an approximation to the mathematical ideal, but also much larger and more cumbersome than the mathematical description was. What is it about mathematics that makes it so much easier to communicate and reason about than nature itself?
• My opinion: mathematics is purely a system of symbolic manipulation that exists only in our minds. The fact that it applies to reality is sort of nice, but mathematics doesn't need reality to exist: it's a purely rule-based system.
– user935
Commented Sep 2, 2018 at 20:52
• I do not quite follow the puzzlement. Why should there be any correlation between processing simplicity and occurrence in nature? Nature is complex, schematic rules are not, that is what makes them easy to process. Evolution would select for the simplest possible schemes that come close enough to survive. Commented Sep 2, 2018 at 21:03
• I'm not sure if you can call a mathematical concept precise. I think precision is more of a measure how well a concept matches reality or how well a measurement translates reality to some conceptual world, but saying mathematical things are of higher precision than things we actually see seems difficult to me. Commented Sep 2, 2018 at 21:27
• It was Hegel’s point that once we recognize a limit we are already past the limit. Such is the nature of Spirit. Commented Oct 19, 2022 at 2:22
• Indeed just think about that your brain is just a tiny tiny part of nature, so why it seems your brain can contemplate and further understand the whole entire world containing it? How strange and magic?... Commented Oct 20, 2022 at 5:49
Just as in religion there is a leap of faith then in science there is a leap of understanding.
The essential notion that physical theory relies upon is that of a universal order; given that understanding, induction are merely attempts to work out the nature, detail and relationships of this universal order.
Pierre Duhem, in his Aim & Structure of Physical Theory pt. 2 ch. 3 "Mathematical Deduction & Physical Theory" is worth reading. §3 gives "an example of a mathematical deduction that can never be utilized" in a physical theory, which is quoted in ch. 5 of the free Chaos film.* In other words: There are mathematical deductions that do not correspond to anything in the physical world.
*The entire film is worth watching, as its other chapters relate to your question, too.
Your question includes a misconception about how we learn about the physical world. Induction is supposedly a process that allows us to use observations to arrive at theories and show that they are true, or probably true or good or something like that. But theories are accounts of how and why something happens and no finite set of observations is equivalent to such an account, so there is no way to get a theory from observations.
We actually create knowledge by noticing problems with our current ideas, guessing solutions to those problems and then criticising the guesses until only one is left and it has no known criticisms. We also learn about maths in the same way - by guessing and criticism.
To learn about mathematical objects we construct physical systems that model those objects according to our best guess about the physical systems and the mathematical objects. In some respects those models and the abstractions they model won't match perfectly. The mathematical objects we are good at modelling are those for which it is possible to construct arbitrarily accurate simulations. For example, you can construct an arbitrarily accurate simulation of a circle by using string of a particular width and making the circle larger to satisfy whatever constraint you want to put on the width of the border relative to the size of the circle. But there is a large subset of mathematics we can't model because no physical system can model them, these include uncomputable functions.
If you want to understand the issues involved better see "Proofs and Refutations" by Lakatos and "The Fabric of Reality" by David Deutsch, chapters 3-7,10. You can discuss them here http://fallibleideas.com/discussion.
The first “light bearer”: man, Prometheus (who has self-consciousness and the ability to reflect) could theoretically bring the whole universe to consciousness and reason, given enough time. (Hegel).
The tragedy is that man the light-bearer has also one foot in the grave (as Freud saw it and for Freud death was final) Freud recognized this.
So history carries reason forward.
• I am actually not so optimistic about “reason” and man as a creature. But once we reflect upon Spinoza and Hegel it is hard not to get a little excited Forgive me. Commented Oct 19, 2022 at 2:55
• In 1889, Plekhanov reported, while he and Pavel Aksel’rod were visiting Engels, the latter stated that “old Spinoza was really right in considering thought and extension as two attributes of the same substance” (Plekhanov 1956–58, II, 360) link.springer.com/article/10.1007/s11212-021-09410-9 Commented Oct 19, 2022 at 3:36
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View the step-by-step solution to:
# A 10-year bond has face value (redemption value) \$150,000 and quarterly coupons of 4%.
A 10-year bond has face value (redemption value) \$150,000 and quarterly coupons of 4%. Consider the time right after the 12th coupon has been paid, when the yield is 6%.
(b)Compute the price of the bond if the yield were to increase by 1 basis point (a basis point is 1/100 of 1%). What is the absolute value of the difference between that price, and your answer to part a)?
(d)Based only on your answer to b), approximately how many basis points (bp) would the yield have to move in order for the bond to increase in value by \$477.10?
(Answer as a positive integer.)
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Categories
# cuantos numeros tiene powerball
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## CГіmo Reclamar Ganancias de Powerball
Existen diferentes mГ©todos para reclamar tus ganancias de Powerball dependiendo de la suma de tu premio. Las reglas para recoger tus ganancias se aplican a la cantidad que has ganado y normalmente a varios juegos de loterГa a la vez, incluyendo loterГas estatales especГficas. Sin embargo, como Powerball cuenta con categorГas de premios, la forma de reclamar tus ganancias dependerГЎ por completo del nГєmero de bolas que hayas acertado.
Si has jugado en lГnea, el proveedor de loterГas con el que hayas jugado serГЎ el que te comunique tu premio mediante correo electrГіnico, y no Lottoamerica.com. Es importante que luego verifiques tu premio entrando en tu cuenta y comprobando los nГєmeros con los del sorteo, ya que las alertas por correo electrГіnico son un tipo de estafa comГєn dentro del mundo de las loterГas.
Si has comprado tu boleto en una tienda, deberГЎs firmar el reverso de tu boleto para confirmar que es tuyo en caso de extravГo o robo. Guarda el boleto en un lugar seguro hasta que llegue el momento de reclamar tu premio. Todos los premios de Powerball se deben reclamar en el mismo estado en el que se comprГі el boleto, por lo que es aconsejable no llevar el boleto contigo si te vas de vacaciones.
### Reclamar tus ganancias
Los premios de menos de 600 \$ normalmente se pueden reclamar desde cualquier vendedor de loterГas autorizado. Antes de acudir a un vendedor de loterГa especГfico a reclamar tu premio, quizГЎ te interese comprobar si dicho vendedor podrГЎ abonГЎrtelo, ya que algunas tiendas, dependiendo de su polГtica o de la de su personal, tienen lГmites de pago de 100 \$ o menos.
En este caso, puedes enviar por correo a tu oficina de loterГas estatal tu boleto premiado firmado junto con un formulario de reclamaciГіn de premios rellenado y un documento de identificaciГіn personal. Sin embargo, deberГЎs comprobar si tu estado ofrece este servicio antes de enviarle el boleto por correo.
Este mГ©todo para reclamar tus ganancias se aplica a todos los boletos premiados que hayan acertado 4 nГєmeros principales o menos. Si has decidido jugar con la opciГіn Power Play y esta multiplica el premio por 10 (para botes de menos de 150 000 000 \$), entonces, tendrГЎs que utilizar el mГ©todo a continuaciГіn para reclamar tus ganancias, ya que estas superan el lГmite de 600 \$.
Los premios de mГЎs de 600 \$ se pueden reclamar en una oficina de distrito local, en un centro de reclamaciones de premios o en la sede central de loterГas. TambiГ©n puedes reclamar tus ganancias por correo postal, enviando a tu oficina de loterГas estatal tu boleto premiado firmado junto con un formulario de reclamaciГіn de premios rellenado. Algunos estados pueden tener lГmites para la cantidad de dinero que se puede reclamar por esta vГa, asГ que es conveniente que te pongas en contacto con tu oficina de loterГas local para obtener mГЎs informaciГіn.
Si has ganado un bote, deberГЎs ponerte en contacto con la sede central de loterГas de tu estado o jurisdicciГіn para organizar la recogida del premio en persona. AllГ se encargarГЎn de comprobar el boleto y de decidir el mГ©todo de pago.
Si has jugado en lГnea, no necesitas ir a ningГєn sitio para recoger tu premio. Todos los premios se pagarГЎn automГЎticamente en tu cuenta en lГnea, desde donde los podrГЎs transferir a tu cuenta bancaria o volver a usar en otros sorteos. Si has ganado un premio significativo, el proveedor de loterГa en lГnea puede podrГa ponerse en contacto contigo directamente para realizar una transferencia bancaria directa de tus ganancias.
Los boletos ganadores no duran para siempre y cada estado tiene un perГodo de reclamaciГіn especГfico para que puedas cobrar tus premios. Comprueba tu estado en la tabla a continuaciГіn para saber de cuГЎnto tiempo dispones para reclamar tu premio:
PerГodo para Reclamar Premios de Powerball por Estado
JurisdicciГіn Periodo de ReclamaciГіn
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## Intermediate Algebra (6th Edition)
We are given the expression $4(6n^{2}-3)-3(8n^{2}+4)$. We can use the distributive property and then combine like terms to simplify. $4\times(6n^{2}-3)-3\times(8n^{2}+4)=24n^{2}-12-24n^{2}-12$ $=(24n^{2}-24n^{2})-(12+12)=(0)-(24)=-24$
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# Cosider the system of differential equationsx_1'=2x_1+x_2x_2'=4x_1-x_2Convert this system to a second order differential equations and solve this second order differential equations
Cosider the system of differential equations
$$x_1'=2x_1+x_2$$
$$x_2'=4x_1-x_2$$
Convert this system to a second order differential equations and solve this second order differential equations
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Consider the given system f equations as $$\begin{cases}x_1'=2x_1+x_2\\x_2'=4x_1-x_2\end{cases}$$
Solve the first equation $$x_1'=2x_1+x_2$$ for $$x_2$$ and obtained as $$x_2=x_1'-2x_1$$
Substitude $$x_2=x_1'-2x_1$$ in the equation $$x_2'=4x_1-x_2$$
$$x_2'=4x_1-x_2$$
$$(x_1'-2x_1)'=4x_1-(x_1'-2x_1)$$
$$x_1''-2x_1'=4x_1-x_1'+2x_1'$$
$$x_1''-x_1'-6x_1=0$$
Thus the second order differential equation is $$x_1''-x_1'-6x_1'=0$$
Consider the equation $$x_1''-x_1'-6x_1=0$$
The characteristics equation of the second order differential equation is
$$m^2-m-6=0$$
Solve the equation $$m^2-m-6=0$$ as follows.
$$m^2-m-6=0$$
$$m^2-3m+2m-6=0$$
$$m(m-3)+2(m-3)=0$$
$$(m-3)(m+2)=0$$
$$m-3=0$$ or $$m+2=0$$
$$m=3$$ or $$m=-2$$
Thus, the general solution of the second order differential equations is obtained as below.
$$y=c_1e^{m_1x}+c_2e^{m_2x}$$
$$=c_1e^{-2x}+c_2e^{3x}$$
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# Local Linear Approximation vs Linearization
1. Nov 10, 2013
### Lebombo
Are Local Linear Approximation, Linear Approximation, and Linearization all the same thing?
Question is, I learned about something called Local Linear Approximation in Calc 1. Now in Calc 2, the topic of Linearization from Calc 1 was mentioned. But I never did anything that was referred to as Linearization. The closest sounding topic was Local Linear Approximation. So are they the same exact topic just referred to with different names?
2. Nov 10, 2013
### HallsofIvy
Perhaps a slight difference. When you have a "local linear approximation", what is "local" is already decided. That is, you have a specific point at which you approximate the curve by a line. The "linearization" of a function requires a choice of the point at which you will linearize. "Linearization" of the same function at two different "x" values will give two different linear functions.
Once you have chosen that point at which to linearize the function, then linearization about that point is the local linear approximation.
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Posted by Andrew Zaske on Tuesday, July 23, 2019
# Stop Motion or High Definition?
It was a rough week for the whole team working on a new customer’s new machine design. Finally, it’s Friday night and you sit down on the couch to relax with your friends. You made some popcorn and just popped in a DVD recommended by your parents. They told you it’s a classic and might be a bit slow but it has a good story.
The movie starts and after 30 seconds, your friends are giving you confused looks. The movie is very difficult to watch because it is filmed in stop motion. The images on the screen are frozen and jump from one to the next. Sometimes it is half a second between shifts and sometimes it can be several seconds. You can see the evening disintegrating and quickly move to switch out the movie to ‘The Fast and the Furious’ for some non-stop action.
The following Monday, your engineering manager is talking about another new project where they are planning on using pneumatic cylinders for a factory automation project. You know the customer has requested an updated design that allows more flexibility and better data trending of performance. Last fall, while attending a trade show in Chicago, you saw some screw-driven electric linear actuators. The exhibitor claimed that electric linear motion control could provide infinite positioning flexibility and real time knowledge of where the electric actuator was at all times.
While chewing on your sandwich at lunch, a few synapses fire off in your brain and you link Friday night at the movies with the new customer project. Pneumatic cylinders with switches at the end of stroke are equivalent to the classic stop motion movie. You get a discrete data point at two points in time, but your knowledge is frozen in time until the next switch is triggered. Electric actuators with a servo motor can give you real time data on position and speed at any time. The electric actuators are similar to ‘The Fast and the Furious’ where the action is continuous.
You knock on your boss’s office door and he invites you to sit down. You talk through the analogy of the stop motion filming technique and the comparison between pneumatic/fluid-power versus electric linear motion control. He asks you a few tough questions and then directs you to go do some more research. He promises you that if the details check out, you can design propose the new system with the electric actuator to the end customer.
Next step? You get the facts. Since Tolomatic manufactures both pneumatic cylinders and electric rod actuators, you check out their whitepaper comparing performance of both technologies from motion control and data collection, to force, velocity and total cost of ownership (TCO).
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XP is just a number PerlMonks
Comment on
Need Help??
This method can be used when a function has a loop, since loops can be implemented using recursion. This is great because it facilitates making iterators from functions where the call to the visitor is not at the end and from functions with multiple calls to the visitor.
For example, let's create a fibonacci generator. A simple implementation is:
```sub fibonacci {
my (\$visitor) = @_;
my (\$i, \$j) = (0, 1);
\$visitor->() for \$i;
for (;;) {
\$visitor->() for \$j;
(\$i, \$j) = (\$j, \$i+\$j);
}
}
Replace the loop with recursion.
```sub fibonacci {
my (\$visitor) = @_;
my (\$i, \$j) = (0, 1);
\$visitor->() for \$i;
_fibonacci(\$i, \$j);
}
sub _fibonacci {
my (\$visitor, \$i, \$j) = @_;
\$visitor->() for \$j;
_fibonacci(\$j, \$i+\$j);
}
Convert for make_iter.
```sub fibonacci {
my (\$i, \$j) = (0, 1);
return ( \\$i, sub { _fibonacci(\$i, \$j) } );
}
sub _fibonacci {
my (\$i, \$j) = @_;
return ( \\$j, sub { _fibonacci(\$j, \$i+\$j) } );
}
Try it out
```{
my \$iter = make_iter(\&fibonacci);
for (0..15) {
my (\$n) = \$iter->()
or last;
print("\$n ");
}
print("\n");
}
```0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610
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## Thursday, 20 February 2014
### Algebra... Different "Faces" of Algebra (II)
The perimeter of a rectangle is (6x + 5y) cm.
Suggest 2 possible dimensions of the rectangle.
e.g.
Length of rectangle = 2x + 2y cm
Breadth of rectangle = x + 0.5y cm
Check:
Perimeter = 2 (2x + 2y) + 2 (x + 0.5y) cm
= 4x + 4y + 2x + y cm
= 6x + 5y cm
#### 1 comment:
1. 1st possible dimensions
Length of rectangle=3x + 3x
Check:
Perimeter= 2(3x) + 2(2.5y)
= (6x + 5y)cm
2nd possible dimensions
Length of rectangle= 2x + 1.5y
Breadth of rectangle= 2x + 2y
Check
Perimeter= 2(2x + 1.5y) + (2x + 2y)
= 4x + 3y + 2x + 2y
= (6x + 5y)cm
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# Jan23 - CSE262: Programming Languages Instructor: Dr. Liang...
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CSE262: Programming Languages Instructor: Dr. Liang Cheng Department of Computer Science and Engineering P.C. Rossin College of Engineering & Applied Science Lehigh University January 23, 2007
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Instructor: Dr. Liang Cheng CSE262: Programming Languages 01/23/07 Outline ± Recap ± Compiler/Compilation in a nutshell ± BNF without ambiguity ± Extended BNF ± Summary and homework
Instructor: Dr. Liang Cheng CSE262: Programming Languages 01/23/07 Language Users <program> <stmts> <stmt> const a <var> = <expr> <var> b <term> + <term> BNF Functionality ± Describing Lists ± Grammar & Derivation ± Parse Trees ± Avoiding Ambiguity <program> => <stmts> => <stmt> => <var> = <expr> => a = <expr> => a = <term> + <term> => a = <var> + <term> => a = b + <term> => a = b + const <program> <stmts> <stmts> <stmt> | <stmt> ; <stmts> <stmt> <var> = <expr> <var> a | b | c | d <expr> <term> + <term> | <term> - <term> <term> <var> | const
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Instructor: Dr. Liang Cheng CSE262: Programming Languages 01/23/07 Language Designers ± Design a BNF grammar for a language that could express a one-digit number, an addition of two one-digit numbers, or a subtraction of two one-digit numbers ± <expr> <term> + <term> | <term> - <term> | <term> ± <term> 0 | 1 | 2 | … | 9
Instructor: Dr. Liang Cheng CSE262: Programming Languages 01/23/07 Language Implementers ± A recursive-descent parser ± Language implementation directly following the BNF grammar ± <expr> <term> + <term> | <term> - <term> | <term> ± <term> 0 | 1 | 2 | … | 9 ± Extra credit homework question ± Pseudo code void expr() { void term() { term(); match( int_literal ); if( token== plus_op } or token== minus_op ) { match(token); void match(expectedToken) { term(); if(token==expectedToken) } getNextToken(); else error(); else error(); }}
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Instructor: Dr. Liang Cheng
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## This note was uploaded on 08/06/2008 for the course CSE 262 taught by Professor Cheng during the Spring '07 term at Lehigh University .
### Page1 / 24
Jan23 - CSE262: Programming Languages Instructor: Dr. Liang...
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http://www.instructables.com/member/Jerry66/
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# Jerry66
• Hi-He hardened the entire chisel when he quenched at its non-magnetic state (cherry red). It is now too brittle to strike with a hammer. He then softened the polished steel by heating up the blunt end and watches the color changes (tempering). When the tip, or just before the tip, reaches a straw color, he quenches it to stop the heat from conducting further. This allows the chisel to be hit with a hammer (on the softened blunt end) without it breaking. From the cutting tip, the steel becomes progressively softer as it approaches the blunt end. The tip is as hard as it ever was, and there is nothing left to do to it with regards to heat treatment I hope that makes sense..
• Jerry66 followed ZackH191 week ago
• Jerry66 followed Magnetic Games, JColvin91 and Plasmana1 month ago
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• Jerry66 followed Plasanator1 month ago
• To convert F to C, subtract 32 and then multiply by 5/9. To convert C to F, add 32 and multiply by 9/5. If that's not cumbersome, I don't know what is; In the US, that is taught in the 5th grade (I think). By summer break, we all forget about it.
Ewe ar eggzactlie korwrecked!
• Jerry66 followed Makersauce2 months ago
• I agree, and I'm an old fart from America! Who can't multiply by 10, 100, or 1,000? Water freezing at O C. and water boiling at 100 C. Easy! As far as how something is spelled (spelt), or pronounced, I don't think it matters. Spilt milk! Or is it spilled milk? Who cares? My dog's name is Phydeaux.
• By using the word intuitively, I meant that Americans can pretty much tell how long a foot is by holding their hands apart. They can come close to knowing when there is a pound of something in their hand. I did not mean they know how many feet in a mile or any other kind of conversion. The metric system makes total sense, but Americans will not intuitively know when they are holding 1 kilogram in their hands etc. They, like the Brits, will just have to get used to it.
• Jerry66 followed tanner_tech2 months ago
• My young American grandson has trouble pronouncing aluminum, but he has no problems pronouncing aluminium. Being a 70 year old American, I can remember going to the "theatre" as a child . It was spelled that way in the U.S. then. To me, theatre seems the correct way to spell it, because that is what I was taught. In any case, I find the whole discussion to be unnecessary. We all believe that we are correct, and this is directly related to where we were born. Neither is correct, nor incorrect. In the end, all of us know what the other is trying to say. It just bugs us somehow??? Now, the metric system is another thing entirely! Americans intuitively know what a pound feels like, or how long a foot or mile is. However, it is much easier to multiply by a 1 with zeros...see more »My young American grandson has trouble pronouncing aluminum, but he has no problems pronouncing aluminium. Being a 70 year old American, I can remember going to the "theatre" as a child . It was spelled that way in the U.S. then. To me, theatre seems the correct way to spell it, because that is what I was taught. In any case, I find the whole discussion to be unnecessary. We all believe that we are correct, and this is directly related to where we were born. Neither is correct, nor incorrect. In the end, all of us know what the other is trying to say. It just bugs us somehow??? Now, the metric system is another thing entirely! Americans intuitively know what a pound feels like, or how long a foot or mile is. However, it is much easier to multiply by a 1 with zeros, than 12, or 16, etc. We just have to get used to it, just as the UK had to do.
• Jerry66 followed ganuganu and The King of Random2 months ago
• Jerry66 commented on harisanonorg's instructable How to Build Circuits2 months ago
For me, conventional current flow confuses things. I learned electron flow in the USAF in the early sixties. The concept of electron flow makes more sense to me because electrons actually carry a negative charge, which is a tangible. An electrically positive charge is really just a point where there are fewer electrons (a void) than the negative point. I agree that unless you are talking about actual earth ground, it tells us little. I like to think of charges as less negative, or more positive. For ex., -5VDC is positive relative to -10VDC. We had several different grounds, such as earth, chassis, vase, circuit, utility, etc. They were not always at the same charge electrically (Scary on vacuum tube circuits). I am not sure what the military teaches now.
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• Jerry66 commented on PaulGetson's instructable Quick and Easy Jacobs Ladder4 months ago
You really cannot blame just one aspect of electric shock resulting in death. I (Current, aka amperage) times E or V (Voltage) = Watts (power). Lightning is very high voltage, but low current, so the overall power in watts is low, but it can still kill you. 120 volts at 10 amps = 1,200 watts of power and can also kill you! Think of voltage as pressure and current (Amps) as the tangible stuff like water or BBs. A BB with lots of pressure driving it will do much damage. A 2 pound cannon ball (Amps) with a relatively small amount of pressure (Volts) will also do some damage. The overall power (Watts) can be the same for both the BB and the cannon ball. Think of the damage a micro meteor traveling at 25,000 plus MPH can do to a satellite, as will a 2 pound mass going 25 MPH.
• Jerry66 commented on hubertofliege's instructable Walk-in open air chicken coop4 months ago
Definitely agree with other posters on burying wire at least 18". Chicken wire comes in 1" hole size also. Put some of that, or 1/2" hardware cloth, over the existing 2" wire from the ground up to 3 feet. Weasels can easily squeeze through 2", snakes too. The bottom sill should be pressure treated or redwood. Not sure why you turned the studs on edge?
