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https://en.academic.ru/dic.nsf/enwiki/2164284/Submarine | 1,606,408,454,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188800.15/warc/CC-MAIN-20201126142720-20201126172720-00582.warc.gz | 279,835,734 | 13,029 | # Submarine depth ratings
Submarine depth ratings
Depth ratings are primary design parameters and measures of a submarine's ability to operate underwater. The depths to which submarines can dive are limited by the strengths of their hulls. As a first order approximation, each 10 meters (33 feet) of depth puts another atmosphere (1 bar, 14.7 psi, 100 kPa) of pressure on the hull, so at 300 meters (1,000 feet), the hull is supporting thirty atmospheres (30 bar, 441 psi, 3,000 kPa) of water pressure. (Note: The one atmosphere of air pressure at sea level is balanced by the roughly one atmosphere maintained inside the sub, so it does not normally strain the hull).
Design depth is the nominal depth listed in the submarine's specifications. From it the designers calculate the thickness of the hull metal, the boat's displacement, and many other related factors. Since the designers incorporate margins of error in their calculations, crush depth of an actual vessel should be slightly deeper than its design depth.
Test depth is the maximum depth at which a submarine is permitted to operate under normal peacetime circumstances, and is tested during sea trials. The test depth is set at two-thirds of the design depth for United States Navy submarines, while the Royal Navy sets test depth slightly deeper than half (4/7ths) of the design depth, and the German Navy sets it at exactly one-half of design depth.[1]
The maximum operating depth[2] (popularly called the never-exceed depth) is the maximum depth at which a submarine is allowed to operate under any (e.g. battle) conditions.
Crush depth, officially called collapse depth[2], is the submerged depth at which a submarine's hull will collapse due to pressure. This is normally calculated; however, it is not always accurate. Submarines from many nations in World War II reported being forced through crush depth, due to flooding or mechanical failure, only to have the water pumped out, or the failure repaired, and succeed in surfacing again. One of the most popular stories of this occurring was the story of U-96, in the movie Das Boot. Note that these reports are not necessarily verifiable, and popular misunderstanding of the difference between test depth and collapse depth can confuse the discussion. (Planesman error sometimes causes submarines to exceed test depth by a few feet or meters during trials; note that a one-degree up-bubble on an Ohio-class boat indicates that the stern is some ten feet or three meters deeper than the bow.)
World War II German U-boats generally had collapse depths in the range of 200 to 280 meters (660 to 920 feet).[citation needed] Modern nuclear attack submarines like the American Seawolf class are estimated to have a test depth of 490 m (1,600 ft),[1] which would imply (see above) a collapse depth of 730 m (2,400 ft).
## References
1. ^ a b Federation of American Scientists (8 December 1998). "Run Silent, Run Deep". Military Analysis Network. Retrieved 10 May 2010.
2. ^ a b Department of Defense (19 August 2009) (PDF). Joint Publication 1-02: Dictionary of Military and Associated Terms. Retrieved 10 May 2010.
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https://elteoremadecuales.com/arrows-impossibility-theorem/ | 1,680,004,460,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948858.7/warc/CC-MAIN-20230328104523-20230328134523-00114.warc.gz | 281,083,610 | 26,109 | # Arrow's impossibility theorem
Arrow's impossibility theorem Arrow's impossibility theorem, the general possibility theorem or Arrow's paradox is an impossibility theorem in social choice theory that states that when voters have three or more distinct alternatives (options), no ranked voting electoral system can convert the ranked preferences of individuals into a community-wide (complete and transitive) ranking while also meeting the specified set of criteria: unrestricted domain, non-dictatorship, Pareto efficiency, and independence of irrelevant alternatives. The theorem is often cited in discussions of voting theory as it is further interpreted by the Gibbard–Satterthwaite theorem. The theorem is named after economist and Nobel laureate Kenneth Arrow, who demonstrated the theorem in his doctoral thesis and popularized it in his 1951 book Social Choice and Individual Values. The original paper was titled "A Difficulty in the Concept of Social Welfare".[1] In short, the theorem states that no rank-order electoral system can be designed that always satisfies these three "fairness" criteria: If every voter prefers alternative X over alternative Y, then the group prefers X over Y. If every voter's preference between X and Y remains unchanged, then the group's preference between X and Y will also remain unchanged (even if voters' preferences between other pairs like X and Z, Y and Z, or Z and W change). There is no "dictator": no single voter possesses the power to always determine the group's preference.
Cardinal voting electoral systems are not covered by the theorem, as they convey more information than rank orders.[2][3] However, Gibbard's theorem shows that strategic voting remains a problem.
The axiomatic approach Arrow adopted can treat all conceivable rules (that are based on preferences) within one unified framework. In that sense, the approach is qualitatively different from the earlier one in voting theory, in which rules were investigated one by one. One can therefore say that the contemporary paradigm of social choice theory started from this theorem.[4] The practical consequences of the theorem are debatable: Arrow has said "Most systems are not going to work badly all of the time. All I proved is that all can work badly at times."[5] Contents 1 Statement 1.1 Independence of irrelevant alternatives (IIA) 2 Formal statement of the theorem 3 Informal proof 3.1 Part one: There is a "pivotal" voter for B over A 3.2 Part two: The pivotal voter for B over A is a dictator for B over C 3.3 Part three: There exists a dictator 4 Interpretations 4.1 Remark: Scalar rankings from a vector of attributes and the IIA property 5 Alternatives based on functions of preference profiles 5.1 Infinitely many individuals 5.2 Limiting the number of alternatives 5.3 Pairwise voting 5.4 Domain restrictions 5.5 Relaxing transitivity 5.6 Relaxing assumption IIA 5.7 Relaxing the Pareto criterion 5.8 Relaxing the Dictatorship prohibition 5.9 Social choice instead of social preference 6 Other alternatives 7 See also 8 References 9 Further reading 10 External links Statement The need to aggregate preferences occurs in many disciplines: in welfare economics, where one attempts to find an economic outcome which would be acceptable and stable; in decision theory, where a person has to make a rational choice based on several criteria; and most naturally in electoral systems, which are mechanisms for extracting a governance-related decision from a multitude of voters' preferences.
The framework for Arrow's theorem assumes that we need to extract a preference order on a given set of options (outcomes). Each individual in the society (or equivalently, each decision criterion) gives a particular order of preferences on the set of outcomes. We are searching for a ranked voting electoral system, called a social welfare function (preference aggregation rule), which transforms the set of preferences (profile of preferences) into a single global societal preference order. Arrow's theorem says that if the decision-making body has at least two members and at least three options to decide among, then it is impossible to design a social welfare function that satisfies all these conditions (assumed to be a reasonable requirement of a fair electoral system) at once: Non-dictatorship The social welfare function should account for the wishes of multiple voters. It cannot simply mimic the preferences of a single voter. Unrestricted domain, or universality For any set of individual voter preferences, the social welfare function should yield a unique and complete ranking of societal choices. Thus: It must do so in a manner that results in a complete ranking of preferences for society. It must deterministically provide the same ranking each time voters' preferences are presented the same way. Independence of irrelevant alternatives (IIA) The social preference between x and y should depend only on the individual preferences between x and y (pairwise independence). More generally, changes in individuals' rankings of irrelevant alternatives (ones outside a certain subset) should have no impact on the societal ranking of the subset. For example, if candidate x ranks socially before candidate y, then x should rank socially before y even if a third candidate z is removed from participation. (See Remarks below.) Monotonicity, or positive association of social and individual values If any individual modifies his or her preference order by promoting a certain option, then the societal preference order should respond only by promoting that same option or not changing, never by placing it lower than before. An individual should not be able to hurt an option by ranking it higher. Non-imposition, or citizen sovereignty Every possible societal preference order should be achievable by some set of individual preference orders. This means that the social welfare function is surjective: It has an unrestricted target space.
A later (1963)[6] version of Arrow's theorem replaced the monotonicity and non-imposition criteria with: Pareto efficiency, or unanimity If every individual prefers a certain option to another, then so must the resulting societal preference order. This, again, is a demand that the social welfare function will be minimally sensitive to the preference profile.
This later version is more general, having weaker conditions. The axioms of monotonicity, non-imposition, and IIA together imply Pareto efficiency, whereas Pareto efficiency (itself implying non-imposition) and IIA together do not imply monotonicity.
Independence of irrelevant alternatives (IIA) The IIA condition has three purposes (or effects):[7] Normative Irrelevant alternatives should not matter. Practical Use of minimal information. Strategic Providing the right incentives for the truthful revelation of individual preferences. Though the strategic property is conceptually different from IIA, it is closely related.
Arrow's death-of-a-candidate example (1963, page 26)[6] suggests that the agenda (the set of feasible alternatives) shrinks from, say, X = {a, b, c} to S = {a, b} because of the death of candidate c. This example is misleading since it can give the reader an impression that IIA is a condition involving two agenda and one profile. The fact is that IIA involves just one agendum ({x, y} in case of pairwise independence) but two profiles. If the condition is applied to this confusing example, it requires this: Suppose an aggregation rule satisfying IIA chooses b from the agenda {a, b} when the profile is given by (cab, cba), that is, individual 1 prefers c to a to b, 2 prefers c to b to a. Then, it must still choose b from {a, b} if the profile were, say: (abc, bac); (acb, bca); (acb, cba); or (abc, cba).
In different words, Arrow defines IIA as saying that the social preferences between alternatives x and y depend only on the individual preferences between x and y (not on those involving other candidates).
Formal statement of the theorem Let A be a set of outcomes, N a number of voters or decision criteria. We shall denote the set of all full linear orderings of A by L(A).
A (strict) social welfare function (preference aggregation rule) is a function {displaystyle F:mathrm {L(A)} ^{N}to mathrm {L(A)} } which aggregates voters' preferences into a single preference order on A.[8] An N-tuple (R1, …, RN) ∈ L(A)N of voters' preferences is called a preference profile. In its strongest and simplest form, Arrow's impossibility theorem states that whenever the set A of possible alternatives has more than 2 elements, then the following three conditions become incompatible: Unanimity, or weak Pareto efficiency If alternative a is ranked strictly higher than b for all orderings R1 , …, RN, then a is ranked strictly higher than b by F(R1, R2, …, RN). (Unanimity implies non-imposition.) Non-dictatorship There is no individual, i whose strict preferences always prevail. That is, there is no i ∈ {1, …, N} such that for all (R1, …, RN) ∈ L(A)N and all a and b, when a is ranked strictly higher than b by Ri then a is ranked strictly higher than b by F(R1, R2, …, RN). Independence of irrelevant alternatives For two preference profiles (R1, …, RN) and (S1, …, SN) such that for all individuals i, alternatives a and b have the same order in Ri as in Si, alternatives a and b have the same order in F(R1, …, RN) as in F(S1, …, SN). Informal proof Based on two proofs appearing in Economic Theory.[9][10] For simplicity we have presented all rankings as if ties are impossible. A complete proof taking possible ties into account is not essentially different from the one given here, except that one ought to say "not above" instead of "below" or "not below" instead of "above" in some cases. Full details are given in the original articles.
We will prove that any social choice system respecting unrestricted domain, unanimity, and independence of irrelevant alternatives (IIA) is a dictatorship. The key idea is to identify a pivotal voter whose ballot swings the societal outcome. We then prove that this voter is a partial dictator (in a specific technical sense, described below). Finally we conclude by showing that all of the partial dictators are the same person, hence this voter is a dictator.
Part one: There is a "pivotal" voter for B over A Part one: Successively move B from the bottom to the top of voters' ballots. The voter whose change results in B being ranked over A is the pivotal voter for B over A.
Say there are three choices for society, call them A, B, and C. Suppose first that everyone prefers option B the least: everyone prefers A to B, and everyone prefers C to B. By unanimity, society must also prefer both A and C to B. Call this situation profile 0.
On the other hand, if everyone preferred B to everything else, then society would have to prefer B to everything else by unanimity. Now arrange all the voters in some arbitrary but fixed order, and for each i let profile i be the same as profile 0, but move B to the top of the ballots for voters 1 through i. So profile 1 has B at the top of the ballot for voter 1, but not for any of the others. Profile 2 has B at the top for voters 1 and 2, but no others, and so on.
Since B eventually moves to the top of the societal preference, there must be some profile, number k, for which B moves above A in the societal rank. We call the voter whose ballot change causes this to happen the pivotal voter for B over A. Note that the pivotal voter for B over A is not, a priori, the same as the pivotal voter for A over B. In part three of the proof we will show that these do turn out to be the same.
Also note that by IIA the same argument applies if profile 0 is any profile in which A is ranked above B by every voter, and the pivotal voter for B over A will still be voter k. We will use this observation below.
Part two: The pivotal voter for B over A is a dictator for B over C In this part of the argument we refer to voter k, the pivotal voter for B over A, as the pivotal voter for simplicity. We will show that the pivotal voter dictates society's decision for B over C. That is, we show that no matter how the rest of society votes, if Pivotal Voter ranks B over C, then that is the societal outcome. Note again that the dictator for B over C is not a priori the same as that for C over B. In part three of the proof we will see that these turn out to be the same too.
Part two: Switching A and B on the ballot of voter k causes the same switch to the societal outcome, by part one of the argument. Making any or all of the indicated switches to the other ballots has no effect on the outcome.
In the following, we call voters 1 through k − 1, segment one, and voters k + 1 through N, segment two. To begin, suppose that the ballots are as follows: Every voter in segment one ranks B above C and C above A. Pivotal voter ranks A above B and B above C. Every voter in segment two ranks A above B and B above C.
Then by the argument in part one (and the last observation in that part), the societal outcome must rank A above B. This is because, except for a repositioning of C, this profile is the same as profile k − 1 from part one. Furthermore, by unanimity the societal outcome must rank B above C. Therefore, we know the outcome in this case completely.
Now suppose that pivotal voter moves B above A, but keeps C in the same position and imagine that any number (even all!) of the other voters change their ballots to move B below C, without changing the position of A. Then aside from a repositioning of C this is the same as profile k from part one and hence the societal outcome ranks B above A. Furthermore, by IIA the societal outcome must rank A above C, as in the previous case. In particular, the societal outcome ranks B above C, even though Pivotal Voter may have been the only voter to rank B above C. By IIA, this conclusion holds independently of how A is positioned on the ballots, so pivotal voter is a dictator for B over C.
Part three: There exists a dictator Part three: Since voter k is the dictator for B over C, the pivotal voter for B over C must appear among the first k voters. That is, outside of segment two. Likewise, the pivotal voter for C over B must appear among voters k through N. That is, outside of Segment One.
In this part of the argument we refer back to the original ordering of voters, and compare the positions of the different pivotal voters (identified by applying parts one and two to the other pairs of candidates). First, the pivotal voter for B over C must appear earlier (or at the same position) in the line than the dictator for B over C: As we consider the argument of part one applied to B and C, successively moving B to the top of voters' ballots, the pivot point where society ranks B above C must come at or before we reach the dictator for B over C. Likewise, reversing the roles of B and C, the pivotal voter for C over B must be at or later in line than the dictator for B over C. In short, if kX/Y denotes the position of the pivotal voter for X over Y (for any two candidates X and Y), then we have shown kB/C ≤ kB/A ≤ kC/B.
Now repeating the entire argument above with B and C switched, we also have kC/B ≤ kB/C.
Therefore, we have kB/C = kB/A = kC/B and the same argument for other pairs shows that all the pivotal voters (and hence all the dictators) occur at the same position in the list of voters. This voter is the dictator for the whole election.
Interpretations Although Arrow's theorem is a mathematical result, it is often expressed in a non-mathematical way with a statement such as no voting method is fair, every ranked voting method is flawed, or the only voting method that isn't flawed is a dictatorship.[11] These statements are simplifications of Arrow's result which are not universally considered to be true. What Arrow's theorem does state is that a deterministic preferential voting mechanism—that is, one where a preference order is the only information in a vote, and any possible set of votes gives a unique result—cannot comply with all of the conditions given above simultaneously.
Various theorists have suggested weakening the IIA criterion as a way out of the paradox. Proponents of ranked voting methods contend that the IIA is an unreasonably strong criterion. It is the one breached in most useful electoral systems. Advocates of this position point out that failure of the standard IIA criterion is trivially implied by the possibility of cyclic preferences. If voters cast ballots as follows: 1 vote for A > B > C 1 vote for B > C > A 1 vote for C > A > B then the pairwise majority preference of the group is that A wins over B, B wins over C, and C wins over A: these yield rock-paper-scissors preferences for any pairwise comparison. In this circumstance, any aggregation rule that satisfies the very basic majoritarian requirement that a candidate who receives a majority of votes must win the election, will fail the IIA criterion, if social preference is required to be transitive (or acyclic). To see this, suppose that such a rule satisfies IIA. Since majority preferences are respected, the society prefers A to B (two votes for A > B and one for B > A), B to C, and C to A. Thus a cycle is generated, which contradicts the assumption that social preference is transitive.
So, what Arrow's theorem really shows is that any majority-wins electoral system is a non-trivial game, and that game theory should be used to predict the outcome of most voting mechanisms.[12] This could be seen as a discouraging result, because a game need not have efficient equilibria; e.g., a ballot could result in an alternative nobody really wanted in the first place, yet everybody voted for.
Remark: Scalar rankings from a vector of attributes and the IIA property The IIA property might not be satisfied in human decision-making of realistic complexity because the scalar preference ranking is effectively derived from the weighting—not usually explicit—of a vector of attributes (one book dealing with the Arrow theorem invites the reader to consider the related problem of creating a scalar measure for the track and field decathlon event—e.g. how does one make scoring 600 points in the discus event "commensurable" with scoring 600 points in the 1500 m race) and this scalar ranking can depend sensitively on the weighting of different attributes, with the tacit weighting itself affected by the context and contrast created by apparently "irrelevant" choices. Edward MacNeal discusses this sensitivity problem with respect to the ranking of "most livable city" in the chapter "Surveys" of his book MathSemantics: making numbers talk sense (1994).
Alternatives based on functions of preference profiles In an attempt to escape from the negative conclusion of Arrow's theorem, social choice theorists have investigated various possibilities ("ways out"). This section includes approaches that deal with aggregation rules (functions that map each preference profile into a social preference), and other functions, such as functions that map each preference profile into an alternative.
Since these two approaches often overlap, we discuss them at the same time. What is characteristic of these approaches is that they investigate various possibilities by eliminating or weakening or replacing one or more conditions (criteria) that Arrow imposed.
Infinitely many individuals Several theorists (e.g., Fishburn[13] and Kirman and Sondermann[14]) point out that when one drops the assumption that there are only finitely many individuals, one can find aggregation rules that satisfy all of Arrow's other conditions.
However, such aggregation rules are practically of limited interest, since they are based on ultrafilters, highly non-constructive mathematical objects. In particular, Kirman and Sondermann argue that there is an "invisible dictator" behind such a rule.[14] Mihara[15][16] shows that such a rule violates algorithmic computability.[17] These results can be seen to establish the robustness of Arrow's theorem.[18] On the other hand, the ultrafilters (indeed, constructing them in an infinite model relies on the axiom of choice) are inherent in finite models as well (with no need of the axiom of choice). They can be interpreted as decisive hierarchies, with the only difference that the hierarchy's top level - Arrow's dictator - always exists in a finite model but can be unattainable (= missing) in an infinite hierarchy. In the latter case, the "invisible dictator" is nothing else but the infinite decisive hierarchy itself. If desired, it can be complemented with a limit point, which then becomes a "visible dictator". Since dictators are inseparable from decisive hierarchies, the Dictatorship prohibition automatically prohibits decisive hierarchies, which is much less self-evident than the Dictatorship prohibition.[19][20][21] See also paragraph "Relaxing the Dictatorship prohibition".
Limiting the number of alternatives When there are only two alternatives to choose from, May's theorem shows that only simple majority rule satisfies a certain set of criteria (e.g., equal treatment of individuals and of alternatives; increased support for a winning alternative should not make it into a losing one). On the other hand, when there are at least three alternatives, Arrow's theorem points out the difficulty of collective decision making. Why is there such a sharp difference between the case of less than three alternatives and that of at least three alternatives?
Nakamura's theorem (about the core of simple games) gives an answer more generally. It establishes that if the number of alternatives is less than a certain integer called the Nakamura number, then the rule in question will identify "best" alternatives without any problem; if the number of alternatives is greater or equal to the Nakamura number, then the rule will not always work, since for some profile a voting paradox (a cycle such as alternative A socially preferred to alternative B, B to C, and C to A) will arise. Since the Nakamura number of majority rule is 3 (except the case of four individuals), one can conclude from Nakamura's theorem that majority rule can deal with up to two alternatives rationally. Some super-majority rules (such as those requiring 2/3 of the votes) can have a Nakamura number greater than 3, but such rules violate other conditions given by Arrow.[22] Pairwise voting A common way "around" Arrow's paradox is limiting the alternative set to two alternatives. Thus, whenever more than two alternatives should be put to the test, it seems very tempting to use a mechanism that pairs them and votes by pairs. As tempting as this mechanism seems at first glance, it is generally far from satisfying even Pareto efficiency, not to mention IIA. The specific order by which the pairs are decided strongly influences the outcome. This is not necessarily a bad feature of the mechanism. Many sports use the tournament mechanism—essentially a pairing mechanism—to choose a winner. This gives considerable opportunity for weaker teams to win, thus adding interest and tension throughout the tournament. This means that the person controlling the order by which the choices are paired (the agenda maker) has great control over the outcome. In any case, when viewing the entire voting process as one game, Arrow's theorem still applies.
Domain restrictions Another approach is relaxing the universality condition, which means restricting the domain of aggregation rules. The best-known result along this line assumes "single peaked" preferences.
