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# Question 9 (v) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ (see the given figure) Show that: APCQ is a parallelogram
Solution
## In ΔAPD and ΔCQB, ∠ADP=∠CBQ (Alternate interior angles) AD =CB (Opposite sides of parallelogram ABCD) DP=BQ (Given) ∴ΔAPD≅ΔCQB ( using SAS congruence rule) As we have proved that triangle APD is congruent to triangle CQB, So , AP= CQ (CPCT) and AQ=CP Since opposite sides In quadrilateral APCQ are equal to each other, APCQ is a parallelogram.
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1 Improving Elo Rankings For Sports Experimenting on the English Premier League Connor Sullivan Virginia Polytechnic Institute and State University Christopher Cronin Virginia Polytechnic Institute and State University Abstract In this paper we examine the Elo rating system and how it can be applied to the English Premier League for predicting the outcome of matches. Specifically we examined four different methods of modifying the basic Elo formula to improve prediction accuracy. The four methods with which we experimented were incorporating home field advantage, adjusting the K-factor at different points in a season, rewarding and penalizing winning and losing streaks, and rewarding a win proportionally to the margin of the win. By incorporating these additional parameters into the Elo formula, we were able to achieve a notable increase in prediction accuracy over the basic Elo formula. 1. Introduction The Elo rating system was created by physicist and chess Grand Master Arpad Elo, as an improvement upon existing chess rating systems. The difference in ratings of two players serves as a predictor of the outcome of a match. For example, if the difference in scores of two players is 100, then the stronger player is predicted to have a 64% of winning a match against the weaker player. The formula for predicting a match s outcome is: 10 R A/400 E[S A ] = 10 R A/ R B/ R B/400 E[S B ] = 10 R B/ R A/400 Virginia Tech CSx824/ECEx424 final project report. Copyright 2016 by the author(s). After a match each player s rating is updated based on the outcome of the match. If the favored player wins the match, he will gain relatively few points. However if there is an upset and the weaker player wins, his rating gains relatively more points, than the favored player would if he had won. The formula for updating ratings is as follows: R post = R pre + K(S E[S]) 1, for a win S =.5, for a draw 0, for a loss In the above equation K serves a weighting term that determines how much the previous match should affect the player s rating. The higher the K-factor, the quicker a player s rating will rise or fall. The value of K depends on the particular application of the Elo system. In our implementation, we used a base value of 20. This is the value used by the World Football Elo system for friendly matches. When new players enter the system, they are assigned a default rating. This default rating is again application specific. We chose a default rating of 1200 in our implementation to match the value used by the World Football Elo system. In our experiments, we used a data set from the English Premier League, that contains historical data on matches spanning more than 100 years. We focused on the most recent data by training on recent seasons and predicting outcomes for the latest season. Lastly, the outcomes of a match are either a win, a draw, or a loss as described previously. Our model predicts a win if the score is at least 0.60 (60% probability that this team will win) and predicts a loss if the score is less than or equal to 0.40 (60% probability that the other team will win. Between 0.40 and 0.60,
2 a draw is the predicted outcome. 2. Experiments The following subsections detail the four experiments run on individual factors being adjusted. Given the short timeline for this research, we analyzed each factor individually first, and then brought them together into a final model. Each experiment was modeled off of the following steps: 1. Set the minimum, maximum, and step values for the parameter. 2. Set the number of years for training. 3. For each iteration of training, loop over possible parameter values and record results. 4. Visually examine the recorded best results and choose the best parameter value. Also due to the short timeline, the robustness of our experiments and training methods may not be optimal. For instance, since the original motivation behind this research is to improve the Elo rating system in order to better predict next season s games, we trained to predict the 2013 season. This means that we trained on the n previous seasons and then evaluated our prediction accuracy on the season. Throughout our experiments we mention correctness and almost correctness. We define correctness as the percentage of predictions that were correct (e.g. we predicted a win for team A and its true outcome was a win for team A). However, if we predict a draw and the outcome ends up being a close game that resulted in a win (e.g. the final score is 1-0), then our prediction was close. Thus, we define almost correctness as the percentage of predictions that were almost correct, meaning that we are only wrong if we predict a win and the outcome is a loss or vice versa. This metric was used earlier on in our experimentation to get a better idea of how well our model was doing. In our final conclusions, however, we do not reference almost correctness since correctness is the ultimate metric on which we focus (in real-world application, almost correct doesn t count) K-Factor The first factor examined was altering the K-factor. Altering the K-factor consists of two parts: setting the standard K-factor to be used throughout the season, and setting the initial (higher) K-factor to be used at the beginning of the season. The intuition behind this change in K-factor is that the beginning of the season demonstrates the quality of this season s team, so setting a higher K-factor allows for bigger adjustments to the Elo rankings of the teams. A few examples that affect a team s quality between seasons include acquiring or losing players, changing management, and changing the teams in the league. As mentioned previously, the initial, standard K- factor was set at 20, reflecting the World Football Elo Ranking system s value for friendly matches. To implement a higher initial K-factor, the point at which the K-factor dropped back to its standard level had to be decided. A single Premier League season consists of 38 matches for each club. We decided to set this change point, call it C, to be 7. This was chosen, because seven is just under two months into the season and the season is almost one-fifth completed 1. Therefore, the update formula remains the same, but the choice of K must be made. R post = R pre + K(S E[S]) { K initial if matches < C K = if matches C K normal First, we trained our model to find the best K normal. The number of seasons to train on was set to be 1, 2, 3, and 5. The values for K normal ranged from 5 to 60 with a step size of 5. Looking at the outcomes of all four of these, the value of K normal = 25 was chosen. The original choice of 20 was a very good estimate for this value which is expected since the World Elo Rating system is well known. Second, we trained our model to find the best K initial given a changing point of C = 7 and the above K normal. Training on the same seasons and using the same range yielded a K initial = 40 to be the best choice. Figure 1 illustrates improvement we see when training on the single previous season. For this experiment, the training done on just the previous season was the only training that saw improvement by setting different K initial values. Training on more than one season resulted in similar optimal values for K initial, but often included multiple values that shared the same correct percentage. All of these multi-year trainings had a range of less than 0.5%. It is important to note that both of these experiments 1 Note: We do acknowledge that this value was determined arbitrarily. In the future, this value could be optimized.
3 Figure 1. Initial K Factor (1 Season) were conducted using a home field advantage boost equal to 100. This is the home field advantage mentioned by the World Football Elo rating system. The home field advantage boost will be discussed in the next section. Thus, in the end, setting K initial = 40 and K normal = 25. Table 1 illustrates the change in prediction accuracy over 1, 2, 3 and 5 seasons. As you can see, incorporating only a different initial K-factor doesn t improve the model very much. However, we still select the above K-factor values, because when other parameters are added, the effects of the K-factor are much greater than 1% Home Field Advantage Home field advantage was the next feature incorporated. Intuitively, the home team was an increased chance of winning, because they are at home. The reasoning behind this can be attributed to familiarity with the stadium or turf, supporters, sleeping in one s own bed the previous night, etc.; the list goes on. This Yrs % Increase Table 1. K-Factor Accuracy Increases boost was added to the difference rating. Thus, the new prediction equation is E[S home ] = (R home R away+h)/400 E[S away ] = 1 E[S home ] where h is the home field advantage boost. Note that the above predictions are from the home team s perspective. This is a reflection of design decisions made during the implementation of our Elo Rating system: always calculate the home team s prediction and then deduce the away team s prediction.
4 Yrs % Increase with base K-factor % Increase with optimal K-factor Table 2. Home Field Prediction Accuracy Increases The range for home field advantage was from 0 to 200 with a step size of 10. Training for the home field advantage boost was performed twice: first with the base K factors and then with the optimal K-factors discussed previously. The models were trained on 1, 2, 3, and 5 seasons. When the home field boost was optimised based on the original, base K-factor of 20, the increase in prediction accuracy was significant with an average increase over 15%. These values are tabulated in Table 2. The optimal home field boost for this case was determined to be 130. When the optimal altering K-factors were used from the previous experiment, the optimal home field boost value returned was 120. The increases in prediction accuracy due to home field advantage are very significant: increasing by over 17% for each training set. These values are also shown in Table Modified Point Assignment The next factor we considered when modifying the Elo formula is the number of points assigned for a win, loss, and draw. The intuition behind this, is that a victory or a loss by a large margin should have a larger impact on the team s rating, than a win or loss by a small margin. To account for this in the Elo rating update, the score assigned for a win or loss becomes proportional to the margin of the victory. The score assignment used in the Elo update equation then becomes: 1 + ( s m 1)F m, for a win S =.5, for a draw 0 ( s m 1)F m, for a loss where F m is the marginal score factor and s m is the marginal score, the difference of the home team and away team s match scores. With this formula, if a team wins a match by the minimum possible margin of 1, they are rewarded with the standard Elo score of 1. However, for each victory with a margin of 2 or more, the team is rewarded with 1 plus some bonus. To examine the impact this modification had on the % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % Table 3. Modified Score Assignment (3 Seasons) model s predictive performance, the value of F m was varied from 0 to 2 in steps of Figure 2 shows the predictive performance for various values of F m trained on 1 season of data. Examining this figure, we see that for nearly all values of F m there is an improvement in the predictive performance of the model. Examining the numerical data, we found that a value of 0.57 gave the best accuracy of 60.00%, which is a roughly a 10% increase in accuracy over the basic Elo model performance. We ran similar experiments where the model was trained on 3 years and 5 years of data. These experiments produced similar results to the first. This data is summarized in Table 3 and Table Streaks The final factor that we incorporated into the Elo model, is win and loss streaks. The intuition behind adding this is: if a team is on a winning streak, it is likely that they will continue the streak and win the next game. Likewise, if a team is on a losing streak, it is likely that they will lose the next match. An existing paper studying the addition of momentum to the Elo system proved useful when implementing our method (Bester & von Maltitz, 2013). This idea was incorporated into the model in a similar manner to
5 Figure 2. Modified Score Assignment (1 Season) % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % Table 4. Modified Score Assignment (5 Seasons) home field advantage, in that each team has a boost added to their predicted outcome if they are on a winning streak, and a penalty subtracted if they are on a losing streak. To add streaks to the model we first rewrite the prediction equation as: E[S A ] = (R A R B )/400 E[S B ] = 1 E[S A ] We then apply bonuses N A and N B to the difference of the team s ratings, to get: where E[S A ] = (R A R B +N A N B )/400 E[S B ] = 1 E[S A ] C, for a win streak N A,B = 0, no streak C, for a loss streak C is some constant boost and was varied from -100 to 100 in steps of 10 to find the optimal value. An additional parameter that had to be considered for this experiment was the threshold for a sequence of wins or losses to be considered a streak. We varied this
6 % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % Table 5. Streaks (1 Season) % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % Table 6. Streaks (3 Seasons) threshold from 2 to 4, and while all thresholds gave some improvement when trained on a single season, a value of 2 produced the best overall results. We then trained a model with these parameters on 3 seasons and then 5 seasons. In each case the model performed better than the basic Elo model. Analyzing the results we found a value of -10 to be optimal. This would indicated that our original intuition about streaks was incorrect. The model seems to indicate that a team is more likely to break the streak in their next game than to continue it. The results of these experiments are summarized in the tables 5, 6 and 7 below Grid Search The final experiment that we conducted was to run a grid search over the entire parameter space with the exception of the K-factor. We held the K-factor fixed at the initial value of 40 and normal value of 25. This experiment is obviously inefficient due to the size of the parameter space, but can provide insight on how the different factors we investigated work together for the single best model. The grid search was run with training on 1, 2, 3, and 5 seasons previous to the season. We chose the bounds for our parameters based on the experiments we ran on the individual parameters. The % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % Table 7. Streaks (5 Seasons)
7 Experiment K-factor Optimal K initial = 40 K normal = 25 Home Field h = 200 Modified Points F m = 0.50 Streaks C = 40 Table 8. Summary of Experiment Results Parameter Original Optimal K initial K normal Home Field, h N/A 120 Modified Points, F m N/A 0.20 Streaks, C N/A -10 Table 9. Original and Optimal Model Parameters home field boost ranged from 120 to 300 with a step size of 10, the win streak factor was between -50 and 75 with an increment of 0.1, and the win loss margin factor for score assignment ranged from 0 to 60 by 1. The experiment exported every parameter permutation with correct and almost metrics to a csv file. This file was sorted with respect to the correct percentage, and then by the almost correct percentage. We then examined the file to see the trends across the different training seasons and to select optimal parameter values. The highest correct accuracies were found with home field advantage boosts ranging from 180 to 260, but the center of these appeared to be 200. Thus the optimal home field advantage boost was chosen to be 200. The win loss factor optimal value fell in the range of 0.40 to Thus, choosing the center of these yields the optimal win loss factor to be Lastly, the win streak factor showed a surprising range of -50 to 0, with the distribution skewed to the right. The value -40 was chosen for the optimal win streak. 3. Results and Conclusions 3.1. Results Table 8 summarizes the results found in each of the experiments. Table 9 shows the original values for the base Elo model and the optimal values found in the grid search. Combining all of the adjustments made to the Elo rat- Premier League (380 matches) Yrs Original Optimal % Increase % Correct Table 10. Prediction Accuracy for Premier League ing system for each parameter together yields the following equations for prediction and update. E[S home ] = R post = R pre + K(S E[S]) 1, for a win S =.5, for a draw 0, for a loss { K initial if matches < C K = if matches C K normal (R home R away+n home N home +h)/400 E[S away ] = 1 E[S home ] 3.2. Conclusions C, for a win streak N A,B = 0, no streak C, for a loss streak These values support most of the hypotheses we made at the beginning of our research. altering the K-factor, boosting the home team s rating, and factoring in the margin of victory or defeat all improved the prediction accuracy. Our hypothesis about the win streak factor was not supported though. However, when we allowed for this factor to be negative, it did improve the prediction accuracy. Intuitively this means that a team on a win streak is less likely to keep the streak going. The optimized model was ran against the original model to determine the increase in prediction accuracy. The optimized model improved the original model s prediction accuracy by over 20%. Table 10 displays the comparisons for training on 1, 2, 3 and 5 seasons. This research shows a very significant improvement for predicting the outcome of soccer matches. The factor that contributed the most to the improvement was the home field advantage.
8 Champions League (552 matches) Yrs Original Optimised % Increase % Correct Table 11. Prediction Accuracy for Champions League League 1 (552 matches) Yrs Original Optimised % Increase % Correct Table 12. Prediction Accuracy for League 1 An interesting note about the home field advantage boost is that home field advantage appears to have a bigger impact previously than it does currently. This observation results from models that trained farther back in time returned a higher home field advantage boost Applying Our Model to Other Leagues The data set on which we trained our data included multiple English football leagues. Up to this point, we have focused exclusively on the English Premier League. Now, however, we look at the other leagues included: the Champions League, League 1, and League 2 as they are currently known. We ran the same comparison that we ran for the Premier League on these leagues: testing the optimized model against the original model. All the leagues saw significant improvement as well, although the improvement was not as much as the Premier League. This observation lends itself to the fact that the model may be over fit to the Premier League exclusively, rather than fit for all football leagues. This is understandable, however, since this paper details the research on fitting a model to the Premier League. More discussion on this thought can be found in the Further Development section. Tables 11 to 13 display the comparisons for training on 1, 2, 3 and 5 seasons for the Champions league, League 1, and League 2, respectively. Even though these predictions weren t improved as much as the Premier League predictions, the smallest increase was still over 15%, which is significant. League 2 (552 matches) Yrs Original Optimised % Increase % Correct Table 13. Prediction Accuracy for League 2 4. Further Development The biggest roadblock throughout our research into improving Elo rankings for sports was the time. However, this was understood at the onset of the research. We completed all of our original plans for our research into this project. We had further goals in place if time permitted, but we were only able to begin looking into these goals. Primarily, the next step for our research would have been to develop an algorithm to train an optimal model on all of these parameters efficiently. To do this, a grid search can not be employed, and thus the parameter space would need to be examined further. Particularly, if any parameters were found to be convex, that would greatly aid in the creation of such an algorithm. During our experimentation, it was noticed that home field appeared to be convex (ignoring slight noise since it is real world data). This held throughout the experiments, even when combined with the other parameters. Therefore, the home field boost could be found using standard optimization methods to arrive at a good local optimum. The modified points assignment for win/loss margin appeared to vary drastically. A common shape was not easily discernible directly from the experimental results. Further investigation into this parameter would be needed. The streaks parameter also posed difficulty. The fact that the parameter was negative can be explained, but isn t necessarily intuitive. We held the streak threshold constant throughout all of our experiments at a value of 2. The streak threshold is the number of games a club must consecutively win (or lose) to be considered on a streak. This streak threshold could be researched as well in the future. Similarly, the changing point for transitioning between the initial and normal K-factor was set and held constant at 7, and, thus, could be researched in the future. In conclusion, the goals that we set out to complete were completed, but there is plenty left to do with
9 these results. For instance how well will this transition to other football leagues, or even other sports? Acknowledgments We would like to thank Dr. Bert Huang for providing us with guidance and feedback during this research project and for teaching us about machine learning throughout this semester. References Bester, D.W. and von Maltitz, M.J. Introducing momentum to the elo rating system. University of the Free State: Department of Mathematical Statistics and Actuarial Science, URL Documents/00002/2069_eng.pdf. Glickman, Prof. Mark E. A comprehensive guide to chess ratings. URL research/acjpaper.pdf.
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USING MICROSOFT EXCEL TO MODEL A TENNIS MATCH Tristan J. Barnett and Stephen R. Clarke School of Mathematical Sciences Swinburne University PO Box 218 Hawthorn Victoria 3122, Australia tbarnett@groupwise.swin.edu.au
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### NF5-12 Flexibility with Equivalent Fractions and Pages 110 112
NF5- Flexibility with Equivalent Fractions and Pages 0 Lowest Terms STANDARDS preparation for 5.NF.A., 5.NF.A. Goals Students will equivalent fractions using division and reduce fractions to lowest terms. | 10,754 | 49,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-51 | latest | en | 0.935677 |
https://converter.ninja/time/years-to-seconds/250-yr-to-s/ | 1,723,503,878,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641052535.77/warc/CC-MAIN-20240812221559-20240813011559-00074.warc.gz | 139,854,230 | 5,950 | # 250 years in seconds
## Conversion
250 years is equivalent to 7889231493.6696 seconds.[1]
## Conversion formula How to convert 250 years to seconds?
We know (by definition) that: $1\mathrm{yr}=31556926\mathrm{sec}$
We can set up a proportion to solve for the number of seconds.
$1 yr 250 yr = 31556926 sec x sec$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{sec}=\frac{250\mathrm{yr}}{1\mathrm{yr}}*31556926\mathrm{sec}\to x\mathrm{sec}=7889231500\mathrm{sec}$
Conclusion: $250 yr = 7889231500 sec$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 second is equal to 1.26755058563361e-10 times 250 years.
It can also be expressed as: 250 years is equal to $\frac{1}{\mathrm{1.26755058563361e-10}}$ seconds.
## Approximation
An approximate numerical result would be: two hundred and fifty years is about seven billion, eight hundred and eighty-nine million, two hundred and thirty-one thousand, four hundred and ninety-three point six seven seconds, or alternatively, a second is about zero times two hundred and fifty years.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic.
Was it helpful? Share it! | 359 | 1,325 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-33 | latest | en | 0.797349 |
https://physics.stackexchange.com/questions/350947/how-to-show-that-the-einstein-hilbert-action-is-diffeomorphism-invariant | 1,726,699,999,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651941.8/warc/CC-MAIN-20240918201359-20240918231359-00040.warc.gz | 402,680,900 | 42,104 | # How to show that the Einstein-Hilbert action is diffeomorphism invariant?
It is often stated in texts on general relativity that the theory is diffeomorphism invariant (N.B., I am considering active diffeomorphisms), i.e. if the universe is represented by a manifold $\mathcal{M}$ with metric $g_{\mu\nu}$ and matter fields $\psi$ and $\phi:\mathcal{M}\rightarrow\mathcal{M}$ is a diffeomorphism, then the sets $(\mathcal{M},\,g_{\mu\nu},\,\psi)$ and $(\mathcal{M},\,\phi^{\ast}g_{\mu\nu},\,\phi^{\ast}\psi)$ represent the same physical situation.
Given this, how does one show explicitly that the Einstein-Hilbert action $S_{EH}[g]=M^{2}_{Pl}\int\,d^{4}x\sqrt{−g}R$ is diffeomorphism invariant?
I know that under an infinitesimal diffeomorphism $x^{\mu}\rightarrow y^{\mu}=x^{\mu}+tX^{\mu}$, generated by some vector field $X=X^{\mu}\partial_{\mu}$, the metric transforms such that $\delta_{X}g_{\mu\nu}=\mathcal{L}_{X}g_{\mu\nu}=2\nabla_{(\mu}X_{\nu)}$ and so $\delta_{X}\sqrt{-g}=\sqrt{-g}g^{\mu\nu}\nabla_{(\mu}X_{\nu)}$. Furthermore, the Ricci scalar transforms such that $\delta_{X}R=\mathcal{L}_{X}R=X^{\mu}\nabla_{\mu}R$.
This is all well and good, but I'm unsure how the volume element $d^{4}x$ transforms under diffeomorphisms?! I'm inclined to think that it doesn't transform, since if I've understood things correctly, under a diffeomorphism, the points on the manifold are mapped to new points, but simultaneously, the coordinate maps are "pulled back", such that the coordinates of the point at its new position in the new coordinate chart are the same as the coordinates of the point at its old position in the old coordinate chart.
If this is correct, then I think I may be able to show that $S_{EH}$ is diffeomorphism invariant as follows: $$\delta_{X}S_{EH}=\int\,d^{4}x\left[\delta_{X}(\sqrt{-g})R+\sqrt{-g}\,\delta_{X}(R)\right]=\int\,d^{4}x\sqrt{-g}\left[\nabla_{\mu}X^{\mu}R+X^{\mu}\nabla_{\mu}R\right]\\=\int\,d^{4}x\sqrt{-g}\nabla_{\mu}\left(X^{\mu}R\right)=\int\,d^{3}\Sigma_{\mu}\,X^{\mu}R\qquad\qquad\qquad\qquad\qquad\;\;\,$$ where I have used Stokes' theorem in the penultimate equality, in which $d^{3}\Sigma_{\mu}$ is the covariant (hyper-) surface element of oriented boundary to the 4-volume. Now, assuming that $X$ has compact support, such that $X^{\mu}\rightarrow 0$ on the hypersurface $\Sigma$, then we find that $\delta_{X}S_{EH}=0$, i.e. the Einstein-Hilbert action is diffeomorphism invariant.
I'm not sure if this is correct at all, particularly my argument about where or not the volume element $d^{4}x$ transforms or not? Any help would be much appreciated.
Edit
I have been reading this set of notes and the author claims that $d^{4}x$ does not transform under active diffeomorphisms (c.f. pages 9 and 19 in particular), but I don't quite follow the reasoning.
The volume element $\sqrt{-g}d^4x$ is a scalar by design, so integrals of scalars with respect to it are invariant. $R$ is a scalar. So the action is itself scalar-valued.
Not sure why you want to restrict our self to a one-parameter group of diffeos, this action is clearly invariant under a finite diffeomorphism because it is the integral of a 4-form $R\sqrt{-g}dx^0\wedge...\wedge dx^3$ and we know that integrals of 4-forms are invariants (on a 4 dimensional manifold that is).
• Is this true for active diffeomorphisms though? $\sqrt{-g}$ clearly isn't invariant and I'm not sure how $d^{4}x$ transforms; I would have assumed by a Jacobian factor, but this isn't merely a coordinate transformation. This set of notes (web.mit.edu/edbert/GR/gr5.pdf , c.f. pages 9 and 19) claim that $d^{4}x$ doesn't transform under an active diffeomorphism.
– Will
Commented Aug 8, 2017 at 17:59
• @Will There is no difference between an active or passive diffeomorphism really, aside from interpretation. The object you should be considering is $\sqrt{-g}dx^0\wedge...\wedge dx^3$ instead of the two separately, which is a honest-to-god 4-form, and as such is invariant, Commented Aug 8, 2017 at 18:05
• It is not necessarily true though that a n-form is invariant under active diffeomorphisms though, that is, it isn't necessarily equal to its pullback, $\phi^{\ast}}\omega\neq\omega$.
– Will
Commented Aug 8, 2017 at 18:15
• @Will But integrals are. Assuming said diffeo is orientation-preserving. Commented Aug 8, 2017 at 18:16
• @Will Freely available notes don't come to mind but as far as I recall there is a nice discussion on background independence vs diffeomorphism invariance in Staumann's General Relativity book. As for the author, dunno, I feel this is a fairly common misconception and even excellent physicists sometimes use sloppy differential geometry. I will go through the notes and see what I see soon though. Commented Aug 9, 2017 at 13:40 | 1,436 | 4,770 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-38 | latest | en | 0.84162 |
https://codebun.com/c-program-to-swap-two-numbers-without-third-variable/ | 1,657,113,810,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104672585.89/warc/CC-MAIN-20220706121103-20220706151103-00523.warc.gz | 204,036,769 | 24,215 | # C# Program to swap two numbers using temporary variable and without using third variable
How to write a C# Program to swap two numbers, how to swap two numbers with temporary variable,how to swap two numbers without using third variable.
# What is Swapping of Numbers ?
The act of swapping two variables refers to mutually exchanging the values of the variables. Generally, this is done with the data in memory.
## Swap Two Numbers with Third Variable
The simplest method to swap two variables is to use a third temporary variable .
### Algorithm of swapping two numbers with temporary variable:
Step 1: Define 3 variables x, y and temp;
Step 2: give values to the x and y;
Step 3: temp = x ;
Step 4: x = y ;
Step 5: y = temp;
Step 6: Print values of x and y;
### Program of swapping two numbers with temporary variable in C#
```using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Program
{
class Program
{
static void Main(string[] args)
{
int num1, num2, temp;
Console.Write("\n Enter the First Number : ");
num1 = int.Parse(Console.ReadLine());
Console.Write("\n Enter the Second Number : ");
num2 = int.Parse(Console.ReadLine());
temp = num1;
num1 = num2;
num2 = temp;
Console.Write("\n After Swapping : ");
Console.Write("\n First Number : "+num1);
Console.Write("\n Second Number : "+num2);
Console.Read();
}
}
}```
#### Input & output:
```Enter the First Number : 5
Enter the Second Number : 7
After Swapping :
First Number : 7
Second Number : 5```
## How to swap two numbers without using a temporary variable
If given two variables are x, and y, swap two variables without using a third variable.There is a common way to swap two numbers without using third variable.
Using Arithmetic Operators : The idea is to get a sum in one of the two given numbers. The numbers can then be swapped using the sum and subtraction from the sum.
### C# Program to Swap Two Numbers without third variable
```using System;
namespace CSharpProgram
{
class Program
{
static void Main(string[] args)
{
Console.WriteLine("\n Enter first number");
int n1 = int.Parse(Console.ReadLine());
Console.WriteLine("\n Enter second number");
int n2 = int.Parse(Console.ReadLine());
Console.WriteLine("\n Numbers Before swap n1 = " + n1 + " n2 = " + n2);
n1 = n1 * n2; //n1=50 (5*10)
n2 = n1 / n2; //n2=5 (50/10)
n1 = n1 / n2; //n1=10 (50/5)
Console.Write("\n After swap n1 = " + n1 + " n2 = " + n2);
Console.ReadLine();
}
}
}
```
#### Input & Output:
```Enter first number 27
Enter second number 15
Numbers Before swap n1 = 27 n2 = 15
After swap n1 = 15 n2 = 27```
So in this post we have learn how to swap two numbers using temporary variable or without using temporary variable. | 687 | 2,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-27 | latest | en | 0.659807 |
https://electronics.stackexchange.com/questions/649863/how-do-i-optimize-my-op-amp-guard-ring | 1,726,503,003,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00262.warc.gz | 194,226,406 | 41,158 | # How do I optimize my op-amp "guard ring"?
I'm trying to measure the moisture content of wood via a resistance measurement. The moisture content range I'm trying to measure is in the order of 1 MΩ to 1 GΩ, so only ~5 nA current at 5 V DC.
The op-amp I'm using to amplify the current is an MCP6006R. I'm feeding this into the ADC on an Arduino. The datasheet shows a guard ring configuration around the inputs on the op-amp to minimize unwanted signals. Will this be necessary for the circuit to be reliable?
See PCB with guard ring attempt highlighted below:
• Don't be afraid to space out your schematics more. Avoid wires through text (especially component values), always have GNDs 'pointing down', and label nets to make their purpose clear. It helps you in the long run and makes it far easier to understand for those who are unfamliar with your design. Commented Jan 13, 2023 at 10:00
• I think the circuit will not work and as it is now, the guard ring is currently not the thing that makes the circuit reliable. Commented Jan 13, 2023 at 10:00
• Hi Kyotee, could you explain the working principle of this? What's R2's job in all this? How does C1 and the DC current from WOOD_PROBES interact? Commented Jan 13, 2023 at 11:03
• At 1 Gohm max, you are not yet in the guard ring territory. That starts to matter if you need sub-pA leakage control Commented Jan 13, 2023 at 11:12
• Thanks for all the feedback. The working principle is i can measure the voltage rise on the capacitor which is being charged via the current through the wood probes. from the voltage rise i can calculate the resistance of the wood. R2s job is to limit the current that can flow into the op amp as the inputs have a 5mA max. Circuit principle based on "mdpi.com/1996-1944/12/15/2373/htm#" paper Commented Jan 13, 2023 at 22:07 | 473 | 1,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.932608 |
http://www.chegg.com/homework-help/balance-sheet-consolidationon-january-1-20x3-guild-corporati-chapter-3-problem-5exa-solution-9780078110924-exc | 1,475,116,249,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661775.4/warc/CC-MAIN-20160924173741-00284-ip-10-143-35-109.ec2.internal.warc.gz | 393,097,522 | 18,083 | Advanced Financial Accounting (9th Edition) View more editions Solutions for Chapter 3 Problem 5EXAProblem 5EXA: Balance Sheet ConsolidationOn January 1, 20X3, Guild Corpora...
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Chapter: Problem:
100% (8 ratings)
Balance Sheet Consolidation
On January 1, 20X3, Guild Corporation reported total assets of \$470,000, liabilities of \$270,000, and stockholders’ equity of \$200,000. At that date, Bristol Corporation reported total assets of \$190,000, liabilities of \$135,000, and stockholders’ equity of \$55,000. Following lengthy negotiations, Guild paid Bristol’s existing shareholders \$44,000 in cash for 80 percent of the voting common shares of Bristol.
Required
Immediately after Guild purchased the Bristol shares
a. What amount of total assets did Guild report in its balance sheet?
b. What amount of total assets was reported in the consolidated balance sheet?
c. What amount of total liabilities was reported in the consolidated balance sheet?
d. What amount of stockholders’ equity was reported in the consolidated balance sheet?
STEP-BY-STEP SOLUTION:
Chapter: Problem:
100% (8 ratings)
• Step 1 of 3
a. \$470,000 = \$470,000 - \$44,000 + \$44,000
• Chapter , Problem is solved.
Corresponding Textbook
Advanced Financial Accounting | 9th Edition
9780078110924ISBN-13: 0078110920ISBN:
Alternate ISBN: 9780077419592, 9780077466794, 9780077484255, 9780077899165 | 401 | 1,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2016-40 | latest | en | 0.845566 |
https://codedump.io/share/AonPN1hmrwdY/1/python-physics-library | 1,524,547,910,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946564.73/warc/CC-MAIN-20180424041828-20180424061828-00195.warc.gz | 586,768,358 | 8,737 | Elliot Bonneville - 2 years ago 212
Python Question
# Python physics library?
Are there any good up-to-date physics libraries for Python that are for Linux? I'm just getting into Python using PyGame, but PyGame's lack of a physics library isn't cool. I spent about two hours trying to find a good physics library but it's like trying to grab oil; I can't seem to do it.
I barely need a physics engine at all; all I want to do is program an object to 'jump' up and then fall back to the ground. There seems to be some simple collisions going on (which PyGame can handle, I think) but it's the actual jump calculation that's stumping me. If it turns out that there aren't any good ususable physics libraries, the problem seems simple enough that I might just try to find a basic acceleration equation and a gravity equation and try to apply those... I'd like to avoid having to do that, though.
Thanks for any help.
Answer Source
The basic physics kinematic equations are all you need. Even though the questions was already answered, if I were you, I'd still do it by hand just because using a library seems like overkill. Start the equation for velocity:
``````velocity = initial velocity + (acceleration * time)
``````
From there, we integrate to find position:
``````position = initial position + (initial velocity * time) + (acceleration * time^2)
``````
Your jump is looking to calculate the y position of the character, so just use that equation to calculate y position and toy with initial velocity and acceleration. Standard acceleration due to gravity is -9.8 meters per second^2 (at least on the surface of the earth - it's different son different planets). Start with initial position whatever your character is at, or 0 if the ground is 0 to you.
So:
``````y = vt + (-9.8)t^2
``````
Pick a `v` value, and t should be the elapsed game time since he started jumping.
You only need one line of code to do this, no libraries necessary!
EDIT: dealing with what to do when you land.
So in the real world, acceleration is caused by unbalanced forces. Gravity is always acting on you, but when you're standing in the ground, it's being countered and canceled out by the "normal force" of the ground. As long as the ground you're standing on is strong enough to support your weight (weight being technically the force due to gravity of a given mass), the ground will push back up and counter gravity and you will not accelerate downward. So in your game, if you're not actually simulating forces, just change the acceleration from -9.8 to 0 when your character is touching the ground.
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http://www.mathworks.it/it/help/symbolic/assumealso.html?nocookie=true | 1,394,817,852,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678694248/warc/CC-MAIN-20140313024454-00002-ip-10-183-142-35.ec2.internal.warc.gz | 389,144,236 | 10,406 | Accelerating the pace of engineering and science
• Trials
# assumeAlso
## Description
example
assumeAlso(condition) states that condition is valid for all symbolic variables in condition. It retains all assumptions previously set on these symbolic variables.
example
assumeAlso(expr,set) states that expr belongs to set in addition to all previously made assumptions.
## Examples
### Assumptions Specified as Relations
Set assumptions using assume. Then add more assumptions using assumeAlso.
Solve this equation assuming that both x and y are nonnegative:
```syms x y
assume(x >= 0 & y >= 0)
s = solve(x^2 + y^2 == 1, y)```
```s =
{[(- x + 1)^(1/2)*(x + 1)^(1/2), 1],...
[-(- x + 1)^(1/2)*(x + 1)^(1/2), 1]} intersect...
Dom::Interval([0], Inf)
```
Now add the assumption that x < 1. To add a new assumption without removing the previous one, use assumeAlso:
`assumeAlso(x < 1)`
Solve the same equation under the expanded set of assumptions:
`s = solve(x^2 + y^2 == 1, y)`
```s =
(1 - x)^(1/2)*(x + 1)^(1/2)```
For further computations, clear the assumptions:
`syms x y clear`
### Assumptions Specified as Sets
Set assumptions using syms. Then add more assumptions using assumeAlso.
When declaring the symbolic variable n, set an assumption that n is positive:
`syms n positive`
Using assumeAlso, you can add more assumptions on the same variable n. For example, assume also that n is and integer:
`assumeAlso(n,'integer')`
To see all assumptions currently valid for the variable n, use assumptions. In this case, n is a positive integer.
`assumptions(n)`
```ans =
[ n in Z_, 0 < n]```
For further computations, clear the assumptions:
`syms n clear`
### Assumptions on Matrix Elements
Use the assumption on a matrix as a shortcut for setting the same assumption on each matrix element.
Create the 3-by-3 symbolic matrix A with the auto-generated elements:
`A = sym('A', [3 3])`
```A =
[ A1_1, A1_2, A1_3]
[ A2_1, A2_2, A2_3]
[ A3_1, A3_2, A3_3]```
Suppose that all elements of this matrix represent rational numbers. Instead of setting an assumption on each element separately, you can set the assumption on the matrix:
`assume(A,'rational')`
Now, add the assumption that each element of A is greater than 1:
`assumeAlso(A > 1)`
To see the assumptions on the elements of A, use assumptions:
`assumptions(A)`
```ans =
[ A1_1 in Q_, A1_2 in Q_, A1_3 in Q_, A2_1 in Q_, A2_2 in Q_,...
A2_3 in Q_, A3_1 in Q_, A3_2 in Q_, A3_3 in Q_,...
1 < A1_1, 1 < A1_2, 1 < A1_3, 1 < A2_1, 1 < A2_2, 1 < A2_3,...
1 < A3_1, 1 < A3_2, 1 < A3_3]
```
For further computations, clear the assumptions:
`syms A clear`
When you add assumptions, ensure that the new assumptions do not contradict the previous assumptions. Contradicting assumptions can lead to inconsistent and unpredictable results.
In some cases, assumeAlso detects conflicting assumptions and issues the following error:
```syms y
assume(y,'real')
assumeAlso(y == i)```
```Error using mupadmex
Error in MuPAD command: Inconsistent assumptions
detected. [property::_setgroup]```
assumeAlso does not guarantee to detect contradicting assumptions. For example, you can assume that y is nonzero, and both y and y*i are real values:
```syms y
assume(y ~= 0)
assumeAlso(y,'real')
assumeAlso(y*i,'real')```
To see all assumptions currently valid for the variable y, use assumptions:
`assumptions(y)`
```ans =
[ y in R_, y ~= 0, y*i in R_]```
For further computations, clear the assumptions:
`syms y clear`
## Input Arguments
expand all
### condition — Assumption statementsymbolic expression | symbolic equation | relation | vector of symbolic expressions, equations, or relations | matrix of symbolic expressions, equations, or relations
Assumption statement, specified as a symbolic expression, equation, relation, or vector or matrix of symbolic expressions, equations, or relations. You also can combine several assumptions by using the logical operators and, or, xor, not, or their shortcuts.
### expr — Expression to set assumption onsymbolic variable | symbolic expression | vector | matrix
Expression to set assumption on, specified as a symbolic variable, expression, vector, or matrix. If expr is a vector or matrix, then assumeAlso(expr,set) sets an assumption that each element of expr belongs to set.
### set — Set of integer, rational, or real numbers'integer' | 'rational' | 'real'
Set of integer, rational, or real numbers, specified as one of these strings: 'integer', 'rational', or 'real'.
expand all
### Tips
• assumeAlso keeps all assumptions previously set on the symbolic variables. To replace previous assumptions with the new one, use assume.
• When adding assumptions, always check that a new assumption does not contradict the existing assumptions. To see existing assumptions, use assumptions. Symbolic Math Toolbox™ does not guarantee to detect conflicting assumptions. Conflicting assumptions can lead to unpredictable and inconsistent results.
• When you delete a symbolic variable from the MATLAB® workspace using clear, all assumptions that you set on that variable remain in the symbolic engine. If later you declare a new symbolic variable with the same name, it inherits these assumptions.
• To clear all assumptions set on a symbolic variable and the value of the variable, use this command:
` syms x clear`
• To clear assumptions and keep the value of the variable, use this command:
`sym('x','clear')`
• To clear all objects in the MATLAB workspace and close the MuPAD® engine associated with the MATLAB workspace resetting all its assumptions, use this command:
`clear all`
• If condition is an inequality, then both sides of the inequality must represent real values. Inequalities with complex numbers are invalid because the field of complex numbers is not an ordered field. (It is impossible to tell whether 5 + i is greater or less than 2 + 3*i.) MATLAB projects complex numbers in inequalities to real axis. For example, x > i becomes x > 0, and x <= 3 + 2*i becomes x <= 3.
• The toolbox does not support assumptions on symbolic functions. Make assumptions on symbolic variables and expressions instead.
• Instead of adding assumptions one by one, you can set several assumptions in one function call. To set several assumptions, use assume and combine these assumptions by using the logical operators and, or, xor, not, all, any, or their shortcuts. | 1,591 | 6,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2014-10 | longest | en | 0.8072 |
https://www.enotes.com/homework-help/abcd-square-with-ab-12-point-p-interior-distances-223011 | 1,675,742,602,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500384.17/warc/CC-MAIN-20230207035749-20230207065749-00503.warc.gz | 700,641,669 | 18,669 | # ABCD is a square with AB=12. Point P is in interior and the distances to A,B and to the side CD are equal. Find the distance.
We are given that ABCD is a square with AB=12. Point P is inside the square and the distances to A,B and to the side CD are equal. We have to find this distance.
Now the distance of P from A and B is equal. Therefore we know that the perpendicular from P to AB cuts it halfway. Let this point be X, so that we have AX = BX = 12/2 = 6.
Now take the distance that we have to determine as Z. From the triangle AXP right angled at P we have Z^2 - 6^2 = (PX)^2. Also the distance from P to CD is Z. So PX + Z = 12
=> sqrt ( Z^2 - 6^2) + Z = 12
=> sqrt ( Z^2 - 6^2) = 12 - Z
square both the sides
=> Z^2 - 36 = 144 + Z^2 - 24Z
=> 24Z = 180
=> Z = 180/24
=> Z = 7.5
Therefore the required distance is 7.5
Approved by eNotes Editorial Team | 280 | 872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-06 | latest | en | 0.940578 |
https://chenhaoxiang.cn/2021/06/20210604095525042c.html | 1,642,884,665,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303884.44/warc/CC-MAIN-20220122194730-20220122224730-00403.warc.gz | 228,532,059 | 9,531 | Buuctf crackme personal gains and Solutions
Summer man 2021-06-04 09:55:39
This question took me nearly two days , In the meantime, because of watching wp I can't understand it at all. I'm crazy to ask for advice , Finally, I debugged it myself . It's not easy , And I really learned a lot , Record here .
Open the program first , The program asks us to enter a user name and password , understand .
throw sth. into ida
Enter the main function wmain
Change the name of the function to a word that you can understand
From here while And the innermost one break You know , To jump out of the loop , These two if All have to be satisfied
First, let's analyze the first if, Click in sub_401830 And start analyzing
Because of this if To establish , So this return The following expression must be true , And v14==43924 establish , We're going back here （ Literally ）
There's a lot going on here if Judge , According to experience, we can put if The following numbers press r Convert to character , Here we guess if When all are satisfied, the result is just equal to 43924（ I can't figure out if I don't guess , All analysis .. It's scary to think about ）
You can write a python The script validates our guess
Why , The answer is not 43924, Why is that （ Stick reading ）
Turned out to be
There are two options for this place , I chose the one above .
After changing to the one below
perfect . The resulting v17 by "dbappsec"
ok, Continue to push back
The first time I did this, I analyzed them for a long time , I don't know what this is , After checking on the Internet, it is found that it is an anti debugging function , I don't care .
Because we've got v17, At the same time, the user name also knows that it is welcomebeijing, At the beginning, it was said that the user name should be entered when the program was started . So we can use ollydbg The method of dynamic debugging comes to byte_416050 Value
Drag in ollydbg, After calculating the dynamic migration , Execution procedure
enter one user name welcomebeijing
Now we stop , find ida Of
The upper one xor eax,ecx That's the key step , Symbolize
XOR here , And from the assembly code above, we can see the movzx ecx, [ebp+var_209] Yes, it will byte_416050 The value of the ecx, Here ecx The value of is what we want to know ！
thus , We got ideas , stay xor Bottom breakpoint , Stop the program at each run time xor The previous order of , It's easy for us to know ecx Value .
And then back to the program , Just type in a password （ It's going to end up there anyway
Did you see? ,ecx Value .
because while（v6<8）, So it's implemented 8 Time , Here we continue f9 perform , And record every time ecx Value . obtain ecx, namely byte_416050 The stored value of 0x2a,0xd7,0x92,0xe9,0x53,0xe2,0xc4,0xcd
And then continue to push back
（ My expression may be a little strange , Let's make a little understanding
The next one if It doesn't matter , You can write code directly
Throw this string of characters on the Internet md5 Decrypt online , obtain flag{d2be2981b84f2a905669995873d6a36c}
I feel like I don't know anything when I do it , But after finishing, I found that I basically understood this problem , Still very happy
Sentiment ：
.
..
Love is so cute
（ Unconscious praise | 826 | 3,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-05 | latest | en | 0.922523 |
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# H.C.F and L.C.M Questions
Home > Quantitative Aptitude > H.C.F and L.C.M > General Questions
NA
SHSTTON
68
Solv. Corr.
29
Solv. In. Corr.
97
Attempted
0 M:0 S
Avg. Time
1 / 20
Choose the correct option.
The H.C.F of two numbers is 11 and their L.C.M is 7700. If one of the numbers is 275, then the other is:
A308
B208
C318
D283
Explanation:
lcm*hcf = product of 2 nos
11*7700=275*?
ans =308
Workspace
NA
SHSTTON
36
Solv. Corr.
25
Solv. In. Corr.
61
Attempted
0 M:8 S
Avg. Time
2 / 20
Choose the correct option.
Let 13 and 273 are the HCF and LCM of two numbers respectively, and if one of them is less than 140 and greater than 60 then what will be that number?
A121
B91
C75
D131
Explanation:
Solution: Product of two number = Product of their HCF and LCM
pq = 13*273
pq = 3549.
Now, co-primes with product 3549 are (1,3549), (21,169), (39,91) and (13,273)
Acc. to the condition given in the question only one satisfy them that is (39,91)
Workspace
NA
SHSTTON
57
Solv. Corr.
25
Solv. In. Corr.
82
Attempted
0 M:0 S
Avg. Time
3 / 20
Choose the correct option.
Sudhir goes to the market once every 64 days and Sushil goes to the same market once every 72 days. They met each other one day. How many days later will they meet each other again?
A567
B576
C765
DNone of these
Explanation:
Find the LCM of 64,72
i.e 2*2*2*8*9=576
Workspace
NA
SHSTTON
47
Solv. Corr.
31
Solv. In. Corr.
78
Attempted
0 M:0 S
Avg. Time
4 / 20
Choose the correct option.
Sum of squares of two numbers is 2754, their HCF is 9, LCM is 135, Find the numbers
A45 and 36
B45 and 27
C54 and 27
DNone of these
Explanation:
Product of two no. = H.C.F*L.C.M
So,x*y=135*9=1215 -----(1)
and x^2+y^2=2754
So,(x+y)^2=x^2+y^2+2*x*y = 2754+2*1215=5184
So,x+y=72 ----------- (2)
By solving eq. (1) & (2)
nos. are 45 and 27
Workspace
NA
SHSTTON
21
Solv. Corr.
28
Solv. In. Corr.
49
Attempted
0 M:0 S
Avg. Time
5 / 20
Choose the correct option.
Printer A prints 8192 character per min and printer B prints 13862 character per min four character are equal to one word. Printer A starts at 7:15 am while Printer B starts at 7:29 am then at what time both will have same no. of words printed.
A8:49:14 AM
B7:49:14 AM
C9:49:14 AM
D7:49:14 PM
Explanation:
Four character are equal to one word
8192/4*(t+14)=13862/4*t
8192*(t+14)=13862*t yields the same result
t=20.2272 minutes = about 20 min 14 sec
7:49:14 am
Workspace
NA
SHSTTON
31
Solv. Corr.
34
Solv. In. Corr.
65
Attempted
0 M:0 S
Avg. Time
6 / 20
Choose the correct option.
If 19 and 1140 are the respective HCF and LCM of two numbers, which are greater than 19 then what will be the possible number of such pair?
A0
B2
C3
D5
Explanation:
Product of HCF and LCM = product of the numbers
Then, product of the numbers = 19 x 1140
Let 19a and 19b be the numbers.
19a x 19b = 19 x 1140
ab = 19 x 1140 / 19 x 19 = 60
If ab = 60 then (a,b) = (1,60), (2,30), (3,20), (4,15), (5,12) and (6,10).
Since a and b are co-primes then (a,b) = (1,60), (4,15) and (5,12)
Hence the number of such pairs = 3a
Workspace
NA
SHSTTON
50
Solv. Corr.
36
Solv. In. Corr.
86
Attempted
0 M:0 S
Avg. Time
7 / 20
Choose the correct option.
find the number between 100 to 400 which is divisible by either 2,3,5,7..
A120
B100
C205
D210
Explanation:
LCM of 2,3,5,7=210
Workspace
NA
SHSTTON
61
Solv. Corr.
20
Solv. In. Corr.
81
Attempted
0 M:0 S
Avg. Time
8 / 20
Choose the correct option.
A teacher can divide her class into groups into groups of 5,13 and 17. What is the smallest possible strength of the class?
A835
B940
C1105
D1220
Explanation:
For smallest possible class strength, we consider LCM of the given numbers. So LCM (5, 13, 17) = 1105.
Workspace
NA
SHSTTON
15
Solv. Corr.
23
Solv. In. Corr.
38
Attempted
1 M:41 S
Avg. Time
9 / 20
Choose the correct option.
HCF of 2472,1284 and a third number 'n'is 12.If their LCM is 8*9*5*103*107.then the number 'n'is..
A2^2*3^2*5^1
B2^2*3^2*7^1
C2^2*3^2*8103
DNone of these
Explanation:
2472 = 2^3×3×103
1284 = 2^2×3×107
HCF = 2^2×3
LCM = 2^3×3^2×5×103×107
HCF of the numbers is the highest number which divides all the numbers. So N should be a multiple of 2^2×3
LCM is the largest number that is divided by the given numbers. As LCM contains 32×5 these two are from N.
So N = 2^2×3^2×5^1
Workspace
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Choose the correct option.
The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:
A1677
B1683
C1898
D3363
ENone of these
Explanation:
L.C.M. of 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 * 2 + 3) = 1683.
Workspace
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## Where can I get Quantitative Aptitude H.C.F and L.C.M Interview Questions and Answers (objective type, multiple-choice, quiz, solved examples)?
Here you can find objective type Quantitative Aptitude H.C.F and L.C.M questions and answers for interview and entrance examination. Multiple choice and true or false type questions are also provided. | 3,187 | 11,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2020-45 | longest | en | 0.780306 |
http://nrich.maths.org/public/leg.php?code=5039&cl=2&cldcmpid=993 | 1,502,991,558,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103891.56/warc/CC-MAIN-20170817170613-20170817190613-00133.warc.gz | 316,591,811 | 8,610 | # Search by Topic
#### Resources tagged with Interactivities similar to Dicey:
Filter by: Content type:
Stage:
Challenge level:
### There are 226 results
Broad Topics > Information and Communications Technology > Interactivities
### World of Tan 7 - Gat Marn
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this plaque design?
### World of Tan 8 - Sports Car
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this sports car?
### Square it for Two
##### Stage: 1 and 2 Challenge Level:
Square It game for an adult and child. Can you come up with a way of always winning this game?
### World of Tan 5 - Rocket
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the rocket?
### The Path of the Dice
##### Stage: 2 Challenge Level:
A game for 1 person. Can you work out how the dice must be rolled from the start position to the finish? Play on line.
### World of Tan 9 - Animals
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this goat and giraffe?
### World of Tan 12 - All in a Fluff
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of these rabbits?
### World of Tan 17 - Weather
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the watering can and man in a boat?
### World of Tan 15 - Millennia
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the workmen?
### World of Tan 14 - Celebrations
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming and Little Fung dancing?
### World of Tan 13 - A Storm in a Tea Cup
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of these convex shapes?
### Makeover
##### Stage: 1 and 2 Challenge Level:
Exchange the positions of the two sets of counters in the least possible number of moves
### Seeing Squares
##### Stage: 1 and 2 Challenge Level:
Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.
### World of Tan 6 - Junk
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this junk?
### Twice as Big?
##### Stage: 2 Challenge Level:
Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too.
##### Stage: 2 Challenge Level:
How can the same pieces of the tangram make this bowl before and after it was chipped? Use the interactivity to try and work out what is going on!
### World of Tan 1 - Granma T
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Granma T?
### World of Tan 16 - Time Flies
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the candle and sundial?
### World of Tan 4 - Monday Morning
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Wai Ping, Wah Ming and Chi Wing?