• Jerry66 followed ArduinoDeXXX4 months ago
• One method uses a paper clip bent to form the shape of the number 7. Poke a tiny hole in the end of the egg, and then insert the short leg of the wire into the egg and then spin the long leg using a drill, or? The hole won't leak egg during boiling.
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• Jerry66 followed Phil B6 months ago
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## Transcription
3 WIND B C A Prepared with Memory-Map Figure 3. MM-Nav plot of sample route. Two tacks to close hauled weather (wind from due North) then some 39 off the wind, which would probably eliminate most of the leeway. To correct a DR leg such as A to B for leeway, you plot it the same length but just rotated downwind by the leeway angle. Here the uncorrected leg is at 045T, the corrected leg at = 052T. You have to know what the leeway is for each wind speed and other conditions for your vessel. Usually, it would be the same for both tacks in a given condition, but you might learn that even this is not true if the waves are not running in line with the wind. Note that we can use the small angle rule to estimate how much C will be off of B for 7. The distance A to B is 5.4 nmi. The rule tells us 6 is one tenth offset, which would be 0.54 nmi, but we are a bit higher (7/6)x0.54 = 0.63 nmi. Checking the chart you see the distance from B to C is about 0.68, which is consistent with the approximations.
4 Figure 1. Sailboat wakes... not really showing much! The two left are essentially straight aft. It is light air and they are both cracked off the wind, so we would expect them to be straight. If we were to detect any leeway visually, they would have to be closer to the wind and the wind stronger. In that case, the two lines shown would be leaning somewhat off to the left, upwind. We might still see a few degrees in the ECS track however, even when not at all visible by eye. The top right case is a candidate for showing the effect, but not for certain, it would not be this big. These are race boats and their heading is changing often to match puffs of wind, so we may have caught them off course a bit but this is sort of what it would look like. It would not be this big however (shown is about 17 ) these are high tech boats in what is still relatively light air. They will not slip more than 3 or 4 in these conditions. If you do get the rare chance to actually see this visually, keep in mind that a finger width at arm s length is about 2, which is a crude way to estimate the effect Small angle Rule. A right triangle of 6 has ratio of sides = 1:10, and this can be scaled up to 18 or down to 1, which if often useful in many aspects of marine navigation.
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## 19.13 Additional remarks on Grothendieck abelian categories
In this section we put some results on Grothendieck abelian categories which are folklore.
Lemma 19.13.1. Let $\mathcal{A}$ be a Grothendieck abelian category. Let $F : \mathcal{A}^{opp} \to \textit{Sets}$ be a functor. Then $F$ is representable if and only if $F$ commutes with colimits, i.e.,
$F(\mathop{\mathrm{colim}}\nolimits _ i N_ i) = \mathop{\mathrm{lim}}\nolimits F(N_ i)$
for any diagram $\mathcal{I} \to \mathcal{A}$, $i \in \mathcal{I}$.
Proof. If $F$ is representable, then it commutes with colimits by definition of colimits.
Assume that $F$ commutes with colimits. Then $F(M \oplus N) = F(M) \times F(N)$ and we can use this to define a group structure on $F(M)$. Hence we get $F : \mathcal{A} \to \textit{Ab}$ which is additive and right exact, i.e., transforms a short exact sequence $0 \to K \to L \to M \to 0$ into an exact sequence $F(K) \leftarrow F(L) \leftarrow F(M) \leftarrow 0$ (compare with Homology, Section 12.7).
Let $U$ be a generator for $\mathcal{A}$. Set $A = \bigoplus _{s \in F(U)} U$. Let $s_{univ} = (s)_{s \in F(U)} \in F(A) = \prod _{s \in F(U)} F(U)$. Let $A' \subset A$ be the largest subobject such that $s_{univ}$ restricts to zero on $A'$. This exists because $\mathcal{A}$ is a Grothendieck category and because $F$ commutes with colimits. Because $F$ commutes with colimits there exists a unique element $\overline{s}_{univ} \in F(A/A')$ which maps to $s_{univ}$ in $F(A)$. We claim that $A/A'$ represents $F$, in other words, the Yoneda map
$\overline{s}_{univ} : h_{A/A'} \longrightarrow F$
is an isomorphism. Let $M \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ and $s \in F(M)$. Consider the surjection
$c_ M : A_ M = \bigoplus \nolimits _{\varphi \in \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(U, M)} U \longrightarrow M.$
This gives $F(c_ M)(s) = (s_\varphi ) \in \prod _\varphi F(U)$. Consider the map
$\psi : A_ M = \bigoplus \nolimits _{\varphi \in \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(U, M)} U \longrightarrow \bigoplus \nolimits _{s \in F(U)} U = A$
which maps the summand corresponding to $\varphi$ to the summand corresponding to $s_\varphi$ by the identity map on $U$. Then $s_{univ}$ maps to $(s_\varphi )_\varphi$ by construction. in other words the right square in the diagram
$\xymatrix{ A' \ar[r] & A \ar@{..>}[r]_{s_{univ}} & F \\ K \ar[r] \ar[u]^{?} & A_ M \ar[u]^\psi \ar[r] & M \ar@{..>}[u]_ s }$
commutes. Let $K = \mathop{\mathrm{Ker}}(A_ M \to M)$. Since $s$ restricts to zero on $K$ we see that $\psi (K) \subset A'$ by definition of $A'$. Hence there is an induced morphism $M \to A/A'$. This construction gives an inverse to the map $h_{A/A'}(M) \to F(M)$ (details omitted). $\square$
Proof. Let $M_ i$, $i \in I$ be a family of objects of $\mathcal{A}$ indexed by a set $I$. The functor $F = \prod _{i \in I} h_{M_ i}$ commutes with colimits. Hence Lemma 19.13.1 applies. $\square$
Remark 19.13.3. In the chapter on derived categories we consistently work with “small” abelian categories (as is the convention in the Stacks project). For a “big” abelian category $\mathcal{A}$ it isn't clear that the derived category $D(\mathcal{A})$ exists because it isn't clear that morphisms in the derived category are sets. In general this isn't true, see Examples, Lemma 110.61.1. However, if $\mathcal{A}$ is a Grothendieck abelian category, and given $K^\bullet , L^\bullet$ in $K(\mathcal{A})$, then by Theorem 19.12.6 there exists a quasi-isomorphism $L^\bullet \to I^\bullet$ to a K-injective complex $I^\bullet$ and Derived Categories, Lemma 13.31.2 shows that
$\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(K^\bullet , L^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I^\bullet )$
which is a set. Some examples of Grothendieck abelian categories are the category of modules over a ring, or more generally the category of sheaves of modules on a ringed site.
Lemma 19.13.4. Let $\mathcal{A}$ be a Grothendieck abelian category. Then
1. $D(\mathcal{A})$ has both direct sums and products,
2. direct sums are obtained by taking termwise direct sums of any complexes,
3. products are obtained by taking termwise products of K-injective complexes.
Proof. Let $K^\bullet _ i$, $i \in I$ be a family of objects of $D(\mathcal{A})$ indexed by a set $I$. We claim that the termwise direct sum $\bigoplus _{i \in I} K^\bullet _ i$ is a direct sum in $D(\mathcal{A})$. Namely, let $I^\bullet$ be a K-injective complex. Then we have
\begin{align*} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(\bigoplus \nolimits _{i \in I} K^\bullet _ i, I^\bullet ) & = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(\bigoplus \nolimits _{i \in I} K^\bullet _ i, I^\bullet ) \\ & = \prod \nolimits _{i \in I} \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet _ i, I^\bullet ) \\ & = \prod \nolimits _{i \in I} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(K^\bullet _ i, I^\bullet ) \end{align*}
as desired. This is sufficient since any complex can be represented by a K-injective complex by Theorem 19.12.6. To construct the product, choose a K-injective resolution $K_ i^\bullet \to I_ i^\bullet$ for each $i$. Then we claim that $\prod _{i \in I} I_ i^\bullet$ is a product in $D(\mathcal{A})$. This follows from Derived Categories, Lemma 13.31.5. $\square$
Remark 19.13.5. Let $R$ be a ring. Suppose that $M_ n$, $n \in \mathbf{Z}$ are $R$-modules. Denote $E_ n = M_ n[-n] \in D(R)$. We claim that $E = \bigoplus M_ n[-n]$ is both the direct sum and the product of the objects $E_ n$ in $D(R)$. To see that it is the direct sum, take a look at the proof of Lemma 19.13.4. To see that it is the direct product, take injective resolutions $M_ n \to I_ n^\bullet$. By the proof of Lemma 19.13.4 we have
$\prod E_ n = \prod I_ n^\bullet [-n]$
in $D(R)$. Since products in $\text{Mod}_ R$ are exact, we see that $\prod I_ n^\bullet [-n]$ is quasi-isomorphic to $E$. This works more generally in $D(\mathcal{A})$ where $\mathcal{A}$ is a Grothendieck abelian category with Ab4*.
Lemma 19.13.6. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor of abelian categories. Assume
1. $\mathcal{A}$ is a Grothendieck abelian category,
2. $\mathcal{B}$ has exact countable products, and
3. $F$ commutes with countable products.
Then $RF : D(\mathcal{A}) \to D(\mathcal{B})$ commutes with derived limits.
Proof. Observe that $RF$ exists as $\mathcal{A}$ has enough K-injectives (Theorem 19.12.6 and Derived Categories, Lemma 13.31.6). The statement means that if $K = R\mathop{\mathrm{lim}}\nolimits K_ n$, then $RF(K) = R\mathop{\mathrm{lim}}\nolimits RF(K_ n)$. See Derived Categories, Definition 13.34.1 for notation. Since $RF$ is an exact functor of triangulated categories it suffices to see that $RF$ commutes with countable products of objects of $D(\mathcal{A})$. In the proof of Lemma 19.13.4 we have seen that products in $D(\mathcal{A})$ are computed by taking products of K-injective complexes and moreover that a product of K-injective complexes is K-injective. Moreover, in Derived Categories, Lemma 13.34.2 we have seen that products in $D(\mathcal{B})$ are computed by taking termwise products. Since $RF$ is computed by applying $F$ to a K-injective representative and since we've assumed $F$ commutes with countable products, the lemma follows. $\square$
The following lemma is some kind of generalization of the existence of Cartan-Eilenberg resolutions (Derived Categories, Section 13.21).
Lemma 19.13.7. Let $\mathcal{A}$ be a Grothendieck abelian category. Let $K^\bullet$ be a filtered complex of $\mathcal{A}$, see Homology, Definition 12.24.1. Then there exists a morphism $j : K^\bullet \to J^\bullet$ of filtered complexes of $\mathcal{A}$ such that
1. $J^ n$, $F^ pJ^ n$, $J^ n/F^ pJ^ n$ and $F^ pJ^ n/F^{p'}J^ n$ are injective objects of $\mathcal{A}$,
2. $J^\bullet$, $F^ pJ^\bullet$, $J^\bullet /F^ pJ^\bullet$, and $F^ pJ^\bullet /F^{p'}J^\bullet$ are K-injective complexes,
3. $j$ induces quasi-isomorphisms $K^\bullet \to J^\bullet$, $F^ pK^\bullet \to F^ pJ^\bullet$, $K^\bullet /F^ pK^\bullet \to J^\bullet /F^ pJ^\bullet$, and $F^ pK^\bullet /F^{p'}K^\bullet \to F^ pJ^\bullet /F^{p'}J^\bullet$.
Proof. By Theorem 19.12.6 we obtain quasi-isomorphisms $i : K^\bullet \to I^\bullet$ and $i^ p : F^ pK^\bullet \to I^{p, \bullet }$ as well as commutative diagrams
$\vcenter { \xymatrix{ K^\bullet \ar[d]_ i & F^ pK^\bullet \ar[l] \ar[d]_{i^ p} \\ I^\bullet & I^{p, \bullet } \ar[l]_{\alpha ^ p} } } \quad \text{and}\quad \vcenter { \xymatrix{ F^{p'}K^\bullet \ar[d]_{i^{p'}} & F^ pK^\bullet \ar[l] \ar[d]_{i^ p} \\ I^{p', \bullet } & I^{p, \bullet } \ar[l]_{\alpha ^{p p'}} } } \quad \text{for }p' \leq p$
such that $\alpha ^ p \circ \alpha ^{p' p} = \alpha ^{p'}$ and $\alpha ^{p'p''} \circ \alpha ^{pp'} = \alpha ^{pp''}$. The problem is that the maps $\alpha ^ p : I^{p, \bullet } \to I^\bullet$ need not be injective. For each $p$ we choose an injection $t^ p : I^{p, \bullet } \to J^{p, \bullet }$ into an acyclic K-injective complex $J^{p, \bullet }$ whose terms are injective objects of $\mathcal{A}$ (first map to the cone on the identity and then use the theorem). Choose a map of complexes $s^ p : I^\bullet \to J^{p, \bullet }$ such that the following diagram commutes
$\xymatrix{ K^\bullet \ar[d]_ i & F^ pK^\bullet \ar[l] \ar[d]_{i^ p} \\ I^\bullet \ar[rd]_{s^ p} & I^{p, \bullet } \ar[d]^{t^ p} \\ & J^{p, \bullet } }$
This is possible: the composition $F^ pK^\bullet \to J^{p, \bullet }$ is homotopic to zero because $J^{p, \bullet }$ is acyclic and K-injective (Derived Categories, Lemma 13.31.2). Since the objects $J^{p, n - 1}$ are injective and since $F^ pK^ n \to K^ n \to I^ n$ are injective morphisms, we can lift the maps $F^ pK^ n \to J^{p, n - 1}$ giving the homotopy to a map $h^ n : I^ n \to J^{p, n - 1}$. Then we set $s^ p$ equal to $h \circ \text{d} + \text{d} \circ h$. (Warning: It will not be the case that $t^ p = s^ p \circ \alpha ^ p$, so we have to be careful not to use this below.)
Consider
$J^\bullet = I^\bullet \times \prod \nolimits _ p J^{p, \bullet }$
Because products in $D(\mathcal{A})$ are given by taking products of K-injective complexes (Lemma 19.13.4) and since $J^{p, \bullet }$ is isomorphic to $0$ in $D(\mathcal{A})$ we see that $J^\bullet \to I^\bullet$ is an isomorphism in $D(\mathcal{A})$. Consider the map
$j = i \times (s^ p \circ i)_{p \in \mathbf{Z}} : K^\bullet \longrightarrow I^\bullet \times \prod \nolimits _ p J^{p, \bullet } = J^\bullet$
By our remarks above this is a quasi-isomorphism. It is also injective. For $p \in \mathbf{Z}$ we let $F^ pJ^\bullet \subset J^\bullet$ be
$\mathop{\mathrm{Im}}\left( \alpha ^ p \times (t^{p'} \circ \alpha ^{pp'})_{p' \leq p} : I^{p, \bullet } \to I^\bullet \times \prod \nolimits _{p' \leq p} J^{p', \bullet } \right) \times \prod \nolimits _{p' > p} J^{p', \bullet }$
This complex is isomorphic to the complex $I^{p, \bullet } \times \prod _{p' > p} J^{p, \bullet }$ as $\alpha ^{pp} = \text{id}$ and $t^ p$ is injective. Hence $F^ pJ^\bullet$ is quasi-isomorphic to $I^{p, \bullet }$ (argue as above). We have $j(F^ pK^\bullet ) \subset F^ pJ^\bullet$ because of the commutativity of the diagram above. The corresponding map of complexes $F^ pK^\bullet \to F^ pJ^\bullet$ is a quasi-isomorphism by what we just said. Finally, to see that $F^{p + 1}J^\bullet \subset F^ pJ^\bullet$ use that $\alpha ^{p + 1p} \circ \alpha ^{pp'} = \alpha ^{p + 1p'}$ and the commutativity of the first displayed diagram in the first paragraph of the proof.
We claim that $j : K^\bullet \to J^\bullet$ is a solution to the problem posed by the lemma. Namely, $F^ pJ^ n$ is an injective object of $\mathcal{A}$ because it is isomorphic to $I^{p, n} \times \prod _{p' > p} J^{p', n}$ and products of injectives are injective. Then the injective map $F^ pJ^ n \to J^ n$ splits and hence the quotient $J^ n/F^ pJ^ n$ is injective as well as a direct summand of the injective object $J^ n$. Similarly for $F^ pJ^ n/F^{p'}J^ n$. This in particular means that $0 \to F^ pJ^\bullet \to J^\bullet \to J^\bullet /F^ pJ^\bullet \to 0$ is a termwise split short exact sequence of complexes, hence defines a distinguished triangle in $K(\mathcal{A})$ by fiat. Since $J^\bullet$ and $F^ pJ^\bullet$ are K-injective complexes we see that the same is true for $J^\bullet /F^ pJ^\bullet$ by Derived Categories, Lemma 13.31.3. A similar argument shows that $F^ pJ^\bullet /F^{p'}J^\bullet$ is K-injective. By construction $j : K^\bullet \to J^\bullet$ and the induced maps $F^ pK^\bullet \to F^ pJ^\bullet$ are quasi-isomorphisms. Using the long exact cohomology sequences of the complexes in play we find that the same holds for $K^\bullet /F^ pK^\bullet \to J^\bullet /F^ pJ^\bullet$ and $F^ pK^\bullet /F^{p'}K^\bullet \to F^ pJ^\bullet /F^{p'}J^\bullet$. $\square$
Remark 19.13.8. Let $\mathcal{A}$ be a Grothendieck abelian category. Let $K^\bullet$ be a filtered complex of $\mathcal{A}$, see Homology, Definition 12.24.1. For ease of notation denote $K$, $F^ pK$, $\text{gr}^ pK$ the object of $D(\mathcal{A})$ represented by $K^\bullet$, $F^ pK^\bullet$, $\text{gr}^ pK^\bullet$. Let $M \in D(\mathcal{A})$. Using Lemma 19.13.7 we can construct a spectral sequence $(E_ r, d_ r)_{r \geq 1}$ of bigraded objects of $\mathcal{A}$ with $d_ r$ of bidgree $(r, -r + 1)$ and with
$E_1^{p, q} = \mathop{\mathrm{Ext}}\nolimits ^{p + q}(M, \text{gr}^ pK)$
If for every $n$ we have
$\mathop{\mathrm{Ext}}\nolimits ^ n(M, F^ pK) = 0 \text{ for } p \gg 0 \quad \text{and}\quad \mathop{\mathrm{Ext}}\nolimits ^ n(M, F^ pK) = \mathop{\mathrm{Ext}}\nolimits ^ n(M, K) \text{ for } p \ll 0$
then the spectral sequence is bounded and converges to $\mathop{\mathrm{Ext}}\nolimits ^{p + q}(M, K)$. Namely, choose any complex $M^\bullet$ representing $M$, choose $j : K^\bullet \to J^\bullet$ as in the lemma, and consider the complex
$\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , I^\bullet )$
defined exactly as in More on Algebra, Section 15.71. Setting $F^ p\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , I^\bullet ) = \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , F^ pI^\bullet )$ we obtain a filtered complex. The spectral sequence of Homology, Section 12.24 has differentials and terms as described above; details omitted. The boundedness and convergence follows from Homology, Lemma 12.24.13.
Remark 19.13.9. Let $\mathcal{A}$ be a Grothendieck abelian category. Let $M, K$ be objects of $D(\mathcal{A})$. For any choice of complex $K^\bullet$ representing $K$ we can use the filtration $F^ pK^\bullet = \tau _{\leq -p}K^\bullet$ and the discussion in Remark 19.13.8 to get a spectral sequence with
$E_1^{p, q} = \mathop{\mathrm{Ext}}\nolimits ^{2p + q}(M, H^{-p}(K))$
This spectral sequence is independent of the choice of complex $K^\bullet$ representing $K$. After renumbering $p = -j$ and $q = i + 2j$ we find a spectral sequence $(E'_ r, d'_ r)_{r \geq 2}$ with $d'_ r$ of bidegree $(r, -r + 1)$, with
$(E'_2)^{i, j} = \mathop{\mathrm{Ext}}\nolimits ^ i(M, H^ j(K))$
If $M \in D^-(\mathcal{A})$ and $K \in D^+(\mathcal{A})$ then both $E_ r$ and $E'_ r$ are bounded and converge to $\mathop{\mathrm{Ext}}\nolimits ^{p + q}(M, K)$. If we use the filtration $F^ pK^\bullet = \sigma _{\geq p}K^\bullet$ then we get
$E_1^{p, q} = \mathop{\mathrm{Ext}}\nolimits ^ q(M, K^ p)$
If $M \in D^-(\mathcal{A})$ and $K^\bullet$ is bounded below, then this spectral sequence is bounded and converges to $\mathop{\mathrm{Ext}}\nolimits ^{p + q}(M, K)$.
Remark 19.13.10. Let $\mathcal{A}$ be a Grothendieck abelian category. Let $K \in D(\mathcal{A})$. Let $M^\bullet$ be a filtered complex of $\mathcal{A}$, see Homology, Definition 12.24.1. For ease of notation denote $M$, $M/F^ pM$, $\text{gr}^ pM$ the object of $D(\mathcal{A})$ represented by $M^\bullet$, $M^\bullet /F^ pM^\bullet$, $\text{gr}^ pM^\bullet$. Dually to Remark 19.13.8 we can construct a spectral sequence $(E_ r, d_ r)_{r \geq 1}$ of bigraded objects of $\mathcal{A}$ with $d_ r$ of bidgree $(r, -r + 1)$ and with
$E_1^{p, q} = \mathop{\mathrm{Ext}}\nolimits ^{p + q}(\text{gr}^{-p}M, K)$
If for every $n$ we have
$\mathop{\mathrm{Ext}}\nolimits ^ n(M/F^ pM, K) = 0 \text{ for } p \ll 0 \quad \text{and}\quad \mathop{\mathrm{Ext}}\nolimits ^ n(M/F^ pM, K) = \mathop{\mathrm{Ext}}\nolimits ^ n(M, K) \text{ for } p \gg 0$
then the spectral sequence is bounded and converges to $\mathop{\mathrm{Ext}}\nolimits ^{p + q}(M, K)$. Namely, choose a K-injective complex $I^\bullet$ with injective terms representing $K$, see Theorem 19.12.6. Consider the complex
$\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , I^\bullet )$
defined exactly as in More on Algebra, Section 15.71. Setting
$F^ p\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , I^\bullet ) = \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet /F^{-p + 1}M^\bullet , I^\bullet )$
we obtain a filtered complex (note sign and shift in filtration). The spectral sequence of Homology, Section 12.24 has differentials and terms as described above; details omitted. The boundedness and convergence follows from Homology, Lemma 12.24.13.