Duncan Black has shown that if there is only one dimension on which every individual has a "single-peaked" preference, then all of Arrow's conditions are met by majority rule. Suppose that there is some predetermined linear ordering of the alternative set. An individual's preference is single-peaked with respect to this ordering if he has some special place that he likes best along that line, and his dislike for an alternative grows larger as the alternative goes further away from that spot (i.e., the graph of his utility function has a single peak if alternatives are placed according to the linear ordering on the horizontal axis). For example, if voters were voting on where to set the volume for music, it would be reasonable to assume that each voter had their own ideal volume preference and that as the volume got progressively too loud or too quiet they would be increasingly dissatisfied. If the domain is restricted to profiles in which every individual has a single peaked preference with respect to the linear ordering, then simple[23] aggregation rules, which include majority rule, have an acyclic (defined below) social preference, hence "best" alternative.[24] In particular, when there are odd number of individuals, then the social preference becomes transitive, and the socially "best" alternative is equal to the median of all the peaks of the individuals (Black's median voter theorem[25]). Under single-peaked preferences, the majority rule is in some respects the most natural voting mechanism.
One can define the notion of "single-peaked" preferences on higher-dimensional sets of alternatives. However, one can identify the "median" of the peaks only in exceptional cases. Instead, we typically have the destructive situation suggested by McKelvey's Chaos Theorem:[26] for any x and y, one can find a sequence of alternatives such that x is beaten by x1 by a majority, x1 by x2, up to xk by y.
Relaxing transitivity By relaxing the transitivity of social preferences, we can find aggregation rules that satisfy Arrow's other conditions. If we impose neutrality (equal treatment of alternatives) on such rules, however, there exists an individual who has a "veto". So the possibility provided by this approach is also very limited.
First, suppose that a social preference is quasi-transitive (instead of transitive); this means that the strict preference {displaystyle succ } ("better than") is transitive: if {displaystyle xsucc y} and {displaystyle ysucc z} , then {displaystyle xsucc z} . Then, there do exist non-dictatorial aggregation rules satisfying Arrow's conditions, but such rules are oligarchic.[27] This means that there exists a coalition L such that L is decisive (if every member in L prefers x to y, then the society prefers x to y), and each member in L has a veto (if she prefers x to y, then the society cannot prefer y to x).
Second, suppose that a social preference is acyclic (instead of transitive): there do not exist alternatives {displaystyle x_{1},ldots ,x_{k}} that form a cycle ( {displaystyle x_{1}succ x_{2},;x_{2}succ x_{3},;ldots ,;x_{k-1}succ x_{k},;x_{k}succ x_{1}} ). Then, provided that there are at least as many alternatives as individuals, an aggregation rule satisfying Arrow's other conditions is collegial.[28] This means that there are individuals who belong to the intersection ("collegium") of all decisive coalitions. If there is someone who has a veto, then he belongs to the collegium. If the rule is assumed to be neutral, then it does have someone who has a veto.
Finally, Brown's theorem left open the case of acyclic social preferences where the number of alternatives is less than the number of individuals. One can give a definite answer for that case using the Nakamura number. See limiting the number of alternatives.
Relaxing assumption IIA There are numerous examples of aggregation rules satisfying Arrow's conditions except IIA. The Borda rule is one of them. These rules, however, are susceptible to strategic manipulation by individuals.[29] See also Interpretations of the theorem above.
Relaxing the Pareto criterion Wilson (1972)[30] shows that if an aggregation rule is non-imposed and non-null, then there is either a dictator or an inverse dictator, provided that Arrow's conditions other than Pareto are also satisfied. Here, an inverse dictator is an individual i such that whenever i prefers x to y, then the society prefers y to x.
Amartya Sen offered both relaxation of transitivity and removal of the Pareto principle.[31] He demonstrated another interesting impossibility result, known as the "impossibility of the Paretian Liberal" (see liberal paradox for details). Sen went on to argue that this demonstrates the futility of demanding Pareto optimality in relation to voting mechanisms.
Relaxing the Dictatorship prohibition Andranik Tangian (2010) introduced measures of dictator's "representativeness", for instance, the "popularity index" defined as the average size of the social group whose pairwise preferences are shared (= represented) by the dictator, averaged over all pairs of alternatives and all preference profiles. It was shown that there always exist "good" Arrow's dictators who on the average represent a majority.[32] Since they are rather representatives of the society - like democratically elected presidents - there are no self-evident reasons to prohibit them. Restricting the notion of dictator to "bad" ones only, i.e. those who on the average represent a minority, Arrow's axioms were proven to be consistent.[20][21] Social choice instead of social preference In social decision making, to rank all alternatives is not usually a goal. It often suffices to find some alternative. The approach focusing on choosing an alternative investigates either social choice functions (functions that map each preference profile into an alternative) or social choice rules (functions that map each preference profile into a subset of alternatives).
As for social choice functions, the Gibbard–Satterthwaite theorem is well-known, which states that if a social choice function whose range contains at least three alternatives is strategy-proof, then it is dictatorial.
As for social choice rules, we should assume there is a social preference behind them. That is, we should regard a rule as choosing the maximal elements ("best" alternatives) of some social preference. The set of maximal elements of a social preference is called the core. Conditions for existence of an alternative in the core have been investigated in two approaches. The first approach assumes that preferences are at least acyclic (which is necessary and sufficient for the preferences to have a maximal element on any finite subset). For this reason, it is closely related to relaxing transitivity. The second approach drops the assumption of acyclic preferences. Kumabe and Mihara[33] adopt this approach. They make a more direct assumption that individual preferences have maximal elements, and examine conditions for the social preference to have a maximal element. See Nakamura number for details of these two approaches.
Other alternatives Arrow originally rejected cardinal utility as a meaningful tool for expressing social welfare,[34] and so focused his theorem on preference rankings, but later stated that a cardinal score system with three or four classes "is probably the best".[2] Arrow's framework assumes that individual and social preferences are "orderings" (i.e., satisfy completeness and transitivity) on the set of alternatives. This means that if the preferences are represented by a utility function, its value is an ordinal utility in the sense that it is meaningful so far as the greater value indicates the better alternative. For instance, having ordinal utilities of 4, 3, 2, 1 for alternatives a, b, c, d, respectively, is the same as having 1000, 100.01, 100, 0, which in turn is the same as having 99, 98, 1, .997. They all represent the ordering in which a is preferred to b to c to d. The assumption of ordinal preferences, which precludes interpersonal comparisons of utility, is an integral part of Arrow's theorem.
For various reasons, an approach based on cardinal utility, where the utility has a meaning beyond just giving a ranking of alternatives, is not common in contemporary economics. However, once one adopts that approach, one can take intensities of preferences into consideration, or one can compare (i) gains and losses of utility or (ii) levels of utility, across different individuals. In particular, Harsanyi (1955)[35] gives a justification of utilitarianism (which evaluates alternatives in terms of the sum of individual utilities), originating from Jeremy Bentham. Hammond (1976)[36] gives a justification of the maximin principle (which evaluates alternatives in terms of the utility of the worst-off individual), originating from John Rawls.
Not all voting methods use, as input, only an ordering of all candidates.[37] Methods which don't, often called "rated" or "cardinal" (as opposed to "ranked", "ordinal", or "preferential") electoral system, can be viewed as using information that only cardinal utility can convey. In that case, it is not surprising if some of them satisfy all of Arrow's conditions that are reformulated.[38] Range voting is such a method.[5][39] Whether such a claim is correct depends on how each condition is reformulated.[40] Other rated electoral system which pass certain generalizations of Arrow's criteria include approval voting and majority judgment. Note that Arrow's theorem does not apply to single-winner methods such as these, but Gibbard's theorem still does: no non-defective electoral system is fully strategy-free, so the informal dictum that "no electoral system is perfect" still has a mathematical basis.[41] Finally, though not an approach investigating some kind of rules, there is a criticism by James M. Buchanan, Charles Plott, and others. It argues that it is silly to think that there might be social preferences that are analogous to individual preferences.[42] Arrow (1963, Chapter 8)[43] answers this sort of criticism seen in the early period, which come at least partly from misunderstanding. | 7,143 | 33,890 | {"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-2023-14 | latest | en | 0.947429 |
https://os.mbed.com/users/soramimi/code/Clock/diff/7974d199ed39/main.cpp/ | 1,713,928,512,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818999.68/warc/CC-MAIN-20240424014618-20240424044618-00749.warc.gz | 390,752,297 | 8,408 | Simply Clock
Revision:
1:7974d199ed39
Parent:
0:7e901efc73e7
--- a/main.cpp Fri Mar 25 16:54:13 2011 +0000
+++ b/main.cpp Mon Mar 28 21:49:06 2011 +0000
@@ -13,36 +13,36 @@
unsigned long convert_gregorian_to_julian(int year, int month, int day)
{
- if (month < 3) {
- month += 9;
- year--;
- } else {
- month -= 3;
- }
- year += 4800;
- int c = year / 100;
- return c * 146097 / 4 + (year - c * 100) * 1461 / 4 + (153 * month + 2) / 5 + day - 32045;
+ if (month < 3) {
+ month += 9;
+ year--;
+ } else {
+ month -= 3;
+ }
+ year += 4800;
+ int c = year / 100;
+ return c * 146097 / 4 + (year - c * 100) * 1461 / 4 + (153 * month + 2) / 5 + day - 32045;
}
void convert_julian_to_gregorian(unsigned long j, int *year, int *month, int *day)
{
- int y, m, d;
- y = (j * 4 + 128179) / 146097;
- d = (j * 4 - y * 146097 + 128179) / 4 * 4 + 3;
- j = d / 1461;
- d = (d - j * 1461) / 4 * 5 + 2;
- m = d / 153;
- d = (d - m * 153) / 5 + 1;
- y = (y - 48) * 100 + j;
- if (m < 10) {
- m += 3;
- } else {
- m -= 9;
- y++;
- }
- *year = y;
- *month = m;
- *day = d;
+ int y, m, d;
+ y = (j * 4 + 128179) / 146097;
+ d = (j * 4 - y * 146097 + 128179) / 4 * 4 + 3;
+ j = d / 1461;
+ d = (d - j * 1461) / 4 * 5 + 2;
+ m = d / 153;
+ d = (d - m * 153) / 5 + 1;
+ y = (y - 48) * 100 + j;
+ if (m < 10) {
+ m += 3;
+ } else {
+ m -= 9;
+ y++;
+ }
+ *year = y;
+ *month = m;
+ *day = d;
}
void display(int year, int month, int day, int hour, int minute, int second)
@@ -57,7 +57,7 @@
eth.setup();
-
+
double last = 0;
@@ -66,8 +66,10 @@
ntp.setTime(server);
+ } else {
+ wait(0.1);
}
-
+
time_t t = time(0);
double s = t + 2440588.0 * 24 * 60 * 60; // chronological julian second
@@ -86,12 +88,10 @@
display(year, month, day, hour, minute, second);
- if (hour == 4 && minute == 0 && second == 0) {
+ if (minute == 59 && second == 30) { | 871 | 1,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-18 | latest | en | 0.244067 |
http://www.enotes.com/homework-help/after-12-00-when-hands-clock-180-degree-each-other-223377 | 1,477,409,834,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720154.20/warc/CC-MAIN-20161020183840-00112-ip-10-171-6-4.ec2.internal.warc.gz | 436,102,330 | 10,702 | # After 12:00, when are the hands of the clock at 180 degree to each other for the first time?
Asked on by alicesmith
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
We need to find what the time is when the hands of the clock are at 180 degree to each other for the first time after 12:00. Let this be after x minutes. Now we know that the minute hand moves (360/60)*x degrees in x minutes. The hour hand moves (360/12*60)*x degrees in a minute. Now when the two hands are at 180 degree to each other it means that (360/60)*x - (360/12*60)*x = 180
canceling 180
=> 2x/60 - 2x/12*60 = 1
=> x (1/ 30 – 1/360) = 1
=> x = (1/ 30 – 1/360) ^-1
=> x = 32.727 min
Therefore the two hands are at 180 degree to each other at 32.72 minutes past 12:00.
krishna-agrawala | College Teacher | (Level 3) Valedictorian
Posted on
Every hour the minute hand of the clock moves one full circle or 360 degree. During the same period the hour hand moves by 1/12th of a complete circle, or 30 degrees.
Thus every 1 hour after 12:00 the net angle between the hour hand and minute hand increases by:
360 - 30 = 330 degrees
Therefore:
The time taken for the angle between hour and minute hand to increase from by 180 degrees from 0 to 180 degrees is given by:
Time for the angle to increase by 180 degrees = 180*(1/330)
= 6/11 hours.
= (6/11)*60 = 32.7272 minutes = 32 minutes and 44 seconds
Therefore:
Time after 12:00 when hands of clock are at 180 degrees to each others = 12:32:44
neela | High School Teacher | (Level 3) Valedictorian
Posted on
We know that the hour hand rotates one full round in 12 hours or 12*60 = 720 minutes. So the hour hand covers 360 degrees in 720 minutes. Therefore the hour hand moves with a speed of 1/2 degrre per minute.
The minute hand moves 360 degree every hour. So the speed of the minute hand per minute is 360/60 = 5 degree per minute.
Therefore difference in the angle of hour hand and minute hand per minute = speed of minute hand/minute - speed of hour hand per minute = 5-1/2 deg/min = 4.5 degree per minute.
Therefore it takes 180 deree/ 4.5 = 40 mintes for the angle to be 180 degree for the first time between the hour hand and the minute hand.
We’ve answered 317,705 questions. We can answer yours, too. | 680 | 2,295 | {"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} | 4.15625 | 4 | CC-MAIN-2016-44 | latest | en | 0.862495 |
https://shop.luckylittlelearners.com/product/2nd-grade-geometry-exit-tickets/ | 1,632,674,234,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057882.56/warc/CC-MAIN-20210926144658-20210926174658-00075.warc.gz | 543,577,260 | 71,453 | 2nd Grade Geometry Exit Tickets - Lucky Little Learners
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Exit Tickets are a great tool to quickly assess your students' understanding of a concept. Exit tickets should take no longer than a couple of minutes to complete and they provide the teacher with valuable information to help guide their instruction.
These 2nd Grade Geometry Exit Tickets are part of a discounted bundle that you can purchase HERE!
This set of exit tickets cover the following 2nd grade geometry concepts:
CCSS.MATH.CONTENT.2.G.A.1
Recognize and draw shapes having specified attributes, such as a given number of angles or a given number of equal faces. Identify triangles, quadrilaterals, pentagons, hexagons, and cubes.
CCSS.MATH.CONTENT.2.g.a.2
Partition a rectangle into rows and columns of same-size squares and count to find the total number of them.
CCSS.MATH.CONTENT.2.G.A.3
Partition circles and rectangles into two, three, or four equal shares, describe the shares using the words halves, thirds, half of, a third of, etc., and describe the whole as two halves, three thirds, four fourths. Recognize that equal shares of identical wholes need not have the same shape.
There are 15 exit tickets in this set (5 different exit tickets per skill). Each page of exit tickets has 2 copies of the same exit ticket to save on paper. Each exit ticket has 3-6 problems on each sheet. The standard is listed in the top right corner of every exit ticket.
Reviews
There are no reviews yet. | 363 | 1,585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-39 | longest | en | 0.906088 |
https://www.techiedelight.com/replace-element-array-product-every-element-without-using-division-operator/ | 1,553,477,581,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203547.62/warc/CC-MAIN-20190325010547-20190325032547-00040.warc.gz | 902,462,320 | 24,965 | # Replace each element of the array with product of every other element without using division operator
Given an array of integers, replace each element of the array with product of every other element in the array without using division operator.
For example,
Input: { 1, 2, 3, 4, 5 }
Output: { 120, 60, 40, 30, 24 }
Input: { 5, 3, 4, 2, 6, 8 }
Output: { 1152, 1920, 1440, 2880, 960, 720 }
Naive approach would be to calculate product of all elements in the left sub-array and right sub-array for each element of the array. The time complexity of above solution is O(n2).
We can solve this problem in linear time by using two auxiliary arrays left[] and right[] where left[i] stores the product of all elements in the sub-array A[0..i-1] and right[i] stores the product of all elements in sub-array A[i+1..n-1]. Now for each element A[i], we simply replace it with product of its left-subarray and right-subarray (i.e. A[i] = left[i] * right[i]).
## C
Output:
1152 1920 1440 2880 960 720
## Java
Output:
The time complexity of above solution is O(n) and auxiliary space used by the program is O(n).
We can use recursion to solve this problem in linear time and constant space. The idea is to recursively calculate the product of all elements in the right sub-array and pass left-subarray product in function arguments.
## C
Output:
1152 1920 1440 2880 960 720
## Java
Output:
[1152, 1920, 1440, 2880, 960, 720]
The time complexity of above solution is O(n) and auxiliary space used by the program is O(1) (if we ignore recursion stack space, no auxiliary space is used).
(3 votes, average: 4.67 out of 5)
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Another O(n) approach without recursion.
-> Find product of right of each number from left and store it in array1.
-> Find product of left of each number from right and store it in array2.
-> Now, multiple array1 and array2 at each index to get the final array (array2 can be avoided by taking care of the multiplication inline)
Guest
There was just a post on HN to a site that went over this exact problem. Essentially, you do two linear scans to calculate the product of every number before an index, and to calculate the product of every number after an index. Then, for each index, the answer is just multiplying those two numbers you found for that index.
Guest
How about this?
– Calculate the product of all array elements using recursion.
– Upon the return from recursion, update array elements starting from the last element.
– At the same time, keep track of the product of elements that are returned from recursive calls.
https://ideone.com/VDwtBA | 717 | 2,832 | {"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} | 3.265625 | 3 | CC-MAIN-2019-13 | latest | en | 0.799973 |
https://electronics.stackexchange.com/questions/245978/can-i-power-multiple-devices-from-a-single-power-supply | 1,723,735,892,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641299002.97/warc/CC-MAIN-20240815141847-20240815171847-00639.warc.gz | 181,881,917 | 42,260 | # Can I power multiple devices from a single power supply?
I'm planning on building a device that has the following requirements:
• 4x devices that are rated as 5v (1.5 amps normal load, however during high loads can require upto 2 amps)
• 1x raspberry pi
I've sourced a power supply that takes 240 AC mains supply and provides 5v @ 8 Amps.
My plan is to connect all the devices in parallel.
My questions are as follows:
• Is this power supply a good idea?
• When the system is idle, will this be an issue, i.e. the devices are not drawing maximum current?
Please advise!
EDIT
This is the kind of power supply I'm planning on using, I don't know if that helps?
http://www.ebay.co.uk/itm/DC-12V-24V-5V-Universal-Regulated-Switching-Power-Supply-for-LED-Strip-CCTV-UK-/291087849273?var=&hash=item43c6301739:m:m6INReCcQQadil-AVWF6qsQ
• There is nothing to go on here to work out if the power supply might cause problems. Ripple voltage, minimal load current, sensitivity of devices to ripple. Tolerance of power supply output, maximum and minimum operating voltages of devices etc.. Commented Jul 14, 2016 at 11:45
• so, in theory would this work? all devices are suppose to be 5v. I'm just concerned about the current. Commented Jul 14, 2016 at 11:48
• There is NO theory. Commented Jul 14, 2016 at 11:52
• Sure there is ;-) You only need to solve this equation: preposterousuniverse.com/blog/wp-content/uploads/2013/01/… then when W = 0 it won't work but when W = 1 it will. Commented Jul 14, 2016 at 12:16
• @FakeMoustache BUT, the Higgs is a Boojum, you see. aND ALSO HERE. Commented Jul 14, 2016 at 15:04
## 2 Answers
4x devices that are rated as 5v (1.5 amps normal load, however during high loads can require up to 2 amps)
1x raspberry pi
I've sourced a power supply that takes 240 AC mains supply and provides 5v @ 8 Amps.
The RPI can see up to 1 Amp draw, or higher. Some take 0.5A by themselves with nothing plugged in.
So you have a system that takes 6.5 Amps normally, but can jump to 9 Amps during high load.
8 Amps is less than 9 Amps.
### So No, that supply isn't a good choice for this.
Is this power supply a good idea?
It is generally a good idea to design a constant voltage power supply with more current capability than you need. This will generally allow the supply to run cooler, and it also allows for the possibility you underestimated your actual needs.
When the system is idle, will this be an issue, i.e. the devices are not drawing maximum current?
This is entirely down to the design of your power supply circuit. An 8 A supply could be designed that does not lose regulation (meaning the output voltage goes out of spec) under low currents down to a few mA. But if you made a very very bad design, it could possibly lose regulation at 1 A or higher. If you just designed a voltage divider and told your boss it was a voltage regulator, it could lose regulation at 7.9 A load.
As the comments point out, there are numerous other considerations that determine whether your power supply is appropriate. The current capability being 8 A and the actual load being 2-3 A is not going to be a problem if the regulator circuit is at all well designed. | 838 | 3,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-33 | latest | en | 0.935204 |
https://gmatclub.com/forum/how-many-four-digit-numbers-divisible-by-4-can-be-made-with-4013.html | 1,498,554,653,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321306.65/warc/CC-MAIN-20170627083142-20170627103142-00266.warc.gz | 762,866,755 | 44,159 | It is currently 27 Jun 2017, 02:10
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How many four digit numbers divisible by 4 can be made with
Author Message
Director
Joined: 03 Jul 2003
Posts: 652
How many four digit numbers divisible by 4 can be made with [#permalink]
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06 Jan 2004, 17:05
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
How many four digit numbers divisible by 4 can be made with the digits 1, 2, 3, 4, 5 if the repetition of digits is not allowed?
Director
Joined: 14 Oct 2003
Posts: 583
Location: On Vacation at My Crawford, Texas Ranch
Show Tags
06 Jan 2004, 17:16
Any number (regardless of number of digits) with the last 2 digits that are divisble by 4 is divisble by 4
out of 1,2,3,4,5
xx12, xx24, xx32, xx52
thus 3!x4 = 24
Director
Joined: 03 Jul 2003
Posts: 652
Show Tags
06 Jan 2004, 17:19
I missed 24. How one can make sure that he/she gets all the numbers
divided by 4 in these kind of questions. These kind of questions tend to
take more time for me. Any suggestions?
Director
Joined: 14 Oct 2003
Posts: 583
Location: On Vacation at My Crawford, Texas Ranch
Show Tags
06 Jan 2004, 17:31
I missed 24. How one can make sure that he/she gets all the numbers
divided by 4 in these kind of questions. These kind of questions tend to
take more time for me. Any suggestions?
Just know the characteristics of some base numbers - that will save you some time.
e.g., multiples of 2,3,4,5,..etc.