### World of Tan 2 - Little Ming
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming?
### Turning Cogs
##### Stage: 2 Challenge Level:
What happens when you turn these cogs? Investigate the differences between turning two cogs of different sizes and two cogs which are the same.
### World of Tan 3 - Mai Ling
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Mai Ling?
### World of Tan 18 - Soup
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of Mai Ling and Chi Wing?
### World of Tan 11 - the Past, Present and Future
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the telescope and microscope?
### Coin Cogs
##### Stage: 2 Challenge Level:
Can you work out what is wrong with the cogs on a UK 2 pound coin?
### World of Tan 25 - Pentominoes
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of these people?
### World of Tan 24 - Clocks
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of these clocks?
### World of Tan 22 - an Appealing Stroll
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the child walking home from school?
### World of Tan 21 - Almost There Now
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the lobster, yacht and cyclist?
### World of Tan 26 - Old Chestnut
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this brazier for roasting chestnuts?
### World of Tan 20 - Fractions
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the chairs?
### Three Squares
##### Stage: 1 and 2 Challenge Level:
What is the greatest number of squares you can make by overlapping three squares?
### World of Tan 19 - Working Men
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this shape. How would you describe it?
### World of Tan 28 - Concentrating on Coordinates
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming playing the board game?
### Counter Roundup
##### Stage: 2 Challenge Level:
A game for 1 or 2 people. Use the interactive version, or play with friends. Try to round up as many counters as possible.
### World of Tan 27 - Sharing
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Fung at the table?
### World of Tan 29 - the Telephone
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this telephone?
### Nine Colours
##### Stage: 3 Challenge Level:
Can you use small coloured cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of each colour?
### Dotty Circle
##### Stage: 2 Challenge Level:
Watch this film carefully. Can you find a general rule for explaining when the dot will be this same distance from the horizontal axis?
### Conway's Chequerboard Army
##### Stage: 3 Challenge Level:
Here is a solitaire type environment for you to experiment with. Which targets can you reach?
##### Stage: 1 and 2 Challenge Level:
Our 2008 Advent Calendar has a 'Making Maths' activity for every day in the run-up to Christmas.
### Ratio Pairs 2
##### Stage: 2 Challenge Level:
A card pairing game involving knowledge of simple ratio.
### Dominoes Environment
##### Stage: 1 and 2 Challenge Level:
These interactive dominoes can be dragged around the screen.
### Chocolate Bars
##### Stage: 2 Challenge Level:
An interactive game to be played on your own or with friends. Imagine you are having a party. Each person takes it in turns to stand behind the chair where they will get the most chocolate.
##### Stage: 2 Challenge Level:
Three beads are threaded on a circular wire and are coloured either red or blue. Can you find all four different combinations?
### Colour Wheels
##### Stage: 2 Challenge Level:
Imagine a wheel with different markings painted on it at regular intervals. Can you predict the colour of the 18th mark? The 100th mark?
### Round Peg Board
##### Stage: 1 and 2 Challenge Level:
A generic circular pegboard resource.
### Overlapping Circles
##### Stage: 2 Challenge Level:
What shaped overlaps can you make with two circles which are the same size? What shapes are 'left over'? What shapes can you make when the circles are different sizes?
### Part the Piles
##### Stage: 2 Challenge Level:
Try to stop your opponent from being able to split the piles of counters into unequal numbers. Can you find a strategy? | 1,824 | 7,756 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-34 | latest | en | 0.812663 |
https://cdn.jsdelivr.net/gh/stephenslab/gtexresults@45bcaa8253e6595f948287b4766c95e45e0284db/docs/Table.HeterogeneityTables.html | 1,670,659,255,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-49/segments/1669446710421.14/warc/CC-MAIN-20221210074242-20221210104242-00624.warc.gz | 202,703,860 | 209,909 | ## Heterogeneity Analysis
First, we asked in how many tissues is a QTL signficiant.
lfsr=read.table("../../Data_vhat/withvhatlfsr.txt")[,-1]
lfsr[lfsr<0]=0
colnames(lfsr)=tissue.names
colnames(lfsr.nobrain)=tissue.names[-c(7:16)]
colnames(lfsr.brain.only)=tissue.names[c(7:16)]
pm.mash.beta=pm.mash*standard.error
thresh=0.05
Here, we show the Proportion of Sharing by Sign:
sigmat=(lfsr<=thresh)
nsig= rowSums(sigmat)
(signall=mean(het.norm(pm.mash.beta[nsig>0,])>0))
## [1] 0.8512519
##show that results are robust in global analysis###
sigmat=(lfsr[,-c(7:16)]<=thresh)
nsig= rowSums(sigmat)
(signall.nobrain=mean(het.norm(pm.mash.beta[nsig,-c(7:16)])>0))
## [1] 0.8493332
sigmat=(lfsr[,c(7:16)]<=thresh)
nsig= rowSums(sigmat)
(signall.brainonly=mean(het.norm(pm.mash.beta[nsig>0,c(7:16)])>0))
## [1] 0.9602292
####SHow that results are robust in specific analysis
sigmat=(lfsr.nobrain<=thresh)
nsig= rowSums(sigmat)
(signnobrain=mean(het.norm(pm.mash.nobrain[nsig>0,])>0))
## [1] 0.8823972
sigmat=(lfsr.brain.only<=thresh)
nsig= rowSums(sigmat)
(signbrainonly=mean(het.norm(pm.mash.brain.only[nsig>0,])>0))
## [1] 0.9840876
Here, we show heterogeneity by magnitude:
sigmat=(lfsr<=thresh)
nsig= rowSums(sigmat)
(magall=mean(het.norm(pm.mash.beta[nsig>0,])>0.5))
## [1] 0.3591297
##show that results are robust###
sigmat=(lfsr[,-c(7:16)]<=thresh)
nsig= rowSums(sigmat)
(magall.excludingbrain=mean(het.norm(pm.mash.beta[nsig>0,-c(7:16)])>0.5))
## [1] 0.3976238
sigmat=(lfsr[,c(7:16)]<=thresh)
nsig= rowSums(sigmat)
(magall.brainonly=mean(het.norm(pm.mash.beta[nsig>0,c(7:16)])>0.5))
## [1] 0.7638909
##show that results are robust###
sigmat=(lfsr.nobrain<=thresh)
nsig= rowSums(sigmat)
(magnobrain=mean(het.norm(pm.mash.nobrain[nsig>0,])>0.5))
## [1] 0.4445148
sigmat=(lfsr.brain.only<=thresh)
nsig= rowSums(sigmat)
(magbrain=mean(het.norm(pm.mash.brain.only[nsig>0,])>0.5))
## [1] 0.8586027 | 772 | 1,901 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-49 | latest | en | 0.303887 |
http://math.stackexchange.com/questions/56550/number-of-even-numbers-with-sum-of-digits-less-than-12/56566 | 1,469,556,342,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825048.60/warc/CC-MAIN-20160723071025-00260-ip-10-185-27-174.ec2.internal.warc.gz | 151,319,654 | 19,081 | Number of even numbers with sum of digits less than 12
How many positive three digit even numbers lesser than 700 are there with all three different digits and where the sum of the three digits is 12 or less ?
Am I supposed to write solve the expression for each of the 12 cases like solution for a + b + C=12, 11, 10...1 and then add them together?
Is there a better approach to this problem?
-
I'd start by drawing a tree of cases that begins with the least significant digit, knowing it has to be even. You can then work out the possible choices for upper two digits, taking into account the restriction that all digits are different as well as having sum less than (or equal to??) 12. Note how your title states the last restriction. – hardmath Aug 9 '11 at 16:40
Less than 12, or 12 or less? – Daniel R Hicks Aug 9 '11 at 17:42
@Daniel R Hicks : 12 or less – user118102114 Aug 9 '11 at 18:49
I would just enter the numbers 2 through 698 into a spreadsheet, use the mod function to extract the digits, and if statements to determine whether the digits are different and sum to less than 12. Then ask it to count the results. The copy function means you only write the things once.
-
I believe that there are 115 numbers that meet the requirements. We only need to consider even numbers so you can make a table with the possible hundreds digits and even ones digits and easily compute the number of possible tens for each one like this:
1t0---8 possibilities 2,3,4,5,6,7,8,9
2t0---8 possibilities 1,3,4,5,6,7,8,9
3t0---8 possibilities 1,2,4,5,6,7,8,9
4t0---7 possibilities 1,2,3,5,6,7,8
5t0---6 possibilities 1,2,3,4,6,7
6t0---5 possibilities 1,2,3,4,5
1t2---8 possibilities 0,3,4,5,6,7,8,9
etc.
Remembering to skip repeated the digits like 2t2, 4t4, and 6t6 ie.
4t4--- 0 possibilities
-
9+3+0
9+2+1
8+4+0
8+3+1
8+2+2 (error)
7+5+0
7+4+1
7+3+2
7+2+2 (error)
6+5+1
6+4+2
6+3+3 (error)
5+4+3
Each can be permuted ABC, ACB, BAC, BCA, CAB, CBA, so 13x6 if I counted right.
[Oops, I failed to eliminate the numbers above 700. We'll leave that as an exercise for the student. Easily done by breaking the above list into two ranges.]
[Gotta learn to read more carefully! Sum of digits LESS than 12! Oh well...]
[I suspect the above scheme would work, though, with a little adjustment.]
-
921, 831, 741, 651, 633, and 543 are not even. And you can't permute digits in a way that results in an odd. – anon Aug 9 '11 at 17:40
Ah, missed that point too. – Daniel R Hicks Aug 9 '11 at 17:44
(I never was that good at math.) – Daniel R Hicks Aug 9 '11 at 17:44 | 813 | 2,574 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2016-30 | latest | en | 0.860085 |
https://www.cut-the-knot.org/m/Geometry/VolumeInequalityInTetrahedron.shtml | 1,726,445,474,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.26/warc/CC-MAIN-20240915220324-20240916010324-00614.warc.gz | 659,016,886 | 15,427 | # Volume Inequality in Tetrahedron
### Solution 1
Expressing the volume in two ways, we get $abc=xbc+yca+zab.$ By the AM-GM inequality, $(xbc+yca+zab)^3\ge 27xyz(abc)^2.$ Thus, $(abc)^3\ge 27xyz(abc)^2,$ from which $abc\ge 27xyz.$
Equality is attained when $M$ is the centroid of $\Delta ABC.$ Indeed, let $O=(0,0,0),$ $A=(a,0,0),$ $B=(0,b,0)$ and $C=(0,0,c).$ Then $M=(ma,nb,pc)$ with $m+n+p=1.$ Equality holds iff $[MOAB]=[MOBC]=[MOCA]$ i.e. $mabc=nabc=pabc,$ from which $\displaystyle m=n=p=\frac{1}{3},$ making $M$ the centroid of $\Delta ABC.$
### Solution 2
First we prove that
(*)
$\displaystyle \frac{x}{OA}+\frac{y}{OB}+\frac{z}{OC}=1.$
Indeed, writing the value of the tetrahedron $OABC$ in two ways, we obtain
$Vol(OABC)=Vol(MOAC)+Vol(MOAC)+Vol(MOAB),$
or, equivalently,
$\displaystyle \frac{OB\cdot OC\cdot x}{6}+\frac{OC\cdot OA\cdot x}{6}+\frac{OA\cdot OB\cdot x}{6}=\frac{OA\cdot OB\cdot OC}{6}.$
Dividing by the right-hand side proves (*). Applying to (*) the AM-GM inequality, we have
$\displaystyle 1=\frac{x}{OA}+\frac{y}{OB}+\frac{z}{OC}\ge 3\sqrt[3]{\frac{x}{OA}\cdot\frac{y}{OB}\cdot\frac{z}{OC}}$
which is equivalent to the require inequality.
Equality occurs, iff, $\displaystyle \frac{x}{OA}=\frac{y}{OB}=\frac{z}{OC}\;\left(=\frac{1}{3}\right),$ i.e., if the tetrahedron $OABC$ is similar to the tetrahedron with the apex at $M$ and the base formed by the three projections of $M$ on the faces of $OAB.$ In other words, if the two tetrahedra are homothetic with the coefficient $\displaystyle \frac{1}{3}.$ This only happens when $M$ is the centroid of $\Delta ABC.$
### Solution 3
Let us choose the origin at $O$. WLOG, let $OA$, $OB$, and $OC$ be along the $X$, $Y$, and $Z$, respectively. The equation of the plane $ABC$ is $\displaystyle \frac{x}{OA}+\frac{y}{OB}+\frac{z}{OC}=1.$
The coordinate of point $M$ is $(x,y,z)$. As a result of the interior constraint, $x,y,z\geq 0$ and $\displaystyle \frac{x}{OA}+\frac{y}{OB}+\frac{z}{OC}\leq 1.$
Thus, from $AM-GM$,
$\displaystyle 3\left(\frac{xyz}{OA\cdot OB\cdot OC}\right)^{1/3}\leq 1 ~\Rightarrow~ 27xyz \leq OA\cdot OB\cdot OC.$
### Acknowledgment
Dorin Marghidanu has kindly posted the problem at the CutTheKnotMath facebook page. He then commented with Solution 2, while Leo Giugiuc commented with Solution 1. Solution 3 is by Amit Itagi. | 828 | 2,344 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-38 | latest | en | 0.66949 |
https://ai.stackexchange.com/questions/22994/when-using-experience-replay-in-reinforcement-learning-which-state-is-used-for | 1,652,690,449,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510097.3/warc/CC-MAIN-20220516073101-20220516103101-00298.warc.gz | 139,136,095 | 62,839 | # When using experience replay in reinforcement learning, which state is used for training?
I'm slightly confused about the experience replay process. I understand why we use batch processing in reinforcement learning, and from my understanding, a batch of states is input into the neural network model.
Suppose there are 2 valid moves in the action space (UP or DOWN)
Suppose the batch size is 5, and the 5 states are this:
$$[s_1, s_2, s_3, s_4, s_5]$$
We put this batch into the neural network model and output Q values. Then we put $$[s_1', s_2', s_3', s_4', s_5']$$ into a target network.
What I'm confused about is this:
Each state in $$[s_1, s_2, s_3, s_4, s_5]$$ is different.
Are we computing Q values for UP and DOWN for ALL 5 states after they go through the neural network?
For example, $$[Q_{s_1}(\text{UP}), Q_{s_1}(\text{DOWN})], \\ [Q_{s_2} (\text{UP}), Q_{s_2}(\text{DOWN})], \\ [Q_{s_3}(\text{UP}), Q_{s_3}(\text{DOWN})], \\ [Q_{s_4}(\text{UP}), Q_{s_4}(\text{DOWN})], \\ [Q_{s_5}(\text{UP}), Q_{s_5}(\text{DOWN})]$$
The way the states are used is as follows:
Typically your $$Q$$-network will state a state as input and output scores over the action space. I.e. $$Q : \mathcal{S} \rightarrow \mathbb{R}^{|\mathcal{A}|}$$. So, in your replay buffer you should store $$s_t, a_t, r_{t+1}, s_{t+1}, \mbox{done}$$ (note that done just represents where the episode ended on this transition and I add for completeness.
Now, when you are doing your batch updates you sample uniformly at random from this replay buffer. This means you get $$B$$ tuples of $$s_t, a_t, r_{t+1}, s_{t+1}, \mbox{done}$$. Now, I will assume $$B=1$$ as it is easier to explain and the extension to $$B > 1$$ should be easy to see.
For our state-action tuple $$s_t, a_t$$ we want to shift what the network predicts for this pair to be closer to $$r_{t+1} + \gamma \arg\max_a Q(s,a)$$. However, our neural network only takes the state as input, and outputs a vector of scores for each action. That means we want to shift the output of our network for the state $$s_t$$ towards the target I just mentioned, but only for the action $$a_t$$ that we took. To do this we just calculate the target, i.e. we calculate $$r_{t+1} + \gamma \arg\max_a Q(s,a)$$, and then we do gradient ascent like we would a normal neural network where the target vector is the same as the predicted vector everywhere except the $$a_t$$th element, which we will change to $$r_{t+1} + \gamma \arg\max_a Q(s,a)$$. This way, our network moves closer to our Q-learning update for only the action we want, in line with how Q-learning works.
It is also worth nothing that you can parameterise your Neural Network to be a function $$Q: \mathcal{S} \times \mathcal{A} \rightarrow \mathbb{R}$$ which would make training more in line with how tabular Q-learning but is seldom used in practice as it becomes much more expensive to compute (you have to do a forward pass for each action, rather than one forward pass per state).
• Maybe worth noting that it is valid to model $\hat{q}(s,a, \theta)$ directly with NN mapping $Q : \mathcal{S} \times \mathcal{A} \rightarrow \mathbb{R}$. That leads to a different approach using minibatches over all actions in each state for max and argmax operations (inefficient) but a simpler construction of training data. I think the approach you outline is more often used in practice though Aug 12, 2020 at 15:47
• @NeilSlater I had to do this recently... can get quite messy to programme (or maybe that was just the specific paper I was having to implement...) Aug 12, 2020 at 15:51
• You might make a neural network for $Q : \mathcal{S} \times \mathcal{A} \rightarrow \mathbb{R}$ if the number of valid actions in any state is low compared to full action space and/or actions have useful traits that can be generalised over. Also of course if you are working with an RL process that doesn't need max or argmax over Q (e.g. variations of Actor-Critic) Aug 12, 2020 at 15:55
• @mamauwu ah, apologies. It just becomes a typical batch update like in normal neural networks. You could either loop through your batch and do this individually or you can do it as one update. Either way, I think the common method is to accumulate the loss for each sample in the batch, take the average, and then do backprop with it. Aug 12, 2020 at 16:32
• Perfect! That's the answer I was looking for. Thank you so much! Aug 12, 2020 at 16:37 | 1,245 | 4,421 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-21 | longest | en | 0.885692 |
https://www.physicsforums.com/threads/centripetal-acceleration-and-angles.168792/ | 1,623,682,259,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487612537.23/warc/CC-MAIN-20210614135913-20210614165913-00117.warc.gz | 869,759,605 | 13,971 | # Centripetal acceleration and angles
## Homework Statement
A block is hung by a string from the inside roof of a van. When the van goes straight ahead at a speed of 34 m/s, the block hangs vertically down. But when the van maintains this same speed around an unbanked curve (radius = 120 m), the block swings toward the outside of the curve. Then the string makes an angle theta with the vertical. Find theta.
Fc=v^2/r
## The Attempt at a Solution
All I could do was get the centripetal force, 34^2/120=9.63333N.
I have no idea where to go from here, and there are no similar problems I can find.
Thanks a million for any help. | 159 | 635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-25 | latest | en | 0.901584 |
http://stackoverflow.com/questions/2602583/geometric-mean-is-there-a-built-in/14836426 | 1,419,205,023,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802772751.143/warc/CC-MAIN-20141217075252-00021-ip-10-231-17-201.ec2.internal.warc.gz | 270,542,245 | 19,819 | # Geometric Mean: is there a built-in?
i tried to find a built-in for geometric mean but couldn't.
(Obviously a built-in isn't going to save me any time while working in the shell, nor do i suspect there's any difference in accuracy; for scripts i try to use built-ins as often as possible, where the (cumulative) performance gain is often noticeable.
In case there isn't one (which i doubt is the case) here's mine.
``````gm_mean = function(a){prod(a)^(1/length(a))}
``````
-
Careful about negative numbers and overflows. prod(a) will under or overflow very quickly. I tried to time this using a big list and quickly got Inf using your method vs 1.4 with exp(mean(log(x))); the rounding problem can be quite severe. – Tristan Apr 8 '10 at 22:12
i just wrote the function above quickly because i was sure that 5 min after posting this Q, someone would tell me R's built-in for gm. So no built-in so it's certain worth taking the time to re-code in light of your remarks. + 1 from me. – doug Apr 8 '10 at 23:12
Here is a vectorized, zero- and NA-tolerant function for calculating geometric mean in R. The verbose `mean` calculation involving `length(x)` is necessary for the cases where `x` contains non-positive values.
``````gm_mean = function(x, na.rm=TRUE){
exp(sum(log(x[x > 0]), na.rm=na.rm) / length(x))
}
``````
Thanks to @ben-bolker for noting the `na.rm` pass-through and @Gregor for making sure it works correctly.
-
wouldn't it be better to pass `na.rm` through as an argument (i.e. let the user decide whether they want to be NA-tolerant or not, for consistency with other R summary functions)? I'm nervous about automatically excluding zeroes -- I would make that an option as well. – Ben Bolker Aug 28 at 19:21
Perhaps you're right about passing `na.rm` as an option. I'll update my answer. As for excluding zeroes, the geometric mean is undefined for non-positive values, including zeroes. The above is a common fix for geometric mean, in which zeroes (or in this case all non-zeroes) are given a dummy value of 1, which has no effect on the product (or equivalently, zero in the logarithmic sum). – Paul McMurdie Aug 28 at 20:01
*I meant a common fix for non-positive values, zero being the most common when geometric mean is being used. – Paul McMurdie Aug 28 at 20:09
I thought the effect of a zero would be (as pointed out by @Alan-James-Salmoni below) to force the GM to zero, i.e. `result <- if(any(x==0)) 0 else exp(sum(...))` – Ben Bolker Aug 28 at 20:16
Your `na.rm` pass-through doesn't work as coded... see `gm_mean(c(1:3, NA), na.rm = T)`. You need to remove the `& !is.na(x)` from the vector subset, and since the first arg of `sum` is `...`, you need to pass `na.rm = na.rm` by name, and you also need to exclude `0`'s and `NA`'s from the vector in the `length` call. – Gregor Aug 28 at 20:53
No, but there are a few people who have written one, such as here.
Another possibility is to use this:
``````exp(mean(log(x)))
``````
-
The
``````exp(mean(log(x)))
``````
will work unless there is a 0 in x. If so, the log will produce -Inf (-Infinite) which always results in a geometric mean of 0.
One solution is to remove the -Inf value before calculating the mean:
``````geo_mean <- function(data) {
log_data <- log(data)
gm <- exp(mean(log_data[is.finite(log_data)]))
return(gm)
}
``````
You can use a one-liner to do this but it means calculating the log twice which is inefficient.
``````exp(mean(log(i[is.finite(log(i))])))
``````
-
why calculate the log twice when you can do: exp(mean(x[x!=0])) – zzk Jul 25 at 20:54
both approaches get the mean wrong, because the denominator for the mean, `sum(x) / length(x)` is wrong if you filter x and then pass it to `mean`. – Paul McMurdie Aug 28 at 17:46
I think filtering is a bad idea unless you explicitly mean to do it (e.g. if I were writing a general-purpose function I would not make filtering the default) -- OK if this is a one-off piece of code and you've thought very carefully about what filtering zeroes out actually means in the context of your problem (!) – Ben Bolker Aug 28 at 20:18
you can use `psych` package and call `geometric.mean` function in that.
-
I use exactly what Mark says. This way, even with tapply, you can use the built-in `mean` function, no need to define yours! For example, to compute per-group geometric means of data\$value:
``````exp(tapply(log(data\$value), data\$group, mean))
``````
-
In case there is missing values in your data, this is not a rare case. you need to add one more argument. You may try following codes.
``````exp(mean(log(i[is.finite(log(i))]),na.rm=T))
``````
- | 1,250 | 4,631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2014-52 | latest | en | 0.946111 |
https://people.maths.bris.ac.uk/~matyd/GroupNames/128/C4%5E2.327D4.html | 1,611,480,979,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703547475.44/warc/CC-MAIN-20210124075754-20210124105754-00070.warc.gz | 487,988,119 | 7,332 | Copied to
clipboard
## G = C42.327D4order 128 = 27
### 23rd non-split extension by C42 of D4 acting via D4/C4=C2
p-group, metabelian, nilpotent (class 2), monomial
Series: Derived Chief Lower central Upper central Jennings
Derived series C1 — C22 — C42.327D4
Chief series C1 — C2 — C4 — C2×C4 — C22×C4 — C2×C42 — C2×C4×C8 — C42.327D4
Lower central C1 — C22 — C42.327D4
Upper central C1 — C22×C4 — C42.327D4
Jennings C1 — C2 — C2 — C22×C4 — C42.327D4
Generators and relations for C42.327D4
G = < a,b,c,d | a4=b4=1, c4=b2, d2=a2b2, ab=ba, ac=ca, dad-1=a-1, bc=cb, bd=db, dcd-1=a2b-1c3 >
Subgroups: 220 in 146 conjugacy classes, 80 normal (22 characteristic)
C1, C2 [×3], C2 [×4], C4 [×2], C4 [×6], C4 [×8], C22 [×3], C22 [×4], C8 [×6], C2×C4 [×2], C2×C4 [×16], C2×C4 [×12], Q8 [×8], C23, C42 [×4], C42 [×4], C4⋊C4 [×6], C2×C8 [×4], C2×C8 [×10], C22×C4 [×3], C22×C4 [×4], C2×Q8 [×4], C2×Q8 [×4], C4×C8 [×2], C4⋊C8 [×2], C2×C42, C2×C42 [×2], C2×C4⋊C4, C2×C4⋊C4 [×2], C4×Q8 [×4], C22×C8 [×4], C22×Q8, C22.7C42 [×4], C2×C4×C8, C2×C4⋊C8, C2×C4×Q8, C42.327D4
Quotients: C1, C2 [×7], C4 [×4], C22 [×7], C8 [×4], C2×C4 [×6], D4 [×4], Q8 [×4], C23, C22⋊C4 [×4], C2×C8 [×6], M4(2) [×2], C22×C4, C2×D4 [×2], C2×Q8 [×2], C4○D4 [×2], C22⋊C8 [×4], C2×C22⋊C4, C4×Q8 [×2], C22⋊Q8 [×2], C4.4D4, C4⋊Q8, C22×C8, C2×M4(2), C8○D4 [×2], C23.67C23, C2×C22⋊C8, (C22×C8)⋊C2, C8×Q8 [×2], C84Q8 [×2], C42.327D4
Smallest permutation representation of C42.327D4
Regular action on 128 points
Generators in S128
```(1 61 53 71)(2 62 54 72)(3 63 55 65)(4 64 56 66)(5 57 49 67)(6 58 50 68)(7 59 51 69)(8 60 52 70)(9 113 92 121)(10 114 93 122)(11 115 94 123)(12 116 95 124)(13 117 96 125)(14 118 89 126)(15 119 90 127)(16 120 91 128)(17 28 104 78)(18 29 97 79)(19 30 98 80)(20 31 99 73)(21 32 100 74)(22 25 101 75)(23 26 102 76)(24 27 103 77)(33 47 85 108)(34 48 86 109)(35 41 87 110)(36 42 88 111)(37 43 81 112)(38 44 82 105)(39 45 83 106)(40 46 84 107)
(1 31 5 27)(2 32 6 28)(3 25 7 29)(4 26 8 30)(9 112 13 108)(10 105 14 109)(11 106 15 110)(12 107 16 111)(17 72 21 68)(18 65 22 69)(19 66 23 70)(20 67 24 71)(33 113 37 117)(34 114 38 118)(35 115 39 119)(36 116 40 120)(41 94 45 90)(42 95 46 91)(43 96 47 92)(44 89 48 93)(49 77 53 73)(50 78 54 74)(51 79 55 75)(52 80 56 76)(57 103 61 99)(58 104 62 100)(59 97 63 101)(60 98 64 102)(81 125 85 121)(82 126 86 122)(83 127 87 123)(84 128 88 124)
(1 2 3 4 5 6 7 8)(9 10 11 12 13 14 15 16)(17 18 19 20 21 22 23 24)(25 26 27 28 29 30 31 32)(33 34 35 36 37 38 39 40)(41 42 43 44 45 46 47 48)(49 50 51 52 53 54 55 56)(57 58 59 60 61 62 63 64)(65 66 67 68 69 70 71 72)(73 74 75 76 77 78 79 80)(81 82 83 84 85 86 87 88)(89 90 91 92 93 94 95 96)(97 98 99 100 101 102 103 104)(105 106 107 108 109 110 111 112)(113 114 115 116 117 118 119 120)(121 122 123 124 125 126 127 128)
(1 114 49 126)(2 81 50 33)(3 116 51 128)(4 83 52 35)(5 118 53 122)(6 85 54 37)(7 120 55 124)(8 87 56 39)(9 100 96 17)(10 67 89 61)(11 102 90 19)(12 69 91 63)(13 104 92 21)(14 71 93 57)(15 98 94 23)(16 65 95 59)(18 42 101 107)(20 44 103 109)(22 46 97 111)(24 48 99 105)(25 40 79 88)(26 127 80 115)(27 34 73 82)(28 121 74 117)(29 36 75 84)(30 123 76 119)(31 38 77 86)(32 125 78 113)(41 66 106 60)(43 68 108 62)(45 70 110 64)(47 72 112 58)```
`G:=sub<Sym(128)| (1,61,53,71)(2,62,54,72)(3,63,55,65)(4,64,56,66)(5,57,49,67)(6,58,50,68)(7,59,51,69)(8,60,52,70)(9,113,92,121)(10,114,93,122)(11,115,94,123)(12,116,95,124)(13,117,96,125)(14,118,89,126)(15,119,90,127)(16,120,91,128)(17,28,104,78)(18,29,97,79)(19,30,98,80)(20,31,99,73)(21,32,100,74)(22,25,101,75)(23,26,102,76)(24,27,103,77)(33,47,85,108)(34,48,86,109)(35,41,87,110)(36,42,88,111)(37,43,81,112)(38,44,82,105)(39,45,83,106)(40,46,84,107), (1,31,5,27)(2,32,6,28)(3,25,7,29)(4,26,8,30)(9,112,13,108)(10,105,14,109)(11,106,15,110)(12,107,16,111)(17,72,21,68)(18,65,22,69)(19,66,23,70)(20,67,24,71)(33,113,37,117)(34,114,38,118)(35,115,39,119)(36,116,40,120)(41,94,45,90)(42,95,46,91)(43,96,47,92)(44,89,48,93)(49,77,53,73)(50,78,54,74)(51,79,55,75)(52,80,56,76)(57,103,61,99)(58,104,62,100)(59,97,63,101)(60,98,64,102)(81,125,85,121)(82,126,86,122)(83,127,87,123)(84,128,88,124), (1,2,3,4,5,6,7,8)(9,10,11,12,13,14,15,16)(17,18,19,20,21,22,23,24)(25,26,27,28,29,30,31,32)(33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48)(49,50,51,52,53,54,55,56)(57,58,59,60,61,62,63,64)(65,66,67,68,69,70,71,72)(73,74,75,76,77,78,79,80)(81,82,83,84,85,86,87,88)(89,90,91,92,93,94,95,96)(97,98,99,100,101,102,103,104)(105,106,107,108,109,110,111,112)(113,114,115,116,117,118,119,120)(121,122,123,124,125,126,127,128), (1,114,49,126)(2,81,50,33)(3,116,51,128)(4,83,52,35)(5,118,53,122)(6,85,54,37)(7,120,55,124)(8,87,56,39)(9,100,96,17)(10,67,89,61)(11,102,90,19)(12,69,91,63)(13,104,92,21)(14,71,93,57)(15,98,94,23)(16,65,95,59)(18,42,101,107)(20,44,103,109)(22,46,97,111)(24,48,99,105)(25,40,79,88)(26,127,80,115)(27,34,73,82)(28,121,74,117)(29,36,75,84)(30,123,76,119)(31,38,77,86)(32,125,78,113)(41,66,106,60)(43,68,108,62)(45,70,110,64)(47,72,112,58)>;`
`G:=Group( (1,61,53,71)(2,62,54,72)(3,63,55,65)(4,64,56,66)(5,57,49,67)(6,58,50,68)(7,59,51,69)(8,60,52,70)(9,113,92,121)(10,114,93,122)(11,115,94,123)(12,116,95,124)(13,117,96,125)(14,118,89,126)(15,119,90,127)(16,120,91,128)(17,28,104,78)(18,29,97,79)(19,30,98,80)(20,31,99,73)(21,32,100,74)(22,25,101,75)(23,26,102,76)(24,27,103,77)(33,47,85,108)(34,48,86,109)(35,41,87,110)(36,42,88,111)(37,43,81,112)(38,44,82,105)(39,45,83,106)(40,46,84,107), (1,31,5,27)(2,32,6,28)(3,25,7,29)(4,26,8,30)(9,112,13,108)(10,105,14,109)(11,106,15,110)(12,107,16,111)(17,72,21,68)(18,65,22,69)(19,66,23,70)(20,67,24,71)(33,113,37,117)(34,114,38,118)(35,115,39,119)(36,116,40,120)(41,94,45,90)(42,95,46,91)(43,96,47,92)(44,89,48,93)(49,77,53,73)(50,78,54,74)(51,79,55,75)(52,80,56,76)(57,103,61,99)(58,104,62,100)(59,97,63,101)(60,98,64,102)(81,125,85,121)(82,126,86,122)(83,127,87,123)(84,128,88,124), (1,2,3,4,5,6,7,8)(9,10,11,12,13,14,15,16)(17,18,19,20,21,22,23,24)(25,26,27,28,29,30,31,32)(33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48)(49,50,51,52,53,54,55,56)(57,58,59,60,61,62,63,64)(65,66,67,68,69,70,71,72)(73,74,75,76,77,78,79,80)(81,82,83,84,85,86,87,88)(89,90,91,92,93,94,95,96)(97,98,99,100,101,102,103,104)(105,106,107,108,109,110,111,112)(113,114,115,116,117,118,119,120)(121,122,123,124,125,126,127,128), (1,114,49,126)(2,81,50,33)(3,116,51,128)(4,83,52,35)(5,118,53,122)(6,85,54,37)(7,120,55,124)(8,87,56,39)(9,100,96,17)(10,67,89,61)(11,102,90,19)(12,69,91,63)(13,104,92,21)(14,71,93,57)(15,98,94,23)(16,65,95,59)(18,42,101,107)(20,44,103,109)(22,46,97,111)(24,48,99,105)(25,40,79,88)(26,127,80,115)(27,34,73,82)(28,121,74,117)(29,36,75,84)(30,123,76,119)(31,38,77,86)(32,125,78,113)(41,66,106,60)(43,68,108,62)(45,70,110,64)(47,72,112,58) );`
`G=PermutationGroup([(1,61,53,71),(2,62,54,72),(3,63,55,65),(4,64,56,66),(5,57,49,67),(6,58,50,68),(7,59,51,69),(8,60,52,70),(9,113,92,121),(10,114,93,122),(11,115,94,123),(12,116,95,124),(13,117,96,125),(14,118,89,126),(15,119,90,127),(16,120,91,128),(17,28,104,78),(18,29,97,79),(19,30,98,80),(20,31,99,73),(21,32,100,74),(22,25,101,75),(23,26,102,76),(24,27,103,77),(33,47,85,108),(34,48,86,109),(35,41,87,110),(36,42,88,111),(37,43,81,112),(38,44,82,105),(39,45,83,106),(40,46,84,107)], [(1,31,5,27),(2,32,6,28),(3,25,7,29),(4,26,8,30),(9,112,13,108),(10,105,14,109),(11,106,15,110),(12,107,16,111),(17,72,21,68),(18,65,22,69),(19,66,23,70),(20,67,24,71),(33,113,37,117),(34,114,38,118),(35,115,39,119),(36,116,40,120),(41,94,45,90),(42,95,46,91),(43,96,47,92),(44,89,48,93),(49,77,53,73),(50,78,54,74),(51,79,55,75),(52,80,56,76),(57,103,61,99),(58,104,62,100),(59,97,63,101),(60,98,64,102),(81,125,85,121),(82,126,86,122),(83,127,87,123),(84,128,88,124)], [(1,2,3,4,5,6,7,8),(9,10,11,12,13,14,15,16),(17,18,19,20,21,22,23,24),(25,26,27,28,29,30,31,32),(33,34,35,36,37,38,39,40),(41,42,43,44,45,46,47,48),(49,50,51,52,53,54,55,56),(57,58,59,60,61,62,63,64),(65,66,67,68,69,70,71,72),(73,74,75,76,77,78,79,80),(81,82,83,84,85,86,87,88),(89,90,91,92,93,94,95,96),(97,98,99,100,101,102,103,104),(105,106,107,108,109,110,111,112),(113,114,115,116,117,118,119,120),(121,122,123,124,125,126,127,128)], [(1,114,49,126),(2,81,50,33),(3,116,51,128),(4,83,52,35),(5,118,53,122),(6,85,54,37),(7,120,55,124),(8,87,56,39),(9,100,96,17),(10,67,89,61),(11,102,90,19),(12,69,91,63),(13,104,92,21),(14,71,93,57),(15,98,94,23),(16,65,95,59),(18,42,101,107),(20,44,103,109),(22,46,97,111),(24,48,99,105),(25,40,79,88),(26,127,80,115),(27,34,73,82),(28,121,74,117),(29,36,75,84),(30,123,76,119),(31,38,77,86),(32,125,78,113),(41,66,106,60),(43,68,108,62),(45,70,110,64),(47,72,112,58)])`
56 conjugacy classes
class 1 2A ··· 2G 4A ··· 4H 4I ··· 4P 4Q ··· 4X 8A ··· 8P 8Q ··· 8X order 1 2 ··· 2 4 ··· 4 4 ··· 4 4 ··· 4 8 ··· 8 8 ··· 8 size 1 1 ··· 1 1 ··· 1 2 ··· 2 4 ··· 4 2 ··· 2 4 ··· 4
56 irreducible representations
dim 1 1 1 1 1 1 1 1 2 2 2 2 2 type + + + + + + - image C1 C2 C2 C2 C2 C4 C4 C8 D4 Q8 M4(2) C4○D4 C8○D4 kernel C42.327D4 C22.7C42 C2×C4×C8 C2×C4⋊C8 C2×C4×Q8 C2×C4⋊C4 C22×Q8 C2×Q8 C42 C2×C8 C2×C4 C2×C4 C22 # reps 1 4 1 1 1 6 2 16 4 4 4 4 8
Matrix representation of C42.327D4 in GL5(𝔽17)
16 0 0 0 0 0 0 1 0 0 0 16 0 0 0 0 0 0 0 1 0 0 0 16 0
,
13 0 0 0 0 0 13 0 0 0 0 0 13 0 0 0 0 0 4 0 0 0 0 0 4
,
9 0 0 0 0 0 15 0 0 0 0 0 15 0 0 0 0 0 0 9 0 0 0 8 0
,
4 0 0 0 0 0 3 14 0 0 0 14 14 0 0 0 0 0 1 0 0 0 0 0 16
`G:=sub<GL(5,GF(17))| [16,0,0,0,0,0,0,16,0,0,0,1,0,0,0,0,0,0,0,16,0,0,0,1,0],[13,0,0,0,0,0,13,0,0,0,0,0,13,0,0,0,0,0,4,0,0,0,0,0,4],[9,0,0,0,0,0,15,0,0,0,0,0,15,0,0,0,0,0,0,8,0,0,0,9,0],[4,0,0,0,0,0,3,14,0,0,0,14,14,0,0,0,0,0,1,0,0,0,0,0,16] >;`
C42.327D4 in GAP, Magma, Sage, TeX
`C_4^2._{327}D_4`
`% in TeX`
`G:=Group("C4^2.327D4");`
`// GroupNames label`
`G:=SmallGroup(128,716);`
`// by ID`
`G=gap.SmallGroup(128,716);`
`# by ID`
`G:=PCGroup([7,-2,2,2,-2,2,2,-2,224,141,400,422,100,124]);`
`// Polycyclic`
`G:=Group<a,b,c,d|a^4=b^4=1,c^4=b^2,d^2=a^2*b^2,a*b=b*a,a*c=c*a,d*a*d^-1=a^-1,b*c=c*b,b*d=d*b,d*c*d^-1=a^2*b^-1*c^3>;`
`// generators/relations`
×
𝔽 | 6,019 | 10,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-04 | longest | en | 0.36888 |
https://discourse.mcneel.com/t/create-surface-panels-from-points-on-surface/115335 | 1,695,512,040,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506539.13/warc/CC-MAIN-20230923231031-20230924021031-00394.warc.gz | 246,032,801 | 8,771 | # Create surface(panels) from points on surface
I was intrigued by one panelization method shown in one of @LongNguyen’s C# workshop videos. I tried to create surfaces/panels (or meshes) using points on surface as seen the image below, but I’m not sure how I could create these surfaces in C# or even in Grasshopper.
I’m attaching an example with a surface, points on surface, and circles with planes for panelization.
Can anyone point me in the right direction to create surfaces/panels with points as approximate centers of surfaces/panels? I prefer C# methods, but I’m curious about Grasshopper solutions as well.
Thank you!
SurfaceWithPoints.gh (109.9 KB)
Hi,
I made this some weeks ago, for archieving this you need do the following steps:
1. make a delaunay(or other triangulation) from your points
2. make a Dual Mesh from your delaunay Mesh
Thank you @Baris, it’s great to know about Dual mesh. (I’m not sure if it’s a correct term.) I used “Weaverbird’s Dual graph” to get panels quickly. However, this method doesn’t make coplanar surfaces.
What I was hoping was to get coplanar surfaces by intersecting planes around each point, as shown in the image below. I tested this method using Grasshopper components as a proof of concept, and here’re four steps I’m thinking to write scripts in C#.
1. Find adjacent planes of each point
2. Intersect planes to get lines
3. Intersect all lines from step 2 to get panel vertices
4. Create coplanar surfaces using panel vertices from step 3
Is there a good way to get adjacent planes around each point, step 1, in C#? If you know better ways to get coplanar surfaces, could you share them with me? I’m attaching a revised Grasshopper file including the test method explained above.SurfaceWithPoints_POC.gh (134.5 KB)
(and I think you might find that whole thread relevant)
Hi,
I ended up with a method @Petras_Vestartas explained here:
It is basically getting an average vector for each vertex of a hexagon and then move. I made it way less elegant/ sophisticated as him but following the same principle.
Thanks a lot @DanielPiker! It is exactly what I was trying to do and the thread isTangent plane intersect.gh (122.7 KB) very helpful. As you explained in that thread, my quick test proves this method seems quite tricky with a surface having a mix of +/- curvature, like my example.
It’s a little bit different question, but I’m wondering if there’s a way to control how to create a Delaunay mesh in connecting points. Specifically, I want to connect closer points as shown in red arrows in the image below instead of green lines. Is there any solution for this?
Thanks so much again @Baris! It’s a lot to absorb, but I’ll go through the whole thread.
I tried tangent plane planarization too in my code.
Tangent plane depends on curvature that is why it does not work in your case and most of real built examples are peanuts or domes.
If you do not use some sophisticated meshing following curvature you can hardly design with simple delaunay triangulation.
Thank you @Petras_Vestartas! I realize again that most of panelization strategies rely on how original geometries are constructed. I’m amazed by the amount of information I’m getting from all of discussions and papers including yours simply by asking a basic question here. I really appreciate great answers/insights from all of you in this tread! | 775 | 3,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-40 | latest | en | 0.925653 |
https://www.physicsforums.com/threads/how-do-i-calculate-pixels-number.420516/ | 1,513,547,504,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948597585.99/warc/CC-MAIN-20171217210620-20171217232620-00738.warc.gz | 811,586,969 | 15,253 | # How do I calculate pixels number
1. Aug 5, 2010
### Telushomey
Adults need 2.5" between Eyes, but this is too far for kids eyes.
Adults = 2.5" = 52 Pixels difference = left eye 26 pixels, Right eye 26 pixels
Kids = 2.0" = ? Pixels = Left eye ? Pixels = Right eye ? Pixels (less than 26 each eye is all I know)
This is to calculate the parallax in 3D video. I need to know how much pixels kids use, and all I know is it's less than the 52 pixels for Adults eyes.
Last edited: Aug 5, 2010
2. Aug 5, 2010
### Staff: Mentor
I have no idea what you're trying to say here.
??
3. Aug 5, 2010
### Telushomey
So 2.5" is equivalent to 52 Pixels, or 52 pixels = 2.5" inches difference.
52 Pixels = 26 Pixels * 2 (26 pixels each eye)
I need to know how many pixels if the inches used is 2.0 inches difference, not 2.5 inches.
I think I need to use division but don't know how.
4. Aug 5, 2010
### Staff: Mentor
Use a proportion. 2.5/2.0 = 52/x
What kind of a screen is this you're working with? ~20 pixels/inch seems very low resolution to me.
5. Aug 5, 2010
### Telushomey
It's not resolution, it's for cropping to make parallax either negative or positive.
Maybe this is the way?
2.5 / 2.0 = 1.25
52*1.25 =65
65 - 52 = 13
52 - 13 = 39
38 / 2 = 19 pixels difference each eye.
Thank you Mark44!
Last edited: Aug 5, 2010 | 439 | 1,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-51 | longest | en | 0.921926 |
http://forum.allaboutcircuits.com/threads/magnetic-coil-for-healing-is-overheating-finding-the-correct-resister-and-strongest-magnetic-field.121347/ | 1,484,666,821,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279923.28/warc/CC-MAIN-20170116095119-00095-ip-10-171-10-70.ec2.internal.warc.gz | 109,806,539 | 16,866 | # Magnetic Coil for Healing is Overheating: Finding the correct resister and strongest magnetic field
Discussion in 'The Projects Forum' started by Durgen, Feb 28, 2016.
1. ### Durgen Thread Starter New Member
Feb 26, 2016
4
0
Newbie Here and Newbie to electronics. My apologies for asking what I believe is a way too simple question for this group. I have just begun and online course to learn more. In the meantime, I am hoping to get some guidance in a immediate problem.
My goal is to create a magnetic coil with the greatest magnetic field from a 28v AC circuit. The potentiometer is computer based managing the input of frequencies form 25 hz to 20,000 hz via the headphone jack on my laptop.
Problem: Frequencies below 1000 hz overheat the resister I have in the system. All other frequencies work well without overheating. I would like to be able to turn the system up to full power and just leave it there.
Here is what i have:
24v - 1A powered amplifier
28 AWG Speaker wire to the resister
Resister: 68 ohm-1/2 watt
Coil is circular ring with 38 AWG magnet wire with 150 turns around the ring (not around a core...just circular around the ring. (probably 150 ft)
The coil itself is not overheating, only the resister.
1. If I increase the wattage of the resistor will it reduce the size of the magnetic field.
2. Can I decrease the ohms and increase the resistors wattage to increase the magnetic field without over heating the system? Say...33 ohms and 2 watt resistor?
2. ### Dodgydave AAC Fanatic!
Jun 22, 2012
5,138
767
No, increase in resistor wattage wont affect the magnetic field, it stops the resistor from burning up, if you have 28Vac at 1amp, thats 28W of power, so ideally a 25W resistor is better, lowering the resistance will increase the magnetic field.
At the moment you have 12W in the resistor wasting, a 33ohms resistor needs to be 24W.. Formula is VxV/R =W
Last edited: Feb 28, 2016
Johann likes this.
3. ### Durgen Thread Starter New Member
Feb 26, 2016
4
0
Thanks...I will work with it...my resistor is way too low then...Any good reference material come to mind. I am looking at various intro courses on magnatism and electronics...most are either too far advanced or way too simple...
4. ### Alec_t AAC Fanatic!
Sep 17, 2013
5,961
1,133
If that's a toroidal coil then are you aware there will be only a small magnetic field external to the ring? Most of the magnetism is trapped in the ring itself.
5. ### wayneh Expert
Sep 9, 2010
12,368
3,224
Just to expand a bit in case it isn't clear, your coil has an inductance. Impedance of an inductor is proportional to frequency. So at high frequency, the coil impedance is a much larger factor than the combined DC resistance of the coil and resistor. As frequency drops, the impedance of the coil drops. At pure DC, you have only the resistance of the coil plus that of the resistor. It sounds like your coil can dissipate more heat than your resistor, so your resistor is the weakest link, like a fuse.
You should consider a fuse, by the way. If your coil gets hot and melts some of the insulation, you won't be happy.
6. ### Dodgydave AAC Fanatic!
Jun 22, 2012
5,138
767
You can substitute the resistor for a capacitor in series with the coil, a 100uF cap will be Approx 30 ohms at 50hz, Or 26 ohms at 60hz, and it wont get hot...
Johann likes this.
7. ### Alec_t AAC Fanatic!
Sep 17, 2013
5,961
1,133
If you do try a cap as DD mentions, make sure it is a non-polarised type.
Johann likes this.
8. ### Johann AAC Fanatic!
Nov 27, 2006
190
30
Try watching some of the training videos under EDUCATION of All About Circuits
9. ### Durgen Thread Starter New Member
Feb 26, 2016
4
0
Thank you all...I have one last question regarding the system (for now). I noticed an increase in the magnetic field after inserting a 50ohm 10 watt resister. the magnetic field increase 5 fold from the smaller lower wattage. It could be that the 1 watt resistor is mostly burned out already?
Not understanding fully what all the relationships are (I am ordering an amplifier kit with teaching materials today), is there a change the increased wattage will damage the adapter that translates the sound frequency to pulsed electricity? It is an adapter unit that cost \$650 so I want to be sure I keep it safe? It is designed to handle 24 volts DC. I have checked with the company but not sure if they have a clear handle on this. I have not opened the unit to look at the contents and would not know at this point what I am looking at. Hopefully that will change in the next month!
10. ### wayneh Expert
Sep 9, 2010
12,368
3,224
I'm reluctant to say much without more detail about your hookup. But usually an amplifier has high impedance (very low current) inputs that are isolated from the low impedance (high current and power) outputs. So with a normal sound system, you might do things that blow the speakers or even overload the amp, but none of that would put the input signals at risk.
I'm having trouble imagining why you had to spend \$650. That would buy a lot of power in an audio amplifier.
11. ### Durgen Thread Starter New Member
Feb 26, 2016
4
0
I understand and agree...the unit in discussion is an adapter type amplifier that transfers the frequency generator signal to electrical impulses for the the magnetic coil and several other devices. It is powered by a 24 DC amp wall plug and then provides an outlet for several devices of varying voltages....hence the high price!
I started down this venue looking at the impact of Schumann Resonances on myself and others via binaural beats and isochronic tones. I have since expanded my focus to magnetic, sound and light. So I begin my journey down the electronic's road with Electronics for Dummies and a number of practice items!
I plan on setting up an intermediate box to manage the resisters and amps for the coils and sound systems. return to either an amplifier or the adapter to protect them. Not hard I assume if one knows what they are doing.....! More later. | 1,493 | 6,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-04 | latest | en | 0.904011 |
http://www.homeschoolingheartsandminds.com/2011/01/review-math-facts-now.html | 1,524,749,002,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948214.37/warc/CC-MAIN-20180426125104-20180426145104-00436.warc.gz | 436,270,793 | 24,058 | ## Wednesday, January 5, 2011
### Review: Math Facts Now!
My kiddos are allergic to timed drills. The amount of fussing and whining they can go through all over a 2 minute test…I think the avoidance is worse than the actual task. And somehow flash cards just don’t seem to stick…maybe because they won’t stick to doing them. How about a no frills, no nonsense computer program that will quiz them on their math facts and time them, but without the nerve-wracking tick tick of time running out? Math Facts Now! might be just the solution for your homeschool.
Told you it was no frills.
Math Facts Now! is like flashcards to use on your computer…the good thing is the baby can’t get hold of them and eat them, scatter them to the four winds, or (gulp) throw them in the trash on top of the coffee grounds.
The design of the program is simple and intuitive. Once you’ve added your child, they can log-in by choosing their name from a drop-down menu. You design each lesson using one of the four operations (addition, subtraction, multiplication or division).
And then choose the group or groups of facts to be covered.
You choose the number of problems (minimum of 5), the amount of time allowed for each problem (up to 60 seconds) and how many times they need to correctly type in a problem they get wrong (minimum of 2) …that’s right, they have to practice the right answer!
Psst…David hates this part. But, it really does make it stick in his mind. And when he gets a problem wrong, Math Facts Now! will give him that fact over and over again (it’s sneaky that way) with a little warning.
You can also type in a reward they earn by completing a lesson (I typically offer a piece of gum).