Remark 19.13.11. Let $\mathcal{A}$ be a Grothendieck abelian category. Let $M, K$ be objects of $D(\mathcal{A})$. For any choice of complex $M^\bullet$ representing $M$ we can use the filtration $F^ pM^\bullet = \tau _{\leq -p}M^\bullet$ and the discussion in Remark 19.13.8 to get a spectral sequence with
$E_1^{p, q} = \mathop{\mathrm{Ext}}\nolimits ^{2p + q}(H^ p(M), K)$
This spectral sequence is independent of the choice of complex $M^\bullet$ representing $M$. After renumbering $p = -j$ and $q = i + 2j$ we find a spectral sequence $(E'_ r, d'_ r)_{r \geq 2}$ with $d'_ r$ of bidegree $(r, -r + 1)$, with
$(E'_2)^{i, j} = \mathop{\mathrm{Ext}}\nolimits ^ i(H^{-j}(M), K)$
If $M \in D^-(\mathcal{A})$ and $K \in D^+(\mathcal{A})$ then $E_ r$ and $E'_ r$ are bounded and converge to $\mathop{\mathrm{Ext}}\nolimits ^{p + q}(M, K)$. If we use the filtration $F^ pM^\bullet = \sigma _{\geq p}M^\bullet$ then we get
$E_1^{p, q} = \mathop{\mathrm{Ext}}\nolimits ^ q(M^{-p}, K)$
If $K \in D^+(\mathcal{A})$ and $M^\bullet$ is bounded above, then this spectral sequence is bounded and converges to $\mathop{\mathrm{Ext}}\nolimits ^{p + q}(M, K)$.
Lemma 19.13.12. Let $\mathcal{A}$ be a Grothendieck abelian category. Suppose given an object $E \in D(\mathcal{A})$ and an inverse system $\{ E^ i\} _{i \in \mathbf{Z}}$ of objects of $D(\mathcal{A})$ over $\mathbf{Z}$ together with a compatible system of maps $E^ i \to E$. Picture:
$\ldots \to E^{i + 1} \to E^ i \to E^{i - 1} \to \ldots \to E$
Then there exists a filtered complex $K^\bullet$ of $\mathcal{A}$ (Homology, Definition 12.24.1) such that $K^\bullet$ represents $E$ and $F^ iK^\bullet$ represents $E^ i$ compatibly with the given maps.
Proof. By Theorem 19.12.6 we can choose a K-injective complex $I^\bullet$ representing $E$ all of whose terms $I^ n$ are injective objects of $\mathcal{A}$. Choose a complex $G^{0, \bullet }$ representing $E^0$. Choose a map of complexes $\varphi ^0 : G^{0, \bullet } \to I^\bullet$ representing $E^0 \to E$. For $i > 0$ we inductively represent $E^ i \to E^{i - 1}$ by a map of complexes $\delta : G^{i, \bullet } \to G^{i - 1, \bullet }$ and we set $\varphi ^ i = \delta \circ \varphi ^{i - 1}$. For $i < 0$ we inductively represent $E^{i + 1} \to E^ i$ by a termwise injective map of complexes $\delta : G^{i + 1, \bullet } \to G^{i, \bullet }$ (for example you can use Derived Categories, Lemma 13.9.6). Claim: we can find a map of complexes $\varphi ^ i : G^{i, \bullet } \to I^\bullet$ representing the map $E^ i \to E$ and fitting into the commutative diagram
$\xymatrix{ G^{i + 1, \bullet } \ar[r]_\delta \ar[d]_{\varphi ^{i + 1}} & G^{i, \bullet } \ar[ld]^{\varphi ^ i} \\ I^\bullet }$
Namely, we first choose any map of complexes $\varphi : G^{i, \bullet } \to I^\bullet$ representing the map $E^ i \to E$. Then we see that $\varphi \circ \delta$ and $\varphi ^{i + 1}$ are homotopic by some homotopy $h^ p : G^{i + 1, p} \to I^{p - 1}$. Since the terms of $I^\bullet$ are injective and since $\delta$ is termwise injective, we can lift $h^ p$ to $(h')^ p : G^{i, p} \to I^{p - 1}$. Then we set $\varphi ^ i = \varphi + h' \circ d + d \circ h'$ and we get what we claimed.
Next, we choose for every $i$ a termwise injective map of complexes $a^ i : G^{i, \bullet } \to J^{i, \bullet }$ with $J^{i, \bullet }$ acyclic, K-injective, with $J^{i, p}$ injective objects of $\mathcal{A}$. To do this first map $G^{i, \bullet }$ to the cone on the identity and then apply the theorem cited above. Arguing as above we can find maps of complexes $\delta ' : J^{i, \bullet } \to J^{i - 1, \bullet }$ such that the diagrams
$\xymatrix{ G^{i, \bullet } \ar[r]_\delta \ar[d]_{a^ i} & G^{i - 1, \bullet } \ar[d]^{a^{i - 1}} \\ J^{i, \bullet } \ar[r]^{\delta '} & J^{i - 1, \bullet } }$
commute. (You could also use the functoriality of cones plus the functoriality in the theorem to get this.) Then we consider the maps
$\xymatrix{ G^{i + 1, \bullet } \times \prod \nolimits _{p > i + 1} J^{p, \bullet } \ar[r] \ar[rd] & G^{i, \bullet } \times \prod \nolimits _{p > i} J^{p, \bullet } \ar[r] \ar[d] & G^{i - 1, \bullet } \times \prod \nolimits _{p > i - 1} J^{p, \bullet } \ar[ld] \\ & I^\bullet \times \prod \nolimits _ p J^{p, \bullet } }$
Here the arrows on $J^{p, \bullet }$ are the obvious ones (identity or zero). On the factor $G^{i, \bullet }$ we use $\delta : G^{i, \bullet } \to G^{i - 1, \bullet }$, the map $\varphi ^ i : G^{i, \bullet } \to I^\bullet$, the zero map $0 : G^{i, \bullet } \to J^{p, \bullet }$ for $p > i$, the map $a^ i : G^{i, \bullet } \to J^{p, \bullet }$ for $p = i$, and $(\delta ')^{i - p} \circ a^ i = a^ p \circ \delta ^{i - p} : G^{i, \bullet } \to J^{p, \bullet }$ for $p < i$. We omit the verification that all the arrows in the diagram are termwise injective. Thus we obtain a filtered complex. Because products in $D(\mathcal{A})$ are given by taking products of K-injective complexes (Lemma 19.13.4) and because $J^{p, \bullet }$ is zero in $D(\mathcal{A})$ we conclude this diagram represents the given diagram in the derived category. This finishes the proof. $\square$
Lemma 19.13.13. In the situation of Lemma 19.13.12 assume we have a second inverse system $\{ (E')^ i\} _{i \in \mathbf{Z}}$ and a compatible system of maps $(E')^ i \to E$. Then there exists a bi-filtered complex $K^\bullet$ of $\mathcal{A}$ such that $K^\bullet$ represents $E$, $F^ iK^\bullet$ represents $E^ i$, and $(F')^ iK^\bullet$ represents $(E')^ i$ compatibly with the given maps.
Proof. Using the lemma we can first choose $K^\bullet$ and $F$. Then we can choose $(K')^\bullet$ and $F'$ which work for $\{ (E')^ i\} _{i \in \mathbf{Z}}$ and the maps $(E')^ i \to E$. Using Lemma 19.13.7 we can assume $K^\bullet$ is a K-injective complex. Then we can choose a map of complexes $(K')^\bullet \to K^\bullet$ corresponding to the given identifications $(K')^\bullet \cong E \cong K^\bullet$. We can additionally choose a termwise injective map $(K')^\bullet \to J^\bullet$ with $J^\bullet$ acyclic and K-injective. (To do this first map $(K')^\bullet$ to the cone on the identity and then apply Theorem 19.12.6.) Then $(K')^\bullet \to K^\bullet \times J^\bullet$ and $K^\bullet \to K^\bullet \times J^\bullet$ are both termwise injective and quasi-isomorphisms (as the product represents $E$ by Lemma 19.13.4). Then we can simply take the images of the filtrations on $K^\bullet$ and $(K')^\bullet$ under these maps to conclude. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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# Counting question
1. Dec 10, 2013
### Miike012
How many 8 letter words have 3 vowels constants do not repeat.
Is it
5^3*26*25*24*23*22 ?
2. Dec 10, 2013
### Staff: Mentor
Don't you have to also factor in order too:
act vs cat vs some nonsense words like tca, cta ...
cafe vs face
3. Dec 10, 2013
### Miike012
Another question how do I know when to use the combination equation or permutation equation?
For instance given the example which is posted above are the following strings the same?
A E I M N P Q R
and
E A I M N P Q R
They both contain the same three vowels and the only thing I changed was the placement of the first and second terms
4. Dec 10, 2013
### Miike012
so I should add 8! ?
5. Dec 10, 2013
### Miike012
I'm not sure that would work, for instance
A1 B A2
and
A2 B A1
are the same.
I almost feel like I should add (8 choose 3) instead of 8!
6. Dec 10, 2013
### Staff: Mentor
I think your consonants are unique but your vowels might repeat so you'd also have to factor that in too.
7. Dec 11, 2013
### haruspex
You mean multiply by 8!?
For the vowels you have 53, so you've already counted all vowel orders.
Similarly, for the consonants, you used 26P5, not 26C5, so you have those ordered.
All that you are missing is integrating the vowels with the consonants. How many ways are there of doing that?
8. Dec 11, 2013
### vela
Staff Emeritus
Remember that the vowels are part of the 26 letters. There are only 21 consonants.
9. Dec 11, 2013
### Staff: Mentor
I missed that, thanks.
10. Dec 11, 2013
How true
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Anonymous
Not applicable
## Total of population in 100% bar chart
Is there a way to show part and whole in 100% bar charts? My example is that I have a total number of employees, and then I have the number of them that left. I want to show the total as the top bar (with gender or race as the legend) and then the departures as the second bar (also with either gender or race as the legend). I can't wrap my head around how to structure this to get it to work.
EmpID Active Jan 1 Departure 1 Active no 2 Active no 3 Active yes 4 no no 5 Active no 6 Active yes 7 Active yes 8 Active no 9 no no
The top bar would be 7, the bottom would be 3. They would both be 100% bar charts.
1 ACCEPTED SOLUTION
Anonymous
Not applicable
Hi @Anonymous ,
Based on your description, I did a test, the table for testing is as follows.
1. Create a new calculated table.
Test1 =
var total_Measure = CALCULATE(COUNTROWS('CASE1'),'CASE1'[Active Jan 1]="Active")
var Departure_Measure = CALCULATE(COUNTROWS('CASE1'),'CASE1'[Departure]="yes")
return
ALLSELECTED(CASE1[gender]),
"departure_table",'CASE1'[Departure_Measure],
"total",'CASE1'[total_Measure]
)
1. Unpivot “departure_table” and “total” columns.
Test2 =
UNION(
SELECTCOLUMNS('Test1',"gender",'Test1'[gender],"attribute","departure","value",'Test1'[departure_table]),
SELECTCOLUMNS('Test1',"gender",'Test1'[gender],"attribute","total","value",'Test1'[total])
)
1. Create a 100% stacked bar chart.
Result:
Hope that's what you were looking for.
Best Regards,
Yuna
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
4 REPLIES 4
Anonymous
Not applicable
Hi @Anonymous ,
Based on your description, I did a test, the table for testing is as follows.
1. Create a new calculated table.
Test1 =
var total_Measure = CALCULATE(COUNTROWS('CASE1'),'CASE1'[Active Jan 1]="Active")
var Departure_Measure = CALCULATE(COUNTROWS('CASE1'),'CASE1'[Departure]="yes")
return
ALLSELECTED(CASE1[gender]),
"departure_table",'CASE1'[Departure_Measure],
"total",'CASE1'[total_Measure]
)
1. Unpivot “departure_table” and “total” columns.
Test2 =
UNION(
SELECTCOLUMNS('Test1',"gender",'Test1'[gender],"attribute","departure","value",'Test1'[departure_table]),
SELECTCOLUMNS('Test1',"gender",'Test1'[gender],"attribute","total","value",'Test1'[total])
)
1. Create a 100% stacked bar chart.
Result:
Hope that's what you were looking for.
Best Regards,
Yuna
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Anonymous
Not applicable
@Anonymous Yuna, I think that's exactly what I need! Thank you! I'm not entirely sure how it will work with my table structure/data connection, but at least this gets me in the right direction. Much appreciated!
Super User
@Anonymous , In power bi you have two100% stacked bars (bar and column bar), Please try that.
https://www.tutorialgateway.org/create-100-stacked-bar-chart-in-power-bi/
Anonymous
Not applicable
What I need is one chart that will show the aggregate of the total along with the breakout of individual groups.
As another example, let's say I have sales per office. I would have bars for New York, Los Angeles, and Chicago. But then I also want a single bar at the top with all offices. Is there any way to get that total bar at the top?
Back to my original example of employee data, I want all of the employees on the top bar, then just the ones that departed on the bottom. The departures would be included in the top bar count (since they were employees on Jan 1). The point is to show ratio of the departures to the ratio of the total.
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# ch106 - CHAPTER CHECKLIST
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ECO2013 Fall 2003 Cobbe Chapter 10 Page 1 of 11 < When you have completed your study of this chapter, you will be able to C H A P T E R C H E C K L I S T Define and calculate the economic growth rate, and explain the implications of sustained growth. 1 Identify the main sources of economic growth. Review some theories of economic growth that try to explain why growth rates vary over time and across countries. 2 3 Describe some policies that might speed economic growth. 4 10.1 THE BASICS OF ECONOMIC GROWTH Economic growth is a sustained expansion of production possibilities measured as the increase in real GDP produced over a given period. < Calculating Growth Rates Economic growth rate The rate of change of real GDP expressed as a percentage per year. 10.1 THE BASICS OF ECONOMIC GROWTH To calculate this growth rate, we use the formula: Growth of real GDP = Real GDP in current year Real GDP in previous year x 100 Real GDP in previous year For example, if real GDP in the current year is \$8.4 trillion and if real GDP in the previous year was \$8.0 trillion, then the growth rate of real GDP is : Growth of real GDP = \$8.4 trillion – \$8.0 trillion \$8.0 trillion x 100 = 5 percent. 10.1 THE BASICS OF ECONOMIC GROWTH The standard of living ultimately depends on real GDP per person. Real GDP per person Real GDP divided by the population. Any individual ’s standard of living depends on the distribution of GDP as well as average real GDP per person, but the average standard of living can be measured by real GDP per person, i.e. real GDP divided by the population. 10.1 THE BASICS OF ECONOMIC GROWTH For example, to calculate this growth rate, of GDP per person, we use the same formula as before, replacing real GDP with real GDP per person. Suppose, for example, that in the current year, when real GDP is \$8.4 trillion, the population is 202 million. Then real GDP per person is \$8.4 trillion divided by 202 million, which equals \$41,584. And suppose that in the previous year, when real GDP was \$8.0 trillion, the population was 200 million. Then real GDP per person in that year was \$8.0 trillion divided by 200 million, which equals \$40,000.
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ECO2013 Fall 2003 Cobbe Chapter 10 Page 2 of 11 10.1 THE BASICS OF ECONOMIC GROWTH Use these two values of real GDP per person in the growth formula to calculate the growth rate of real GDP per person. It is: Growth rate of real GDP per person \$41,584 – \$40,000 \$40,000 x 100 = 4 percent. = 10.1 THE BASICS OF ECONOMIC GROWTH The growth rate of real GDP per person can also be calculated by using the approximation, very accurate for growth rates less than about 10% per year, Growth of real GDP per person Growth rate of real GDP Growth rate of population = Growth of population 202 million – 200 million 200 million x 100 = 1 percent. = 10.1 THE BASICS OF ECONOMIC GROWTH
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Solutions
Calculus Early Transcendentals Single Variable
# Calculus Early Transcendentals Single Variable (10th Edition) View more editions Solutions for Chapter 9.4
• 5071 step-by-step solutions
• Solved by publishers, professors & experts
• iOS, Android, & web
Over 90% of students who use Chegg Study report better grades.
May 2015 Survey of Chegg Study Users
Chapter: Problem:
100% (6 ratings)
SAMPLE SOLUTION
Chapter: Problem:
100% (6 ratings)
• Step 1 of 5
(a)
Consider the following series,
,
The objective is to find the sum of the series.
Rewrite the series as,
The series is a convergent geometric series
And the series is also a convergent geometric series
Find the sum.
Therefore, the series is convergent and its sum is.
• Step 2 of 5
(b)
Consider the following series,
The objective is to find the sum of the series.
• Step 3 of 5
Rewrite the series as,
• Step 4 of 5
The series is a convergent geometric series
Now,
So that the partial sum of the series is,
Thus,
Since the limit of partial sum has been expressed in the closed form and the limit as exists, the series is convergent and converges to 1.
• Step 5 of 5
Find the sum.
=
=
Therefore, the given series is convergent and its sum is.
Corresponding Textbook
Calculus Early Transcendentals Single Variable | 10th Edition
9780470647684ISBN-13: 047064768XISBN: Authors:
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## 16220
16,220 (sixteen thousand two hundred twenty) is an even five-digits composite number following 16219 and preceding 16221. In scientific notation, it is written as 1.622 × 104. The sum of its digits is 11. It has a total of 4 prime factors and 12 positive divisors. There are 6,480 positive integers (up to 16220) that are relatively prime to 16220.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 11
• Digital Root 2
## Name
Short name 16 thousand 220 sixteen thousand two hundred twenty
## Notation
Scientific notation 1.622 × 104 16.22 × 103
## Prime Factorization of 16220
Prime Factorization 22 × 5 × 811
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 8110 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 16,220 is 22 × 5 × 811. Since it has a total of 4 prime factors, 16,220 is a composite number.
## Divisors of 16220
1, 2, 4, 5, 10, 20, 811, 1622, 3244, 4055, 8110, 16220
12 divisors
Even divisors 8 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 34104 Sum of all the positive divisors of n s(n) 17884 Sum of the proper positive divisors of n A(n) 2842 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 127.358 Returns the nth root of the product of n divisors H(n) 5.70725 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 16,220 can be divided by 12 positive divisors (out of which 8 are even, and 4 are odd). The sum of these divisors (counting 16,220) is 34,104, the average is 2,842.
## Other Arithmetic Functions (n = 16220)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 6480 Total number of positive integers not greater than n that are coprime to n λ(n) 1620 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1883 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 6,480 positive integers (less than 16,220) that are coprime with 16,220. And there are approximately 1,883 prime numbers less than or equal to 16,220.
## Divisibility of 16220
m n mod m 2 3 4 5 6 7 8 9 0 2 0 0 2 1 4 2
The number 16,220 is divisible by 2, 4 and 5.
• Arithmetic
• Abundant
• Polite
## Base conversion (16220)
Base System Value
2 Binary 11111101011100
3 Ternary 211020202
4 Quaternary 3331130
5 Quinary 1004340
6 Senary 203032
8 Octal 37534
10 Decimal 16220
12 Duodecimal 9478
20 Vigesimal 20b0
36 Base36 cik
## Basic calculations (n = 16220)
### Multiplication
n×i
n×2 32440 48660 64880 81100
### Division
ni
n⁄2 8110 5406.67 4055 3244
### Exponentiation
ni
n2 263088400 4267293848000 69215506214560000 1122675510800163200000
### Nth Root
i√n
2√n 127.358 25.3134 11.2853 6.95041
## 16220 as geometric shapes
### Circle
Diameter 32440 101913 8.26517e+08
### Sphere
Volume 1.78748e+13 3.30607e+09 101913
### Square
Length = n
Perimeter 64880 2.63088e+08 22938.5
### Cube
Length = n
Surface area 1.57853e+09 4.26729e+12 28093.9
### Equilateral Triangle
Length = n
Perimeter 48660 1.13921e+08 14046.9
### Triangular Pyramid
Length = n
Surface area 4.55682e+08 5.02905e+11 13243.6
## Cryptographic Hash Functions
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# Re-writting an Equation
1. Jul 16, 2009
### EngWiPy
1. The problem statement, all variables and given/known data
Hello,
I have the following equation:
$$f(x)=\sum_{r=m}^{M_B}\,\sum_{i=0}^{M_B-r}\,\sum_{j=0}^{r+i}\,\sum_{k=0}^{j(N_B-1)}(-1)^{i+j}{M_B\choose r}{M_B-r\choose i}{r+i\choose j}\left(\frac{x}{C}\right)^k\,e^{jx/C}$$
and I want to write it in the form of $$f(x)=1+R(x)$$
2. Relevant equations
$$m$$ will be any number from 1 up to $$M_B$$, and $$f(x)=1-e^{x/C}$$ for $$M_B=N_B=1$$.
3. The attempt at a solution
I tried to extract the following parameters to have the 1 value:
$$r=M_B, i=0, j=0, k=0$$
but after that I don't know what to do by the indices. It does not seem straightforward as one can see.
Regards
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## SpinOffs
A blog on what's new, notable, and next in turbomachinery
# Specific Speed Performance Effects
By Mark R. Anderson
Jan 18, 2019
Share
In my last blog, Specific Speed Demystified, I covered the mathematical definition of specific speed and how it relates to flow and work coefficient. The concept of specific speed has been a guiding principle for radial turbomachinery design for many years. Use of specific speed has been most heavily emphasized by Balje in his famous textbook and early publications. In these works, he laid out several graphs that are still widely used today.
The first figure shows islands of potential efficiency for turbines as a function of specific speed and specific diameter, a parameter related to specific speed.
Image from https://barber-nichols.com/media/tools-resources/
The next two plots are more simplified figures from Balje expressing potential efficiency as a function of specific speed alone. The first plot is for radial turbines and the second is for radial compressors.
Images from Introduction to Turbomachinery by David Japikse and Nicholas C Baines
The physical explanation of the poor performance at low specific speed is that long narrow passages generate significant viscous losses that degrade efficiency. The degradation at high specific speed is generally attributed to large curvature in the geometry. Indeed, one of the driving forces for shifting to axial machines at high specific speeds is the need to reduce this effect.
Though widely used, it’s easy to overestimate the effect of specific speed. The figure below plots several different radial turbine designs and clearly shows just how varied the performance can be across the specific speed.
Above Image from Axial and Radial Turbines by
Hany Moustapha, Mark F. Zelesky, Nicholas C. Baines, and David Japikse
These results clearly show that the lines of peak performance as a function of specific speed are much, much blurrier than the sharp lines of Balje’s plots would imply.
The value of specific speed is not in the prediction of performance, but rather as a rough guide to the machine shape needed that will tend to give better performance. Concepts text book, Introduction to Turbomachinery, describes specific speed as, “slightly coarse, but never the less useful, tool for selecting the type of turbomachine required…”.
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# How do you convert from Fahrenheit to radians?
## How do you convert from Fahrenheit to radians?
Multiply the number of degrees by π/180. Since you know this, all you have to do is multiply the number of degrees you’re working with by π/180 to convert it to radian terms.
## What is 60f when converted to Rankine?
Fahrenheit to Rankine conversion table
Fahrenheit (°F) Rankine (°R)
50 °F 509.67 °R
60 °F 519.67 °R
70 °F 529.67 °R
80 °F 539.67 °R
What does R mean in temperature?