Like I said - multiples of 4 - the last two digits will always be multiples of 4.
Multiples of 2 will always be even (obviously)
Multiples of 5 will always end in 0 or 5.
Multiples of 10 will always end in 0.
Senior Manager
Joined: 30 Aug 2003
Posts: 322
Location: dallas , tx
Show Tags
08 Jan 2004, 14:30
okay,,this is the problem i was referrign to...
_________________
shubhangi
Director
Joined: 14 Oct 2003
Posts: 583
Location: On Vacation at My Crawford, Texas Ranch
Show Tags
08 Jan 2004, 14:39
shubhangi wrote:
okay,,this is the problem i was referrign to...
Shubanghi,
Yes it shouldn't be 3! per se just (3x2) - which conveniently is the same as 3!
thus for xx12 3 letters for the first number x 2 letters that we can choose from the numbers that are left.
08 Jan 2004, 14:39
Display posts from previous: Sort by | 803 | 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} | 3.953125 | 4 | CC-MAIN-2017-26 | longest | en | 0.924369 |
https://binary-notes.ru/calculation-of-powerset/ | 1,679,980,666,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948765.13/warc/CC-MAIN-20230328042424-20230328072424-00387.warc.gz | 169,145,692 | 40,215 | # Calculation of PowerSet on C#
This text explains a binary algorithm to calculate a power set of any sequence. It has a sample code in C#.
The Wikipedia text can tell you in details about powerset algorithm. Personally it has a clean mathematical description which is hard to transform into programming code. Especially if you try apply it to some more complex case than just a sequence of elements. Further in text I’ll try to explain the basics with a human-friendly language and propose a generic solution to apply to any case.
From the Wikipedia text you should notice that a whole number of sets for a sequence of N elements is 2N (or the binary left shift of 1 to N). Let’s take a sequence of 5 elements. The number of all subsets is 32 (binary 100000). In order to find subsets you can use binary representation of the index of a subset. For example, the subset #3 has binary representation as 000011. Each set bit corresponds to an index in the main set. The corresponding subset for #3 is the sequence {0,1}. Let’s take another number – #16 (001000). The corresponding sequence is {3}. Index of each of subset elements can be obtained by evaluating indexes of set bits.
The whole example in you can find in my GitHub repository with tests.
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## Speeding up calculation of an average in continuous processesSpeeding up calculation of an average in continuous processes
Processing big sets of data can cause a low performance due to inefficient calculations. In order to overcome this issue we should rewrite calculations in a way which allows getting | 350 | 1,652 | {"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-2023-14 | latest | en | 0.87261 |
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In economics an isocost line shows all combinations of inputs which cost the same total amount The isocost line is combined with the isoquant map to determine the optimal production point at any given level of output. Specifically, the point. Isocost-isoquant analysis: theory of production: The production function: a figure known as an isoquant diagram (Figure 1). In the graph, goldsmith-hours per. Isoquants: An isoquant (equal quantity) is a curve that shows the combinations of certain inputs such as Labor (L) and Capital (K) that will produce a certain.
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Isoquant and Isocost Lines (With Diagram) | Economics
Then an outlay of Rs. That is why, in the case of constant returns to scale, the production function is homogeneous of degree one. To see this, consider an example. A line joining tangency points of isoquants and isocosts with input prices held constant is called the expansion path. But combination S is preferred to combination B, being on the higher portion of isoquant IQ 1. In other words when the quantity of labour is increased by LR and RS, the output declines from to and to In it, two factors capital and labour replace two commodities of consumption.
To explain the law, capital is taken as a fixed factor and labour as a variable factor. The firm is in equilibrium at point P where the isoquant curve is tangent to the isocost line CL.
Explain a firm’s equilibrium with the help of isoquants and isocost line.
Views Read View source View history. The different tax authorities and their functions are outlined below: In fact, in between the units of output,etc. The slope depends on the prices of factors of production and the amount of money which the firm spends on the factors.
I JUCA PIRAMA DE GONALVES DIAS PDF
All these economies help isoquany increasing the returns to scale more than proportionately. This is known as the stage of diminishing returns.
There are two essential or second order conditions for the equilibrium of jsoquant firm. There being perfect competition, intensive bidding raises wages, rent and interest.
As we move along an isoquant downward to the right, each point on it represents the substitution of labour for capital. For the two production inputs labour and capital, with fixed unit costs of the inputs, the equation of the isocost line is.
The slope of iso cost line indicates the ratio of the factor prices. Increasing returns to scale also result from specialisation and division of labour. Isoquaant arrive at the conclusion that a firm will find it profitable to produce only in the second stage of the law of variable proportions for it will be uneconomical to produce in the regions to the left or right of the ridge lines which form the first stage and the third stage of the law respectively.
On the other hand, combination A is preferred to R, the former being on the higher portion of the isoquant IQ. A profit maximisation firm faces two choices of optimal combination of factors inputs: If izoquant were relatively more expensive, the isocost lines would be steeper in Fig.
The producer must have sufficient capacity to buy necessary factor inputs to be able to reach its desired production level. When the amount of money spent by the firm changes, the isocost line may shift but its slope remains the same.
DEFINITIVE ANTLR REFERENCE PDF
According to Hibbdon, Economic Rent is the difference between the actual payment to a factor and its supply price You must be logged in to post a comment. Isoquant indicates various combinations of two factors of production which give the same level of output per unit of time.
Isocost – Wikipedia
The dotted segments of an isoquant are the waste- bearing segments. This is an isoquant. In the upper dotted portion, more capital and in the lower dotted portion more labour than necessary is employed. Self-Assessment is a system under which the taxpayer is required to declare the basis of his assessment e. isouqant
An isocost line shows the alternative quantities of two factors viz. Some items of equipment or some activities have a minimum size and cannot be divided into smaller units. Such a situation leads to increasing marginal returns.
ISO QUANT AND ISOCOST – WikiEducator
The firm also maximises its profits by maximising its output, given its cost outlay and the prices of the two factors. An isoquant shows the various combination of two inputs that can be used to produce a specific level of output. To treble output, units of both factors are trebled. On the other hand, where these ridge lines cut the isoquants, the marginal product of the inputs is zero. | 1,069 | 4,911 | {"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-2020-16 | latest | en | 0.90896 |
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Get Full Access to Conceptual Physics - 12 Edition - Chapter 1 - Problem 10e
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# Scientists call a theory that unites many ideas in a
ISBN: 9780321909107 29
## Solution for problem 10E Chapter 1
Conceptual Physics | 12th Edition
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Problem 10E
Scientists call a theory that unites many ideas in a simple way “beautiful.” Are unity and simplicity among the criteria of beauty outside of science? Support your answer.
Step-by-Step Solution:
Step 1 of 3
Solution 10E Something that is beautiful for a scientist may or may not be the same for an audience outside of science. A scientist is well aware of scientific principles, theories etc. But same may not be true for a person not studying science. Therefore, for a theory to be understood by common people it has be simply described to make it attractive. For example, let us consider the topic of an object sinking or floating in water. A scientist will easily conclude that an object having a density higher than that of water will sink and an object having a lesser density than water will float. But a common person may have difficulty in understanding without proper examples. An iron ball may be taken and dropped into a bucket of water. It will sink in water as it has more density. Similarly, ice floats in water as it has density lower than that of water. Such examples will make things look simple to a person outside of science. Uniting many ideas into a theory may also look beautiful to a common person. But those scientific ideas has to be easily understandable or has to be explained with proper examples. When such ideas are united to make a theory, it may seem beautiful to a person having less understanding of science. For example, to explain Newton’s first law of motion we can present some simple examples like pushing a book on a table to a particular direction or stopping a moving ball to change their states of rest or motion. Uniting both examples, the theory of Newton’s first law of motion can be defined. This make things simpler to understand and that is where the beauty in this approach may lie for a common person to understand scientific theories.
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# MCQ: Simple Interest - 1 - SSC CGL MCQ
Test Description
## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ: Simple Interest - 1
MCQ: Simple Interest - 1 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The MCQ: Simple Interest - 1 questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ: Simple Interest - 1 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ: Simple Interest - 1 below.
Solutions of MCQ: Simple Interest - 1 questions in English are available as part of our Quantitative Aptitude for SSC CGL for SSC CGL & MCQ: Simple Interest - 1 solutions in Hindi for Quantitative Aptitude for SSC CGL course. Download more important topics, notes, lectures and mock test series for SSC CGL Exam by signing up for free. Attempt MCQ: Simple Interest - 1 | 15 questions in 15 minutes | Mock test for SSC CGL preparation | Free important questions MCQ to study Quantitative Aptitude for SSC CGL for SSC CGL Exam | Download free PDF with solutions
MCQ: Simple Interest - 1 - Question 1
### Shivam lent out Rs. 50X for Y years at Y% per annum simple interest. If he received Rs. 32X as interest, what is the value of Y?
Detailed Solution for MCQ: Simple Interest - 1 - Question 1
Y2 = 64
Y = 8
Hence, Option B is correct.
MCQ: Simple Interest - 1 - Question 2
### Madhav borrows Rs. 15000 at the rate of 24% per annum simple interest and Bhuvan borrows Rs. 24000 at the rate of 12% per annum simple interest. In how many years will their amounts of debt be equal?
Detailed Solution for MCQ: Simple Interest - 1 - Question 2
Let the time = T
15,000 + 3,600T = 24,000 + 2,880T
720T = 9,000
T = 12.50
So the answer = 12.50 years
Hence, Option B is correct.
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MCQ: Simple Interest - 1 - Question 3
### Manoj lent out Rs. 20,000 at 8% per annum on simple interest. If he received Rs. 40,000 as interest, for how much time did he lent out?
Detailed Solution for MCQ: Simple Interest - 1 - Question 3
T = 25 years
Hence, Option D is correct.
MCQ: Simple Interest - 1 - Question 4
Saurabh lent out some money at 12.50% per annum simple interest for two years and the same amount at 15% per annum simple interest for two years. If he received a combined interest of Rs. 3,300, how much amount did he lend out in each case?
Detailed Solution for MCQ: Simple Interest - 1 - Question 4
Let the sum in each case = Rs. X
55X = 3,300 × 100
X = Rs. 6,000
Hence, Option A is correct.
MCQ: Simple Interest - 1 - Question 5
A sum of Rs. 24,000 amounts to Rs. 32,000 in 4 years at a certain rate of simple interest per annum. To what does a sum of Rs. 6,000 amount at the same rate in 9 years?
Detailed Solution for MCQ: Simple Interest - 1 - Question 5
Interest in first case = 32,000 – 24,000 = 8,000
Hence, Option D is correct.
MCQ: Simple Interest - 1 - Question 6
Aman lent out Rs. 4000 to Mohan at X% per annum simple interest and Rs. 4000 to Ajit at Y% per annum simple interest. If at the end of 3 years, he received Rs. 60 more from Mohan than from Ajit, what is the difference between X and Y?
Detailed Solution for MCQ: Simple Interest - 1 - Question 6
The difference is created by the difference of the rate of interest.
X – Y = 0.50
Hence, Option A is correct.
MCQ: Simple Interest - 1 - Question 7
When the annual rate of simple interest increases from X% to (X + 3)% , a man’s yearly interest increases by Rs. 630. Find the principal amount.
Detailed Solution for MCQ: Simple Interest - 1 - Question 7
3% of the principal amount = 630
Hence, Option D is correct.
MCQ: Simple Interest - 1 - Question 8
Shri deposited Rs. 8,000 which amounted to Rs. 9,200 in 3 years at X% p.a. simple interest. Had the rate been “X + 2”% p.a., what would be the amount in 3 years?
Detailed Solution for MCQ: Simple Interest - 1 - Question 8
So the answer = 9200 + 480 = Rs. 9680
Hence, Option C is correct.
MCQ: Simple Interest - 1 - Question 9
A man lent out Rs. 1,200 and Rs. 1,500 for 3 years and 2 years respectively at 5% p.a. and at 6% p.a.. What amount did he take as interest?
Detailed Solution for MCQ: Simple Interest - 1 - Question 9
Hence, Option C is correct.
MCQ: Simple Interest - 1 - Question 10
A man lent out Rs. 2,000 on simple interest for 8 years at the rate of 5% per annum. What is the earned interest?
Detailed Solution for MCQ: Simple Interest - 1 - Question 10
Hence, Option D is correct.
MCQ: Simple Interest - 1 - Question 11
Monty lent out Rs. 12,500 to Lakhan at M% per annum simple interest and Rs. 12,500 to Ram at N% per annum simple interest. If at the end of 2 years, he received Rs. 600 less from Lakhan than from Ram, what is the difference between M and N?
Detailed Solution for MCQ: Simple Interest - 1 - Question 11
The difference is created by the difference of the rate of interest.
Hence, Option B is correct.
MCQ: Simple Interest - 1 - Question 12
A man lent out Rs. X on simple interest for 6 years at the rate of 10% per annum. If he received interest of Rs. 4,800, what is the value of X?
Detailed Solution for MCQ: Simple Interest - 1 - Question 12
X = Rs. 8,000
Hence, Option A is correct.
MCQ: Simple Interest - 1 - Question 13
A lends Rs. 4800 to B and Rs. ‘X’ to C for the same time at the same rate of interest of 12.50% per annum. If after 4 years A received a combined interest of Rs. 4800 from both of them, what is the value of ‘X’?
Detailed Solution for MCQ: Simple Interest - 1 - Question 13
Total interest of 4 years = 4 × 12.50% = 50% of total principal
So the answer = Rs.9,600 – Rs. 4,800 = Rs. 4,800
Hence, Option C is correct.
MCQ: Simple Interest - 1 - Question 14
Aisha lent out Rs. 15,000 on simple interest for 84 months at the rate of 20% per annum. What amount would she receive as interest?
Detailed Solution for MCQ: Simple Interest - 1 - Question 14
Hence, Option B is correct.
MCQ: Simple Interest - 1 - Question 15
Rajdeep lent out a combined amount of Rs. 72,000 such that simple interest in 2 years at 10% per annum on the first part is the same as the simple interest in 2 years at 15% per annum on the second part. What is the interest in 2 years at 10% per annum on the first part?
Detailed Solution for MCQ: Simple Interest - 1 - Question 15
Interest on the first part = 2 × 10% = 20%
Interest on the second part = 2 × 15% = 30%
Let the first part = X
Second part = (72,000 – X)
20% of X = 30% of (72,000 – X)
50% of X = 21,600
X = 43,200
So the answer = 20% of 43,200 = Rs. 8,640
Hence, Option D is correct.
## Quantitative Aptitude for SSC CGL
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## Quantitative Aptitude for SSC CGL
314 videos|170 docs|185 tests | 2,094 | 7,132 | {"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-2024-38 | latest | en | 0.840174 |
https://www.physicsforums.com/threads/mechanical-efficiency-of-a-ramp.493808/ | 1,571,462,917,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986688826.38/warc/CC-MAIN-20191019040458-20191019063958-00365.warc.gz | 1,054,720,561 | 18,760 | # Archived Mechanical Efficiency of a Ramp
#### iRamie
1. Homework Statement
We have to find the efficiency of a ramp for a lab. We are given an inclined plane, a wooden block and a few masses. We are also given a newton spring scale.
2. Homework Equations
efficiency = (Eout/Ein)*100%
Eg = mgh
Work = FD
3. The Attempt at a Solution
Is the efficiency equal to Eg/Work?
Because the only work i am putting in is me pushing the wooden block up the ramp. That will be the Ein. The Eout will be the gravitational potential energy of the object?
How would i calculate the efficiency of the ramp?
Related Introductory Physics Homework Help News on Phys.org
#### Nahyo
Substitute the Eg value for the Eout value and substitute the work value into the Ein value. Perform the calculation.
#### late347
The ramp is what is called a non-isolated mechanical system.
According to my own high school physics textbook
$W_{intake}= pushing work$ (I'm thinking that this is probably correct, but you have laboratory so why not test it out?)
$W_{benefit}= E_{pot.~at ~highpoint}$ (probably true as well...)
$W_{wasted} = W_{friction}$ (definitely true)
Wintake=Wbenefit+Wwasted
Wintake-Wwasted= Wbenefit
ramp efficiency η = $\frac { W_{benefit}}{W_{intake}}$
Initially static friction will be overcome by the pushing action, and afterwards sliding friction will cause work. Or in other words, friction requires more force in order to be overcome by the ramp-user who pushes the box.
You would have less work required, if the ramp were frictionless. But of course in real-life ramp you most likely have some friction, so therefore more work is required to push the box along the ramp.
Overall, you were on the correct track to solving the problem. The practical problem for you is to simply figure out how you can find out the correct values for Wintake, Wbenefit and Wwasted by using measurement and calculation in the lab.
This is my understanding of the situation and anybody more experrienced is welcome to criticize.
η
= $\frac {W_{benefit}}{W_{input}}$
#### ehild
Homework Helper
Measure both the height and the length of the ramp (from the front side of the block to the top. Move the block along the ramp slowly, with constant speed so as the spring scale shows a constant force. Read that force: it is F. The work done is Wintake=FD where D is the length of the ram. You know Wbenefit= mgh, so you can calculate the efficiency
$\eta =\frac {W_{benefit}}{W_{input}}$.
#### late347
Measure both the height and the length of the ramp (from the front side of the block to the top. Move the block along the ramp slowly, with constant speed so as the spring scale shows a constant force. Read that force: it is F. The work done is Wintake=FD where D is the length of the ram. You know Wbenefit= mgh, so you can calculate the efficiency
$\eta =\frac {W_{benefit}}{W_{input}}$.
how should you measure the force in practical terms with newton-scale (spring scale)?
Especially if you are in the lab alone, and you do all work with only your own two hands?
If the pulling force has different direction in the vector compared to the actual angle in which the distance is travelled that could be problematic.
Therefore the pulling should occur at same angle as the ramp itself.
Its probably easier to use the scale with pulling.
If you pull at some constant speed, then I reckon the force reading should stay at a stable value prety much?
#### ehild
Homework Helper
how should you measure the force in practical terms with newton-scale (spring scale)?
Especially if you are in the lab alone, and you do all work with only your own two hands?
If the pulling force has different direction in the vector compared to the actual angle in which the distance is travelled that could be problematic.
Therefore the pulling should occur at same angle as the ramp itself.
Its probably easier to use the scale with pulling.
If you pull at some constant speed, then I reckon the force reading should stay at a stable value prety much?
Yes, you attach the block to the spring scale and pull the scale parallel with the ramp, so as the block moves with constant speed. Usually such a ramp is on a lab table and fixed to it, so you need only one hand, like in the picture
"Mechanical Efficiency of a Ramp"
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7.
# Multiply Mixed Fractions
- Grade 5 Math
Taking their understanding of multiplying fractions further, fifth graders are ready to multiply mixed numbers. This fun multiplying mixed fractions game goes a step ahead of your regular worksheets. Children begin by estimating the product of two mixed numbers. They then multiply mixed numbers by converting them to fractions.
What’s inside?
- Estimate the product of two mixed numbers.
- Convert mixed numbers to improper fractions and then multiply the two.
Real-World Application
We come across multiplication of mixed numbers ever so often. When preparing a liter of mango smoothie, say we need 2¾ cups of mango juice. To prepare 2½ liters of mango smoothie, we’ll need 2½ × 2¾ cups of mango smoothie.
What’s next?
Now that fifth graders are comfortable with multiplying fractions, they can proceed to dividing fractions by a whole number and dividing whole by a unit fraction.
### Common Core Alignment
5.NF.6Solve real world problems involving multiplication of fractions and mixed numbers, e.g., by using visual fraction models or equations to represent the problem. | 297 | 1,462 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-04 | latest | en | 0.849426 |
https://web2.0calc.com/questions/help_432 | 1,723,343,312,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640843545.63/warc/CC-MAIN-20240811013030-20240811043030-00610.warc.gz | 476,566,742 | 5,623 | +0
# Help
0
232
1
30 apples and 35 pears in Store X costs \$33.50. 24 apples and 24 pears in Store Y costs \$22.80. The price of the pears in both stores are the same, but each apple in Store X is 10 cents more than each apple in Store Y. What is the price of an apple in Store Y?
Jul 16, 2022
### 1+0 Answers
#1
+129805
+1
30 (A +.10) + 35P = 33.50 → 30A + 3 + 35P = 33.50 → 30A + 35P = 30.50 (1)
24A + 24P = 22.80 (2)
Multiply (1) by 24 and (2) by - 35
720A + 840P = 732
-840A - 840P = -798 add these
-120 A = -66
A = 66/120 = .55 = price of apple in store Y
Jul 18, 2022 | 281 | 628 | {"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} | 4.03125 | 4 | CC-MAIN-2024-33 | latest | en | 0.396667 |
https://www.tes.com/teaching-resource/simplifying-expressions-cootie-catchers-11281156 | 1,534,845,846,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221218101.95/warc/CC-MAIN-20180821092915-20180821112915-00033.warc.gz | 1,025,513,179 | 30,582 | # Simplifying Expressions Cootie Catchers
Created by
ScienceSpot
Simplifying Expressions: These simplifying expressions cootie catchers are a great way for students to have fun while they practice their skills with simplifying expressions. How to Play and Assembly Instructions are included.
There are 2 cootie catchers in this product, each one having 8 problems for a total of 16 problems. Step-by-Step answers are provided for each problem.
*** Enjoy this Lesson? Do a search for my other Cootie Catchers: ***
Arrays
Balancing Equations
Coins
Estimating Sums
Expanded Form
Fact Families
Greater Than Less Than
Long Division
Multiplication Word Problems
Number Bonds Word Problems
Skip Counting
Subtraction
Telling Time
Balancing Equations
Capacity
Comparing Decimals
Decimals
Elapsed Time
Expanded Form
Exponents
Fact Families
Factors & Multiples
Fractions
Fractions on a Number Line
Greater Than Less Than
Greatest Common Factors
Least Common Multiple
Long Division
Mean, Median, Mode, & Range
Metric Measurement
Mixed Numbers
Multiplication
Number Patterns
Order of Operations
Percents
Place Value
Prime & Composite Numbers
Prime Factorization
Probability
Properties of Mult.