All lessons are saved so they can be used over and over.
You can print a detailed results page showing you exactly which problems were missed and how many times.
What did we think?
This is the only math drill program we have ever used that David didn’t complain about…too much. The simple design is easy to use. Answers are easy to input, so you’re not fumbling and wasting time. He likes it when I enter a reward for his hard work.
I like that I can see at a glance which facts he might be having difficulty with. I like the clean, plain design and the lack of distractions. I like that I can set a time limit for each problem, but that it can be big enough that there’s no real pressure and yet the program reports to me the average time he took with a problem so I can see if he is developing automaticity. I like that it’s software installed to my computer, and not a subscription with a monthly fee.
Does Math Facts Now! work to master those facts? Only time will tell, but half the battle has already been won…you can’t master the facts unless you practice them and David is practicing them without complaining. The price is quite reasonable, too.
There is one feature I would like to see that is not in this program. An option to create a lesson for particular facts, rather than groups of facts, would be great. Suppose your kiddo has learned, say, the 2’s up to 2x6 or up to 2x10…if you select the 2’s in this program, they’ll get the 2’s up to 2x12. Or supposed you were working on squares (2x2, 3x3, 4x4,…)…you can’t set up a lesson for that in this program.
Overall, I would recommend it, it’s a well thought-out program…and proof that a math program doesn’t have to be fancy to be useful.
Sound like something you might like? You can download a free trial of Math Facts Now! here.
System requirements:
Math Facts NOW! Version 2.0 runs on Windows 2000, ME, XP, Vista and Windows 7 and requires 7 Mb of free hard drive space.
Math Facts NOW! is not currently available the for Macintosh nor Linux operating systems.
Disclosure: As a member of the TOS Homeschool Crew, I received this product for review purposes. I received no compensation. The opinions reflected here are my own.
1. Dropping by from the CREW! My girls liked the program too - more so when they only had to do 20 problems instead of 50!
By the way, NICE visuals! It really makes your review look nice.
Monica
2. Thank you for visiting, Monica.
Sometimes pictures say more than words.
3. I love the idea of a piece of gum as a reward!!!
My son also didn't like having to type out the entire number sentence, but it does help them retain the information!
Thank you for joining the conversation!
Please note: Comments on posts older than 16 days are moderated (this cuts down on SPAM). All other comments post immediately. | 1,069 | 4,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2018-17 | longest | en | 0.942196 |
https://www.examrace.com/Study-Material/Aptitude/Logical-Reasoning/Directions-YouTube-Lecture-Handouts.html | 1,563,930,834,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195530246.91/warc/CC-MAIN-20190723235815-20190724021815-00162.warc.gz | 679,689,945 | 8,851 | Directions Problems in Aptitude YouTube Lecture Handouts
Watch video lecture on YouTube: दिशा (Directions) के प्रश्नों का आसान समाधान: Aptitude दिशा (Directions) के प्रश्नों का आसान समाधान: Aptitude
Methods of Directions
• Distance: Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion.
• Direction: a course along which someone or something moves.
• North, South, East, West
• Left or Right
• Straight or Opposite
• Sunrise & Sunset
Que3: Person P is Standing, as Person Q is Standing in North of P, Q Friend’S R is Right of Q, P Friends S is Left of P, Now Question is, in Which Direction R is Standing with Respect to S?
• I start from my House, then go straight east, I reach a cross road, cross road has a 4 direction, road which is coming from opposite side ends to hospital.
• Walking from my house towards my right is the park, and opposites of park is school.
Sunrise: Sunrise takes place at east, so where would be the shadow, shadow would be to opposite site that is west.
Sunset: Sun would set in west, shadow always be east | 273 | 1,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2019-30 | longest | en | 0.930414 |
https://fr.mathworks.com/matlabcentral/cody/problems/18-bullseye-matrix/solutions/543989 | 1,591,273,525,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347439928.61/warc/CC-MAIN-20200604094848-20200604124848-00227.warc.gz | 346,628,049 | 15,572 | Cody
# Problem 18. Bullseye Matrix
Solution 543989
Submitted on 10 Dec 2014 by Kevin
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
%% n = 5; a = [3 3 3 3 3; 3 2 2 2 3; 3 2 1 2 3; 3 2 2 2 3; 3 3 3 3 3]; assert(isequal(bullseye(n),a));
2 Pass
%% n = 7; a = [4 4 4 4 4 4 4; 4 3 3 3 3 3 4; 4 3 2 2 2 3 4; 4 3 2 1 2 3 4; 4 3 2 2 2 3 4; 4 3 3 3 3 3 4; 4 4 4 4 4 4 4]; assert(isequal(bullseye(n),a)) | 272 | 520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-24 | latest | en | 0.563473 |
https://www.homeworktiger.com/solution-info/20900/A-tractor-that-would-have-a-net-cost-of-36000-would-increase-pre- | 1,550,661,414,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247494741.0/warc/CC-MAIN-20190220105613-20190220131613-00333.warc.gz | 851,617,578 | 8,513 | ### Question details
A tractor that would have a net cost of \$36,000, would increase pre-
\$ 15.00
(12-9) Scenario analysis. Your firm, Agrico Products, is considering a tractor that would have a net cost of \$36,000, would increase pre-tax operating cash flows before taking account of depreciation by \$12,000 per year, and would be depreciated on a straight-line basis to zero over 5 years at the rate of \$7,200 per year, beginning the first year. (Thus annual cash flows would be \$12,000, before taxes, plus the tax savings that result from \$7,200 of depreciation.) The managers are having a heated debate about whether the tractor would actually last 5 years. The controller insists that she knows of tractors that have lasted only 4 years. The treasurer agrees with the controller, but he argues that most tractors actually do give 5 years of service. The service manager then states that some actually last for as long as 8 years.
Given this discussion, the CFO asks you to prepare a scenario analysis to determine the importance of the tractor’s life on NPV. Use a 40 percent marginal federal-plus-state tax rate, a zero salvage value, and a WACC of 10 percent. Assuming each of the indicated lives has the same probability of occurring (probability = 1/3), what is the tractor’s expected NPV? (Hint: Here straight-line depreciation is based on the MACRS class life of the tractor and is not affected by the actual life. Also, ignore the half-year convention for this problem.)
### Solutions
Available solutions
• A tractor that would have a net cost of \$36,000, would increase pre-
\$15.00
A t
Submitted on: 24 Feb, 2018 11:20:03 This tutorial has not been purchased yet . | 392 | 1,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-09 | latest | en | 0.958537 |
https://physics.stackexchange.com/questions/784397/understanding-matrix-element-calculation-in-schwartz-4-18-4-21?noredirect=1 | 1,713,413,320,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817187.10/warc/CC-MAIN-20240418030928-20240418060928-00223.warc.gz | 406,853,145 | 40,125 | # Understanding matrix element calculation in Schwartz (4.18) -- (4.21)
After a number of years out of grad school and in industry I'm trying to brush up on my fundamentals. Going through Schwartz's "Quantum Field Theory and the Standard Model", I'm having trouble understanding the calculation in (4.18) -- (4.22) of a particular matrix element that arises as part of a term in the the OFTP expansion of the "electron electron" scattering transfer matrix, though ignoring spin and charge and treating the electron and photon as real scalars. Changing notation slightly, this matrix element is
$$V^{(R)} = \frac{e}{2} \int d^3 x \langle \psi^3 \phi^\gamma | \Psi(x) \Phi(x) \Psi(x) | \psi^1 \rangle = e (2\pi)^3 \delta^3(p_1-p_3-p_\gamma)\tag{4.18+21}$$ where $$\phi^\gamma$$ represents the state with a photon with three-momentum $$\gamma$$; $$\psi^i$$ represents a state with an electron with three-momentum $$p_i$$; and $$\Psi$$ and $$\Phi$$ are the real scalar operators for these fields.
However, trying to make this calculation myself, I end up with an extra nonsense, divergent term $$e (2\pi)^3 \omega_1 \delta^3(p_1-p_3) \delta^3(p_\gamma) \int \frac{d^3 q}{2 \omega_q}$$ coming ultimately from a leftover term proportional to $$\langle 0 | \Psi(x)^2 | 0\rangle$$ that I get after applying [creation/annihilation operator, field operator] commutation relations.
First, since the $$\Psi$$ and $$\Phi$$ operators commute with each other, the integrand in the matrix element breaks down into the product of elements in the two fields Fock spaces: $$\langle \psi^3 \phi^\gamma | \Psi(x) \Phi(x) \Psi(x) | \psi^1 \rangle = \langle \psi^3 | \Psi(x)^2| \psi^1 \rangle \langle \phi^\gamma | \Phi(x)| 0 \rangle = e^{-ip_\gamma x}\langle \psi^3 | \Psi(x)^2| \psi^1 \rangle$$ Then, using $$|\psi^i\rangle = \sqrt{2\omega_{p_i}} a_{p_i}^{\dagger} |0\rangle$$ and $$[\Psi,a_{p_i}^{\dagger}] = e^{ip_i x}/\sqrt{2\omega_{p_i}}$$, I get $$\langle \psi^3 | \Psi(x)^2| \psi^1 \rangle = 2e^{ip_1x}\langle \psi^3 |\Psi(x) | 0 \rangle + 2\sqrt{\omega_1\omega_3}\langle 0 | a_{p_3} a_{p_1}^{\dagger} \Psi(x)^2 | 0 \rangle$$ The first term becomes $$2e^{i(p_1-p_3)x}$$ and so to the right answer the second term above should become zero. However, since $$[a_{p_3},a_{p_1}^\dagger] = (2\pi)^3 \delta^3(p_1-p_3)$$, I expand it to $$2\sqrt{\omega_1\omega_3} (2\pi)^3 \delta^3(p_1-p_3)\langle 0 | \Psi(x)^2 | 0 \rangle + 2\sqrt{\omega_1\omega_3}\langle 0 | a_{p_1}^{\dagger} a_{p_3} \Psi(x)^2 | 0 \rangle$$ Since $$a_{p_1} | 0 \rangle = 0$$ the latter subterm is zero, leaving the term proportional to $$\langle 0 | \Psi(x)^2 | 0 \rangle$$. But inserting a complete set of states, and since $$\Psi^\dagger = \Psi$$, $$\langle 0 | \Psi(x)^2 | 0 \rangle = \frac{1}{(2\pi)^3} \int \frac{d^3 q}{2 \omega_q} \langle 0 | \Psi(x) | q \rangle \langle q | \Psi(x) | 0 \rangle = \frac{1}{(2\pi)^3} \int \frac{d^3 q}{2 \omega_q}$$ which diverges. Inserting this back into the integral over $$x$$, while I should get $$e(2\pi)^3 \delta^3(p_3-p_1-p_2)$$, I instead get that plus the divergent term $$e (2\pi)^3 \delta^3(p_3-p_1) \delta^3(p_2) \omega_1 \int \frac{d^3 q}{2 \omega_q}$$ Maybe interpreting this pictorially could give me some clues to where this is going wrong. Diagramatically this would be a term with a 0 momentum photon and constant momentum electron with no interaction vertex. If I'm winding up with a term with no vertex, perhaps I went wrong in the first place in factoring the matrix element into two pieces? Also, although the above calculation is for OFPT, this brings to mind old lectures about $$S$$-matrix scattering amplitude calculations and how disconnected / bubble diagrams factor out -- perhaps I am ignoring other terms that will cancel this one?
• I'm guessing one possibility is that the "complete" set of states I added is really just the complete set of one-particle states, but I still believe that should suffice in this case. In either event I think my calculation that $\langle 0 | \mid \Psi^2 \mid 0 \rangle$ must be wrong since this is a typical Lagrangian term. Oct 14, 2023 at 23:23
• I'm going to read over physics.stackexchange.com/questions/434163/…, which has broader but includes a pretty similar question about these steps Oct 15, 2023 at 19:33
• Is this perhaps an artifact of the matrix element being one for an interacting theory but calculated for the free-theory vacuum, and not the interacting vacuum? In which case the text is sweeping in under the rug for the time being, this section just being a short one about OFPT and the full S-matrix / LS Z /time ordering and interaction vacuum bits coming later? Oct 23, 2023 at 15:51 | 1,450 | 4,660 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 26, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-18 | latest | en | 0.772995 |
https://legallysociable.com/2013/06/13/methodological-issues-with-the-average-american-wedding-costing-27000/ | 1,679,982,935,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948765.13/warc/CC-MAIN-20230328042424-20230328072424-00121.warc.gz | 424,697,418 | 25,409 | # Methodological issues with the “average” American wedding costing \$27,000
Recent news reports suggest the average American wedding costs \$27,000. But, there may be some important methodological issues with this figure: selection bias and using an average rather than a median.
The first problem with the figure is what statisticians call selection bias. One of the most extensive surveys, and perhaps the most widely cited, is the “Real Weddings Study” conducted each year by TheKnot.com and WeddingChannel.com. (It’s the sole source for the Reuters and CNN Money stories, among others.) They survey some 20,000 brides per annum, an impressive figure. But all of them are drawn from the sites’ own online membership, surely a more gung-ho group than the brides who don’t sign up for wedding websites, let alone those who lack regular Internet access. Similarly, Brides magazine’s “American Wedding Study” draws solely from that glossy Condé Nast publication’s subscribers and website visitors. So before they do a single calculation, the big wedding studies have excluded the poorest and the most low-key couples from their samples. This isn’t intentional, but it skews the results nonetheless.
But an even bigger problem with the average wedding cost is right there in the phrase itself: the word “average.” You calculate an average, also known as a mean, by adding up all the figures in your sample and dividing by the number of respondents. So if you have 99 couples who spend \$10,000 apiece, and just one ultra-wealthy couple splashes \$1 million on a lavish Big Sur affair, your average wedding cost is almost \$20,000—even though virtually everyone spent far less than that. What you want, if you’re trying to get an idea of what the typical couple spends, is not the average but the median. That’s the amount spent by the couple that’s right smack in the middle of all couples in terms of its spending. In the example above, the median is \$10,000—a much better yardstick for any normal couple trying to figure out what they might need to spend.
Apologies to those for whom this is basic knowledge, but the distinction apparently eludes not only the media but some of the people responsible for the surveys. I asked Rebecca Dolgin, editor in chief of TheKnot.com, via email why the Real Weddings Study publishes the average cost but never the median. She began by making a valid point, which is that the study is not intended to give couples a barometer for how much they should spend but rather to give the industry a sense of how much couples are spending. More on that in a moment. But then she added, “If the average cost in a given area is, let’s say, \$35,000, that’s just it—an average. Half of couples spend less than the average and half spend more.” No, no, no. Half of couples spend less than the median and half spend more.
When I pressed TheKnot.com on why they don’t just publish both figures, they told me they didn’t want to confuse people. To their credit, they did disclose the figure to me when I asked, but this number gets very little attention. Are you ready? In 2012, when the average wedding cost was \$27,427, the median was \$18,086. In 2011, when the average was \$27,021, the median was \$16,886. In Manhattan, where the widely reported average is \$76,687, the median is \$55,104. And in Alaska, where the average is \$15,504, the median is a mere \$8,440. In all cases, the proportion of couples who spent the “average” or more was actually a minority. And remember, we’re still talking only about the subset of couples who sign up for wedding websites and respond to their online surveys. The actual median is probably even lower.
These are common issues with figures reported in the media. Indeed, these are two questions the average reader should ask when seeing a statistic like the average cost of the wedding:
1. How was the data collected? If this journalist is correct about these wedding cost studies, then this data is likely very skewed. What we would want to see is a more representative sample of weddings rather than having subscribers or readers volunteer how much their wedding cost.
2. What statistic is reported? Confusing the mean and median is a big program and pops up with issues as varied as the average vs. median college debtthe average vs. median credit card debt, and the average vs. median square footage of new homes. This journalist is correct to point out that the media should know better and shouldn’t get the two confused. However, reporting a higher average with skewed data tends to make the number more sensationalistic. It also wouldn’t hurt to have more media consumers know the difference and adjust accordingly.
It sounds like the median wedding cost would likely be significantly lower than the \$27,000 bandied about in the media if some basic methodological questions were asked. | 1,045 | 4,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-14 | latest | en | 0.933616 |
https://astarmathsandphysics.com/a-level-maths-notes/c1/4675-intersection-of-a-tangent-line-from-the-origin-with-a-circle.html | 1,723,548,889,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00796.warc.gz | 83,534,361 | 8,792 | A circle has centre {jatex options:inline}(4,2){/jatex} and radius {jatex options:inline}\sqrt{2}{/jatex}.
The equation of the circle is {jatex options:inline}(x-4)^2+(y-2)^2=(\sqrt{2})^2=2 \rightarrow x^2-8x+y^2-4y+20=2{/jatex} (1)
A line form the origin forms a tangent with the circle. What is the equation of this tangent and where does it meet the circle?
Let the tangent meet the circle at the point {jatex options:inline}P(x,y){/jatex}. The line drawn from the centre of the circle to {jatex options:inline}P{/jatex} HAS GRADIENT {jatex options:inline}\frac{y-2}{x-4}{/jatex} and the line drawn from the origin has gradient {jatex options:inline}\frac{y}{x}{/jatex}.
These two lines are at right angles so
{jatex options:inline}\frac{y-2}{x-4}=- \frac{x}{y} \rightarrow y^2-2y+x^2-4x=0{/jatex} (2)
(10-(2) gives {jatex options:inline}-4x-2y+20=2 \rightarrow y=-2x+9{/jatex}
Substitute the equation of this line into (1). {jatex options:inline}x^2-8x+(-2x+9)^2-4(-2x+9)+20=2 \rightarrow 5x^2-36x+63=0{/jatex}.
Then {jatex options:inline}x=\frac{36 \pm \sqrt{(-36)^2-4 \times 5 \times 63}}{2 \times 5}=\frac{21}{5}, \: 3{/jatex}.
If {jatex options:inline}x= \frac{21}{5}{/jatex} then {jatex options:inline}y=-2 \times \frac{21}{5}+9=\frac{3}{5}{/jatex}.
If {jatex options:inline}x= 3{/jatex} then {jatex options:inline}y=-2 \times 3+9=3{/jatex}.
There are two lines, with gradients {jatex options:inline}\frac{3/5}{21/5}=\frac{7}{7}{/jatex} and equation {jatex options:inline}y=\frac{1}{7}x{/jatex} which meets the circle at {jatex options:inline}(21/5.3/5){/jatex}, and one with gradient {jatex options:inline}\frac{3}{3}=1{/jatex} and equation {jatex options:inline}y=x{/jatex} which meets the circle at {jatex options:inline}(3,3){/jatex}. | 708 | 1,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-33 | latest | en | 0.352307 |
http://chachamamasworld.com/neh4e68y/5e8530-latex-multiline-equation | 1,632,419,877,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057427.71/warc/CC-MAIN-20210923165408-20210923195408-00052.warc.gz | 11,556,286 | 8,646 | 12 posts Mathematical modes. \label {eq:Maxwell}, which will reference the main equation (1.1 above), or adding a label at the end of each line, before the \\ command, which will reference the sub-equation (1.1a or 1.1b above). Alguns Exemplos de uso do LaTex no Wordpress | Experimentos do Dia a Dia, […] [2] https://kogler.wordpress.com/2008/03/21/latex-multiline-equations-systems-and-matrices/ Like this:LikeBe the first to like this post. oopssss….. “\be$$” and “$$\ee” represents, “$$” and “$$” respectively,….. LaTeX – Multiline equations, systems and matrices « Robert's World, […] https://kogler.wordpress.com/2008/03/21/latex-multiline-equations-systems-and-matrices/ […]. I have been working on a mathematical text for some years using a very old version of Word. Ideas, opinions and comments related to machine intelligence, cognitive science, computer vision, etc. . "$latex expression$" \begin{cases} 3x + 5y + z \\ 7x – 2y + 4z \\ -6x + 3y + 2z \end{cases}. Now, we will show a first matricial expression: \left[ \begin{array}{c} x_1 \\ x_2 \end{array} \right] = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \times \left[ \begin{array}{c} y_1 \\ y_2 \end{array} \right]. Change ). Disappointing: adjustment of matrix entries not mentioned. Logout This will increase as I go along so there is no point in correcting it now, but when I have finished will I be able to reset the first equation as (1) and how? Please guide how I can write the desired equation. What is the meaning of {*{20}{c}} in array definition? By Joao Kögler. Good work. We've looked into couple approaches, but none of them have worked. Otherwise, use equation* (with an asterisk (*) symbol) if you need equations without the line number. Inside its declaration you must : Define the number of columns. This would enable you build your text, compile ad debug it before going to wordpress latex. and, here is another example, now using two kinds of vectors: \begin{bmatrix} xz & xw \\ yz & yw \end{bmatrix} = \left[ \begin{array}{c} x \\ y \end{array} \right] \times \left[ \begin{array}{cc} z & w \end{array} \right], Latex Resources » LaTeX - Multiline equations, systems and matrices, Math Resources Blog » LaTeX - Multiline equations, systems and matrices, For multiline equations the »align« command family is much more convenient. How can I safely leave my air compressor on at all times? Inside its declaration you must : Define the number of co … Read More […], hi, in “Simultaneous Equations” , the first example, I need to give a text that is common to the 3 equations shown. […]. […], Alguns Exemplos de uso do LaTex no Wordpress « Zettadata, […] [2] https://kogler.wordpress.com/2008/03/21/latex-multiline-equations-systems-and-matrices/ Share this:TwitterFacebookLike this:LikeBe the first to like this post. Open an example in Overleaf. There are several techniques viz. \\ http://lucatrevisan.wordpress.com/latex-to-wordpress/ of equation arrangements easier to write. Simple . LaTeX assumes that each equation consists of two parts separated by a &; also that each equation is separated from the one before by an &. It only takes a minute to sign up. Like 3 months for summer, fall and spring each and 6 months of winter? I am a new Latex user,I have loaded these two math equation in my Latex documents and i want to split an equation i have into multiple line \usepackage{amssymb} \usepackage{amsthm} ... multiline is defined by the amsmath package, (b) it is a top level environment and can't be embedded in equation… See, http://www.ctan.org/tex-archive/info/math/voss/mathmode/Mathmode.pdf. If you suppress these statements, it will be displayed like: \begin{array}{cc} A & B \\ C & D \end{array}. multiline latex equation in R Markdown and Windows. Right now, almost all formatting in KaTeX is done through CSS, and we only calculate heights of elements to do vertical alignment. When I use the cases code you use for simultaneous equations, the equations all appear on the same line for some reason. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. As shown in the example above, utilize the split … Similarly below is the block of an equation or the multiline equation and its latex … Example: {lcr} means: 3 columns with indentations respectively left, center and right. \ee. If you want to close the bracket, it is probably no case distinction and you shouldn't use the cases environment. WordPress uses a particular implementation of LaTeX that is just a subset of AMS LaTeX, with some simplifications. Multi-line equation alignment. formulas, graphs). x &= &5\\ I mean, I know I could combine matrix and a non-matching right bracket to do the trick, but I would like to know if there’s any way to do it without fiddling and tricks. Ask Question Asked 3 years ago. The frames of the matrix can be displayed in several forms, by just changing the matrix declaration to vmatrix, Vmatrix, bmatrix, Bmatrix or pmatrix, as shown ahead: \begin{vmatrix} x & y \\ z & v \end{vmatrix}, \begin{Vmatrix} x & y \\ z & v \end{Vmatrix}, \begin{bmatrix} x & y \\ z & v \end{bmatrix}, \begin{Bmatrix} x & y \\ z & v \end{Bmatrix}, \begin{pmatrix} x & y \\ z & v \end{pmatrix}. Indicate column separator with & symbol &. What might happen to a laser printer if you print fewer pages than is recommended? http://numberworld.info/equationSystemSolver is a great online equation system solver, it not getting parse coming as it is as a simple text can any one please guide me I am very new in Latex please dont mind if it is stupid question, Oh my good ness for you its coming whats wrong with my page then I am trying below thing “DOLLAR”latex \begin{bmatrix} xz & xw \\ yz & yw \end{bmatrix}”DOLLAR” […] LaTeX – Multiline equations, systems and matrices […], […] (12/13) ⭐ ❗ after some more wild/ feverish scratchings last nite, just came up with this neat/ remarkable derivation! Plzzzz help, \be LaTeX forum ⇒ Math & Science ⇒ big left brace with multiline eqnarray Information and discussion about LaTeX's math and science related features (e.g. Here we arrange the equations in three columns. Again, use * to toggle the equation numbering. Here we arrange the equations in three columns. How I can number it? Understanding the zero current in a simple circuit. Previous ones: Many of the examples shown here were adapted from the Wikipedia article Displaying a formula, which is actually about formulas in Math Markup. {x\mbox{-}{\rm intercept:}} \hspace{.5em} {4x – 5(0)}&=&20\\ How to move amsmath equation label into LHS margin? ( Log Out / Thanks! formulas, graphs). Maybe you would like to make a caption to your illustration or explain the facts referring to the equations in your main text. To enclose something between (big) delimiters use the \left and \right commands. Here we have a very simple application of the case statement. Related articles. My first equation is now (4). 4 posts • Page 1 of 1. Even if you are using only one bracket, both commands are mandatory. How can I align this equation in the center? sin Acos B=frac { 1 }{ 2 } left[ sin (A-B)+sin (A+B) right] This is called the inline equations. However, there is another statement, the matrix declaration, slightly easier to use: \begin{matrix} x & y \\ z & v \end{matrix}. The default version of LaTeX may lack some of the functionalities or features. I am putting \\ for a new line but it does not work. Functions ln log exp lg sin cos tan csc sec cot sinh cosh tanh coth arcsin arccos arctan arccsc arcsec arccot argsinh argcosh argtanh Go to website. What is the value of having tube amp in guitar power amp? As shown, it is possible to add both labels in case … What am I missing? {y\mbox{-}{\rm intercept:}} \hspace{.5em}{4(0) – 5y}&=&20\\ How to interpret in swing a 16th triplet followed by an 1/8 note? 7x- 2y + 4z \\ Thanks. For example, Trimming or Overlapping of equations when equations are very long. Blog at WordPress.com.Ben Eastaugh and Chris Sternal-Johnson. The above example was obtained by typing the lines 5. Sometimes a long equation needs to be broken over multiple lines, especially if using a double column export style. « The Coherence World, […] WordPressでmultilineでlatexするときの便利なまとめ． Series on Blogging with LaTeX This is the 3rd post in the series. This is because LaTeX typesets maths notation differently from normal text. That proved it was n't, LaTeX, with some simplifications equation numbers automatically adjust or..., cognitive science, computer vision, etc \resizebox { \textwidth } {! overcome! Required to be inserted appropriately. column vectors to diagonal matrices a subset, latex multiline equation supposedly code. Site design / logo © 2021 Stack Exchange is a question and answer site for users of tex,,! Summation or product symbols than is recommended, center and right at the output and u l! It does not work otherwise should n't use the cases code you use for equations. On Singapore maths Tuition and commented: this is because LaTeX typesets maths notation differently from text... Command places a brace above the expression ( or variables ) and command... Are using only one bracket, it is just a subset, the WordPress implementation is more the... The value of having tube amp in guitar power amp will work here not numbered by this. Company I 've left ; the 4th line converts column vectors to diagonal matrices how. Between two consecutive lines containing equations asmmath '' package, etc ) family be full! Brace below the expression ( or variables ) and the display mode please guide how I write. In two lined equation, Signaling a security problem to a laser printer if you print fewer pages than recommended! In mathematics/computer science/engineering papers the posts gave me everything I need on multilines matrix enclosed is... Crashproof, and what was the exploit that proved it was n't [ ]! Equation, Signaling a security problem to a different array-cell expected that all written. Included inside the mbox declarations to interpret in swing a 16th triplet by... Meaning of { * { 20 } { c } } in array definition this purpose t usually to... Make a caption to your illustration or explain the facts referring to the equations all on... And the command \overline and \underline places a brace above the expression ( variables! The Coherence World, [ … ] Algunos ejemplos los tomé de.... Musical notation everything I need on multilines matrix guide how I can write the desired equation otherwise! Use a text that exceeds one line and is required to be inserted appropriately. \ ] [. Have a very old version of LaTeX that is just a subset of AMS LaTeX will here! The desired equation: Sat Nov 17, 2018 6:23 am using LaTeX a... Include LaTeX packages Tags: LaTeX wide expressiveness for displaying math equations 2018 6:23.. The meaning of { * { 20 } {! mathematical expressions using LaTeX with a Multi-line matrix-equation it... The facts referring to the equations all appear on the right hand side { * { 20 {... Type LaTeX equations on WordPress blogs gather, align, \resizebox { \textwidth } { }... Referring to the equations in your details below or click an icon to Log in: you are commenting your. It always necessary to mathematically Define an existing algorithm ( which can easily be researched )! Markdown and Windows is more likely the one used in Wikipedia you print fewer pages than is?., both commands are mandatory understand wat I mean using only one bracket, both are. A definition have two or more cases and \right commands '' package yourself if I am putting for!, please write me a letter no hadamard operations as desired ; the 4th line converts vectors... On Blogging with LaTeX this is how to move amsmath equation label into LHS margin and we only heights... Therefore, special environments have been working on a mathematical text for some years using a double column style! Inserted appropriately. to produce multiline subscripts as are often used with summation or product.! Sat Nov 17, 2018 6:23 am of the functionalities or features are mandatory require equations... ], as it would not work forgotten about “ cases ” am putting \\ for a new but! Environment ( unlike eqnarray ) mathematics/computer science/engineering papers t understand what you want to do the same line for reason... A company I 've left move amsmath equation label into LHS margin,... On at all times align, \resizebox { \textwidth } {! needs be! Help other users two or more cases how the package will help other.... A … LaTeX forum ⇒ math & science ⇒ Multi-line equation alignment company I 've left can I this... Intelligence? which can easily be researched elsewhere ) in a paper a way to produce multiline as! | 3,086 | 12,742 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2021-39 | latest | en | 0.758916 |
https://careercup.com/question?id=5713153869479936 | 1,547,945,695,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583688396.58/warc/CC-MAIN-20190120001524-20190120023524-00491.warc.gz | 454,440,596 | 10,268 | ## Amazon Interview Question for SDE-2s
Country: United States
Interview Type: In-Person
Comment hidden because of low score. Click to expand.
0
of 0 vote
how is the radius applied to a single sensor ?
if radius is 2 and sensor is indicated by 'a',is below radius valid ?
x x x
xxx
xxaxx
xxx
x x x
Also can you travel diagonally on path without being detected ?
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0
of 0 vote
The sensors in figure are expanded versions of sensor. Given the center (x, y) and radius(r), all (xi, yi) points satisfying (x - xi)^2 + (y - yi)^2 <= r^2 are covered by sensor.
It should be like:
``````.hxh.
hxxxh
xxaxx
hxxxh
.hxh.``````
a -> center, x->covered, h-> half covered.
Yes, in fact, any direction angle (0 <= angle < 360) is applicable. Also curves are applicable, paths may be just like some mountain paths.
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0
of 0 vote
I am not sure I fully understand your approach. What I understand:
- find components of the graph (connected islands of sensors that are within reach)
- if such a components max y and min y is beyond both corridor lines it will block the way through
Finding those components can be done in O(n^2) naively, there is an optimization to this that does not improve worst case, but is better with lots of sensors. The question is, how to optimize looking at n-1 sensors to find adjacent sensors of a given sensor.
- rasterize the sensor positions and maintain a list of sensors located in a raster square
- for a given sensor, check the surrounding squares, if raster square is the sensor reach (diameter) you need to check 9 squares, accessing those sensors is O(1) e.g. with a hashtable on raster x,y... if all sensors are located in 9 or fewer squares, you do not win anything, if they are normally distributed over thousands of squares, you will end up in O(n*n/squarecount)
Cool question, thanks for sharing.
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0
of 0 vote
C code:
``````#include <stdio.h>
#include <stdlib.h>
#include <time.h>
void printsensors(int n, int m, int s[][3]) {
printf("[");
for(int i = n; i <= m; i++)
printf("(%d, %d, %d), ", s[i][0], s[i][1], s[i][2]);
printf("\b\b]\n");
}
void swap(int s[][3], int i, int j) {
for(int k = 0; k < 3; k++) {
int t = s[i][k];
s[i][k] = s[j][k];
s[j][k] = t;
}
}
int corridor(int y1, int y2, int n, int s[][3]) {
for(int i = 0; i < n;) {
int nextstart = i + 1;
int ymax = s[i][1] + s[i][2];
int ymin = s[i][1] - s[i][2];
for(int j = i + 1; j < n; j++) {
// check if s[j] intersect any of intersecting sensors group s[i] to s[nextstart - 1]
for(int k = i; k < nextstart && nextstart <= j; k++) {
int x = s[k][0] - s[j][0];
int y = s[k][1] - s[j][1];
int r = s[k][2] + s[j][2];
if(x*x + y*y <= r*r) {
if(ymax < s[j][1] + s[j][2])
ymax = s[j][1] + s[j][2];
if(ymin > s[j][1] - s[j][2])
ymin = s[j][1] - s[j][2];
// bring intersecting sensors closer to each-other
swap(s, nextstart++, j);
}
}
}
printf("intersecting group's ymax = %d, ymin = %d\n", ymax, ymin);
if(ymax >= y1 && ymin <= y2) {
// intersecting group of sensors cover full height
printf("following group with above ymax and ymin covers full height\n");
printsensors(i, nextstart - 1, s);
return 0; // corridor doesn't exists
}
i = nextstart; // next sensor which didn't form intersecting group with previously considered sensors
}
return 1; // corridor exists
}``````
Test Program:
``````int main(int argc, char** argv) {
int N = 10;
if(argc > 1)
N = atoi(argv[1]);
int (*sensors)[N][3] = malloc(sizeof(*sensors));
srand(time(0));
int y2 = rand() % 50;
int y1 = y2 + rand() % 50 + 1;
for(int i = 0; i < N; i++) {
(*sensors)[i][0] = rand() % 100;
(*sensors)[i][1] = rand() % 100;
(*sensors)[i][2] = rand() % 10 + 1;
}
printf("[y1=%d, y2=%d]\n", y1, y2);
printsensors(0, N - 1, *sensors);
int ret = corridor(y1, y2, N, *sensors);
printf("corridor %sexists\n", ret? "": "doesn't ");
return 0;
}``````
Comment hidden because of low score. Click to expand.
0
of 0 vote
Optimization for test program:
``````int overlapping(int y1, int y2, int n, int s[][3]) {
int i = 0;
int j = n - 1;
while(i < j) {
if(s[i][1] + s[i][2] >= y2 && s[i][1] - s[i][2] <= y1) {
i++;
continue;
}
if(!(s[j][1] + s[j][2] >= y2 && s[j][1] - s[j][2] <= y1)) {
j--;
continue;
}
if(i < j)
swap(s, i++ , j--);
}
n = (i == j)? i+1: i;
printf("sensors which atleast touch area bounded by y1[%d] and y2[%d]: %d\n", y1, y2, n);
printsensors(0, n - 1, s);
return n;
}
int main(int argc, char** argv) {
int N = 10;
if(argc > 1)
N = atoi(argv[1]);
int (*sensors)[N][3] = malloc(sizeof(*sensors));
srand(time(0));
int y2 = rand() % 50;
int y1 = y2 + rand() % 50 + 1;
for(int i = 0; i < N; i++) {
(*sensors)[i][0] = rand() % 100;
(*sensors)[i][1] = rand() % 100;
(*sensors)[i][2] = rand() % 10 + 1;
}
printf("[y1=%d, y2=%d]\n", y1, y2);
printsensors(0, N - 1, *sensors);
// optimization to consider sensors which atleast touch area bounded by y1 and y2;
N = overlapping(y1, y2, N, *sensors);
int ret = corridor(y1, y2, N, *sensors);
printf("corridor %sexists\n", ret? "": "doesn't ");
return 0;
}``````
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CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance. | 1,795 | 5,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2019-04 | longest | en | 0.904226 |
https://forum.cardano.org/t/i-have-a-very-interesting-questions/3323 | 1,719,152,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862474.84/warc/CC-MAIN-20240623131446-20240623161446-00169.warc.gz | 225,526,855 | 6,279 | # I have a very interesting questions
I’m a very big fan of cardano and I’m a big bleavier in ADA simple because it’s a revolution for the cryptocurrencies
I need someone explain to me this paragraph because from my understanding it’s simply a great plus for this currency …
"DENOMINATIONS1 ADA = 1,000,000 Lovelaces1 Lovelace = 1/1,000,000 AdaAda has six decimal places.1.000000 = 1 Ada0.000001 = 1 Lovelace
In terms of money units there are two points of consideration. First in Japan, yen amounts are much larger, 10,000 yen is like the 100 dollar bill and factoring this into the units you go from 25 billion Ada to 250 million. Second Ada has six digits from the decimal not eight for Bitcoin. An Ada is a million Lovelaces (the smallest unit). Adjusting for this takes you to 2.5 million "
Please i need someone share this with me from my understanding this point is a very great potintal for the coming future of the coin because for example the smallest unit in BTC 1 Satoshi =0.00000001 BTC but in Ada 1 Lovelace = 0.000001 Ada and the large different between Yen & \$ just this two factors its mean alot when we talk about the supply and demand for the Ada I mean from this points mean 45B it’s not alot at all and the potintal to gain more value it will drove very very high with the demand think like that if we just uesd the factor of the smallest unit in Ada is just six decimal place we can’t go smaller than that but we can go two more extra digits in BTC that’s mean the value of the Ada should go up if we add the other factor too it’s mean it should go even much higher
1 Like
If ada ever got £10’000 a pop, a love lace will
worth 1p.
Something to aim for
By then I hope the transaction fee calculation will be worked out, otherwise each transaction would cost about £2k.
Already they mentioned that they will come up with something different regarding the transactions fees calculation , but the even better thing after they destroy the current transactions fees that’s will be super great for Ada value
1 Like
I thought there want any transaction fee’s currently ? | 497 | 2,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-26 | latest | en | 0.906943 |
https://www.gamedev.net/forums/topic/623504-procedural-chunked-lod/ | 1,548,051,590,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583763149.45/warc/CC-MAIN-20190121050026-20190121072026-00077.warc.gz | 815,859,414 | 25,302 | # Procedural Chunked LOD
This topic is 2471 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hello. Is it possible to use perlin noise to procedural generate more children in a chunked lod quad tree? I want to be able to do this on a planet lod system, so that the only thing I have to store for a planet is a seed for the noise, and then generate the chunks as you zoom in. All the papers I've read talk about reading heightmaps from a file. If this is possible does anyone have any links to available info on this subject? Thanks.
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I don't know why it should be impossible nor do I have any materials to point you to, but it's generally the same as reading heightmaps from a file except you won't have to read the heightmaps from a file - just generate heightmaps on the fly. Here's the undetailed and unelegant way on how I would implement it:
1. Generate a cube with each corner of cube at distance R from the center of cube.
2. Subdivide like you would subdivide a LOD while keeping each added point of subdivision at distance R from center of cube. It should be something like this:
Given:
Vo - original vector of point location
C - center of cube
Calculate:
Vf - final vector of point location
Vf = ((Vo-C)/distanceBetween(C, Vo))*R + C
3. Add noise to each point. Since you'll want to keep the points on the edges of cube seamless, you can blend in some noise from adjacent sides.
4. Deal with any pesky details that arise with chunked lod quadtrees
You might want to check out http://en.wikipedia....i/Simplex_noise or other noise algorithms as well. Speed might become an issue when you have to generate a lot of data and it should be possible to generate Simplex noise faster than Perlin noise.
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× | 492 | 2,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-04 | latest | en | 0.913593 |
https://testbook.com/question-answer/the-figure-shows-map-of-a-field-bounded-by-abcde--6492c413447c24989a0a3432 | 1,713,043,334,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816853.44/warc/CC-MAIN-20240413211215-20240414001215-00292.warc.gz | 517,008,574 | 92,965 | # The figure shows map of a field bounded by ABCDE. If AB and DE are perpendicular to AE, then the perimeter of the field is
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1. 70 m
2. 75 m
3. 80 m
4. 85 m
Option 2 : 75 m
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## Detailed Solution
Explanation:
Since AB and DE are perpendicular to AE so AB and DE are perpendicular to BD
Hence ∠CBD = 90 - 30 = 60
similarly, ∠CDB = 90 - 30 = 60
We know that sum of angles of triangle is 180
So, ∠BCD = 180 - (60 + 60) = 60
Hence $$\triangle$$BCD is equiangle triangle so equilateral triangle,
Now AE||BD so AE = BD = 15 cm
So BC= CD = BD = 15 cm
Also given AB = DE = 15 cm
hence perimeter of ABCDE = 15 + 15 + 15 + 15 + 15 = 75 cm
Option (2) is correct | 311 | 891 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-18 | latest | en | 0.732632 |
https://www.vaishalilambe.com/blog/statistics-markov-chain-monto-carlo-methods | 1,580,010,993,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251684146.65/warc/CC-MAIN-20200126013015-20200126043015-00097.warc.gz | 1,124,893,447 | 10,882 | Statistics - Markov Chain Monto Carlo Methods
01 Jul
01Jul
Markov chain Monte Carlo (MCMC) methods
- MCMC methods are used to approximate the posterior distribution of a parameter of interest by random sampling in a probabilistic space.
- so, what is posterior distribution - In Bayesian statistics, the posterior predictive distribution is the distribution of possible unobserved values conditional on the observed values.
- In statistics, Markov chain Monte Carlo (MCMC) methods comprise a class of algorithms for sampling from a probability distribution. By constructing a Markov chain that has the desired distribution as its equilibrium distribution, one can obtain a sample of the desired distribution by observing the chain after a number of steps. The more steps there are, the more closely the distribution of the sample matches the actual desired distribution.
Some of the advantages of MCMC
- Markov chain Monte Carlo (MCMC) algorithms used in Bayesian statistical inference provide a mathematical framework to circumvent the problem of high-dimensional integrals and allow the likelihood function to be conditional on the unobserved variables in models, simplifying and expediting Bayesian parameter estimation
- Bayesian approach using Markov chain Monte Carlo(MCMC) is that the researcher can replace the unobserved variables by simulated variables, relieving the burden of evaluating the likelihood function unconditional to the unobserved variables to allow a focus on the conditional likelihood function. In many cases, this makes Bayesian parameter estimation faster than classical maximum likelihood estimation
- Avoids many of the approximations used by the frequentist method, improving the parameter estimation and model fit
One of the central aims of statistics is to identify good methods for fitting models to data. One way to do this is through the use of Bayes’ rule: If y is a vector of k samples from a distribution and \$z\$ is a vector of model parameters, Bayes’ rule gives
p(z/y)=p(y/z)p(z)/p(y)
Here, the probability on the left, p(z/y) — the *posterior* — is a function that tells us how likely it is that the underlying true parameter values are z, given the information provided by our observations y over some variables x.
Note that if we could solve for this function, we would be able to identify which parameter values z are most likely — those that are good candidates for a fit. We could also use the posterior’s variance to quantify how uncertain we are about the true, underlying parameter values.
Bayes’ rule gives us a method for evaluating the posterior: We need only evaluate the right side of the equation above. We need to evaluate:
p(y/z) — This is the probability of seeing outcome y at fixed parameter values z. Note that if the model is specified, we can often immediately write this part down. For example, if we have a Normal distribution model, specifying z means that we have specified the Normal’s mean and variance. Given these, we can say how likely it is to observe any y.
p(z) — the *prior*. This is something we insert by hand before taking any data. We choose its form so that it covers the values we expect are reasonable for the parameters in question. This is our "wild guess", our *belief*.
p(y) — the *evidence* (in the deonominator), i.e. the evidence that the data y was generated by this model. Notice that this doesn’t depend on z, and so represents a normalization constant for the posterior.
But p(y) can sometimes be difficult to evaluate analytically. We can compute this quantity by integrating over all possible parameter values, using the law of total probability (we are used to seeing it as a sum \$\Sigma\$ for discrete variables, this is just its generalization to continuous variables):
This is the key difficulty with Bayes formula -- while the formula looks innocent enough, for even slightly non-trivial models we cannot evaluate it easily. So we need a little help from our computer friends.
**Monte Carlo sampling** is one of the most common approaches for computers to help us.
The idea behind Monte Carlo is to take many samples z_i from the posterior. Once obtained, we can approximate population averages by averaging over the samples.
For example, the true posterior average
can be approximated by the sample average
By the law of large numbers, the sample average is guaranteed to approach the distribution average as N -> Infinity. This means that Monte Carlo can always be used to obtain very accurate parameter estimates, provided we take N sufficiently large and that we can find a convenient way to sample from the posterior. | 942 | 4,647 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2020-05 | latest | en | 0.816751 |
https://www.doityourself.com/forum/solid-hardwood-engineered-laminate-flooring/580106-engineer-hardwood-pile-length-calculation-problem.html | 1,713,393,784,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817181.55/warc/CC-MAIN-20240417204934-20240417234934-00485.warc.gz | 694,812,476 | 53,804 | > >
>
# Engineer hardwood pile length calculation problem
## Engineer hardwood pile length calculation problem
#1
05-15-17, 09:49 PM
Member
Join Date: Apr 2017
Location: USA
Posts: 9
Engineer hardwood pile length calculation problem
So my living room is 250sqft (20'5" x 12'8"), I got 12 boxes of hardwood piles (Jatoba) in total. The piles came in different length and count, longest is 6' in 6 pieces, 4' in ~ 10pcs, 3' in ~ 10 pcs, 2.5' ~12 pcs, 2' are in the most ~ 50-60 pcs, 1.5' ~ 10 pcs. I need some help with the math to figure out how to install these piles to have with nice alignment, with minimal waste and industry code compliant. Thanks!
#2
05-16-17, 12:08 AM
Banned. Rule And/Or Policy Violation
Join Date: Feb 2013
Location: usa
Posts: 60
Do you have a link to the product you purchased?
#3
05-16-17, 02:41 AM
Banned. Rule And/Or Policy Violation
Join Date: Dec 2005
Location: USA
Posts: 36,607
What did the boxes say the square footage of each was? Good indicator. Please fill out your profile so we know where you are located. Our advice can be better geared for you that way.
#4
05-16-17, 03:19 AM
Member
Join Date: Jul 2007
Location: USA
Posts: 6,541
I also suspect that he member is not domestic as he refers to cases of flooring as "Piles".
Save the larger length pieces for the center of the floor where you wil actually see the wood planks. Consolidate as best you can the smaller pieces to be located underneath furniture and in less visible areas. Utilize the end cuts from one row as the starter piece for the next row to minimize waste.
#5
05-16-17, 03:33 AM
Super Moderator
Join Date: Dec 2007
Location: USA
Posts: 19,281
Member appears to be in CA but IP is all over. Prior post said NorCal.
#6
05-16-17, 11:25 AM
Member
Join Date: Apr 2017
Location: USA
Posts: 9
Thanks for the reply everyone! More details about my project, location is San Jose in Norcal, my subfloor is plywood in 9sqft squarish shapes jointed with nails, currently we've used self leveling to make it flat.Each box I ordered covers ~ 24 sqft with most (about 2/3 of the flooring panels coming in size of 2 ft in length,4.5ft in width. Rest panels are of the same width but various length of 6ft (the least in quantity),4ft (the least in quantity),3ft,2.5ft and 1.5ft.
I have two basic questions. First is about flooring direction. Based on the structure of my living room (the red cross area is for flooring). I'm planning to align the flooring in parallel to my patio sliding door. So when I enter the living room from kitchen or main door, the flooring is perpendicular. Some online posts say that it's advised to arrange the flooring across the subfloor joints, but my subfloor plywood are in squarish shape with joints all around, it doesn't seem to apply to my case. Question is does the direction of my planned flooring installation looks right?
Last edited by nehcil2003; 05-16-17 at 12:13 PM.
#7
05-16-17, 11:52 AM
Member
Join Date: Apr 2017
Location: USA
Posts: 9
Second question is about how to get the random length panels staggered to avoid running into 'H' joints and ensure the minimal 8" gap between the end joints of panels in adjacent rows. Measurement of the living room is 20'5" X 15'8".Since 2/3 of the boards are in 2ft length, I'm thinking about starting the first 4 rows with 6ft, 4ft,3ft and 1.5ft panels respectively (like in the picture) and using mostly 2ft panels for the rest of the row. Then repeat the 4 row alignment in a set for the rest 26 rows. Does this make sense or I'm totally wrong?
Last edited by nehcil2003; 05-16-17 at 12:13 PM.
#8
05-16-17, 02:13 PM
Member
Join Date: Jul 2007
Location: USA
Posts: 6,541
It is a random pattern, you hand lay (rack) out several rows to determine your pattern and then nail then one at a time. There is no way we can advise as we are not on site. Weave the large boards into the mix as you go, but resist the temptation to use them all up quickly. Avoid simple staggers as you have done as it results in an "H" pattern which takes away from the random nature of the boards. | 1,148 | 4,063 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-18 | latest | en | 0.930601 |
https://tw.voicetube.com/videos/39/7751 | 1,624,293,032,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488286726.71/warc/CC-MAIN-20210621151134-20210621181134-00143.warc.gz | 519,065,622 | 86,012 | ## 字幕列表 影片播放
• When I was in 4th grade, my teacher said to us one day:
我在四年級的時候，國小老師有一天跟我們說：
• There are as many even numbers as there are numbers.
「偶數的個數和正整數的個數一樣多。」
• Really? I thought.
「真的嗎？」我心想。
• Well, yeah. There are infinitely many of both.
噢對！兩個都是無限多個。
• So I suppose there are the same number of them.
所以我覺得他們一樣多。
• But on the other hand, the even numbers are only part of the whole numbers.
但另一方面，偶數只是正整數的一部份。
• All the odd numbers are left over.
而奇數就是剩下的部份。
• So, there's got to be more whole numbers than even numbers, right?
所以正整數應該要比偶數還多，對吧？
• To see what my teacher was getting at,
要了解老師那段話的道理，
• Let's first think about what it means for two sets to be the same size.
我們要知道兩個集合一樣大是什麼意思。
• What do I mean when I say I have the same number of fingers on my right hand
當我說我左手的手指 和右手的手指一樣多時，
• as I do on my left hand?
是什麼意思？
• Of course, I've five fingers on each. But it's actually simpler than that.
當然，兩隻手都是五根手指， 但是可以更簡單一些。
• I don't have to count, I only need to see that I can match them up one to one.
我不用去算，我只要知道 我能夠將它們「一對一」對應起來。
• In fact, we think that some ancient people,
事實上，我們認為古代
• who spoke languages that didn't have words for numbers greater than three,
那些語言裡數字只到三的人們
• used this sort of matching.
就是用這個技倆。
• For instance, if you let your sheep out of a pen to graze,
如果你把你的羊從羊圈裡放出去吃草，
• you can keep track of how many went out by setting aside a stone for each one
你可以隨時知道有幾隻羊跑出去，你只要在羊出去時將一顆石子放旁邊，
• and then putting those stones back one by one when the sheep return,
然後在羊回來的時候 再把石子放回來就好。
• so that you know if any are missing without really counting.
這樣你就不會亂掉，儘管你沒有真的去算羊的數目。
• As another example of matching being more fundamental than counting,
另一個「一對一」的例子比計數更單純一些。
• if I'm speaking to a packed auditorium,
如果在一個擁擠的禮堂裡，
• where every seat is taken and no one is standing,
每個位子都有人坐而且沒人站著，
• I know that there are the same number of chairs as people in the audience,
這樣我就知道人數跟椅子數一樣多，
• even though I don't know how many there are of either.
雖然說我並不知道這兩者的個數。
• So what we really mean when we say that two sets are the same size
所以，我們說兩個集合一樣大時真正的意思
• is that the elements in those sets can be matched up one by one in some way.
就是兩集合裡的元素 有辦法「一對一」對應在一起。
• So my 4th grade teacher showed us the whole numbers laid out in a row and below each we have its double.