Learn about this topic in these articles: Fahrenheit scale is called the Rankine (°R) scale. These scales are related by the equations K = °C + 273.15, °R = °F + 459.67, and °R = 1.8 K. Zero in both the Kelvin and Rankine scales is at absolute zero.
### What is the Fahrenheit equivalent of 500 R?
Rankine to Fahrenheit conversion table
Rankine (°R) Fahrenheit (°F)
290 °R -169.67 °F
300 °R -159.67 °F
400 °R -59.67 °F
500 °R 40.33 °F
### What is the Fahrenheit temperature of 40 C?
104°F
Celsius to Fahrenheit Conversion Chart
Celsius Fahrenheit
40°C 104°F
50°C 122°F
60°C 140°F
70°C 158°F
What is 378 R converted to Kelvin?
Convert 378 Rankine to Kelvin
378 Rankine (Ra) 210.000 Kelvin (K)
1 Ra = 0.555556 K 1 K = 1.800000 Ra
#### Which temperature is the warmest?
The current official highest registered air temperature on Earth is 56.7 °C (134.1 °F), recorded on 10 July 1913 at Furnace Creek Ranch, in Death Valley in the United States.
#### What is a function in R?
What are functions? A key feature of R is functions. Functions are “self contained” modules of code that accomplish a specific task. Functions usually take in some sort of data structure (value, vector, dataframe etc.), process it, and return a result.
How to convert Fahrenheit to degrees in degrees Rankine?
The temperature T in degrees Rankine (°R) is equal to the temperature T in degrees Fahrenheit (°F) plus 459.67: T (°R) = T (°F) + 459.67. Example. Convert 68 degrees Fahrenheit to degrees Rankine: T (°R) = 68°F + 459.67 = 527.67 °R. Fahrenheit to Rankine conversion table
## Which is the correct formula for a temperature conversion calculator?
Calculator Use Temperature conversions are performed by using a formula, which differs depending on the two temperature scales you are converting between. For example, to convert 50 degrees Celsius (centigrade) to Fahrenheit, we plug our numbers into the formula as shown below: F = C * 9/5 + 32 F = 50 * 9/5 + 32
## Which is colder 0 k or 1 °R?
Thus, a temperature of 0 K (−273.15 °C; −459.67 °F) is equal to 0 °R, and a temperature of −458.67 °F equal to 1 °R. Some important temperatures relating the Rankine scale to other temperature scales are shown in the table below.
Which is greater 0 °F or 459.67 °R?
0 °F = 459.67 °R. The temperature T in degrees Rankine (°R) is equal to the temperature T in degrees Fahrenheit (°F) plus 459.67: T (°R) = T (°F) + 459.67.
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# How many pages will a 25000 word book be once printed
## How many pages is a 30000 word book?
How many words per page?Word CountPages (single-spaced)Pages (double-spaced)12,500 WORDS25 PAGES50 PAGES15,000 WORDS30 PAGES60 PAGES20,000 WORDS40 PAGES80 PAGES25,000 WORDS50 PAGES100 PAGES
## How many pages should a book be?
For a short and snappy non-fiction/educational type book, aim for between 100 – 150 pages, based on approximately 367 words per page using Arial font size 12 with a paragraph spacing of 1.5. This would equate to 36,700 – 55,000 words in total.
## How many words is a 5×8 book page?
5.25×8”: 260 words/page. 5.5×8. 5”: 290 words/page. 6×9”: 340 words/page.
## How many Microsoft Word pages equal a book page?
By default, a MS Word is set at 8.5 x 11 inches with 1 inch margin on four sides. If you write in 12 point Times New Roman font, that’ll be about 500 – 550 words per page. So a 175 paperback pages novel will be about 100 Microsoft Word pages, i.e. one Microsoft word page is approximately equivalent to 1.75 novel pages.
## Is 20 000 words enough for a book?
This means that you cannot fit many words on a printed page and keep it readable. … This trim size is good for only 200 to 250 words per printed book page. This translates into just 20,000 to 25,000 words for a 100 page book, 30,000 to 42,500 words for a 150-page book, perhaps just 40,000 words for a 200-page book.
## Is 30000 words enough for a book?
Big, epic stories get anywhere from 120,000 to 200,000 words.” But, he also mentions that “The Wizard of Oz was 40,000 words. The Old Man and the Sea was about 25 to 30,000 words, tops.” … If you’re working on a novel-length book, aim for 50,000 words at the very least — but it’s better to aim for 90,000.
You might be interested: Often asked: What does utc mean?
## How many pages is 40000 words?
40,000 words is 80 pages single spaced, 160 pages double spaced. 50,000 words is 100 pages single spaced, 200 pages double spaced.
## How many pages is 90000 words?
Here’s how 90,000 words (which is 360 pages at 250 words per page) breaks down by act: Act One: 90 pages (22,500 words)7 мая 2018 г.
## What is a good amount of chapters for a book?
Just how many chapters should you include in your novel? Most novels have between 10 to 12 chapters, but that’s not set in stone. You can have two chapters or 200 — it all depends on how comfortable you are with experimenting. Consider your dear reader.
## Is 15000 words enough for a book?
Children’s books are 10,000 to 15,000. Mysteries and young adult books run between 40,000 and 80,000 words. … Nonfiction books also vary tremendously in word count. Short and punchy books are 40,000 to 60,000 words, but many nonfiction books are lengthy: 80,000 words and up.
## Is 50 000 words enough for a novel?
NOVEL: A manuscript over 40,000 words is considered to be a novel. However, very few novels these days are as short as that. Generally a 50,000-word novel would be the minimum word count. Most novels are between 60,000 and 100,000 words.
70.0 pages
## Is 100 pages too short for a book?
Book Length by the Numbers
A typical manuscript page (8.5×11 paper, 1-inch margins, standard 11- or 12-point font, doublespaced—like you would type in Word) is considered to be 250 words. So a 25,000-word manuscript is about 100 pages. A 50,000-word manuscript is 200 pages.
You might be interested: What is the book of life in the bible
## How many pages is 100000 word?
It depends on the font you are using, of course, but in general, 250-300 words per page. Therefore, a 55,000 word book should be about 200 manuscript pages. A 100,000 word book would be about 400. Editors like 12 point font.
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How to Calculate ROI (Return on Investment)
You may have asked…why should I switch over to an engineered compressed air product if my system already works? Or…How can your products be much different?
Manufacturing has always been an advocate for cost savings, where they even have job positions solely focused on cost savings. Return on Investment (ROI) is a metric they look toward to help make good decisions for cost savings. The term is used to determine the financial benefits associated with the use of more efficient products or processes compared to what you are currently using. This is like looking at your homes heating costs and then changing out to energy efficient windows and better insulation. The upfront cost might be high but the amount of money you will save over time is worth it.
How is ROI calculated? It is very simple to calculate out the potential savings of using an EXAIR Intelligent Compressed Air® Product. We have easy to use calculators on our websites Resources where filling in a few blanks will result in an ROI when switching to a EXAIR product! Here they Are, Calculators.
I’ll go ahead and break down the simple ROI calculations for replacing open blow offs with an EXAIR Super Air Nozzle:
• ¼” Copper Pipe consumes 33 SCFM at 80 psig (denoted below as CP)
• A Model 1100 ¼” Super Air Nozzle can be used to replace and only uses 14 SCFM at 80 psig (denoted below as EP)
Calculation:
(CP air consumption) * (60 min/hr) * (8 hr/day) * (5 days/week) * (52 weeks/year) = SCF used per year for Copper Pipe
(33) * (60) * (8) * (5) * (52) = 4,118,400 SCF
(EP air consumption) * (60 min/hr) * (8 hr/day) * (5 days/week) * (52 weeks/year) = SCF used per year for EXAIR Product
(14) * (60) * (8) * (5) * (52) = 1,747,200 SCF
Air Savings:
SCF used per year for Copper Pipe – SCF used per year for EXAIR Product = SCF Savings
4,118,400 SCF – 1,747,200 SCF = 2,371,200 SCF in savings
If you know the facilities cost to generate 1,000 SCF of compressed air you can calculate out how much this will save. If not, you can use \$0.25 to generate 1,000 SCF which is the value used by the U.S. Department of Energy to estimate costs.
Yearly Savings:
(SCF Saved) * (Cost / 1000 SCF) = Yearly Savings
(2,371,200 SCF) * (\$0.25 / 1000 SCF) = \$592.80 annual Savings
With the simple investment of \$42 (as of date published) you can calculate out the time it will take to pay off the unit.
Time Until payoff:
(Yearly Savings) / (5 days/week * 52 weeks/year) = Daily Savings
(\$592.80/year) / (5 days/week * 52 weeks/year) = \$2.28 per day
(Cost of EXAIR Unit) / (Daily Savings) = Days until product has been paid off
(\$42) / (\$2.28/day) = 17.9 days
As you can see it doesn’t have to take long for the nozzle to pay for itself, and then continue to contribute toward your bottom line.
If you have any questions about compressed air systems or want more information on any of EXAIR’s products, give us a call, we have a team of Application Engineers ready to answer your questions and recommend a solution for your applications.
Jordan Shouse
Application Engineer
Send me an Email
Find us on the Web
Hand Holding money Image from Pictures of Money Creative Commons license
Benefits of Engineered Air Jets and Nozzles
EXAIR Nozzles and Jets are designed and manufactured to take advantage of the Coanda (wall attachment of a high velocity fluid) effect which can amplify the airflow up to 25 times. The compressed air exits through the small holes on the nozzle which entrains the surrounding air. The effect from this is a high volume, high velocity blast using less compressed air. EXAIR manufactures many sizes and styles of air nozzles from the smallest, but quite powerful M4 x 0.5 thread Atto Super Air Nozzles to our largest 1-1/4 NPT Super Air Nozzle. We also offer Flat Super Air Nozzles, and the Back Blow style nozzle for cleaning out tubes, pipes, channels or holes from 1/4″ to 16″ in diameter.
The Air Jets are 1/8 NPT threads and blow air out at a right angle from the inlet. They produce a vacuum on the larger diameter side which pulls in surrounding ambient air into the total output flow. Air Jets are available in brass or Type 303 stainless steel. You can choose from a fixed flow style or an adjustable flow style to provide flexibility for your applications. The adjustable flow models have a clear micrometer air gap indicator to assure consistent and accurate results.
All of our Air Nozzles and Jets are engineered to meet or exceed OSHA Standard 1910.24(b) for 30 PSIG dead end pressure, they cannot be dead-ended as there is always a route for the air to escape. In addition, our products are going to meet the OSHA Standard CFR 29 – 1910.95(a) for allowable noise exposure levels.
EXAIR’s Swivel Fittings make it easy to adjust the position of the Air Nozzles and Air Jets. The fittings allow for movement of 25° form the center axis for a total movement of 50°. There are nine different models available and all of them are made from stainless steel
If you would like to discuss blow off, noise levels, dead end pressure or any of EXAIR’s Intelligent Compressed Air® usage solutions, I would enjoy hearing from you…give me a call.
Jordan Shouse
Application Engineer
Send me an email
Find us on the Web
Two Important Safety Factors When Choosing Air Nozzles
At EXAIR, we have a statement, “Safety is everyone’s responsibility”. And we also manufacture safe compressed air products. In the United States, we have an organization called Occupational Safety and Health Administration, OSHA, that enforces directives for safe and healthy working environments. They do training, outreach programs, and educational assistance for manufacturing plants. They will also enforce these directives with heavy fines for violations. The two most common violations with compressed air are air guns and blow-off devices are described in 29CFR 1910.242(b) for dead-end pressure/chip shielding and 29CFR 1910.65(a) for maximum allowable noise exposure.
Here is an example of a nozzle that is dangerous. As you can see, there is only one opening where the air can come out from the nozzle. Other types of nozzles that would fall into this same group would include copper tube, extensions, and open pipes.
They are dangerous as the compressed air cannot escape if it is blocked with your body or skin. If operated above 30 PSIG (2 bar), these nozzles could create an air embolism within the body which can cause bodily harm or death. This is a hazard which can be avoided by using EXAIR Super Air Nozzles and Safety Air Guns. The nozzles are designed with fins which allows the air to escape and not be blocked by your skin. So, you can use the EXAIR Super Air Nozzles safely even above 30 PSIG (2 bar).
To counteract the dead-end pressure violation, some nozzle manufacturers create a hole through the side of the nozzle (Reference photo above). This will allow for the compressed air to escape, but, now the issue is noise level. With an “open” hole in the nozzle, the compressed air is very turbulent and very loud. The National Institute for Occupational Safety and Health, NIOSH, states that 70% to 80% of all hearing loss within a manufacturing plant is caused by compressed air. OSHA created a chart to show the maximum allowable noise exposure. This chart shows the time and noise limits before requiring hearing protection. The EXAIR Super Air Nozzles, Super Air Knives, Super Air Amplifiers are designed to have laminar flow which is very quiet. As an example, the model 1210 Safety Air Gun has a sound level of only 74 dBA; well under the noise exposure limit for 8 hours.
NIOSH created an overview of how to handle hazards in the workplace. They call it the Hierarchy of Controls to best protect workers from dangers. The most effective way is by eliminating the hazard or substituting the hazard. The least effective way is with Personal Protective Equipment, or PPE. For unsafe compressed air nozzles and guns, the proper way to reduce this hazard is to substitute it with an engineered solution.
One of the last things that companies think about when purchasing compressed air products is safety. Loud noises and dead-end pressure can be missed or forgotten. To stop any future fines or additional personal protective equipment (PPE), it will be much cheaper to purchase an EXAIR product. And with the Hazard Hierarchy of Controls, the first method is to remove any hazards. The last method for control is to use PPE. In the middle of the hierarchy is for an engineered solution. EXAIR products are that engineered solution. If you would like to improve the safety in your facility with your current blow-off devices, an Application Engineer can help you.
John Ball
Application Engineer
Email: johnball@exair.com
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https://www.crazy-numbers.com/en/42359
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Discover a lot of information on the number 42359: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
Mathematical properties of 42359
Is 42359 a prime number? Yes
Is 42359 a perfect number? No
Number of divisors 2
List of dividers 1, 42359
Sum of divisors 42360
Prime factorization 42359
Prime factors 42359
How to write / spell 42359 in letters?
In letters, the number 42359 is written as: Forty-two thousand three hundred and fifty-nine. And in other languages? how does it spell?
42359 in other languages
Write 42359 in english Forty-two thousand three hundred and fifty-nine
Write 42359 in french Quarante-deux mille trois cent cinquante-neuf
Write 42359 in spanish Cuarenta y dos mil trescientos cincuenta y nueve
Write 42359 in portuguese Quarenta e dois mil trezentos cinqüenta e nove
Decomposition of the number 42359
The number 42359 is composed of:
1 iteration of the number 4 : The number 4 (four) is the symbol of the square. It represents structuring, organization, work and construction.... Find out more about the number 4
1 iteration of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2
1 iteration of the number 3 : The number 3 (three) is the symbol of the trinity. He also represents the union.... Find out more about the number 3
1 iteration of the number 5 : The number 5 (five) is the symbol of freedom. It represents change, evolution, mobility.... Find out more about the number 5
1 iteration of the number 9 : The number 9 (nine) represents humanity, altruism. It symbolizes generosity, idealism and humanitarian vocations.... Find out more about the number 9
Other ways to write 42359
In letter Forty-two thousand three hundred and fifty-nine
In roman numeral
In binary 1010010101110111
In octal 122567
In US dollars USD 42,359.00 (\$)
In euros 42 359,00 EUR (€)
Some related numbers
Previous number 42358
Next number 42360
Next prime number 42373
Mathematical operations
Operations and solutions
42359*2 = 84718 The double of 42359 is 84718
42359*3 = 127077 The triple of 42359 is 127077
42359/2 = 21179.5 The half of 42359 is 21179.500000
42359/3 = 14119.666666667 The third of 42359 is 14119.666667
423592 = 1794284881 The square of 42359 is 1794284881.000000
423593 = 76004113274279 The cube of 42359 is 76004113274279.000000
√42359 = 205.81302193982 The square root of 42359 is 205.813022
log(42359) = 10.65393619226 The natural (Neperian) logarithm of 42359 is 10.653936
log10(42359) = 4.6269456988479 The decimal logarithm (base 10) of 42359 is 4.626946
sin(42359) = -0.7871976641602 The sine of 42359 is -0.787198
cos(42359) = -0.61670076823426 The cosine of 42359 is -0.616701
tan(42359) = 1.2764661643184 The tangent of 42359 is 1.276466
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https://justaaa.com/civil-engineering/1029503-how-i-can-calculate-rb-angles-if-i-dont-know
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Question
# how i can calculate RB angles if i dont know where is laying but i know...
how i can calculate RB angles if i dont know where is laying but i know the vertical and WCB angles
however i do not have any graph
If you have whole circle bearing with their vertical direction then you obviously calculate the reduced bearing. First of all we have to draw vertical direction as North direction, from there we can have to plot whole circle bearing. If whole circle bearing is lesser than 180° then we have to reduce 180 from them, then we got reduced bearing. For example if WCB = 130° then reduced bearing = 180- 130= 50°, i.e, S50°E, similarly if WCB is greater than 180° then we have to reduce 180 ° from them to find RB. If WCB is greater than 270° then we to reduce 360° from them to deduct reduced bearing. For example if whole circle bearing= 300° , then Reduced bearing = 360- 300= 60°, i.e, N60° W.
#### Earn Coins
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https://nickadamsinamerica.com/christmas-math-worksheets-coloring-pages/19aa936d6f3ef355df7b879b1edbd322/
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# 19aa936d6f3ef355df7b879b1edbd322
By . Worksheet. At Monday, October 04th 2021, 02:14:12 AM.
This section contains all of the graphic previews for the Differential Equations Worksheets. We have slope field, separable equation, differential equation, and exponential growth worksheets for your use. These Calculus worksheets are a good resource for students in high school.
When we add numbers that need to be carried over we can only carry over a digit into the right spot. For example, when we add the 4 and 9 from the problem on the board we have to carry over the answer (13) into the correct place value spots.
These mixed problems worksheets may be configured for adding and subtracting 2, 3, and 4 digit problems in a vertical format. For the subtraction problems you may select some regrouping, no regrouping, all regrouping, or subtraction across zero. You may select up to 30 mixed problems per worksheet.
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http://www.enotes.com/homework-help/how-do-you-factor-x-2-7-436516
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# How do you factor x^2-7?
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
Use the relation a^2 - b^2 = (a - b)(a + b)
`x^2 - 7`
= `x^2 - (sqrt 7)^2`
=> `(x - sqrt 7)(x + sqrt 7)`
It should be noted that x^2 - 7 does not have rational factors.
The expression `x^2 - 7 = (x - sqrt 7)(x + sqrt 7)`
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https://hub.jmonkeyengine.org/t/how-to-make-a-lego-block/6035
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# How to Make a Lego Block?
I'm trying to create a Lego block in jMonkey, but I'm having trouble figuring out how to go about doing this. I'm pretty new to jMonkey (and 3D graphics in general), so I'd greatly appreciate some help.
I want to draw a three-cylinder Lego block, but I'm also hoping to be able to customize it to other sizes.
My initial idea was to extend the Trimesh class and specify the appropriate vertex, normal, texture, and index data, but thanks to my lack of understanding and inexperience with this, I'm running into some problems.
The Lego block would essentially be a box with the bottom exposed, showing the bottom three cylinders. I could probably figure out how to draw a box without a bottom face (mostly by just examining the Box class), but how would I incorporate the vertices for the three cylinders? The Lego block's vertex data would consist of the points of the box part of the Lego as well as radial samples and axis samples for the cylinder part of the Lego. How can I combine this vertex data together, as well as create the proper placement for each cylinder of the Lego block?
If I understand correctly, index data specifies (in threes) the indices of what vertices create triangles. I understand vertex, normal, and index data, but how does texture data work? I looked at the Box class, and I wasn't able to figure out why each of the box's 6 planes has (1,0), (0,0), (0,1), and (1,1) for its texture data. Can anyone explain this to me please or point me to a tutorial of some sort that explains this?
My second idea would be to create a custom Lego class that simply utilizes the Box and Cylinder classes. However, that would need a few customizations, as I'd have to get rid of the bottom plane of the Box, and each Cylinder would need a closed top and open bottom. Would this be the simpler way to go, or should I not consider it because of performance issues or anything like that? Any tips as to how to go about this approach, such as whether or not to use the SharedNode class to render the cylinders?
I'm not sure which approach to use, and I'm very new at this, so any help would be greatly appreciated. If I could be pointed to a tutorial on making custom objects in jMonkey, that'd be great, too.
I strongly recommend modelling the brick in something like blender, 3D Studio Max or Maya, exporting it in some format jme can read (collada, obj, md5, …) and just loading the model into the application.
An alternative, if you want to support parametric blocks, is to generate the vertex and index data into arrays in memory, and then create the TriMesh object with the vertex, normal and index data arrays as FloatBuffer/IntBuffer arguments. See the constructor for TriMesh(). You can generate triangles for the walls, cylinders, etc, all using math. For example, you could write a class something like:
```public class BlockData { TriMesh theTriMesh; public BlockData(int width, int height) { float[] vertices = new float[ number of vertices for the given size * 3 ]; float[] normals = new float[ number of vertices for the given size * 3 ]; int[] indices = new int[ number of triangles for the given size * 3 ]; generateData(width, height, vertices, normals, indices); theTriMesh = new TriMesh(vertices, normals, null, null, indices); } public TriMesh getTriMesh() { return theTriMesh; } } ```
The trick is, of course, to write the "generateData" function and by extension the function that calculates the needed amount of data. If you don't yet know how meshes are put together of vertex data and indices that generate triangles, then I suggest you go about finding a basic 3D mesh tutorial first. Then you can google for something like "parametric geometry generation."
If that's too much work for you, then model each size brick in a modeler like Blender, MilkShape, DeleD, 3D Studio Max or Maya, and import it as a model (.obj, .ase or similar format). The draw-back is that you can't make bricks of arbitrary size; you have to model each.
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http://nrich.maths.org/5997/note?nomenu=1
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## Sorting Shapes
These shapes can be sorted in many different ways.
In this box there is a collection of triangles:
In this box there is a collection of small blue shapes:
Now you can make your own. Think of a good name for each of your collections.
Full screen version
If you can see this message Flash may not be working in your browser
Please see http://nrich.maths.org/techhelp/#flash to enable it.
### Why do this problem?
This problem provides an opportunity for children to sort and categorise, both of which are important mathematical processes. This open activity challenges children to find their own categories and then name them. This means it is a good way to introduce specific vocabulary associated with properties of shape, or to remind children of relevant mathematical language.
### Possible approach
You could begin by gathering the class together, perhaps on the floor where everyone can see, and showing them a set of 2D shapes. (These can be any set that you have, they don't have to match those in the interactivity!) Gather some of them together and ask the children what your group of shapes has in common. Talk with them about different ways to describe the set of shapes and try some other groups. Alternatively (or in addition), you could use the interactivity with the whole class for a similar purpose.