Rounding
Word Problems
Customary Measurements
Fractions: Add & Sub, Decimals, & Percents
Fractions: Equivalent & Reducing Fractions
Greater Than Less Than
Improper Fractions & Mixed Numbers
Integers
Operations with Fractions
Rational Numbers
Ratios
Simple Interest
♦ Geometry:
3D Shapes
Angle Pair Relationships
Area of a Circle & Composite Figures
Circumference of a Circle
Missing Angles
Perimeter
Polygons
Pythagorean Theorem
Surface Area of Rectangular Prisms
Volume & Surface Area of Cylinders
Volume of Cones, Rectangular Prisms, & Triangular Prisms
♦ Algebra:
Absolute Value
Combining Like Terms
Distributive Property
Equations
Evaluating Expressions
Inequalities
Linear Equations
Polynomials
Proportions
Scientific Notation
Simplifying Expressions
Slope
System of Equations
Two Step & Multi Step Equations
Writing Expressions
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### Info
Created: May 19, 2016
Updated: Feb 22, 2018
jpg, 76 KB
Cover-Page
jpg, 124 KB
Version-1
jpg, 129 KB
Version-2
Report a problem | 552 | 2,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2018-34 | longest | en | 0.781376 |
https://www.dataunitconverter.com/kilobyte-per-minute-to-pebibyte-per-hour | 1,702,073,687,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100779.51/warc/CC-MAIN-20231208212357-20231209002357-00109.warc.gz | 820,237,935 | 16,657 | # kB/Min to PiB/Hr → CONVERT Kilobytes per Minute to Pebibytes per Hour
info 1 kB/Min is equal to 0.000000000053290705182007513940334320064 PiB/Hr
Input Kilobytes per Minute (kB/Min) - and press Enter.
You are converting .
Sec
Min
Hr
Day
Sec
Min
Hr
Day
S = Second, M = Minute, H = Hour, D = Day
## Kilobytes per Minute (kB/Min) Versus Pebibytes per Hour (PiB/Hr) - Comparison
Kilobytes per Minute and Pebibytes per Hour are units of digital information used to measure storage capacity and data transfer rate.
Kilobytes per Minute is a "decimal" unit where as Pebibytes per Hour is a "binary" unit. One Kilobyte is equal to 1000 bytes. One Pebibyte is equal to 1024^5 bytes. There are 1,125,899,906,842.624 Kilobyte in one Pebibyte. Find more details on below table.
Kilobytes per Minute (kB/Min) Pebibytes per Hour (PiB/Hr)
Kilobytes per Minute (kB/Min) is a unit of measurement for data transfer bandwidth. It measures the number of Kilobytes that can be transferred in one Minute. Pebibytes per Hour (PiB/Hr) is a unit of measurement for data transfer bandwidth. It measures the number of Pebibytes that can be transferred in one Hour.
## Kilobytes per Minute (kB/Min) to Pebibytes per Hour (PiB/Hr) Conversion Formula and Steps
kB/Min to PiB/Hr Calculator Tool allows you to easily convert from Kilobytes per Minute (kB/Min) to Pebibytes per Hour (PiB/Hr). This converter uses the below formula and steps to perform the conversion.
The formula of converting the Kilobytes per Minute (kB/Min) to Pebibytes per Hour (PiB/Hr) is represented as follows :
diamond CONVERSION FORMULA PiB/Hr = kB/Min x 1000 ÷ 10245 x 60
Source Data Unit Target Data Unit
Equal to 1000 bytes
(Decimal Unit)
Equal to 1024^5 bytes
(Binary Unit)
The conversion from Data per Minute to Hour can be calculated as below.
x 60
x 60
x 24
Data
per
Second
Data
per
Minute
Data
per
Hour
Data
per
Day
÷ 60
÷ 60
÷ 24
Now let us apply the above formula and see how to manually convert Kilobytes per Minute (kB/Min) to Pebibytes per Hour (PiB/Hr). We can further simplify the formula to ease the calculation.
FORMULA
Pebibytes per Hour = Kilobytes per Minute x 1000 ÷ 10245 x 60
STEP 1
Pebibytes per Hour = Kilobytes per Minute x 1000 ÷ (1024x1024x1024x1024x1024) x 60
STEP 2
Pebibytes per Hour = Kilobytes per Minute x 1000 ÷ 1125899906842624 x 60
STEP 3
Pebibytes per Hour = Kilobytes per Minute x 0.0000000000008881784197001252323389053344 x 60
STEP 4
Pebibytes per Hour = Kilobytes per Minute x 0.000000000053290705182007513940334320064
Example : If we apply the above Formula and steps, conversion from 1 Kilobytes per Minute (kB/Min) to Pebibytes per Hour (PiB/Hr) will be processed as below.
1. = 1 x 1000 ÷ 10245 x 60
2. = 1 x 1000 ÷ (1024x1024x1024x1024x1024) x 60
3. = 1 x 1000 ÷ 1125899906842624 x 60
4. = 1 x 0.0000000000008881784197001252323389053344 x 60
5. = 1 x 0.000000000053290705182007513940334320064
6. = 0.000000000053290705182007513940334320064
7. i.e. 1 kB/Min is equal to 0.000000000053290705182007513940334320064 PiB/Hr.
Note : Result rounded off to 40 decimal positions.
You can use above formula and steps to convert Kilobytes per Minute to Pebibytes per Hour using any of the programming language such as Java, Python or Powershell.
### Unit Definitions
#### Kilobyte
A Kilobyte (kB) is a decimal unit of digital information that is equal to 1000 bytes (or 8,000 bits) and commonly used to express the size of a file or the amount of memory used by a program. It is also used to express data transfer speeds and in the context of data storage and memory, the binary-based unit of kibibyte (KiB) is used instead.
arrow_downward
#### Pebibyte
A Pebibyte (PiB) is a binary unit of digital information that is equal to 1,125,899,906,842,624 bytes (or 9,007,199,254,740,992 bits) and is defined by the International Electro technical Commission(IEC). The prefix 'pebi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'petabyte' (PB). It is widely used in the field of computing as it more accurately represents the storage size of high end servers and data storage arrays.
## Excel Formula to convert from Kilobytes per Minute (kB/Min) to Pebibytes per Hour (PiB/Hr)
Apply the formula as shown below to convert from 1 Kilobytes per Minute (kB/Min) to Pebibytes per Hour (PiB/Hr).
A B C
1 Kilobytes per Minute (kB/Min) Pebibytes per Hour (PiB/Hr)
2 1 =A2 * 0.0000000000008881784197001252323389053344 * 60
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Kilobytes per Minute (kB/Min) to Pebibytes per Hour (PiB/Hr) Conversion
You can use below code to convert any value in Kilobytes per Minute (kB/Min) to Kilobytes per Minute (kB/Min) in Python.
kilobytesperMinute = int(input("Enter Kilobytes per Minute: "))
pebibytesperHour = kilobytesperMinute * 1000 / (1024*1024*1024*1024*1024) * 60
print("{} Kilobytes per Minute = {} Pebibytes per Hour".format(kilobytesperMinute,pebibytesperHour))
The first line of code will prompt the user to enter the Kilobytes per Minute (kB/Min) as an input. The value of Pebibytes per Hour (PiB/Hr) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Pebibytes(PiB) are there in a Kilobyte(kB)?expand_more
There are 0.0000000000008881784197001252323389053344 Pebibytes in a Kilobyte.
#### What is the formula to convert Kilobyte(kB) to Pebibyte(PiB)?expand_more
Use the formula PiB = kB x 1000 / 10245 to convert Kilobyte to Pebibyte.
#### How many Kilobytes(kB) are there in a Pebibyte(PiB)?expand_more
There are 1125899906842.624 Kilobytes in a Pebibyte.
#### What is the formula to convert Pebibyte(PiB) to Kilobyte(kB)?expand_more
Use the formula kB = PiB x 10245 / 1000 to convert Pebibyte to Kilobyte.
#### Which is bigger, Pebibyte(PiB) or Kilobyte(kB)?expand_more
Pebibyte is bigger than Kilobyte. One Pebibyte contains 1125899906842.624 Kilobytes.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation. | 1,911 | 6,153 | {"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-2023-50 | latest | en | 0.650578 |
https://mymathtutors.com/factors-of-quadratic/radical-expressions/how-to-solve-functions-in.html | 1,627,622,279,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153931.11/warc/CC-MAIN-20210730025356-20210730055356-00047.warc.gz | 419,209,588 | 12,668 | Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
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Solve for:
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Number of inequalities to solve: 23456789
Ineq. #1:
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• holt middle school math sheets on triangles | 1,037 | 4,436 | {"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-2021-31 | latest | en | 0.89496 |
https://discussions.unity.com/t/chase-script-makes-my-sprite-thin-and-blurry/369219 | 1,723,276,107,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00088.warc.gz | 157,703,558 | 7,005 | Chase script makes my sprite thin and blurry
I’m using this script I found here to make enemies chase my player.
``````using UnityEngine;
using System.Collections;
//[RequireComponent(typeof(CharacterController))]
public class Chaser : MonoBehaviour {
public float speed = 20.0f;
public float minDist = 1f;
public Transform target;
// Use this for initialization
void Start ()
{
// if no target specified, assume the player
if (target == null) {
if (GameObject.FindWithTag ("Player")!=null)
{
target = GameObject.FindWithTag ("Player").GetComponent<Transform>();
}
}
}
// Update is called once per frame
void Update ()
{
if (target == null)
return;
// face the target
transform.LookAt(target);
//get the distance between the chaser and the target
float distance = Vector3.Distance(transform.position,target.position);
//so long as the chaser is farther away than the minimum distance, move towards it at rate speed.
if(distance > minDist)
transform.position += transform.forward * speed * Time.deltaTime;
}
// Set the target of the chaser
public void SetTarget(Transform newTarget)
{
target = newTarget;
}
}
``````
It works, but it makes my enemy sprite all thin and blurry. Why?
Is this project in 2D? You should take a look at the enemy’s transform while the game is running to see if any rotation has been applied to the X or Y axes. `transform.LookAt` will align the objects `forward` vector in the direction of the target. However, `forward` is typically in the Z direction, even for 2D objects. This will end up applying rotation on X and Y which will cause the sprite to rotate away from the camera.
You may want to try using Quaternion.LookRotation to ensure that direction of `forward` does not change and that rotation is applied only to the Z axis.
``````Vector3 direction = target.position - transform.position;
Quaternion rotation = Quaternion.LookRotation(new Vector3(0, 0, 1), direction);
transform.rotation = rotation;
``````
Note that LookRotation will create a rotation to align the objects “up” to point at the target direction, so you may need to change the base sprite to be rotated accordingly.
Thanks. I put your code in and it worked, except I had to change it to (0, 0, 0) because the sprites were created looking up. | 507 | 2,261 | {"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-2024-33 | latest | en | 0.773082 |
https://convertoctopus.com/587-deciliters-to-fluid-ounces | 1,632,429,110,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057447.52/warc/CC-MAIN-20210923195546-20210923225546-00336.warc.gz | 229,769,320 | 7,857 | ## Conversion formula
The conversion factor from deciliters to fluid ounces is 3.3814022558919, which means that 1 deciliter is equal to 3.3814022558919 fluid ounces:
1 dL = 3.3814022558919 fl oz
To convert 587 deciliters into fluid ounces we have to multiply 587 by the conversion factor in order to get the volume amount from deciliters to fluid ounces. We can also form a simple proportion to calculate the result:
1 dL → 3.3814022558919 fl oz
587 dL → V(fl oz)
Solve the above proportion to obtain the volume V in fluid ounces:
V(fl oz) = 587 dL × 3.3814022558919 fl oz
V(fl oz) = 1984.8831242086 fl oz
The final result is:
587 dL → 1984.8831242086 fl oz
We conclude that 587 deciliters is equivalent to 1984.8831242086 fluid ounces:
587 deciliters = 1984.8831242086 fluid ounces
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 fluid ounce is equal to 0.00050380800149063 × 587 deciliters.
Another way is saying that 587 deciliters is equal to 1 ÷ 0.00050380800149063 fluid ounces.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that five hundred eighty-seven deciliters is approximately one thousand nine hundred eighty-four point eight eight three fluid ounces:
587 dL ≅ 1984.883 fl oz
An alternative is also that one fluid ounce is approximately zero point zero zero one times five hundred eighty-seven deciliters.
## Conversion table
### deciliters to fluid ounces chart
For quick reference purposes, below is the conversion table you can use to convert from deciliters to fluid ounces
deciliters (dL) fluid ounces (fl oz)
588 deciliters 1988.265 fluid ounces
589 deciliters 1991.646 fluid ounces
590 deciliters 1995.027 fluid ounces
591 deciliters 1998.409 fluid ounces
592 deciliters 2001.79 fluid ounces
593 deciliters 2005.172 fluid ounces
594 deciliters 2008.553 fluid ounces
595 deciliters 2011.934 fluid ounces
596 deciliters 2015.316 fluid ounces
597 deciliters 2018.697 fluid ounces | 565 | 2,066 | {"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-2021-39 | latest | en | 0.721848 |
https://www.weegy.com/?ConversationId=ALJVMAPO&Link=i&ModeType=2 | 1,632,586,022,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057687.51/warc/CC-MAIN-20210925142524-20210925172524-00576.warc.gz | 1,056,509,591 | 10,240 | Angelo earns \$2,080 each month. His total deductions are 30% of his pay. How much is deducted from his pay each month? A. \$62.40 B. \$69.33 C. \$624.00 D. \$693.33
Angelo earns \$2,080 each month. His total deductions are 30% of his pay. The amount deducted from his pay each month is \$624.00. \$2080*30% = \$624.00.
s
Question
Updated 322 days ago|11/7/2020 3:25:17 PM
Edited by Sting [11/7/2020 3:25:15 PM], Confirmed by Sting [11/7/2020 3:25:16 PM]
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 684 | 2,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-39 | latest | en | 0.855631 |
http://mathhelpforum.com/calculus/98342-solved-integration-two-variables.html | 1,481,336,433,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542932.99/warc/CC-MAIN-20161202170902-00281-ip-10-31-129-80.ec2.internal.warc.gz | 174,702,127 | 10,948 | # Thread: [SOLVED] integration with two variables
1. ## [SOLVED] integration with two variables
Hi im just trying to prove to myself that I can solve this integration. It is infact the distribution function to the exponential distribution derived from the gamma distribution when $\alpha=1$
$F_{x}(x)=\int_0^x(\lambda).exp[-(\lambda)x]dx=1-exp[-(\lambda)x]$
Im not sure how this is done. Two variables are chi and x. Could somebody show me the working for this, perhaps explain how the chi variable is dealt here.
Thanks.
2. Originally Posted by i_zz_y_ill
$F_{x}(x)=\int_0^x\chi e^{-\chi x}dx=1-e^{-\chi x}$
Is this what you meant ?
In that case, use an integration by parts
chi will be like a constant, that's all
3. could you show me please i dont quite get this?
4. If $\lambda$ is just a constant, then it should be this:
$-(\lambda)^2(exp[-(\lambda)x]-1)$ which gives me the right answer when i take the minus inside the brackets but where does the $\lambda^2$go from here?
5. ok If lambda is just a constant surely you can just take it out the brackets giving an answer of $\lambda[-exp[-\lambda.x]+1]$ why is there a lambda still there?
6. oh right yeah iman idiot gotitnow!
7. I hope it's okay, I wasn't connected and didn't understand much of your modifications
(and it was a chi, not a lambda...but it doesn't matter) | 363 | 1,341 | {"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": 0, "img_math": 0, "codecogs_latex": 7, "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-2016-50 | longest | en | 0.901043 |
https://mathsnoproblem.com/mathematics/primary-key-stage1-and-key-stage-2/maths-no-problem-2a-workbook.html | 1,597,026,065,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738603.37/warc/CC-MAIN-20200810012015-20200810042015-00122.warc.gz | 387,660,875 | 15,484 | BOOKSTORE
# Maths — No Problem! Workbook 2A
ISBN: 9781910504031
278 Page
Availability: In stock
£8.49
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## Details
### This workbook is for the first half of year 2
Maths — No Problem! is a series of textbooks and workbooks written to meet the requirements of the 2014 English National Curriculum.
The focus of the series is on teaching to mastery. This research-based approach emphasises problem solving and utilises pupils' core competencies to develop a relational understanding of mathematical concepts.
In 2017, Maths — No Problem! textbooks were the only resource to meet the Department of Education’s standard for excellence in maths mastery teaching. The good news? Your school might be eligible for match funding. Learn More
Consultant and Author: Dr. Yeap Ban Har
UK Consultant: Dr. Anne Hermanson
Authors: Dr. Foong Pui Yee, Lim Li Gek Pearlyn, Wong Oon Hua
To experience the full value of the Maths — No Problem! programme why not register for a FREE trial? (Available for teachers only).
• Chapter 1: Numbers to 100
• Lesson 1: Counting to 100
• Lesson 2: Place Value
• Lesson 3: Comparing Numbers
• Lesson 4: Number Bonds
• Lesson 5: Number Patterns
• Lesson 6: Number Patterns
• Chapter 2: Addition and Subtraction
• Lesson 5: Adding with Renaming
• Lesson 6: Adding with Renaming
• Lesson 7: Simple Subtracting
• Lesson 8: Simple Subtracting
• Lesson 9: Simple Subtracting
• Lesson 10: Simple Subtracting
• Lesson 11: Subtracting with Renaming
• Lesson 12: Subtracting with Renaming
• Lesson 13: Addition of Three Numbers
• Chapter 3: Multiplication of 2, 5 and 10
• Lesson 1: Multiplication As Equal Groups
• Lesson 2: 2 Times Table
• Lesson 3: 2 Times Table
• Lesson 4: 5 Times Table
• Lesson 5: 5 Times Table
• Lesson 6: 10 Times Table
• Lesson 7: 10 Times Table
• Lesson 8: Multiplying by 2, 5 and 10
• Lesson 9: Multiplying by 2, 5 and 10
• Lesson 10: Solving Word Problems
• Chapter 4: Multiplication and Division of 2, 5 and 10
• Lesson 1: Grouping
• Lesson 2: Sharing
• Lesson 3: Dividing by 2
• Lesson 4: Dividing by 5
• Lesson 5: Dividing by 10
• Lesson 6: Multiplication and Division
• Lesson 7: Solving Word Problems
• Lesson 8: Odd and Even Numbers
• Revision 1
• Chapter 5: Length
• Lesson 1: Measuring Length in Metres
• Lesson 2: Measuring Length in Centimetres
• Lesson 3: Comparing Length in Metres
• Lesson 4: Comparing Length in Centimetres
• Lesson 5: Comparing the Length of Lines
• Lesson 6: Solving Word Problems
• Lesson 7: Solving Word Problems
• Lesson 8: Solving Word Problems
• Chapter 6: Mass
• Lesson 1: Measuring Mass in Kilograms
• Lesson 2: Measuring Mass in Grams
• Lesson 3: Measuring Mass in Grams
• Lesson 4: Comparing Masses of Two Objects
• Lesson 5: Comparing the Mass of Three Objects
• Lesson 6: Solving Word Problems
• Lesson 7: Solving More Word Problems
• Chapter 7: Temperature | 836 | 2,903 | {"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-2020-34 | latest | en | 0.81874 |
https://brainly.ph/question/234168 | 1,485,006,164,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281084.84/warc/CC-MAIN-20170116095121-00505-ip-10-171-10-70.ec2.internal.warc.gz | 796,043,620 | 10,120 | # A class has 40 pupils. 2/5 of them are boys, how many are girls? 5 parts =40 1 part=? 2 part=? 3 part=?
1
by fidelity6
2015-09-15T20:37:31+08:00
Girls: 24
Boys: 16
First, multiply 2/5 by 40.
2/5 X 40/1 = 80/5
Then, divide it.
80 ÷ 5 = 16
*The answer will be the number of boys.
Subtract the number of boys by the total number of students which is 40.
40 - 16 = 24
*The answer will be the number of girls.
Checking:
24 + 16 = 40
40 = 40
-KookEin
Hi! Sad to say but, I don't understand the thing with the 'parts'. Good luck with it. :)
Oh yeah. Sorry. There is a big box the teacher drew for us. Like 1/2 of the pizza but using box as parts. Again, Thank you so much!
No problem, always feel free to ask. :)
Ahhhh. Cnan you help me again about " Arman had 240 eggs..." Thank you this is important to me. | 278 | 810 | {"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} | 4.0625 | 4 | CC-MAIN-2017-04 | latest | en | 0.927747 |
https://rukodelkino.com/en/cutting/curtains/smad-parecida.html | 1,585,663,003,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500482.27/warc/CC-MAIN-20200331115844-20200331145844-00388.warc.gz | 643,535,993 | 10,963 | Подписывайтесь на еженедельный дайджест с новыми и самыми популярными статьями.
#### The proportion of mid-and shoulder Smad, peracid etc.
1 - mid -2 - shoulder h - depth Smad ...
#### Useful tips
To achieve the most appropriate and proportionate implementation of pelmet must follow some rules to avoid costly mistakes and rework.In...
#### The calculation of the length (L) Smad on the roof
Smad 1 = 3/3 Smad 2 = 3/3 + 3/3 = 6/3 Overlap 1/3 6/3 -...
#### Asymmetrical swags (one-way)
Construction: 1) Find the direction of the thread at an angle of 45°. We estimated the bend. 2) Perpendicular to...
#### Swag with a perpendicular shoulder
Construction:1) draw to the fold of the fabric perpendicular, on which lay off 1/2 of the mid-Smad, value (1) -...
#### Equilateral swag
Construction:1) draw to the fold of the fabric perpendicular, on which lay off 1/2 of the mid-svaga value (1) -...
#### Equilateral peracid
Construction: 1) draw to the fold of the fabric perpendicular. Which is delayed 1/2 of the mid-parecida value (1) -...
#### Mechanical swag
Construction: 1) draw to the fold of the fabric perpendicular. Lay off on the perpendicular value (1) - 1/2 mid...
#### Smad, parecida, asymmetric Smad and other elements of the pelmet
For the manufacture of the most beautifully and correctly vypolnennykh elements lambrequin (Smad, Smad asymmetrical, parecido, asymmetric parecido, Smad transition...