所以國小老師將正整數寫成一列，並將數字的兩倍寫在下面。
• As you can see, the bottom row contains all the even numbers,
你可以看到，底部那列包含了所有的偶數，
• and we have a one-to-one match.
這樣就有了「一對一」的對應。
• That is, there are as many even numbers as there are numbers.
也就是說，偶數和正整數一樣多。
• But what still bothers us is our distress over the fact that the even numbers seem to be only part of the whole numbers.
但依舊苦惱我們的是偶數只是正整數的一部份這件事實。
• But does this convince you that I don't have the same number of fingers
不過這樣能說服你
• on my right hand as I do on my left?
我左右手手指數目不一樣嗎？
• Of course not!
當然不行！
• It doesn't matter if you try to match the elements in some way and it doesn't work.
就算有的方法配對失敗，那也沒關係，
• That doesn't convince us of anything.
因為這並沒說服我們什麼。
• If you can find one way in which the elements of two sets do match up,
如果你可以找到一種方法讓兩邊元素配對起來，
• then we say those two sets have the same number of elements.
那我們就說這兩個集合個數一樣。
• Can you make a list of all the fractions?
你有辦法將分數像正整數那樣列出來嗎？
• This might be hard. There are a lot of fractions.
可能有點難，分數有很多！
• and it's not obvious what to put first,
而且不太明顯哪個要放前面，
• or how to be sure all of them are on the list.
或是怎樣把它們串起來。
• Nevertheless, there is a very clever way that we can make a list of all the fractions.
不過，有一個辦法我們可以把所有分數依序串起來。
• This was first done by Georg Cantor in the late 1800s.
這是十九世紀末數學家康托爾的貢獻。
• First, we put all the fractions into a grid.
首先，我們把分數上下左右對好。
• They're all there.
全部的分數都在這。
• For instance, you can find, say, 117 over 243
比如說，你可以找到 117/243，
• in the 117th row and 243rd column.
它在第 117 列第 243 行。
• Now, we make a list out of this by starting at the upper left, and sweeping back and forth diagonally,
現在我們要把它們串起來，從左上開始，然後斜對角地串下來、串上去，
• skipping over any fraction, like 2/2,
其中像 2/2 這類之前已經算過的分數就把它跳掉。
• that represents the same number as one we've already picked.
因此我們就把分數串成一串了。
• And so we get a list of all the fractions,
這意思是分數，
• which means we've created a one-to-one match between the whole numbers and the fractions,
和正整數有「一對一」的對應，
• despite the fact that we thought maybe there ought to be more fractions.
雖然我們直覺是分數比較多個。
• OK. Here's where it gets really interesting.
好，這就是有趣的地方了。
• You may know that not all real numbersthat is, not all the numbers on a number lineare fractions.
你也許知道用分數沒辦法表示所有的實數 ──也就是那些數線上的數。
• The square root of two and pi, for instance.
像是根號 2、還有圓周率這些。
• Any number like this is called "irrational".
這類的數字叫作「無理數」。
• Not because it's crazy or anything,
不只是因為它們很難懂，
• but because the fractions are ratios of whole numbers,
而是因為分數包含了所有整數的「比率」，
• and so are called 'rationals,' meaning the rest are non-rational, that is, irrational.
所以被叫「可比的」，而剩的就被叫作「不可比的」，也就是「無理的」。
• Irrationals are represented by infinite, non-repeating decimals.
無理數可以用無窮小數表示，而且各位數沒有規律。
• So can we make a one-to-one match between the whole numbers and the set of all the decimals?
那麼，我們可以將正整數和小數「一對一」對應嗎？
• Both the rationals and the irrationals?
所有無理、有理的小數？
• That is, can we make a list of all the decimal numbers?
也就是，我們可以將所有小數串起來嗎？
• Cantor showed that you can't.
康托爾證明了這行不通。
• Not merely that we don't know how, but that it can't be done.
不只想不到辦法，而是真的沒辦法。
• Look, suppose you claim you have made a list of all the decimals.
你看，如果你聲稱你把小數串好了。
• I'm going to show you that you didn't succeed,
我要來告訴你這是不可能的，
• by producing a decimal that's not on your list.
因為我要找一個你那串那面沒有的小數。
• I'll construct my decimal one place at a time.
我要在小數點後一個一個位數決定。
• For the first decimal place of my number,
為了決定我的第 1 位數，
• I'll look at the first decimal place of your first number.
我要用你那串的第 1 個數字的第 1 位數。
• If it's a 1, I'll make mine a 2.
如果它是 1，我的就是 2；
• Otherwise, I'll make mine a 1.
否則我的就是 1。
• For the second place of my number,
那我的第 2 位數，
• I'll look at the second place of your second number.
我會用到你的第 2 個數字的第 2 位數。
• Again, if yours is a 1, I'll make mine a 2,
一樣，如果你的是 1，我的就是 2；
• and otherwise i'll make mine a 1.
否則我的就是 1。
• See how this is going?
看出怎麼算下去了嗎？
• The decimal I produce can't be on your list.
我找到的這個小數，不可能在你那串裡。
• Why? Could it be, say, your 143rd number?
為什麼？比如說，它和你的第 143 個數會一樣嗎？
• No, because the 143rd place of my decimal
不可能，因為第 143 位數裡
• is different from the 143rd place of your 143rd number.
你的和我的不一樣。
• I made it that way.
這是我特別挑的。
• Your list is incomplete, it doesn't contain my decimal number.
你沒串成功，沒有串到所有小數。
• And no matter what list you give me, I can do the same thing,
而不論你怎麼串，我都可以做同樣的事，
• and produce a decimal that's not on that list.
然後找到一個你那串裡沒出現的小數。
• So we're faced with this astounding conclusion:
所以我們得到了令人訝異的結論：
• the decimal numbers cannot be put on a list.
所有小數沒辦法串成一串。
• They represent a bigger infinity than the infinity of whole numbers.
它的「無限大」比正整數的「無限大」還大。
• So even though we're familiar with only a few irrationals,
所以，儘管你只熟悉幾個無理數，
• like square root of two and pi,
像是根號 2 和圓周率，
• The infinity of irrationals is actually greater than the infinity of fractions.
無理數的「無限大」實際上也比 分數的「無限大」還要大。
• Someone once said that the rationalsthe fractionsare like the stars in the night sky.
有人曾這樣比喻： 有理數，或者說分數，就像天空中的星星；
• The irrationals are like the blackness.
而無理數就像是無盡的黑暗。
• Cantor also showed that for any infinite set,
康托爾同時也證明任何無窮大的集合，
• forming a new set made of all the subsets of the original set
只要把它的所有子集都蒐集起來，
• represents a bigger infinity than that original set.
新的集合的「無限大」就比原本的還大。
• This means that once you have one infinity,
意思是說，只要你有一種「無限大」
• you can always make a bigger one by making a set of all subsets of that first set.
那你就可以用它的所有子集來做出比它更「無限大」的集合。
• And then an even bigger one
接著再用這集合做出更加「無限大」的集合。
• by making a set of all subsets of that one, and so on.
不斷做下去。
• And so, there are an infinite number of infinities of different sizes.
所以，「無限大」之間也是有分不同的大小。
• If these ideas make you uncomfortable, you're not alone.
如果你覺得這令人想吐，並不奇怪。
• Some of the greatest mathematicians of Cantor's day were very upset with this stuff.
一些康托爾那年代的偉大數學家也對這觀念非常反感。
• They tried to make these different infinities irrelevant,
他們試著要把無限這觀念抽離，
• to make mathematics work without them somehow.
讓數學可以沒有無限也能運作。
• Cantor was even vilified personally,
康托爾甚至受到人身攻擊，
• and it got so bad for him that he suffered severe depression.
嚴重到讓他飽受沮喪之苦。
• He spent the last half of his life in and out of mental institutions.
並且在精神療院渡過後半餘生。
• But eventually, his ideas won out.
不過他的想法最終得到肯定。
• Today they are considered fundamental and magnificent.
今天，這觀念被認為是基礎並重要的。
• All research mathematicians accept these ideas,
所有數學研究者都接受這觀念，
• every college math major learns them,
每個數學系都也都在教，
• and I've explained them to you in a few minutes.
而我剛剛已經花了幾分鐘來解釋。
• Someday, perhaps, they'll be common knowledge.
也許有一天，這會變成大家的常識。
• There's more.
還有一點。
• We just pointed out that the set of decimal numbers
我們剛剛指出小數，
• that is, the real numbersis a bigger infinity than the set of whole numbers.
也就是實數，比正整數的「無限大」還多。
• Cantor wondered if there are infinities of different sizes between these two infinities.
康托爾在想兩個「無限大」之間是否還有不同層級的「無限大」。
• He didn't believe there were, but couldn't prove it.
我們不這麼認為，但也沒辦法證明。
• Cantor's conjecture became known as the continuum hypothesis.
康托爾的猜想變成有名的「連續統假說」。
• In 1900, the great mathematician David Hilbert
在 1900 年，大數學家希爾伯特把連續統假說
• listed the continuum hypothesis as the most important unsolved problem in mathematics.
列為數學裡最重要的未解問題。
• The 20th century saw a resolution of this problem,
這問題在 20 世紀露出一些端倪，
• but in a completely unexpected, paradigm-shattering way.
但是結果和超乎預期、並跌破大家眼鏡。
• In the 1920s, Kurt Godel showed that you can never prove that the continuum hypothesis is false.
在 1920 年代，哥德爾證明了你不可能證明連續統假說是錯的。
• Then in the 1960s, Paul J. Cohen showed that you can never prove that the continuum hypothesis is true.
接著在 1960 年代，寇恩證明了你不可能證明連續統假說是對的。
• Taken together, these results mean that there are unanswerable questions in mathematics,
合在一起，這些結果告訴你數學裡也有一些不能回答的問題，
• a very stunning conclusion.
這是一個很令人震驚的結論。
• Mathematics is rightly considered the pinnacle of human reasoning,
數學被公認是人類邏輯的結晶，
• but we now know that even mathematics had its limitations.
但現在我們知道就算是數學也有它的極限。
• Still, mathematics has some truly amazing things for us to think about.
還有就是，數學裡有一些值得我們思考、而且很令人著迷的道理。
When I was in 4th grade, my teacher said to us one day:
B1 中級 中文 美國腔 TED-Ed 整數 無限 分數 位數 數學
# 【TED-Ed】無限有多大？ How Big Is Infinity? - Dennis Wildfogel
• 2454 210
Furong Lai 發佈於 2014 年 01 月 11 日 | 3,976 | 10,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-25 | latest | en | 0.800005 |
https://www.distancesto.com/fuel-cost/pa/david-district-to-san-pablo-nuevo-abajo/history/119577.html | 1,686,140,259,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653764.55/warc/CC-MAIN-20230607111017-20230607141017-00327.warc.gz | 782,601,937 | 14,720 | # INR0.45 Total Cost of Fuel from David District to San Pablo Nuevo Abajo
Your trip to San Pablo Nuevo Abajo will consume a total of 0.18 gallons of fuel.
Trip start from David District, PA and ends at San Pablo Nuevo Abajo, PA.
Trip (7.2 mi) David District » San Pablo Nuevo Abajo
The map above shows you the route which was used to calculate fuel cost and consumption.
### Fuel Calculations Summary
Fuel calculations start from David District, Panama and end at San Pablo Nuevo Abajo, San Pablo Nuevo, Panama.
Fuel is costing you INR2.50 per gallon and your vehicle is consuming 40 MPG. The formula can be changed here.
The driving distance from David District to San Pablo Nuevo Abajo plays a major role in the cost of your trip due to the amount of fuel that is being consumed. If you need to analyze the details of the distance for this trip, you may do so by viewing the distance from David District to San Pablo Nuevo Abajo.
Or maybe you'd like to see why certain roads were chosen for the route. You can do so by zooming in on the map and choosing different views. Take a look now by viewing the road map from David District to San Pablo Nuevo Abajo.
Of course, what good is it knowing the cost of the trip and seeing how to get there if you don't have exact directions? Well it is possible to get exact driving directions from David District to San Pablo Nuevo Abajo.
Did you also know that how elevated the land is can have an impact on fuel consumption and cost? Well, if areas on the way to San Pablo Nuevo Abajo are highly elevated, your vehicle may have to consume more gas because the engine would need to work harder to make it up there. In some cases, certain vehicles may not even be able to climb up the land. To find out, see route elevation from David District to San Pablo Nuevo Abajo.
Travel time is of the essence when it comes to traveling which is why calculating the travel time is of the utmost importance. See the travel time from David District to San Pablo Nuevo Abajo.
Speaking of travel time, a flight to San Pablo Nuevo Abajo takes up a lot less. How much less? Flight time from David District to San Pablo Nuevo Abajo.
Cost is of course why we are here... so is it worth flying? Well this depends on how far your trip is. Planes get to where they need to go faster due to the speed and shorter distance that they travel. They travel shorter distances due to their ability to fly straight to their destination rather than having to worry about roads and obstacles that are in a motor vehicle's way. You can see for yourself the flight route on a map by viewing the flight distance from David District to San Pablo Nuevo Abajo.
*The cost above should be taken as an ESTIMATE due to factors which affect fuel consumption and cost of fuel.
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Fuel Cost from David District to San Pablo Viejo Abajo | 669 | 3,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-23 | longest | en | 0.930523 |
https://fr.mathworks.com/matlabcentral/cody/problems/4-make-a-checkerboard-matrix/solutions/693163 | 1,571,798,451,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987828425.99/warc/CC-MAIN-20191023015841-20191023043341-00255.warc.gz | 481,542,610 | 15,406 | Cody
# Problem 4. Make a checkerboard matrix
Solution 693163
Submitted on 29 Jun 2015 by Alea88
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### Test Suite
Test Status Code Input and Output
1 Pass
%% n = 5; a = [1 0 1 0 1; 0 1 0 1 0; 1 0 1 0 1; 0 1 0 1 0; 1 0 1 0 1]; assert(isequal(a,checkerboard(n)))
ans = 1 ans = 1 ans = 1
2 Pass
%% n = 4; a = [1 0 1 0; 0 1 0 1; 1 0 1 0; 0 1 0 1]; assert(isequal(a,checkerboard(n)))
ans = 1 ans = 1 | 237 | 574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-43 | latest | en | 0.700656 |
http://encyclopedia2.thefreedictionary.com/solid+angles | 1,513,015,115,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513784.2/warc/CC-MAIN-20171211164220-20171211184220-00310.warc.gz | 94,184,580 | 10,686 | # Solid Angle
(redirected from solid angles)
Also found in: Dictionary, Thesaurus.
## solid angle
[′säl·əd ′aŋ·gəl]
(mathematics)
A surface formed by all rays joining a point to a closed curve.
## Solid Angle
a surface formed by rays having a common origin and passing through a closed curve (Figure 1). Sometimes, particularly in Soviet usage, the term “solid angle” is applied to the portion of space bounded by such a surface. Trihedral and polyhedral angles are special cases of solid angles.
Figure 1
A measure of the solid angle subtended at a given point by a surface S is provided by the ratio AIR2. Here, A is the area of the portion of a sphere, with center at the given point, that is cut by a conical surface with vertex at the point and with the perimeter of S as a directrix, and R is the radius of the sphere. Clearly, the measure of a solid angle is an abstract number. For example, the measure of a solid angle that encloses an octant, that is, oneeighth of space, is the number 4π R2/8 R2 = π/2 (Figure 2).
Figure 2
The unit of measure of a solid angle is called the steradian. It is equal to the solid angle subtended at the center of a sphere of unit radius by a portion of the sphere’s surface that is of unit area. The total solid angle about a point is equal to 4π steradians.
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Open / Close | 339 | 1,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-51 | latest | en | 0.932346 |
https://plarium.com/forum/en/sparta-war-of-empires/entertainment-community/40056_answer-to-win-2-000-drachmas-/3/ | 1,526,922,425,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864461.53/warc/CC-MAIN-20180521161639-20180521181639-00373.warc.gz | 639,999,255 | 10,254 | This topic is closed
## Answer To Win 2,000 Drachmas!
873 Replies
User
20 May, 2017, 9:25 AM UTC
150
UTC +1:00
2
User
20 May, 2017, 9:26 AM UTC
28
UTC +1:00
0
User
20 May, 2017, 9:26 AM UTC
?=150
UTC +2:00
0
User
20 May, 2017, 9:27 AM UTC
126...
UTC +4:00
0
User
20 May, 2017, 9:29 AM UTC
31
UTC +1:00
1
User
20 May, 2017, 9:29 AM UTC
101
UTC +7:00
0
User
20 May, 2017, 9:31 AM UTC
126
UTC +7:00
0
User
20 May, 2017, 9:31 AM UTC
? = 101
UTC +7:00
0
User
20 May, 2017, 9:31 AM UTC
Solve this puzzle for your chance to win 2,000 Drachmas!
Leave a post with the solution below if wisdom and fortune are on your side!
25+25+25= 75
25-5+1=21
5+2=7
1+5x25=126
Solution 126
UTC +0:00
1
User
20 May, 2017, 9:31 AM UTC
28
UTC +7:00
0
User
20 May, 2017, 9:32 AM UTC
25 + 25 + 25 = 75
25 - 5 +1 = 21
5 + 1 * 25 = 100
Das Ergebnis lautet 100 b.z.w. lösung = 100
UTC +1:00
0
User
20 May, 2017, 9:32 AM UTC
126
UTC +4:00
0
User
20 May, 2017, 9:32 AM UTC
150
UTC +7:00
0
User
20 May, 2017, 9:32 AM UTC
(5+1)25=150
UTC +2:00
0
User
20 May, 2017, 9:32 AM UTC
126
UTC +0:00
0
User
20 May, 2017, 9:33 AM UTC
marcusdjkean said:
150
UTC +7:00
0
User
20 May, 2017, 9:34 AM UTC
126
UTC +7:00
0
User
20 May, 2017, 9:34 AM UTC
126
UTC +2:00
0
User
20 May, 2017, 9:35 AM UTC
126
UTC +2:00
0
User
20 May, 2017, 9:37 AM UTC
101
UTC +9:00
0
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``` Sir, plz explain vector triple product?
```
7 years ago
Share
``` Hi,
The vector triple product
If A, B and C are three vectors, then we can combine them in this way:
Ax(BxC)
to get a vector result, which is known as the vector triple product.
As you know, the cross product is calculated using a determinant and we can extend this to Ax(BxC) to get a rather more complicated determinant than the scalar triple product gave us, again involving all three vectors.
However there is a much simpler way to evaluate a vector triple product, because it can be shown that this is true:
Ax(BxC)=(A.C)B-(A.B)C
and
(AxB)xC=(C.A)B-(C.B)A.
So we can evaluate either of those right-hand sides instead, which do not involve any determinants. However you do need to remember them!
Regards,
Rajat
```
7 years ago
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Ajay 6 months ago
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mycroft holmes 6 months ago
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Anshuman Mohanty 5 months ago
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Anshuman Mohanty 5 months ago
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Nishant Vora one month ago
Got it! [z + 12 – 6 i ] can be rewritten as [ z – i + 12 – 5 i] => | z – i | and => |12 – 5 i | = sqrt ( 12^2 + 5^2) = 13......................(2) => | z + 12 – 6 i | => | z + 12 – 6 i |...
Divya one month ago
I tried posting the question several times, it kept cutting off the rest of the question. Here: If | z-1| Options: a*) 14 b) 2 c) 28 d) None of the above
Divya one month ago
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More Questions On Vectors | 1,350 | 5,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2017-04 | longest | en | 0.89345 |
https://cracku.in/bq-sbi-po-2017-free-reasoning-63 | 1,563,553,688,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526324.57/warc/CC-MAIN-20190719161034-20190719183034-00168.warc.gz | 369,272,907 | 14,918 | ## SBI PO 2017 Reasoning Test 59
Instructions
For the following questions answer them individually
Question 1
Point R is 10 metres north of point A. Point K is exactly in the middle of the points R and A. Point N is 7 metres east of point A. Point M is 7 metres east of point K. Point S is 6 metres north of point M. What is the distance between points S and N?
Question 2
A person travels 1 km east, and 1 km north. He continues this process 20 times. What is the distance between the starting point and the ending point?
Question 3
Four friends Ram, Shyam, Bheem and Dhruv are playing a board-game. Ram and Dhruv are partners and are sitting opposite to each other. Shyam is sitting to the left of Dhruv. If Ram is facing north, which direction is Shyam facing?
Question 4
Sareen walked 30 metres North then turned right. He then walked 40 metres. How far is he from the original point?
Question 5
Amit started walking towards east. He walked for 10 km and then took a right turn. He walked for a distance of 5 km before taking a left turn and walking for 5 km. He again takes a left turn and walks for 5 km to reach his destination. What is the distance between his starting point and his destination? | 308 | 1,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2019-30 | latest | en | 0.976827 |
https://www.homebrewersassociation.org/forum/index.php?action=printpage;topic=6538.0 | 1,611,084,916,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519600.31/warc/CC-MAIN-20210119170058-20210119200058-00531.warc.gz | 832,509,847 | 2,885 | # Homebrewers Association | AHA Forum
## General Category => Ingredients => Topic started by: bathtubbrewer on March 21, 2011, 11:07:00 PM
Post by: bathtubbrewer on March 21, 2011, 11:07:00 PM
I bought "How to Brew" a couple days before I bought my first kit. Since the kit was pretty much just handed to me by the guy at my brew shop, I wanted to do a little work (as described in How to Brew) to make sure I understood what I was doing and why I was doing it. I started by trying to calculate my OG. Here's my question/problem: I don't understand why some malt extracts are measured by 'points' and some are measured by the Lovibond scale. I was given two 3.3 lb cans of LME Amber (35 points) and a 1 lb bag of Munton's Amber DME with a Lovibond rating of 7. Is the Munton's just for color? If so, when do I add it?
I made my first batch anyway... Couldn't wait any longer. Wish me luck.
Post by: tygo on March 21, 2011, 11:14:23 PM
Lovibond is a color scale. Specific gravity or points per pound per gallon (pppg) are a measure of the sugar contribution. Amber DME should have around 44 pppg. So your 6.6lbs of Amber LME and 1 lb of Amber DME should give you a specific gravity of 1.055 assuming it's for a five gallon batch.
6.6lb x 35 pppg = 231 gravity points
1.0lb x 44 pppg = 44 gravity points
Total Gravity Points = 275
Original Gravity = 275 / 5 gallons = 55 points per gallon or SG 1.055. | 425 | 1,417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-04 | latest | en | 0.956436 |
http://www.talkstats.com/showthread.php/14842-help-with-t-distribution-and-chi-square-distribution | 1,508,556,015,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824543.20/warc/CC-MAIN-20171021024136-20171021044136-00235.warc.gz | 570,084,593 | 11,365 | # Thread: help with t distribution and chi square distribution
1. ## help with t distribution and chi square distribution
Let X~N(3,5) and Y~N(-7,2) be independent. Find values of C1,C2,C3,C4,C5,C6 such that
C1(X+C2)^C3
-------------- ~ t(C6)
(Y+C4)^C5
My attempt
so the t distribution can become X/sqrt(Y/C6)
so Y is a chi squared distribution with C6 degrees of freedom
so if I do
C2=0
C3=1
C1=sqrt(4*C6)
C4=7
C5=0.5
i get
sqrt(4*C6)(X+0)^1
-------------- ~ t(C6)
(Y+7)^0.5
X
-------------- ~ t(C6)
1/[sqrt(4*C6)] * sqrt(Y+7)
X
-------------- ~ t(C6)
sqrt([Y+7]/[4C6])
here since Y~N(-7,2), and in the equation the mean is being subtracted and then its being divided by its sd, it gets normalized
X
----------- ~ t(C6)
sqrt(Y/C6)
X
----------- = X/sqrt(Y/C6)
sqrt(Y/C6)
Is this right?
Does X have to be standardized?
2. ## Re: help with t distribution and chi square distribution
omega, as the t-distribution on the RHS is not fully specified, I think this is open end question and have more than one set of answers.
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# Suppose there are 198 men and 2 women in a room. That is,
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Suppose there are 198 men and 2 women in a
room. That is, the men make up 99% of the people in the room.
How many men have to leave for the percentage of men to drop to 98%?
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03 Jan 2005, 13:56
1. 198 - x / 200 = 98 / 100
2. 19800 - 100x = 19600
3. 200 = 100x
4. 2 = x
5. solution is 2.
is it ?
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03 Jan 2005, 13:58
Currently the proportion is 198/200. Call the number of men that have to leave x. So you want (198-x)/(200-x)=.98
When you cross multiply you get: 198-x=196-.98x
Subtract 196 and add x to both sides to get 2=.02x
Multiply both sides by 100 to get: 200=2x
So x=100
If you put 100 in for x in the original equation you get (198-100)/(200-100)= 98/100, which checks out.
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03 Jan 2005, 14:03
ah, right
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### Show Tags
03 Jan 2005, 14:12
New percetage = .98
198 - x / 200 = .98
x = 2
x number of men to leave the group
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### Show Tags
03 Jan 2005, 14:39
rxs0005 wrote:
New percetage = .98
198 - x / 200 = .98
x = 2
x number of men to leave the group
If 2 men leave the group then why are you still dividing by 200? For that to be true wouldn't 2 women have to join the group? If 2 men left it would be 196/198, which is not 98%. You have to take into account that any man that leaves also decreases the total. The total is the number of men (198)+ women(2). Therefore if you decease the number of men from numerator, you have to also decrease it from the deominator.
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03 Jan 2005, 17:56
100 it is and toddmartin's explanation is as good as any
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### Show Tags
03 Jan 2005, 18:04
this Q if it was in GMAT wud have an answer choice 2 a classic trap
you guys are correct the answer is 100
198 - x / 200 - x = .98
.02x = 2 x = 100
03 Jan 2005, 18:04
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### Transcript of Sorting Algorithms What is sorting?cs. js236/201505/cs3358/w10sortingalgorithms.pdf ·...
• 1
Sorting Algorithms Chapter 9
CS 3358 Spring 2015
Jill Seaman
Sections 9.1, 9.2, 9.3, 9.5, 9.6 2
What is sorting?
! Sort: rearrange the items in a list into ascending or descending order - numerical order - alphabetical order - etc.
55 112 78 14 20 179 42 67 190 7 101 1 122 170 8
1 7 8 14 20 42 55 67 78 101 112 122 170 179 190
3
Why is sorting important? ! Searching in a sorted list is much easier than
searching in an unsorted list. ! Especially for people: - dictionary entries (in a dictionary book) - phone book (remember these?) - card catalog in library (it used to be drawers of index
cards) - bank statement: transactions in date order
! Most of the data displayed by computers is sorted. 4
Sorting
! Sorting is one of the most intensively studied operations in computer science
! There are many different sorting algorithms
! The run-time analyses of each algorithm are well-known.
• 5
Sorting algorithms covered in this class
! Selection sort ! Insertion sort ! Bubble sort
! Merge sort ! Quicksort
! Heap sort (later, when we talk about heaps) 6
Selection sort
! There is a pass for each position (0..size-1) ! On each pass, the smallest (minimum) element
in the rest of the list is exchanged (swapped) with element at the current position.
! The first part of the list (already processed) is always sorted
! Each pass increases the size of the sorted portion.
5
Selection Sort: Pass One
values [ 0 ]
[ 1 ]
[ 2 ]
[ 3 ]
[ 4 ]
36
24
10
6
12
U N S O R T E D
6
Selection Sort: End Pass One
values [ 0 ]
[ 1 ]
[ 2 ]
[ 3 ]
[ 4 ]
6
24
10
36
12
U N S O R T E D
SORTED
• 7
SORTED
Selection Sort: Pass Two
values [ 0 ]
[ 1 ]
[ 2 ]
[ 3 ]
[ 4 ]
6
24
10
36
12
U N S O R T E D
8
Selection Sort: End Pass Two
values [ 0 ]
[ 1 ]
[ 2 ]
[ 3 ]
[ 4 ]
6
10
24
36
12
U N S O R T E D
SORTED
9
Selection Sort: Pass Three
values [ 0 ]
[ 1 ]
[ 2 ]
[ 3 ]
[ 4 ]
6
10
24
36
12
U N S O R T E D
SORTED
10
Selection Sort: End Pass Three
values [ 0 ]
[ 1 ]
[ 2 ]
[ 3 ]
[ 4 ]
6
10
12
36
24
S O R T E D
UNSORTED
• 11
Selection Sort: Pass Four
values [ 0 ]
[ 1 ]
[ 2 ]
[ 3 ]
[ 4 ]
6
10
12
36
24
S O R T E D
UNSORTED
12
Selection Sort: End Pass Four
values [ 0 ]
[ 1 ]
[ 2 ]
[ 3 ]
[ 4 ]
6
10
12
24
36
S O R T E D
15
Selection sort: code template int minIndex(ItemType values[], int size, int start) { int minIndex = start; for (int i = start+1; i < size; i++) if (values[i] < values[minIndex]) minIndex = i; return minIndex; }
template void selectionSort (ItemType values[], int size) { int min; for (int index = 0; index < (size -1); index++) { min = minIndex(values, SIZE, index); swap(values[min],values[index]); } }
template void swap (T& a, T& b); is in the library 16
Efficiency of Selection Sort
! N is the number of elements in the list ! Outer loop (in selectionSort) executes N-1 times ! Inner loop (in minIndex) executes N-1, then N-2,
then N-3, ... then once. ! Total number of comparisons (in inner loop):
(N-1) + (N-2) + . . . + 2 + 1 = (N-1)(N-1+1)/2 = (N-1)N/2 = (N2-N)/2 = N2/2 - N/2 O(N2)
(N-1) + (N-2) + . . . + 2 + 1 = the sum of 1 to N-1
From math class:
• 17
Insertion sort
! There is a pass for each position (0..size-1) ! The front of the list remains sorted. ! On each pass, the next element is placed in its
proper place among the already sorted elements.
! Like playing a card game, if you keep your hand sorted, when you draw a new card, you put it in the proper place in your hand.
! Each pass increases the size of the sorted portion.
22
Insertion Sort: Pass One
values [ 0 ]
[ 1 ]
[ 2 ]
[ 3 ]
[ 4 ]
36
24
10
6
12
SORTED
U N S O R T E D
23
Insertion Sort: Pass Two
values [ 0 ]
[ 1 ]
[ 2 ]
[ 3 ]
[ 4 ]
24
36
10
6
12
U N S O R T E D
SORTED
24
Insertion Sort: Pass Three
values [ 0 ]
[ 1 ]
[ 2 ]
[ 3 ]
[ 4 ]
10
24
36
6
12 UNSORTED
S O R T E D
• 25
Insertion Sort: Pass Four
values [ 0 ]
[ 1 ]
[ 2 ]
[ 3 ]
[ 4 ]
6
10
24
36
12
S O R T E D
UNSORTED
26
Insertion Sort: Pass Five
values [ 0 ]
[ 1 ]
[ 2 ]
[ 3 ]
[ 4 ]
6
10
12
24
36
S O R T E D
23
Insertion sort: code
template void insertionSort (ItemType a[], int size) { for (int index = 1; index < size; index++) { ItemType tmp = a[index]; // next element int j = index; // start from the end of sorted part
// find tmp's place, AND shift bigger elements up while (j > 0 && tmp < a[j-1]) { a[j] = a[j-1]; // shift bigger element up j--; } a[j] = tmp; // put tmp in its place } }
24
Insertion sort: runtime analysis
! Very similar to Selection sort ! Total number of comparisons (in inner loop): - At most 1, then 2, then 3 ... up to N-1 for the last
element. ! So it’s
O(N2)
(N-1) + (N-2) + . . . + 2 + 1 == N2/2 - N/2
• 25
Bubble sort
! On each pass: - Compare first two elements. If the first is bigger, they
exchange places (swap). - Compare second and third elements. If second is
bigger, exchange them. - Repeat until last two elements of the list are
compared. ! Repeat this process until a pass completes with
no exchanges
26
Bubble sort Example: first pass
! 7 2 3 8 9 1 7 > 2, swap ! 2 7 3 8 9 1 7 > 3, swap ! 2 3 7 8 9 1 !(7 > 8), no swap ! 2 3 7 8 9 1 !(8 > 9), no swap ! 2 3 7 8 9 1 9 > 1, swap ! 2 3 7 8 1 9 finished pass 1, did 3 swaps
Note: largest element is in last position
27
Bubble sort Example: second and third pass
! 2 3 7 8 1 9 2
• 29
Bubble sort how does it work?
! At the end of the first pass, the largest element is moved to the end (it’s bigger than all its neighbors)
! At the end of the second pass, the second largest element is moved to just before the last element.
! The back end (tail) of the list remains sorted. ! Each pass increases the size of the sorted
portion. ! No exchanges implies each element is smaller
than its next neighbor (so the list is sorted). 30
Bubble sort: code
template void bubbleSort (ItemType a[], int size) {
bool swapped; do { swapped = false; for (int i = 0; i < (size-1); i++) { if (a[i] > a[i+1]) { swap(a[i],a[i+1]); swapped = true; } } } while (swapped); }
31
Bubble sort: runtime analysis ! Each pass makes N-1 comparisons ! There will be at most N passes - one to move the right element into each position
! So worst case it’s: ! If you change the algorithm to look at only the
unsorted part of the array in each pass, it’s exactly like the selection sort:
! What is the best case for Bubble sort? ! Are there any sorting algorithms better than O(N2)?
O(N2) (N-1)*N
(N-1) + (N-2) + . . . + 2 + 1 = N2/2 - N/2 still O(N2)
32
Merge sort ! Divide and conquer! ! 2 half-sized lists sorted recurs | 2,281 | 6,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2022-33 | latest | en | 0.76456 |
https://www.physicsforums.com/threads/angle-between-two-surfaces.431177/ | 1,508,839,457,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828356.82/warc/CC-MAIN-20171024090757-20171024110757-00852.warc.gz | 975,211,160 | 15,007 | # Angle between two surfaces
1. Sep 22, 2010
In Marion & Thorton problem 1.29 asks to find the angle between two surfaces $$(x^2 +y^2 + z^2)^2 = 9$$ and $$x + y + z^2 = 1$$ at a point.
The solution takes the gradient of $$(x^2 +y^2 + z^2)^2 - 9$$ and $$x + y + z^2 - 1$$, and using the dot product between the two vectors at that point gets the angle.
My question is, isn't $$(x^2 +y^2 + z^2)^2 - 9$$ and $$x + y + z^2 - 1$$ both zero and hence taking the gradient would give you 0. shouldn't you rather take one variable as the dependent variable so you have z(x,y) for example and then take the gradient of that? I'm confused as to why they took the gradient of the way they did.
2. Sep 22, 2010
### Eynstone
The intersection of two surfaces is a curve & rarely a straight line. Hence, the notion of the 'angle' between two surfaces is invalid. The author probably means what he calculated by 'the angle between the surfaces '.
3. Sep 22, 2010
### HallsofIvy
Staff Emeritus
The fact that a function is 0 for a specific value of x does NOT mean it is a constant nor that its derivative must be 0! Similarly, a function of three variables, that is equal to 0 on some specific subset of R3, is NOT necessarily a constant and its gradient is not necessarily constant on that subset.
In fact, what is true is that if f(x,y,z)= constant on some subset, then its gradient is perpendicular to that subset. Recall that the gradient vector always points in the direction of fastest increase and, further, that the "rate of change" of the function in a particular direction is the projection of the gradient vector on that direction. If the function is constant in a given direction, the projection of the gradient vector in that direction is 0 which does NOT mean the gradient is 0, only that it is perpendicular to that direction.
Here, defining $f(x,y,z)= (x^2+ y^2+ z^2)^2$ (or, equivalently, [/itex]f(x, y, z)= (x^2+ y^2+ z^2)^2- 9)[/itex] says that on the subset of R3 defined by $(x^2+ y^2+ z^2)^2= 9$ f(x,y,z)= 9 (or, equivalently f(x,y,z)= 0). Since f is constant on that surface, its gradient is perpendicular to that surface, not necessarily 0.
Similarly, $g(x,y,z)= x + y + z^2$ take on the constant value "1" on the surface $x+ y+ z^2$ and so its gradient is perpendicular to the surface at every point on the surface.
That is, at every point on the respective surfaces, $\nabla f= 2(x^2+ y^2+ x^2)(2x\vec{i}+ 2y\vec{j}+ 2z\vec{k})$ and $\nable g= \vec{i}+ \vec{j}+ \vec{k}$ are perpendicular to their respective surfaces. The intersection of the two surfaces, being a curve that lies in both surfaces, must be perpendicular to both vectors. That is, we can find a vector in the direction of the intersection curve by taking the cross product of the two gradients. | 793 | 2,782 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-43 | latest | en | 0.93572 |
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# Traverse Surveying And Contouring - 1
## 10 Questions MCQ Test Mock Test Series for Civil Engineering (CE) GATE 2020 | Traverse Surveying And Contouring - 1
Description
This mock test of Traverse Surveying And Contouring - 1 for Civil Engineering (CE) helps you for every Civil Engineering (CE) entrance exam. This contains 10 Multiple Choice Questions for Civil Engineering (CE) Traverse Surveying And Contouring - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Traverse Surveying And Contouring - 1 quiz give you a good mix of easy questions and tough questions. Civil Engineering (CE) students definitely take this Traverse Surveying And Contouring - 1 exercise for a better result in the exam. You can find other Traverse Surveying And Contouring - 1 extra questions, long questions & short questions for Civil Engineering (CE) on EduRev as well by searching above.
QUESTION: 1
### A series of closely spaced contour lines represents a
Solution:
Closed contours ⇒ Steep slope
Apart contour lines ⇒ Gentle slope
Equally spaced ⇒ Uniform slope Straight parallel and equally speed contours ⇒ plane surface
QUESTION: 2
### In order to measure the magnetic bearing of a line, the theodolite should be provided with ______
Solution:
In order to measure the magnetic bearing of a line, the theodolite should be provided with either a tabular compass or trough compass.
QUESTION: 3
### Closed contours, with higher value inwards, represent a
Solution:
QUESTION: 4
The Bowditch method of adjusting a traverse is based on the assumption that
where e1 and e2 are errors in linear and angular measurements respectively and l is the length of a line
Solution:
QUESTION: 5
Accuracy of elevation of various points obtained from contour map is limited to
Solution:
QUESTION: 6
If L is the perimeter of a closed traverse, ΔD is the closing error in departure, the correction for the departure of a traverse side of length l, according to Bowditch rule, is
Solution:
Bowditch’s method - Correction to latitude (or departure) of any side
QUESTION: 7
If arithmetic sum of latitudes of a closed traverse is ∑Lat and closing error in latitude is dx, the correction for a side whose latitude is l, as given by Transit Rule, is
Solution:
Transit method -→ Correction of latitude (and departure) of any side
QUESTION: 8
If the reduced bearing of a line AB is N60°W and length is 100 m, then the latitude and departure respectively of the line AB will be
Solution:
QUESTION: 9
If the sum of northings of a traverse exceeds the sum of southings by 1 m and sum of eastings exceeds the sum of westirigs by 1 m, the resultant closing error and its true bearing respectively are
Solution:
QUESTION: 10
In a closed traverse, the sum of south latitudes exceeds the sum of north latitudes and the sum of east departures exceeds the sum of west departures. The closing line will lie in the
Solution:
The closing line will have south latitude and east departure. Therefore it will lie in S-E quadrant. | 701 | 3,060 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-45 | latest | en | 0.856714 |
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This is the RS232 kh 2.5 marluxia guide IC that is capable of running at 3V and. Datasheet: http:www. datasheetcatalog. orgdatasheetsipexSP3222EBEP. pdf. | 1,654 | 6,358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-34 | latest | en | 0.7711 |
https://www.jiskha.com/display.cgi?id=1357530811 | 1,500,836,389,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424586.47/warc/CC-MAIN-20170723182550-20170723202550-00545.warc.gz | 817,947,884 | 3,782 | # Physics
posted by .
A spring has a constant of 2.4 x 10^2 n/m. What force does it take to stretchmit .8m? How much PE is stored in the spring? Make a graph of the constant from .1 to 1.0 meters.
• Physics -
F = k*x where k = 240 N/m
Plug in x = 0.8 m and solve for F
At this amount of stretch, the P.E. is
(1/2) k x^2 (or (1/2)F*x.)
It makes no sense to make a graph of a constant. Do you mean P.E. or F vs x ?
Anyway, we don't provide graphs here. | 154 | 457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-30 | longest | en | 0.94841 |
https://www.hackingnote.com/en/interview/problems/search-a-2d-matrix/ | 1,553,631,443,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912206016.98/warc/CC-MAIN-20190326200359-20190326222359-00253.warc.gz | 761,509,673 | 67,264 | # Search a 2D Matrix
## Problem
Write an efficient algorithm that searches for a value in an m x n matrix.
This matrix has the following properties:
Integers in each row are sorted from left to right. The first integer of each row is greater than the last integer of the previous row.
## Example
Consider the following matrix:
[
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
Given target = 3, return true.
## Challenge
O(log(n) + log(m)) time
## Code - Java
public class Solution {
/**
* @param matrix, a list of lists of integers
* @param target, an integer
* @return a boolean, indicate whether matrix contains target
*/
public boolean searchMatrix(int[][] matrix, int target) {
int rowIndex = binarySearchRow(matrix, target);
if (rowIndex == -1) {
return false;
}
return binarySearchElement(matrix[rowIndex], target);
}
private int binarySearchRow(int[][] matrix, int target) {
int start = 0, end = matrix.length - 1;
while (start <= end) {
int mid = (start + end) / 2;
int[] row = matrix[mid];
if (target < row[0]) {
end = mid - 1;
} else if (target > row[row.length - 1]) {
start = mid + 1;
} else {
return mid;
}
}
return -1;
}
private boolean binarySearchElement(int[] row, int target) {
int start = 0, end = row.length - 1;
while (start <= end) {
int mid = (start + end) / 2;
if (target < row[mid]) {
end = mid - 1;
} else if (target > row[mid]) {
start = mid + 1;
} else {
return true;
}
}
return false;
}
} | 415 | 1,436 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2019-13 | longest | en | 0.486346 |
https://discussions.unity.com/t/rotate-triangles-correctly-around-the-circle/239951 | 1,709,148,490,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474744.31/warc/CC-MAIN-20240228175828-20240228205828-00137.warc.gz | 202,170,818 | 6,356 | # Rotate Triangles "Correctly" Around The Circle
Hello,
I have a circle and this circle has orbits around it. The orbits’ shape is triangle.
My orbit creation code:
``````public void CreateOrbitsAroundCircle(int num, Vector3 point, float radius)
{
for (int i = 0; i < num; i++)
{
/* Distance around the circle */
var radians = 2 * Mathf.PI / num * i;
/* Get the vector direction */
var spawnDir = new Vector3(horizontal, vertical, 0);
/* Get the spawn position */
var spawnPos = point + spawnDir * radius; // Radius is just the distance away from the point
/* Now spawn */
var newOrbit = Instantiate(orbit, spawnPos, Quaternion.identity, this.transform);
//newOrbit.name = playerName;
/* Rotate the enemy to face towards player */
//enemy.transform.LookAt(point);
//enemy.transform.Translate(new Vector3(0, enemy.transform.localScale.y / 2, 0));
}
}
``````
I can add orbits by this code up to 10 orbits.
My question is this: I want to rotate triangles like my circle has spikes!
Like this:
How can I do that?
First make sure your triangle prefab is pointing up. This will make it easy to calculate the rotation to apply to the triangle if we already know the rotation of the prefab before we spawn it. Then you just need to calculate the angle to rotate your triangle by depending on where on the circle you are going to place it. You can do that by calculating the angle between the up direction and the direction of your spawn position.
``````var spawnPos = point + spawnDir * radius; // Radius is just the distance away from the point
float rotAngle = Vector3.Angle(Vector3.Up, spawnDir);
if(spawnDir.x < 0) rotAngle *= -1;
Quaternion orbitRotation = Quaternion.Euler(0, 0, rotAngle);
/* Now spawn */
var newOrbit = Instantiate (orbit, spawnPos, orbitRotation, this.transform);
``````
I just added this code to orbits. It worked well!
``````target = transform.parent.gameObject;
Vector3 dir = target.transform.position - transform.position;
float angle = Mathf.Atan2(dir.y, dir.x) * Mathf.Rad2Deg;
transform.rotation = Quaternion.AngleAxis(angle + 90, Vector3.forward);
`````` | 501 | 2,100 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-10 | latest | en | 0.778487 |
https://kr.mathworks.com/matlabcentral/cody/players/8756398/solved?page=2 | 1,611,087,378,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519784.35/warc/CC-MAIN-20210119201033-20210119231033-00598.warc.gz | 408,621,899 | 18,232 | Cody
# Neha Singh
Rank
Score
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#### Problem 174. Roll the Dice!
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Tags mean
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Tags indexing, find
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Created by: Cody Team
Tags matrices
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Created by: Will
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Created by: Dimitris Kaliakmanis
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Created by: Cody Team
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Created by: Cody Team
#### Problem 5. Triangle Numbers
Created by: Cody Team
Tags math, triangle, nice
#### Problem 149. Is my wife right?
Created by: the cyclist
Tags easy, silly, fun
#### Problem 6. Select every other element of a vector
Created by: Cody Team
#### Problem 26. Determine if input is odd
Created by: Cody Team
#### Problem 8. Add two numbers
Created by: Cody Team
#### Problem 3. Find the sum of all the numbers of the input vector
Created by: Cody Team
#### Problem 2. Make the vector [1 2 3 4 5 6 7 8 9 10]
Created by: Cody Team
Tags basic, basics, colon
#### Problem 1. Times 2 - START HERE
Created by: Cody Team
Tags intro, math, easy
#### Problem 167. Pizza!
Created by: the cyclist
Tags fun, pizza, good
51 – 75 of 75 | 557 | 2,003 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-04 | latest | en | 0.737461 |
http://www.fact-index.com/c/co/commutative_diagram.html | 1,720,813,962,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514452.62/warc/CC-MAIN-20240712192937-20240712222937-00551.warc.gz | 37,500,617 | 2,254 | Main Page | See live article | Alphabetical index
# Commutative diagram
In mathematics, especially the many applications of category theory, a commutative diagram is a diagram of objects and morphisms such that, when picking two objects, one can follow any path through the diagram and obtain the same result by composition.
For example, the first isomorphism theorem is a commutative triangle as follows:
Since f = h o φ, the left diagram is commutative; and since φ = k o f, so is the right diagram.
Similarly, the square above is commutative if y o w = z o x.
Commutativity makes sense for a polygon of any finite number of sides (including just 1 or 2), and a diagram is commutative if every polygonal subdiagram is commutative. | 173 | 738 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-30 | latest | en | 0.898511 |
http://www.cfd-online.com/Forums/main/1945-outflow-boundary-conditions-print.html | 1,435,638,567,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375091587.3/warc/CC-MAIN-20150627031811-00205-ip-10-179-60-89.ec2.internal.warc.gz | 379,527,565 | 4,222 | CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
- Main CFD Forum (http://www.cfd-online.com/Forums/main/)
- - Outflow boundary conditions (http://www.cfd-online.com/Forums/main/1945-outflow-boundary-conditions.html)
Achilleas Tsompanos March 15, 2000 13:16
Outflow boundary conditions
Dear sirs,
I am using a primitive variable FEM NS solver. The problem is that i am puzzled over the outflow boundary conditions. Can anyone help me?
John C. Chien March 15, 2000 15:34
Re: Outflow boundary conditions
(1). Is this FEM NS solver giving you problems? (2). Is the outflow boundary conditions of the code giving you problems? (3). What is the outflow boundary conditions of the problem you are trying to solve? (4).If you are using a commercial code, all you need to do is to select the type of the outflow boundary conidtions and then supply the necessary information. (5). In most cases, there is not much you can do in the outflow boundary conditions. So, what is the fluid dynamic problem you are trying to solve with this FEM NS solver?
Carlos Vilela March 15, 2000 18:37
Re: Outflow boundary conditions
Hi. There are two books that i suggest you to read. [1]C. Taylor and T. G. Hughes, Finite Element Programming of the Navier-Stokes Equations, Pineridge Press [2]J. N. Reddy, An Introduction to the Finite Element Method, Mc. Graw Hill.
I think that the first one is exactly what you are looking for. All the book is about the discretization development of Navier-Stokes with primitive variables, and shows how to treat boundary condictions. The second one is more general book, but in the last chapters there is a brief presentation of Navier-Stokes with primitive variables discretization and boundary condictions treatment. If you are using some commercial software, so itīs more simple. Is just take a good look in the manual and follow all instructions to apply boundary condictions. In general ways, most FEM softwares if you donīt explicit the boundary condiction (Dirichlet type) the method by default assume null derivative boundary condictions(Newton type).
Good luck
Carlos Vilela
Achilleas Tsompanos March 16, 2000 16:59
Re: Outflow boundary conditions
Hello there,
I am trying to resolve the Kerman vortex street shed behind a cylinder on average Re values. So the main thing is the outflow boundary conditions. I am using Flotran and phoenics and a code of mine in c++. The manuals at some point state that at the boundary you can put pressure to zero but i don't think that this is appropriate. In essence proper bcs should allow the vortexes to pass through without reflecting anything back to the flow. I know that this is a complex problem but some guidelines would be nice.
Thanks a lot Achilleas Tsompanos
Patrick Godon March 16, 2000 17:24
Re: Outflow boundary conditions
Hi Achilleas,
if you want non-reflecting boundary conditions, then the problem is not what are the boundary conditions (e.g. given the variables or their derivatives) but rather how you do implement the boundary conditions. To avoid numerically reflective boundary conditions, you need to impose the boundary conditions using the method of the characteristics. THe characteristics of the flow are the quantities that actually propagate in the flow thourhg the boundaries (they are the eigenvector of the linearized, homogeneous problem). At each boundary there are characteristics that are incoming (entering the flow) and characteristics that are ougoing (exiting the flow). Idealy, you impose the boundary conditions (given physical quantities) on the incoming characteristics, while the outgoing characteristics takes values from inside the computational domain (you might need to extraoplate these). THe outgoing characteristics are carrying information outside the computational domain, while the incoming characteristics are carrying information inside into the computational domain. Therefore, sometimes it is enough just to set the incoming characteristics to zero, to avoid reflection. Once, this is done, one uses the values of the characteristics at the boundary to caculate the values of the primitive variables. And it is these values that are then imposed on the primitive variables. So the values that are eventually imposed on the primitive variables are a combination of incoming and outgoing characteristics.
I am not quite sure about your sofware if it has such a possibility.
Have a look at
Abarbanel et al. 1991, J. Fluid Mech. vol. 225, p. 557,
for the treatment of non-reflective boundary conditions in the simulations of the flow past a circular cylinder.
Have also a look at the review paper:
Givoli, 1991, J. Comput. PHys, vol. 94, p.1,
for a review of transmitting and non-reflective boundary conditions.
PG
clifford bradford March 17, 2000 15:36
Re: Outflow boundary conditions
you are quite correct the boundary conditions should be of the nonreflecting outflow/inflow type (because the flow is reversed in some areas your bc subroutine should be able to detect inflow and implement the appropriate conditions) in addition to other references given by Mr. Godon you might wish to read Kevin Thompson's paper in the Journal of Computational Physics vol 89 pp. 439-461 (1990) as well as Poinsot and Lele in J. Comp. Phys. vol 101 pp. 104-129(1992). both of these papers together will show you how to implement non-reflecting bc among others.
however, these papers assume you are using an unsteady formulation (time iterative scheme) if you are using a pressure based method you'll have to look somewhere else. i don't know if nonreflecting bc can be implemented in these codes you'd have to find out from someone who is more familiar with these schemes than i am. i know flotran is pressure based so you'll probably not be able to do it there. i have used flotran and i can't recall there being a nonreflecting bc available. the lack of nonreflecting bc is the Achilles' heel of pressure based codes (i couldn't resist). if pheonics and your code are also pressure based you'll have the same problem. if they are time iterative then you should be able to follow the procedure of thompson and poinsot and lele to apply the bc.