Ideally, children should have a set of shapes or logic blocks to handle in pairs so that they can move and sort physically. However, if you have access to a computer suite, you might want to ask learners to work in pairs at a screen. (If you do have sets of old Dienes' Logiblocks, then you'll notice they come in different thicknesses too, which adds another variable.) Listen out for children who are clearly justifying their reasons for grouping particular shapes and encourage (or introduce) useful vocabulary.
As a plenary, you could ask some pairs of children to share one of their sets for all to see on the whiteboard. Alternatively, you could challenge pairs to hold up a shape which fits your description, for example "show me a small, red circle".
You could try a similar activity with 3D shapes, too.
### Key questions
What is the same about these two blocks?
Can we find others that could go with them?
What could we call this collection?
### Possible extension
A challenging extension to this activity could be to ask children to draw another different shape which goes with their set.
### Possible support
Some children might need prompting to sort particular groups to begin with, for example "find all the squares". However, as this is an open task, they have the scope to make it as straightforward or difficult as they like, so some children may well surprise you!
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http://money.cnn.com/calculator/retirement/retirement-need/?source=yahoo_hosted
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# Will you have enough to retire?
Methodology
This calculator estimates how much you'll need to save for retirement. To make sure you're thinking about the long haul, we assume you'll live to age 92. But you could live to be 100 or incur large medical bills early on in retirement that may raise your costs even further. Social Security is factored into these calculations, but other sources of income, such as pensions and annuities, are not. All calculations are pre-tax.
The results offer a general idea of how much you'll need and are not intended to be investment advice. The results are presented in both future dollars (at retirement) and today's dollars, which is calculated using an inflation rate of 2.3%.
How we calculate your savings goal
First, we determine what your income will be at the time you retire by growing your current income at an annual rate of 3.8% (the inflation rate of 2.3%, plus the salary growth rate of 1.5%). We then assume you can live comfortably off of 85% of your pre-retirement income. So if you earn \$100,000 the year you retire, we estimate you will need \$85,000 during the first year of retirement. For each subsequent year, we increase your income need by 2.3% to keep up with inflation. We then factor in Social Security by subtracting your estimated benefits (more on that below) since that income will reduce the amount you will need to save.
The second step is to calculate the total savings you will need at the time you retire, in order to generate enough income for each year of retirement. To do this, we determine what it would cost to purchase a fixed income annuity, with inflation-adjusted payments, using a discount rate (or rate of return) of 6%. The cost to purchase this hypothetical annuity is your target savings goal.
How we calculate the amount you will save
To figure out how much you will save by the time you retire, we first estimate your future income by growing your current income at a rate of 3.8% (the inflation rate of 2.3%, plus the salary growth rate of 1.5%). Then, we determine what the sum of your annual contributions will be between now and retirement. We assume your current savings and future contributions are invested and will earn an average annual rate of return of 6%.
How we estimate Social Security benefits
We estimate your Social Security benefits based on the assumption that you will have worked at least 35 years and will start collecting benefits at age 67. For most people who are working today, that's considered full retirement age. If you plan on retiring after age 67, we assumed the benefits are invested (along with your savings) and grown at the same average rate of return of 6%. We use your estimated pre-retirement income to calculate your estimated annual Social Security benefits, based on current benefit formulas and accounting for inflation. To better understand your actual Social Security benefits, please visit www.ssa.gov.
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https://www.jiskha.com/questions/560132/prove-that-2sin-pi-6-cosec-7pi-6cos-pi-3-3-2
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# trignometry
Prove That 2sin^pi/6 + cosec^7pi/6cos^pi/3 = 3/2
1. 👍 1
2. 👎 0
3. 👁 179
1. Your expression makes no sense
We use the ^ symbol as an exponent sign.
so your 2 sin^π/6 would be
2(sin)^(π/6) which is meaningless
if you simply meant:
2 sin(π/6) + cosec(7π/6)cos(π/3)
that would be
2(1/2) + (-2)(1/2) = 0
1. 👍 0
2. 👎 0
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# Physics 434, 2012: Lecture 14
Back to the main Teaching page.
Back to Physics 434, 2012: Information Processing in Biology. Good lecture notes on the subject are in Bialek's draft textbook, version 2011.
## Warmup question
1. E. coli and many other bacteria exhibit a phenomenon of persistence, so elegantly studied in Balaban et al., 2004. Briefly bacteria may choose to switch from a growing to a non-growing phenotype, when they are less sensitive to effects of antibiotics and persist through antibiotic applications. This is advantageous when antibiotics are applied, but they don't grow in the persistent state, making it a loosing choice when there are no antibiotics. Leaving aside the issue of how they actually switch (we will study a related problem a bit later in class), can you predict what should the bacterial strategy be in choosing whether to divide or not to divide? Which considerations should enter the bacterial decision?
## Main Lecture
### Population biology
• Once switched, a bacterium remains committed to the state for a while. We will discretize the time and consider the bacterium in making a choice 0 (growing) and 1 (persistent) once per time step. The time step is much longer than the switching time and the division/death time of a single bacterium.
• Each time step does (1) or doesn't (0) have an application of an antibiotic, and the bacterium can be in the persistent (1) or the growing state (0).
• For the state of bacteria and the world ${\displaystyle [bw]=\{[00],[01],[10],[11]\}}$, at the end of the time period, the number of the bacteria will be multiplied by the growth rate of the bacteria, which we call ${\displaystyle \alpha _{bw}}$. We choose ${\displaystyle \alpha _{00}=\alpha ^{(0)}}$ -- growth in unhindered conditions; ${\displaystyle \alpha _{01}=0}$ -- growing bacteria die in antibiotic world; ${\displaystyle \alpha _{10}=\alpha _{11}=1}$ -- persistent bacteria persist in the world that either has or doesn't have the antibiotics.
• We suppose that ${\displaystyle P(w=1)=p}$ (antibiotic probability), and ${\displaystyle P(w=0)=1-p=q}$.
• If the bacterium can sense presence/absence of antibiotics and switch appropriately, then for ${\displaystyle \alpha ^{(0)}>1}$ the bacterial growth rate in a single time step is ${\displaystyle {\bar {\alpha }}=\alpha ^{(0)}(1-n_{i})+n_{i}}$, where ${\displaystyle n_{i}=(0,1)}$ is a random variable describing if antibiotics is present or absent in this time step. Over many trials, the population if ${\displaystyle K_{0}}$ individuals at the beginning will grow as ${\displaystyle K(N)=K_{0}\prod _{i=1}^{N}\left[\alpha ^{(0)}(1-n_{i})+n_{i}\right]=K_{0}\prod _{i=1}^{N}2^{\left[(1-n_{i})\log _{2}\alpha ^{(0)}+n_{i}\right]}=K_{0}2^{\Lambda _{\rm {best}}N}}$.
• The growth rate ${\displaystyle \Lambda _{\rm {best}}={\frac {1}{N}}\sum \left[(1-n_{i})\log _{2}\alpha ^{(0)}+n_{i}\right]}$. By the central limit theorem, ${\displaystyle {\frac {1}{N}}\sum n_{i}\to p}$. Thus ${\displaystyle \Lambda \to (1-p)\log _{2}\alpha ^{(0)}+p}$. This is the fastest possible growth rate. However, as argued by Kussel and Leibler (2005), sensing the environment is resource-consuming and might not be useful in all situations, and the bacteria might instead prefer to switch randomly, hedging their bets, rather than sensing.
• Suppose that the bacterium is going to switch randomly at the beginning of every time step. The fraction ${\displaystyle f}$ of the population will be in the persistent state, and ${\displaystyle 1-f}$ in non-persistent. The actual state of the world will be ${\displaystyle n_{i}}$. Then the growth rate of the colony at time step ${\displaystyle i}$ is ${\displaystyle \alpha _{i}=\alpha ^{(0)}(1-f)(1-n_{i})+f=\left[\alpha ^{(0)}(1-f)+f\right](1-n_{i})+fn_{i}=2^{(1-n_{i})\log _{2}\left[\alpha ^{(0)}(1-f)+f\right]+n_{i}\log _{2}f}}$. After ${\displaystyle N}$ time steps, the whole population is ${\displaystyle K(N)=K_{0}\prod _{i=1}^{N}2^{(1-n_{i})\log _{2}\left[\alpha ^{(0)}(1-f)+f\right]+n_{i}\log _{2}f}\equiv K_{0}2^{\Lambda N}}$.
• ${\displaystyle \Lambda ={\frac {1}{N}}\sum _{i}\left\{(1-n_{i})\log _{2}\left[\alpha ^{(0)}(1-f)+f\right]+n_{i}\log _{2}f\right\}}$. By central limit theorem ${\displaystyle {\frac {1}{N}}\sum n_{i}\to p}$. Thus ${\displaystyle \Lambda \to (1-p)\log _{2}\left[\alpha ^{(0)}(1-f)+f\right]+p\log _{2}f}$.
• To find the optimum fraction of bacteria that should choose to be persistent at any time, we maximize the growth rate ${\displaystyle \Lambda }$. We do ${\displaystyle \partial \Lambda /\partial f}$. Solving the ensuing equation, we get for the optimal ${\displaystyle f}$ ${\displaystyle f_{0}=\min \left\{{\frac {p\alpha ^{(0)}}{\alpha ^{(0)}-1}},1\right\}}$.
• ${\displaystyle f_{0}}$ is proportional to ${\displaystyle p}$ -- special case of matching, see Gallistel et al. 2001.
• When ${\displaystyle \alpha ^{(0)}<1/(1-p)}$, that is, the maximum growth is not too fast, ${\displaystyle f_{0}=1}$ -- it makes sense not to hedge bets and always stay in the persistent state.
• Plugging the result into the growth rate formula (for large ${\displaystyle \alpha ^{(0)}}$), we get:${\displaystyle \Lambda _{\rm {max}}=\Lambda _{\rm {best}}-S[p]-p(1-\log _{2}(\alpha ^{(0)}/(1-\alpha ^{(0)})}$. Note that the growth rate decreases by ${\displaystyle S[p]}$, the amount of what the bacterium doesn't know about the world. The amount of information about the outside world is actually important in the evolutionary sense: if you know the world, you multiply faster.
### Building effective models
• Information theory provides a measure for characterization of quality of input output relations. But in addition, due to the data processing inequality, it also provides ways of unambiguously reducing dimensionality of the modeled biological system.
• (Indeed, say we have a large-dimensional signal ${\displaystyle {\vec {s}}}$ and response ${\displaystyle {\vec {r}}}$. There's a certain mutual information between these ${\displaystyle I[{\vec {s}};{\vec {r}}]=I_{0}}$. If we propose a reduction of the signal and response to ${\displaystyle f=f({\vec {s}}),g=g({\vec {r}})}$, then ${\displaystyle I[f;g]\leq I[{\vec {s}};{\vec {r}}]}$ by the data processing inequality.
• We can, for example, solve the problem like: Which inputs are informative of the outputs (and hence need to be accounted for in a model)? We omit different subsets of the inputs ${\displaystyle s_{i}}$, calculate ${\displaystyle I[{\vec {s}}-s_{i};{\vec {r}}]=I_{\not i}}$, and calculate the error due to omitting this signal ${\displaystyle \Delta _{i}={\frac {I_{0}-I_{\not i}}{I_{0}}}}$. Those components that have a small ${\displaystyle \Delta _{\not i}}$ can be safely neglected. This type of analysis can be used, for example, to understand which features of the neural code are important. In the subsequent presentation by Farhan, we will hear about using this trick to understand if high precision of neural spikes is important or not.
• The ARACNE algorithm.
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Topic: A box contains 4 red chips and 2 blue chips. If two chips are
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A garage has a stock of side mirrors. The ratio of right side mirrors to left side mirrors is 5:3. Iris, a garage worker, attaches pairs of the left and right mirrors with an average with an adhesive tape until no more pairs can be made. If 30 right mirrors are left unpaired, how many left and righ...
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## due in 3 hours 1050 word count
You are starting your own Internet business. You decide to form a company that will sell cookbooks online. Justcookbooks.com is scheduled to launch 6 months from today. You estimate that the annual cost of this business will be as follows:
Part I
Deliverable Length: 1 graph plus calculations
You must give up your full-time job, which paid \$50,000 per year, and you worked part-time for half of the year.
The average retail price of the cookbooks will be \$30, and their average cost will be \$20.
Assume that the equation for demand is Q = 40,000 – 500P, where
Q = the number of cookbooks sold per month
P = the retail price of books.
Show what the demand curve would look like for price between \$25 and \$35.
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https://math.stackexchange.com/questions/3290827/square-root-of-a-derivative-operator
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# "Square root" of a derivative operator?
I have been studying derivatives as operators on a function. Specifically, how we may write, for a function $$f(x)$$
$$\frac{df}{dx}$$
as
$$Df$$
where $$D$$ denotes $$\frac{d}{dx}$$. I've seen how successive application of an operator is seen as multiplying the operators together to give rise to a new operator. For example
$$\frac{d^2f}{dx^2}$$
can be written as $$D^2f$$.
What I have been pondering is this: Consider a certain operator $$\Phi$$, which has the property that
$$\Phi^2f=Df$$
i.e the $$\Phi$$ operator acts like a sort of "square root" of the derivative operator. This would imply that
$$\Phi f = h$$
and
$$\Phi h = g$$
where $$g(x)$$ and $$h(x)$$ are functions and that
$$\frac{df}{dx}=g$$
My question is, has such an operator been studied before? Does it have a unique name? I do not necessarily belive $$\Phi$$ to be unique, perhaps many unique operators may fill the role of $$\Phi$$. But can we determine if such an operator is unique or not?
• You mean this? en.wikipedia.org/wiki/Fractional_calculus Jul 12, 2019 at 10:46
• @MattiP. wow thank you for that link. Can you please somehow convert it into an answer so I can accept it? Jul 12, 2019 at 10:47
• Jul 12, 2019 at 11:08
• There is a nice channel on youtube by Dr. Peyam. He has several videos related to the fractional calculus and the reasoning behind defining some of the concepts. Here is a playlist for reference Jul 12, 2019 at 14:08
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# maths
what fraction of 30p is 5p?
1. 👍
2. 👎
3. 👁
4. ℹ️
5. 🚩
1. 5/30
Divide the numerator and denominator by 5 to find the fraction in its lowest terms.
5/30 = 1/6
1. 👍
2. 👎
3. ℹ️
4. 🚩
👤
Ms. Sue
2. What fraction of 30p is 5p
1. 👍
2. 👎
3. ℹ️
4. 🚩
3. £2 is 30p
1. 👍
2. 👎
3. ℹ️
4. 🚩
4. Wow amazing website. I TOTALLY RECOMMEND THIS AMAZING WEBSITE!! If you haven't checked out this website, try it out NOW!!!Btw, im not a bot ^~^
1. 👍
2. 👎
3. ℹ️
4. 🚩
5. Dong het it
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2. 👎
3. ℹ️
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6. Sorry dont get it
Thanks
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## Similar Questions
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# Treatise on Geometry and Trigonometry: For Colleges, Schools and Private Students
Van Antwerp, Bragg & Company, 1868 - Geometry - 420 pages
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### Contents
LOGICAL TERMS 9 PROBLEMS 17 PART SECOND PLANE GEOMETRY 28 PERPENDICULAR AND OBLIque Lines 38 CHAPTER IV 52 TANGENTS 68 POSITIONS OF TWO CIRCUMFERENCES 78 CHAPTER V 85
PYRAMIDS 222 MEASURE OF VOLUME 232 SIMILAR POLYEDRONS 239 CHAPTER XI 245 SPHERES 250 SPHERICAL AREAS 261 SPHERICAL VOLUMES 271 PART FOURTH TRIGONOMETRY 277
EQUALITY OF TRIANGLES 93 SIMILAR TRIANGLES 101 CHAPTER VI 119 MEASURE OF AREA 128 EQUIVALENT SURFACES 135 CHAPTER VII 143 REGULAR POLYGONS 151 ISOPERIMETRY 159 RECTIFICATION OF THE CIRCUMFERENCE 172 DIEDRAL ANGLES 185 TRIEDRALS 195 POLYEDRALS 209
CONSTRUCTION AND USE OF TABLES 296 RIGHT ANGLED TRIANGLES 302 CHAPTER XIII 314 RIGHT ANGLED SPHERICAL TRIangles 324 CHAPTER XIV 334 383 1 722 8 52 2 17 8 58 11 122 56
### Popular passages
Page 98 - If two triangles have two sides of the one equal to two sides of the...
Page 182 - ... the plane at equal distances from the foot of the perpendicular, are equal...
Page 141 - The square described on the hypothenuse of a right-angled triangle is equivalent to the sum of the squares described on the other two sides.
Page 91 - Conversely, if two angles of a triangle are equal, the sides opposite them are also equal, and the triangle is isosceles.
Page 84 - If a circle have any number of equal chords, what is the locus of their points of bisection? 21. If any point, not the center, be taken in a diameter of a circle, of all the chords which can pass through that point, that one is the least which is at right angles to the diameter. 22. If from any point there extend two lines tangent to a circumference, the angle contained by the tangents is double the angle contained by the line joining the points of contact and the radius extending to one of them....
Page 117 - ABC, so that DE shall be equal to the difference of BD and CE. 22. In a given circle, to inscribe a triangle similar to a given triangle. 23. In a given circle, find the locus of the middle points of those chords which pass through a given point. 25. If a line bisects an exterior angle of a triangle, it divides the base produced into segments ^A which are proportional to the adjacent sides.
Page 307 - The sum of any two sides of a triangle is to their difference, as the tangent of half the sum of the angles opposite to those sides, to the tangent of half their difference.
Page 237 - The volume of any prism is equal to the product of its base by its altitude. Let V denote the volume, B the base, and H the altitude of the prism DA'.
Page 233 - The volume of a rectangular parallelepiped is equal to the product of its three dimensions.
Page 126 - Theorem. — Two parallelograms are equal when two adjacent sides and the included angle in the one, are respectively equal to those parts in the other.
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# PARABOLAS Topic 7.2.
## Presentation on theme: "PARABOLAS Topic 7.2."— Presentation transcript:
PARABOLAS Topic 7.2
Definition The set of all points in a plane that are the same distance from a given point called the focus and a given line called the directrix.
Writing linear equation in parabolic form
GOAL: Turn general form Standard form
Writing linear equation in parabolic form
Start with Group the two x-terms Pull out the constant with x2 from the grouping Complete the square of the grouping **Look back to Topic 6.3 for help** Write the squared term as subtraction so that you end with standard form
Group x-terms Pull out GCF Complete the Square **Remember that whatever you add in the grouping must be subtracted from the c-value** Factor and simplify
Summary of Parabolas in Standard Form
Equation Axis of symmetry x = h y = k Vertex (h, k) Focus Directrix Direction of opening Up: a>0, Down: a<0 Right: a>0, Left: a<0 Latus Rectum
Graph of prior example
You Try!! Write the following equation in parabolic form. Then identify all of the parts. Vertex: (5, -32) Focus: (5.25, -32) Axis of Symmetry: Latus Rectum: 1 X=5 Direction of Opening: Right Directrix: x= 4.75
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https://health.icalculator.com/army-body-fat-calculator.html
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# Army Body Fat Percentage Calculator
The Army Body Fat Percentage Calculator is an essential tool in the military used to assess the fitness level of personnel by calculating the percentage of their body fat. This tutorial elucidates the concept, calculations, and formulas involved, interesting facts, and its applications outside the military.
🖹 Normal View 🗖 Full Page View Male Female Waist Size Inches Neck Size Inches Hip Size Inches Height Inches
Army Body Fat = %
An interesting fact about this calculator is that it was developed to ensure that personnel meet the fitness and readiness standards of the Army. It helps the Army maintain an optimal level of combat readiness by ensuring that personnel do not exceed the maximum allowable body fat percentages.
The formula used by the Army Body Fat Percentage Calculator is based on the U.S. Army body fat standards and considers factors such as gender, age, weight, and measurements of the neck, waist, and hips (for women). For men, the formula is:
Body Fat Percentage = 86.010 × log10(abdomen - neck) - 70.041 × log10(height) + 36.76
For women, the formula is:
Body Fat Percentage = 163.205 × log10(waist + hip - neck) - 97.684 × log10(height) - 78.387
Beyond the military, these calculations are relevant in health and fitness industries to assess and monitor the body composition of individuals. The calculation of body fat percentage is crucial in designing appropriate fitness and diet plans.
For instance, let's consider a male soldier who is 70 inches tall with a neck measurement of 16 inches and an abdominal measurement of 36 inches. Using these measurements in the formula:
Body Fat Percentage = 86.010 × log10(36 - 16) - 70.041 × log10(70) + 36.76 = 18.6%
Col. Pauline Santiano, a prominent figure in military health and fitness, has excelled in implementing and enforcing the Army's physical fitness and body composition standards. Her dedication and efforts have led to the enhancement of physical readiness and health among Army personnel.
## Health Calculators
You may also find the following Health Calculators useful.
## Use of the Health and Medical Calculators
Please note that the Army Body Fat Calculator is provided for your personal use and designed to provide information and information relating to the calculations only. The Army Body Fat Calculator should not be used for you to self-diagnose conditions, self-medicate or alter any existing medication that you are currently prescribed by your Doctor. If the Army Body Fat Calculator produces a calculation which causes you concern, please consult your Doctor for support, advice and further information.
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http://superuser.com/questions/147683/how-to-plot-survey-results-on-excel
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# How to plot survey results on Excel?
I would like to plot the average of these data samples:
``````1. Contribution to the project
2. Affiliation with project owners
3. Level of expertise
4. Learning Curve
Yes No 3 2
No No 3 2
No No 4 3
No No 5 3
No No 3 3
Yes Yes 5 4
No No 4 3
No No 3 2
No Yes 2 4
No No 5 2
No No 2 2
No No 3 3
...
``````
I know how to do the average of each numeric column, but I have a few questions for you:
• If the column contains non numeric values (e.g. "no answer") the average doesn't work. Is there a workaround for this?
• How can I count the yes/no results and plot only a percentage?
-
If the column contains non numeric values (e.g. "no answer") the average doesn't work. Is there a workaround for this?
I assume you're doing an average like this: `=AVERAGE(C1:C13)`. If so, the formula will simply skip that cells that have text in them. So, in your example, you have 13 rows, the average formula will be dividing by 12, rather than 13. You can check this by replacing the "No answer" with a 0:
``````AVERAGE(C1:13) = 2.75 (has "No Answer")
AVERAGE(C1:13) = 2.54 (has a 0)
``````
So it's up to you whether you want to count the people who didn't answer as zeros, or just not include them in your statistics.
How can I count the yes/no results and plot only a percentage?
This one is easy - you'll use Excel's `COUNTIF` formula.