#### Swag, rolling in de ruff
Construction: 1) draw the intended fold of the fabric (at an angle of 45°) perpendicular. Which pushes the...
#### Mechanical peracid, rolling in de ruff
Designations: l1 - the value of the right shoulder of peracid on the sketch (30 cm) to build...
#### Double swag with bell
Construction: 1) draw to the fold of the fabric perpendicular, on which lay off the magnitude of AB = 25...
#### Double swag with bell - mechanical
Construction: 1) From the edge of the fabric, the warp threads are 1/2-delayed values of the lower SAG Smad -...
#### Classic kokila (straight)
In the model used: Smad with a perpendicular from the shoulder, de frill tselnokrajnimi bell "Classic...
#### Classic kokila (from one point)
Item of pelmet is usually performed on the lining, with a thin, transparent fabric - processed, double lapped seam in...
#### Kokila on the bias
Kokila on the bias Corner kokila on the bias ...
#### Hard lambrequin. Manufacturer Bando
This stripe fabric is doubled spec. material: thermo bandeau, adhesive, bandeau, phleselin, dublerin, prolamin, glue crinoline, thermal synthetic padding, wood... | 683 | 2,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-16 | longest | en | 0.648709 |
http://openstudy.com/updates/50f20ed6e4b0694eaccf469c | 1,448,645,628,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398449793.41/warc/CC-MAIN-20151124205409-00242-ip-10-71-132-137.ec2.internal.warc.gz | 172,941,624 | 15,434 | ## Ben103 2 years ago How do you graph x>-13/2
1. geerky42
|dw:1358040778330:dw|
2. Azteck
It depends whether you want to graph that on a number plane or a number line.
3. Azteck
Usually if it just says graph x>-13/2, then it's a on a number plane, not a number line.
4. geerky42
It can be on number line since there is only one variable. Just because the variable is x does not mean we should use number plane. But again, you're right, it does depend. What subject are you on? @Ben103
5. geerky42
Ben103?
6. Ben103
algebra 2
7. Ben103
how would you graph x<-13/4
8. Azteck
For you @Ben103 , it's a number line then. So @geerky42 got you covered.
9. Ben103
would it be going left opened circle?
10. Ben103
?
11. geerky42
I think you should use number plane. Algebra 2 usually focuses on inequality, which usually have two variables. |dw:1358041581217:dw| Does this help you? Need any more examples?
12. Ben103
so its a closed circle?
13. Ben103
going right?
14. geerky42
For number line, use open circle, for plane, use dashed line. ">" only mean greater than, so it does not include the number itself. Yes it goes right. Is it clear?
15. Ben103
its closed circle no?
16. Ben103
ok opened circle going right. correct?
17. geerky42
I just use dot so you know it's -13/2. Just use dashed line. If you use number line, then no, it's not closed circle, it's opened. Which one do you use?
18. geerky42
Don't use circle if you use plane. Just dashed line.
19. Ben103
its x<-13/4
20. Ben103
and it goes to the right correct?
21. geerky42
x<-13/4 goes left.
22. geerky42
< goes left, > goes right.
23. Ben103
ok. And does the line start before the -4 or after it?
24. Ben103
*-5 i mean.
25. Azteck
In graphs, all you have to put are important features. you don't need to write a 4 or a 5. Just write -13/4. Unrelevant features does not give you extra marks.
26. Azteck
-13/2* | 584 | 1,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2015-48 | longest | en | 0.872152 |
https://oeis.org/A238678 | 1,669,972,017,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710900.9/warc/CC-MAIN-20221202082526-20221202112526-00056.warc.gz | 463,712,994 | 3,918 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A238678 Sixth prime p such that (p+n)^2+n is prime but (p+j)^2+j is not prime for all 0
53, 97, 71, 199, 787, 1019, 401, 1741, 1103, 1447, 10453, 1283, 1223, 9631, 1021, 12109, 15361, 1913, 14723, 6397, 26513, 2789, 25603, 21491, 6689, 87103, 10247, 8597, 254911, 12007, 71453, 47521, 37529, 39971, 109147, 59453, 12791, 256147, 59611, 78317 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Alois P. Heinz, Table of n, a(n) for n = 1..100 CROSSREFS Row n=6 of A238086. Sequence in context: A096697 A033234 A266845 * A142296 A180520 A139959 Adjacent sequences: A238675 A238676 A238677 * A238679 A238680 A238681 KEYWORD nonn AUTHOR Alois P. Heinz, Mar 02 2014 STATUS approved
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Last modified December 1 23:44 EST 2022. Contains 358485 sequences. (Running on oeis4.) | 498 | 1,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2022-49 | latest | en | 0.735297 |
http://www.barnesandnoble.com/w/calculate-with-confidence-deborah-c-gray-morris/1100099474?ean=9780323056298 | 1,455,441,786,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701171770.2/warc/CC-MAIN-20160205193931-00064-ip-10-236-182-209.ec2.internal.warc.gz | 288,857,527 | 45,674 | # Calculate with Confidence / Edition 5
ISBN-10: 0323056296
ISBN-13: 9780323056298
Pub. Date: 09/14/2009
Publisher: Elsevier Health Sciences
This popular text covers the ratio and proportion, formula, and dimensional analysis methods offering a step-by-step approach to the calculation and administration of drug dosages. With over 2,000 practice problems, Gray Morris focuses on enhancing the learning experience of nursing students at all curricular levels by making content clinically applicable.
… See more details below
## Overview
This popular text covers the ratio and proportion, formula, and dimensional analysis methods offering a step-by-step approach to the calculation and administration of drug dosages. With over 2,000 practice problems, Gray Morris focuses on enhancing the learning experience of nursing students at all curricular levels by making content clinically applicable. Calculate with Confidence, 6th Edition addresses the increasing responsibility of the nurse in medication administration, prioritizes client safety, and reflects the current scope of practice.
• Tips for Clinical Practice
boxes
call attention to information critical to math calculation and patient safety.
Safety Alert boxes highlight issues that may lead to mediation errors and empower you to identify actions that must be taken to avoid calculation errors
Chapter review problems test all major topics presented in the chapter.
Separate basic math review test allows you to assess and evaluate your understanding of basic math material covered in Unit 1, directing you to review chapters if you miss any of these test questions.
Pre-test basic math review tests help you assess your basic math skills and identify areas of strength and weakness in competency of basic math.
Comprehensive unit on basic math review offers complete coverage of basic math: roman numerals, fractions, decimals, ratio and proportion, and percentages.
• NEW! Integration of QSEN information related to patient safety
in the Medication Administration chapter and throughout text.
NEW! NCLEX-style questions on Evolve help prepare you for the NCLEX-RN Examination.
NEW! Content additions and updates includes word problems involving dosages, Critical Thinking Scenarios, a discussion of the concepts regarding safety issues with medication administration, plus significant updates in the insulin, critical care and IV chapters.
NEW! Reorganization of Answer Key features answers and the work to practice problems at the end of each chapter rather than in the back of the book.
## Product Details
ISBN-13:
9780323056298
Publisher:
Elsevier Health Sciences
Publication date:
09/14/2009
Edition description:
Older Edition
Pages:
720
Product dimensions:
8.50(w) x 10.80(h) x 1.10(d)
## Related Subjects
Unit One: Math Review
Pre-Test
1. Roman Numerals
2. Fractions
3. Decimals
4. Ratio and Proportion
5. Percentages
Post-Test
Unit Two: Systems of Measurement
6. Metric System
7. Apothecary and Household Systems
8. Converting Within and Between Systems
9. Additional Conversions Useful in the Health Care Setting
Unit Three: Methods of Administration and Calculation
11. Understanding and Interpreting Medication Orders
12. Medication Administration Records and Drug Distribution Systems
14. Dosage Calculation Using the Ratio and Proportion Method
15. Dosage Calculation Using the Formula Method
16. Dosage Calculation Using the Dimensional Analysis Method
Unit Four: Oral and Parenteral Dosage Forms, Insulin and Pediatric Dosage Calculations
17. Calculation of Oral Medications
18. Parenteral Medications
19. Reconstitution of Solutions
20. Insulin
Unit Five: Intravenous, Heparin, and Critical Care Calculations and Pediatric Dosage Calculations
22. Intravenous Calculations
23. Heparin Calculations
24. Critical Care Calculations
25. Pediatric and Adult Dosage Calculation Based on Weight
Comprehensive Post-Test
Bibliography
Index
## Customer Reviews
Average Review:
Write a Review
and post it to your social network
Calculate With Confidence 4.5 out of 5 based on 0 ratings. 4 reviews.
Anonymous More than 1 year ago
I bought this book for school. It is very easy to read as well as to understand. They have step by step explanations (kind of), and they also have "rules" to remember throughout the textbook. Practice questions help too! I haven't used the "evolve" website since the book is more than helpful.
Skippyboy More than 1 year ago
The solving is helping me a lot.IV problems too.
Anonymous More than 1 year ago | 960 | 4,554 | {"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-2016-07 | latest | en | 0.860717 |
http://www.jiskha.com/display.cgi?id=1266309253 | 1,498,586,554,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321497.77/warc/CC-MAIN-20170627170831-20170627190831-00253.warc.gz | 572,839,194 | 3,898 | # Chemistry
posted by .
If 720 mL of 0.458 M aqueous Ba(OH)2 and 720 mL of 0.431 M aqueous HClO4 are reacted stoichiometrically according to the balanced equation, how many milliliters of 0.458 M aqueous Ba(OH)2 remain? Round your answer to 3 significant figures.
Ba(OH)2(aq) + 2HClO4(aq) → Ba(ClO4)2(aq) + 2H2O(l)
• Chemistry -
Make sure the equation is balanced.
moles Ba(OH)2 = M x L.
moles HClO4 = M x L.
From the question, I assume Ba(OH)2 is in excess and HClO4 is the limiting reagent. If so, use the coefficients in the balanced equation to convert moles HClO4 to moles Ba(OH)2 used. The difference between initial moles Ba(OH)2 and moles Ba(OH)2 used gives moles Ba(OH)2 remaining. Then M = moles/L. You know moles and M, calculate L remaining. | 251 | 758 | {"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-2017-26 | latest | en | 0.825729 |
https://reposcope.com/man/en/3/cacos | 1,632,859,531,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060882.17/warc/CC-MAIN-20210928184203-20210928214203-00006.warc.gz | 508,994,056 | 2,926 | Linux repositories inspector
# cacos(3)
2019-03-06
Aliases: cacosf(3), cacosf(3), cacosf(3), cacosf(3), cacosf(3), cacosf(3), cacosf(3), cacosf(3), cacosf(3), cacosf(3), cacosl(3), cacosl(3), cacosl(3), cacosl(3), cacosl(3), cacosl(3), cacosl(3), cacosl(3), cacosl(3), cacosl(3)
### manpages-dev
Manual pages about using GNU/Linux for development
### man-pages
Linux kernel and C library user-space interface documentation
## NAME
cacos, cacosf, cacosl - complex arc cosine
## SYNOPSIS
#include <complex.h>
double complex cacos(double complex z);
float complex cacosf(float complex z);
long double complex cacosl(long double complex z);
## DESCRIPTION
These functions calculate the complex arc cosine of z. If y = cacos(z), then z = ccos(y). The real part of y is chosen in the interval [0,pi].
One has:
``` cacos(z) = -i * clog(z + i * csqrt(1 - z * z))
```
## VERSIONS
These functions first appeared in glibc in version 2.1.
## ATTRIBUTES
For an explanation of the terms used in this section, see attributes(7).
Interface Attribute Value Thread safety MT-Safe
## CONFORMING TO
C99, POSIX.1-2001, POSIX.1-2008.
## EXAMPLE
#include <complex.h> #include <stdlib.h> #include <unistd.h> #include <stdio.h>
int main(int argc, char *argv[]) {
double complex z, c, f;
double complex i = I;
if (argc != 3) {
fprintf(stderr, "Usage: %s <real> <imag>\n", argv[0]);
exit(EXIT_FAILURE);
}
z = atof(argv[1]) + atof(argv[2]) * I;
c = cacos(z);
printf("cacos() = %6.3f %6.3f*i\n", creal(c), cimag(c));
f = -i * clog(z + i * csqrt(1 - z * z));
printf("formula = %6.3f %6.3f*i\n", creal(f), cimag(f));
exit(EXIT_SUCCESS); } | 547 | 1,631 | {"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} | 2.546875 | 3 | CC-MAIN-2021-39 | latest | en | 0.371008 |
https://socratic.org/questions/how-do-you-integrate-1-e-x-1-e-x | 1,558,682,288,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257553.63/warc/CC-MAIN-20190524064354-20190524090354-00031.warc.gz | 614,591,531 | 5,770 | # How do you integrate (1 + e^x )/(1 - e^x)?
Apr 19, 2018
$\int \frac{1 + {e}^{x}}{1 - {e}^{x}} \mathrm{dx} = x - 2 \ln \left\mid 1 - {e}^{x} \right\mid + C$
#### Explanation:
Substitute:
$t = {e}^{x}$
$\mathrm{dt} = {e}^{x} \mathrm{dx}$
$\mathrm{dx} = \frac{\mathrm{dt}}{t}$
so:
$\int \frac{1 + {e}^{x}}{1 - {e}^{x}} \mathrm{dx} = \int \frac{1 + t}{1 - t} \frac{\mathrm{dt}}{t}$
Use partial fractions decomposition:
$\frac{1 + t}{t \left(1 - t\right)} = \frac{A}{t} + \frac{B}{1 - t}$
$\frac{1 + t}{t \left(1 - t\right)} = \frac{A \left(1 - t\right) + B t}{t \left(1 - t\right)}$
$1 + t = A - A t + B t$
$\left\{\begin{matrix}A = 1 \\ - A + B = 1\end{matrix}\right.$
$\left\{\begin{matrix}A = 1 \\ B = 2\end{matrix}\right.$
so:
$\int \frac{1 + {e}^{x}}{1 - {e}^{x}} \mathrm{dx} = \int \frac{\mathrm{dt}}{t} + 2 \int \frac{\mathrm{dt}}{1 - t}$
$\int \frac{1 + {e}^{x}}{1 - {e}^{x}} \mathrm{dx} = \ln \left\mid t \right\mid - 2 \ln \left\mid 1 - t \right\mid + C$
and undoing the substitution:
$\int \frac{1 + {e}^{x}}{1 - {e}^{x}} \mathrm{dx} = x - 2 \ln \left\mid 1 - {e}^{x} \right\mid + C$ | 518 | 1,112 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2019-22 | longest | en | 0.230906 |
https://plati.market/itm/6-1-4-the-solution-of-the-problem-of-the-collection-of/2054392?lang=en-US | 1,596,994,248,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738562.5/warc/CC-MAIN-20200809162458-20200809192458-00478.warc.gz | 449,632,248 | 13,139 | # 6.1.4 The solution of the problem of the collection of
Affiliates: 0,01 \$how to earn
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Content: 6_1_4.png 49,19 kB
Loyalty discount! If the total amount of your purchases from the seller TerMaster more than:
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## Product description
Homogeneous wire, bent at right angles, is suspended as shown in Fig. Determine the ratio l1 / l2, where l2 is the length of a segment in a horizontal position, if OK = 0,2l2. | 161 | 521 | {"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-2020-34 | latest | en | 0.754931 |
https://au.mathworks.com/matlabcentral/cody/problems/8-add-two-numbers/solutions/1086192 | 1,579,732,974,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250607596.34/warc/CC-MAIN-20200122221541-20200123010541-00081.warc.gz | 324,529,469 | 15,569 | Cody
# Problem 8. Add two numbers
Solution 1086192
Submitted on 21 Dec 2016 by Ashley Jacob
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
a = 1; b = 2; c_correct = 3; assert(isequal(add_two_numbers(a,b),c_correct))
2 Pass
a = 17; b = 2; c_correct = 19; assert(isequal(add_two_numbers(a,b),c_correct))
3 Pass
a = -5; b = 2; c_correct = -3; assert(isequal(add_two_numbers(a,b),c_correct)) | 164 | 517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-05 | latest | en | 0.672913 |
https://cs.stackexchange.com/questions/86056/polynomial-time-problems-with-provably-high-degree-time-complexity/86066 | 1,585,555,814,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496669.0/warc/CC-MAIN-20200330054217-20200330084217-00487.warc.gz | 422,183,145 | 31,389 | # Polynomial time problems with provably high degree time complexity?
For any integer $k$, does there exist a decision problem in $\textbf P$ that can be proven to require $\Omega(n^k)$ steps?
• Yes. This follows from the time hierarchy theorem. – Yuval Filmus Dec 29 '17 at 22:19
• Is the proof constructive? Can you find a concrete problem for any $k$? – jnalanko Dec 29 '17 at 22:23
• The problem is deciding whether a given Turing machine halts on a given string of length $n$ in time $n^k$. – Yuval Filmus Dec 29 '17 at 22:24
• A more concrete example is, given a graph, to determine whether or not there exists a $k$-clique for a constant $k$. (observe that this is essentially a weaker version of the time-hierarchy theorem) – quicksort Dec 30 '17 at 9:08
• @quicksort I'm pretty sure that the complexity of k-clique is an open poroblem – Ariel Dec 30 '17 at 16:25
The time hierarchy theorem, proved by diagonalization, shows that for every reasonable function $f$ there exists a problem solvable in $O(f(n))$ but not in $o(f(n)/\log f(n))$. One such problem is, given a Turing machine and in input $x$, to determine whether the machine halts in $x$ within $f(|x|)$ steps. In your case, you can choose $f(n) = n^k \log n$ to obtain a problem solvable in $O(n^k \log n)$ but not in $o(n^k)$. | 377 | 1,300 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-16 | latest | en | 0.808091 |
https://library.fiveable.me/ap-physics-e-m/unit-4/biot%E2%80%93savart-law-amp%C3%A8res-law/study-guide/HChNR7z7O4ql5SF8879S | 1,713,905,085,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818740.13/warc/CC-MAIN-20240423192952-20240423222952-00313.warc.gz | 315,770,167 | 79,764 | or
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# 4.4 Biot–Savart Law and Ampère’s Law
Peter Apps
Peter Apps
As promised in section 4.2, we're going to look at how we can define internal magnetic fields for wires of any configuration. There are two laws that help us to tackle this: Biot-Savart Law and Ampère’s Law.
## Biot-Savart Law
The Biot-Savart Law lets us determine the magnetic field in a region of space that is caused by current in a wire. To solve this, we break up the wire into sections of length dl, each of which causes a small magnetic field dB.
Image From phys.libretexts.org
\hat{r} is the unit vector and is equal to
, and μ_o is the permeability of free space. This describes how well magnetic fields can propagate through space, similarly to how we used ε_0 to describe how well electric fields can propagate in Unit 1.
We often only care about the magnitude of the field, since we can usually find the direction of the field using the right-hand curl rule. Ignoring the directional aspect we find that:
### Common Geometries for using Biot-Savart Law
• Vertical components of the magnetic field cancel due to symmetry
• (In)Finite Straight Wire
• Because the angle will change with respect to θ, we need to define r in terms of d and x
• If the wire is infinite in length, the integral becomes: (which is the equation we used in 4.2)
## Ampère's Law
Ampère's Law is the magnetic equivalent of Gauss' Law. This can be derived from the Biot-Savart law (if you're interested in how). We'll create a path around the object we care about, and then integrate to determine the enclosed current.
Let's use Ampere's Law to look at a pair of coaxial cables with currents running through both. We'll calculate the magnetic field at 4 different radii.
Image from web.iit.edu/
• r < a
• Some fraction of the current I_o is enclosed by our path.
• a < r < b
• We now can enclose the entire current in the in inner cable.
• b < r < c
• We enclose all of the inner current, and need to subtract some portion of the outer current (since they're moving in different directions)
• r > c
• We encircle both cables, so I_enc = 0 and B = 0
## Solenoids
Solenoids create a special case where we need to apply Ampere's Law. A solenoid is basically a bunch of loops of wire tightly wound together. Its purpose is to produce a strong magnetic field as a current is passed through the wire. Every gas-powered car uses a solenoid to push a piston which starts the engine running. Without that 'starter' we'd be using hand cranks circa 1900.
Image from wikipedia.org
To apply Ampere's Law, we define our path as a rectangle enclosing a certain number of loops, N, in the solenoid. The current that is enclosed is equal to the current in the wire times N. If we add up the dl from the path, we'll get the total length of the solenoid. Your reference tables already use N as the # of charge carriers per volume, so we'll end up using n in our equation, where n = N/L.
As you might have guessed, there's a right-hand rule for solenoids, too. Curl your fingers in the direction of the (conventional) current, and your thumb points in the direction of the magnetic field.
Image from wikipedia.org
## Practice Questions
1)
Image from apclassroom.collegeboard.org
The single, circular wire loop of radius R shown above carries a current I that produces a magnetic field B at the center of the loop. If the current remains constant while the loop is enlarged to a radius of 2R, what happens to the magnetic field at the center?
Using Biot-Savart Law
2)
Image from collegeboard.org
A solenoid is used to generate a magnetic field. The solenoid has an inner radius a, length l, and N total turns of wire. A power supply, not shown, is connected to the solenoid and generates current I, as shown in the figure on the left above. The x-axis runs along the axis of the solenoid. Point P is in the middle of the solenoid at the origin of the xyz-coordinate system, as shown in the cutaway view on the right above. Assume l >> a.
(a) Select the correct direction of the magnetic field at point P. Justify your selection.
_ +x-direction, _ -x-direction, _ +y-direction, _ -y-direction, _ +z-direction, _ -z-direction
(b) i. On the cutaway view, clearly draw an Amperian loop that can be used to determine the magnetic field at point P at the center of the solenoid.
ii. Use Ampere's law to derive an expression for the magnetic field strength at point P. Express your answer in terms of I, l, N, a, and physical constants, as appropriate.
a) Use the Right Hand Rule to determine the direction of the induced current. (In this case for a solenoid, curve your fingers in the direction of the current.) Your fingers curl clockwise around the solenoid and your thumb points to the right (which is the positive x direction).
b)
Image from collegeboard.org
ii. Similar to our derivation above. Remember n=N/l and because we're only getting part of the solenoid not the entire things, multiply by h not l to find the total enclosed current.