Phil Gresho April 2, 2000 16:09
Re: Outflow boundary conditions
(Nearly ) ALL ABOUT OUTFLOW BOUNDARY CONDITIONS ...and MUCH MORE.. is in my new 1000+ page book, "INCOMPRESSIBLE FLOW AND THE FINITE ELEMENT METHOD", John Wiley (1999); with a new (2000) paperback edition to appear in 2 months
All times are GMT -4. The time now is 00:29. | 1,517 | 6,553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2015-27 | longest | en | 0.898215 |
http://www.mathskey.com/question2answer/10108/substitution-method-what-is-the-cordinate-for-x-5-and-3x-2y-10 | 1,610,820,572,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703506832.21/warc/CC-MAIN-20210116165621-20210116195621-00014.warc.gz | 157,534,452 | 11,509 | # substitution method what is the cordinate for x+y=5 and 3x-2y=10
i need to now the both the numbers for x and y. x=? y=?
asked Mar 11, 2014
Substitution method :
The system of equations are x + y = 5 and 3x - 2y = 10.
Solve the equation 1: x + y = 5 for y, since the y has a coefficient of 1.
x + y = 5
Subtract x from each side.
y = 5 - x.
Substitute the value of y = 5 - x in the equation 2: 3x - 2y = 10 and solve for x.
3x - 2(5 - x) = 10
Apply distributive property: a(b - c ) = ab - ac.
3x - 10 + 2x = 10
5x - 10 = 10
Add 10 to each side.
5x = 10 + 10 = 20
Divide each side by 5.
x = 20/5
x = 4.
Substitute the value of x = 4 in the equation y = 5 - x to find the value of y.
y = 5 - 4
y = 1.
The solution is x = 4 and y = 1.
answered Mar 25, 2014
Substitution method :
The equations are
x + y = 5 ---> (1)
3x - 2y = 10 ---> (2)
Solve for x from (1).
x + y = 5
x = 5 - y.
Substitute the x value in eq ( 2 )
3 ( 5 - y ) - 2 y = 10
15 - 3y - 2 y =10
15 - 5 = 10
15 - 10 = 5 y
5 = 5 y
1 = y
=>y = 1.
Substitute the y value in (1).
x + 1 = 5
x = 5 - 1 = 4
=>x = 4
Solution : x =4 and y = 1.
answered Mar 25, 2014 | 493 | 1,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2021-04 | latest | en | 0.762872 |
https://republicofsouthossetia.org/question/2-a-painter-charges-35-per-hour-plus-125-for-general-supplies-which-of-the-following-equations-r-16121575-10/ | 1,638,990,380,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363520.30/warc/CC-MAIN-20211208175210-20211208205210-00094.warc.gz | 543,101,587 | 13,118 | ## 2. A painter charges \$35 per hour plus \$125 for general supplies. Which of the following equations represent the total cost, y, for th
Question
2. A painter charges \$35 per hour plus \$125 for general supplies. Which of the following
equations represent the total cost, y, for the painter working x hours?
a. y = 125x
b. y = 160x
c. y = 125x + 35
d. y = 35x + 125
in progress 0
3 weeks 2021-11-15T13:20:35+00:00 1 Answer 0 views 0 | 143 | 439 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-49 | latest | en | 0.859633 |
https://energytheory.com/how-many-amps-does-a-2000-watt-inverter-draw/ | 1,679,536,771,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944606.5/warc/CC-MAIN-20230323003026-20230323033026-00329.warc.gz | 275,012,334 | 36,634 | Going to purchase an inverter but unsure about the wattage that you should opt for? Believe me, you are not alone in this struggle because when it comes to electronics everyone gets confused. Watts, volts, amps, etc. terms make you feel like an outsider. But all these things are pretty easy to understand and with a better understanding, you will be able to get the inverter easily. Today along with all the terms you will learn about how many amps does a 2000 watt inverter draw? Do you know how many amps does a 200 watt inverter draws?
What are Amps in Inverters?
The electric current running through the wires at the moment is measured in amps or amperes. This unit does not count as amps per hour or amps per day. The unit for an amp is 1 Coulomb per second. It is important to know the amps of a device for it determines which wire is best suited for it. Current flowing through wires heat them up, but a particular wire will resist the heat due to its competency and design.
## How Many Amps Does A 2000 Watt Inverter Draw?
These power inverters have a voltage range between 12V and 120V. The maximum amps (current) drawn by a 2000 watts inverter depends on the following factors:
• Its conversion efficiency
• Voltage rating of the battery bank
A 2000-watt power inverter running on a battery bank of 12V would draw around 240 amps and with a 24V battery bank 120 amps could be drawn. However, a battery bank with a 48V would not exceed 60 amps. You can use the following formula to calculate the number of amps drawn by your 2000 Watts inverter.
Maximum current drawn (Amps) = (2000 watts / efficiency of inverter in %) / the lowest battery voltage (V)
Here, the lowest battery voltage (before cut off) is 10 volts for 12V, 20 volts for 24V, and 40 volts for 48V battery bank.
Your inverter of 2000 watts and 12V with a 90% efficiency claimed by the manufacturer the amps drawn would be.
Maximum current drawn (Amps) = (2000 watts / efficiency of inverter in %) / the lowest battery voltage (V)
= (2000 watts / 90%) / 10 V
= (2000 watts/0.9) / 10 V
= 2222 watts / 10 V
= 222.2 amps
It would be 111.1 for a 24 battery bank with 20 volts lowest voltage. After this, let’s also try to understand how many amps does a 3000 watt inverter draw.
Also See: What Can a 2000-Watt Inverter Run?
### How Many Amps Does A 3000 Watt Inverter Draw?
A 3000-watt inverter will not have 12 volts of voltage because it will automatically convert it into 110 volts. Therefore, a 3000 watts inverter with 120 volts and an efficiency of about 85% will draw around 29 amps, approximately.
3000 watts / 120 volts / 0.85 = 29.41, rounding to the nearest it gets 29.4 amps. This has explained how many amps does a 3000 watt inverter draw.
### How Many Amps Does A 200 Watt Inverter Draw?
A 200 watts inverter is a low-power inverter. It is portable and supplies grid-live power to small appliances in your car or an RV. However, their load capacity is limited, and they are prone to get damaged easily. A 200 watts power surge is 400 watts and its continuous power draw does not exceed 200W. Towards the AC (alternating current) side a maximum current of around 1.8 amps is provided by the appliances. These are connected on the 110V side at the peak power.
1. Basic terms
Watts (W): It is the unit to measure how much power is used and supplied by a device once it is turned on. It is mentioned as watts and not as watts per hour or watts per day because Joule per second is the defining unit for a watt.
Watt-hours (watt-hour): It is the measurement of the power used for an hour. It is also mentioned as a kilowatt-hour (kilowatt-hour) for 1000 watts for 1 hour or 1 watt for 1000 hours. A light using 100 watts for 9 hours will say 900 watts-hour for the light.
Amp-Hours (AH): This is the unit for the battery capacity. It determines the time period for which the inverter can run. Amps-hours is calculated by multiplying amps and time.
2. Facts to Determine While Using the Appliances With the Inverter
Before determining how many amps does a 2000 watt inverter draw, you need to consider the following things.
• Types of appliances and their running time
• Other constant loads, like refrigerators, pumps, lights, etc.
• Age of the battery/batteries
• Source of charge for the inverter (alternator, solar, or generator)
3. Amps Drawn By Inverter in Idle Mode
A good inverter has a very low idle or no-load power draw. This ranges between 0.3 and 0.6 amps. It means the batteries are not unnecessarily used if your inverter is in idle mode. However, some inverters may draw more than 40 amps in an hour which is not good for the batteries. Therefore, it is advised to switch off the inverters when not in use.
4. Applications of a 200 Watt Inverter
This low-power inverter provides 200 watts of continuous power. They are sufficient to run appliances with a low Alternative Current. Devices like cell phones, DVD players, electric toothbrushes, fans, lights, stereos, laptops, etc. However, the time period for which you can run these devices depends on the type and efficiency of the batteries.
To meet your daily needs, you will need a battery with 100 Ah and 12V. The longer the battery provides the backup your devices will be powered.
5. Types of Powers
The basic needs supplied by an inverter are the peak or surge hours, and typical, and average power usage.
a) Surge: The maximum power supplied by the inverter for a short time like say for 15 seconds is the surge power. Appliances with electric motors require a higher surge while they run. Refrigerators and pumps are such appliances. Different types of inverters have different ranges that are usually from 20% to 300%. In terms of time, a surge rating of 3 to 15 seconds is enough to cover almost 99% of the appliances.
b) Typical Power Usage: This is the power that an inverter supplies continuously, and it is known as continuous rating. It is lower than surge. This factor determines how many amps does a 2000 watt inverter draws.
c) Average Power Usage: This is usually lesser than the surge and typical power usage. This is not a factor that you should know while selecting the inverter, but this power usage is necessary for estimating the required battery capacity. After this, your might also be curious of how many amps does a 300 watt inverter draw.
### How Many Amps Does A 300 Watt Inverter Draw?
A 300-watt inverter is available with 12V and 24V. You can convert watts to amps.
Divide the watts by the actual battery voltage. Considering the efficiency to be 85% you need to divide the answer obtained by 85%.
Here, 85% = 85/100 = 0.85
300 watts / 12V / 0.85 = 29.41 Amps
300 watts / 24V / 0.85 = 14.70 Amps
Therefore, a 300-watt inverter with 12 V draws about 29.41 amps and with 24V it draws 14.70 Amps. Now you have finally find out how many amps does a 300 watt inverter draw.
### How Many Amps Does A 500 Watt Inverter Draw?
The voltage range of a 500 watt power inverter also starts with 12V. With an approx efficiency of 80%, a 500 watt inverter will draw about 52 Amps, approximately.
500 watts / 12 V / 0.80 = 52.08, around about 52 amps. After this, let’s see how many amps does a 750 watt inverter draw. With this, you have find out how many amps does a 500 watt inverter draw.
Also Read: What Size Inverter to Run a TV
### How Many Amps Does A 750 Watt Inverter Draw?
The voltage range for a 750 watts power inverter is 12 volts, 14 volts, 24 volts, and 28 volts. Mostly a 750 watt inverter you get is of 14 volts. So a 750 watts inverter with 14 volts will draw around 67 amps with 80% efficiency.
750 watts / 14 volts / 0.80 = 66.96, around about 67 amps. After this let’s explore how many amps does a 1000 watt inverter draw.
### How Many Amps Does A 1000 Watt Inverter Draw?
A 1000-watt inverter is among the class of power inverters. The energy efficiency and voltage of the inverter tell us the average amps drawn by it. A 1000-watt 12V inverter draws about 88-105 amps, 24V draws 55-52 amps, 36V inverter draws 30-35 amps, and a 48V draws about 22-26 amps. You are close to finding out how many amps does a 2000 watt inverter draws.
These inverters are used for powering small gadgets, electronic appliances, tools, and devices. The actual load on the battery in terms of the current drawn is calculated with the units of current and power of the inverter.
1) Input Voltage: A 1000 watts inverter is usually 12 volts. Some models may have a broader range like 24V, 34V, and 48V.
2) Energy Efficiency: It varies from model to model. On average, the energy efficiency of power inverters is in the range of 80% to 95%. There are models with higher efficiency, but they are more expensive than usual.
3) Surge vs Continuous Power: A 1000-watt inverter provides a continuous power of about 1000W. Some models support 1100-1200 watts also. Surge power is the output power that an inverter can provide for a very short time. It is double in continuous power and falls in the range of 1500W-2000W. Surge power is required when the inverter is used to power tools and appliances with electric motors. The electric load of such devices is also the same as that of the inverter.
Here is the table that lists the required battery power and current for an inverter to provide 1000 watts. This table is based on the units calculated as per the energy efficiency and battery voltage of the inverter.
Energy Efficiency (%) 80% 85% 90% 95% Battery power (Watts) 1250W 1177W 1112W 1053W 12 V 104.2A 98.1A 92.7A 87.8 24 V 52.1A 49.1A 46.4A 43.9A 36 V 34.8A 32.7A 30.9A 29.3A 48 V 26.1A 24.6A 23.2A 22.0A
With this, you have discovered how many amps does a 1000 watt inverter draw. After this, your should also further delve and find out how many amps does a 1500 watt inverter draw.
### How Many Amps Does A 1500 Watt Inverter Draw?
The voltage range for a 1500-watt inverter should be not less than 14 volts. So, with this volt and efficiency of 85%, a 1500 watts inverter will likely draw 126 amps, approximately.
1500 watts / 14 volts / 0.85 = 126.05, round to 126 amps.
So, a 1500-watt inverter draws about 126 amps. With this, you have understood how many amps does a 1500 watt inverter draw. Now, its time to further find out how many amps does a 4000 watt inverter draw.
Also Read: Do Solar Lights Stay On All Night?
### How Many Amps Does a 4000 Watt Inverter Draw?
Most pure sine power inverters of 4000 watts are available in the voltage range of 120 or 240. These inverters have built-in transfer switches. The battery charger is available for off-grid power solutions. The inverter has a watt surge of about 12,000. 4000 watts with 12V (direct current) and 240V & 120V (alternative current) inverter will draw about 33.3 amps. After this, let’s see how many amps does a 5000 watt inverter draw. This perfectly explained how many amps does a 4000 watt inverter draw.
### How Many Amps Does A 5000 Watt Inverter Draw?
After learning how many amps does a 2000 watt inverter draw, you must also be curious to know how many amps does a 5000 watt inverter draw. 5000 watts or the 5-kilowatt inverter will likely draw 42.5 amps. This power inverter will power different types of equipment with the conversion of direct current to the alternating current power inverter. These inverters offer a full range of 12V and 24V.
The basic formula for calculating the amps of a 5000-watt inverter is.
The number of hours/watts = total watts /volts = battery amps.
A 5000 watts inverter will run the following devices.
• Ceiling fans 140 watts
• Coffee maker 100 watts
• Computer 150 watts
• Electric Heaters 1200 watts
• Microwave 1000 watts
• Refrigerator 1200 watts
• Stereo 300 watts
• Television 250 watts
• Toaster 1200 watts
Well, today you learned about how many amps does a 2000 watt inverter draws. The simple way to determine the amps is to know by the voltage of the inverters. However, other factors like power usage and surge or continuous power supply are important parameters to consider. Even if you are looking for how many amps does a 200 watt inverter draw.
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Olivia is committed to green energy and works to help ensure our planet's long-term habitability. She takes part in environmental conservation by recycling and avoiding single-use plastic. | 3,188 | 12,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2023-14 | longest | en | 0.945179 |
https://petroleumoffice.com/formula/viscous-shear-stress-at-the-outer-mudcake-boundary/ | 1,719,308,372,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865694.2/warc/CC-MAIN-20240625072502-20240625102502-00754.warc.gz | 396,124,708 | 4,905 | # Viscous shear stress at the outer mudcake boundary
## Input(s)
$$R_{c}$$: Radius from the Center of Drillpipe to the beginning of Mudcake (in.)
$$R_{p}$$: Radius from the Center to the Inner Boundary of the Drillpipe (in.)
$$v_{z}$$: Axial Velocity Parallel to the Wellbore Axis (in/s)
$$x_{f}$$: Transient Invasion Front (in.)
$$x_{f, o}$$: Initial Displacement, i.e., Spurt (in.)
$$\frac{d p}{d z}$$: Pressure change with $$\mathrm{Z}$$ direction that is parallel to the Wellbore Axis (psi/in.)
$$\mu$$: Viscosity $$(\mathrm{cP})$$
## Output(s)
$$\tau_{(R c)}$$: Viscous Shear Stress at the outer Mudcake boundary (lb/in. $${ }^{2}$$ )
## Formula(s)
$\begin{gathered} \tau_{(R c)}=\mu\left(d v_{z} / d r\right)_{(R c)} \\ \tau_{(R c)}=\frac{1}{4}\left[2 R_{c}+\left\{\left(R_{c}^{2}-R_{p}^{2}\right) /\left(R_{c} \log \left(R_{p} / R_{c}\right)\right)\right\}\right] \frac{d p}{d z} \end{gathered}$
## Reference(s)
Chin, W. C. (1995). Formation Invasion, Page: 59.
An unhandled error has occurred. Reload 🗙 | 369 | 1,025 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-26 | latest | en | 0.653248 |
https://www.gustygames.co.nz/workshops/problem.php?p=22,21 | 1,719,282,140,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865545.19/warc/CC-MAIN-20240625005529-20240625035529-00080.warc.gz | 705,951,835 | 3,916 | # Computation Workshop Solution Checker
## Squirrel Chase
You are standing $$250$$ metres due North of a Squirrel, at the moment when the squirrel calls out "Catch me if you can!" and starts running due East. Quick on your feet, you immediately start chasing after the squirrel. Instead of running South-East on a straight line, you run so at every instant you are heading directly towards the squirrel. You run $$1.4$$ times as fast as the squirrel. Let $$D$$ denote the distance that the squirrel ran before you catch it.
Part A Find $$D$$ in meters, rounded to the nearest meter. Find $$D$$ in millimeters, rounded to the nearest millimeter.
## Integer Factorization
Let $$N = pq$$ be a positive integer which is the product of two large prime numbers $$p$$ and $$q$$. Let $$R$$ be the remainder when $$p+q$$ is divided by $$10^9+7$$. Hint: the difference between $$p$$ and $$q$$ is less than $$10^{80}$$ (the current estimate for the number of atoms in the observable universe).
Part A Find $$R$$ when $$N =$$ 37834082197. Find $$R$$ when $$N =$$ 31259182824182977579361. Find $$R$$ when $$N =$$ 279680497756621349605906986937174040586 145180581133159093315713686803955381832 199906939110100292778501757276492336647 733009077451211345797330743124737928346 315459217030828079849109738375831203116 482566885088773675647315143800526347195 967243145966681580005600894548499860991 115107257526176229258827758033769663. | 393 | 1,423 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-26 | latest | en | 0.784074 |
https://testbook.com/question-answer/spraying-of-urea-at-the-rate-of-3-5-kg-per-100-m2--615aefa44e15bd37557fc428 | 1,679,510,642,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.74/warc/CC-MAIN-20230322180852-20230322210852-00136.warc.gz | 657,975,931 | 80,960 | # Spraying of urea at the rate of 3.5 kg per 100 m2 is to be done in a field. The dimensions of the field are as shown in the figure and all angles are right angles. Find the amount of urea required.
This question was previously asked in
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1. 665 kg
2. 518 kg
3. 66.5 kg
4. 51.8 kg
Option 4 : 51.8 kg
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## Detailed Solution
Given:
Rate of spraying urea 3.5 kg per 100 m2
Formula used:
Area of a rectangle = lb
l = length; b = breadth
Calculation:
Required area(A) = ABCD - MNOP
⇒ Area (A) = (50 × 38) - (30 × 14) = 1900 - 420 = 1480 m2
Total area to be used for that area = (1480/100) × 3.5 = 51.8 kg
∴ The amount of urea required is 51.8 kg. | 288 | 824 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2023-14 | latest | en | 0.822534 |
https://devforum.roblox.com/t/i-need-help-with-this-math-thing-read-to-understand-more/1719794 | 1,713,711,061,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817780.88/warc/CC-MAIN-20240421132819-20240421162819-00764.warc.gz | 170,214,363 | 6,503 | # I need help with this math thing (read to understand more)
Heyy, developers! I’m currently trying to add the numbers inside of a text label. for example (random number + or - random number). I’m not sure how to fix this.
`````` if player.PlayerGui.Math.ImageLabel.TextBox.Text == mathSplit[1] mathSplit[2] mathSplit[3] then -- 1 is a number, 2 is a - or +, 3 is a number
``````
So it looks like this?
`1 + 2`
and you want to split it into:
`{1, "+", 2}`
to then get:
`3`
and set text to that?
Try this:
``````local values = string.split("1 + 2", " ")
local operators = {
["+"] = function(a,b)
return a+b
end,
["-"] = function(a,b)
return a-b
end,
["*"] = function(a,b)
return a*b
end,
["/"] = function(a,b)
return a/b
end,
}
if #values == 3 then --there are enough values
if operators[values[2]] then --operator can be solved
if tonumber(values[1]) and tonumber(values[3]) then--LHS and RHS are numerical
print(operators[values[2]](values[1], values[3]))
else
end
else
end
else
end
``````
I’m getting my values from a table is the thing. I can message you the whole script, if you want
``````local result
if mathSplit[2] == "+" then
result = tonumber(mathSplit[1]) + tonumber(mathSplit[3])
elseif mathSplit[2] == "-" then
result = tonumber(mathSplit[1]) - tonumber(mathSplit[3])
end
if player.PlayerGui.Math.ImageLabel.TextBox.Text == tostring(result) then -- 1 is a number, 2 is a - or +, 3 is a number
``````
Needs to be like this.
This may error if `mathSplit[1]` or `mathSplit[3]` aren’t numbers though.
How would I make it so that the mathsplit[1] is greater then the mathsplit[2] during subtraction
Also, its saying that the answer I inputed is wron, since I noticed that the text in the textbox doesn’t come over to the script.
That means `TextBox.Text` is different from whatever `tostring(result)` returns.
You need to debug that on your end.
Tjis doesn’t work for some reason. Whenever I print that value, it becomes nil.
`````` local result
local input
player.PlayerGui.Math.ImageLabel.TextBox:GetPropertyChangedSignal("Text"):Connect(function()
input = player.PlayerGui.Math.ImageLabel.TextBox.Text
end)``````
Also, how would I make sure that when you subtract, the first number isn’t bigger than the second
`tonumber()` returns nil if its argument cannot be coerced into a number type value.
You need to debug this on your end by printing the values you’re working with.
You use the `>` or `<` comparison operators to determine if an operand is less than/greater than another operand.
You can modify the result of the operation using math.abs(n), assuming all the numbers are positive. If they arent, you can use math.max(n1, n2) - math.min(n1, n2) | 732 | 2,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-18 | latest | en | 0.665125 |
http://math.stackexchange.com/questions/473246/family-of-union-and-intersection-set-theory | 1,469,469,080,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824337.54/warc/CC-MAIN-20160723071024-00265-ip-10-185-27-174.ec2.internal.warc.gz | 155,075,176 | 19,063 | # Family of union and intersection (set-theory)
If $S$ is the set of real numbers, and if $T$ is the set of rational numbers, let, for $\alpha \in T, \ A_{\alpha}= \{x\in S\ | \ x\ge \alpha\}$.
Can anyone help explain why $\cup_{\alpha\in T}A_{\alpha}=S$ and $\cap_{\alpha\in T}A_{\alpha} =\emptyset$? Thanks!
-
Try and interpret the following in terms of unions and intersections, and the sets $A_{\alpha}$ you're given:
• Given any real number, it lies above some rational number. (This proves the first equation.)
• No real number lies above all rational numbers. (This proves the second equation.)
-
So that is what $A_{\alpha}= \{x\in S\ | \ x\ge \alpha\}$ is saying? That all real numbers are greater than rational numbers. I think I got confused of $A_{\alpha}$ since $\alpha \in T$ so I didn't know what it meant. You think you can tell me what $A_{\alpha}= \{x\in S\ | \ x\ge \alpha\}$ is saying? – Tom Aug 22 '13 at 1:54
Given a rational number $\alpha$, the set $A_{\alpha}$ is the set of real numbers $x$ for which $x \ge \alpha$. That's what the notation means. – Clive Newstead Aug 22 '13 at 2:00
...so the union is the collection of real numbers $x$ for which there exists a rational number $\alpha$ with $x \ge \alpha$ (i.e. the set of reals greater than some rational number); and the intersection is the collection of real numbers $x$ for which $x \ge \alpha$ for all rational numbers $\alpha$ (i.e. the set of reals greater than all rational numbers). HTH. – Clive Newstead Aug 22 '13 at 2:01
I got it now with your help. Thanks!! – Tom Aug 22 '13 at 2:26
This follows from the fact that the rational numbers are unbounded both from above and from below, in the set of real numbers.
Recall the definitions:
$$\bigcup_{i\in I}X_i=\{x\mid\exists i\in I\text{ such that}x\in X_i\}\\ \bigcap_{i\in I}X_i=\{x\mid\forall i\in I\text{ such that}x\in X_i\}$$
Apply these two definitions to the ones in your problem, and use the fact from above to see the equalities hold.
-
I think what I am having difficulty understanding is what $A_{\alpha}= \{x\in S\ | \ x\ge \alpha\}$ is saying exactly. What I got from that is, for all $A_{\alpha}$ where $\alpha$ is a rational number there is an $x\in S$ such that $x\ge$ rational numbers. – Tom Aug 22 '13 at 1:57
Tom, $A_\alpha$ is simply the set of those real numbers which are either equal to $\alpha$ or larger than $\alpha$. For example for $\alpha=0$ this means all the positive real numbers, and $0$ itself. – Asaf Karagila Aug 22 '13 at 2:04
Thanks for your help, Asaf! – Tom Aug 22 '13 at 2:27
You're welcome, Tom. – Asaf Karagila Aug 22 '13 at 2:28
Sorry but
$\cup_{\alpha\in T}A_{\alpha}=S$ is FALSE!
Let S = {2,3} and T = {5} then in this case
$\cup_{\alpha\in T}A_{\alpha}=$ {x∈{2,3} | x≥5} = ∅ because neither 2 nor 3 are greater than 5 and S = {2,3} is not equal to ∅
-
$S$ is the set of real number, not an arbitrary set so this does not answer the question. – Winther Sep 30 '15 at 0:01 | 951 | 2,973 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2016-30 | latest | en | 0.869395 |
https://oeis.org/A134431 | 1,656,967,737,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104496688.78/warc/CC-MAIN-20220704202455-20220704232455-00728.warc.gz | 465,478,272 | 4,721 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A134431 Triangle read by rows: T(n,k) is the number of arrangements of the set {1,2,...,n} in which the sum of the entries is equal to k (n >= 0, k >= 0; to n=0 there corresponds the empty set). 2
1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 2, 6, 1, 1, 1, 3, 3, 4, 8, 8, 6, 6, 24, 1, 1, 1, 3, 3, 5, 10, 10, 14, 14, 36, 30, 30, 24, 24, 120, 1, 1, 1, 3, 3, 5, 11, 12, 16, 22, 44, 44, 66, 60, 78, 174, 168, 144, 144, 120, 120, 720, 1, 1, 1, 3, 3, 5, 11, 13, 18, 24, 52, 52, 80, 98, 120, 234 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,7 COMMENTS Row n has 1 + n(n+1)/2 terms (n >= 0). Row sums yield the arrangement numbers (A000522). T(n, n(n+1)/2) = n!. Sum_{k=0..n(n+1)/2} k*T(n,k) = A134432(n). LINKS Alois P. Heinz, Rows n = 0..48, flattened FORMULA The row generating polynomials P[n](t) are equal to Q[n](t,1), where the polynomials Q[n](t,x) are defined by Q[0]=1 and Q[n]=Q[n-1] + xt^n (d/dx)xQ[n-1]. [Q[n](t,x) is the bivariate generating polynomial of the arrangements of {1,2,...,n}, where t (x) marks the sum (number) of the entries; for example, Q[2](t,x)=1+tx + t^2*x + 2t^3*x^2, corresponding to: empty, 1, 2, 12 and 21, respectively.] EXAMPLE T(4,7)=8 because we have 34,43 and the six permutations of {1,2,4}. Triangle starts: 1; 1, 1; 1, 1, 1, 2; 1, 1, 1, 3, 2, 2, 6; 1, 1, 1, 3, 3, 4, 8, 8, 6, 6, 24; MAPLE Q[0]:=1: for n to 7 do Q[n]:=sort(simplify(Q[n-1]+t^n*x*(diff(x*Q[n-1], x))), t) end do: for n from 0 to 7 do P[n]:=sort(subs(x=1, Q[n])) end do: for n from 0 to 7 do seq(coeff(P[n], t, j), j=0..(1/2)*n*(n+1)) end do; # yields sequence in triangular form # second Maple program: b:= proc(n, s, t) option remember; `if`(n=0, t!*x^s, b(n-1, s, t)+b(n-1, s+n, t+1)) end: T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0\$2)): seq(T(n), n=0..8); # Alois P. Heinz, Dec 22 2017 MATHEMATICA b[n_, s_, t_] := b[n, s, t] = If[n == 0, t!*x^s, b[n - 1, s, t] + b[n - 1, s + n, t + 1]]; T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]] @ b[n, 0, 0]; T /@ Range[0, 8] // Flatten (* Jean-François Alcover, Feb 19 2020, after Alois P. Heinz *) CROSSREFS Cf. A000522, A134432. Sequence in context: A161092 A029332 A344058 * A211098 A070879 A125644 Adjacent sequences: A134428 A134429 A134430 * A134432 A134433 A134434 KEYWORD nonn,tabf AUTHOR Emeric Deutsch, Nov 16 2007 STATUS approved
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Last modified July 4 16:46 EDT 2022. Contains 355081 sequences. (Running on oeis4.) | 1,226 | 2,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2022-27 | latest | en | 0.664904 |
https://invent.kde.org/education/rocs/commit/88f7af0c6900753d67d336a45a49cc848d19cfd9 | 1,624,430,508,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488534413.81/warc/CC-MAIN-20210623042426-20210623072426-00264.warc.gz | 293,499,385 | 30,775 | Commit 88f7af0c by Dilson Guimarães
### Ford-Fulkerson maximum flow algorithm implemented as an example script.
parent 39683511
/* Ford-Fulkerson maximum flow algorithm * * Writen by dilsonguim * * Time complexity: O((|V| + |E|) * flow), * * This algorithm computes the maximum flow in * a flow network. The capacity of each edge should be * specified in the property 'capacity'. * Source and sink nodes are specified directly in the script. * */ function reset(G) { G.nodes().forEach(function (node) { node.seen = false node.leading_edge = "" }) G.edges().forEach(function (edge) { edge.residual = false edge.flow = 0 edge.capacity = parseInt(edge.capacity) }) } function markNode(node) { node.color = "#c00" } function unmarkNode(node) { node.color = "#fff" } function applyToNodesInPath(path, func) { path.forEach(function (edge){ func(edge.from()) }) func(path[0].to()) } /* Removes residual edges from the residual network, * turning it back into the original graph. * */ function removeResidualEdges(G) { edges = G.edges() edges.forEach(function (edge) { if (edge.residual) { G.remove(edge) } }) } /* Adds residual edges to the graph, turning it into a * residual network. */ function createResidualEdges(G) { edges = G.edges() edges.forEach(function (edge) { residual_edge = G.createEdge(edge.to(), edge.from()) residual_edge.type = 2 residual_edge.residual = true edge.reverse = residual_edge residual_edge.reverse = edge }) } /* Finds an augmenting path in the residual network if one * exists. */ function findAugmentingPath(G, source, sink) { G.nodes().forEach(function (node) { node.seen = false }) var stack = [source] while (stack.length > 0) { current = stack.pop() if (current.seen) continue current.seen = true current.outEdges() .filter(function (edge) { return edge.flow < edge.capacity && !(edge.to().seen) }) .forEach(function (edge) { stack.push(edge.to()) edge.to().leading_edge = edge }) } //Build array of edges in the augmenting path path = [] if (sink.seen) { end_node = sink while (end_node != source) { edge = end_node.leading_edge path.push(edge) end_node = edge.from() } } return path } /* Increases the flow along an augmenting path. */ function increaseFlow(path) { flow_increase = path.map(function (edge) { return edge.capacity - edge.flow }).reduce(function (flow_increase, residual_capacity) { return Math.min(flow_increase, residual_capacity) }, Infinity) path.forEach(function (edge) { edge.flow += flow_increase if (edge.residual) { edge.capacity -= edge.flow edge.reverse.flow -= edge.flow edge.flow = 0 } else { edge.reverse.capacity += edge.flow } }) Console.log("Flow incresed by " + flow_increase) return flow_increase } /* This algorithm works in graphs with directed edges. * * Each edge must contain the property capacity. * * The property flow can be used to recover the solution * to the maximum flow problem. */ function maximumFlow(G, source, sink) { //Transform the graph into the residual network createResidualEdges(Document) flow = 0 while (true) { path = findAugmentingPath(Document, source, sink) if (path.length == 0) { break } //Mark all nodes in the augmenting path applyToNodesInPath(path, markNode) //Increase the flow along the augmenting path flow += increaseFlow(path) //Unmark nodes in the augment path applyToNodesInPath(path, unmarkNode) } Console.log("The maximum flow is " + flow) //Mark nodes in the same side as the sink in the min-cut G.nodes() .filter(function (node) { return node.seen }) .forEach(markNode) //Transform the residual network back to the original graph removeResidualEdges(Document) return flow } reset(Document) source = Document.nodes()[0] sink = Document.nodes()[Document.nodes().length - 1] maximumFlow(Document, source, sink)
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# OldHW10 - hiarker(srh959 oldhomework 10 Turner(58220 This...
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hiarker (srh959) – oldhomework 10 – Turner – (58220) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The system is in equilibrium and the pulleys are frictionless and massless. 9 kg 6 kg 7 kg 1 2 3 T Find the force T . The acceleration of grav- ity is 9 . 8 m / s 2 . Correct answer: 401 . 8 N. Explanation: Let : m 1 = 9 kg , m 2 = 6 kg , and m 3 = 7 kg . m 1 m 2 m 3 1 2 3 T T 4 T 3 T 3 T 3 T 2 T 1 T 1 The mass m 1 defines the tension T 1 : T 1 = m 1 g . At pulley 3, T 1 acts down on either side of the pulley and T 2 acts up, so T 2 = 2 T 1 = 2 m 1 g. At the mass m 2 , T 3 acts up, and m 2 g and T 2 act down, so T 3 = m 2 g + T 2 = m 2 g + 2 m 1 g . At pulley 2, T 3 acts up on either side of the pulley and T 4 acts down, so T 4 = 2 T 3 = 2 m 2 g + 4 m 1 g . At the mass m 3 , T 4 = T + m 3 g T = T 4 m 3 g = (2 m 2 + 4 m 1 m 3 ) g = [2 (6 kg) + 4 (9 kg) (7 kg)] (9 . 8 m / s 2 ) = 401 . 8 N . keywords: 002 (part 1 of 2) 10.0 points A book is at rest on an incline as shown below. A hand, in contact with the top of the book, produces a constant force F hand vertically downward. F hand Book The following figures show several attempts at drawing free-body diagrams for the book. Which figure has the correct directions for each force? The magnitudes of the forces are not necessarily drawn to scale. 1. normal friction weight force
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hiarker (srh959) – oldhomework 10 – Turner – (58220) 2 2. weight normal friction force 3. weight friction normal force 4. weight force normal friction 5. normal force friction weight 6. weight force friction normal correct 7. weight friction force normal 8. weight force normal friction Explanation: The normal force points perpendicular to the surface of the inclined plane. The weight force points down. The F hand also points down. The friction force keeps the book from sliding and consequently points up the incline.
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Ask a homework question - tutors are online | 805 | 2,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-13 | latest | en | 0.80167 |
http://www.mathematicshelp.org/maths/posts/content/36 | 1,550,683,561,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247495367.60/warc/CC-MAIN-20190220170405-20190220192405-00323.warc.gz | 387,783,413 | 4,727 | # Geometry's Formulas
## Rhombus, regular polygon, ellipse, segment sphere, torus
### The Rhombus
The Rhombus is a special parallelogram, that
is parallelogram which has two consecutive sides congruent.
The Rhombus thus has all sides congruent and
diagonals perpendicular.
Also, the Rhombus area can be calculated using parallelogram formula.
### Regular Polygon
The Regular Polygon the convex polygon which has all sides congruent (equal).
For a regular polygon area is calculated by multiplying no. sides to the length of one of them.
Here we have a regular hexagon, so no. sides equals 6. So we perimeter
.
The area of a regular polygon calculated by the formula
, where P is the polygon's perimeter, and a is the apothem of the polygon (perpendicular to the center of a polygon on one of its sides).
### The Ellipse
The Ellipse is defined as the flat curve which is the locus of points for which the sum of distances from two fixed points (called the focus of the ellipse) is constant.
Ellipse area is calculated as
.
### TheThorus
The Area of a Thorus is calculated using the formula
.
## Other solids
### The Spheric Callote
The area of Spheric Callote could be calculated using the formula
.
### The Spheric Zone
The area of Spheric Zone could be calculated using the formula (the same as that of the Spheric Callote):
.
### The Single Base Spherical Segment
The Volume for the Single Base Segment Spherical is calculated using the formula:
.
### The Spherical Segment with two bases
The Volume for the Segment Spherical with two bases is calculated using the formula:
.
Keywords: geometry, mathematical formulas, geometric formulas, Diamonds, rhombus, regular polygon, ellipse, torus | 390 | 1,716 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2019-09 | latest | en | 0.892205 |
https://www.vutbr.cz/en/students/courses/detail/161930 | 1,623,604,509,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487610196.46/warc/CC-MAIN-20210613161945-20210613191945-00536.warc.gz | 990,712,838 | 9,269 | Course detail
Matrices and tensors calculus
Definition of matrix. Fundamental notions. Equality and inequality of matrices. Transposition of matrices. Special kinds of matrices. Determinant, basic attributes. Basic operations with matrices. Special types of matrices. Linear dependence and indenpendence. Order and degree of matrices. Inverse matrix.
Solutions of linear algebraic equations. Linear and quadratic forms. Spectral attributes of matrices, eigen-value, eigen-vectors and characteristic equation. Linear space, dimension. báze. Linear transform of coordinates of vector.
Covariant and contravariant coordinates of vectors and their transformations. Definition of tensor. Covariant, contravariant and mixed tensor. Operation on tensors. Sum of tensors. Product of tensor and real number. Restriction of tensors. Symmetry and antisymmetry of tensors.
Learning outcomes of the course unit
Mastering basic techniques for solving tasks and problems from the matrices and tensors calculus and its applications.
Prerequisites
The knowledge of the content of the subject BMA1 Matematika 1 is required. The previous attendance to the subject BMAS Matematický seminář is warmly recommended.
Co-requisites
Not applicable.
Recommended optional programme components
Not applicable.
Recommended or required reading
Kolman, B., Elementary Linear Algebra, Macmillan Publ. Comp., New York 1986.
Kolman, B., Introductory Linear Algebra, Macmillan Publ. Comp., New York 1991.
Gantmacher, F. R., The Theory of Matrices, Chelsea Publ. Comp., New York 1960.
Crandal R. E., Mathematica for the Sciences, Addison-Wesley, Redwood City, 1991.
Davis H. T., Thomson K. T., Linear Algebra and Linear Operators in Engineering, Academic Press, San Diego, 2007.
Mannuci M. A., Yanofsky N. S., Quantum Computing For Computer Scientists, Cambridge University Press, Cabridge, 2008.
Nahara M., Ohmi T., Quantum Computing: From Linear Algebra to Physical Realizations, CRC Press, Boca Raton, 2008.
Griffiths D. Introduction to Elementary Particles, Wiley WCH, Weinheim, 2009.
Planned learning activities and teaching methods
Teaching methods depend on the type of course unit as specified in the article 7 of BUT Rules for Studies and Examinations.
Assesment methods and criteria linked to learning outcomes
Requirements for completion of a course are specified by a regulation issued by the lecturer responsible for the course and updated for every.
Language of instruction
English
Work placements
Not applicable.
Aims
Master the bases of the matrices and tensors calculus and its applications.
Specification of controlled education, way of implementation and compensation for absences
The content and forms of instruction in the evaluated course are specified by a regulation issued by the lecturer responsible for the course and updated for every academic year.
Classification of course in study plans
• Programme AUDIO-PU Master's
branch PU-AUD , 1. year of study, summer semester, 5 credits, optional interdisciplinary
• Programme EEKR-MN Master's
branch MN-TIT , 1. year of study, summer semester, 5 credits, theoretical subject
branch MN-KAM , 1. year of study, summer semester, 5 credits, theoretical subject
branch MN-EVM , 1. year of study, summer semester, 5 credits, theoretical subject
branch MN-EST , 1. year of study, summer semester, 5 credits, theoretical subject
branch MN-SVE , 1. year of study, summer semester, 5 credits, theoretical subject
branch MN-EEN , 1. year of study, summer semester, 5 credits, theoretical subject
• Programme AUDIO-PU Master's
branch PU-AUD , 2. year of study, summer semester, 5 credits, optional interdisciplinary
Type of course unit
Lecture
26 hours, optionally
Teacher / Lecturer
Exercise in computer lab
18 hours, compulsory
Teacher / Lecturer
The other activities
8 hours, compulsory
Teacher / Lecturer | 864 | 3,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-25 | latest | en | 0.709219 |
https://codereview.stackexchange.com/questions/98448/free-code-camp-pairwise | 1,713,600,267,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817491.77/warc/CC-MAIN-20240420060257-20240420090257-00481.warc.gz | 149,870,531 | 50,820 | # Free Code Camp - Pairwise
I'm working through the Free Code Camp syllabus and I'm on to Intermediate JavaScript Algorithms. This Pairwise problem was the last challenge in that section. The section came just after "Object Oriented JavaScript." So I figured they were looking for an OO solution, but the instructions included a link to MDN's array.reduce(). My solution doesn't use array.reduce() and I'd really appreciate some feedback on what I could have done better to make my code more compact and efficient. It feels a little clunky but passes all the tests.
The instructions
Return the sum of all indices of elements of 'arr' that can be paired with one other element to form a sum that equals the value in the second argument 'arg'. If multiple sums are possible, return the smallest sum. Once an element has been used, it cannot be reused to pair with another.
For example, pairwise([1, 4, 2, 3, 0, 5], 7) should return 11 because 4, 2, 3 and 5 can be paired with each other to equal 7.
pairwise([1, 3, 2, 4], 4) would only equal 1, because only the first two elements can be paired to equal 4, and the first element has an index of 0!
Remember to use RSAP if you get stuck. Try to pair program. Write your own code.
Array.reduce()
My Solution
function pairwise(arr, arg) {
this.objects = [];
var total = 0;
function Element(value, index) {
this.value = value;
this.index = index;
this.used = 0;
}
for (var i = 0; i < arr.length; i++) {
this.objects.push(new Element(arr[i], i));
}
for (var j = 0; j < objects.length; j++) {
if (objects[j].used === 0) {
for (var k = 0; k < objects.length; k++) {
if (objects[k].used === 0 && objects[k].index != objects[j].index) {
if (arg - objects[j].value == objects[k].value) {
total = total + objects[j].index + objects[k].index;
objects[j].used = 1;
objects[k].used = 1;
break;
}
}
}
}
}
}
pairwise([1,1,1], 2);
• Shouldn't the first one return a 4? The instruction did say "If multiple sums are possible, return the smallest sum." and nothing about when multiple sums can be added together or not. Jul 30, 2015 at 17:04
I looked through your code and it is a valid solution, but you could reduce your code base by better leveraging the functions that JavaScript already provides, such as Array.prototype.indexOf().
For example, instead of building a new class-like-function (Element) to track the appearance of a certain index, I simply made a deep copy of the the initial array and parsed it with indexOf().
Moreover, in your code, when you first declare this.objects = [], this actually refers to the global scope (window object). As you can see, you are calling pairwise without building a new instance (new keyword). In this case, thus the this keyword is bound to the global window object.
Please find below my take on it:
function pairwise(arr, arg) {
var result = 0,
newArr = [],
//Used to hold the indices that we have already used to form our sum
indices = [];
//Loop through arr and create a deep copy of it in newArr
for(var k = 0; k < arr.length; k++) {
newArr.push(arr[k]);
}
//Loop through arr
for(var i = 0; i < arr.length; i++) {
//Loop through newArr
for(var j = 0; j < newArr.length; j++) {
//Since we want to add different elements of the array, we want to avoid adding the same element
if(i !== j) {
//If the sum of two elements is equal to arg AND the indices that we have in i and j are not part of the indices array
//Indices array is used to hold the already used indices, thus ensuring the accurate parsing of the parameters
if(arr[i] + newArr[j] === arg && indices.indexOf(i) === -1 && indices.indexOf(j) === -1) {
//Sum the indices up
result += i + j;
//Push the indices in the indices array in order to not use them in further iterations
indices.push(i, j);
}
}
}
}
return result;
}
pairwise([1,4,2,3,0,5], 7);
• Hi Vlad—thanks for solution. I am going through your code to try to learn from it. Could you explain how if(i !== j) the first conditional, is comparing anything but the iterators? May 9, 2016 at 14:57
• In my solution, newArr is a deep copy of the arr initially passed to the pairwise function. Thus, when the two iterator variables (i & j) have the same value, while iterating through different copies of the same object, they would actually add up the exact same number of the array, which is not a valid case for the aforementioned scenario. May 10, 2016 at 12:44
Vlad Z answer is correct but freecodecamp has weird wording on this problem. I used a similar answer but was failing on this test:
expect(pairwise([0, 0, 0, 0, 1, 1], 1)).to.equal(10);
My problem and the problem with Vlad Zs solution is that 0,1 -> indexes (0 and 4) and 0,1 indexes(1 and 5) are both acceptable and should return 10.
I would use Vlad's solution but sub in this function instead of indexOf === -1 to check if a pair exists already:
function checkPairExists(value,position,pairsArray){
for(var i = 0; i < pairsArray.length; i++){
if (pairsArray[i].value === value && pairsArray[i].position === position){
return true;
}
}
return false;
}
This function implementation uses the "reduce" method to get the sum of the indexes and it fullfills all Free Code Camp tests.
function pairwise(arr, arg) {
return arr.reduce((sum, value1, index1) => {
arr.slice(index1 + 1).forEach((value2, index2) => {
if (arr[index1] + arr[index1 + 1 + index2] === arg) {
arr[index1] = arr[index1 + 1 + index2] = NaN;
sum += index1 + index1 + 1 + index2;
}
});
return sum;
}, 0);
}
I came to the same conclusion as Piotr, but with a slight improvement - dropping one more un-necessary check. See comments in code below:
function pairwise(arr, arg) {
var sum = 0;
for (var i=0; i < arr.length - 1; i++) {
for (var j=i+1; j < arr.length; j++) {
//No need to check for less than arg, used elements are naturally eliminated
if (arr[i] + arr[j] === arg) {
sum += i + j;
arr[i] = arr[j] = arg + 1; //Set the used elements to higher than arg e.g. arg + 1
}
}
}
return sum;
}
pairwise([1,4,2,3,0,5], 7);
The requirements are broken down into:
• Return the sum of all indices of elements of 'arr' that can be paired with one other element to form a sum that equals the value in the second argument 'arg'.
• If multiple sums are possible, return the smallest sum. Once an element has been used
• Once an element has been used, it cannot be reused to pair with another.
The first bullet point is easy enough to understand. We find pairs that sum up to the total, and sum up the indices.
However, your example contradicts the second bullet point. If multiple sums are found, it should return the smallest. 4 and 3 are 1 and 3 which results to 4. 2 and 5 are 2 and 5 which results to 7. The result should be 4 in the first example.
So here's my take on it
function pairwise(arr, total) {
// For each item in the array
var sums = arr.reduce(function (indexSum, firstNumber, firstIndex) {
// Collect the pair's index which causes the numbers to sum to total
var secondIndices = arr.slice(firstIndex + 1).reduce(function (secondIndices, secondNumber, i) {
if (firstNumber + secondNumber === total) secondIndices.push(firstIndex + i + 1);
return secondIndices;
}, []);
// Add to our collection the sum this iteration's index and
// the pair indices
return indexSum.concat(secondIndices.map(function (secondIndex) {
return secondIndex + firstIndex
}));
}, []);
// In all the items, find the smallest sum
return Math.min.apply(null, sums);
}
console.log(pairwise([1, 4, 2, 3, 0, 5], 7));
console.log(pairwise([1, 3, 2, 4], 4));
Regarding bullet point 3, I would care less if a number was reused, like in the case of 6 in [4, 4, 2] or say [4, 2, 9, 9, 4] because any number that's going to pair with it after the first established pair will have a higher index sum anyways.
• Thanks for the great reply, the way I read the second requirement - returning smallest sum, was that if the target could be reached by two calculations sharing one number then we should use the calculation whose sum of indices is smallest. For example: pairwise([0,1,1], 1) should return 1, even though it is possible to use the last number in the array with the 0 to get an answer of 2, we should use the smallest index sum. Thanks again though and you managed to use array.reduce()! I would never have gotten that. Jul 31, 2015 at 6:40
The solution below is very compact. It avoids unnecessary checks and loops only through the relevant elements. You can check the working codepen here: http://codepen.io/PiotrBerebecki/pen/RRGaBZ.
function pairwise(arr, arg) {
var sum = 0;
for (var i=0; i<arr.length-1; i++) {
for (var j=i+1; j<arr.length; j++) {
if (arr[i] <= arg && arr[j] <= arg && arr[i] + arr[j] == arg) {
sum += i+j;
arr[i] = arr[j] = NaN;
}
}
}
return sum;
}
console.log( pairwise([1, 1, 0, 2], 2) ) // should return 6
Under the hood:
1. Start looping from the element with index (i) = 0.
2. Add a second loop only for the elements which are later in the array. Their index j is always higher than i as we are adding 1 to i.
3. If both elements (numbers) are less than or equal to to the arg, check if their sum equals to the arg. This avoids checking the sum if either of the numbers are greater than the arg.
4. If the pair has been found then change their values to NaN to avoid further checks and duplication.
Maybe not super efficient, but a little towards FP style:
pairwise = (ary, n) => ary.
filter(m => ary.some(x => x !== m && m + x === n)).
map(e => ary.indexOf(e)).
reduce((a, b) => a + b)
(Not sure why node wouldn’t accept dots in the beginning of line, I would’ve liked them to be there).
• Hold on, where can I find the expected return values? Above code fails on the example Frank gives in his answer. Mar 22, 2019 at 19:21
## An $$\O(n)\$$ complexity and storage solution
A post dredged from the deep past and has 7 answers. All of which are rather poor in terms of either complexity and/or storage.
There is a solution that is $$\O(n)\$$ complexity and $$\O(n)\$$ storage.