In a new set of cells, you'll want to count each yes or no. For your example, I put my yes in K2 and No in K3. You'll want to use the `COUNTIF` like so:
``````For the Yes row (K2): =COUNTIF(A:A, "Yes")
For the No row (K3): =COUNTIF(A:A, "No")
``````
Then just use a pie chart referencing the countif cells, and it should look like this:
-
+1 Nice. I was figuring how to go about the Yes/No condition. Although, in OpenOffice. – Sathya Jun 1 '10 at 17:33
Great answer. I would be tempted to adapt the countif to count "yes" and "not yes" to make sure to catch any oddities that might occur in the data now or in the future (such as "no answer"). So for example using =COUNTIF(A:A, "<>Yes") should work. Only problem then of course is if someone enters "Y" instead of "Yes" it still does not count as a yes (no worse than the current method, but possibly more misleading). – AdamV Jun 1 '10 at 20:52
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https://www.askiitians.com/forums/Engineering-Entrance-Exams/a-sonometer-string-and-a-tuning-fork-when-sounded_84403.htm
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A sonometer string and a tuning fork when sounded together gives 6 beats/sec, whether the length of sonometer string is 85 cm or 100 cm. the frequency of tuning fork is (a) 262 Hz (b) 256 Hz (c) 260 Hz (d) 234 Hz
Arun
25750 Points
4 years ago
Probable frequencies of tuning fork be
n+4
or
n4
Frequency of sonometer wire
n1l
n+4n4=10095
or
95(n+4)=100(n4)
or
95n+380=100n400
or
5n=780
or
n=156
Vikas TU
14149 Points
4 years ago
Letn = Frequency of the tuning fork
n1=Frequency of the wire corresponding to length of L1 (39.5 cm)
n2=Frequency of the wire corresponding to length of L2 (40.5 cm)
The frequency of wire is inversely proportional to its length
Hence,n1>n2
So, n1/n2=n+4 / n−4 [ since number of beats = 4]
But, n1/n2=L2/L1
n+4 / n−4 = 40.5/39.5
40.5n−162=39.5n+158
n=320 HzAns:Frequency of the tuning fork=320 Hz
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# Nibbles to Kilobits - 324 Nibble to kbit Conversion
Input Nibble - and press Enter.
Nibble
You are converting Nibble to Kilobit, a decimal unit.
verified_user RESULT =
324 Nibble = 1.296 kbit
( Equal to 1.296E+0 kbit )
content_copy
Calculated as → 324 x 4 ÷ 1000... - View details
## Nibble to Kilobit (kbit) Conversion Formula and Steps
Nibble to kbit Calculator Tool allows you to easily convert from Nibble to Kilobit (kbit). This converter uses the below formula and steps to perform the conversion.
The formula of converting the Nibble to Kilobit (kbit) is represented as follows :
diamond CONVERSION FORMULA kbit = Nibble x 4 ÷ 1000
Source Data Unit Target Data Unit
Nibble
Equal to 4 bits
(Basic Unit)
Kilobit
Equal to 1000 bits
(Decimal Unit)
Now let us apply the above formula and see how to manually convert Nibble to Kilobit (kbit). We can further simplify the formula to ease the calculation.
FORMULA
Kilobits = Nibbles x 4 ÷ 1000
STEP 1
Kilobits = Nibbles x 0.004
If we apply the above Formula and steps, conversion from 324 Nibble to Kilobit (kbit) will be processed as below.
1. = 324 x 4 ÷ 1000
2. = 324 x 0.004
3. = 1.296
4. i.e. 324 Nibble is equal to 1.296 kbit.
Note : Result rounded off to 40 decimal positions.
You can use above formula and steps to convert Nibbles to Kilobits using any of the programming language such as Java, Python or Powershell.
### Unit Definitions
#### Nibble
A Nibble is a unit of digital information that consists of 4 bits. It is half of a byte and can represent a single hexadecimal digit. It is used in computer memory and data storage and sometimes used as a basic unit of data transfer in certain computer architectures.
arrow_downward
#### Kilobit
A Kilobit (kb or kbit) is a decimal unit of digital information that is equal to 1000 bits. It is commonly used to express data transfer speeds, such as the speed of an internet connection and to measure the size of a file. In the context of data storage and memory, the binary-based unit of Kibibit (Kibit) is used instead.
## Excel Formula to convert from Nibble to Kilobit (kbit)
Apply the formula as shown below to convert from 324 Nibble to Kilobit (kbit).
A B C
1 Nibble Kilobit (kbit)
2 324 =A2 * 0.004
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Nibble to Kilobit (kbit) Conversion
You can use below code to convert any value in Nibble to Nibble in Python.
nibbles = int(input("Enter Nibbles: "))
kilobits = nibbles * 0.004
print("{} Nibbles = {} Kilobits".format(nibbles,kilobits))
The first line of code will prompt the user to enter the Nibble as an input. The value of Kilobit (kbit) is calculated on the next line, and the code in third line will display the result.
## Similar Conversions & Calculators
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## Future value calculator time
Future Value Calculator - The value of an asset or cash at a specified date in the future that is equivalent in value to a specified sum today.
Iteration - by calculating the future value for different values of interest rate or time , one gradually can converge on the solution. Financial calculator or Inflation Calculator, Future Value Calculator helps you calculate the future value of money based on the Inflation rate. eg You can calculate the value of 1 lakh You can use the future value of money calculator to compute the amount which is to be invested, manually as well but it is time-consuming. To efficiently and 10 Nov 2015 Several financial planning calculators are available on the web. The interesting point is that your investment grew over four times in 20 years. It is important to know what will be the future value of, say, today's Rs 10,000, The future cost calculator computes the value by just inputting figures such as present cost, time period, and inflation rate. Steps in using future cost calculator. The Our FV calculator computes the future value of your investment based on the rate of interest on the principal amount, time period. To understand the future value Calculate the Future Value of your Initial and Periodic Investments with you to budget and be disciplined with your spending habits, and many times it can this investment calculator will help you find out the future value of your investment .
## 5 Mar 2018 According to the time value of money, a dollar in hand today is worth more than a dollar received at a certain point in the future. That's because
Future value (FV) calculator is an online investment return value estimation tool to calculate future time value of money or asset. Generally the asset value is calculated in equivalent value of money. Using the future value calculator. This calculator can help you calculate the future value of an investment or deposit given an initial investment amount, the nominal annual interest rate and the compounding period. Optionally, you can specify periodic contributions or withdrawals and how often these are expected to occur. Future Value Calculator Use this calculator to determine the future value of an investment which can include an initial deposit and a stream of periodic deposits. Javascript is required for this calculator. Present Value Calculator. This present value calculator can be used to calculate the present value of a certain amount of money in the future or periodical annuity payments.
### Future Value Calculator This future value calculator forecasts the FV of a present value/deposit or investment over a specific period of time, with or without annuity payments (ordinary / due). There is in depth information about this indicator below the tool.
1 Apr 2016 So how do we tackle the question of value over time? Future Value. Let's take our \$1,000 today and see what that might be worth in a year's time Calculate future value (FV) based on present value (PV), rate of return (R), and time (t) in years with present value amortization table. Use this Present Value Calculator to get the present value PV. Indicate the future value, the interest rate, number of years n and the type of compounding. Let k k k be the number of times the money is compounded in a year. For example, for 15 Nov 2019 The present value calculator estimates what future money is worth now. That is to say, the present value of \$120 if your time-frame is 3 years Understanding the calculation of present value can help you set your retirement saving goals and app, or any other financial calculator app that you can download to your smartphone or tablet. Retirement and the Unknown Time Horizon.
### 15 Nov 2019 The present value calculator estimates what future money is worth now. That is to say, the present value of \$120 if your time-frame is 3 years
Calculate Future Value. To help you in calculating the sum of money you would receive if you invest an amount now at an assumed compounded rate for a “N”. Total number of payments periods. “I/Y”. Annual interest rate. “PV”. Present Value. “FV”. Future Value. “PMT”. Payment amount. “?” Down arrow on calculator \$100 in a year. Your \$1,000 now can become \$1,100 in a year's time. We say the Present Value of \$1,100 next year is \$1,000 calculator exponent button . 1 Apr 2016 So how do we tackle the question of value over time? Future Value. Let's take our \$1,000 today and see what that might be worth in a year's time Calculate future value (FV) based on present value (PV), rate of return (R), and time (t) in years with present value amortization table. Use this Present Value Calculator to get the present value PV. Indicate the future value, the interest rate, number of years n and the type of compounding. Let k k k be the number of times the money is compounded in a year. For example, for
## Present value is the value right now of some amount of money in the future. It's based upon the best risk-free interest rate you could get now for the time period
You can use the future value of money calculator to compute the amount which is to be invested, manually as well but it is time-consuming. To efficiently and 10 Nov 2015 Several financial planning calculators are available on the web. The interesting point is that your investment grew over four times in 20 years. It is important to know what will be the future value of, say, today's Rs 10,000, The future cost calculator computes the value by just inputting figures such as present cost, time period, and inflation rate. Steps in using future cost calculator. The
Time value of money calculators to determine relative worth, present value of money versus future value of money. Calculate present value of lump sum and investments, and future value of investments given interest earned and inflation variables.
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# Having trouble getting triangle to speed up?
I’m trying to get the triangle to speed up after it hits the wall, but I’m unsure how to do that.
``````//variables
float xa=30;float ya=75;float xb=58;float yb=20;
float xc=86;float yc=75;float xSpeed=10;float ySpeed=10;
void setup(){
size(800,600);//set up screen size 800 pixels by 600 pixels
}
void draw (){
background (0);//refreshing the background
fill (225,225,225);
textSize(18);
text("Cliking the mouse on the background might slow down the object",110,20);
Triangle();
}
void Triangle(){//user defined function for Triangle
fill(random(255),random(255),random(255));//selects a radom mix of red, blue and green
triangle (xa,ya,xb,yb,xc,yc);
//have all of the points moving
xa=xa+xSpeed;
xb=xb+xSpeed;
xc=xc+xSpeed;
ya=ya+ySpeed;
yb=yb+ySpeed;
yc=yc+ySpeed;
if (xa>width || xa<0 || xb>width || xb<0 || xc>width || xc<0){ //if any of the X values goes out of bounds, reverse the speed
xSpeed=xSpeed*-1;
}
if (ya>height || ya<0 || yb>height || yb<0 || yc>height || yc<0 ){ //if any of the Y values goes out of bounds, reverse the speed
ySpeed=ySpeed*-1;
}
}
void mouseReleased() {//when mouse is released then pressed there is a radom chance that the triangle will slow down
if (.5<random(1)){
xSpeed=1;
ySpeed=1;
}
else{
xSpeed=10;
ySpeed=10;
}}
``````
1 Like
Hi ant,
You almost already have all the code for that
Since you want that event to appear after you hit the walls, you need to detect when the triangle hits the wall.
``````if (xa>width || xa<0 || xb>width || xb<0 || xc>width || xc<0){
xSpeed=xSpeed*-1;
}
if (ya>height || ya<0 || yb>height || yb<0 || yc>height || yc<0 ){
ySpeed=ySpeed*-1;
}
``````
Now, your new speed depends on the old one. So if the previous speed was slow, the new one will also be slow. But all you really care about the old speed is its direction so you can inverse it.
So let’s just keep the sign and put the speed we want instead:
`````` if (xa>width || xa<0 || xb>width || xb<0 || xc>width || xc<0){
xSpeed = 10 * (xSpeed/abs(xSpeed)) * -1;
ySpeed = 10 * (ySpeed/abs(ySpeed));
}
if (ya>height || ya<0 || yb>height || yb<0 || yc>height || yc<0 ){
ySpeed = 10 * (ySpeed/abs(ySpeed)) * -1;
xSpeed = 10 * (xSpeed/abs(xSpeed));
}
``````
1 Like
Would there be a way to continuously make it faster each time it hits the wall not just the initial hit of the wall?
Sure!
Here I put 10 as the speed but you can put any numbers that you want. And especially you can put here a variable. If so, you can increase the value of that variable every time you hit the wall and get the desired effect.
But I’ll leave it as an exercise for you
1 Like
Alright, I got the triangle to speed up and slow down when hitting different axis, but now my code for mouse released when you can stop the triangle sometimes the triangle gets stuck on the edges or disappears completely.
``````//variables
float xa=30;float ya=75;float xb=58;float yb=20;
float xc=86;float yc=75;float xSpeed=5;float ySpeed=5;
void setup(){
size(800,600);//set up screen size 800 pixels by 600 pixels
}
void draw (){
background (0);//refreshing the background
fill (225,225,225);
textSize(18);
text("Cliking the mouse on the background might slow down the object",110,20);
Triangle();
}
void Triangle(){//user defined function for Triangle
fill(random(255),random(255),random(255));//selects a radom mix of red, blue and green
triangle (xa,ya,xb,yb,xc,yc);
//have all of the points moving
xa=xa+xSpeed;
xb=xb+xSpeed;
xc=xc+xSpeed;
ya=ya+ySpeed;
yb=yb+ySpeed;
yc=yc+ySpeed;
if (xa>width || xa<0 || xb>width || xb<0 || xc>width || xc<0){//if any of the X values goes out of bounds, reverse the speed
//when the ball hit the x-axis it speeds up
xSpeed = 17 * (xSpeed/abs(xSpeed)) * -1;
ySpeed = 17 * (ySpeed/abs(ySpeed));
}
if (ya>height || ya<0 || yb>height || yb<0 || yc>height || yc<0 ){//if any of the Y values goes out of bounds, reverse the speed
//when ball hits the y-axis it slows down
ySpeed = 11 * (ySpeed/abs(ySpeed)) * -1;
xSpeed = 11 * (xSpeed/abs(xSpeed));
}
}
void mouseReleased() {//when mouse is released then pressed there is a radom chance that the triangle stop
if (.5<random(1)){
xSpeed=0;
ySpeed=0;
}
else{
xSpeed=5;
ySpeed=5;
}}
``````
1 Like
It is because of the way the logic is handled.
Let’s imagine we are in this case:
You triangle is moving toward the bottom right and the position is updated:
The bottom edge get below the screen so we enter your if statement and reverse the xSpeed. So if nothing happen, the following frame will look like this:
But instead, let’s imagine that the user clicked the mouse at the right time so we enter the mousePressed event. What happen is that your speed is override so the next frame will instead look like this:
Now the bottom edge is still below the screen so again we enter your if statement and reverse the xSpeed and in the next frame we update the triangle position. But because we went a bit lower that we should have to in the previous step, the up motion is not enough for the triangle to get above the screen edge:
So now, because we couldn’t get high enough to get above the screen, we still enter your if statement and we still reverse the xSpeed. This time it makes the triangle go down:
Then the speed is reverse again so you go up, then down again, then up, then down… in an endless loop.
2 Likes
So is there a way to make the triangle stop and start using mouseReleased or because of the reverse it’s not possible?
Everything is possible, and there are plenty of ways to overcome your problem. It really just depends on how you want to tackle the problem and which behavior you want to have.
For example, if you want a realistic bouncing, your logic is too simple and with big speed their might be some issues. The reason is that you are allowing the triangle to get out of the screen for a while. If the speed is too high (let’s say higher than the dimension of your triangle) then you might get some frame where the triangle is completely out of the screen.
One way to quickly solve your problem is to prevent the user to modify the speed if the triangle is outside of the screen. So you just need to copy your if statement inside the mousePressed event. (You can also use a global boolean variable to save some computation time but it’s not really useful here),
Sorry I’m a bit confused what if statement and where would it go in the mousePressed event.
What you want to do in mousePressed is:
``````if the shape is outside the screen:
do nothing;
else:
I can stop or change the speed of my triangles
``````
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Easyphotoeditor.org
Easyphotoeditor.org
## The Domino Effect
Domino is a word that describes a chain reaction in which one event sets off other events, like dominoes falling over each other. A good example of the Domino Effect is someone making a change to their lifestyle, such as cutting back on sedentary activities, which can then lead to a shift in other habits. For example, a person who cuts back on watching TV may also start eating healthier food as a side effect.
A domino is a game piece with an arrangement of spots on its face. Dominoes are cousins to playing cards and are one of the oldest tools for game play. They can be used to play a wide variety of games, including scoring and blocking games. Some of these games use the same rules as card games, while others require more skill and patience.
To win a domino game, a player must empty his hand and block opponents’ play. Normally, players do this by choosing dominoes from their boneyard that have matching numbers to the ones already played on the table. Once the player has done this, he must play that domino, usually by placing it so that its exposed ends touch other adjacent dominoes that have a number showing on them (one’s touching two’s and two’s touching three’s). Often, the points are awarded based on the number of dots exposed on each end of the laying domino.
Another way to play domino is by choosing a specific total to be reached. The first player to reach that total wins the game. A player can score by laying dominoes edge to edge, so that the exposed numbers match (one’s touching two’s and so on). Some players also place doubles at right angles to each other, in which case only the exposing edges count (one’s and two’s for example).
Dominoes are sometimes used to create art. For example, artists can draw a plan for their work and then use different colored dominoes to make it look 3D. This is a great way to show how a story works out and how it might end. This is something that many people do when they are writing their books, as it helps them keep track of the plot points and shows them how each part of the story relates to the next.
Whether you compose your novel off the cuff or take your time planning it out, it’s important to think about the Domino Effect. Plotting a novel is all about what happens next, and thinking about how your characters react to each other will help you develop a compelling narrative.
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http://plus.maths.org/content/searching-missing-truth
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# Searching for the missing truth
Many people like mathematics because it gives definite answers. Things are either true or false, and true things seem true in a very fundamental way.
Hugh Woodin.
The fact is, however, that mathematics has been moving on somewhat shaky philosophical ground for some time now. With the work of the logician Kurt Gödel and others in the 1930s it became clear that there are limits to the power of mathematics to pin down truth. In fact, it's possible to build different versions of mathematics in which certain statements are true or false depending on your preference. This raises the possibility that mathematics is little more than a game in which we choose the rules to suit our purpose. Good-bye to the beautiful Platonic notion of an eternal and independent mathematical truth.
But there may be some ground for optimism. At the International Congress of Mathematicians last year Plus talked to the mathematician Hugh Woodin, who believes that although many distinct mathematical "universes" are possible, one may soon be singled out as what you might consider the "right one".
### Numbers from nothing
Woodin works in set theory, an area that lies at the very foundation of mathematics. A set in mathematics is just a collection of objects — what these objects are doesn't matter. They could be numbers, symbols, triangles, or a mixture of them all. Sets are written using curly brackets, for example the set containing the numbers 1,2,3 is written as The members of a set are called its elements. Importantly, a set can itself be an element of another set. For example, if you consider the contents of your shopping trolley as a set, then the bag of oranges you bought is itself a collection of things and therefore a set.
A set is a collection of objects.
That the harmless notion of a set should be so fundamental to mathematics may seem surprising, but it turns out that pretty much all mathematical objects can be described in the language of sets.
As an example, think of the natural numbers 0, 1, 2, … . If you met an alien who had no conception of numbers, but understood sets, you could define the numbers recursively as follows.
• 0 is the empty set, denoted , that is the set with nothing in it.
• 1 is the set containing only the number 0 (which we've just defined).
• 2 is the set consisting of the two previous numbers.
• 3 is
• Etc
• In general, the number is the set containing all the previously defined numbers.
This hierarchical definition immediately gives you the ordering of the numbers, as well as a notion of what it means to add 1 to a given number: you simply go one up in the hierarchy. These two notions are enough to give you all of arithmetic, since addition and multiplication just correspond to repeatedly adding 1s and subtraction and division are their reverse. Thus, the arithmetic of the natural numbers can be built, literally, from nothing, using only the notion of sets. And as it turns out, virtually every other mathematical object can be constructed from sets too.
Perhaps the most important property of sets is that they give us a glimpse of the infinite. As the mathematician Georg Cantor noted in the late nineteenth century, to see if two sets and have the same number of elements, you don't need to count them. All you need to do is match each element of with exactly one element of , so that no elements in have the same partner in . If you can do that without any left-overs in either set, then the two sets have the same size, or as mathematicians say, the same cardinality.
Georg Cantor.
Now take two infinite sets, for example the set containing all the natural numbers and the set containing all the even numbers. Each element of can be matched with exactly one of as follows:
N 1 2 3 … n … E 2 4 6 … 2n …
So even though the two sets are infinite, and one is contained in the other, we can say that they have the same cardinality.
What Cantor also showed, however, is that it is impossible to find an exact matching between the natural numbers and the real numbers (the reals are all the numbers that lie along the number line). There are "more" real numbers than natural ones because however you match natural numbers with reals, there will always be reals left over.
This suggests that there are two different types of infinity, one bigger than the other. The first, the infinity of the natural numbers, is what is called a countable infinity. The second infinity is called the continuum. The question of whether there's an infinity "in-between" these two turns out to be a can of worms to which we'll return later.
Cantor didn't stop at these two types of infinity, but actually defined a whole hierarchy of them, each bigger than the one before. He called these infinities cardinal numbers and even found a way of doing arithmetic with them. Each cardinal number measures the size of infinite sets with certain properties. (You can find out more about this construction in the Plus article Cantor and Cohen: Infinite investigators.)
Ever since Cantor's work, mathematicians have extended his gallery of infinity, adding ever larger infinite monsters. The structure that emerges is truly compelling. "The remarkable thing is that there are many different ways in which one might approach formulating a hierarchy of large cardinals," explains Woodin, "but all the approaches end up with the same hierarchy. There are very deep theorems saying that one notion of infinity is exactly measured by another. That in part justifies the claim that the hierarchy of infinity is the fundamental core of set theory."
### Formal mathematics
Spurred on, in part, by the abstract power of sets, mathematicians at the beginning of the twentieth century thought themselves on the verge of achieving an ancient dream: to put all of mathematics on a watertight logical footing and show that it contains no contradictions. This may sound strange, since mathematics is the most rigorous of disciplines. But the fact is that it's full of hidden assumptions and leaps of faith. For example, in our definition of the natural numbers above we've implicitly assumed that there is such a thing as the empty set.
Cantor realised that there are more real numbers than natural numbers. This was just the beginning of a hierarchy of infinities.
To eliminate such grey areas from mathematics, you need to build it up as a formal system. You need a clear statement of the assumptions you're prepared to accept without proof — these are called your axioms. You also need to be clear about the rules of logical inference you consider valid, for example rules like "if x=y and y=z, then x=z". You then only consider a statement to be true if you can deduce it from the axioms using your rules of inference.
Set theory seemed to provide the perfect setting for such a formal system. Virtually all mathematical objects can be defined in the language of sets and the notion of a set is simple enough to come up with a relatively short and sharp set of axioms. This mathematicians duly did. Ernst Zermelo and Abraham Fraenkel came up with a set of eight axioms, known as the ZF axioms, which include the assertion that the empty set exists, as well as other rather intuitive statements such as "two sets are equal if and only if they have the same members" (click here for a complete list of the ZF axioms). The set of axioms most commonly used today is made up of the ZF axioms together with the rather intriguing axiom of choice. They are known as the ZFC axioms.
### Incomplete mathematics
The axiomatic dream was shattered in the 1930s, in particular by two results proved by the mathematician Kurt Gödel. A consequence of Gödel's famous incompleteness theorems is that within any formal mathematical system that describes the arithmetic of the natural numbers there will always be statements that can't be proved either true or false from the axioms. As Gödel put it in a letter to Zermelo, "For every formal system there are statements which are expressible within the system, but which may not be decided from the axioms of that system." (You can read more about the incompleteness theorems in the Plus article Gödel and the limits of logic.)