# Key Terms to Review (9)
Amperian Loop
: An Amperian loop is an imaginary closed loop used to analyze the magnetic field around a current-carrying conductor or through a specific region in space. It helps determine the direction and magnitude of the magnetic field using Ampere's law.
Biot-Savart Law
: The Biot-Savart law is an equation that describes how an infinitesimal current element produces a magnetic field at any given point in space. It provides a way to calculate the magnitude and direction of this magnetic field.
Center of a Loop
: The center of a loop refers to the point at which the magnetic field produced by the loop is strongest.
Coaxial Cables
: Coaxial cables are cables consisting of two concentric conductors separated by insulation, often used for transmitting high-frequency electrical signals with low loss.
Current
: Current refers to the flow of electric charge in an electric circuit. It represents the rate at which charges pass through a given cross-sectional area of a conductor and is measured in amperes (A).
Infinite Straight Wire
: An infinite straight wire refers to an idealized wire with no end that carries an electric current.
Permeability of Free Space
: A physical constant denoted by μ₀, representing the ability of free space to allow magnetic field lines to pass through it. It has a value of approximately 4π x 10⁻⁷ T·m/A.
Right-hand Curl Rule
: A method used to determine the direction of induced currents or magnetic fields resulting from changing magnetic flux. It states that if you curl your right hand around a wire or loop in the direction of current flow, your thumb points in the direction of the magnetic field.
Solenoid
: A solenoid is a coil of wire wound tightly in the shape of a cylinder, often with an iron core inside, used to generate strong magnetic fields when electric current flows through it.
# 4.4 Biot–Savart Law and Ampère’s Law
Peter Apps
Peter Apps
As promised in section 4.2, we're going to look at how we can define internal magnetic fields for wires of any configuration. There are two laws that help us to tackle this: Biot-Savart Law and Ampère’s Law.
## Biot-Savart Law
The Biot-Savart Law lets us determine the magnetic field in a region of space that is caused by current in a wire. To solve this, we break up the wire into sections of length dl, each of which causes a small magnetic field dB.
Image From phys.libretexts.org
\hat{r} is the unit vector and is equal to
, and μ_o is the permeability of free space. This describes how well magnetic fields can propagate through space, similarly to how we used ε_0 to describe how well electric fields can propagate in Unit 1.
We often only care about the magnitude of the field, since we can usually find the direction of the field using the right-hand curl rule. Ignoring the directional aspect we find that:
### Common Geometries for using Biot-Savart Law
• Vertical components of the magnetic field cancel due to symmetry
• (In)Finite Straight Wire
• Because the angle will change with respect to θ, we need to define r in terms of d and x
• If the wire is infinite in length, the integral becomes: (which is the equation we used in 4.2)
## Ampère's Law
Ampère's Law is the magnetic equivalent of Gauss' Law. This can be derived from the Biot-Savart law (if you're interested in how). We'll create a path around the object we care about, and then integrate to determine the enclosed current.
Let's use Ampere's Law to look at a pair of coaxial cables with currents running through both. We'll calculate the magnetic field at 4 different radii.
Image from web.iit.edu/
• r < a
• Some fraction of the current I_o is enclosed by our path.
• a < r < b
• We now can enclose the entire current in the in inner cable.
• b < r < c
• We enclose all of the inner current, and need to subtract some portion of the outer current (since they're moving in different directions)
• r > c
• We encircle both cables, so I_enc = 0 and B = 0
## Solenoids
Solenoids create a special case where we need to apply Ampere's Law. A solenoid is basically a bunch of loops of wire tightly wound together. Its purpose is to produce a strong magnetic field as a current is passed through the wire. Every gas-powered car uses a solenoid to push a piston which starts the engine running. Without that 'starter' we'd be using hand cranks circa 1900.
Image from wikipedia.org
To apply Ampere's Law, we define our path as a rectangle enclosing a certain number of loops, N, in the solenoid. The current that is enclosed is equal to the current in the wire times N. If we add up the dl from the path, we'll get the total length of the solenoid. Your reference tables already use N as the # of charge carriers per volume, so we'll end up using n in our equation, where n = N/L.
As you might have guessed, there's a right-hand rule for solenoids, too. Curl your fingers in the direction of the (conventional) current, and your thumb points in the direction of the magnetic field.
Image from wikipedia.org
## Practice Questions
1)
Image from apclassroom.collegeboard.org
The single, circular wire loop of radius R shown above carries a current I that produces a magnetic field B at the center of the loop. If the current remains constant while the loop is enlarged to a radius of 2R, what happens to the magnetic field at the center?
Using Biot-Savart Law
2)
Image from collegeboard.org
A solenoid is used to generate a magnetic field. The solenoid has an inner radius a, length l, and N total turns of wire. A power supply, not shown, is connected to the solenoid and generates current I, as shown in the figure on the left above. The x-axis runs along the axis of the solenoid. Point P is in the middle of the solenoid at the origin of the xyz-coordinate system, as shown in the cutaway view on the right above. Assume l >> a.
(a) Select the correct direction of the magnetic field at point P. Justify your selection.
_ +x-direction, _ -x-direction, _ +y-direction, _ -y-direction, _ +z-direction, _ -z-direction
(b) i. On the cutaway view, clearly draw an Amperian loop that can be used to determine the magnetic field at point P at the center of the solenoid.
ii. Use Ampere's law to derive an expression for the magnetic field strength at point P. Express your answer in terms of I, l, N, a, and physical constants, as appropriate.
a) Use the Right Hand Rule to determine the direction of the induced current. (In this case for a solenoid, curve your fingers in the direction of the current.) Your fingers curl clockwise around the solenoid and your thumb points to the right (which is the positive x direction).
b)
Image from collegeboard.org
ii. Similar to our derivation above. Remember n=N/l and because we're only getting part of the solenoid not the entire things, multiply by h not l to find the total enclosed current.
# Key Terms to Review (9)
Amperian Loop
: An Amperian loop is an imaginary closed loop used to analyze the magnetic field around a current-carrying conductor or through a specific region in space. It helps determine the direction and magnitude of the magnetic field using Ampere's law.
Biot-Savart Law
: The Biot-Savart law is an equation that describes how an infinitesimal current element produces a magnetic field at any given point in space. It provides a way to calculate the magnitude and direction of this magnetic field.
Center of a Loop
: The center of a loop refers to the point at which the magnetic field produced by the loop is strongest.
Coaxial Cables
: Coaxial cables are cables consisting of two concentric conductors separated by insulation, often used for transmitting high-frequency electrical signals with low loss.
Current
: Current refers to the flow of electric charge in an electric circuit. It represents the rate at which charges pass through a given cross-sectional area of a conductor and is measured in amperes (A).
Infinite Straight Wire
: An infinite straight wire refers to an idealized wire with no end that carries an electric current.
Permeability of Free Space
: A physical constant denoted by μ₀, representing the ability of free space to allow magnetic field lines to pass through it. It has a value of approximately 4π x 10⁻⁷ T·m/A.
Right-hand Curl Rule
: A method used to determine the direction of induced currents or magnetic fields resulting from changing magnetic flux. It states that if you curl your right hand around a wire or loop in the direction of current flow, your thumb points in the direction of the magnetic field.
Solenoid
: A solenoid is a coil of wire wound tightly in the shape of a cylinder, often with an iron core inside, used to generate strong magnetic fields when electric current flows through it. | 3,262 | 13,836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-18 | latest | en | 0.875925 |
https://www.physicsforums.com/threads/converting-2nd-order-ode-to-1st-order-ode.55582/ | 1,618,406,272,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077818.23/warc/CC-MAIN-20210414125133-20210414155133-00475.warc.gz | 1,051,119,526 | 15,917 | # Converting 2nd Order ODE to 1st Order ODE
The position function x(t) in a certain nonlinear system is described by the second order ODE:
< equation.gif >
Transform this ODE into a pair of first order ODEs for x1=x and x2=dx/dt. (Note that x2 represents the velocity in this system.)
I've thought about calculating the homogeneous equation and then the particular integral, but (a) how is this done with a sin(dx/dt) on the RHS and (b) how does this yield a pair of solutions? Or is there another way to go about it?
#### Attachments
• equation.gif
1 KB · Views: 486
krab
Show us what happens when you substitute x2=dx/dt. Isn't that a 1st order equation? What's the other one? Hint: it is given in this very post.
dextercioby
Homework Helper
krab said:
Show us what happens when you substitute x2=dx/dt. Isn't that a 1st order equation? What's the other one? Hint: it is given in this very post.
Krab,u should have given him the result and let him strive to find the solution for the messy equation he gets:
$$\frac{dx_{1}}{dt}=x_{2}$$
$$\frac{dx_{2}}{dt}=-\int x_{2}dt- \alpha\sin x_{2}$$.
That should fully answer your problem.If u want to solve the second equation,try numerical methods.It is a nonlinear first order integro-differential equation with constant coefficients.Impossible to solve analytically.For almost all cases.
Last edited:
Thanks, I guess I was expecting the problem to be harder than that . Yeah, the second part of the question involves evaluation by the Runge-Kutta method.
Last edited:
krab
$$\frac{dx_1}{dt}=x_2$$
$$\frac{dx_2}{dt}=-x_1- \alpha\sin x_2$$ | 442 | 1,594 | {"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} | 3.921875 | 4 | CC-MAIN-2021-17 | latest | en | 0.913397 |
https://ask.includehelp.com/t/how-to-solve-this-string-matrix-problem/241 | 1,590,871,317,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347410352.47/warc/CC-MAIN-20200530200643-20200530230643-00392.warc.gz | 260,323,616 | 5,422 | # How to solve this string matrix problem
A matrix of characters schematically represents a swamp. The swamp is composed of muddy areas, represented with the ‘.’ character, and rocky areas, represented with the ‘’ character. The matrix size can be defined using #define, however it should be no greater than 25 rows and 80 columns.
Example of swamp:
**.
.
.…**.
.
.
.....
..
Write a C program that searches a path in the swamp, from the left to the right, without jumps, only including consecutive rocky areas. Suppose that each rocky area can have at most one other rocky area on its right (there are no branches), i.e., either on the same row, or in the previous row, or in the following one. The program shall print the row sequence of the path (the columns are implicit – there shall be a rocky area for each column), or report that no path exists.
For example: in picture
Path: 2 3 4 3 4 3 2 3 4 3 4
Hint: in a preliminary version, use a predefined matrix of strings and test the program; then modify the program and read the swamp from the keyboard input (in the future it would be possible to read it from a file).
Thanks
Hi @ali , following is my solution. kindly comment if you find anything wrong. I haven’t checked many testcases.
Also, I have returned only path, there can be many possible paths. My program is able to print all paths possible, but needs to be modified. Modification is commented within code.
Hope it helps. Any improvement will be appreciated.
`````` #include <stdio.h>
#define ROW 25
#define COL 80
char arr[ROW][COL];
int vis[COL],store[COL];
int issafe(int vis[],int curr,int curc,int r,int c){
//printf("%c\n",arr[curr][curc]);
if(curr<0 || curr>=r || curc<0 || curc>=c || arr[curr][curc]=='.')
return 0;
return 1;
}
int findpath(int vis[],int store[],int curr,int curc,int r,int c){
//base case
if(curc==c){
//store[curc]=curr;
printf("The path can be: ");
for(int i=0;i<c;i++){
printf("%d ",store[i]);
}
printf("\n");
return 1;
}
if(issafe(vis,curr,curc,r,c)){
vis[curc]=1;
store[curc]=curr;
//printf("%d\n",store[curc]);
if(findpath(vis,store,curr,curc+1,r,c))
return 1;
if(findpath(vis,store,curr+1,curc+1,r,c))
return 1;
if(findpath(vis,store,curr-1,curc+1,r,c))
return 1;
vis[curc]=0;
store[curc]=0;
return 0;
}
else{
return 0;
}
}
int main()
{
// FILE *fptr;
// fptr = fopen("input.txt", "r");
// if (fptr == NULL)
// {
// printf("Cannot open file \n");
// exit(0);
// }
int r,c;
//printf("Enter number of rows and column\n");
scanf("%d %d",&r,&c);
for(int i=0;i<r;i++){
scanf(" %[^\n]",arr[i]);
}
// for(int i=0;i<r;i++){
// for(int j=0;j<c;j++){
// printf("%c ",arr[i][j]);
// }
// printf("\n");
// }
int flag=0;
for(int i=0;i<r;i++){
for(int j=0;j<c;j++)
vis[j]=0;
for(int j=0;j<c;j++)
store[j]=0;
if(findpath(vis,store,i,0,r,c)){
flag=1;
break; //don't break here, if you need all possible paths
}
}
if(flag==0)
printf("No path there\n");
return 0;
}
``````
1 Like | 868 | 2,966 | {"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} | 2.671875 | 3 | CC-MAIN-2020-24 | latest | en | 0.831317 |
http://www.jiskha.com/display.cgi?id=1337124846 | 1,498,592,501,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321536.20/warc/CC-MAIN-20170627185115-20170627205115-00533.warc.gz | 556,722,741 | 4,203 | # physics
posted by .
Three charges sit on the vertices of an equilateral triangle, the side of which are 30.0 cm long. If the charges are A = +4.0 C, B = +5.0 C, and C = +6.0 C (clockwise from the top vertex), find the force on each charge.
• physics -
Charge A(q1)
F12 = k•q1•q2/a².
F13 = k•q1•q3/a².
F(A) = sqrt(F12² + F13² - 2•F12•F13•cos 120º).
Charge B(q2)
F21 = k•q1•q2/a².
F23 = k•q2•q3/a².
F(B) = sqrt(F21²+F23²-2•F21•F23•cos120º).
Charge C (g3),
F31 = k•q1•q3/a².
F32 = k•q2•q3/a².
F(C) = sqrt(F31² + F32² - 2•F31•F32•cos 120º).
• physics -
How did you get F12 and F13 ??
• physics -
physics - sara, Friday, August 17, 2012 at 10:45pm
How did you get F12 and F13 ??
F12 = k•q1•q2/a²=9•10^9•4•10^-6•5•10^-6/0.09=2 N
F13 = k•q1•q3/a² = 9•10^9•4•10^-6•6•10^-6/0.09=2.4 N
F(A) = sqrt(F12² + F13² - 2•F12•F13•cos 120º)=
=sqrt[4+5.76-2•2•2.4(-0.5)] =3.82 N | 440 | 875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-26 | latest | en | 0.580698 |
https://www.yemialadeworld.com/how-do-you-find-the-least-common-factor-of-a-polynomial/ | 1,669,780,977,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710719.4/warc/CC-MAIN-20221130024541-20221130054541-00728.warc.gz | 1,087,082,441 | 14,283 | # How do you find the least common factor of a polynomial?
## How do you find the least common factor of a polynomial?
All you need to do is get the factors of each polynomial, multiply the common and remaining terms to get the L.C.M.
## What is the LCM of a set of polynomials?
To find the lowest common multiple (L.C.M.) of polynomials, we first find the factors of polynomials by the method of factorization and then adopt the same process of finding L.C.M.
How do you find the lowest common denominator of a polynomial fraction?
To find the LCD of two rational expressions, we factor the expressions and multiply all of the distinct factors. For instance, if the factored denominators were (x+3)(x+4) ( x + 3 ) ( x + 4 ) and (x+4)(x+5) ( x + 4 ) ( x + 5 ) , then the LCD would be (x+3)(x+4)(x+5) ( x + 3 ) ( x + 4 ) ( x + 5 ) .
How do you use least common denominator?
Explanation: To find the least common denominator, list out the multiples of both denominators until you find the smallest multiple that is shared by both. Because 28 is the first shared multiple of 4 and 7, it must be the least common denominator for these two fractions.
### What is a least common denominator and how do you determine the least common denominator in rational expressions?
The easiest common denominator to use will be the least common denominator or LCD. The LCD is the smallest multiple that the denominators have in common. To find the LCD of two rational expressions, we factor the expressions and multiply all of the distinct factors.
### How do you find a least common denominator?
To find the least common denominator, list out the multiples of both denominators until you find the smallest multiple that is shared by both. Because 20 is the first shared multiple of 4 and 5, it must be the least common denominator for these two fractions.
Who invented LCM?
mathematician Euclid
… algorithm, procedure for finding the greatest common divisor (GCD) of two numbers, described by the Greek mathematician Euclid in his Elements (c. 300 bc). The method is computationally efficient and, with minor modifications, is still used by computers.
What is least common multiple of polynomials?
Least Common Multiple (LCM) of Polynomials The Least Common Multiple of two or more algebraic expressions is the expression of lowest degree (or power) such that the expressions exactly divide it. Least Common Multiple (LCM) of Polynomials
## How do you factor a polynomial with a common factor?
When the terms of a polynomial have a common factor, the distributive law, is used to factor the polynomial. One factor is the greatest common factor of all the terms of the polynomial. The other factor is the entire quotient, obtained by dividing each term of the polynomial by the common factor; that is,
## What is the LCM of the given polynomial?
The LCM contains the highest number of each of the different factors in the given polynomials. The first polynomial has z to the first power; the second polynomial has z to the second power. So only a z^2 is required.
Which polynomial contains the highest number of factors?
The LCM contains the highest number of each of the different factors in the given polynomials. The first polynomial has z to the first power; the second polynomial has z to the second power. So only a z^2 is required. Comment on Judith Gibson’s post “The LCM contains the high…” | 758 | 3,403 | {"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} | 4.65625 | 5 | CC-MAIN-2022-49 | latest | en | 0.92509 |
https://mathoverflow.net/questions/122557/what-manifolds-are-boundaries-of-euclidian-spaces | 1,708,955,661,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474660.32/warc/CC-MAIN-20240226130305-20240226160305-00784.warc.gz | 382,082,863 | 26,748 | # What manifolds are boundaries of euclidian spaces ?
I would like to know if there are compact (n-1)-manifolds $N$ that are not spheres but such that there is a manifold with boundary $M$ which satisfies the following two properties:
• $\partial M\cong N$
• $M-\partial M\cong \mathbb{R}^n$
I am primarily interested in this question in the category of smooth manifolds but I would be interested to know the answer in the topological case as well.
$N$ has to be a homotopy-sphere. So as long as it's dimension isn't $4$, there's a proof that it has to be the standard $S^{n-1}$.
These arguments appear in the Kosinski book on smooth manifolds. The basic idea goes like this.
1) $N$ is simply connected. There's the inclusion map $N \to M$, but if $p \in int(M)$ then there's also a retraction map $M \setminus \{p\} \to N$. Provided $n \geq 3$ removal of a point does not affect the fundamental group.
2) $N$ is a homology sphere by Alexander/Poincare duality of the pair $(M,N)$. Part (1) technically gives us this as well but this argument works even if $M$ is not contractible.
So by the Whitehead theorem, $N$ is a homotopy sphere. Moreover, $M$ is a contractible manifold whose boundary is a homotopy sphere. So $M$ is a disc by the h-cobordism theorem, and the disc has a unique smooth structure (not known in dimension $5$ still).
So that resolves all cases except $dim(N)=4$.
The answer is in fact given in the question Misha links to in his comment. For completeness, I wanted to give the details in a comment, but that became too long, so I turned it into this answer. The idea is the same as in Ryan Budney's answer, but the arguments are a bit more detailed and direct. I do apologize for the repetition. Please upvote Misha's comment, the answer he links to, and Ryan Budney's answer.$\newcommand{\RR}{\mathbb{R}} \newcommand{\int}{\operatorname{int}}$
Edit: I have slightly refined the argument below so that it goes through when the interior of the smooth manifold $M$ is only homeomorphic to $\RR^{n+1}$, not necessarily diffeomorphic. By the way, this weaker assumption makes no difference when $n\neq 3$, as the uniqueness (up to diffeomorphism) of differentiable structures on $\RR^{n+1}$ implies that the interior of $M$ is then actually diffeomorphic to $\RR^{n+1}$. So the easier, more elementary argument below assuming the interior of $M$ is diffeomorphic to $\RR^{n+1}$ is enough when $n\neq 3$.
The final answer is: if a compact smooth manifold $M$ has interior homeomorphic to $\RR^{n+1}$, then its boundary $\partial M$ is diffeomorphic to $S^n$, as long as $n>4$. In dimension $n=4$, we get only a homeomorphism between the boundary and $S^n$ by using Freedman's h-cobordism theorem instead of the smooth h-cobordism theorem in the argument below. We still get a diffeomorphism in dimension $n=3$, by the recently proved original Poincaré conjecture.
The argument goes as follows. It is simpler when the interior is actually diffeomorphic to $\RR^{n+1}$, so let us assume that at first. Let $M$ be a compact smooth manifold with interior diffeomorphic to $\RR^{n+1}$. Then removing the interior of the closed unit disc $D$ from $\RR^{n+1}$ (which we identify with the interior of $M$), we get a smooth cobordism $N=M\setminus(\int D)$ between $\partial M$ and $\partial D=S^n$ (the boundary of the removed disc).
On the one hand, the inclusion of $\partial D=S^n$ into $N$ is a homotopy equivalence, as $N$ is actually homotopy equivalent to $N\setminus \partial M=\RR^{n+1}\setminus(\int D)$ (since $\partial M$ has a collaring in $M$ and thus in $N$).
On the other hand, the inclusion of $\partial M$ into $N$ is also a homotopy equivalence. Again, note that $\partial M$ has a collar in $N$, and the complement of this collar is compact in the interior of $M$, i.e. in $\RR^{n+1}$. So by pushing off to infinity radially within $\RR^{n+1}\setminus(\int D)$ (via a suitable homotopy with bounded support starting at the identity of $\RR^{n+1}\setminus(\int D)$), we obtain a deformation of $N$ into the collar of $\partial M$ which fixes $\partial M$ pointwise. Then the collar of $\partial M$ deformation retracts onto $\partial M$, and these two deformations together give a deformation retraction of $N$ onto $\partial M$.