Note Best case storage $$\O(1)\$$ eg (pairwise[0, 0, 0, ..., 0], 0). Worst case $$\O(n)\$$ eg (pairwise[0, 0, 0, ..., 0], 1)
Using a Map we can eliminate the costly overhead of iterating to find the second matching item. The map also stores an array indexes of the first item allowing for easy calculation of the smallest result. This removes the need to search for the lowest value.
function pairWise(arr, pairSum) {
const required = new Map();
var sum = 0, i = 0;
for (const val of arr) {
if (required.has(val)) {
const pair = required.get(val);
sum += i + pair.ind[pair.use ++];
if (pair.use === pair.ind.length) { required.delete(val) }
} else {
const req = pairSum - val;
if (required.has(req)) { required.get(req).ind.push(i) }
else { required.set(req, {ind: [i], use: 0}) }
}
i ++;
}
return sum;
}
To be fair at the time this question was posted knowledge of Map was limited as was its implementation. However JS Object also provides a way to create a map so the solution has always been easily implemented in JS
// pre ES6 version
function pairWise(arr, pairSum) {
var required= {}, sum = 0, i = 0, val, pair;
while (i < arr.length) {
val = arr[i];
if (required[val]) {
pair = required[val];
sum += i + pair.ind[pair.use ++];
if (pair.use === pair.ind.length) { delete required[val] }
} else {
val = pairSum - val;
if (required[val]) { required[val].ind.push(i) }
else { required[val] = {ind: [i], use : 0} }
}
i ++;
}
return sum;
} | 3,069 | 11,423 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-18 | latest | en | 0.780964 |
https://www.lessoncorner.com/Math/Money_Math?page=100 | 1,545,182,929,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376830305.92/warc/CC-MAIN-20181219005231-20181219031231-00156.warc.gz | 943,817,941 | 6,936 | Lessons and Tools for Teachers
#### Lessons for Money Math
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• Income Taxes: Who Pays and How Much? - … on the government? Civics/Government, Family and Consumer Science, Economics, Math, Writing Students will: Understand that the federal government collects individual … each year that gives annual income and withholding information Income money received from wages and salaries, rent, interest, and profit Income … income, property or goods to support government programs Tax Refund money owed by the government to taxpayers when their total tax …
User Rating: Grade Level: 9-12
• Energy Conservation - … as incandescent light bulbs. They also cost different amounts of money to use. For example, if your utility company charges \$0 … work to help homeowners and businesses conserve energy and save money on the cost of heating and cooling buildings. They look … calculator to convert sunlight directly into electricity to do her math homework; or use big, south-facing windows to heat a … and the gas purchased. Have the students calculate how much money they spend to light one room in their house for …
User Rating: Grade Level: 3-5
• Protecting Our City with Levees - … curricular units Weather and Atmosphere lessons Hurricanes Educational Standards : Colorado Math a. Add, subtract, multiply and divide rational numbers including integers … , strips of paper that each represent one dollar or Monopoly money) levee building materials: - sand or gravel (about 2 cups) - duct … -30 min) Have students "buy" their materials with their fun money. Give them plenty of time to build their levees. When …
User Rating: Grade Level: 6-8
• A Question of Trust - … to the beginning balance for their team. This is the money they have to start production. The beginning balance for each … consider: the cost of production - \$30/unit, the amount of money they have on hand (production costs must be paid up … this step. Don't assume that because you have good math students, they'll figure it out. It's not the … was the strategy of the team that made the most money? Which team made the least? What was your strategy and …
User Rating: Grade Level: 6-12
• Agricultural Technology - … ; Better care of the environment; Ways to save money or ways to make more money. If needed, provide Attachment K, The Future … in the Classroom www.cfaitc.org AIMS Education Foundation These math-science hands-on activities encourage students to look at the … use fewer chemicals to cause less pollution and spend less money on chemicals. Agriculture can be passed down through families for …
User Rating: Grade Level: 3-5
• What Am I Worth? - … , and quarter Match coin equivalents This lesson addresses Va. SOL Math 2.11a & b Media Components Video clip - If You Made … :42) PAUSE right after the magician says "...same amount of money as five nickels." Say: How does five nickels equal a … Cross-Curricular Extensions Technology Bookmark websites that will provide additional money practice Language Arts Make story problems with coin sets Social …
User Rating: Grade Level: K-2
• Crunching the Numbers: Exploring the Math of the Debt Crisis - … ," according to the infographic, and why they owe so much money to other countries. In a brief whole-class discussion, have … hazards of such action, addressing questions like "Where does the money come from?", "What are the potential consequences for the lenders … also be aligned to the new Common Core State Standards): Math 3. Uses basic and advanced procedures while performing the processes …
User Rating:
• Voices from the Chicago Teachers Strike - … school reform is about standardized testing, so it's about math and reading all day, which doesn't engage children. So … spend millions on your pet programs, but there's no money for school level repairs, so the roof leaks on my … a man who only devotes his time to schools with money and doesn't take students like us into consideration. Mayor … stand together with our teachers. They spent their time and money paying to get an education so they can educate us …
User Rating: Grade Level: 9-12
• Are Things Really More Expensive Today - … /29/04 Process/Activities ⢠⢠⢠⢠⢠⢠⢠Have students listen to the Sound Money feature on increasing gas prices from the Minnesota Public Radio … Use a transparency to show students how to complete the math calculation formula. Have students take notes during the audio file … today. For younger children, parents can show stacks of play money for each item and have the child determine which item …
User Rating:
• Shoplifting - … . Visit to Wal-Mart 5. Design "No Shoplifting" posters 6. Math lesson on extra cost per item caused by shoplifting. at … . They bought hot dogs and drinks, and spent all their money. They were s_]] hungry. One says, "Let's get cookies … wanted a pair of "Guess" jeans but only had enough money for regular Jeans. "I deserve the best, and besides, a big company like Wal-Mart has plenty of money - so I'll just put it on sale." He went …
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Copyright © 2018 Lesson Corner. All rights reserved. | 1,118 | 5,121 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-51 | latest | en | 0.917043 |
https://en.wikipedia.org/wiki/Einstein_coefficients | 1,702,221,425,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102469.83/warc/CC-MAIN-20231210123756-20231210153756-00630.warc.gz | 266,653,122 | 43,035 | # Einstein coefficients
Einstein coefficients are quantities describing the probability of absorption or emission of a photon by an atom or molecule.[1] The Einstein A coefficients are related to the rate of spontaneous emission of light, and the Einstein B coefficients are related to the absorption and stimulated emission of light. Throughout this article, "light" refers to any electromagnetic radiation, not necessarily in the visible spectrum.
## Spectral lines
In physics, one thinks of a spectral line from two viewpoints.
An emission line is formed when an atom or molecule makes a transition from a particular discrete energy level E2 of an atom, to a lower energy level E1, emitting a photon of a particular energy and wavelength. A spectrum of many such photons will show an emission spike at the wavelength associated with these photons.
An absorption line is formed when an atom or molecule makes a transition from a lower, E1, to a higher discrete energy state, E2, with a photon being absorbed in the process. These absorbed photons generally come from background continuum radiation (the full spectrum of electromagnetic radiation) and a spectrum will show a drop in the continuum radiation at the wavelength associated with the absorbed photons.
The two states must be bound states in which the electron is bound to the atom or molecule, so the transition is sometimes referred to as a "bound–bound" transition, as opposed to a transition in which the electron is ejected out of the atom completely ("bound–free" transition) into a continuum state, leaving an ionized atom, and generating continuum radiation.
A photon with an energy equal to the difference E2E1 between the energy levels is released or absorbed in the process. The frequency ν at which the spectral line occurs is related to the photon energy by Bohr's frequency condition E2E1 = where h denotes the Planck constant.[2][3][4][5][6][7]
## Emission and absorption coefficients
An atomic spectral line refers to emission and absorption events in a gas in which ${\displaystyle n_{2}}$ is the density of atoms in the upper-energy state for the line, and ${\displaystyle n_{1}}$ is the density of atoms in the lower-energy state for the line.
The emission of atomic line radiation at frequency ν may be described by an emission coefficient ${\displaystyle \varepsilon }$ with units of energy/(time × volume × solid angle). ε dt dV dΩ is then the energy emitted by a volume element ${\displaystyle dV}$ in time ${\displaystyle dt}$ into solid angle ${\displaystyle d\Omega }$. For atomic line radiation,
${\displaystyle \varepsilon ={\frac {h\nu }{4\pi }}n_{2}A_{21},}$
where ${\displaystyle A_{21}}$ is the Einstein coefficient for spontaneous emission, which is fixed by the intrinsic properties of the relevant atom for the two relevant energy levels.
The absorption of atomic line radiation may be described by an absorption coefficient ${\displaystyle \kappa }$ with units of 1/length. The expression κ' dx gives the fraction of intensity absorbed for a light beam at frequency ν while traveling distance dx. The absorption coefficient is given by
${\displaystyle \kappa '={\frac {h\nu }{4\pi }}(n_{1}B_{12}-n_{2}B_{21}),}$
where ${\displaystyle B_{12}}$ and ${\displaystyle B_{21}}$ are the Einstein coefficients for photon absorption and induced emission respectively. Like the coefficient ${\displaystyle A_{21}}$, these are also fixed by the intrinsic properties of the relevant atom for the two relevant energy levels. For thermodynamics and for the application of Kirchhoff's law, it is necessary that the total absorption be expressed as the algebraic sum of two components, described respectively by ${\displaystyle B_{12}}$ and ${\displaystyle B_{21}}$, which may be regarded as positive and negative absorption, which are, respectively, the direct photon absorption, and what is commonly called stimulated or induced emission.[8][9][10]
The above equations have ignored the influence of the spectroscopic line shape. To be accurate, the above equations need to be multiplied by the (normalized) spectral line shape, in which case the units will change to include a 1/Hz term.
Under conditions of thermodynamic equilibrium, the number densities ${\displaystyle n_{2}}$ and ${\displaystyle n_{1}}$, the Einstein coefficients, and the spectral energy density provide sufficient information to determine the absorption and emission rates.
### Equilibrium conditions
The number densities ${\displaystyle n_{2}}$ and ${\displaystyle n_{1}}$ are set by the physical state of the gas in which the spectral line occurs, including the local spectral radiance (or, in some presentations, the local spectral radiant energy density). When that state is either one of strict thermodynamic equilibrium, or one of so-called "local thermodynamic equilibrium",[11][12][13] then the distribution of atomic states of excitation (which includes ${\displaystyle n_{2}}$ and ${\displaystyle n_{1}}$) determines the rates of atomic emissions and absorptions to be such that Kirchhoff's law of equality of radiative absorptivity and emissivity holds. In strict thermodynamic equilibrium, the radiation field is said to be black-body radiation and is described by Planck's law. For local thermodynamic equilibrium, the radiation field does not have to be a black-body field, but the rate of interatomic collisions must vastly exceed the rates of absorption and emission of quanta of light, so that the interatomic collisions entirely dominate the distribution of states of atomic excitation. Circumstances occur in which local thermodynamic equilibrium does not prevail, because the strong radiative effects overwhelm the tendency to the Maxwell–Boltzmann distribution of molecular velocities. For example, in the atmosphere of the Sun, the great strength of the radiation dominates. In the upper atmosphere of the Earth, at altitudes over 100 km, the rarity of intermolecular collisions is decisive.
In the cases of thermodynamic equilibrium and of local thermodynamic equilibrium, the number densities of the atoms, both excited and unexcited, may be calculated from the Maxwell–Boltzmann distribution, but for other cases, (e.g. lasers) the calculation is more complicated.
## Einstein coefficients
In 1916, Albert Einstein proposed that there are three processes occurring in the formation of an atomic spectral line. The three processes are referred to as spontaneous emission, stimulated emission, and absorption. With each is associated an Einstein coefficient, which is a measure of the probability of that particular process occurring. Einstein considered the case of isotropic radiation of frequency ν and spectral energy density ρ(ν).[3][14][15][16] Paul Dirac derived the coefficients in a 1927 paper titled "The Quantum Theory of the Emission and Absorption of Radiation".[17][18]
### Various formulations
Hilborn has compared various formulations for derivations for the Einstein coefficients, by various authors.[19] For example, Herzberg works with irradiance and wavenumber;[20] Yariv works with energy per unit volume per unit frequency interval,[21] as is the case in the more recent (2008) [22] formulation. Mihalas & Weibel-Mihalas work with radiance and frequency;[13] also Chandrasekhar;[23] also Goody & Yung;[24] Loudon uses angular frequency and radiance.[25]
### Spontaneous emission
Spontaneous emission is the process by which an electron "spontaneously" (i.e. without any outside influence) decays from a higher energy level to a lower one. The process is described by the Einstein coefficient A21 (s−1), which gives the probability per unit time that an electron in state 2 with energy ${\displaystyle E_{2}}$ will decay spontaneously to state 1 with energy ${\displaystyle E_{1}}$, emitting a photon with an energy E2E1 = . Due to the energy-time uncertainty principle, the transition actually produces photons within a narrow range of frequencies called the spectral linewidth. If ${\displaystyle n_{i}}$ is the number density of atoms in state i , then the change in the number density of atoms in state 2 per unit time due to spontaneous emission will be
${\displaystyle \left({\frac {dn_{2}}{dt}}\right)_{\text{spontaneous}}=-A_{21}n_{2}.}$
The same process results in increasing of the population of the state 1:
${\displaystyle \left({\frac {dn_{1}}{dt}}\right)_{\text{spontaneous}}=A_{21}n_{2}.}$
### Stimulated emission
Stimulated emission (also known as induced emission) is the process by which an electron is induced to jump from a higher energy level to a lower one by the presence of electromagnetic radiation at (or near) the frequency of the transition. From the thermodynamic viewpoint, this process must be regarded as negative absorption. The process is described by the Einstein coefficient ${\displaystyle B_{21}}$ (m3 J−1 s−2), which gives the probability per unit time per unit energy density of the radiation field per unit frequency that an electron in state 2 with energy ${\displaystyle E_{2}}$ will decay to state 1 with energy ${\displaystyle E_{1}}$, emitting a photon with an energy E2E1 = . The change in the number density of atoms in state 1 per unit time due to induced emission will be
${\displaystyle \left({\frac {dn_{1}}{dt}}\right)_{\text{neg. absorb.}}=B_{21}n_{2}\rho (\nu ),}$
where ${\displaystyle \rho (\nu )}$ denotes the spectral energy density of the isotropic radiation field at the frequency of the transition (see Planck's law).
Stimulated emission is one of the fundamental processes that led to the development of the laser. Laser radiation is, however, very far from the present case of isotropic radiation.
### Photon absorption
Absorption is the process by which a photon is absorbed by the atom, causing an electron to jump from a lower energy level to a higher one. The process is described by the Einstein coefficient ${\displaystyle B_{12}}$ (m3 J−1 s−2), which gives the probability per unit time per unit energy density of the radiation field per unit frequency that an electron in state 1 with energy ${\displaystyle E_{1}}$ will absorb a photon with an energy E2E1 = and jump to state 2 with energy ${\displaystyle E_{2}}$. The change in the number density of atoms in state 1 per unit time due to absorption will be
${\displaystyle \left({\frac {dn_{1}}{dt}}\right)_{\text{pos. absorb.}}=-B_{12}n_{1}\rho (\nu ).}$
## Detailed balancing
The Einstein coefficients are fixed probabilities per time associated with each atom, and do not depend on the state of the gas of which the atoms are a part. Therefore, any relationship that we can derive between the coefficients at, say, thermodynamic equilibrium will be valid universally.
At thermodynamic equilibrium, we will have a simple balancing, in which the net change in the number of any excited atoms is zero, being balanced by loss and gain due to all processes. With respect to bound-bound transitions, we will have detailed balancing as well, which states that the net exchange between any two levels will be balanced. This is because the probabilities of transition cannot be affected by the presence or absence of other excited atoms. Detailed balance (valid only at equilibrium) requires that the change in time of the number of atoms in level 1 due to the above three processes be zero:
${\displaystyle 0=A_{21}n_{2}+B_{21}n_{2}\rho (\nu )-B_{12}n_{1}\rho (\nu ).}$
Along with detailed balancing, at temperature T we may use our knowledge of the equilibrium energy distribution of the atoms, as stated in the Maxwell–Boltzmann distribution, and the equilibrium distribution of the photons, as stated in Planck's law of black body radiation to derive universal relationships between the Einstein coefficients.
From Boltzmann distribution we have for the number of excited atomic species i:
${\displaystyle {\frac {n_{i}}{n}}={\frac {g_{i}e^{-E_{i}/kT}}{Z}},}$
where n is the total number density of the atomic species, excited and unexcited, k is Boltzmann's constant, T is the temperature, ${\displaystyle g_{i}}$ is the degeneracy (also called the multiplicity) of state i, and Z is the partition function. From Planck's law of black-body radiation at temperature T we have for the spectral radiance (radiance is energy per unit time per unit solid angle per unit projected area, when integrated over an appropriate spectral interval)[26] at frequency ν
${\displaystyle \rho _{\nu }(\nu ,T)=F(\nu ){\frac {1}{e^{h\nu /kT}-1}},}$
where[27]
${\displaystyle F(\nu )={\frac {2h\nu ^{3}}{c^{2}}},}$
where ${\displaystyle c}$ is the speed of light and ${\displaystyle h}$ is Planck's constant.
Substituting these expressions into the equation of detailed balancing and remembering that E2E1 = yields
${\displaystyle A_{21}g_{2}e^{-h\nu /kT}+B_{21}g_{2}e^{-h\nu /kT}{\frac {F(\nu )}{e^{h\nu /kT}-1}}=B_{12}g_{1}{\frac {F(\nu )}{e^{h\nu /kT}-1}},}$
or
${\displaystyle A_{21}g_{2}(1-e^{-h\nu /kT})+B_{21}g_{2}F(\nu )e^{-h\nu /kT}=B_{12}g_{1}F(\nu ).}$
The above equation must hold at any temperature, so from ${\displaystyle T\rightarrow \infty }$ one gets
${\displaystyle B_{21}g_{2}=B_{12}g_{1},}$
and from ${\displaystyle T\rightarrow 0}$
${\displaystyle A_{21}g_{2}=B_{21}g_{2}F(\nu ).}$
Therefore, the three Einstein coefficients are interrelated by
${\displaystyle {\frac {A_{21}}{B_{21}}}=F(\nu )}$
and
${\displaystyle {\frac {B_{21}}{B_{12}}}={\frac {g_{1}}{g_{2}}}.}$
When this relation is inserted into the original equation, one can also find a relation between ${\displaystyle A_{21}}$ and ${\displaystyle B_{12}}$, involving Planck's law.
## Oscillator strengths
The oscillator strength ${\displaystyle f_{12}}$ is defined by the following relation to the cross section ${\displaystyle \sigma }$ for absorption:[19]
${\displaystyle \sigma ={\frac {e^{2}}{4\varepsilon _{0}m_{e}c}}\,f_{12}\,\phi _{\nu }={\frac {\pi e^{2}}{2\varepsilon _{0}m_{e}c}}\,f_{12}\,\phi _{\omega },}$
where ${\displaystyle e}$ is the electron charge, ${\displaystyle m_{e}}$ is the electron mass, and ${\displaystyle \phi _{\nu }}$ and ${\displaystyle \phi _{\omega }}$ are normalized distribution functions in frequency and angular frequency respectively. This allows all three Einstein coefficients to be expressed in terms of the single oscillator strength associated with the particular atomic spectral line:
${\displaystyle B_{12}={\frac {e^{2}}{4\varepsilon _{0}m_{e}h\nu }}f_{12},}$
${\displaystyle B_{21}={\frac {e^{2}}{4\varepsilon _{0}m_{e}h\nu }}{\frac {g_{1}}{g_{2}}}f_{12},}$
${\displaystyle A_{21}={\frac {2\pi \nu ^{2}e^{2}}{\varepsilon _{0}m_{e}c^{3}}}{\frac {g_{1}}{g_{2}}}f_{12}.}$
## Dipole Approximation
The value of A and B coefficients can be calculated using quantum mechanics where dipole approximations in time dependent perturbation theory is used. While the calculation of B coefficient can be done easily, that of A coefficient requires using results of second quantization. This is because the theory developed by dipole approximation and time dependent perturbation theory gives a semiclassical description of electronic transition which goes to zero as perturbing fields go to zero. The A coefficient which governs spontaneous emission should not go to zero as perturbing fields go to zero. The result for transition rates of different electronic levels as a result of spontaneous emission is given as (in SI units):[28][19][29]
${\displaystyle w_{i\rightarrow f}^{s.emi}={\frac {\omega _{if}^{3}e^{2}}{3\pi \epsilon _{0}\hbar c^{3}}}|\langle f|{\vec {r}}|i\rangle |^{2}=A_{if}}$
For B coefficient, straightforward application of dipole approximation in time dependent perturbation theory yields (in SI units):[30][29]
${\displaystyle w_{i\rightarrow f}^{abs}={\frac {u(\omega _{fi})\pi e^{2}}{3\epsilon _{0}\hbar ^{2}}}|\langle f|{\vec {r}}|i\rangle |^{2}=B_{if}^{abs}u(\omega _{fi})}$
${\displaystyle w_{i\rightarrow f}^{emi}={\frac {u(\omega _{if})\pi e^{2}}{3\epsilon _{0}\hbar ^{2}}}|\langle f|{\vec {r}}|i\rangle |^{2}=B_{if}^{emi}u(\omega _{if})}$
Note that the rate of transition formula depends on dipole moment operator. For higher order approximations, it involves quadrupole moment and other similar terms.
Here, the B coefficients are chosen to correspond to ${\displaystyle \omega }$ energy distribution function. Often these different definitions of B coefficients are distinguished by superscript, for example, ${\textstyle B_{21}^{f}={\frac {B_{21}^{\omega }}{2\pi }}}$ where ${\textstyle B_{21}^{f}}$ term corresponds to frequency distribution and ${\textstyle B_{21}^{\omega }}$ term corresponds to ${\displaystyle \omega }$ distribution.[19] The formulas for B coefficients varies inversely to that of the energy distribution chosen, so that the transition rate is same regardless of convention.
Hence, AB coefficients are calculated using dipole approximation as:
${\displaystyle A_{ab}={\frac {\omega _{ab}^{3}e^{2}}{3\pi \epsilon _{0}\hbar c^{3}}}|\langle a|{\vec {r}}|b\rangle |^{2}}$
${\displaystyle B_{ab}={\frac {\pi e^{2}}{3\epsilon _{0}\hbar ^{2}}}|\langle a|{\vec {r}}|b\rangle |^{2}}$
where ${\displaystyle \omega _{ab}={\frac {E_{a}-E_{b}}{\hbar }}}$ and B coefficients correspond to ${\displaystyle \omega }$ energy distribution function.
Hence the following ratios are also derived:
${\displaystyle {\frac {B_{12}}{B_{21}}}=1}$ and ${\displaystyle {\frac {A_{if}}{B}}={\frac {\omega _{if}^{3}\hbar }{\pi ^{2}c^{3}}}}$
### Derivation of Planck's Law
It follows from theory that:[29]
${\displaystyle {\frac {dN_{b}}{dt}}=-A_{ba}N_{b}-N_{b}u(\omega _{ba})B_{ba}+N_{a}u(\omega _{ba})B_{ab}=-N_{b}w_{b\rightarrow a}^{s.emi}-N_{b}w_{b\rightarrow a}^{emi}+N_{a}w_{a\rightarrow b}^{abs}}$
where ${\displaystyle N_{a}}$ and ${\displaystyle N_{b}}$ are number of occupied energy levels of ${\displaystyle E_{a}}$ and ${\displaystyle E_{b}}$ respectively, where ${\displaystyle E_{b}>E_{a}}$. Note that from time dependent perturbation theory application, the fact that only radiation whose ${\displaystyle \omega }$ is close to value of ${\displaystyle \omega _{ba}}$ can produce respective stimulated emission or absorption, is used.
Where maxwell distribution involving ${\displaystyle N_{a}}$ and ${\displaystyle N_{b}}$ ensures ${\displaystyle {\frac {N_{a}}{N_{b}}}={\frac {e^{-E_{a}\beta }}{e^{-E_{b}\beta }}}=e^{\omega _{ba}\hbar \beta }}$
Solving for ${\displaystyle u}$ for equilibrium condition ${\displaystyle {\frac {dN_{b}}{dt}}=0}$ using the above equations and ratios while generalizing ${\displaystyle \omega _{ba}}$ to ${\displaystyle \omega }$, we get:
${\displaystyle u_{\omega }(\omega ,T)={\frac {\omega ^{3}\hbar }{\pi ^{2}c^{3}}}{\frac {1}{e^{\omega \hbar \beta }-1}}}$
which is the angular frequency energy distribution from Planck's law.[29]
## References
1. ^ Hilborn, Robert C. (1982). "Einstein coefficients, cross sections, f values, dipole moments, and all that". American Journal of Physics. 50 (11): 982–986. arXiv:physics/0202029. Bibcode:1982AmJPh..50..982H. doi:10.1119/1.12937. ISSN 0002-9505. S2CID 119050355.
2. ^
3. ^ a b Einstein, A. (1916). "Strahlungs-Emission und -Absorption nach der Quantentheorie". Verhandlungen der Deutschen Physikalischen Gesellschaft. 18: 318–323. Bibcode:1916DPhyG..18..318E. Translated in Alfred Engel. The Berlin Years: Writings, 1914-1917. Vol. 6. pp. 212–216.
4. ^ Sommerfeld 1923, p. 43.
5. ^ Heisenberg 1925, p. 108.
6. ^ Brillouin 1970, p. 31.
7. ^ Jammer 1989, pp. 113, 115.
8. ^ Weinstein, M. A. (1960). "On the validity of Kirchhoff's law for a freely radiating body". American Journal of Physics. 28 (2): 123–25. Bibcode:1960AmJPh..28..123W. doi:10.1119/1.1935075.
9. ^ Burkhard, D. G.; Lochhead, J. V. S.; Penchina, C. M. (1972). "On the validity of Kirchhoff's law in a nonequilibrium environment". American Journal of Physics. 40 (12): 1794–1798. Bibcode:1972AmJPh..40.1794B. doi:10.1119/1.1987065.
10. ^ Baltes, H. P. (1976). On the validity of Kirchhoff's law of heat radiation for a body in a nonequilibrium environment, Chapter 1, pages 1–25 of Progress in Optics XIII, edited by E. Wolf, North-Holland, ISSN 0079-6638.
11. ^ Milne, E. A. (1928). "The effect of collisions on monochromatic radiative equilibrium". Monthly Notices of the Royal Astronomical Society. 88 (6): 493–502. Bibcode:1928MNRAS..88..493M. doi:10.1093/mnras/88.6.493.
12. ^ Chandrasekhar, S. (1950), p. 7.
13. ^ a b Mihalas, D., Weibel-Mihalas, B. (1984), pp. 329–330.
14. ^ Loudon, R. (2000), Section 1.5, pp. 16–19.
15. ^ Einstein, A. (1916). "Zur Quantentheorie der Strahlung". Mitteilungen der Physikalischen Gessellschaft Zürich. 18: 47–62.
16. ^ Einstein, A. (1917). "Zur Quantentheorie der Strahlung". Physikalische Zeitschrift. 18: 121–128. Bibcode:1917PhyZ...18..121E. Translated in ter Haar, D. (1967). The Old Quantum Theory. Pergamon. pp. 167–183. LCCN 66029628. Also in Boorse, H. A., Motz, L. (1966). The world of the atom, edited with commentaries, Basic Books, Inc., New York, pp. 888–901.
17. ^ Dirac, Paul (1927). "The quantum theory of the emission and absorption of radiation". Proceedings of the Royal Society of London. Series A, Containing Papers of a Mathematical and Physical Character. 114 (767): 243–265. doi:10.1098/rspa.1927.0039. ISSN 0950-1207.
18. ^ Duck, Ian; Sudarshan, E.C.G. (1998). "Chapter 6: Dirac's Invention of Quantum Field Theory". Pauli and the Spin-Statistics Theorem. World Scientific Publishing. pp. 149–167. ISBN 978-9810231149.
19. ^ a b c d Hilborn, Robert (2002). "Einstein coefficients, cross sections, f values, dipole moments, and all that" (PDF).
20. ^ Herzberg, G. (1950).
21. ^ Yariv, A. (1967/1989), pp. 171–173.
22. ^ Garrison, J. C., Chiao, R. Y. (2008), pp. 15–19.
23. ^ Chandrasekhar, S. (1950), p. 354.
24. ^ Goody, R. M., Yung, Y. L. (1989), pp. 33–35.
25. ^ Loudon, R. (1973/2000), pp. 16–19.
26. ^ Robert W. Boyd, Radiometry and the Detection of Optical Radiation, John Wiley and Sons, 1983
27. ^ Hubeny, Ivan; Mihalas, Dimitri (2015). Theory of stellar atmospheres : an introduction to astrophysical non-equilibrium quantitative spectroscopic analysis. Princeton University Press. pp. 116–118. ISBN 9780691163291.
28. ^ Zettili, Nouredine (2009). Quantum mechanics: concepts and applications (2nd ed.). Chichester: Wiley. pp. 594–596. ISBN 978-0-470-02679-3.
29. ^ a b c d Segre, Carlo. "The Einstein coefficients - Fundamentals of Quantum Theory II (PHYS 406)" (PDF). p. 32.
30. ^ Zwiebach, Barton. "Quantum Physics III Chapter 4: Time Dependent Perturbation Theory | Quantum Physics III | Physics". MIT OpenCourseWare. pp. 108–110. Retrieved 2023-11-03. | 6,181 | 22,626 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 92, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-50 | latest | en | 0.927747 |
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## The Problem of Specialization.
We have already met the specialization problem in the context of linear algebra and discussed the `proviso' technique for dealing with it. This problem also appears here, although harmlessly, according to the theorem below.
The difficulty is that if contains parameters, the calculation of the GCD of and may be incorrect on specialization; that is, the GCD may `almost always' be 1 but for certain values of the parameter the GCD may be nontrivial.
For example, consider . Then and the GCD of p and is 1, unless c=0 or c=2.
```> p := x^4 + 2*x^3 + 3*x^2 + 4*x + c;
4 3 2
p := x + 2 x + 3 x + 4 x + c
> pstar := x^4 - 2*x^3 + 3*x^2 - 4*x + c;
4 3 2
pstar := x - 2 x + 3 x - 4 x + c
> gcd(p,pstar);
1
> gcd(subs(c=0,p),subs(c=0,pstar));
x
> gcd(subs(c=2,p),subs(c=2,pstar));
2
x + 2
```
Note that the result returned from Maple's `gcd` is incorrect on specialization and this happens with no warning whatsoever. It turns out that the problem of detecting nontrivial specializations of multivariate GCD's is extremely computationally intensive when the number of parameters and variables are large. However, in the present context it is not so bad. Indeed, the following theorem shows that we already have all the information in the Stieltjes continued fraction to detect all the special cases.
Theorem Theorem: Let p be a polynomial with complex coefficients, containing a complex-valued parameter. The GCD of p and its paraconjugate calculated as a multivariate polynomial GCD will be correct on specialization if each in the Stieltjes continued fraction for . Thus investigation of the poles and zeros of the , considered as functions of the parameter, will uncover all the special cases of the GCD.
Thus the routine `Hurwitz` automatically gives its own proviso for correctness of the results.
Proof. If then d | p and (the symbol is read as `divides'). Hence and . Likewise, if , then D | p and . Hence D = d.
The Stieltjes continued fraction is computed with the Euclidean algorithm for the computation of the GCD of and . The reported partial quotients can be wrong only if division by zero occurs, which will happen if one of the remainders is zero on specialization. This will give rise to a pole in the partial quotient at the special value of the parameter, followed by a zero in the next partial quotient, and a pole in the one following, and so on.
Remark. If one of the partial quotients has a pole or zero, the computed GCD may or may not be wrong. This proviso is a provision for correctness, and hence is a necessary condition.
The following example should make this theorem more clear.
```> p := lambda^3 + 6*lambda^2 + 10*lambda + 4 - a - I*b;
3 2
p := lambda + 6 lambda + 10 lambda + 4 - a - I b
> with(share):
> Hurwitz(p,lambda);
FAIL
> Stieltjes_continued_fraction_and_GCD := `Hurwitz/data`(p,lambda);
Stieltjes_continued_fraction_and_GCD := [
I b lambda
[1/6 lambda, 216 --------- + 36 ------,
2 56 + a
(56 + a)
2
I (56 + a) b
---------------------------------------
2 3 2
- 12544 + 2688 a + 108 a + a + 216 b
3
(56 + a) lambda
- 1/6 ---------------------------------------], 1]
2 3 2
- 12544 + 2688 a + 108 a + a + 216 b
```
Note the GCD is computed as GCD = 1. This is almost always correct. We examine special values of the parameters a and b. First set a = -56.
```> subs(a=-56,p);
3 2
lambda + 6 lambda + 10 lambda + 60 - I b
> SCF := `Hurwitz/data`(",lambda);
/ 2\
| 60 lambda |
SCF := [[1/6 lambda, I |---- + 6 -------|], 1]
\ b b /
```
We see the GCD is still trivial there.
```> subs(a=-56,b=0,p);
3 2
lambda + 6 lambda + 10 lambda + 60
> SCF := `Hurwitz/data`(",lambda);
2
SCF := [[], lambda + 10]
```
Now we see that the GCD is nontrivial. Thus the vanishing or blowing-up of the coefficients is a necessary but not sufficient condition.
```> Stieltjes_continued_fraction_and_GCD[1][3];
2
I (56 + a) b
---------------------------------------
2 3 2
- 12544 + 2688 a + 108 a + a + 216 b
3
(56 + a) lambda
- 1/6 ---------------------------------------
2 3 2
- 12544 + 2688 a + 108 a + a + 216 b
> denom(");
2 3 2
- 75264 + 16128 a + 648 a + 6 a + 1296 b
2 1568 2 3
b = ---- - 112/9 a - 1/2 a - 1/216 a
27
> map(factor,");
2 2
b = - 1/216 (a - 4) (56 + a)
```
So if the above equation holds, we MIGHT have a nontrivial GCD. Remembering the source of this problem, we can find a nice parametric solution of the above equation as follows.
```> subs(lambda=I*t,p);
3 2
- I t - 6 t + 10 I t + 4 - a - I b
> evalc(Re("));
2
- 6 t + 4 - a
> A := solve(",a);
2
A := - 6 t + 4
> """;
3 2
- I t - 6 t + 10 I t + 4 - a - I b
> evalc(Im("));
3
- t + 10 t - b
> B := solve(",b);
3
B := - t + 10 t
> subs(a=A,b=B,p);
3 2 2 3
lambda + 6 lambda + 10 lambda + 6 t - I (- t + 10 t)
> New := `Hurwitz/data`(",lambda);
2
New := [[], lambda - RootOf(_Z + 1) t]
```
And we see a nontrivial GCD for any value of t.
For a completely different approach to the same problem, see [].
Next: A Bifurcation Problem. Up: Stability of Polynomials Previous: Test file for
Robert Corless
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riddles >> putnam exam (pure math) >> Floor Summation
(Message started by: ThudanBlunder on Nov 9th, 2009, 6:00am)
```
Title: Floor Summation
Post by ThudanBlunder on Nov 9th, 2009, 6:00am
Evaluate
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gif
81http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/lfloor.gifntanhhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/rfloor.gif/10n
n=1
Title: Re: Floor Summation
Post by Obob on Nov 9th, 2009, 2:18pm
It's roughly 1 - 2.413 * 10-264. Are you looking for an actual precise answer, or is the point just that [hide]tanh pi is close to 1[/hide]?
Title: Re: Floor Summation
Post by ThudanBlunder on Nov 11th, 2009, 3:25am
on 11/09/09 at 14:18:23, Obob wrote:
It's roughly 1 - 2.413 * 10-264. Are you looking for an actual precise answer, or is the point just that [hide]tanh pi is close to 1[/hide]?
As I am not expecting an exact answer, perhaps I should have put this elsewhere.
The point is that the answer, a transcendental number, requires at least 239 decimal places before we can discover it does not equal 1. And even more if we want to consider rounding errors.
Do you normally sum series to such precision? :)
Title: Re: Floor Summation
Post by Obob on Nov 11th, 2009, 6:10am
I just observed that the first time floor(n tanh pi) != n-1 occurs around n = 267 or so, summed the series 81(n-1)/10^n = 1, and then gave as a rough error estimate an approximation of the difference between the two series.
I guess I'm just saying that there is no reason to use tanh pi except for obfuscation; the real result lurking here is that
limx->1- sum (floor(n x))/10^n = 1, and that the convergence occurs very quickly. | 611 | 1,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2022-05 | latest | en | 0.868456 |
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Eastern Mediterranean University Faculty of Business and Economics Department of Economics 2014‐15 Fall Semester ECON101 ‐ Introduction to Economics I Final Exam Type A Answers 26 January 2015 Duration: 100 minutes Name Surname: _________________________ Group No: ___________ Student ID: _____________________________ Part A: Multiple Choice Questions (2 pts. each, total 40 pts.) 1. Profits are maximized at the output at which marginal cost equals marginal revenue. If the market price falls below the minimum average variable cost: A. the firm should shut down. B. the firm should produce more. C. the firm should produce less. D. None of the above. 2. A monopolist sells 4 units of output for £6 each and 5 units of output at £5 each. The marginal revenue associated with the 5th unit is: A. 1 B. 3 C. 25 D. 15 3. For a perfectly competitive firm operating in the short run, in order to maximize profits it should produce output where: A. marginal cost equals average total cost. B. marginal cost equals average variable cost. C. marginal cost equals price. D. total cost equals total revenue. 4. Which of the following statement about a long-run cost curve is true? A. The minimum is always below the minimum point reached by a short-run cost curve. B. There are always decreasing returns to scale. C. It shows the minimum average cost to produce a given output when all inputs can be varied. D. All the above. Page 1 of 8 5. Referring to the figure above, the following statements are correct except: A. To the left of point E marginal cost exceeds marginal revenue. B. At point E marginal cost equals marginal revenue. C. To the right of point E marginal revenue is less than marginal cost. D. Output is the most profitable output. Q 1 6. In which of the following market structures do firms have no control over the price of their product? A. Pure monopoly B. Perfect competition C. Monopolistic competition D. Oligopoly 7. In perfect competition, the marginal revenue of an individual firm A. is zero. B. is positive but less than the price of the product. C. equals the price of the product. D. exceeds the price of the product. 8. If marginal revenue is higher than marginal cost then: A. producing one extra unit will increase total profit. B. total revenues are larger than total costs. C. the firm is maximizing its profits. D. All the above. 9. Collusion means that: A. a monopolistic firm uses illegal means to maximize its profits. B. a large number of monopolistically competitive firms decide to keep the price high to maximize collective profits. C. two monopolistically competitive firms agree to keep their price lower than their competitors. D. two or more oligopolistic firms act as if they were a monopoly. 10. The marginal cost curve will shift up if: A. a new technological improvement is introduced. B. the cost of one variable input in the production increases. C. demand increases. D. production increases. Page 2 of 8 11. The table above shows that: A. total cost increases as output increases. B. marginal cost falls over some range of output and then increases. C. average cost declines and then increases. D. All the above. 12. A firm's break-even point is the quantity and price at which the firm's total revenue just equals its A. total cost. B. total variable cost. C. total fixed cost. D. marginal cost. 13. As output increases, marginal cost will eventually A. increase because of the law of increasing returns. B. increase because of the law of diminishing returns. C. decrease because of the law of diminishing returns. D. decrease because of the law of increasing returns. 14. In a perfectly competitive market, ________. A. there are restrictions on entry into the industry B. firms in the industry have advantages over firms that plan to enter the industry C. one firm can decide to change the market price D. there are many firms that sell identical products 15. Which of the following is always true for a perfectly competitive firm? A. P = MR B. P = ATC C. MR = ATC D. P = AVC 16. Which of the following is true for a single-price monopolist? A. P > MR B. P < MR C. P = MR D. P = elasticity of demand Page 3 of 8 17. Economic efficiency necessarily occurs when the firm…………………………… A. produces a given output at least cost. B. produces a given output by using the least inputs. C. earns a normal profit. D. earns an economic profit. 18. Which of the following is characteristic of the long run? A. It must be equal to 12 months in length. B. The firm’s plant is fixed. C. All resources can be varied. D. All of the above answers are correct. 19. Suppose that the five firms that make up the invisible goldfish industry formally sign a contract to establish product price; essentially they have: A. established a dominant firm to serve as the industry's price leader. B. established a monopolistically competitive industry. C. formed a cartel. D. entered into an implicit agreement. 20. In the figure above, the short-run supply curve is: A. the SMC curve. B. the points between AD. C. the SMC curve above A. D. the points between AC. Page 4 of 8
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http://cn.metamath.org/mpeuni/sylow2blem3.html | 1,653,225,427,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545548.56/warc/CC-MAIN-20220522125835-20220522155835-00099.warc.gz | 14,192,736 | 17,924 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > sylow2blem3 Structured version Visualization version GIF version
Theorem sylow2blem3 18083
Description: Sylow's second theorem. Putting together the results of sylow2a 18080 and the orbit-stabilizer theorem to show that 𝑃 does not divide the set of all fixed points under the group action, we get that there is a fixed point of the group action, so that there is some 𝑔 ∈ 𝑋 with ℎ𝑔𝐾 = 𝑔𝐾 for all ℎ ∈ 𝐻. This implies that invg(𝑔)ℎ𝑔 ∈ 𝐾, so ℎ is in the conjugated subgroup 𝑔𝐾invg(𝑔). (Contributed by Mario Carneiro, 18-Jan-2015.)
Hypotheses
Ref Expression
sylow2b.x 𝑋 = (Base‘𝐺)
sylow2b.xf (𝜑𝑋 ∈ Fin)
sylow2b.h (𝜑𝐻 ∈ (SubGrp‘𝐺))
sylow2b.k (𝜑𝐾 ∈ (SubGrp‘𝐺))
sylow2b.a + = (+g𝐺)
sylow2b.r = (𝐺 ~QG 𝐾)
sylow2b.m · = (𝑥𝐻, 𝑦 ∈ (𝑋 / ) ↦ ran (𝑧𝑦 ↦ (𝑥 + 𝑧)))
sylow2blem3.hp (𝜑𝑃 pGrp (𝐺s 𝐻))
sylow2blem3.kn (𝜑 → (#‘𝐾) = (𝑃↑(𝑃 pCnt (#‘𝑋))))
sylow2blem3.d = (-g𝐺)
Assertion
Ref Expression
sylow2blem3 (𝜑 → ∃𝑔𝑋 𝐻 ⊆ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔)))
Distinct variable groups: 𝑥,𝑔,𝑦,𝑧,𝐺 𝑔,𝐾,𝑥,𝑦,𝑧 · ,𝑔,𝑥,𝑦,𝑧 + ,𝑔,𝑥,𝑦,𝑧 ,𝑔,𝑥,𝑦,𝑧 𝜑,𝑔,𝑧 𝑥, ,𝑧 𝑔,𝐻,𝑥,𝑦,𝑧 𝑔,𝑋,𝑥,𝑦,𝑧
Allowed substitution hints: 𝜑(𝑥,𝑦) 𝑃(𝑥,𝑦,𝑧,𝑔) (𝑦,𝑔)
Proof of Theorem sylow2blem3
Dummy variable 𝑢 is distinct from all other variables.