Kurt Gödel
So what kind of statements are undecidable within the ZFC axioms? We have already encountered one above. As Cantor noted, the infinity of the continuum is "greater" than that of the natural numbers. The assertion that there's no infinity "in-between" the two is known as the continuum hypothesis. It turns out that within the system based on the ZFC axioms, the continuum hypothesis is unprovable. It's neither true nor false, it's simply beyond the limits of ZFC. (You can find out more about the continuum hypothesis in this Plus article.)
This is rather shocking, since naively we might feel that questions like the continuum hypothesis should have an answer. Perhaps, the ZFC axioms are not strong enough and we need to add a few more? We could even add the continuum hypothesis itself as an axiom, in other words we could agree to accept it without proof. (This did in fact happen with the axiom of choice, see the Plus article Cantor and Cohen: Infinite investigators.)
However, this approach is problematic. Firstly, as Gödel's result proves, any new system with whatever extra axioms will also contain undecidable questions. Secondly, there are many, many results which are undecidable in ZFC set theory. Adding axioms all over the place just in order to settle them may not only introduce contradictions, but smacks of cheating.
So what does this mean for mathematical truth? "One can take the view, and some have, that the ubiquity of unsolvable problems in set theory is an indication that set theory is a result of an excess of human imagination and that there's no meaning there. It's just a result of our biological hard-wiring," says Woodin. "I don't think that's true, but until we discover a new civilisation and see if they have the same mathematics as we do, it's hard to tell for sure."
### The missing axiom
There is, however, a middle way, which doesn't involve searching for alien life. The accepted axioms of set theory are themselves products of human invention.They have been chosen because they "feel" natural and reflect our intuition of what sets are and of the nature of infinity. Perhaps there is an extra axiom (or several of them), which complete this intuition in a meaningful way. If we add them, we would still be left with undecidable questions, but perhaps we could settle a large proportion of the avalanche of undecidability that has cascaded over set theory. "One might say this is just a game, you're just choosing the axioms to solve your problem," says Woodin. "But not quite. There are basic fundamental intuitions that frame set theory. If you find an answer based on the extrapolations of those principles, then that's not just a game. It's saying that the ZFC axioms did not capture all of our intuition."
Back in the 1930s Gödel himself came up with an extra axiom, called the axiom of constructibility, which together with the ZFC axioms settled the continuum hypothesis, as well as many other undecidable questions. To be precise, Gödel devised a class of sets that satisfy the ZFC axioms and have an extra property which makes it possible to answer these undecidable questions. The axiom of constructibility asserts that Gödel's constructible universe of sets is in fact all there is: there are no other sets which mess things up.
What's the missing axiom?
Alas, Gödel's universe does not contain most of the infinite sets that arise from the hierarchy begun by Cantor. In plain old ZFC set theory the existence of these large cardinals is undecidable, just as the continuum hypothesis is. But once you add the axiom of constructibility to ZFC, the undecidability goes away: you can prove that most of these large infinities don't exist. This is unacceptable to many set theorists: an axiom which cuts off a large chunk of the fundamental hierarchy seems too restrictive. For this and some other reasons too, Gödel's universe and the axiom of constructibility were dismissed.
But all was not lost. For some time now Woodin and his colleagues have been systematically modifying Gödel's universe to admit larger and larger infinite sets. It's this work that has led Woodin to formulate a notion of an ultimate version of Gödel's universe of sets, which can accommodate all known large cardinals, and to what might be the ultimate missing axiom. The axiom he has in mind involves the existence of certain very large infinities known as Woodin cardinals.
Woodin's optimism that he is indeed on the right path is based on some previous success. For some time, mathematicians had been puzzling over a group of classical questions central to a certain area of maths, that were undecidable under ZFC. It was clear that these questions would be settled if all sets satisfied a property called projective determinacy, but there was no reason why they should. On the face of it, projective determinacy had little to do with the hierarchy of infinity, but the approach of Woodin and his colleagues showed that the two were linked. If you're prepared to accept a certain axiom involving Woodin cardinals, then projective determinacy pops out as a consequence. They were also able to show that their axiom is essentially the only one that can give you projective determinacy, subject to some fairly general structural constraints.
This synergy between seemingly unrelated concepts lends credence to Woodin's approach to finding the ultimate missing axiom. "If this were all just a human enterprise then there's no reason to expect this kind of success. We were searching for new axioms on the basis of our intuition and in this sort of situation there's no reason to believe that the search should be successful. It's a bit like looking for a unicorn. We think we know what a unicorn looks like, but that doesn't mean we're going to find it." If you do find the unicorn, then you must be doing something right.
Success, however, is by no means guaranteed. Whether or not Woodin's ultimate axiom really makes sense is something that depends on open questions that still need to be answered. "It's very speculative," he admits. "But there are a series of conjectures formulated over the last two or three years, which if they go the right way will lead one to this unique conception of the universe of sets. If this happens, then it will settle the continuum hypothesis [as true] and many other problems that are unsolvable. To me, the whole subject is at a critical crossroads."
At the ICM 2010 Woodin made the bold and controversial prediction that his axiom will be "validated on compelling and accepted principles of infinity". But he has already shown that he's not averse to changing his mind in the light of new evidence. A few years ago he was on record as saying he believed the continuum hypothesis could actually be considered false, based on work he was doing then. He's now reversed his opinion.
But does he have any idea as to when his new axiom might be validated? "Impossible to say. It could be a year, it could be a hundred years. I personally think we will know a lot more in a year or two, though some of these conjectures look quite difficult. That's the thing: if you have a conjecture in mathematics, you don't know if it will be settled tomorrow or in a thousand years."
Marianne Freiberger is co-editor of Plus. She interviewed Hugh Woodin, Professor of Mathematics at the University of California, Berkeley, at the International Congress of Mathematicians in August 2010.
If you're up for some technical maths, the book Roads to Infinity by John Stillwell gives a good introduction to the issues discussed in this article. It's not exactly for a general audience, but accessible to anyone with stamina and some grounding in logic and set theory.
You can buy the books and help Plus at the same time by clicking on the link on the left to purchase from amazon.co.uk, and the link to the right to purchase from amazon.com. Plus will earn a small commission from your purchase.
If you're up for a really advanced presentation, you can watch a video of Hugh Woodin's lecture at the International Congress of Mathematicians.
### set theory
This was very interesting, easy to read, and easy to comprehend.
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Question
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Solving time: 6 mins
# Refer to Q.2, Exercise 14.2. What is the empirical probability that an engineer lives:(i) less than from her place of work?(ii) more than or equal to from her place of work?(iii) Within from her place of work?Solution:
## Solutions (2)
The distance (in ) of 40 engineers from their residence to their place of work were found as follows:
Total numbers of engineers =40
(i) Number of engineers living less than from their place of work =9
the probability that an engineer lives less than from her place of work =9 / 40
(ii) Number of engineers living more than or equal to from their place of work
therefore probability that an engineer lives more than or equal to from her place of work =31 / 40
(iii) Number of engineers living within from their place of work =0
therefore the probability that an engineer lives within from her place of work =0 / 40=0
26
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Question Text Refer to Q.2, Exercise 14.2. What is the empirical probability that an engineer lives:(i) less than from her place of work?(ii) more than or equal to from her place of work?(iii) Within from her place of work?Solution: Updated On Apr 9, 2023 Topic Probability Subject Mathematics Class Class 9 Answer Type Text solution:1 Video solution: 1 Upvotes 120 Avg. Video Duration 13 min
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# Improvement of weak type inequality for Hardy-Littlewood Maximal inequality
Let $B(x,R)$ denotes the ball in centered at $x\in \mathbb{R}^n$ with radius $R$. The centered Hardy-Littlewood maximal operator $M$ is defined by $$Mf(x)=\sup_{B(x,R)} \frac{1}{|B(x,R)|} \int_{B(x,R)} |f(y)| dy$$ for every locally integrable function $f \in L^1_{\text{loc}}(\mathbb{R}^n)$. We know that $M$ is of weak type-$(1,1)$, that is, there exist a constant $C>0$ such that $$|\{ x\in \mathbb{R}^n: Mf(x)>\lambda \}| \le \frac{C\|f\|_{L^1}}{\lambda},$$ for every $\lambda>0$ and $f\in L^1(\mathbb{R}^n)$.
My question is that could we improve the right hand side bound in the sense as follows:
There exist a constant $C>0$ such that $$|\{ x\in \mathbb{R}^n: Mf(x)>2\lambda \}| \le \frac{C\|f \chi_{\{|f|>\lambda\}}\|_{L^1}}{\lambda},$$ for every $\lambda>0$ and $f\in L^1(\mathbb{R}^n)$.
Yes, this result is true. And actually you can explicitly write, for $f$ non-negative, $$\mu\{Mf>\alpha\}\leq \frac{3^d}{(1-\delta)\alpha}\int_{\{f>\delta\alpha\}}f\,dx$$ where $d$ denotes the dimension.
For general $f$, write $f=f^+-f^-$
Let $E_\alpha:=\{Mf > \alpha \}$. By definition of maximal function we can choose a ball $B_x$, centered at $x$ for each $x\in E_\alpha$, such that $$\frac{1}{\mu(B_x)}\int_{B_x}|{f}|dx>\alpha,$$ and $\{B_x\}_{x\in E_\alpha}$ covers $E_\alpha$.
Then by Vitali covering theorem, we can choose a sequence of disjoint balls $\{B_k\}$ from $\{B_x\}_{x\in E_\alpha}$ such that $E_\alpha\subset \bigcup_k B_k$ and
$$\mu(E_\alpha)\leq 3^d \sum_k \mu(B_k).$$
Then by we have $$\mu(E_\alpha)\leq 3^d\sum_k \mu(B_k) \leq \sum_k\frac{3^d}{\alpha}\int_{B_k}|{f}|dx\leq \frac{3^d}{\alpha}\int_{\bigcup_kB_k}|{f}|dx$$ Now since $f\geq 0$, for each given $\alpha$, we can choose $0<\delta<1$ such that
$$\{Mf(x)>\alpha\}\subset \bigcup_k B_k\subset \{f(x)>\delta\alpha\}.$$ Thus, we have $$\mu(E_\alpha)\leq \frac{3^d}{\alpha}\int_{\bigcup_kB_k}|{f}|dx \leq \frac{3^d}{\alpha}\int_{\{{f>\delta\alpha}\}}|{f}|dx\leq \frac{3^d}{\alpha(1-\delta)}\int_{\{f>\delta\alpha\}}|{f}|dx.$$
• Could you give the hint to prove this result or the reference? Is it related to weak type-(1,1) and strong type-$(\infty,\infty)$ for $M$? – beginner Nov 16 '14 at 2:10
• Now the answer is complete – spatially Nov 16 '14 at 2:22
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How to select by pos and compare 2 orders at a time? How to make a OrderSelect loop? - page 3
3243
johnnybegoode:
Ya, does not work and only 1 order is close and not both together
Thanks, but is this comparing 1 with 2, then 2 with 3, then 3 with 4?
I need to specifically compare order
1 & 2 only
3 & 4 only
5 & 6 only
..... and so on
other orders should not be compared.
```if (...) {} // do what you need with both
```
`pos++;`
629
qjol:
```void CloseAll()
{
for (int i=0; i<OrdersTotal(); i++)
{
{
int ticketA = OrderTicket();
double profitA = OrderProfit();
for (i++; i<OrdersTotal(); i++)
{
{
int ticketB = OrderTicket();
double profitB = OrderProfit();
i++;
if ((profitA + profitB) > Profit)
{
OrderClose(ticketA,OrderLots(),OrderClosePrice(),5,Violet);
OrderClose(ticketB,OrderLots(),OrderClosePrice(),5,Violet);
}
}
}
}
}
}```
Thanks, but does not work, It closes any profitable orders. It is quite a mess.
EX. I have 10 orders like order 1,2,3,4,5,6,7,8,9,10. I opened them in sequential order, meaning opened order 1 then 2 then follow by 3 ... etc.
I would like to group them in pairs (see below).
1 & 2 (Pair 1) or
3 & 4 (Pair 2) or
5 & 6 (Pair 3) or
7 & 8 (Pair 4) or
9 & 10 (Pair 5)
I would also like a loop to go through the 10 orders over and over again until the addition of any pair is more than Profit.
Pair must be inseparable.
If the addition of any of the 5 pairs is more than Profit, only then close that pair. It should not close 1 & 5, or 1 & 3, or 2 & 3 ... etc.
Do you think that my ticketA/B and profitA/B is updated correctly in the loop as (1 & 2), (3 & 4), (4 & 5) ... etc.?
Or do I need to store ticket number in some sort of an array?
420
is it possible to compare select first order and compare with others all orders with same pairs and order type? if not match do this or break and erase variable, then again start with 2nd order select and compare with others below match with same pairs and order type? and go on like this.
16111
1. ```OrderClose(ticketA,OrderLots(),OrderClosePrice(),5,Violet);
OrderClose(ticketB,OrderLots(),OrderClosePrice(),5,Violet); ```
OCP is the close price of the last selected order. You're trying to close both orders at the same price. You must reselect as price will have moved during the first server call.
2. In the presence of multiple orders (one EA multiple charts, multiple EA's, manual trading) you must count down when closing/deleting/modifying in a position loop. Get in the habit of always counting down. Loops and Closing or Deleting Orders - MQL4 forum
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# Experiment to Find the Focal Length and Power of a Diverging Lens.
Extracts from this document...
Introduction
Joanne Bell
Experiment to Find the Focal Length and Power of a Diverging Lens
Aim
The aim of this experiment is to measure the focal length and power of a diverging lens. Since negative or diverging lenses do not form real images of real objects, an experimental virtual object will need to be set up for the diverging lens by using a converging lens. To do this, place a converging lens in front of the diverging lens. If the converging lens has a shorter focal length than the diverging lens, the two-lens combination will form a real image of a distant object.
Figure 1
When the lenses of individual focal lengths f c and f D, are placed in contact, the effective focal length f of the combination is given by:
Equation 1
Middle
Set up the apparatus as shown in figure 2 but with both the converging and diverging lenses in the lens holder. With the object to lens distance, u, at 6f , move the viewing screen until a clearly focused image can be seen.Measure the image to lens distance and include uncertainty measurements (where the distance can be changed more but the image is still in focus).Repeat procedures 2 and 3 for object to lens distance at 5f, 4f, 3f, 2f, and just over 1f.Solve the lens equation (Equation 1) using the values found in the above procedures to give a more accurate combined focal length.Use Equation 2 to find the power of the combined lens.
Step 3: Finding the focal length and power of the diverging lens:
Conclusion
/p>
50
11
0.02
0.090909
0.110909
9.016393
60
12.4
0.016667
0.080645
0.097312
10.27624
Diverging lens:
u, object to lens (cm) v, image to lens (cm) 1/u 1/v 1/f f (cm) 15 170 0.066667 0.005882 0.072549 13.78378 30 31 0.033333 0.032258 0.065591 15.2459 45 22 0.022222 0.045455 0.067677 14.77612 60 17 0.016667 0.058824 0.07549 13.24675 75 16 0.013333 0.0625 0.075833 13.18681 90 15 0.011111 0.066667 0.077778 12.85714
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# Related AS and A Level Microscopes & Lenses essays
1. ## Physics coursework; Finding the focal length of a lens using a graphical method.
4 star(s)
is appropriate level off sensitivity; any more sensitivity would be unsuitable for the apparatus being used and since I can quite easily judge to 2mm, any less would be too insensitive. ==> This experiment will be carried out in dark conditions to make the bright image easier to see, thereby increasing the accuracy of the experiment.
2. ## In this experiment I will be investigating the efficiency of a motor. I hope ...
The problem however is that glass reflects a proportion of the light. It reflects about 10% of the light this can be reduced by adding a anti-reflection coating to the lens. This increases the transmission of light up to 99% making the image brighter.
1. ## Does the focal length of a lens depend on the colour of light used?
the light source through the image and lens onto the screen, and my meter rule or tape measurer will be measuring these distances as if they were straight. If they are all in line then the image should be projected onto the screen without having to move the screen (I
2. ## Relationship Between U and V For a Convex Lens
Safety I shall take the following safety precautions were taken, to ensure that the experiment was safe to do. They were: * I will put my things out the way, so they don't come in the way of other people (e.g.
1. ## The focal length of a convex lens.
Also, make sure the image distance was measured correctly. In experiment 2, make sure the pin, the lens and the mirror are vertically placed. If you cannot see the image of the pin, the probable reasons are (a) you are too close to the pin, less than the least distance of distinct vision, (b)
2. ## To investigate the relationship between the distance between a lens and an object, and ...
I will set up equipment as shown above and then adjust the distance between the screen and the lens until the image shown on the screen is as clear as possible. I will record the v value for each of the different u values.
1. ## Measuring the focal length of a lens for red and green light- Case Study
do not know, another social drawback may be that some people may feel uncomfortable wearing glasses to help them in helping their eye defects. There are also limitations to how much they can magnify something so if a human's eye is too bad lenses may not be able to fully help them.
2. ## Lenses experiment
So I have hopefully explained why in my experiment I will use a convex lens, but I have not yet said why I believe the convex lens will cause the magnification of the image to decrease as the object moves further away from the focal point of the lens.
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# timeofday
Elapsed time since midnight for datetime arrays
## Description
example
T = timeofday(DT) returns a duration array whose values equal the elapsed time since midnight for each element in DT.
For datetime arrays without time zones, and in most other cases, T is equal to
E = hours(DT.Hour) + minutes(DT.Minute) + seconds(DT.Second)
If DT has its TimeZone property set to a time zone that respects Daylight Saving Time (DST), then timeofday takes DST into account. For more information, see Algorithms.
The output argument T is also equivalent to DT - dateshift(DT,'start','day').
example
[T,D] = timeofday(DT) also returns the date portions of the values in DT as the datetime array D.
The output argument D is equivalent to dateshift(DT,'start','day').
## Examples
collapse all
Create a datetime array.
DT = datetime('now') + hours(1:3)
DT = 1x3 datetime
26-Nov-2022 08:26:53 26-Nov-2022 09:26:53 26-Nov-2022 10:26:53
Calculate the elapsed time since midnight for each input value. Display the times as hours, minutes, and seconds.
T = timeofday(DT)
T = 1x3 duration
08:26:53 09:26:53 10:26:53
Create a datetime array.
DT = datetime('now') + hours(1:3)
DT = 1x3 datetime
26-Nov-2022 09:10:30 26-Nov-2022 10:10:30 26-Nov-2022 11:10:30
Return the date portions of DT as a second datetime array. Because the hour, minute, and second components of D are all set to midnight (or 00:00:00 in 24-hour format), D displays only the dates.
[T,D] = timeofday(DT)
T = 1x3 duration
09:10:30 10:10:30 11:10:30
D = 1x3 datetime
26-Nov-2022 26-Nov-2022 26-Nov-2022
Calculate elapsed times since midnight on a day with a Daylight Saving Time (DST) shift.
Create a datetime array. Set its TimeZone property to a time zone that observes DST. Set the date to a day when a DST shift occurred.
tz = 'America/New_York';
fmt = 'dd-MMM-yyyy HH:mm:ss z';
DT = datetime(2015,3,8,'TimeZone',tz,'Format',fmt) + hours(1:4)
DT = 1x4 datetime
Columns 1 through 2
08-Mar-2015 01:00:00 EST 08-Mar-2015 03:00:00 EDT
Columns 3 through 4
08-Mar-2015 04:00:00 EDT 08-Mar-2015 05:00:00 EDT
Calculate the elapsed times. The DST shift occurred at 02:00 on March 8, 2015 in this time zone. timeofday accounts for the shift for times at or after 02:00 on this date. Set the format of T to display elapsed times in hours.
T = timeofday(DT);
T.Format = 'h'
T = 1x4 duration
1 hr 2 hr 3 hr 4 hr
For comparison, calculate elapsed times using the hour, minute, and second components of DT. This calculation does not account for the DST shift.
E = hours(DT.Hour) + minutes(DT.Minute) + seconds(DT.Second)
E = 1x4 duration
1 hr 3 hr 4 hr 5 hr
Set the times of day in a datetime array according to the times of day in another datetime array.
There are two ways to set the times of day.
• Split the time portions from the first datetime array by using timeofday and add the time portions to the second datetime array.
• Set the Hour, Minute, and Second components of the second datetime array equal to the Hour, Minute, and Second components of the first datetime array.
If you use the first way, you might not always account for Daylight Saving Time (DST) shifts correctly. Only the second way is always correct across any DST shifts.
Create a datetime array. Each element has a different time component.
DT1 = datetime(2015,3,7) + hours(1:4)
DT1 = 1x4 datetime
Columns 1 through 3
07-Mar-2015 01:00:00 07-Mar-2015 02:00:00 07-Mar-2015 03:00:00
Column 4
07-Mar-2015 04:00:00
Create a second datetime array. Each element has the same date and time components.
DT2 = datetime(2015,3,[8 8 8 8])
DT2 = 1x4 datetime
08-Mar-2015 08-Mar-2015 08-Mar-2015 08-Mar-2015
Set the times of day in DT2 according to the times of day in DT1. Because DT1 and DT2 do not have time zones their time components are identical.
DT2 = dateshift(DT2,'start','day') + timeofday(DT1)
DT2 = 1x4 datetime
Columns 1 through 3
08-Mar-2015 01:00:00 08-Mar-2015 02:00:00 08-Mar-2015 03:00:00
Column 4
08-Mar-2015 04:00:00
Create a datetime array with elements that have the TimeZone property set to 'America/New_York'. Because DT3 has a time zone and a DST shift that occurs on March 8, 2015, the time components of DT3 are not the same as those of DT1.
tz = 'America/New_York';
fmt = 'dd-MMM-yyyy HH:mm:ss z';
DT3 = datetime(2015,3,8,'TimeZone',tz,'Format',fmt) + timeofday(DT1)
DT3 = 1x4 datetime
Columns 1 through 2
08-Mar-2015 01:00:00 EST 08-Mar-2015 03:00:00 EDT
Columns 3 through 4
08-Mar-2015 04:00:00 EDT 08-Mar-2015 05:00:00 EDT
Display the elapsed times since midnight. timeofday accounts for the DST shift. The elapsed times show that the times in DT3 are incorrect. The last time in DT3 corresponds to 5:00 a.m. But in DT1 only four hours elapsed since midnight.
T = timeofday(DT3)
T = 1x4 duration
01:00:00 02:00:00 03:00:00 04:00:00
To set the times of day of DT4 correctly regardless of the time zone or the day of year, use the Hour, Minute, and Second properties of DT1.