To modify the above argument when the interior of $M$ is only homeomorphic to $\RR^{n+1}$ via a homeomorphism $\varphi:\int M \to \RR^{n+1}$, make the following changes:
• $N=M\setminus(\int D)$ is now defined to be $M$ minus the interior of a closed disc $D$ smoothly embedded in the interior of $M$. So $N$ is still a smooth cobordism between $\partial M$ and $\partial D$.
• Importantly, note that the image $\varphi(\partial D)$ need not be a standard sphere in $\RR^{n+1}$. However, $\varphi(\partial D)$ is locally flat in $\RR^{n+1}$ (since $\partial D$ is locally flat in $\int M$), so the Shoenflies theorem implies that it is taken to a standard sphere in $\RR^{n+1}$ via some homeomorphism of $\RR^{n+1}$. By modifying $\varphi$, we can then assume that $\varphi(\partial D)$ is the standard sphere $S^n\subset \RR^{n+1}$.
• Since $\varphi(\partial D)=S^n$ is now the standard sphere in $\RR^{n+1}$, we can apply, essentially unchanged, the above two arguments showing that the inclusions of $\partial M$ and $\partial D$ into $N$ are deformation retracts.
In conclusion, $N$ is a smooth h-cobordism between $\partial M$ and $\partial D$, where $\partial D$ is diffeomorphic to the standard sphere $S^n$. Since $S^n$ is simply connected, the h-cobordism theorem implies that $\partial M$ is diffeomorphic to $S^n$ when $n>4$. As stated earlier, in dimension $n=4$, we only get a homeomorphism between $\partial M$ and $S^n$ by Freedman's h-cobordism theorem. In dimension $n=3$ we can use the Poincaré conjecture in dimension 3 (recently proved) to conclude that $\partial M$ is diffeomorphic to $S^3$, since those manifolds are homotopy equivalent to $N$ and thus to each other. In dimensions $n<3$, the classification of closed $2$-manifolds and $1$-manifolds shows we get a diffeomorphism again.
Finally, note that the elementary argument above (before the modification requiring Schoenflies theorem) showing that $N$ is a h-cobordism between $\partial M$ and $\partial D$ holds equally well for a compact topological manifold $M$ whose interior is homeomorphic to $\RR^{n+1}$. Then $\partial M\simeq N\simeq \partial D\cong S^n$. Since the Poincaré conjecture holds topologically in all dimensions, we conclude that the boundary of $M$ is homeomorphic to $S^n$.
• Thanks for a great answer Ricardo. I have accepted Ryan's answer but it's nice to have details. Feb 21, 2013 at 22:52
• @Geoffroy: No problem! Feb 22, 2013 at 1:13 | 1,859 | 6,723 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-10 | latest | en | 0.94533 |
https://www.coursehero.com/file/217478/ans1/ | 1,493,547,180,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917124478.77/warc/CC-MAIN-20170423031204-00030-ip-10-145-167-34.ec2.internal.warc.gz | 866,428,731 | 71,938 | # ans1 - Department of Economics University of California,...
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Department of Economics University of California, Berkeley Spring 2006 Economics 182 Suggested Solutions to Problem Set 1 Problem 1: National income accounts Using the national income identity we replace the known values to find the unknown variables: Y = C + I + G + EX - IM + NFP 100 = 70 + 40 + 20 + 20 - IM + 0 IM = 50 CA = EX - IM + NFP CA = 20 - 50 + 0 CA = –30 We find that imports total 50 and that there is a current account deficit of 30, that is, the country is a net borrower from the rest of the world. In open economies, savings equals investment plus the current account balance. The savings rate is then calculated by dividing savings by output. Replacing we have, S = I + CA S = 40 - 30 S = 10 s S Y = 10% If we breakdown savings into private and government savings and use T=10 we have S p = Y - T - C S p = 20 s p = 20% S g = T - G S g = - 10 s g = –10% S = S p + S g S = 10 s = 10% Problem 2: BoP transactions Recall that every export of a good, service or asset (financial or cash) enters the balance of payments accounts with a positive sign (+), counting as a credit. Likewise, every import of a good, service, or asset (financial or cash) enters the balance of payments with a negative sign (–), counting as a debit. (a) Purchase of Portuguese stock: U.S. asset import – debit in the financial account. Paying via a wire transfer from Wells Fargo to a Portuguese bank: U.S. asset export – credit in the financial account. (b) U.S. company earnings abroad that are not repatriated are credited to the U.S. investment income account. They do not actually have to cross a border. And 1
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firms do have to report these earnings to the U.S. government, so they are ob- served. They enter the current account with a positive sign, as a credit. The purchase of the machine would enter the financial account under foreign di- rect investment (as a debit since it is an acquisition of an asset located abroad), because the U.S. company is expanding its foreign holdings of capital. (c) Car rental: U.S. service export – credit in the current account. Purchase of claim on Australian credit card: U.S. asset import – debit in the financial account. Problem 3: BoP identity Yes, it is possible for a country to have a current account surplus and a balance of payments deficit. Recall that CA + KA + NRFA + SD = OSB where CA is the current account surplus, KA is the capital account balance, NRFA is the non-reserve portion of the financial account balance, SD is the statistical discrep- ancy, and OSB is the official settlements balance or the balance of payments. Therefore, if a country is running a current account surplus of, say, \$100 billion but there is a non- reserve financial account deficit of \$150 billion, then the balance of payments has a \$50 billion deficit. The situation described above could happen in a country undergoing an economic
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## This note was uploaded on 08/01/2008 for the course ECON 182 taught by Professor Kasa during the Spring '08 term at Berkeley.
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Ask a homework question - tutors are online | 886 | 3,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-17 | longest | en | 0.903021 |
https://boldlygo.tech/archive/2024-01-15-type-unification/ | 1,722,773,737,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640398413.11/warc/CC-MAIN-20240804102507-20240804132507-00805.warc.gz | 114,701,546 | 9,941 | # Type unification
### January 15, 2024
A couple of days ago we saw the spec reference the concept of “type unification”. Today we start through that explanation….
### Type unification
Type inference solves type equations through type unification. Type unification recursively compares the LHS and RHS types of an equation, where either or both types may be or contain bound type parameters, and looks for type arguments for those type parameters such that the LHS and RHS match (become identical or assignment-compatible, depending on context).
Let be break up this long paragraph with a quick example.
So we have a type equation `P ➞ A`… let’s imagine that `P`, the type parameter, is defined as `[P ~int|~float64]`. Then `A` represents an argument of `int(3)`. Type unification is the process of comparing the left-hand side (LHS; `~int|~float`) and the right-hand side (RHS; `int(3)`) to look for an identical match. In this case, `~int` and `int(3)` match, to the equation is solved, and the type parameter is `int`.
If the RHS represents an untyped constant, example, let’s say simply the literal `3`, then it’s not an identical match, but an assignment-compatible match, because the untyped constnat `3` is assignable to `int` (and to `float64`, for that matter, but it would default to `int` in this case).
To that effect, type inference maintains a map of bound type parameters to inferred type arguments; this map is consulted and updated during type unification. Initially, the bound type parameters are known but the map is empty. During type unification, if a new type argument A is inferred, the respective mapping `P ➞ A` from type parameter to argument is added to the map. Conversely, when comparing types, a known type argument (a type argument for which a map entry already exists) takes the place of its corresponding type parameter. As type inference progresses, the map is populated more and more until all equations have been considered, or until unification fails. Type inference succeeds if no unification step fails and the map has an entry for each type parameter.
In other words, an internal mapping is kept of all relevant `P ➞ A` values in the equation set. This is important because there may be multiple places where the same `P ➞ A` mapping are relevant, and once it’s solved for one of the places, that solution can be applied to the others (or trigger a type inference failure, if they conflict). Consider:
``````func sum[S ~int|~float64](a, b S) S { return a+b }
func product[P ~int|~float64](a, b P) P { return a*b }
x := product(sum(1, 2), sum(3, 4))
``````
In this case we have three generic function calls in play: Two to `sum()`, and one to `product()`. Our complete set of type parameter equations looks something like this:
```P1 ➞ sum(1, 2)
P2 ➞ sum(3, 4)
S1 ➞ 1
S2 ➞ 2
S3 ➞ 3
S4 ➞ 4
```
And our initial (empty) mapping of type parameters to inferred types something like:
```P ➞ ?
S ➞ ?
```
Type unification then goes through the list of equations, and looks for one it can solve. If we go in the order presented above, the first one that can be solved is `S``1`, because we know the default type of the untyped constnat `1` is `int`, and that is assignable to `S`’s type parameter of `~int|~float`. So the map of inferred types looks like:
```P ➞ ?
S ➞ int
```
If we apply this newly inferred `int` type to all the other instances of `S` in the list of equations above, we see that there are no conflicts. And we are also able to continue to solve the `P``1` and `P``2` equations, because the RHS of the equations is now known also to be `int`.
On the other hand, if our function call looked like this:
``````x := product(sum(1, float54(2)), sum(3, 4))
``````
we’d be in a different situation. We’d solve the `S``1` equation the same way, but when we tried to apply the solution of `int` to `S``2`, we’d see that `int` and `float64` are not assignment-compatible, even though both are assignable to `~int|~float64`, and type inference would fail.
The spec offers its own example, as well:
For example, given the type equation with the bound type parameter `P`
``` [10]struct{ elem P, list []P } ≡A [10]struct{ elem string; list []string }
```
type inference starts with an empty map. Unification first compares the top-level structure of the LHS and RHS types. Both are arrays of the same length; they unify if the element types unify. Both element types are structs; they unify if they have the same number of fields with the same names and if the field types unify. The type argument for `P` is not known yet (there is no map entry), so unifying `P` with string adds the mapping `P ➞ string` to the map. Unifying the types of the list field requires unifying `[]P` and `[]string` and thus `P` and `string`. Since the type argument for `P` is known at this point (there is a map entry for `P`), its type argument string takes the place of `P`. And since `string` is identical to `string`, this unification step succeeds as well. Unification of the LHS and RHS of the equation is now finished. Type inference succeeds because there is only one type equation, no unification step failed, and the map is fully populated.
Quotes from The Go Programming Language Specification Version of August 2, 2023 | 1,326 | 5,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-33 | latest | en | 0.833016 |
https://www.geeksforgeeks.org/gradient-descent-in-linear-regression/ | 1,696,424,763,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511369.62/warc/CC-MAIN-20231004120203-20231004150203-00637.warc.gz | 828,557,293 | 45,856 | Open In App
Related Articles
• Write an Interview Experience
• Machine Learning Algorithms
• Gradient Descent in Linear Regression
# Gradient Descent in Linear Regression
We know that in any machine learning project our main aim relies on how good our project accuracy is or how much our model prediction differs from the actual data point. Based on the difference between model prediction and actual data points we try to find the parameters of the model which give better accuracy on our dataset\, In order to find these parameters we apply gradient descent on the cost function of the machine learning model.
Gradient Descent is an iterative optimization algorithm that tries to find the optimum value (Minimum/Maximum) of an objective function. It is one of the most used optimization techniques in machine learning projects for updating the parameters of a model in order to minimize a cost function.
The main aim of gradient descent is to find the best parameters of a model which gives the highest accuracy on training as well as testing datasets. In gradient descent, The gradient is a vector that points in the direction of the steepest increase of the function at a specific point. Moving in the opposite direction of the gradient allows the algorithm to gradually descend towards lower values of the function, and eventually reaching to the minimum of the function.
### Steps Required in Gradient Descent Algorithm
• Step 1 we first initialize the parameters of the model randomly
• Step 2 Compute the gradient of the cost function with respect to each parameter. It involves making partial differentiation of cost function with respect to the parameters.
• Step 3 Update the parameters of the model by taking steps in the opposite direction of the model. Here we choose a hyperparameter learning rate which is denoted by alpha. It helps in deciding the step size of the gradient.
• Step 4 Repeat steps 2 and 3 iteratively to get the best parameter for the defined model
t ← 0
max_iterations ← 1000
w, b ← initialize randomly
while t < max_iterations do
t ← t + 1
w_t+1 ← w_t − η ∇w_t
b_t+1 ← b_t − η ∇b_t
end
Here max_iterations is the number of iteration we want to do to update our parameter
W,b are the weights and bias parameter
η is the learning parameter aslo denoted by alpha
To apply this gradient descent on data using any programming language we have to make four new functions using which we can update our parameter and apply it to data to make a prediction. We will see each function one by one and understand it
1. gradient_descent – In the gradient descent function we will make the prediction on a dataset and compute the difference between the predicted and actual target value and accordingly we will update the parameter and hence it will return the updated parameter.
2. compute_predictions – In this function, we will compute the prediction using the parameters at each iteration.
3. compute_gradient – In this function we will compute the error which is the difference between the actual and predicted target value and then compute the gradient using this error and training data.
4. update_parameters – In this separate function we will update the parameter using learning rate and gradient that we got from the compute_gradient function.
function gradient_descent(X, y, learning_rate, num_iterations):
Initialize parameters = θ
for iter in range(num_iterations):
predictions = compute_predictions(X, θ)
return θ
function compute_predictions(X, θ):
return X*θ
error = predictions - y
gradient = Xᵀ * error / m
θ = θ - learning_rate ⨉ gradient
In the Machine Learning Regression problem, our model targets to get the best-fit regression line to predict the value y based on the given input value (x). While training the model, the model calculates the cost function like Root Mean Squared error between the predicted value (pred) and true value (y). Our model targets to minimize this cost function.
To minimize this cost function, the model needs to have the best value of θ1 and θ2(for Univariate linear regression problem). Initially model selects θ1 and θ2 values randomly and then iteratively update these value in order to minimize the cost function until it reaches the minimum. By the time model achieves the minimum cost function, it will have the best θ1 and θ2 values. Using these updated values of θ1 and θ2 in the hypothesis equation of linear equation, our model will predict the output value y.
### How do θ1 and θ2 values get updated?
Linear Regression Cost Function:
so our model aim is to Minimize \frac{1}{2m} \sum_{i=1}^{m} (h_\theta(x^{(i)}) – y^{(i)})^2 and store the parameters which makes it minimum.
### Gradient Descent Algorithm For Linear Regression
Gradient descent algorithm for linear regression
-> θj : Weights of the hypothesis.
-> hθ(xi) : predicted y value for ith input.
-> i : Feature index number (can be 0, 1, 2, ......, n).
-> α : Learning Rate of Gradient Descent.
### How Does Gradient Descent Work
Gradient descent works by moving downward toward the pits or valleys in the graph to find the minimum value. This is achieved by taking the derivative of the cost function, as illustrated in the figure below. During each iteration, gradient descent step-downs the cost function in the direction of the steepest descent. By adjusting the parameters in this direction, it seeks to reach the minimum of the cost function and find the best-fit values for the parameters. The size of each step is determined by parameter α known as Learning Rate
In the Gradient Descent algorithm, one can infer two points :
• If slope is +ve : θj = θj – (+ve value). Hence the value of θj decreases.
If slope is +ve in Gradient Descent
• If slope is -ve : θj = θj – (-ve value). Hence the value of θj increases.
If slope is -ve in Gradient Descent
### How To Choose Learning Rate
The choice of correct learning rate is very important as it ensures that Gradient Descent converges in a reasonable time. :
• If we choose α to be very large, Gradient Descent can overshoot the minimum. It may fail to converge or even diverge.
Effect of large alpha value on Gradient Descent
• If we choose α to be very small, Gradient Descent will take small steps to reach local minima and will take a longer time to reach minima.
Effect of small alpha value on Gradient Descent
## Python Implementation of Gradient Descent
At first, we will import all the necessary Python libraries that we will need for mathematical computation and plotting like numpy for mathematical operations and matplotlib for plotting. Then we will define a class Linear_Regression that represents the linear regression model.
We will make a update_coeffs method inside the class to update the coefficients (parameters) of the linear regression model using gradient descent. To calculate the error between the predicted output and the actual output we will make a predict method that will make predictions using the current model coefficients.
For updating and calculating the gradient of the error we will make compute_cost which will apply gradient descent on (mean squared error) between the predicted values and the actual values.
## Python3
# Implementation of gradient descent in linear regressionimport numpy as npimport matplotlib.pyplot as plt class Linear_Regression: def __init__(self, X, Y): self.X = X self.Y = Y self.b = [0, 0] def update_coeffs(self, learning_rate): Y_pred = self.predict() Y = self.Y m = len(Y) self.b[0] = self.b[0] - (learning_rate * ((1/m) * np.sum(Y_pred - Y))) self.b[1] = self.b[1] - (learning_rate * ((1/m) * np.sum((Y_pred - Y) * self.X))) def predict(self, X=[]): Y_pred = np.array([]) if not X: X = self.X b = self.b for x in X: Y_pred = np.append(Y_pred, b[0] + (b[1] * x)) return Y_pred def get_current_accuracy(self, Y_pred): p, e = Y_pred, self.Y n = len(Y_pred) return 1-sum( [ abs(p[i]-e[i])/e[i] for i in range(n) if e[i] != 0] )/n # def predict(self, b, yi): def compute_cost(self, Y_pred): m = len(self.Y) J = (1 / 2*m) * (np.sum(Y_pred - self.Y)**2) return J def plot_best_fit(self, Y_pred, fig): f = plt.figure(fig) plt.scatter(self.X, self.Y, color='b') plt.plot(self.X, Y_pred, color='g') f.show() def main(): X = np.array([i for i in range(11)]) Y = np.array([2*i for i in range(11)]) regressor = Linear_Regression(X, Y) iterations = 0 steps = 100 learning_rate = 0.01 costs = [] # original best-fit line Y_pred = regressor.predict() regressor.plot_best_fit(Y_pred, 'Initial Best Fit Line') while 1: Y_pred = regressor.predict() cost = regressor.compute_cost(Y_pred) costs.append(cost) regressor.update_coeffs(learning_rate) iterations += 1 if iterations % steps == 0: print(iterations, "epochs elapsed") print("Current accuracy is :", regressor.get_current_accuracy(Y_pred)) stop = input("Do you want to stop (y/*)??") if stop == "y": break # final best-fit line regressor.plot_best_fit(Y_pred, 'Final Best Fit Line') # plot to verify cost function decreases h = plt.figure('Verification') plt.plot(range(iterations), costs, color='b') h.show() # if user wants to predict using the regressor: regressor.predict([i for i in range(10)]) if __name__ == '__main__': main()
Output:
100 epochs elapsed
Current accuracy is : 0.9836456109008862
Regression line before gradient descent iteration
Regression line after gradient descent iteration
Accuracy graph for gradient descent on model
Note: Gradient descent sometimes is also implemented using Regularization. | 2,320 | 10,054 | {"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} | 3.28125 | 3 | CC-MAIN-2023-40 | latest | en | 0.82927 |
https://community.shopify.com/c/ecommerce-marketing/facebook-ad-what-are-the-kpi-s-for-a-successful-ad/td-p/510205 | 1,679,472,839,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943750.71/warc/CC-MAIN-20230322051607-20230322081607-00229.warc.gz | 223,280,533 | 56,207 | Solved
Tourist
5 0 0
Hey guys,
Pretty new here, we started our first Shopify store three weeks ago. We started running some FB ads and got some results (Reach 2,611, Impressions: 2,807, Link Clicks: 28, CTR: 1.00%, Cost per content view: \$0.39). Still 0 Orders.... as our site reached its 500 visits. Problem is, we got nothing to compare these figures to. Maybe it is too soon to get any conclusions but could you guys share your views about what are good KPIs for an ad ?
Every comment is welcome here 🙂
Daniel
Accepted Solutions (2)
Accepted Solutions
Shopify Partner
45 2 45
This is an accepted solution.
Let me work through some quick definitions and math so everyone can follow along.
Reach (Facebook reach is the number of unique people who saw your content): 2611
Impressions (How many times 50% or more of your ad (measured in pixels) was on a potential customers screen): 2807
Link Clicks (How many times a potential customer clicked you ad and was directed to your website): 28
CTR - Click Through Rate (The total number of clicks divided by the number of impressions - displayed as a percentage): ~1.00%
Okay, the math starts now.
On average advertisers online can achieve a CTR ranging from 1% to 3% (depending on the product and audience).
That means in order to get 100 people to click the link and reach your store you need to display your ads to between ~3000 - 10,000 people
Once they make it to your store there is additional drop off.
Simply put
1 - 3% of people click the link
Not everyone that views your store buys a product. The average conversion rate (depending on your products and audience) range from 1 - 7%
1 - 7% makes a purchase
1 - 7 people buy a product.
HOWEVER
We also have to measure these numbers from Unique Users. The same person may visit your store multiple times and never make a purchase (increasing your Impressions/link clicks without actually converting.
Industry average KPI's
1 - 3% CTR
1 - 7% Conversion
In many industries competition is quite high. As profit margins go up, the amount an advertiser is willing to spend to acquire a customer is more (legal services, financial products etc.)
Best regards,
Matthew Herchel
2H.Media
Connect with me here:
Shopify Expert
1568 152 934
This is an accepted solution.
When it comes to an ecom store... your KPI is either going to be your CPA & return on ad spend (ROAS) or your LTV of a customer. Everything else is a distraction because your focus should be on building a profitable business. Things like CPC and CTR are a distraction and not worth heavily focusing on.
We help ecommerce & DTC brands create, manage and scale profitable PPC campaigns: TakeSomeRisk.com
P.S. Take my Google Shopping Course and help grow your revenue this month.
Replies 4 (4)
Shopify Partner
45 2 45
This is an accepted solution.
Let me work through some quick definitions and math so everyone can follow along.
Reach (Facebook reach is the number of unique people who saw your content): 2611
Impressions (How many times 50% or more of your ad (measured in pixels) was on a potential customers screen): 2807
Link Clicks (How many times a potential customer clicked you ad and was directed to your website): 28
CTR - Click Through Rate (The total number of clicks divided by the number of impressions - displayed as a percentage): ~1.00%
Okay, the math starts now.
On average advertisers online can achieve a CTR ranging from 1% to 3% (depending on the product and audience).
That means in order to get 100 people to click the link and reach your store you need to display your ads to between ~3000 - 10,000 people
Once they make it to your store there is additional drop off.
Simply put
1 - 3% of people click the link
Not everyone that views your store buys a product. The average conversion rate (depending on your products and audience) range from 1 - 7%
1 - 7% makes a purchase
1 - 7 people buy a product.