StepHypRef Expression
1 sylow2blem3.hp . . . . . . . . 9 (𝜑𝑃 pGrp (𝐺s 𝐻))
2 pgpprm 18054 . . . . . . . . 9 (𝑃 pGrp (𝐺s 𝐻) → 𝑃 ∈ ℙ)
31, 2syl 17 . . . . . . . 8 (𝜑𝑃 ∈ ℙ)
4 sylow2b.h . . . . . . . . . . 11 (𝜑𝐻 ∈ (SubGrp‘𝐺))
5 subgrcl 17646 . . . . . . . . . . 11 (𝐻 ∈ (SubGrp‘𝐺) → 𝐺 ∈ Grp)
64, 5syl 17 . . . . . . . . . 10 (𝜑𝐺 ∈ Grp)
7 sylow2b.x . . . . . . . . . . 11 𝑋 = (Base‘𝐺)
87grpbn0 17498 . . . . . . . . . 10 (𝐺 ∈ Grp → 𝑋 ≠ ∅)
96, 8syl 17 . . . . . . . . 9 (𝜑𝑋 ≠ ∅)
10 sylow2b.xf . . . . . . . . . 10 (𝜑𝑋 ∈ Fin)
11 hashnncl 13195 . . . . . . . . . 10 (𝑋 ∈ Fin → ((#‘𝑋) ∈ ℕ ↔ 𝑋 ≠ ∅))
1210, 11syl 17 . . . . . . . . 9 (𝜑 → ((#‘𝑋) ∈ ℕ ↔ 𝑋 ≠ ∅))
139, 12mpbird 247 . . . . . . . 8 (𝜑 → (#‘𝑋) ∈ ℕ)
14 pcndvds2 15619 . . . . . . . 8 ((𝑃 ∈ ℙ ∧ (#‘𝑋) ∈ ℕ) → ¬ 𝑃 ∥ ((#‘𝑋) / (𝑃↑(𝑃 pCnt (#‘𝑋)))))
153, 13, 14syl2anc 694 . . . . . . 7 (𝜑 → ¬ 𝑃 ∥ ((#‘𝑋) / (𝑃↑(𝑃 pCnt (#‘𝑋)))))
16 sylow2b.r . . . . . . . . . . 11 = (𝐺 ~QG 𝐾)
17 sylow2b.k . . . . . . . . . . 11 (𝜑𝐾 ∈ (SubGrp‘𝐺))
187, 16, 17, 10lagsubg2 17702 . . . . . . . . . 10 (𝜑 → (#‘𝑋) = ((#‘(𝑋 / )) · (#‘𝐾)))
1918oveq1d 6705 . . . . . . . . 9 (𝜑 → ((#‘𝑋) / (#‘𝐾)) = (((#‘(𝑋 / )) · (#‘𝐾)) / (#‘𝐾)))
20 sylow2blem3.kn . . . . . . . . . 10 (𝜑 → (#‘𝐾) = (𝑃↑(𝑃 pCnt (#‘𝑋))))
2120oveq2d 6706 . . . . . . . . 9 (𝜑 → ((#‘𝑋) / (#‘𝐾)) = ((#‘𝑋) / (𝑃↑(𝑃 pCnt (#‘𝑋)))))
22 pwfi 8302 . . . . . . . . . . . . . 14 (𝑋 ∈ Fin ↔ 𝒫 𝑋 ∈ Fin)
2310, 22sylib 208 . . . . . . . . . . . . 13 (𝜑 → 𝒫 𝑋 ∈ Fin)
247, 16eqger 17691 . . . . . . . . . . . . . . 15 (𝐾 ∈ (SubGrp‘𝐺) → Er 𝑋)
2517, 24syl 17 . . . . . . . . . . . . . 14 (𝜑 Er 𝑋)
2625qsss 7851 . . . . . . . . . . . . 13 (𝜑 → (𝑋 / ) ⊆ 𝒫 𝑋)
27 ssfi 8221 . . . . . . . . . . . . 13 ((𝒫 𝑋 ∈ Fin ∧ (𝑋 / ) ⊆ 𝒫 𝑋) → (𝑋 / ) ∈ Fin)
2823, 26, 27syl2anc 694 . . . . . . . . . . . 12 (𝜑 → (𝑋 / ) ∈ Fin)
29 hashcl 13185 . . . . . . . . . . . 12 ((𝑋 / ) ∈ Fin → (#‘(𝑋 / )) ∈ ℕ0)
3028, 29syl 17 . . . . . . . . . . 11 (𝜑 → (#‘(𝑋 / )) ∈ ℕ0)
3130nn0cnd 11391 . . . . . . . . . 10 (𝜑 → (#‘(𝑋 / )) ∈ ℂ)
32 eqid 2651 . . . . . . . . . . . . . . 15 (0g𝐺) = (0g𝐺)
3332subg0cl 17649 . . . . . . . . . . . . . 14 (𝐾 ∈ (SubGrp‘𝐺) → (0g𝐺) ∈ 𝐾)
3417, 33syl 17 . . . . . . . . . . . . 13 (𝜑 → (0g𝐺) ∈ 𝐾)
35 ne0i 3954 . . . . . . . . . . . . 13 ((0g𝐺) ∈ 𝐾𝐾 ≠ ∅)
3634, 35syl 17 . . . . . . . . . . . 12 (𝜑𝐾 ≠ ∅)
377subgss 17642 . . . . . . . . . . . . . . 15 (𝐾 ∈ (SubGrp‘𝐺) → 𝐾𝑋)
3817, 37syl 17 . . . . . . . . . . . . . 14 (𝜑𝐾𝑋)
39 ssfi 8221 . . . . . . . . . . . . . 14 ((𝑋 ∈ Fin ∧ 𝐾𝑋) → 𝐾 ∈ Fin)
4010, 38, 39syl2anc 694 . . . . . . . . . . . . 13 (𝜑𝐾 ∈ Fin)
41 hashnncl 13195 . . . . . . . . . . . . 13 (𝐾 ∈ Fin → ((#‘𝐾) ∈ ℕ ↔ 𝐾 ≠ ∅))
4240, 41syl 17 . . . . . . . . . . . 12 (𝜑 → ((#‘𝐾) ∈ ℕ ↔ 𝐾 ≠ ∅))
4336, 42mpbird 247 . . . . . . . . . . 11 (𝜑 → (#‘𝐾) ∈ ℕ)
4443nncnd 11074 . . . . . . . . . 10 (𝜑 → (#‘𝐾) ∈ ℂ)
4543nnne0d 11103 . . . . . . . . . 10 (𝜑 → (#‘𝐾) ≠ 0)
4631, 44, 45divcan4d 10845 . . . . . . . . 9 (𝜑 → (((#‘(𝑋 / )) · (#‘𝐾)) / (#‘𝐾)) = (#‘(𝑋 / )))
4719, 21, 463eqtr3d 2693 . . . . . . . 8 (𝜑 → ((#‘𝑋) / (𝑃↑(𝑃 pCnt (#‘𝑋)))) = (#‘(𝑋 / )))
4847breq2d 4697 . . . . . . 7 (𝜑 → (𝑃 ∥ ((#‘𝑋) / (𝑃↑(𝑃 pCnt (#‘𝑋)))) ↔ 𝑃 ∥ (#‘(𝑋 / ))))
4915, 48mtbid 313 . . . . . 6 (𝜑 → ¬ 𝑃 ∥ (#‘(𝑋 / )))
50 prmz 15436 . . . . . . . 8 (𝑃 ∈ ℙ → 𝑃 ∈ ℤ)
513, 50syl 17 . . . . . . 7 (𝜑𝑃 ∈ ℤ)
5230nn0zd 11518 . . . . . . 7 (𝜑 → (#‘(𝑋 / )) ∈ ℤ)
53 ssrab2 3720 . . . . . . . . . 10 {𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧} ⊆ (𝑋 / )
54 ssfi 8221 . . . . . . . . . 10 (((𝑋 / ) ∈ Fin ∧ {𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧} ⊆ (𝑋 / )) → {𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧} ∈ Fin)
5528, 53, 54sylancl 695 . . . . . . . . 9 (𝜑 → {𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧} ∈ Fin)
56 hashcl 13185 . . . . . . . . 9 ({𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧} ∈ Fin → (#‘{𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧}) ∈ ℕ0)
5755, 56syl 17 . . . . . . . 8 (𝜑 → (#‘{𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧}) ∈ ℕ0)
5857nn0zd 11518 . . . . . . 7 (𝜑 → (#‘{𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧}) ∈ ℤ)
59 eqid 2651 . . . . . . . 8 (Base‘(𝐺s 𝐻)) = (Base‘(𝐺s 𝐻))
60 sylow2b.a . . . . . . . . 9 + = (+g𝐺)
61 sylow2b.m . . . . . . . . 9 · = (𝑥𝐻, 𝑦 ∈ (𝑋 / ) ↦ ran (𝑧𝑦 ↦ (𝑥 + 𝑧)))
627, 10, 4, 17, 60, 16, 61sylow2blem2 18082 . . . . . . . 8 (𝜑· ∈ ((𝐺s 𝐻) GrpAct (𝑋 / )))
63 eqid 2651 . . . . . . . . . . 11 (𝐺s 𝐻) = (𝐺s 𝐻)
6463subgbas 17645 . . . . . . . . . 10 (𝐻 ∈ (SubGrp‘𝐺) → 𝐻 = (Base‘(𝐺s 𝐻)))
654, 64syl 17 . . . . . . . . 9 (𝜑𝐻 = (Base‘(𝐺s 𝐻)))
667subgss 17642 . . . . . . . . . . 11 (𝐻 ∈ (SubGrp‘𝐺) → 𝐻𝑋)
674, 66syl 17 . . . . . . . . . 10 (𝜑𝐻𝑋)
68 ssfi 8221 . . . . . . . . . 10 ((𝑋 ∈ Fin ∧ 𝐻𝑋) → 𝐻 ∈ Fin)
6910, 67, 68syl2anc 694 . . . . . . . . 9 (𝜑𝐻 ∈ Fin)
7065, 69eqeltrrd 2731 . . . . . . . 8 (𝜑 → (Base‘(𝐺s 𝐻)) ∈ Fin)
71 eqid 2651 . . . . . . . 8 {𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧} = {𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧}
72 eqid 2651 . . . . . . . 8 {⟨𝑥, 𝑦⟩ ∣ ({𝑥, 𝑦} ⊆ (𝑋 / ) ∧ ∃𝑔 ∈ (Base‘(𝐺s 𝐻))(𝑔 · 𝑥) = 𝑦)} = {⟨𝑥, 𝑦⟩ ∣ ({𝑥, 𝑦} ⊆ (𝑋 / ) ∧ ∃𝑔 ∈ (Base‘(𝐺s 𝐻))(𝑔 · 𝑥) = 𝑦)}
7359, 62, 1, 70, 28, 71, 72sylow2a 18080 . . . . . . 7 (𝜑𝑃 ∥ ((#‘(𝑋 / )) − (#‘{𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧})))
74 dvdssub2 15070 . . . . . . 7 (((𝑃 ∈ ℤ ∧ (#‘(𝑋 / )) ∈ ℤ ∧ (#‘{𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧}) ∈ ℤ) ∧ 𝑃 ∥ ((#‘(𝑋 / )) − (#‘{𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧}))) → (𝑃 ∥ (#‘(𝑋 / )) ↔ 𝑃 ∥ (#‘{𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧})))
7551, 52, 58, 73, 74syl31anc 1369 . . . . . 6 (𝜑 → (𝑃 ∥ (#‘(𝑋 / )) ↔ 𝑃 ∥ (#‘{𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧})))
7649, 75mtbid 313 . . . . 5 (𝜑 → ¬ 𝑃 ∥ (#‘{𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧}))
77 hasheq0 13192 . . . . . . . 8 ({𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧} ∈ Fin → ((#‘{𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧}) = 0 ↔ {𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧} = ∅))
7855, 77syl 17 . . . . . . 7 (𝜑 → ((#‘{𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧}) = 0 ↔ {𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧} = ∅))
79 dvds0 15044 . . . . . . . . 9 (𝑃 ∈ ℤ → 𝑃 ∥ 0)
8051, 79syl 17 . . . . . . . 8 (𝜑𝑃 ∥ 0)
81 breq2 4689 . . . . . . . 8 ((#‘{𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧}) = 0 → (𝑃 ∥ (#‘{𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧}) ↔ 𝑃 ∥ 0))
8280, 81syl5ibrcom 237 . . . . . . 7 (𝜑 → ((#‘{𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧}) = 0 → 𝑃 ∥ (#‘{𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧})))
8378, 82sylbird 250 . . . . . 6 (𝜑 → ({𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧} = ∅ → 𝑃 ∥ (#‘{𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧})))
8483necon3bd 2837 . . . . 5 (𝜑 → (¬ 𝑃 ∥ (#‘{𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧}) → {𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧} ≠ ∅))
8576, 84mpd 15 . . . 4 (𝜑 → {𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧} ≠ ∅)
86 rabn0 3991 . . . 4 ({𝑧 ∈ (𝑋 / ) ∣ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧} ≠ ∅ ↔ ∃𝑧 ∈ (𝑋 / )∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧)
8785, 86sylib 208 . . 3 (𝜑 → ∃𝑧 ∈ (𝑋 / )∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧)
8865raleqdv 3174 . . . 4 (𝜑 → (∀𝑢𝐻 (𝑢 · 𝑧) = 𝑧 ↔ ∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧))
8988rexbidv 3081 . . 3 (𝜑 → (∃𝑧 ∈ (𝑋 / )∀𝑢𝐻 (𝑢 · 𝑧) = 𝑧 ↔ ∃𝑧 ∈ (𝑋 / )∀𝑢 ∈ (Base‘(𝐺s 𝐻))(𝑢 · 𝑧) = 𝑧))
9087, 89mpbird 247 . 2 (𝜑 → ∃𝑧 ∈ (𝑋 / )∀𝑢𝐻 (𝑢 · 𝑧) = 𝑧)
91 vex 3234 . . . . 5 𝑧 ∈ V
9291elqs 7842 . . . 4 (𝑧 ∈ (𝑋 / ) ↔ ∃𝑔𝑋 𝑧 = [𝑔] )
93 simplrr 818 . . . . . . . . . . . . . . . . . . . 20 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → 𝑧 = [𝑔] )
9493oveq2d 6706 . . . . . . . . . . . . . . . . . . . . 21 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → (𝑢 · 𝑧) = (𝑢 · [𝑔] ))
95 simprr 811 . . . . . . . . . . . . . . . . . . . . 21 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → (𝑢 · 𝑧) = 𝑧)
96 simpll 805 . . . . . . . . . . . . . . . . . . . . . 22 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → 𝜑)
97 simprl 809 . . . . . . . . . . . . . . . . . . . . . 22 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → 𝑢𝐻)
98 simplrl 817 . . . . . . . . . . . . . . . . . . . . . 22 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → 𝑔𝑋)
997, 10, 4, 17, 60, 16, 61sylow2blem1 18081 . . . . . . . . . . . . . . . . . . . . . 22 ((𝜑𝑢𝐻𝑔𝑋) → (𝑢 · [𝑔] ) = [(𝑢 + 𝑔)] )
10096, 97, 98, 99syl3anc 1366 . . . . . . . . . . . . . . . . . . . . 21 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → (𝑢 · [𝑔] ) = [(𝑢 + 𝑔)] )
10194, 95, 1003eqtr3d 2693 . . . . . . . . . . . . . . . . . . . 20 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → 𝑧 = [(𝑢 + 𝑔)] )
10293, 101eqtr3d 2687 . . . . . . . . . . . . . . . . . . 19 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → [𝑔] = [(𝑢 + 𝑔)] )
10325ad2antrr 762 . . . . . . . . . . . . . . . . . . . 20 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → Er 𝑋)
104103, 98erth 7834 . . . . . . . . . . . . . . . . . . 19 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → (𝑔 (𝑢 + 𝑔) ↔ [𝑔] = [(𝑢 + 𝑔)] ))
105102, 104mpbird 247 . . . . . . . . . . . . . . . . . 18 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → 𝑔 (𝑢 + 𝑔))
1066ad2antrr 762 . . . . . . . . . . . . . . . . . . 19 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → 𝐺 ∈ Grp)
10738ad2antrr 762 . . . . . . . . . . . . . . . . . . 19 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → 𝐾𝑋)
108 eqid 2651 . . . . . . . . . . . . . . . . . . . 20 (invg𝐺) = (invg𝐺)
1097, 108, 60, 16eqgval 17690 . . . . . . . . . . . . . . . . . . 19 ((𝐺 ∈ Grp ∧ 𝐾𝑋) → (𝑔 (𝑢 + 𝑔) ↔ (𝑔𝑋 ∧ (𝑢 + 𝑔) ∈ 𝑋 ∧ (((invg𝐺)‘𝑔) + (𝑢 + 𝑔)) ∈ 𝐾)))
110106, 107, 109syl2anc 694 . . . . . . . . . . . . . . . . . 18 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → (𝑔 (𝑢 + 𝑔) ↔ (𝑔𝑋 ∧ (𝑢 + 𝑔) ∈ 𝑋 ∧ (((invg𝐺)‘𝑔) + (𝑢 + 𝑔)) ∈ 𝐾)))
111105, 110mpbid 222 . . . . . . . . . . . . . . . . 17 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → (𝑔𝑋 ∧ (𝑢 + 𝑔) ∈ 𝑋 ∧ (((invg𝐺)‘𝑔) + (𝑢 + 𝑔)) ∈ 𝐾))
112111simp3d 1095 . . . . . . . . . . . . . . . 16 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → (((invg𝐺)‘𝑔) + (𝑢 + 𝑔)) ∈ 𝐾)
113 oveq2 6698 . . . . . . . . . . . . . . . . . 18 (𝑥 = (((invg𝐺)‘𝑔) + (𝑢 + 𝑔)) → (𝑔 + 𝑥) = (𝑔 + (((invg𝐺)‘𝑔) + (𝑢 + 𝑔))))
114113oveq1d 6705 . . . . . . . . . . . . . . . . 17 (𝑥 = (((invg𝐺)‘𝑔) + (𝑢 + 𝑔)) → ((𝑔 + 𝑥) 𝑔) = ((𝑔 + (((invg𝐺)‘𝑔) + (𝑢 + 𝑔))) 𝑔))
115 eqid 2651 . . . . . . . . . . . . . . . . 17 (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔)) = (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔))
116 ovex 6718 . . . . . . . . . . . . . . . . 17 ((𝑔 + (((invg𝐺)‘𝑔) + (𝑢 + 𝑔))) 𝑔) ∈ V
117114, 115, 116fvmpt 6321 . . . . . . . . . . . . . . . 16 ((((invg𝐺)‘𝑔) + (𝑢 + 𝑔)) ∈ 𝐾 → ((𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔))‘(((invg𝐺)‘𝑔) + (𝑢 + 𝑔))) = ((𝑔 + (((invg𝐺)‘𝑔) + (𝑢 + 𝑔))) 𝑔))
118112, 117syl 17 . . . . . . . . . . . . . . 15 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → ((𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔))‘(((invg𝐺)‘𝑔) + (𝑢 + 𝑔))) = ((𝑔 + (((invg𝐺)‘𝑔) + (𝑢 + 𝑔))) 𝑔))
1197, 60, 32, 108grprinv 17516 . . . . . . . . . . . . . . . . . . 19 ((𝐺 ∈ Grp ∧ 𝑔𝑋) → (𝑔 + ((invg𝐺)‘𝑔)) = (0g𝐺))
120106, 98, 119syl2anc 694 . . . . . . . . . . . . . . . . . 18 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → (𝑔 + ((invg𝐺)‘𝑔)) = (0g𝐺))
121120oveq1d 6705 . . . . . . . . . . . . . . . . 17 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → ((𝑔 + ((invg𝐺)‘𝑔)) + (𝑢 + 𝑔)) = ((0g𝐺) + (𝑢 + 𝑔)))
1227, 108grpinvcl 17514 . . . . . . . . . . . . . . . . . . 19 ((𝐺 ∈ Grp ∧ 𝑔𝑋) → ((invg𝐺)‘𝑔) ∈ 𝑋)
123106, 98, 122syl2anc 694 . . . . . . . . . . . . . . . . . 18 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → ((invg𝐺)‘𝑔) ∈ 𝑋)
12467ad2antrr 762 . . . . . . . . . . . . . . . . . . . 20 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → 𝐻𝑋)
125124, 97sseldd 3637 . . . . . . . . . . . . . . . . . . 19 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → 𝑢𝑋)
1267, 60grpcl 17477 . . . . . . . . . . . . . . . . . . 19 ((𝐺 ∈ Grp ∧ 𝑢𝑋𝑔𝑋) → (𝑢 + 𝑔) ∈ 𝑋)
127106, 125, 98, 126syl3anc 1366 . . . . . . . . . . . . . . . . . 18 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → (𝑢 + 𝑔) ∈ 𝑋)
1287, 60grpass 17478 . . . . . . . . . . . . . . . . . 18 ((𝐺 ∈ Grp ∧ (𝑔𝑋 ∧ ((invg𝐺)‘𝑔) ∈ 𝑋 ∧ (𝑢 + 𝑔) ∈ 𝑋)) → ((𝑔 + ((invg𝐺)‘𝑔)) + (𝑢 + 𝑔)) = (𝑔 + (((invg𝐺)‘𝑔) + (𝑢 + 𝑔))))
129106, 98, 123, 127, 128syl13anc 1368 . . . . . . . . . . . . . . . . 17 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → ((𝑔 + ((invg𝐺)‘𝑔)) + (𝑢 + 𝑔)) = (𝑔 + (((invg𝐺)‘𝑔) + (𝑢 + 𝑔))))
1307, 60, 32grplid 17499 . . . . . . . . . . . . . . . . . 18 ((𝐺 ∈ Grp ∧ (𝑢 + 𝑔) ∈ 𝑋) → ((0g𝐺) + (𝑢 + 𝑔)) = (𝑢 + 𝑔))
131106, 127, 130syl2anc 694 . . . . . . . . . . . . . . . . 17 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → ((0g𝐺) + (𝑢 + 𝑔)) = (𝑢 + 𝑔))
132121, 129, 1313eqtr3d 2693 . . . . . . . . . . . . . . . 16 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → (𝑔 + (((invg𝐺)‘𝑔) + (𝑢 + 𝑔))) = (𝑢 + 𝑔))
133132oveq1d 6705 . . . . . . . . . . . . . . 15 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → ((𝑔 + (((invg𝐺)‘𝑔) + (𝑢 + 𝑔))) 𝑔) = ((𝑢 + 𝑔) 𝑔))
134 sylow2blem3.d . . . . . . . . . . . . . . . . 17 = (-g𝐺)
1357, 60, 134grppncan 17553 . . . . . . . . . . . . . . . 16 ((𝐺 ∈ Grp ∧ 𝑢𝑋𝑔𝑋) → ((𝑢 + 𝑔) 𝑔) = 𝑢)
136106, 125, 98, 135syl3anc 1366 . . . . . . . . . . . . . . 15 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → ((𝑢 + 𝑔) 𝑔) = 𝑢)
137118, 133, 1363eqtrd 2689 . . . . . . . . . . . . . 14 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → ((𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔))‘(((invg𝐺)‘𝑔) + (𝑢 + 𝑔))) = 𝑢)
138 ovex 6718 . . . . . . . . . . . . . . . 16 ((𝑔 + 𝑥) 𝑔) ∈ V
139138, 115fnmpti 6060 . . . . . . . . . . . . . . 15 (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔)) Fn 𝐾
140 fnfvelrn 6396 . . . . . . . . . . . . . . 15 (((𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔)) Fn 𝐾 ∧ (((invg𝐺)‘𝑔) + (𝑢 + 𝑔)) ∈ 𝐾) → ((𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔))‘(((invg𝐺)‘𝑔) + (𝑢 + 𝑔))) ∈ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔)))
141139, 112, 140sylancr 696 . . . . . . . . . . . . . 14 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → ((𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔))‘(((invg𝐺)‘𝑔) + (𝑢 + 𝑔))) ∈ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔)))
142137, 141eqeltrrd 2731 . . . . . . . . . . . . 13 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ (𝑢𝐻 ∧ (𝑢 · 𝑧) = 𝑧)) → 𝑢 ∈ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔)))
143142expr 642 . . . . . . . . . . . 12 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ 𝑢𝐻) → ((𝑢 · 𝑧) = 𝑧𝑢 ∈ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔))))
144143ralimdva 2991 . . . . . . . . . . 11 ((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) → (∀𝑢𝐻 (𝑢 · 𝑧) = 𝑧 → ∀𝑢𝐻 𝑢 ∈ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔))))
145144imp 444 . . . . . . . . . 10 (((𝜑 ∧ (𝑔𝑋𝑧 = [𝑔] )) ∧ ∀𝑢𝐻 (𝑢 · 𝑧) = 𝑧) → ∀𝑢𝐻 𝑢 ∈ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔)))
146145an32s 863 . . . . . . . . 9 (((𝜑 ∧ ∀𝑢𝐻 (𝑢 · 𝑧) = 𝑧) ∧ (𝑔𝑋𝑧 = [𝑔] )) → ∀𝑢𝐻 𝑢 ∈ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔)))
147 dfss3 3625 . . . . . . . . 9 (𝐻 ⊆ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔)) ↔ ∀𝑢𝐻 𝑢 ∈ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔)))
148146, 147sylibr 224 . . . . . . . 8 (((𝜑 ∧ ∀𝑢𝐻 (𝑢 · 𝑧) = 𝑧) ∧ (𝑔𝑋𝑧 = [𝑔] )) → 𝐻 ⊆ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔)))
149148expr 642 . . . . . . 7 (((𝜑 ∧ ∀𝑢𝐻 (𝑢 · 𝑧) = 𝑧) ∧ 𝑔𝑋) → (𝑧 = [𝑔] 𝐻 ⊆ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔))))
150149reximdva 3046 . . . . . 6 ((𝜑 ∧ ∀𝑢𝐻 (𝑢 · 𝑧) = 𝑧) → (∃𝑔𝑋 𝑧 = [𝑔] → ∃𝑔𝑋 𝐻 ⊆ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔))))
151150ex 449 . . . . 5 (𝜑 → (∀𝑢𝐻 (𝑢 · 𝑧) = 𝑧 → (∃𝑔𝑋 𝑧 = [𝑔] → ∃𝑔𝑋 𝐻 ⊆ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔)))))
152151com23 86 . . . 4 (𝜑 → (∃𝑔𝑋 𝑧 = [𝑔] → (∀𝑢𝐻 (𝑢 · 𝑧) = 𝑧 → ∃𝑔𝑋 𝐻 ⊆ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔)))))
15392, 152syl5bi 232 . . 3 (𝜑 → (𝑧 ∈ (𝑋 / ) → (∀𝑢𝐻 (𝑢 · 𝑧) = 𝑧 → ∃𝑔𝑋 𝐻 ⊆ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔)))))
154153rexlimdv 3059 . 2 (𝜑 → (∃𝑧 ∈ (𝑋 / )∀𝑢𝐻 (𝑢 · 𝑧) = 𝑧 → ∃𝑔𝑋 𝐻 ⊆ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔))))
15590, 154mpd 15 1 (𝜑 → ∃𝑔𝑋 𝐻 ⊆ ran (𝑥𝐾 ↦ ((𝑔 + 𝑥) 𝑔)))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 196 ∧ wa 383 ∧ w3a 1054 = wceq 1523 ∈ wcel 2030 ≠ wne 2823 ∀wral 2941 ∃wrex 2942 {crab 2945 ⊆ wss 3607 ∅c0 3948 𝒫 cpw 4191 {cpr 4212 class class class wbr 4685 {copab 4745 ↦ cmpt 4762 ran crn 5144 Fn wfn 5921 ‘cfv 5926 (class class class)co 6690 ↦ cmpt2 6692 Er wer 7784 [cec 7785 / cqs 7786 Fincfn 7997 0cc0 9974 · cmul 9979 − cmin 10304 / cdiv 10722 ℕcn 11058 ℕ0cn0 11330 ℤcz 11415 ↑cexp 12900 #chash 13157 ∥ cdvds 15027 ℙcprime 15432 pCnt cpc 15588 Basecbs 15904 ↾s cress 15905 +gcplusg 15988 0gc0g 16147 Grpcgrp 17469 invgcminusg 17470 -gcsg 17471 SubGrpcsubg 17635 ~QG cqg 17637 pGrp cpgp 17992 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1762 ax-4 1777 ax-5 1879 ax-6 1945 ax-7 1981 ax-8 2032 ax-9 2039 ax-10 2059 ax-11 2074 ax-12 2087 ax-13 2282 ax-ext 2631 ax-rep 4804 ax-sep 4814 ax-nul 4822 ax-pow 4873 ax-pr 4936 ax-un 6991 ax-inf2 8576 ax-cnex 10030 ax-resscn 10031 ax-1cn 10032 ax-icn 10033 ax-addcl 10034 ax-addrcl 10035 ax-mulcl 10036 ax-mulrcl 10037 ax-mulcom 10038 ax-addass 10039 ax-mulass 10040 ax-distr 10041 ax-i2m1 10042 ax-1ne0 10043 ax-1rid 10044 ax-rnegex 10045 ax-rrecex 10046 ax-cnre 10047 ax-pre-lttri 10048 ax-pre-lttrn 10049 ax-pre-ltadd 10050 ax-pre-mulgt0 10051 ax-pre-sup 10052 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3or 1055 df-3an 1056 df-tru 1526 df-fal 1529 df-ex 1745 df-nf 1750 df-sb 1938 df-eu 2502 df-mo 2503 df-clab 2638 df-cleq 2644 df-clel 2647 df-nfc 2782 df-ne 2824 df-nel 2927 df-ral 2946 df-rex 2947 df-reu 2948 df-rmo 2949 df-rab 2950 df-v 3233 df-sbc 3469 df-csb 3567 df-dif 3610 df-un 3612 df-in 3614 df-ss 3621 df-pss 3623 df-nul 3949 df-if 4120 df-pw 4193 df-sn 4211 df-pr 4213 df-tp 4215 df-op 4217 df-uni 4469 df-int 4508 df-iun 4554 df-disj 4653 df-br 4686 df-opab 4746 df-mpt 4763 df-tr 4786 df-id 5053 df-eprel 5058 df-po 5064 df-so 5065 df-fr 5102 df-se 5103 df-we 5104 df-xp 5149 df-rel 5150 df-cnv 5151 df-co 5152 df-dm 5153 df-rn 5154 df-res 5155 df-ima 5156 df-pred 5718 df-ord 5764 df-on 5765 df-lim 5766 df-suc 5767 df-iota 5889 df-fun 5928 df-fn 5929 df-f 5930 df-f1 5931 df-fo 5932 df-f1o 5933 df-fv 5934 df-isom 5935 df-riota 6651 df-ov 6693 df-oprab 6694 df-mpt2 6695 df-om 7108 df-1st 7210 df-2nd 7211 df-wrecs 7452 df-recs 7513 df-rdg 7551 df-1o 7605 df-2o 7606 df-oadd 7609 df-omul 7610 df-er 7787 df-ec 7789 df-qs 7793 df-map 7901 df-en 7998 df-dom 7999 df-sdom 8000 df-fin 8001 df-sup 8389 df-inf 8390 df-oi 8456 df-card 8803 df-acn 8806 df-cda 9028 df-pnf 10114 df-mnf 10115 df-xr 10116 df-ltxr 10117 df-le 10118 df-sub 10306 df-neg 10307 df-div 10723 df-nn 11059 df-2 11117 df-3 11118 df-n0 11331 df-xnn0 11402 df-z 11416 df-uz 11726 df-q 11827 df-rp 11871 df-fz 12365 df-fzo 12505 df-fl 12633 df-mod 12709 df-seq 12842 df-exp 12901 df-fac 13101 df-bc 13130 df-hash 13158 df-cj 13883 df-re 13884 df-im 13885 df-sqrt 14019 df-abs 14020 df-clim 14263 df-sum 14461 df-dvds 15028 df-gcd 15264 df-prm 15433 df-pc 15589 df-ndx 15907 df-slot 15908 df-base 15910 df-sets 15911 df-ress 15912 df-plusg 16001 df-0g 16149 df-mgm 17289 df-sgrp 17331 df-mnd 17342 df-submnd 17383 df-grp 17472 df-minusg 17473 df-sbg 17474 df-mulg 17588 df-subg 17638 df-eqg 17640 df-ga 17769 df-od 17994 df-pgp 17996 This theorem is referenced by: sylow2b 18084
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Find out about the five-term project (January 2014 to July 2015) which NRICH is leading in conjunction with Haringey Council, funded by London Schools Excellence Fund.
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Is problem solving at the heart of your curriculum? In this article for teachers, Lynne explains why it should be.
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In this article for teachers, Lynne explains the difference between 'rich tasks' and 'low threshold high ceiling' tasks, using examples from the website.
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This short article outlines a few activities which make use of interlocking cubes.
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An article for teachers which first appeared in the MA's Equals journal, featuring activities which use counters.
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An outline of 'Everyday Maths', a project run by Bristol University, working with parents of Year 3/4 children.
### ACME Report: Developing Able Young Mathematicians
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Lynne McClure gives an overview of the ACME report 'Raising the bar: developing able young mathematicians', published in December 2012.
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An article describing what LTHC tasks are, and why we think they're a good idea.
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Jenny Piggott reflects on the event held to mark her retirement from the directorship of NRICH, but also on problem solving itself.
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Jennifer Piggott and Steve Hewson write about an area of teaching and learning mathematics that has been engaging their interest recently. As they explain, the word ‘trick’ can be applied to. . . .
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Alf and Tracy explain how the Kingsfield School maths department use common tasks to encourage all students to think mathematically about key areas in the curriculum.
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7 core tips for effective studying
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Doug has just finished the first year of his undergraduate engineering course at Cambridge University. Here he gives his perspectives on engineering.
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Teachers who participated in an NRICH workshop produced some posters suggesting how they might use a tessellation interactivity in a range of situations.
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Ideas to support mathematics teachers who are committed to nurturing confident, resourceful and enthusiastic learners.
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In this article, Jennifer Piggott talks about just a few of the problems with problems that make them such a rich source of mathematics and approaches to learning mathematics.
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Helen Joyce interviews the neuropsychologist Brian Butterworth whose research has shown that we are all born with a "built-in" sense of cardinal number.
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Avril Crack describes how she went about planning and setting up a Maths trail for pupils in Bedfordshire.
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Jenny Murray writes about the sessions she leads in schools for parents to work alongside children on mathematical problems, puzzles and games.
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The second in a series, this article looks at the possible opportunities for children who operate from different intelligences to be involved in "typical" maths problems.
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This article, the first in a series, discusses mathematical-logical intelligence as described by Howard Gardner.
### Path of Discovery Series 3: I Do and I Understand
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Marion Bond recommends that children should be allowed to use 'apparatus', so that they can physically handle the numbers involved in their calculations, for longer, or across a wider ability band,. . . .
### Path of Discovery Series: 2. First Steps
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This article takes a closer look at some of the toys and games that can enhance a child's mathematical learning.
### Path of Discovery Series: 1. Uncertain Beginnings
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Marion Bond suggests that we try to imagine mathematical knowledge as a broad crazy paving rather than a path of stepping stones. There is no one right place to start and there is no one right route. . . . | 1,775 | 7,797 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-04 | longest | en | 0.879205 |
https://physics.meta.stackexchange.com/questions/12620/creating-a-community-wiki-post-for-special-relativity-questions-about-comparing | 1,679,321,040,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943483.86/warc/CC-MAIN-20230320114206-20230320144206-00245.warc.gz | 522,617,838 | 29,845 | # Creating a community wiki post for special relativity questions about comparing events in different reference frames
As I am sure many are aware, we get a fair amount of questions about special relativity that are essentially variances of
Alice sees this event happen at this time over here, and Bob sees this event happen at this time over there, so then how can Alice see this if Bob sees this?
Of course questions like these don't necessarily need to involve an "Alice" and a "Bob", but the questions essentially boil down to how to reconcile time dilation, length contraction, relativity of simultaneity, etc. of various events as seen by various observers. Some (or many?) of these questions get closed/ignored due to quality, clarity, or "homework-ness", but there are also ones that survive and receive answers.
Some recent examples of questions like these, or questions recent posts have been duplicates of:
High Fives on the Relativistic Train of Death
Does Special Relativity Imply Multiple Realities?
Question about Lorentz transformations with time
Special Relativity - Regarding the Simultaneity of Events During the Train Paradox
2 Different Results for Length Contraction
I am sure there are better examples out there, but hopefully the type of question I am am referring to is evident. Questions like these can easily become convoluted in trying to specify when/where events occur in the various frames, what defines the events, etc. But, at least to me, it seems like many of these questions could "easily" be resolved with the appropriate application of Lorentz transformations, rather than some of the introductory equations of time dilation and length contraction.
Additionally, I know there are already great community wiki posts for other SR topics, and these serve as great posts to link to as a duplicate for other posts asking about these topics:
How can time dilation be symmetric?
What is the proper way to explain the twin paradox?
I am wondering if we can make a post like the two above that have a similar purpose.
Therefore, would it be useful to create a community wiki post that covers how to handle "event comparison" questions described above, or are the questions of this type actually not as frequent as I think or already adequately handled so that a community wiki post would not be useful? If one such post would be useful, what would this post look like, both in terms of the "question" that sets up the post and the type of answer(s) that would be ideal for such a post? Additionally, who would be responsible for creating the post/answer? I am by far an expert in SR, so I don't think I could handle creating an answer (even if it is open for the public to edit and make better).
• I think there's some sort of rule that all SR & GR "FAQ" type questions just be written (and answered) by John Rennie. Dec 27, 2019 at 17:27
• @KyleKanos I mean who else would do it? Dec 27, 2019 at 17:34
• Could whoever gave the downvote explain why they think this isn't something that is needed? Dec 28, 2019 at 2:59
• Good idea! I'm very much in favor of this Dec 28, 2019 at 9:14
• The core problem is with new users not searching properly, if at all. So while the idea is not without merit it will not solve the issues of duplication, homeworkness etc. If only people cared to check even the suggested links before submitting a question... but one cannot force a donkey to drink, even it you bring it water. Dec 31, 2019 at 3:06
• @ZeroTheHero I wasn't proposing this to fix those issues. Just a post that more experienced users can link to as a duplicate, just like how many twin paradox posts can be marked as a duplicate of the twin paradox community wiki. Dec 31, 2019 at 3:44
• Right but if you can identify a problem that this would solve then I’d happily support. Don’t get me wrong: there’s a seed of idea here but I’m not sure - short of writing something that new users would not read - how the site is ahead. Maybe this needs to be fleshed out a little more? Dec 31, 2019 at 5:17
This doesn't seem like a good idea to me. The question would have to be extremely broad and vague. The answer would also have to be extremely broad and vague, something like "Use spacetime diagrams and the Lorentz transformation. Don't try to figure out all of special relativity using length contraction and time dilation. It doesn't work that way."
Possibly it would be useful to create a Q&A where the question was something like, "Can special relativity be reduced to length contraction and time dilation?," and the answer was an explanation of why not, with an explanation of spacetime diagrams and the Lorentz transformation. Is this different from what you're proposing?
• Thanks for the thoughts. I'm not proposing anything specific. I just was thinking about the possibility of having something to direct users to who ask questions like the ones I discuss in the question. Dec 29, 2019 at 14:46
While I really like the idea at a naive level, I feel that it's implementation would be difficult at best and perhaps practically impossible.
The proposal is, essentially, to write the definitive introduction to relativistic mechanics - the one that somehow answers everyone's questions in the order in which they might ask them. I think that there are a variety of sources for learning relativistic mechanics and if the people asking questions were that way inclined, they could learn directly from these.
What Physics Stack Exchange does is allow people who do not feel confident with such resources to ask a question in their own way looking for an answer that respects the way in which the question was asked. Even when a question is a duplicate, it is often asked in a slightly different way, and the fact of it being a duplicate is part of the answer provided by the contributors of answers on stack exchange.
Telling people to go read the Wiki, while it sounds like what they should be doing, is likely to be off-putting to someone who is feeling confused about the concept behind the question. It is the one-on-one sense of support that makes stack exchange different from, say, the Wikipedia - which has many excellent articles on introductory mathematics and physics.
• I'm not proposing a post that covers all of introductory relativistic mechanics. Just a post that takes posts that are trying to compare events using time dilation/length contraction and show how Lorentz transformations (1D would be sufficient) are a better, more general way to approach the comparison of various events as viewed in different frames. Dec 31, 2019 at 1:05
• I am with you in intent. But, how do you get people to notice that this generic answer exits? Or is it that we should have the definitive self-answered question on the topic which people direct the original poster to? I suppose if we could get community support for a group of such definitive questions, that might prevent some otherwise wasted effort? Jan 1, 2020 at 23:33
• The point isn't for people to notice it exists, although that would certainly be ideal. It would exist to serve as a question to link to as a duplicate, just like the other community wikis I have linked to in my question. If users fail to see the post, then others could still direct them to it and close their question as a duplicate. Jan 1, 2020 at 23:36
• Okay, I see the idea. But, I remain unconvinced that it is really any different from everyone linking to a popular answer to the duplicated question. I think that the personal service is part of what attracts people to a forum rather than the Wikipedia - which I would not expect us to improve on as far as being a wiki is concerned. Jan 3, 2020 at 0:16
• So you think the community wikis that are already established should not have been made? Jan 3, 2020 at 1:37
• Well, this question has caused me to think a lot on the issue. For something for which there is not another commonly known wiki, I can see a point, but don't feel it's core to what is important about Stack Exchange. For something for which well written wiki's are available, I feel it is not Stack Exchange's role to compete. What Stack Exchange does is give personal service (so to speak). You get to talk to a person, not directed to a FAQ. Jan 6, 2020 at 2:58
• I'm not proposing to do something that reflects the "core of Stack Exchange", nor am I asking for a competition with other sites. It would essentially be just like a normal post you would see on SE, but more free for others to edit as they see fit. Jan 6, 2020 at 3:02
• Could you not do this unilaterally - with encouragment to those with enough reputation points to edit - and then link to this in your answers? If that became popular you would get it by default. And if not, then the question of whether has been answered in a practical sense. Jan 7, 2020 at 4:49
• A friend of mine suggested that perhaps in the answer archetype a list of duplications should be kept. Jan 15, 2020 at 0:17 | 2,001 | 8,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-14 | latest | en | 0.949486 |
http://www.cs.utep.edu/vladik/cs1401.13/test1k.html | 1,511,414,938,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806736.55/warc/CC-MAIN-20171123050243-20171123070243-00557.warc.gz | 349,835,988 | 2,120 | ## CS 1401, Exam #1, MW 12-1:20 version
Date: Wednesday, September 18, 2013
Name (please type legibly, ideally in block letters): ______________________________________________________________________
1. On September 18, 1998, the International Corporation for Assigned Names and Numbers (ICANN) was started, an organization that controls the naming of websites.
• Explain how Java is different from all other programming languages, and how this difference helps to transfer computations from one computer to another.
• What programming language(s) was Java based on?
For extra credit: describe one more event from the history of computing.
```
```
2. For each of the following sequences of symbols, describe which can be valid Java identifiers and which cannot be; if you believe they cannot be, briefly explain why (e.g., "is a reserved word" or "does not start with a letter"):
• ICANN
• double
• 1998
• 18September
• Sept-18-1998
```
```
3. The following formula enables us to compute the area a of a right triangle with sides x and y:
``` 1
a = - xy
2
```
Assuming that x and y are already placed in the corresponding variables of type double, write a Java code statement for assigning the corresponding value to the variable a of type double. Explain, step-by-step, which arithmetic operations will be performed first, which next, etc., and trace the computations on the above example. Describe two different ways to avoid getting 0 as the result of evaluating 1/2. Explain what happens if you simply write xy in your Java code.
```
```
4-5. To register a website for several years, you need to pay a per-year registration fee. For simplicity, let us take \$35 as this fee. Write the main method which asks the user for the URL of the website, asks for how many years we want to register this website, and prints a memo describing the price. For example, if we want to register CS website http://www.cs.utep.edu for 3 years, your program should print the following message:
```From: ICANN
To register your website http://www.cs.utep.edu, you need to pay
\$35 X 3 = \$105.
```
Declare 35 as an integer constant, so that it will be easy to change if needed.
Reminder: to read from the keyboard, you can define the reader as follows:
```Scanner reader = new Scanner(System.in);
```
the header of the main method is:
```public static void main(String[] args){
```
```
```
6. Suppose that would like to add 2 to the number of years. If the number of years is stored in the integer variable years, which of the two lines of code leads to a correct increase:
• years = years + 2.0;
• years = years + 2;
If originally, before each line, we had 3 years, explain what will happen after each of these lines is implemented by Java. What is a clearer way (different from those above) to add 2 to the variable years? | 663 | 2,830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-47 | latest | en | 0.898447 |
https://www.tessshebaylo.com/one-solution-linear-equation-definition/ | 1,723,128,080,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640728444.43/warc/CC-MAIN-20240808124926-20240808154926-00457.warc.gz | 776,698,198 | 14,390 | # One Solution Linear Equation Definition
By | December 11, 2018
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Linear equations definition solving equation wikipedia with zero one or 2x y 2 and 10x 5y 3 have solution systems of explained in two variables solutions to no using determinants when does a
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 467 | 2,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-33 | latest | en | 0.742264 |
http://www.answers.com/topic/hawksley-workman | 1,550,351,506,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481122.31/warc/CC-MAIN-20190216210606-20190216232606-00613.warc.gz | 308,332,681 | 52,580 | ## Results for: Hawksley-workman
In Personal Finance
# What are the 5Cs of credit?
5 C's of Credit refer to the factors that lenders of money evaluate to determine credit worthiness of a borrower. They are the following:. 1. Borrower's CHARACTER. 2. Borrow (MORE)
In Acronyms & Abbreviations
# What does 5c stand for?
The Iphone 5C is Iphone 5Colorful 5c can also stand for thenumber 500 ("c" is the Roman numeral for 100) or for 5 degreesCelsius (centigrade) . +++ . "5c" can not stand fo (MORE)
In Coins and Paper Money
# What animal is on a 5c coin?
There are multiple animals on 5 cent coins depending on the country and time period such as the Buffalo on the US "buffalo nickel", the Beaver on the Canadian nickel, etc.
In Math and Arithmetic
# What is -5c plus 9 and how?
You can't tell a thing about -5c+9 until you know what 'c' is. And every time 'c' changes, -5c+9 changes.
In Volume
# What is 5c in milliliters?
5cc? cc means cubic centimetres which is equal to ml, so 5ml. if you mean cl, then that is equal to 50ml
In Numerical Analysis and Simulation
# What is the answer for 5c equals -75?
The 'answer' is the number that 'c' must be, if 5c is really the same as -75. In order to find out what number that is, you could use 'algebra'. First, write the equatio (MORE)
In Uncategorized
# When did the career of Hawksley Workman take off?
Hawksley Workman's career took off with the release of his second album '(Last Night we were) The Delicious Wolves' in 2001. He released 2 successful singles from the album 'S (MORE)
In iPhone 5
# How many pixels does the iPhone 5c have?
The iPhone 5c is 640 x 1136 pixels. That is about 326 pixels persquare inch (ppi).
In Temperature
# What is minus 5c in Fahrenheit?
(-5) degrees Celsius = 23 degrees Fahrenheit. Formula: [°F] = [°C] à 9 â 5 + 32
In iPhone 5
# How many inches is a iPhone 5c?
The screen is 4" big. The height is 4.9", width is 2.33" and thedepth is 0.35" | 574 | 1,951 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2019-09 | latest | en | 0.92183 |
http://www.verycomputer.com/18_0a8cd5d68aa19cf8_1.htm | 1,582,508,946,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145869.83/warc/CC-MAIN-20200224010150-20200224040150-00158.warc.gz | 235,682,707 | 4,626 | ## Putting together TeX boxes
### Putting together TeX boxes
Hi,
I am facted with the following problem: I have to vertical boxes
(\vbox) of different height. No I want to glue them together in a
horizontal box (\hbox) such that the top lienes of the two boxes are
on the same line (see figure below). Who can I do this, preferably in
plain TeX (but*is OK too)?
Please reply by e-mail. I will summarize to the net!
Figure
+----------------------------+ +-----------------+
| Box 1 | | |
+----------------------------+ | Box 2 |
| |
+-----------------+
Claude
-----------------------------------------------------------------------------
Claude G. Diderich PGP V2.3 public key available
Swiss Federal Institute of Technology, Lausanne -----------------------------
Department of Computer Science Fields of interest:
Computer Science Theory Laboratory - Complexity theory
CH-1015 Lausanne (Switzerland - Europe) - Combinatorial optimization
Phone: (021)/693-52-86 - Parallel computations
-----------------------------------------------------------------------------
### Putting together TeX boxes
Some time ago I posted the following question to the TeX net.
Quote:> I am faced with the following problem: I have to vertical boxes
> (\vbox) of different height. No I want to glue them together in a
> horizontal box (\hbox) such that the top lienes of the two boxes are
> on the same line (see figure below). Who can I do this, preferably in
> plain TeX (but*is OK too)?
> Please reply by e-mail. I will summarize to the net!
> Figure
> +----------------------------+ +-----------------+
> | Box 1 | | |
> +----------------------------+ | Box 2 |
> | |
> +-----------------+
Here are some of the repies I received:
---------------------------------------------------------------------------
If you want an easy*solution, try
\parbox[t]{3in}{ First box here } \hspace{ a little space if needed }%
\parbox[t]{3in}{ Second box here }\\
---------------------------------------------------------------------------
Did you try \hbox{\vtop{stuff}\vtop{stuff}} ?
\vbox aligns with the bottom baseline rather than the top.
---------------------------------------------------------------------------
\line{\vtop{ ... Box 1 ...}\hfill\vtop{... Box 2 ...}}
should do what you want: if the boxes are \vbox's their bottoms
are lined up, but if they are \vtop's their tops are aligned.
---------------------------------------------------------------------------
The best thing to do is use \vtop rather than \vbox. \vtop is exactly
like \vbox, except the reference point is always one line from the top
of the box (off hand, I think it's really at the reference point of the
first box it encloses). When I have this problem, I usually have text
in the \vboxes, so \vtop does what I want.
The second best thing is to use \setbox and fiddle with heights:
% first use the scratch boxes \box0 and \box2 to hold the
% contents of the two boxes in your figure
\setbox0=\vbox{<whatever's in Box 1>}
\setbox2=\vbox{<whatever's in Box 2>}
% then fiddle with the height of box 2
\skip2=\ht 2 % skip 2 has the height of box 2
\ht 2=\ht 0 % set height of box 2 to be the same as box 1
% and finally adjust the depth of box 2 to take up the
% slack. We can't \advance \dp2 by \skip2, incidentally
\advance\skip2 by -\ht 2
\advance\skip2 by \dp 2
\dp 2=\skip2
% you need to be careful that you don't do anything with
% \box0, \box2, or \skip2 while this is going on. If in
% doubt, define registers with \newbox and \newskip
% oh -- assemble in an \hbox
\hbox{\box 0 \hskip 4 pt \box 2}
Hope this helps.
---------------------------------------------------------------------------
In fact only the last reply solved my problem. In fact the two boxes I want
to align contain figures. Therefore aligning on the top baseline doesn't
change the problem.
Claude
-----------------------------------------------------------------------------
Claude G. Diderich PGP V2.3 public key available
Swiss Federal Institute of Technology, Lausanne -----------------------------
Department of Computer Science Fields of interest:
Computer Science Theory Laboratory - Complexity theory
CH-1015 Lausanne (Switzerland - Europe) - Combinatorial optimization
Phone: (021)/693-52-86 - Parallel computations
-----------------------------------------------------------------------------
In summary: I seek a style file for use with TeX which will allow me to
put incidental text in a framed box, perhaps with a slightly shaded
background, and allow the box to float to an appropriate position close
to the text it complements. If in addition the boxes could be numbered
then you would make my day.
Dear all,
I am currently in the process of writing up my thesis.
I would like to include some little pieces of additional
information which is aside to the main text. For example,
it would be nice to have a little biography of James Clerk Maxwell
beside his equations.
I particularly like the way that this sort of thing has been done by
Aki and Richards (seismologists know who these are) but if you read
the NewScientist you will also be familiar with the little boxes
containing extra information.
I could probably just about write a tex style file to do this, but I'd
rather not re-invent the wheel and so I am hoping someone out there will
be able to help me. If someone has a partially working style file I
would be happy to try help make it complete.
Many thanks,
University of Edinburgh | mathematicians and all those who make
Dept of Geology and Geophysics | empty prophecies. The danger already
http://www.glg.ed.ac.uk/~ajsw | exists that mathematicians have made a
phone +44 131 650 8533 | covenant with the devil to darken the
fax +44 131 668 3184 | spirit and confine man in the bonds of
| Hell." -- St. Augustine | 1,470 | 6,257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-10 | latest | en | 0.513172 |
https://math.answers.com/Q/Which_number_is_bigger_78_decimeters_or_10_meters_why | 1,709,611,712,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948217723.97/warc/CC-MAIN-20240305024700-20240305054700-00714.warc.gz | 399,442,743 | 46,203 | 0
# Which number is bigger 78 decimeters or 10 meters why?
Updated: 10/18/2022
Wiki User
9y ago
10 m = 100 dm
100 dm > 78 dm
10 m > 78 dm
Wiki User
9y ago
Earn +20 pts
Q: Which number is bigger 78 decimeters or 10 meters why?
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Related questions
### Is 2 meters bigger the 15 decimeters?
Yes, 15 decimeters is 1.5 meters. There are 10 decimeters per meter.
### Is 4 decimeters bigger than 4 meters?
1 meter is 10 decimeters, so 4 meters would be 40 decimeters. therefore your answer is NO
### Which is bigger 4 meters or 400 decimeters?
400 decimeters. 1 meter equals 10 decimeters.
### 400 decimeters equals how many meters?
A meter has 10 decimeters, so just divide the number of decimeters by 10.A meter has 10 decimeters, so just divide the number of decimeters by 10.A meter has 10 decimeters, so just divide the number of decimeters by 10.A meter has 10 decimeters, so just divide the number of decimeters by 10.
### Which operation should you use to change meters to decimeters?
To change meters to decimeters, you should multiply the number of meters by 10, as there are 10 decimeters in one meter.
### What is bigger decimeter or meters?
Meter is bigger than decimeter. 1 meter = 10 decimeters
### How many meters are in 100 decimeters?
1 Decimeter = 10 meters. So multiply the number of decimeters by 10 to get the number of meters. 10 Decimeters = 100 meters. * * * * * What a load of rubbish! 10 decimetre = 1 metre (not the other way round). So 100 decimetres = 100/10 = 10 metres.
### How many decimeters are in 9.1 meters?
1 meter is 10 decimeters. Then, 9.1 meters is 91 decimeters. To show that, multiply 10 by 9.1 to get 91 decimeters. Formally, 9.1 meters * 10 decimeters / meters = 91 decimeters
### Is decimeter greater than meters?
yes decimeters are indeed bigger or as you might say greater than meters example: 9 decimeters= _ meters 9 times 10 = 90
### How many decimeters are in 10 meters?
1 meter = 10 decimeters There are 10 decimeters in 1 meter or 10/10, which equals 1.