DT4 = datetime(2015,3,[8 8 8 8],'TimeZone',tz,'Format',fmt);
DT4.Hour = DT1.Hour;
DT4.Minute = DT1.Minute;
DT4.Second = DT1.Second;
DT4
DT4 = 1x4 datetime
Columns 1 through 2
08-Mar-2015 01:00:00 EST 08-Mar-2015 03:00:00 EDT
Columns 3 through 4
08-Mar-2015 03:00:00 EDT 08-Mar-2015 04:00:00 EDT
In this time zone, 2:00 a.m. Eastern Standard Time did not exist on March 8, 2015 because the DST shift occurred then. The second element of the result has a time component of 3:00 a.m. Eastern Daylight Time.
## Input Arguments
collapse all
Input dates and times, specified as a datetime array.
## Output Arguments
collapse all
Time portions of input array, returned as a duration array.
Date portions of input array, returned as a datetime array.
## Algorithms
A datetime array can have its TimeZone property set to a time zone that observes Daylight Saving Time (DST). The timeofday function takes DST into account.
• If the input argument DT is a datetime array with no time zone, then the output T is also equal to
E = hours(DT.Hour) + minutes(DT.Minute) + seconds(DT.Second)
• If DT has its TimeZone property set to a time zone that does not observe DST, then T is equal to E.
• If DT has its TimeZone property set to a time zone that observes DST, then timeofday accounts for the DST shift on days when the shift occurs. On those days, for times after the DST shift occurs, T differs from E by the amount of the shift.
## Version History
Introduced in R2014b
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## DEV Community
Abhishek Chaudhary
Posted on
# Single Number III
Given an integer array `nums`, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. You can return the answer in any order.
You must write an algorithm that runs in linear runtime complexity and uses only constant extra space.
Example 1:
Input: nums = [1,2,1,3,2,5]
Output: [3,5]
*Explanation: * [5, 3] is also a valid answer.
Example 2:
Input: nums = [-1,0]
Output: [-1,0]
Example 3:
Input: nums = [0,1]
Output: [1,0]
Constraints:
• `2 <= nums.length <= 3 * 104`
• `-231 <= nums[i] <= 231 - 1`
• Each integer in `nums` will appear twice, only two integers will appear once.
SOLUTION:
``````import bisect
class Solution:
def singleNumber(self, nums: List[int]) -> List[int]:
nums.sort()
op = []
numset = set(nums)
for num in numset:
if bisect.bisect_right(nums, num) == bisect.bisect_left(nums, num) + 1:
op.append(num)
return op
``````
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# chapter5-datapath02 - Manchester Carry Chain Digital IC 1...
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Digital IC Manchester Carry Chain 1
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Digital IC Carry-Skip Adder Carry-ripple is slow through all N stages Carry-skip allows carry to skip over groups of n bits Decision based on n-bit propagate signal Slide 2 in + S 4:1 P 4:1 A 4:1 + S 8:5 P 8:5 A + S 12:9 P 12:9 A 12:9 B 12:9 + S 16:13 B 16:13 C out C 4 1 0 C 0 C 12 1 0 1 0
Digital IC 4/16 bit Block Carry Skip Adder Worst-case delay carry from bit 0 to bit 15 = carry generated in bit 0, ripples through bits 1, 2, and 3, skips the middle two groups (B is the group size in bits), ripples in the last group from bit 12 to bit 15 C i,0 Sum Carry Propagation Setup Sum Carry Propagation Setup Sum Carry Propagation Setup Sum Carry Propagation Setup bits 0 to 3 bits 4 to 7 bits 8 to 11 bits 12 to 15 T add = t setup + B t carry + ((N/B) - 1) t skip +(B-1) t carry + t sum
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Digital IC RCA, Carry Skip Adder Comparison 0 10 20 30 40 50 60 70 8 bits 16 bits 32 bits 48 bits 64 bits RCA CSkA B=2 B=3 B=4 B=5 B=6
Digital IC Carry-Skip PG Diagram Slide 6 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 15:0 14:0 13:0 12:0 11:0 10:0 9:0 8:0 7:0 6:0 5:0 4:0 3:0 2:0 1:0 0:0 16:0 skip t : : : 1: : : i j i k i k k j i j i k k j G G P G P P P 0 : 0 1 : 4 1 : 4 0 : 4 G P G G 0 : 4 5 : 8 5 : 8 0 : 8 G P G G 0 : 8 9 : 12 9 : 0 : G P G G 0 : 13 : 16 : 0 : G P G G
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Digital IC Carry-Skip PG Diagram Slide 7 skip xor 2 1 ( 1) pg AO t t n k t t 0 : 0 1 : 4 1 : 4 0 : 4 G P G G 0 : 4 5 : 8 5 : 8 0 : 8 G P G G 0 : 8 9 : 12 0 : 0 : G P G G 0 : 13 : 16 : 0 : G P G G 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 15:0 14:0 13:0 12:0 11:0 10:0 9:0 8:0 7:0 6:0 5:0 4:0 3:0 2:0 1:0 0:0 16:0
Digital IC Carry-Skip PG Diagram 8 0 , 0 1 , 4 1 , 4 0 , 4 G P G G P.CHAN and M.Schlag,”Analysis and design of CMOS Manchester adders with variable carry- skip” IEEE Trans.Computers ,Vol.39,No.8,Aug.1990,pp.983-992
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Digital IC Variable Group Size 0 3 4 5 6 7 8 9 10 11 12 13 14 15 16 15:0 14:0 13:0 12:0 11:0 10:0 9:0 8:0 7:0 6:0 5:0 4:0 3:0 2:0 1:0 0:0 16:0 Slide 9 Delay grows as O(sqrt(N))
Digital IC 10 C ok fA k B k C 1 G k P k C o k 1 + == A N- 1 , B N- 1 A 1 , B 1 P 1 S 1 ??? S N- 1 P N- 1 C i , N- 1 S 0 P 0 C i ,0 C i ,1 A 0 , B 0
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Digital IC Look-Ahead: Topology 11 C ok G k P k G k1 P C o k 2 + + = C G k P k G P P 1 G 0 P 0 C i0 + + + + = Expanding Lookahead equations: All the way: C o ,3 C i ,0 V DD P 0 P 1 P 2 P 3 G 0 G 1 G 2 G 3
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# 4.3: Body Weight
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Now that we know how to find the volume of a body, we just need to measure body mass in order to find body density. We typically measure the mass of a body by first measuring the weight using a scale, and then calculating mass from the measured weight. Weight is just another name for the force of gravity on an object. In everyday experience, a force (F) is any push or pull on an object. Forces can move objects, deform objects, or both. Often W is used to abbreviate weight, but is also used because it reminds us that an object’s weight and the force of gravity on the object are the same thing. Throughout this book we will learn about other forces, including buoyant force, tension, normal force, friction, and air resistance. We typically represent forces with arrows that point in the direction the force pushes (or pulls). We usually try to make the length of the arrows proportional to how big the forces are, in which case the arrows can be called vectors. The SI unit for weight, and all other forces, is the Newton (N). In the U.S. we often use pounds (lbs) instead of Newtons as our unit of force. One pound is equal to 4.45 Newtons.
## Reinforcement Activity
Draw a picture of yourself jumping on a trampoline (a stick figure will work). Then add an arrow representing gravity acting on you while you are in the air. The arrows should start at your center and point in the direction that the force is pushing or pulling you. Label the forces arrows.
Draw a second figure that is just standing on the trampoline and add arrows to represent the forces acting on the person. Label the forces. (Hint: There are two.) Do you think the lengths of the two arrows should be the same or different? Explain your thought process.
This page titled 4.3: Body Weight is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Lawrence Davis (OpenOregon) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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## The History of Calculating Machines
People have needed to carry out, mathematical calculations as part of their job since the earliest of times. These have mainly been scientists, engineers, mathematicians, accountants and tax collectors. As these calculations became more complex, it was no longer possible to use mental arithmetic, fingers or paper and pencil. To solve these more complex problems involving larger numbers, it was necessary to develop some form of calculating machine. Like most things, when someone invents a machine it is not long before someone else improves on the design, making it difficult to identify all the different types of calculating machines. The list below shows some main inventions created over time to simplify the process of mathematical calculation.
## The Earliest Calculating Machines
The earliest recorded calculating machine was the abacus or counting frame. The name abacus, derived from the Ancient Greek word “Abax” meaning a counting board containing grooves for counters. The records indicate that there were many types of abacus including Mesopotamian, Egyptian, Persian, Greek, Roman, Chinese, Indian, Japanese, Korean, Native American and Russian. The earliest recorded was a dust abacus around 2400 Before Common Era (BCE) used by the Babylonians. It consisted of a stone slab covered with dust or sand, with letters and numbers drawn in the sand. Eventually, addition and subtraction used small pebbles to aid the calculations.
## The Salamis Tablet
The archaeological evidence of a counting tablet found on the Greek island of Salamis in 1846 dating from 300 BCE. In brief, this was a counting board known as the Salamis Tablet, made from marble. However, the calculating machine most people would recognize as an abacus is the Saunpan of Chinese origin, which dates from the 2nd century BCE. Saunpan means calculating tray.
Attribution: Wilhelm Kubitschek: „XII. Die Salaminische Rechentafel“ in: Numismatische Zeitschrift, Bd. 31, Wien, 1899, S. 394 ff, Public domain, via Wikimedia Commons
Salamis Tablet
How to use a ‘Saunpan’ abacus by Pelle Lindblå
## The Roman Abacus
The Roman abacus was in fact similar to the Salamis tablet by moving pebbles on a board. Also covered with a thin layer of wax by the 1st century BCE, and inscribed with a stylus with columns and figures.
How to use a Roman abacus by Dasty Jones
## Antikythera mechanism
Presently, many people consider the Antikythera mechanism to be the world’s first computer. Archaeologist’s discovered it in 1902 in a shipwreck off the coast of the island of Antikythera. This island lies in the straits between the Greek island of Peloponnese and Crete. This clockwork mechanism used in Greece around 150 and 60 BCE, at the present time, thought to be for calculating the lunar and solar cycles. Reconstruction of the mechanism it was based on the geocentric model of the Solar System in which the Earth is in the centre with the Sun and the Moon orbiting the Earth.
Hellenic Republic Ministry of Culture and Tourism
## The medieval counting table
The medieval counting table or Exchequer table, once used to calculate taxes and goods. Unfortunately, there are very few examples of counting tables survive. One exists in the Museum of Notre Dame in Strasbourg. In the UK civil service “Her Majesty’s Exchequer”, the central government accounting process takes its name from the Medieval Exchequer table.
## Scottish Mathematician John Napier
It was the Scottish mathematician John Napier (1550 – 1617), accredited with inventing Napierian logarithms in 1614 who also invented a calculating device known as Napier’s bones. Sometimes known as Napier’s rods, consisted of flat rods, that performed multiplication of numbers between 2 and 9. He described his work on rods In his book “Rabdologia” published in 1617.
## The Gunter Scale
A contemporary of John Napier was Edmund Gunter (1581 – 1626) an English mathematician and clergyman. Gunther created the forerunner to the slide rule, using a rule marked with various scales. One or more of these were logarithmic scales. Gunther, then, used these scales and a pair of dividers, to allow him to carry out mathematical calculations. These included navigational problems.
Another English mathematician and clergyman, William Oughtred (1575-1660) replaced the need for a pair of dividers with a second rule. Sliding the second rule past the first, he was able to achieve the same result as Gunter had with dividers. William Oughtred’s device was the invention of the slide rule that continued in use by engineers until the 1970s.
## The Slide Rule
The basic slide rules are all similar to the image on the left, consisting of three scales. The slide rule consists of two scales fixed together, and a third sliding scale sandwiched between the two fixed scales. When the sliding scale aligns accurately on the fixed scale, the second fixed scale displays the correct answer. A perspex sliding window with a cursor makes reading the results easier. The Museum of HP Calculators page provides a good description of how a modern slide works. The same website also has details of circular and cylindrical slide rules.
Other designs of slide rules have specific tasks. These included designs specifically for carrying out hydraulic engineering calculations. Calculations such as hydraulic force, fluid pressure, piston speed, flow, stroke time and volume and fluid speed. I have included a photograph of two of these hydraulic slide rules on my page about Maths Tools.
## The Pascaline
The Pascaline designed by the French mathematician and physicist Blaise Pascal (1623 – 1662). In 1642, he started designing a calculating machine in order to assist his father as a supervisor of taxes in Rouen. The final version of the machine appeared in 1645 after three years of work. The Pascaline or Pascal’s Arithmetic Machine was the first calculator that went into production commercially. Over 50 of these calculating machines went into manufacture over the following 10 years.
The Rechenuhr a calculating clock, designed in 1623 by Wilhelm Schickard (1592 – 1635) predated the Pascaline. Since there was little knowledge of this clock until the 1950s, the Pascaline had claimed to be the first mechanical calculator. The design of the Pascaline would eventually lead to later calculators such as Gottfried Leibenz wheel in 1671 and Thomas Colmar Arithmometer in 1820.
How the Pascaline works by MechanicalComputing
## Leibniz Wheel
Another calculating machine that expanded on the design of the Pascaline was the Leibniz Wheel. A German mathematician, Gottfried Wilhelm von Leibniz (1646 – 1716) designed and built the Leibniz Wheel in 1673. After he became acquainted with Pascal’s calculating machine, the Pascaline, he decided to improve the design, allowing his machine to carry out multiplication and division.
To perform multiplication and division he designed a stepped drum mechanism or the Leibniz wheel. This was a drum with teeth of varying lengths around the drum. The rotation of a drum allowed teeth to mesh with the teeth on the counting wheel as it moved along the length of the drum and to carry out the calculations. In 1676 Leibniz demonstrated a small working model of an arithmetic machine using several of his Leibniz wheels. However, it wasn’t until 1694 that the manufacture of a full working Stepped Reckoner took place.
The Leibniz wheel mechanism
The Leibniz Stepped Reckoner
## The Jacquard Loom
Joseph Marie Jacquard (1752 – 1834) was a French weaver and merchant. He designed a weaving loom in 1801 that made use of punched cards to control the pattern of the longitudinal warp threads. This would allow unskilled workers to weave complex patterns. He later developed his loom with the punched cards joined to form a continuous loop. The concept of adopting punched cards to store information was in use until the early 1970s.
## The Arithmometer
The industrial revolution highlighted a need for a reliable calculator in the office environment. Charles Xavier Thomas (1785 – 1870) from Colmar in France, later known as Charles Xavier Thomas, de Colmar, invented The Arithmometer in 1820. In 1851, it was the first successful mechanical calculator to go into production commercially. By 1914 copies of the Arithmometer were in manufacture by approximately 20 European companies.
## Babbage Difference Engine
The mathematician, Charles Babbage (1791 – 1871) used a punched card system for the design of his Analytical Engine in 1833. By 1840 the Analytical Engine was almost complete and considered the first general purpose computing machine.
People probably know Charles Babbage better for designing his Difference Engine rather than his Analytical Engine mentioned above. He started work on his first Difference Engine in 1820, designed to calculate polynomial functions and print out the results as a table. The Government cut funding in 1842 due to the cost of producing high quality and precision components for the engine, resulting in abandonment of the project.
Work had already halted on the construction of the engine when Babbage had turned his attention to the Analytical Engine. With the successful completion of the Analytical Engine in 1840, Babbage started work on the design of an improved Difference Engine. In 1849, he had completed the design of the new Difference Engine. Babbage never constructed the Difference Engine during his life. It was only in 1985 that the Museum of Science in London constructed a working engine from his original design drawings.
## Hydraulic Computers
Vladimir Sergeevich Lukianov (1902-1980) a Russian Soviet scientist and doctor of technical services proposed a fundamentally new way of mechanising the calculations of unsteady processes in 1934. 1936 saw the world’s first hydro-mechanical analogue computer produced. The Lukyanova developed the Hydraulic Integrator to solve partial differential equations. The changes in the machine water levels, indicated by marks on graph paper attached to measuring tubes (piezometers), produced a curved graph.
It was not just the Soviet Union that used water in computing. William Philips (1914 – 1975) was an engineer and economist born in New Zealand. After World War II, he studied economics at the London School of Economics. While studying, he used his previous training as an engineer to develop the Monetary National Income Analogue Computer, (MONIAC) in 1949. The Phillips machine consisted of transparent plastic tanks and pipes, fastened to a wooden frame. The computer demonstrated how the UK national economy worked, using coloured water to represent the flow of money within the economy. Although the machine did have limitations, as there was no provision for inflation or swings in the credit cycle.
Phillips Hydraulic Computer (MONIAC)
## Colossus
By the 1940s, electronic digital programmable computers were beginning to take over from the mechanical, clockwork and hydraulic calculators of previous eras. One of the better know computers was the Colossus, developed between 1943 and 1945. The Colossus used vacuum tubes, (thermionic valves) to carry out Boolean and counting operations. During WW II, Tommy Flowers (1905 – 1998) developed Colossus for codebreaking. Allen Coombs (1911 – 1995) was the principle designer for improved Colossus Mark II in 1944.
## The Post-War Years
The post-war years, saw an increase in the number of electronic computers developed for different commercial applications. John Mauchly and J. Presper Eckert designed the Electronic Numerical Integrator and computer (ENIAC) in 1946. Also, the Victoria University of Manchester built The Small Scale Experimental Machine (SSEM) in the same year. The Electronic Discrete Variable Computer (EDVAC) in 1952 was the first mainframe that used magnetic tapes. In 1953 IBM produced the IBM 702, which also used magnetic tapes. Between 1953 and 1969 IBM produced the first mass-produced IBM 650. British Ferranti Ltd. produced the Ferranti Pegasus 1 in 1956, followed in 1959 by the Ferranti Pegasus 2. All these early machines were all mainframe computers. It wasn’t until 1981 that the IBM Personal computer model 5150 made computer technology mainstream.
The 1980s was a major turning point for calculating machines. During this decade, there was the widespread use of desktop business computer’s, and the pocket calculator replaced the slide rule that had been around since the 17th century. More information on calculating machines and modern computer software is available on my page about maths tools.
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0 Members and 1 Guest are viewing this topic.
#### Jan666
• Jr. Member
• Posts: 52
« on: June 18, 2022, 12:37:12 pm »
Hello,
I wanna create a touch controller, a small circle i can move around with a finger inside a bigger circle. But i have no idea, how i set a border radius, that the smaller circle still movable just inside it, any help?
#### Jan666
• Jr. Member
• Posts: 52
##### Re: Touch joystick border radius
« Reply #1 on: June 20, 2022, 02:10:21 pm »
Can samone give me an advise
sf::Vector2i currPos = sf::Touch::getPosition(0, window);
sf::Vector2f mapPos = window.mapPixelToCoords(currPos);
float distance = sqrt((circIn.getPosition().x * circIn.getPosition().x) + (circIn.getPosition().y * circIn.getPosition().y));
angle = atan2(circIn.getPosition().x, circIn.getPosition().y);
angle = angle * 180 / 3.141592654f;
plr = circIn.getGlobalBounds();
if(plr.contains(mapPos)){
sf::Vector2i touch = sf::Touch::getPosition(0, window);
touch.x = Origin.x + 150 * cos(angle);
touch.y = Origin.y + 150 * sin(angle);
circIn.setPosition(touch.x , touch.y);
It doenst work anyway
#### eXpl0it3r
• SFML Team
• Hero Member
• Posts: 10235
##### Re: Touch joystick border radius
« Reply #2 on: June 20, 2022, 03:59:20 pm »
Are you on Android or iOS?
sf::Touch isn't implemented for other OSs right now.
If you have a Joystick controller, you'll need to use the Joystick API instead
Official FAQ: https://www.sfml-dev.org/faq.php
Nightly Builds: https://www.nightlybuilds.ch/
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Dev Blog: https://dev.my-gate.net/
Thor: http://www.bromeon.ch/libraries/thor/
#### Jan666
• Jr. Member
• Posts: 52
##### Re: Touch joystick border radius
« Reply #3 on: June 20, 2022, 04:08:03 pm »
I have android, i wanna build the touch controll by myself just with a circle, but iam not fit in math, sin, cos etc. I just need to know how i can rotate/ move a circle in a radius
#### eXpl0it3r
• SFML Team
• Hero Member
• Posts: 10235
##### Re: Touch joystick border radius
« Reply #4 on: June 20, 2022, 04:25:28 pm »
You take the center of the circle and the center of the joystick position and subtract the two positions, then you can calculate the distance (= sqrt((a - b)^2)) and if that gets too large, you stop moving the circle, or rather you need to move it to the minimum distance from the center.
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#### Jan666
• Jr. Member
• Posts: 52
##### Re: Touch joystick border radius
« Reply #5 on: June 20, 2022, 11:22:48 pm »
Yes i know the theory but i cant get fix it with the code, but thsnks for your reply
#### Jan666
• Jr. Member
• Posts: 52
##### Re: Touch joystick border radius
« Reply #6 on: June 22, 2022, 12:04:52 am »
Okay i got it, can you give me an advise how to stop the joytsick circle when it reach the distance?
if(plr.contains(mapPos)){
diff.x =circOutRectPos.x - circIn.getPosition().x;
diff.y =circOutRectPos.y - circIn.getPosition().y;
distance = sqrt(diff.x * diff.x + diff.y * diff.y);
circIn.setPosition(circIn.getPosition().x, mapPos.x);
circIn.setPosition(circIn.getPosition().y, mapPos.y);
if(distance < 175)
????????
How i can stop the circIn (Circle for touch ), it should slidly move along the radius border (distance),
#### fallahn
• Sr. Member
• Posts: 464
• Buns.
##### Re: Touch joystick border radius
« Reply #7 on: June 22, 2022, 10:40:21 am »
Regular circle-circle collision detection will work (there are plenty of tutorials out there), but instead of making sure the minimum distance between centres is always >= the sum of the radii, you make sure the *maximum* distance is <= large radius minus the small radius
#### Jan666
• Jr. Member
• Posts: 52
##### Re: Touch joystick border radius
« Reply #8 on: June 22, 2022, 12:29:21 pm »
I dont understand, can you explain it in a code example ?
#### Hapax
• Hero Member
• Posts: 3058
• My number of posts is shown in hexadecimal.
##### Re: Touch joystick border radius
« Reply #9 on: June 23, 2022, 10:30:29 pm »
Another option is to instead use a polar vector (or a polar form of one). This defines a vector by angle/direction and length rather than horizontal and vertical distances.
e.g.
• Take the vector (you can still use SFML's Vector2f here) from the centre of the "range" circle to the current location of the "touch" point (by taking the difference: touchPoint - rangeCircle)
• Then, convert that vector to its polar form, giving you a direction and length.
• Clamp the maximum length i.e. if the length is greater than the maximum allowed length, set it to that maximum length.
• Then, convert the vector back from its polar form.
This approach will allow you to drag past the outer circle's border and continue dragging around the outside and the inner circle will keep updating, always pointing towards where the finger is but clamped by the range.
You need to know a little maths for this. Try this wiki: Polar coordinates
If you want to see some code converting to and from polar co-ordinates - or just use these vectors to do it for you - see my library, Plinth's, Vector class implementation.
Selba Ward - SFML drawables
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Rectangular Boundary Collision - Rectangular SAT Collision
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