### How many decimeter are in 8000 meters?
There are 80,000 decimeters in 8000 meters. 8,000 meters x 10 decimeters/1 meters = 80,000 decimeters 1 meter = 10 decimeters
### How do you calculate decimeters into meters?
Divide decimeter by 10 and you get meters. Multiply meters by 10 and you get decimeters. 1 decimeter is 1/10 of a meter. There are 10 decimeters in a meter. | 696 | 2,437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-10 | latest | en | 0.905302 |
http://www.bosspintar.com/blog/kitchenaid-mini-jsah/15f8d7-complete-graph-with-5-vertices | 1,621,326,209,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989756.81/warc/CC-MAIN-20210518063944-20210518093944-00545.warc.gz | 61,610,949 | 9,417 | Next Qn. Definition: Complete. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Example2: Show that the graphs shown in fig are non-planar by finding a subgraph homeomorphic to K 5 or K 3,3. Now, for a connected planar graph 3v-e≥6. It erases all existing edges and edge properties, arranges the vertices in a circle, and then draws one edge between every pair of vertices. C 5. Can a simple graph exist with 15 vertices each of degree 5 ? W 4 Dl{ back to top. There is then only one choice for the last city before returning home. 2 Paths After all of that it is quite tempting to rely on degree sequences as an infallable measure of isomorphism. Substituting the values, we get-3 x 4 + (n-3) x 2 = 2 x 21. a) (n*(n+1))/2 b) (n*(n-1))/2 c) n d) Information given is insufficient View Answer . answered Jan 27, 2018 Salazar. Had it been If the simple graph G has 5 vertices and 7 edges, how many edges does G have ? Ask Question Asked 7 years, 7 months ago. Viewed 425 times 0 $\begingroup$ If a graph has 5 vertices, all of them connected to each other vertex, how many different spanning trees exist? We denote by C n a complete convex geometric graph with n vertices, i.e., a complete geometric graph whose vertices are in convex position (note that all these graphs are weakly isomorphic to each other). The number of edges in a complete bipartite graph is m.n as each of the m vertices is connected to each of the n vertices. Example: Draw the complete bipartite graphs K 3,4 and K 1,5. For convenience, suppose that n is a multiple of 6. Complete Graph: A simple undirected graph can be referred to as a Complete Graph if and only if the each pair of different types of vertices in that graph is connected with a unique edge. Complete Graphs The number of edges in K N is N(N 1) 2. The task is to calculate the total weight of the minimum spanning tree of this graph. True False 1.2) A complete graph on 5 vertices has 20 edges. Question: True Or False: A Complete Graph With Five Vertices Has An Euler Circuit. Consider a complete graph G. n >= 3. a. The bull graph is planar with chromatic number 3 and chromatic index also 3. sage: g. order (); g. size 5 5 sage: g. radius (); g. diameter (); g. girth 2 3 3 sage: g. chromatic_number 3. nC2 = n!/(n-2)!*2! 2n = 36 ∴ n = 18 . (5 points, 1 point for each) True/False Questions 1.1) In a simple graph on n vertices, the degree of a vertex is at most n - 1. B Contains a circuit. Now give an Euler trail through the graph with this new edge by listing the vertices in the order visited. complete graph K4. If we add all possible edges, then the resulting graph is called complete. True False 1.3) A graph on n vertices with n - 1 must be a tree. in Sub. Find the number of cycles in G of length n. b. Any scenario in which one wishes to examine the structure of a network of connected objects is potentially a problem for graph theory. Given an undirected weighted complete graph of N vertices. From each of those, there are three choices. Math. 5K 1 = K 5 D?? Suppose we had a complete graph with five vertices like the air travel graph above. Solution.Every vertex of a graph on n vertices has degree between 0 and n − 1. 5 vertices - Graphs are ordered by increasing number of edges in the left column. View Answer Answer: 6 30 A graph is tree if and only if A Is planar . 1.8.2. The list contains all 34 graphs with 5 vertices. Sum of degree of all vertices = 2 x Number of edges . Answer: b Explanation: Number of ways in which every vertex can be connected to each other is nC2. Active 7 years, 7 months ago. Any help would be appreciated, thanks. Algebra. Chromatic Number . Complete Graph draws a complete graph using the vertices in the workspace. So to properly it, as many different colors are needed as there are number of vertices in the given graph. The given Graph is regular. in Sub. 2 The sum of degrees of all vertices is even, but we can see ∑ v ∈ V deg (v) = 15 × 5 = 75 is odd. the problem is that you counted each edge twice - one time as $(u,v)$ and one time as $(v,u)$ so you need to divide by two, and then you get that you have $\frac {n(n-1)}{2}$ edges in a complete simple graph. My answer 8 Graphs : For un-directed graph with any two nodes not having more than 1 edge. A simple graph G ={V,E} is said to be complete if each vertex of G is connected to every other vertex of G. The complete graph with n vertices is denoted Kn. In our flrst example, Figure 2, we have two connected simple graphs, each with flve vertices. A graph with N vertices can have at max nC2 edges.3C2 is (3!)/((2!)*(3-2)!) Graph with 5 vertices - # of spanning trees. The maximum packing problem of K v with copies of G has been studied extensively for G=K 3,K 4,K 5,K 4 −e and for other specific graphs (see for references). = n(n-1)/2 This is the maximum number of edges an undirected graph can have. You should check that the graphs have identical degree sequences. → Related questions 0 votes. Solution: No, it can’t. Weight sets the weight of an edge or set of edges. Next → ← Prev. From each of those cities, there are two possible cities to visit next. (6) Suppose that we have a graph with at least two vertices. This is intuitive in the sense that, you are basically choosing 2 vertices from a collection of n vertices. Qn. Labeling the vertices v1, v2, v3, v4, and v5, we can see that we need to draw edges from v1 to v2 though v5, then draw edges from v2 to v3 through v5, then draw edges between v3 to v4 and v5, and finally draw an edge between v4 and v5. 1 answer. Select True Or False: The Koenisgburg Bridge Problem Is Not Possible Because Some Of The Vertices In The Graph That Represents The Problem Have An Odd Degree. There is a closed-form numerical solution you can use. => 3. claw ∪ K 1 DJ{ back to top. If a complete graph has n vertices, then each vertex has degree n - 1. There are exactly M edges having weight 1 and rest all the possible edges have weight 0. The bull graph has chromatic polynomial $$x(x - 2)(x - 1)^3$$ and Tutte polynomial $$x^4 + x^3 + x^2 y$$. [ Select] True Of False: The Koenisgburg Bridge Problem Is Not Possible Because An Euler Circuit Cannot Be Completed. with 5 vertices a complete graph can have 5c2 edges => 10 edges . a) True b) False View Answer. If G is a connected graph, then the number of bounded faces in any embedding of G on the plane is equal to A 3 . I The Method of Pairwise Comparisons can be modeled by a complete graph. If you are considering non directed graph then maximum number of edges is $\binom{n}{2}=\frac{n!}{2!(n-2)!}=\frac{n(n-1)}{2}$. In general, the best way to answer this for arbitrary size graph is via Polya’s Enumeration theorem. True False 1.4) Every graph has a spanning tree. The array arr[][] gives the set of edges having weight 1. Definition. In exercises 13-17 determine whether the graph is bipartite. From asking for help elsewhere I was told the formula for the number of subgraphs in a complete graph with n vertices is 2^(n(n-1)/2) In this problem that would give 2^3 = 8. B 4. So you can compute number of Graphs with 0 edge, 1 edge, 2 edges and 3 edges. Notes: ∗ A complete graph is connected ∗ ∀n∈ , two complete graphs having n vertices are isomorphic ∗ For complete graphs, once the number of vertices is From Seattle there are four cities we can visit first. W 4 DQ? What is the number of edges present in a complete graph having n vertices? Vertices in a graph do not always have edges between them. Recently, Zhang and Yin and Ge studied maximum packings of K v with copies of a graph G of five vertices having at least one vertex … Proof. The bull graph has 5 vertices and 5 edges. A complete graph is an undirected graph where each distinct pair of vertices has an unique edge connecting them. u can be any vertex that is not v, so you have (n-1) options for this. claw ∪ K 1 Ds? That is, a graph is complete if every pair of vertices is connected by an edge. D Is completely connected. 5. Hence, for K 5, we have 3 x 5-10=5 (which does not satisfy property 3 because it must be greater than or equal to 6). Show that it is not possible that all vertices have different degrees. K 5 D~{ back to top. comment ← Prev. Then G would've had 3 edges. The complete bipartite graph is an undirected graph defined as follows: . 21-25. The sum of all the degrees in a complete graph, K n, is n(n-1). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … suppose $(v,u)$ is an edge, then v can be any of the vertices in the graph - you have n options for this. Its vertex set is a disjoint union of a subset of size and a subset of size ; Its edge set is defined as follows: every vertex in is adjacent to every vertex in .However, no two vertices in are adjacent to each other, and no two vertices in are adjacent to each other. How many edges are in K15, the complete graph with 15 vertices. In a complete graph, each vertex is connected with every other vertex. Solution: The complete graph K 5 contains 5 vertices and 10 edges. We know that edges(G) + edges(G)=10 so edges(G)=10-7=3. Thus, Total number of vertices in the graph = 18. 1. In a complete graph, every vertex is connected to every other vertex. How many cycles in a complete graph with 5 vertices? Suppose are positive integers. Thus, K 5 is a non-planar graph. P 3 ∪ 2K 1 DN{ back to top. the other hand, the third graph contains an odd cycle on 5 vertices a,b,c,d,e, thus, this graph is not isomorphic to the first two. 5. Graph theory is the study of mathematical objects known as graphs, which consist of vertices (or nodes) connected by edges. Theorem 5 . Consider the graph given above. I This formula also counts the number of pairwise comparisons between N candidates (recall x1.5). The number of isomorphism classes of extendable graphs weakly isomorphic to C n is at least 2 Ω (n 4). 12 + 2n – 6 = 42. 5 Graph Theory Graph theory – the mathematical study of how collections of points can be con-nected – is used today to study problems in economics, physics, chemistry, soci- ology, linguistics, epidemiology, communication, and countless other fields. However, that would be a mistake, as we shall now see. Weights can be any integer between –9,999 and 9,999. Question 1. I Vertices represent candidates I Edges represent pairwise comparisons. A graph G = (V, E) is called a complete bipartite graph if its vertices V can be partitioned into two subsets V 1 and V 2 such that each vertex of V 1 is connected to each vertex of V 2. P 3 ∪ 2K 1 Do? 2n = 42 – 6. In the case of n = 5, we can actually draw five vertices and count. K 5 - e = 5K 1 + e = K 2 ∪ 3K 1 D?O K 5 - e D~k back to top. Complete Graphs- A complete graph is a graph in which every two distinct vertices are joined by exactly one edge. C Is minimally. D 6 . Its radius is 2, its diameter 3, and its girth 3. A basic graph of 3-Cycle. 29 Let G be a simple undirected planar graph on 10 vertices with 15 edges. We are done. (In the figure below, the vertices are the numbered circles, and the edges join the vertices.) The default weight of all edges is 0. Add an edge so the resulting graph has an Euler trail (without repeating an existing edge). Vertices have different degrees 13-17 determine whether the graph = 18. complete K! 3. a vertices each of degree 5, 2 edges and 3 edges complete if every pair of vertices an! N-3 ) x 2 = 2 x 21 vertices with 15 vertices. listing the vertices. 7 months.. Give an Euler trail ( without repeating an existing edge ) edge connecting them given an graph... The left column complete bipartite graphs K 3,4 and K 1,5 examine the structure of a of... The minimum spanning tree of this graph cities to visit next 2 = 2 x 21 is. V, so you have ( n-1 ) /2 this is intuitive in the order visited 3 ∪ 1! The array arr [ ] [ ] [ ] gives the set of edges you (... As we shall now see 1 must be a tree and 9,999 on n vertices with n - must! Explanation: number of cycles in G of length n. b that edges ( ... Extendable graphs weakly isomorphic to C n is n ( n-1 ) options for.! Be any integer between –9,999 and 9,999 x number of vertices in the case of n.., that would be a tree ( recall x1.5 ) False 1.4 ) every graph has n vertices with vertices... Task is to calculate the Total weight of an edge so the resulting graph has an Euler trail ( repeating. On degree sequences as an infallable measure of isomorphism classes of extendable graphs isomorphic. Not v, so you can compute number of edges in the order visited any integer between and. Colors are needed as there are exactly M edges having weight 1 and rest all the degrees in graph! Of edges an undirected graph defined as complete graph with 5 vertices:, is n ( n-1 ) /2 this is in... N is a graph with any two nodes not having more than 1 edge, 2 and... And rest all the degrees in a complete graph, K n is a closed-form numerical solution you can number... Graph above a spanning tree of degree 5 check that the graphs shown in fig are non-planar by finding subgraph! Nc2 = n ( n-1 ) /2 this is intuitive in the case of n,... Sequences as an infallable measure of isomorphism classes of extendable graphs weakly to. By finding a subgraph homeomorphic to K 5 or complete graph with 5 vertices 3,3 Circuit can not be Completed every pair vertices. From Seattle there are three choices with 5 vertices and count, 1 edge, 2 edges and 3.... However, that would be a simple graph G ) =10-7=3 the Total weight an! The list contains all 34 graphs with 5 vertices and 5 edges + (. Follows: each distinct pair of vertices is connected with every other vertex connected an. The weight of the minimum spanning tree would be a tree we add all possible edges have weight.... Our flrst example, figure 2, we can actually draw five vertices and 5 edges choosing... Task is to calculate the Total weight of an edge or set of edges in the figure below, complete! Subgraph homeomorphic to K 5 or K 3,3 should check that the shown! If and only if a is planar a collection of n =,. One choice for the last city before returning home as many different colors are needed there! Of spanning trees between 0 complete graph with 5 vertices n − 1 i edges represent pairwise comparisons one wishes examine... A multiple of 6 is 2, we can visit first ] ]! Distinct pair of vertices in the order visited possible that all vertices have different degrees from each those!, there are three choices DN { back to top the case of n = 5, we x... With n - 1 the degrees in a graph with five vertices like the air travel graph.. G. n > complete graph with 5 vertices 3. a other is nc2 of extendable graphs weakly to. Are in K15, the vertices are the numbered circles, and its girth 3 travel graph above 10... 5 contains 5 vertices ( n-1 ) /2 this is the study of mathematical objects known as,. C n is at least two vertices. ( n-1 ) options for.... Nc2 = n! / ( n-2 )! * 2 edge ) =10-7=3. 5 edges vertices have different degrees vertices is connected to every other vertex minimum spanning tree: number of in... ) x 2 = 2 x number of ways in which every vertex is by! Colors are needed as there are four cities we can visit first undirected weighted complete graph can have edges! If we add all possible edges have weight 0 flrst example, figure,... –9,999 and 9,999 graphs: for un-directed graph with any two nodes having... Mathematical objects known as graphs, each vertex is connected with every other vertex, number! To C n is n ( n-1 ) options for this connected every..., how many edges are in K15, the vertices are the numbered circles, its. Ordered by increasing number of ways in which every vertex is connected by edges degree between 0 and −. Weakly isomorphic to C n is n ( n-1 ) options for this ) connected by edges in our example... If every pair of vertices has degree between 0 and n − 1 of the... Circuit can not be Completed collection of n vertices has an unique edge connecting them it is not Because. The number of isomorphism answer: b Explanation: number of edges as infallable! ( without repeating an existing edge ) have edges between them those, there are exactly M edges weight! K n, is n ( n-1 ) options for this > 3.! 1 DJ { back to top 3, and its girth 3 Show that the shown., there are four cities we can visit first graph do not always have edges them! Graphs with 0 edge, 1 edge: the Koenisgburg Bridge Problem is not v, so you use... 15 vertices. we had a complete graph, each vertex is connected to every other vertex graphs shown fig!, you are basically choosing 2 vertices from a collection of n vertices, then the graph... Can not be Completed which every vertex is connected to every other vertex different degrees tree. Of False: the Koenisgburg Bridge Problem is not possible that all vertices 2! 8 graphs: for un-directed graph with this new edge by listing the vertices in the case n! Or nodes ) connected by edges cities to visit next, each with flve vertices ). Euler trail ( without repeating an existing edge ) different degrees of in... By a complete graph G. n > = 3. a weight sets the weight of edge! Draw the complete bipartite graph is tree if and only if a is planar vertex has degree between and. Visit first the vertices. is potentially a Problem for graph theory using the vertices in the workspace - are! Graph do not always have edges between them graph having n vertices with edges. The resulting graph is called complete: for un-directed graph with any two nodes having. ) a complete graph on n vertices. draw five vertices like the air graph... Edges an undirected graph can have 5c2 edges = > 10 edges ) x 2 = 2 x.... Having more than 1 edge do not always have edges between them scenario in which every vertex connected! Degree n - 1 must be a mistake, as many different are! N candidates ( recall x1.5 ) if and only if a complete graph with 5 vertices and edges. We know that edges ( G ) =10-7=3 weight of the minimum spanning tree of this.! That would be a simple graph G ` has 5 vertices is 2, we actually... Substituting the values, we can visit first always have edges between them x number of edges having 1... Sets the weight of an edge so the resulting graph is an undirected graph defined as follows: is... Structure of a graph on 5 vertices. of edges in K n, is n ( )! For the last city before returning home you have ( n-1 ) options for this objects as... V, so you have ( n-1 ) /2 this is intuitive in figure. Examine the structure of a network of connected objects is potentially a Problem for graph theory is complete graph with 5 vertices of! ∪ 2K 1 DN { back to top and 10 edges 1 must be a simple exist. Each distinct pair of vertices ( or nodes ) connected by edges 1.2 ) a graph is if. Can actually draw five vertices and 5 edges which consist of vertices in the complete graph with 5 vertices 5. Without repeating an existing edge ) graphs the number of edges present in complete! Are two possible cities to visit next the vertices in the left column of... X 21 a mistake, as many different colors are needed as are... It, as many different colors complete graph with 5 vertices needed as there are two possible cities to visit next vertices... That, you are basically choosing 2 vertices from a collection of complete graph with 5 vertices vertices has n! Possible cities to visit next, suppose that we have two connected simple graphs, each with flve vertices )... Any scenario in which one wishes to examine the structure of a graph in which one to... Examine the structure of a graph is tree if and only if a graph... Calculate the Total weight of the minimum spanning tree answer answer: b Explanation: number of pairwise between. Koenisgburg Bridge Problem is not v, so you have ( n-1 ) this. | 5,050 | 19,848 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-21 | latest | en | 0.905132 |
http://betterlesson.com/lesson/resource/2648850/measuring-with-two-different-tools-at-the-same-time-png | 1,488,220,892,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501173405.40/warc/CC-MAIN-20170219104613-00209-ip-10-171-10-108.ec2.internal.warc.gz | 31,340,848 | 21,903 | ## Measuring With Two Different Tools at the Same Time.png - Section 3: Measuring With Non Standard Units
Measuring With Two Different Tools at the Same Time.png
# Tiles and Cubes and Clips, Oh My!
Unit 8: Non Standard Measuring
Lesson 1 of 8
## Objective: SWBAT measure lengths using a variety of units. SWBAT measure accurately.
### Thomas Young
203 Lessons14 new
## Big Idea: Were off to get the measure, the wonderful measure of side! The students will identify the longest side of an object and measure its length with a variety of non standard units.
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### Thomas Young
203 Lessons | 14 new
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http://javaexplorer03.blogspot.com/2016/02/find-element-repeated-more-than-n2-times.html | 1,524,504,936,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946120.31/warc/CC-MAIN-20180423164921-20180423184921-00442.warc.gz | 170,390,359 | 17,784 | ## Tuesday, 16 February 2016
### Find the element repeated more than n/2 times
There is an array (of size N) with an element repeated more than N/2 number of time and the rest of the element in the array can also be repeated but only one element is repeated more than N/2 times. Find the number.
Approach#1
Keep the count of each number in a hash map.
Extra space required for this approach.
Approach#2
Simplest, sort the array and the number at n/2+1th index is the required number.
Time complexity to sort array is: O (nlogn).
Approach#3
Moore’s Voting Algorithm
1. Define two variables majority_elem to keep track of majority element and counter (count).
2. Initially we set the first element of the array as the majority element.
3. Traverse the array:
a. If the current element == majority_elem
Increment count
else
Decrement count
b. If count becomes zero,
Set count = 1
Set majority_elem = current element.
4. Print majority_elem.
array = [1, 2, 3, 4, 5, 5, 5, 5, 5 ]
majority_elem = items[0]
count = 1
for i ß 0 to end {
if (items[i] == majority_elem) {
count += 1;
} else {
count -= 1
}
if (count == 0) {
majority_elem = items[i];
count = 1;
}
}
print(majority_elem)
Note: For boundary condition, Check that the occurrence of element is more than n/2.
Intuition behind the algorithm:
Suppose that you were to have a roomful of people each holding one element of the array. Whenever two people find each other where neither is holding the same array element as the other, the two of them sit down. Eventually, at the very end, if anyone is left standing, there's a chance that they're in the majority, and you can just check that element. As long as one element occurs with frequency at least N/2, you can guarantee that this approach will always find the majority element. | 470 | 1,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-17 | latest | en | 0.887465 |
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# 11.2: From the Center of the Polygon
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Geometry, Chapter 10, Lesson 6.
## Problem 1 – Area of a Regular Pentagon
A regular polygon is a polygon that is equiangular and equilateral. The apothem of a regular polygon is a line segment from the center of the polygon to the midpoint of one of its sides.
Start the Cabri Jr. application by pressing APPS and selecting Cabri Jr. Open the file PENTAGON by pressing Y=\begin{align*}Y=\end{align*}, selecting Open..., and selecting the file. You are given regular pentagon ABCDE\begin{align*}ABCDE\end{align*} with center R\begin{align*}R\end{align*}. You are given the length of CD¯¯¯¯¯¯¯¯\begin{align*}\overline{CD}\end{align*} (side of the polygon), RM¯¯¯¯¯¯¯¯¯\begin{align*}\overline{RM}\end{align*} (apothem), and the area of the polygon.
1. Drag point D\begin{align*}D\end{align*} to 4 different positions and record the data in the table below. The perimeter and apothem multiplied by perimeter will need to be calculated. The perimeter is the number of sides multiplied the length of CD¯¯¯¯¯¯¯¯\begin{align*}\overline{CD}\end{align*}.
Position Apothem (a)\begin{align*}(a)\end{align*} Perimeter (p)\begin{align*}(p)\end{align*} ap\begin{align*}a \cdot p\end{align*} (apothem times perimeter) Area
1
2
3
4
2. Using the table discuss how the area and ap\begin{align*}a \cdot p\end{align*} are related?
## Problem 2 – Area of a Regular Hexagon
In the problem, you will repeat the process from Problem 1 for a regular hexagon. Open the file HEXAGON showing a regular hexagon with center R\begin{align*}R\end{align*}. You are given the length of CD¯¯¯¯¯¯¯¯\begin{align*}\overline{CD}\end{align*} (side of the polygon), RM¯¯¯¯¯¯¯¯¯\begin{align*}\overline{RM}\end{align*} (apothem), and the area of the polygon.
3. Drag point D\begin{align*}D\end{align*} to 4 different positions and record the data in the table. The perimeter and apothem multiplied by perimeter will need to be calculated. The perimeter is the number of sides multiplied by the length of CD¯¯¯¯¯¯¯¯\begin{align*}\overline{CD}\end{align*}.
Position Apothem (a)\begin{align*}(a)\end{align*} Perimeter (p)\begin{align*}(p)\end{align*} ap\begin{align*}a \cdot p\end{align*} (apothem times perimeter) Area
1
2
3
4
4. Using the data you found in both problems, give formula for the area of a regular polygon.
## Problem 3 – Area of a Regular Polygon
Now you will look at the proof of the formula for the area of a regular polygon. Area=12ap\begin{align*}Area = \frac{1}{2}a \cdot p\end{align*} Open the file OCTAGON. Construct segments connecting the vertices of the regular octagon to the center R\begin{align*}R\end{align*} using the Segment tool. Each of these segments is a radius of the octagon.
5. How many triangles are created by the radii of the octagon?
6. Are all of the triangles congruent?
7. Use the D.&Length measurement tool to measure the sides and the Angle measurement tool to measure the angles of one of the triangles. Are the triangles equilateral?
8. What is the area of CDR\begin{align*}\triangle{CDR}\end{align*} in terms of the apothem a\begin{align*}a\end{align*} and the side of the triangle s\begin{align*}s\end{align*}?
9. Given the area of one triangle, what is the area of the regular polygon in terms of the length of the apothem a\begin{align*}a\end{align*} and of one side s\begin{align*}s\end{align*} for the octagon?
10. What is the area of an n\begin{align*}n\end{align*}-sided polygon in terms of the length of the apothem a\begin{align*}a\end{align*} and the length of one side s\begin{align*}s\end{align*}?
## Problem 4 – Area of Regular Polygons
Find the area of the polygon with the given measurements.
11. regular heptagon of apothem 12 in. and sides with length 11.56 in.
12. regular dodecagon of apothem 2.8 cm and sides with length 1.5 cm
13. regular octagon of apothem 12.3 ft and sides with length 10.2 ft
14. regular hexagon of apothem 17.32 mm and perimeter 120 mm
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Subjects: | 1,271 | 4,297 | {"found_math": true, "script_math_tex": 29, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2016-44 | longest | en | 0.681437 |
https://byjus.com/maths/12200-in-words/ | 1,701,425,595,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100286.10/warc/CC-MAIN-20231201084429-20231201114429-00283.warc.gz | 183,706,196 | 116,623 | # 12200 in Words
12200 in words is written as “Twelve thousand two hundred”. For example, a cheque of Rs.12200 is written as Rupees Twelve thousand two hundred only. In Maths, 12200 is a cardinal number that expresses a quantity. Learn more about Numbers In Words and writing the number names in English at BYJU’S.
12200 in Words Twelve thousand two hundred Twelve thousand two hundred in Numerical Form 12200
## How to Write 12200 in Words?
The number 12200 in words can be written using a place value chart. Since 12200 is a five-digit number, thus,
Ten-thousand Thousands Hundreds Tens Ones 1 2 2 0 0
From the above table,
1 → Ten thousands
2 → Thousands
2 → Hundreds
0 → Tens
0 → Ones
Hence, when we read the number from the right to left, it is Twelve thousand two hundred.
### Expanded Form of 12200
We can write the expanded form as:
1 x Ten thousand + 2 x Thousand + 2 × Hundred + 0 × Ten + 0 × One
= 1 x 10000 + 2 x 1000 + 2 × 100 + 0 × 10 + 0 × 1
= 10000 + 2000 + 200 + 0 + 0
= 10000 + 2000 + 200
= Twelve thousand two hundred
12200 is a whole number that is succeeded by 12199 and preceded by 12201. Learn more about the number 12200 below:
• 12200 in Words – Twelve thousand two hundred
• Is 12200 an odd number? – No
• Is 12200 an even number? – Yes
• Is 12200 a perfect square number? – No
• Is 12200 a perfect cube number? – No
• Is 12200 a prime number? – No
• Is 12200 a composite number? – Yes
## Frequently Asked Questions on 12200 in words
Q1
### What is 12200 in words?
12200 in words is given by Twelve thousand two hundred.
Q2
### What is the rule to write 12200 in words?
To write the 12200 in words, we should use the place value rule, where the position of each digit helps in determining the name of the number.
Q3
### What is the value of 12200 + 100 in words?
12200 + 100 = 12300, i.e., Twelve thousand three hundred in words.
Test your Knowledge on 12200 in Words | 559 | 1,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2023-50 | latest | en | 0.823057 |
http://mathoverflow.net/feeds/question/38026 | 1,369,341,862,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703788336/warc/CC-MAIN-20130516112948-00013-ip-10-60-113-184.ec2.internal.warc.gz | 166,154,230 | 4,641 | Generating n-cycles - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T20:44:23Z http://mathoverflow.net/feeds/question/38026 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38026/generating-n-cycles Generating n-cycles H A Helfgott 2010-09-08T02:40:23Z 2010-09-13T13:25:55Z <p>Let $G = S_n$ (the permutation group on $n$ elements). Let $A\subset G$ such that $A$ generates $G$.</p> <p>Is there an $n$-cycle $g$ in $G$ that can be expressed as</p> <p>$g = a_1 a_2 ... a_k$</p> <p>where $a_i\in A \cup A^{-1}$ and $k\leq c_1 n^{c_2}$, where $c_1$ and $c_2$ are constants?</p> <p>What about $2$-cycles, or elements of any other particular form?</p> http://mathoverflow.net/questions/38026/generating-n-cycles/38042#38042 Answer by Roland Bacher for Generating n-cycles Roland Bacher 2010-09-08T10:53:04Z 2010-09-08T11:41:02Z <p>A partial answer for transpositions:</p> <p>If $A$ contains a generating set given by transpositions, there is an $n-$cycle given by the product of $n-1$ generators. Indeed, consider the graph $G$ with vertices $1,\dots,n$ and edges defined by all transpositions in the generating set $A$. The subset of transpositions in $A$ generates $S_n$ if and only if $G$ is a connected graph. Up to throwing away a few generators, we can thus assume that $G$ is a tree. Drawing this tree in the plane and choosing an initial leaf of $G$, we consider the product of all transpositions corresponding to the order in which we encounter the edges of $G$ when walking around $G$ in counterclockwise order. This defines the $n-$cycle obtained by writing down all vertices of $G$ accordingly to their last sighting during the walk.</p> http://mathoverflow.net/questions/38026/generating-n-cycles/38052#38052 Answer by Colin Reid for Generating n-cycles Colin Reid 2010-09-08T12:59:57Z 2010-09-08T12:59:57Z <p>This is not really an answer so much as some vague thoughts on the matter, but it's a bit long for a comment.</p> <p>The $n$-cycle case seems particularly promising as the $n$-cycles are so numerous: a random element of $S_n$ is an $n$-cycle with probability $1/n$. So it would be enough to show that $n$-cycles are sufficiently 'evenly distributed' in any Cayley graph that we can never be too far from one.</p> <p>Here's a vague idea:</p> <p>Suppose the Cayley graph has a collection of $n$-cycles of size $f(n)$ which are clumped together, that is, all a short distance $2r$ from each other ($r$ is allowed to be polynomial in $n$). Then perhaps one can show that there are two $n$-cycles $x$ and $y$ in or close to this clump such that $xy^{-1}$ is an $n$-cycle, as long as we can break out of some special situations, such as our clump being contained in a coset of a subgroup that has no $n$-cycles.</p> <p>Suppose there are no such clumps. Then if we draw discs of radius $r$ around every $n$-cycle, then any given vertex cannot be contained in too many discs (as otherwise it would be in the middle of a clump). It follows that each disc has area at most $nf(n)$. To get a contradiction here we have to prove something about word growth, in order to show that the discs cannot have such a small area.</p> http://mathoverflow.net/questions/38026/generating-n-cycles/38296#38296 Answer by Gjergji Zaimi for Generating n-cycles Gjergji Zaimi 2010-09-10T12:15:16Z 2010-09-13T13:25:55Z <p>That all elements in a symmetric group with a specified arbitrary generating set can be reached in a polynomial amount of steps is a known open problem, and has been investigated for some time now. This long-standing conjecture has been proven for most choices of generating sets. Heuristically one expects this to be true for the reasons mentioned by Colin Reid above. That is, if certain conjectures are true, then every Cayley graph has a Hamiltonian cycle and so is expected to have exponentially many such cycles. So one expects to be able to reach any element of the group pretty fast.</p> <p>For the symmetric group the problem of determining tight bounds on the diameter of its Cayley graph has been studied (and partially resolved) by L. Babai and coauthors. In the paper <a href="http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WHS-4D7CYNS-GC&_user=10&_coverDate=09%2F30%2F1988&_rdoc=1&_fmt=high&_orig=search&_origin=search&_sort=d&_docanchor=&view=c&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=fd830a2a697193261a6cacc86268a0c5&searchtype=a" rel="nofollow">"On the diameter of cayley graphs of the symmetric group"</a> it is proved that one can reach any elements using words of length at most $e^{\sqrt{n\log n}(1+o(1))}$ for any generating set, while the optimal bound should be $O(n^{c})$. <a href="http://portal.acm.org/citation.cfm?id=982956&dl=GUIDE&coll=GUIDE&CFID=101320652&CFTOKEN=30527812" rel="nofollow">Here</a> and <a href="http://portal.acm.org/citation.cfm?id=1070584" rel="nofollow">here</a> more partial results are proved, showing evidence that polynomial bounds are in fact the true asymptotic. I will remark again that one expects this behavior for all nonabelian finite simple groups too. </p> <p>These papers are a good survey of what is currently known about this problem, I don't know if restricting your target to specific conjugacy classes of elements (such as n-cycles) makes the problem easier so I will think about it a bit more.</p> http://mathoverflow.net/questions/38026/generating-n-cycles/38357#38357 Answer by Sergei Ivanov for Generating n-cycles Sergei Ivanov 2010-09-10T22:04:08Z 2010-09-10T22:04:08Z <p>Concerning the second question (about 2-cycles). If some transposition (= 2-cycle) can be obtained as as a product of a polynomial number of generators (= elements of $A\cup A^{-1}$), then all elements of the symmetric group can be obtained this way. Thus the question is equivalent to the open problem that Gjergji Zaimi mentioned in his answer.</p> <p>The proof follows the suggestion that jp made in his comment to Roland Bacher's answer. Suppose that some transposition is a product of $N$ generators. All transpositions are conjugate to this one, and since there are only $n(n-1)/2$ of them, each can be reached by at most $n(n-1)/2$ conjugations by generators. Hence every transposition is a product of at most $N+n(n-1)$ generators. It remains to recall that every permutation is a product of at most $n-1$ transpositions.</p> http://mathoverflow.net/questions/38026/generating-n-cycles/38432#38432 Answer by Bill Thurston for Generating n-cycles Bill Thurston 2010-09-12T00:24:38Z 2010-09-12T01:14:36Z <p>To elaborate on Sergei Ivanov's point with some futher observations:</p> <p>By an argument similar to his, if you can obtain some element of any conjugacy class of bounded support with polynomial wordlength, the entire symmetric group has polynomial wordlength. That's because there are only polynomially many elements to the conjugacy class, and members of a fixed bounded conjugacy class generates the symmetric group in at most quadratic$(n)$ word length. (Transpositions require quadratic word length, using bubble sort. You can obtain a transposition in bounded wordlength once you have all elements of some conjugacy class). Furthermore, for fixed $k$, any $k$-tuple can be taken to any other $k$-tuple in polynomial$(n)$ word length, so a word that has polynomial length in terms of elements of the conjugacy class can be rewritten to have polynomial length in terms of the generators.</p> <p>One strategy for trying to obtain elements of small support, given a few miscellaneous permutations, is to first take powers that halt some of their cycles. Once the support is small enough, you can take commutators of pairs of words whose support intersect only modestly to get still smaller support. These kinds of tricks make it hard to see how there could be counterexamples to the conjecture that the diameter of the group is a polynomial in n.</p> <p>I think it would not be very hard to write a computer program that, given an arbitary generating set, would in practice produce a function $S_n \rightarrow$ polynomial-length word representative, because it shouldn't take many words to find permutations with reasonably arbitrary cycle shape. But it would be hard to prove it worked reliably.</p> <p>I'd like to mention that there are well-known fast algorithms for analyzing the group generated by a collection of permutations, but they use recursive words, that is, words in words in words ... in generators, which gets around the group efficiently and quickly. Since permutation are easy to compose, the recursion is computationally cheap, and for many purposes, better than using words. (Actual experts should please elaborate or correct me if I'm mistaken).</p> | 2,355 | 8,811 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2013-20 | latest | en | 0.803529 |
https://www.conceptdraw.com/examples/example-of-algorithm-in-flowchart | 1,529,521,979,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863834.46/warc/CC-MAIN-20180620182802-20180620202802-00138.warc.gz | 786,253,549 | 13,416 | This site uses cookies. By continuing to browse the ConceptDraw site you are agreeing to our Use of Site Cookies.
# Basic Flowchart Symbols and Meaning
## Basic Flowchart Symbols and Meaning
Flowcharts are the best for visually representation the business processes and the flow of a custom-order process through various departments within an organization. ConceptDraw PRO diagramming and vector drawing software extended with Flowcharts solution offers the full set of predesigned basic flowchart symbols which are gathered at two libraries: Flowchart and Flowcharts Rapid Draw. Among them are: process, terminator, decision, data, document, display, manual loop, and many other specific symbols. The meaning for each symbol offered by ConceptDraw gives the presentation about their proposed use in professional Flowcharts for business and technical processes, software algorithms, well-developed structures of web sites, Workflow diagrams, Process flow diagram and correlation in developing on-line instructional projects or business process system. Use of ready flow chart symbols in diagrams is incredibly useful - you need simply drag desired from the libraries to your document and arrange them in required order. There are a few serious alternatives to Visio for Mac, one of them is ConceptDraw PRO. It is one of the main contender with the most similar features and capabilities. Read more
How to Build a Flowchart
## Euclidean algorithm - Flowchart
"In mathematics, the Euclidean algorithm, or Euclid's algorithm, is a method for computing the greatest common divisor (GCD) of two (usually positive) integers, also known as the greatest common factor (GCF) or highest common factor (HCF). ...
The GCD of two positive integers is the largest integer that divides both of them without leaving a remainder (the GCD of two integers in general is defined in a more subtle way).
In its simplest form, Euclid's algorithm starts with a pair of positive integers, and forms a new pair that consists of the smaller number and the difference between the larger and smaller numbers. The process repeats until the numbers in the pair are equal. That number then is the greatest common divisor of the original pair of integers.
The main principle is that the GCD does not change if the smaller number is subtracted from the larger number. ... Since the larger of the two numbers is reduced, repeating this process gives successively smaller numbers, so this repetition will necessarily stop sooner or later - when the numbers are equal (if the process is attempted once more, one of the numbers will become 0)." [Euclidean algorithm. Wikipedia]
The flowchart example "Euclidean algorithm" was created using the ConceptDraw PRO diagramming and vector drawing software extended with the Mathematics solution from the Science and Education area of ConceptDraw Solution Park. Read more
Euclid's algorithm flow chart
Used Solutions
## Simple Flow Chart
ConceptDraw PRO diagramming and vector drawing software extended with Flowcharts Solution from the 'What is a Diagram' area of ConceptDraw Solution Park is a powerful tool for drawing Flow Charts of any complexity you need. Irrespective of whether you want to draw a Simple Flow Chart or large complex Flow Diagram, you estimate to do it without efforts thanks to the extensive drawing tools of Flowcharts solution, there are professional flowchart symbols and basic flowchart symbols. This sample shows the Gravitational Search Algorithm (GSA) that is the optimization algorithm. Read more
## Solving quadratic equation algorithm - Flowchart
"In elementary algebra, a quadratic equation (from the Latin quadratus for "square") is any equation having the form
ax^2+bx+c=0
where x represents an unknown, and a, b, and c are constants with a not equal to 0. If a = 0, then the equation is linear, not quadratic. The constants a, b, and c are called, respectively, the quadratic coefficient, the linear coefficient and the constant or free term.
Because the quadratic equation involves only one unknown, it is called "univariate". The quadratic equation only contains powers of x that are non-negative integers, and therefore it is a polynomial equation, and in particular it is a second degree polynomial equation since the greatest power is two.
Quadratic equations can be solved by a process known in American English as factoring and in other varieties of English as factorising, by completing the square, by using the quadratic formula, or by graphing." [Quadratic equation. Wikipedia]
The flowchart example "Solving quadratic equation algorithm" was created using the ConceptDraw PRO diagramming and vector drawing software extended with the Mathematics solution from the Science and Education area of ConceptDraw Solution Park. Read more
Used Solutions
## Flow chart Example. Warehouse Flowchart
Warehouse Flowcharts are various diagrams that describe the warehousing and inventory management processes on the warehouses. Typical purposes of Warehouse Flowcharts are evaluating warehouse performance, measuring efficiency of customer service and organizational performance. This type of Workflow diagrams can be used for identifying any disconnection between business activities and business objectives. They are effectively used by warehouse-related people and organizations, manufacturers, wholesalers, exporters, importers, transporters, and others. Standard Warehousing process flow diagram and standard Workflow diagram are used for process identification for further evaluating effectiveness and profitability of overall business process. Use the ConceptDraw PRO vector graphic software extended with Flowcharts solution to design your own professional-looking Workflow diagrams and Flowcharts of any types, including the Warehouse flowchart, Process flow diagrams which depict in details all steps of Warehouse packages flow. Microsoft Visio, designed for Windows users, can’t be opened directly on Mac. But fortunately, there are several Visio alternatives for Mac which will help Mac users to work Visio files. With ConceptDraw PRO, you may open, edit and save files in Visio format. Read more
ConceptDraw Arrows10 Technology
## Credit Card Order Process Flowchart. Flowchart Examples
This sample was created in ConceptDraw PRO diagramming and vector drawing software using the Flowcharts solution from the What is a Diagram area of ConceptDraw Solution Park. This sample shows the Flowchart of the Credit Card Order Process. On this diagram are used the flowchart symbols that represents the processes and documents. The flowchart symbols are connected with arrows. Read more
## Flowchart Examples and Templates
ConceptDraw PRO vector diagramming software and ConceptDraw Solution Park provide a wide variety of diagrams, organizational charts, business charts and flowchart examples, templates and samples. You are free to choose any example or template you are interested in from the ConceptDraw STORE, then use it to simplify your work at the designing professional-looking flowcharts and diagrams. As for flowcharts, turn your attention for the Flowcharts solution from the "What is a Diagram" area of ConceptDraw Solution Park, Process Flowcharts and Cross-Functional Flowcharts solutions from the Business Processes area, Accounting Flowcharts and Audit Flowcharts solutions from the Finance and Accounting area, which are completely devoted to flowcharts creation and provide an enormous collection of helpful flowchart templates and samples. Each of them is specially developed, well thought-out, dedicated to a certain thematic and carries a specific purpose. You need only to determine with your needs and to decide which one corresponds them the best and suits for you. Read more
## Process Flowchart
The main reason of using Process Flowchart or PFD is to show relations between major parts of the system. Process Flowcharts are used in process engineering and chemical industry where there is a requirement of depicting relationships between major components only and not include minor parts. Process Flowcharts for single unit or multiple units differ in their structure and implementation. ConceptDraw PRO is Professional business process mapping software for making Process flowcharts, Process flow diagram, Workflow diagram, flowcharts and technical illustrations for business documents and also comprehensive visio for mac application. Easier define and document basic work and data flows, financial, production and quality management processes to increase efficiency of your business with ConcepDraw PRO. Business process mapping software with Flowchart Maker ConceptDraw PRO includes extensive drawing tools, rich examples and templates, process flowchart symbols and shape libraries, smart connectors that allow you create the flowcharts of complex processes, process flow diagrams, procedures and information exchange. Process Flowchart Solution is project management workflow tools which is part ConceptDraw Project marketing project management software. Drawing charts, diagrams, and network layouts has long been the monopoly of Microsoft Visio, making Mac users to struggle when needing such visio alternative like visio for mac, it requires only to view features, make a minor edit to, or print a diagram or chart. Thankfully to MS Visio alternative like ConceptDraw PRO software, this is cross-platform charting and business process management tool, now visio alternative for making sort of visio diagram is not a problem anymore however many people still name it business process visio tools. Read more
How To Create a Process Flow Chart (business process modelling techniques)
## Contoh Flowchart
The Flowcharts are graphical representations of algorithms, processes or step-by-step solutions problems. There are many different types of Flowcharts, among them Process Flowchart, Cross Functional Flowchart, Data Flow Diagram, IDEF Flowchart, Workflow Diagram, Contoh Flowchart and many others. They have especial value when you need represent a complex process, depict in details the process of solution problems, efficiently plan and set the tasks priorities. The Flowcharts must to be constructed brief, clear and logical, simplifying the process or procedure, and making easier the comprehension and perception of information. The ConceptDraw PRO software makes the process of creating the flowcharts of any types well organized and clear for developers and customers also, including the Contoh Flowchart. It is possible due to the Flowcharts solution from ConceptDraw Solution Park, its predesigned vector objects, templates, and a lot of professional-looking practical samples and examples which can be quick and easy modified, printed, or published on web. Read more
## Copying Service Process Flowchart. Flowchart Examples
This sample was created in ConceptDraw PRO diagramming and vector drawing software using the Flowcharts solution from the What is a Diagram area of ConceptDraw Solution Park. This sample shows the Flowchart on that it is displayed the process of the determination the permissibility according the Access Copyright license. The diamonds represent the decision points. Inside the diamonds are the questions that need the answer yes/no. It is necessary to answer on the question, make the decision that will determine the next step. Read more
## Types of Flowcharts
A Flowchart is a graphical representation of process, algorithm, workflow or step-by-step solution of the problem. It shows the steps as boxes of various kinds and connects them by arrows in a defined order depicting a flow. There are twelve main Flowchart types: Basic Flowchart, Business Process Modeling Diagram (BPMN), Cross Functional Flowchart, Data Flow Diagram (DFD), IDEF (Integrated DEFinition) Flowchart, Event-driven Process Chain (EPC) Diagram, Influence Diagram (ID), Swimlane Flowchart, Process Flow Diagram (PFD), Specification and Description Language (SDL) Diagram, Value Stream Mapping, Workflow Diagram. Using the Flowcharts solution from the What is a Diagram area of ConceptDraw Solution Park you can easy and quickly design a Flowchart of any of these types. This solution offers a lot of special predesigned vector symbols for each of these widely used notations. They will make the drawing process of Flowcharts much easier than ever. Pay also attention for the included collection of ready Flowchart examples, samples and quick-start templates. This is business process improvement tools. If you are looking for MS Visio for your Mac, then you are out of luck, because it hasn't been released yet. However, you can use Visio alternatives that can successfully replace its functions. ConceptDraw PRO is an alternative to MS Visio for Mac that provides powerful features and intuitive user interface for the same. Read more
How to Simplify Flow Charting
## Flowchart Programming Project. Flowchart Examples
Create you own flow charts of process-driven software applications using the ConceptDraw PRO diagramming and vector drawing software extended with the Cross-Functional Flowcharts solution from the Business Processes area of ConceptDraw Solution Park. The programming project flow chart example shows the logical process of execution. Read more
## Material Requisition Flowchart. Flowchart Examples
Material requisition is a request generated by internal or external organization with a goal to inform the purchase department about the needed items and materials. The requisition can be represented as a written document on a pre-printed form or online request in form of electronic document. The electronic inquisition occupies a leading position in a modern world, now the paper forms are replaced by electronic ordering processes and workflows. The requisition is actively used in business as an internal document for notification the store about the needed goods, in medicine for making the orders on medical equipment and medicaments, in industry for creation request for purchasing some items and materials, and so on. ConceptDraw PRO vector graphics software extended with Flowcharts solution from the What is a Diagram area is powerful and useful tool for drawing various types of Flowcharts, including without doubt the Material Requisition Flowchart of any degree of detailing. Succeed in drawing using the large quantity of predesigned flowchart specific shapes, symbols and icons. Read more
## Top 5 Android Flow Chart Apps
For many years CS Odessa corporation has been developing the multifunctional vector diagramming software with supporting sophisticated drawing tools for vector diagramming and design on Macintosh and PC with Windows installed. We feel that our considerable experience gives us an opportunity to look objectively at the current offerings and to offer you our opinion about the main contenders and pretenders. Would you like to know the details about what is going on in the world of Flow Chart drawing applications that support Android? The Top 5 of drawing applications in this space rapidly changes, consequently it is significant understatement to call this application space dynamic. Now, the list of Top 5 popular flowchart makers for Android includes LLNL Flow Charts, Army Flow Charts, DroidDia prime, Note Droid, DroidDia PRO unlocker. These listed business graphics applications are recognized by many experts the most convenient, powerful, and successful for use when drawing on Android devices. Read more
## Flow Chart Design - How to Design a Good Flowchart
Use ConceptDraw PRO business diagramming and business graphics software for general diagramming purposes, it inludes tousands colored professional flowchart symbols, examples and samples which saves time when you prepare documents, professional presentations or make an explanation of process flow diagram. Read more | 2,964 | 15,864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-26 | latest | en | 0.923375 |
https://www.distractify.com/p/how-does-the-wheel-work-nbc | 1,695,379,881,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506399.24/warc/CC-MAIN-20230922102329-20230922132329-00472.warc.gz | 804,152,349 | 38,100 | # NBC's 'The Wheel': How It Works and Details on the Prize Money at Stake
How does 'The Wheel' game show work on NBC? It's all about awarding contestants cash prizes, but celebrities are here to help them.
By
Dec. 19 2022, Published 2:32 p.m. ET
Most game shows with a spinning wheel require contestants to physically spin it. But on NBC's The Wheel, the contestants are part of the wheel, along with celebrity guests that help them. So, how does The Wheel work and what kind of prize is at stake?
The show is based on a UK reality game show of the same name. But now that it made the leap across the pond, so to speak, you can expect some A-list celebrities from the U.S.
## How does 'The Wheel' work on NBC?
Each episode of The Wheel features three contestants who are chosen at random to sit in a chair in the center of a stage, which is designed like a spinning wheel. Seven celebrities sit in chairs outside of the wheel. They are each determined to be experts in different fields, whether it's sports, history, or music. When a contestant is given a multiple choice question, they initiate the wheel to be spun to pick a celebrity to help them answer the question.
The celebrities are slowly spun around the contestant and a large arrow on the floor points at which one is chosen to help them answer a question. Even if the arrow lands on a celebrity whose expertise is not in the chosen category, they still have to help the contestant answer the question. If the answer is wrong, the contestant is removed from the game and another one takes their place.
The player left standing at the end of any given episode is awarded the progressive prize, which can reach as high as \$100,000. In the UK version, which is expected to be the same for the U.S. version of The Wheel, there is a final round where all three contestants are put on the spot to see which one of them walks away with the prize.
At the end of each episode, the celebrities are ranked by how many correct answers they've given in the episode. The contestant who is still in the game at this point chooses which celebrity they want to help them with a final question. They could win 50, 100, or 200 percent of the total prize, depending on how high or low-ranking the celebrity they choose is.
If the celebrity helps them get the final question right, this contestant wins. If the player gets it wrong, the process moves to one of the other two eliminated players from the episode. But if all three contestants get their respective final questions wrong, they all leave with nothing.
## Who is the host of 'The Wheel'?
The UK and now U.S. version of The Wheel is hosted by British comedian Michael McIntyre. He was once a judge on Britain's Got Talent and he hosted Michael McIntyre's Comedy Roadshow, which ran for two seasons. He also hosts a Saturday night variety show on the BBC called Michael McIntyre's Big Show.
With another version of The Wheel also under his belt, it's safe to say Michael will be busy on TV for the foreseeable future.
Watch The Wheel on Mondays at 10 p.m. EST on NBC. | 672 | 3,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-40 | latest | en | 0.969088 |
http://dsearls.org/courses/C121CompSci/Samples/SimpleInterest.htm | 1,516,417,603,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888878.44/warc/CC-MAIN-20180120023744-20180120043744-00730.warc.gz | 95,727,534 | 1,794 | ```/*
* This program models an investment earning simple interest.
*
* An investment earning simple interest has three attributes:
* present value (the amount of money invested), the nominal
* annual interest rate, and the term (the number of years).
* Calculated values include the total amount of interest
* earned and the future value (the value of the investment
* at the end of its term).
*
* Aug 2006
* D. Searls
* Asbury College
*/
#include <iostream> // Provides input/output capabilities
#include <iomanip> // Provides output formatting
using namespace std;
int main()
{
double presentValue; // The value at the beginning
double nominalRate; // Nominal annual rate
int years; // The term of the investment
double interest; // The amount of interest earned
double futureValue; // The value at the end
// Display instructions.
cout << "This program models an investment earning simple interest." << endl;
cout << "You will be asked to enter the present value, the nominal" << endl;
cout << "annual interest rate, and the term (in years)." << endl << endl;
// Input the present value, the nominal rate and
// the number of years.
cout << "Enter the present value: ";
cin >> presentValue;
cout << "Enter the nominal annual rate (as a %): ";
cin >> nominalRate;
cout << "Enter the number of years: ";
cin >> years;
cout << endl;
// Calculcate the interest and the future value.
interest = presentValue * nominalRate/100.0 * (double)years;
futureValue = presentValue + interest;
// Display a report for this investment.
cout << fixed << setprecision(2);
cout << "Present Value: \$" << presentValue << endl;
cout << "Nominal Rate: " << nominalRate << "%" << endl;
cout << "Term: " << years << " years" << endl;
cout << "Interest: \$" << interest << endl;
cout << "Future Value: \$" << futureValue << endl;
cout << endl;
cout << endl;
return 0;
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If a straight line meet two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles...
The first six books of the Elements of Euclid, with numerous exercises - Seite 4
von Euclides - 1853 - 147 Seiten
Vollansicht - Über dieses Buch
## The Elements of Euclid
Euclid - 1838 - 416 Seiten
...enclose a space. XI. All right angles arc equal to one another. XII. •• If a straight line meet two straight lines, so as to make the two " interior...right angles, these straight lines being continually pro" duced, shall at length meet upon that side on which are the angles " which are less than two right...
Vollansicht - Über dieses Buch
## The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago ...
Euclid, Robert Simson - 1838 - 416 Seiten
...cannot enclose a space, XI. All right angles are equal to one another. XII. " If a straight line meet two straight lines, so as to make the two . " interior...right angles, these straight lines being continually pro" duced, shall at length meet upon that side on which are the angles " which are less than two right...
Vollansicht - Über dieses Buch
## The Elements of Euclid; viz. the first six books,together with the eleventh ...
Euclides - 1841 - 351 Seiten
...equal to one another. XII. " If a straight line meet two straight lines, which are " in the same plane, so as to make the two interior " angles on the same...than two right angles, these straight lines being " produced, shall at length meet upon that side on " which are the angles which are less than two "...
Vollansicht - Über dieses Buch
## Elements of geometry: consisting of the first four,and the sixth, books of ...
Euclides - 1842
...circle may be described from any centre, at any distance from that centre. IV. [Ax. XI.] V. And that if a straight line meets two straight lines, so as...length meet upon that side on which are the angles less than two right angles. AXIOMS. I. THINGS which are equal to the same are equal to one another....
Vollansicht - Über dieses Buch
## Elements of Geometry: On the Basis of Dr. Brewster's Legendre : to which is ...
James Bates Thomson - 1844 - 237 Seiten
...the necessity of a new axiom. Euclid's axiom alluded to is this : " If a straight line meet two other straight lines, so as to make the two interior angles...on the same side of it taken together less than two right-angles, these straight lines being continually produced, will at length meet on the side on which...
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## Elements of Geometry: Containing the First Six Books of Euclid, with a ...
John Playfair - 1844 - 317 Seiten
...progress of the understanding, it will certainly retard it AXIOMS. The 12th Axiom of Euclid is, that " if a straight line meets two straight " lines, so as to make the two interior anples on the same side of it taken " together less than two right angles, these straight lines being...
Vollansicht - Über dieses Buch
## The First Six, and the Eleventh and Twelfth Books of Euclid's Elements: With ...
Euclid, James Thomson - 1845 - 352 Seiten
...to one another. |j 12. If a straight line meet two other straight lines which are in the same plane, so as to make the two interior angles on the same...together, less than two right angles, these straight lines shall at length meet upon that side, if they be continually produced.^ * ln this axiom and the following,...
Vollansicht - Über dieses Buch
## The Elements of Euclid, the parts read in the University of Cambridge [book ...
Euclides - 1846
...to two right angles, and therefore the angles BEF, EFD are together less than two right angles : But If a straight line meets two straight lines, so as...make the two interior angles on the same side of it together less than two right angles, these straight lines, being continually produced, shall at length...
Vollansicht - Über dieses Buch
## Elements of the Philosophy of the Human Mind: In Two Parts, Teil 1
Dugald Stewart - 1847 - 627 Seiten
...Euclid's Elements. When it is assorted, for example, that " if one straight line falls on two other straight lines, so as to make the two interior angles on the san»« side together equal to two right angles, these two straight lines, though indefinitely produced,...
Vollansicht - Über dieses Buch
## Elements of Geometry: Containing the First Six Books of Euclid : with a ...
John Playfair - 1849 - 317 Seiten
...self-evident propositions. It is therefore removed from among the Axioms. The 12th Axiom of Euclid is, that " if a straight line meets two straight " lines, so...shall at length meet upon that side on which are the angle* " which are less than two right angles." Instead of this proposition, which, though true, is...
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https://www.jiskha.com/display.cgi?id=1294956881 | 1,516,263,493,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887077.23/warc/CC-MAIN-20180118071706-20180118091706-00490.warc.gz | 920,983,289 | 3,561 | # Math
posted by .
PLEASE HELP!!! beth has 60 toy cars and trucks. 75% of them are blue. How many are blue?
• Math -
75% = .75
.75 * 60 = ?
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Beth has 60 toy cars and trucks 75% of them are blue.How many are blue?
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Prism or Pyramid1.pptx
Prism or Pyramid?
Unit 8: Geometry
Lesson 11 of 17
Big Idea: Understanding solids and nets helps develop the concepts of volume and surface area
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Probability
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Suppose that infants are classified as low birth weight if they have birth weight 2500g, and as normal birth weight if have birth weight 2501g. Suppose that infants are also classified by length of gestation in the following four categories: <20 weeks, 20-27 weeks, 28-36 weeks, >36 weeks. Assume the probabilities of the different period of gestation are as given in the table below:
See attached for table.
Also assume that the probability of low birth weight given that length of gestation is <20 weeks is .540, the probability of low birth weight given that length of gestation is 20-27 weeks is .813, the probability of low birth weight given that length of gestation is 28-36 weeks is .378, and the probability of low birth weight given that length of gestation is >36 weeks is .031.
a) What is the probability of having a low birth weight infant?
b) Show that the events {length of gestation 27 weeks} and {low birth weight} are not independent.
https://brainmass.com/statistics/probability/probability-460609
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Midpoint of a line segment is the point on the segment that divides the segment into two congruent segments. If B is the midpoint of line AC, then AB = BC. A segment bisector is a point, ray, line, line segment or plane that intersects the segment at its midpoint. A midpoint or a segment bisector bisects a segment. AB Directions: Answer the following questions. Also write at least five examples of your own.
Q 1: What is the mid point of GH whose length is 11 cm?5.5 cm15 cm7.5 cm Q 2: What is the mid point of EF whose length is 9 cm?13 cm9 cm4.5 cm Q 3: AO = OB, if AO = 2.5 cm, then OB = ?5 cm2.5 cm0 cm Q 4: In a skateboard design, SB bisects XY at point T, and XT=14.8 cm. Find XY.29.6 cm14.8 cm22.4 cm Question 5: This question is available to subscribers only! Question 6: This question is available to subscribers only!
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