HOWEVER
We also have to measure these numbers from Unique Users. The same person may visit your store multiple times and never make a purchase (increasing your Impressions/link clicks without actually converting.
Industry average KPI's
1 - 3% CTR
1 - 7% Conversion
In many industries competition is quite high. As profit margins go up, the amount an advertiser is willing to spend to acquire a customer is more (legal services, financial products etc.)
Best regards,
Matthew Herchel
2H.Media
Connect with me here:
Shopify Expert
1568 152 934
This is an accepted solution.
When it comes to an ecom store... your KPI is either going to be your CPA & return on ad spend (ROAS) or your LTV of a customer. Everything else is a distraction because your focus should be on building a profitable business. Things like CPC and CTR are a distraction and not worth heavily focusing on.
We help ecommerce & DTC brands create, manage and scale profitable PPC campaigns: TakeSomeRisk.com
P.S. Take my Google Shopping Course and help grow your revenue this month.
Tourist
5 0 0
Exactly the kind of numbers I was looking for, cheers 🙂
My website is https://iamnotredame.com/
FB Pixel says we just reached 1.3K PageViews and 460 Contentviews. No orders yet. So either we continue the FB ads to have some bigger numbers, either we"ll have to change the content
Shopify Expert
1568 152 934
@Dantheman wrote:
Exactly the kind of numbers I was looking for, cheers 🙂
My website is https://iamnotredame.com/
FB Pixel says we just reached 1.3K PageViews and 460 Contentviews. No orders yet. So either we continue the FB ads to have some bigger numbers, either we"ll have to change the content
I would look at how many people add to card and initiated checkout to tell if your ad is working or not. Those will help you tell if people are closing to making a sale.
We help ecommerce & DTC brands create, manage and scale profitable PPC campaigns: TakeSomeRisk.com
P.S. Take my Google Shopping Course and help grow your revenue this month.
Top Contributors
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https://www.pcreview.co.uk/threads/indirect-worksheet-function.3382333/ | 1,660,645,945,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572286.44/warc/CC-MAIN-20220816090541-20220816120541-00263.warc.gz | 809,364,797 | 14,604 | # Indirect worksheet function
R
#### rk0909
All,
I am using the =SUM('Start>>:<<End'!B7) formula to sum data from 50 sheets.
I want to make this formula little dynamic. I want the '7" in the formula to
be a cell link e.g. A1 so that it sums up the field in the cells whose row is
defined by the number input in A!.
I tried using =SUM(INDIRECT("('Start>>:<<End'!B"&A1) but this gives errors.
Any solutions or alternatives to this method.
Thanks much,
RK
H
#### Herbert Seidenberg
Make a list of all your sheets
and name the list FL.
=SUMPRODUCT(SUMIF(A1,A1,INDIRECT(FL&"!B"&A1)))
If the list can be generated by some
algorithm, further automation is possible.
R
#### rk0909
Thanks Herbert.
I am new to lists, could you please explain little bit more on how to create
the list. I looked at excel help and couldn't find a way to create a list
across diff. sheets.
thanks much,
RK
H
#### Herbert Seidenberg
The most straight forward way is to type
the sheet names into 50 adjacent cells
and name those 50 cells FL.
The automated way is to use a macro:
Sub listsheets()
For i = 1 To Worksheets.Count
Cells(i, "A") = Sheets(i).Name
'List starts at A1.
'If you want to start the list at B22 use
'Cells(i + 21, "B") = Sheets(i).Name
Next i
End Sub
If you sheet names have some order to them,
or you are willing to change the names,
then I can give you an easy shortcut. | 390 | 1,383 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-33 | latest | en | 0.816763 |
http://forum.roulette30.com/index.php?PHPSESSID=f2g1uth4bthbais0b9hmniesi4&topic=351.0 | 1,516,332,750,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887729.45/warc/CC-MAIN-20180119030106-20180119050106-00295.warc.gz | 135,355,681 | 8,679 | ### Author Topic: Black Magic (Read 3583 times)
0 Members and 1 Guest are viewing this topic.
#### ignatus
• Mature Member
• Posts: 470
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##### Black Magic
« on: June 09, 2015, 08:06:19 AM »
Very good results from the first tests.
This is 8 numbers bet, -flatbet-
Consider these 4 black diamonds
2-4-6-8
11-13-15-17
20-22-24-26
29-31-33-35
Trigger: Wait for 2 red numbers to hit in sequence, then start the attack and bet the two cold black diamonds (last two NOT hit) for 4 spins, if any black number hit STOP the attack.
for an example
22
32
2
3
14 --Trigger 2 RED in sequence, now start to bet the two cold black diamonds: 11-13-15-17 & 29-31-33-35.
4/4 Games won
#### december
• Veteran Member
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##### Re: Black Magic
« Reply #1 on: June 09, 2015, 08:27:57 AM »
This would be 4 which has the least hits?
#### ignatus
• Mature Member
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##### Re: Black Magic
« Reply #2 on: June 09, 2015, 09:20:38 AM »
This would be 4 which has the least hits?
#### december
• Veteran Member
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##### Re: Black Magic
« Reply #3 on: June 09, 2015, 01:59:06 PM »
Sorry just now I see - two cold black diamonds (last two NOT hit).
So, we invest 4x8=32 in hope to get 28, 20, 12 or 4!
Do you think it would be the same result (as your test) with any other 8 numbers?
#### ignatus
• Mature Member
• Posts: 470
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• Gender:
##### Re: Black Magic
« Reply #4 on: June 09, 2015, 02:54:42 PM »
Sorry just now I see - two cold black diamonds (last two NOT hit).
So, we invest 4x8=32 in hope to get 28, 20, 12 or 4!
Do you think it would be the same result (as your test) with any other 8 numbers?
i tried to bet the two hottest diamonds (the two most recent hit) but that didn't work so well, so the original bet is kept. (the two coldest diamonds).
#### BlueAngel
• I always express my opinion
• Hero Member
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• Do you want truth? You cannot handle the truth!
##### Re: Black Magic
« Reply #5 on: June 09, 2015, 05:07:38 PM »
Why not the red X's too?
I guess you have tried with red X's but you didn't have positive results,right?
So does it make sense to you that the red X's don't work but black O's do?
This pattern on the table layout, black O's and red X's,it's like playing TIC TAC TOE.
There are 4 red X's,2 of them including only red numbers and the other 2 including 4 red and 1 black.
Also there are 4 small black O's (or diamonds) ,the middle dozen has 1 big black X which is including 6 black numbers and 1 big red O which includes 6 numbers.
The smaller X's are red dominated,while the big X in the middle is totally black,the 4 small O's are black,while the big one inside the second dozen is red.
Therefore:
4 BLACK O PLUS 1 BIG RED
4 RED X PLUS 1 BIG BLACK
Why you think exists such symmetrical pattern on the table layout,pure coincidence?
#### ignatus
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##### Re: Black Magic
« Reply #6 on: June 09, 2015, 05:19:41 PM »
it's just easier to play and track the black diamonds, i guess.
#### december
• Veteran Member
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##### Re: Black Magic
« Reply #7 on: June 10, 2015, 08:37:04 AM »
I just start to use Rx - still trial period.
Can you send me or publish Rx code for this system.
If you or somebody have more Rx codes to share, it would be nice help! | 1,078 | 3,488 | {"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-2018-05 | latest | en | 0.851715 |
https://currencyconverts.com/eur/sgd/99226 | 1,660,439,918,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571989.67/warc/CC-MAIN-20220813232744-20220814022744-00271.warc.gz | 210,708,728 | 11,242 | # 99226 EUR to SGD
99,226Euro (€)
=
139,968.20Singapore Dollar (\$)
## 99,226 Euro (EUR) to Singapore Dollar (SGD)
You have converted 99226 Euro (EUR) to Singapore Dollar (SGD). Exchange rate for the conversion is 139968.1956 for 14 August 2022, Sunday. How much is 99226 Euro to Singapore Dollar? 139,968.20 Singapore Dollar's.
## How much Ninety-nine Thousand Two Hundred Twenty-six (99226) Euro in Singapore Dollar?
Today, 99,226 (ninety-nine thousand two hundred twenty-six) Euro are worth 139,968.20 Singapore Dollar. That's because the current exchange rate, to S\$, is 1.411. So, to make Euro to Singapore Dollar conversion, you just need to multiply the amount in € by 1.411.
• Name Euro
• Money99226
• Country Europa
• Symbol
• ISO 4217 EUR
• Name Singapore Dollar
• Money139968.1956
• Country Singapore
• Symbol S\$
• ISO 4217 SGD
This page provides the exchange rate of ninety-nine thousand two hundred twenty-six Euro to Singapore Dollar, sale and conversion rate. We added the list of the most popular conversions for visualization and the history table with exchange rate diagram for 99226 Euro to Singapore Dollar from Sunday, 14/08/2022 till Wednesday, 3/08/2022.
DateEuro (€)Singapore Dollar (S\$)
14 August 2022, Sunday99,226 EUR139,968.196 SGD
13 August 2022, Saturday99,226 EUR139565.437266 SGD
12 August 2022, Friday99,226 EUR139625.36977 SGD
11 August 2022, Thursday99,226 EUR140254.661062 SGD
10 August 2022, Wednesday99,226 EUR139929.695912 SGD
09 August 2022, Tuesday99,226 EUR139616.538656 SGD
08 August 2022, Monday99,226 EUR139346.247032 SGD
07 August 2022, Sunday99,226 EUR139466.508944 SGD
06 August 2022, Saturday99,226 EUR139723.008154 SGD
05 August 2022, Friday99,226 EUR139723.008154 SGD
04 August 2022, Thursday99,226 EUR139895.76062 SGD
03 August 2022, Wednesday99,226 EUR139234.617782 SGD
This graph show how much is 99226 Euro in Singapore Dollar = 139968.1956 SGD, according to actual pair rate equal 1 EUR = 1.4106 SGD. Yesterday this currency exchange rate plummeted on -139,565.4373 and was S\$ 139,565.4373 Singapore Dollars for € 1. On this graph you can see trend of change 99,226 EUR to SGD. And average currency exchange rate for the last week was S\$ 1.4078 SGD for €1 EUR.
``<a href="https://currencyconverts.com/eur/sgd/99226" title="99,226 Euro(EUR) to Singapore Dollar(SGD) Currency Rates Today">99,226 Euro(EUR) to Singapore Dollar(SGD)</a>`` | 713 | 2,407 | {"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-2022-33 | latest | en | 0.75462 |
https://www.myelectrical2015.com/2020/01/laws-of-reflection-and-refraction.html | 1,726,142,025,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00372.warc.gz | 830,760,510 | 20,733 | ## 23/01/2020
### Laws of Reflection and Refraction
• When a ray of light strikes any boundary between two transparent substances, it is divided in to reflected ray and refracted ray.
Law of Reflection
θ1 = θ1
Where
θ1 = Angle of incidence
θ1’ = Angle of reflection
I = Incident ray
R’ = Reflected ray
The incident ray ( I ) , reflected ray ( R’ ) and normal ( N ) are in one plane.
Laws of Refraction
It is known as Snell’s law
( Sin θ1 / Sin θ2 ) = ( µ1 / µ2 )
Where
θ1 = Angle of incidence
θ2 = Angle of refraction
µ1 = Refractive index of medium 1
µ2 = Refractive index of medium 2
The incident ray ( I ) , refracted ray ( R ) and normal ( N ) at the point of incidence are in one place.
You may also like : | 215 | 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} | 2.75 | 3 | CC-MAIN-2024-38 | latest | en | 0.889725 |
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Amortization Schedule Calculator. Determine Principal and Interest Payments As Time Passes
This loan amortization calculator numbers your loan payment and interest expenses at different re re payment periods. Merely input the key amount lent, the size of the loan while the annual rate of interest and also the calculator does the others.
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How exactly to Create, Utilize, and Know Amortization Schedules
Should you determine that loan payment interest that is including principal?
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Amortization Schedules 101
An amortization routine is just a table detailing each payment that is periodic amortizing a loan. Amortization is the method of paying down a debt in the long run through regular re re payments.
Loan re payments contain principal and interest. At the start of the loan term the attention element of each repayment is extremely high since the balance owed regarding the loan is high. Given that principal gets compensated in the loan the proportionate level of each repayment gets paid off until almost the whole payment becomes principal toward the termination of the loan term.
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Drawbacks Of Amortized Loans
Although amortized loans would be the most typical, a number of faculties among these loans you want to take into consideration.
Amortized loans try not to build up much equity in the end that is front of loan. For instance, you will notice this together with your home loan. Even although you’ve been spending your 30-year mortgage for seven years, you are going to nevertheless owe quite a little more for the loan that is original than you may expect. That is since the early repayments are mostly interest.
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How Fast Am I Able To Pay Back an loan that is amortized?
Whenever your loan is amortized, your lender determines your equal monthly premiums so you will probably pay your loan coincident off utilizing the end associated with loan term. Should you want to spend your loan off faster, you’ll want to boost your major payments. Doing this has a few advantages:
• You are going to spend exponentially less interest whenever you boost your major payments. Numerous loan providers permit you to make principal-only repayments in addition to your frequently scheduled re re payments.
• By paying down your loan faster, it is possible to prevent the stress that is included with financial obligation. You never always have to take a faster loan term to cover down the debt faster. Keep in mind you are able to pay back, for instance, a 30-year mortgage in 15 years if you are paying it just like a 15-year home loan. This technique will allow you to steer clear of the anxiety of experiencing which will make a greater payment that is monthly allowing a choice of settling the loan quicker.
For inspiration to incorporate additional principal to your repayments, simply utilize the amortization routine calculator to determine exactly how much interest you will put away. Based on terms, the total amount may be dramatic.
How To Locate Loan Payments And Interest Costs
You’ll find total interest expenses making use of the Amortization Schedule Calculator by scrolling right down to the finish regarding the amortization routine where it shows the grand total both for major payments and interest re payments.
It will likewise show your loan re payment quantity and just how a lot of each re payment goes toward principal and interest. It is possible to determine loan re re re payments for several periods, including month-to-month, quarterly, semi-quarterly, and intervals that are annual.
Summary
An amortization dining dining table is a tremendously of good use device for making monetary decisions. While determining your loan amortization by hand or with the use of spreadsheet is hard, the procedure is easy using this Amortization Schedule Calculator.
Make use of this calculator to prepare your financial troubles payoff and lower your total interest expenses in order to advance from paying down financial obligation to building wide range. | 1,061 | 5,555 | {"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-2021-17 | latest | en | 0.961992 |
https://math.stackexchange.com/questions/859620/how-to-find-the-average-direction-of-a-set-of-vectors | 1,643,323,516,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305317.17/warc/CC-MAIN-20220127223432-20220128013432-00560.warc.gz | 433,057,673 | 33,073 | # How to find the "average" direction of a set of vectors?
I have a series of vector directions and I need to find the "average" direction. I am not looking for the overall direction which would be the sum of the directions and this can't be used in cases where the vectors all cancel out.
I know of spherical linear interpolation but its not really practical when I have to do this for a large set of vectors.
EDIT: I misinterpreted my problem and so here is the updated version, sorry for any inconvenience.
What I have is a set of directions and one of the directions is selected as "leading" (shown as the red line) direction and then I need to find the average direction produced by the remaining vectors (the black ones) and it seems by drawing it out I have figured out one way of how I can do it.
However, Rahul's comment about means of circular quantities does seem to be a much better method which I will be looking into.
The dashed red vector is meant to be the opposite of the red vector.
• By average direction, do you mean an average of the directions without considering their magnitude? In which case you could normalise all the vectors (so that they have the same direction but magnitude 1) and then add them Jul 7 '14 at 23:46
• How would that work if the resulting sum is a vector with zero magnitude? And, yes the magnitude is irrelevant in my case. Jul 7 '14 at 23:49
• I think the point then is that the average direction is not defined - for example if I pull in one direction and you pull in the opposite direction, then there is no total pull, so you can't really say that the average direction is in any particular direction. Jul 7 '14 at 23:50
• Would it not be perpendicular to the two vectors in the case? Jul 7 '14 at 23:56
• Perpendicular in which direction? If we factor in 3 dimensions, that gives infinitely many options (e.g. up, down, at a 30 degree angle to the ground etc) And I'm not convinced perpendicular is the answer anyway. Jul 7 '14 at 23:58 | 459 | 1,994 | {"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-2022-05 | latest | en | 0.968236 |
https://www.shaalaa.com/question-bank-solutions/coordinate-geometry-examples-solutions-find-area-quadrilateral-abcd-whose-vertices-are-a-3-1-b-2-4-c-4-1-d-3-4_50520 | 1,576,014,930,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540529006.88/warc/CC-MAIN-20191210205200-20191210233200-00383.warc.gz | 852,785,701 | 11,066 | Share
# Find the Area of the Quadrilateral Abcd, Whose Vertices Are A(−3, −1), B (−2, −4), C(4, − 1) and D (3, 4). - CBSE Class 10 - Mathematics
ConceptCoordinate Geometry Examples and Solutions
#### Question
Find the area of the quadrilateral ABCD, whose vertices are A(−3, −1), B (−2, −4), C(4, − 1) and D (3, 4).
#### Solution
The given quadrilateral i.e., ABCD whose vertices are A (−3, −1), B (−2, −4), C (4, −1) and D (3, 4) can be drawn as follows:
Here, B is joined with D.
We know that the area of a triangle whose vertices are (x1 , y1 ), (x2 , y2 ) and (x3 , y3 ) is given by
=1/2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]
=1/2[-3(-8)-2(5)+3(3)]
=1/2[24-10+9]
=23/2
=11.5 sq.inits
∴ar(ΔABD)
=1/2[-3(-4-4)+(-2)(4+1)+3(-1+4)]
∴ar (ΔCDB)
=1/2[4(4+4)+3(-4+1)+(-2)(-1-4)]
=1/2[(4xx8)+(3x-3)-2xx(-5)]
=1/2[32-9+10]
=33/2
=16.5 sp.unit
Thus, ar (ABCD) = ar (ΔABD) + ar (ΔCDB) = (11.5 + 16.5) sq units = 28 sq units
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
Solution Find the Area of the Quadrilateral Abcd, Whose Vertices Are A(−3, −1), B (−2, −4), C(4, − 1) and D (3, 4). Concept: Coordinate Geometry Examples and Solutions.
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REQUIRED RESOURCES
Text
Amacher, R., & Pate, J. (2013). Microeconomics principles and policies [Electronic version]. Retrieved from https://content.ashford.edu/
• This text is a Constellation™ course digital materials (CDM) title.
Article
Martin, E. R. (2014, March 27). The ethics of big data (Links to an external site.)Links to an external site.. Forbes. Retrieved from http://www.forbes.com/sites/emc/2014/03/27/the-ethics-of-big-data/#4091edf730c7
Multimedia
DISCUSSION 1
Let’s assume that you own a fast food restaurant and you are faced with many customers each day eating in the restaurant without any tables. Describe the difference between the short run and long run in the example to bringing about more tables for the customers. How is the restaurant able to differentiate between the short run and long run?
Guided Response: Review the discussion board posts of your classmates. Discuss the difference between short run and long run with relation to costs. Respond to at least two of your classmates. Discuss how short run and long run vary in a firm.
DISCUSSION 2
After reading Chapter 8 in the text and viewing the Fixed, Variable, and Marginal Cost (Links to an external site.)Links to an external site. video, address the following in your initial post:
• First, describe several different fixed costs and variable costs associated with operating an automobile.
• Next, assume that you would like to travel from Los Angeles to New York City by either car or plane. Which costs would you take into account in making your decision, fixed costs, variable costs or both? Make sure to explain your analysis in the decision that you have to make.
Guided Response: Review the discussion board posts of your classmates. Analyze the difference in your answer to your peers in response to the different costs that are discussed. Respond to at least two of your classmates. Discuss the different types of costs like fixed, variable, or both
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1. Question : Jiffy Wax Corp. Can sell common stock for \$15 per share and its investors require a 14 % return. However, the administrative or flotation costs associated with selling the stock amount to \$2.40 per share. What is the cost of capital for Jiffy Wax if the corporation raises money by selling preferred stock?
2. Question : Kinslow Manufacturing Company paid a dividend yesterday of \$2.50 per share. The dividend is expected to grow at a constant rate of 5% per year. The price of Kinslow’s common stock today is \$25 per share. If Kinslow decides to issue new common stock, flotation costs will equal \$2.00 per share. Key’s marginal tax rate is 34 %. Based on the above information, the cost of retained earnings is _________
3. Question : Nickel Industries is considering the purchase of a new machine that will cost \$178,000, plus an additional \$12,000 to ship and install. The new machine will have a 5 year useful life and will be depreciated using the straight line method. The machine is expected to generate new sales of \$85,000 per year and is expected to increase operating costs by \$ 10,000 annually. Nickel’s income tax rate is 40%. What is the projected incremental cash flow of the machine or year 1?
4. Question : Nargo Inc. Wants to replace a 7 year old machine with a new machine that is more efficient. The old machine cost \$50,000 when new and has a current book value of \$10,000. Margo can sell the machine to a foreign buyer for \$12,000. Margo’s tax rate is 30%. The effect of the sale of the old machine on the initial outlay for the new machine is ________
5. Question : A capital budgeting project has a net present value of \$10,000 and a modified internal rate of return of 13%. The project's required rate of return is 11 %. The internal rate of return is ______
6. Question : Higgins Office Corp. Plans to maintain its optimal capital structure of 40 percent debt, 10 percent preferred stock, and 50 percent common equity indefinitely. The required return on each component source of capital is as follows: debt 8 percent; preferred stock- 12 percent; common equity- 16 percent. Assuming a 40 percent marginal tax rate, what after tax rate of exchange must Higgins Office Corp. Earn on its investments if the value of the firm is to remain unchanged?
7. Question : Zellar’s, Inc. Is considering two mutually exclusive projects, A and B. Project A costs \$ 75,000 and is expected to generate \$48,000 in year one and \$45,000 in year two. Project B costs \$80,000 and is expected to generate \$34,000 on year one, \$37,000 in year two, \$26,000 in year three, and \$25,000 in year four. Zellar, Inc.’s required rate of return for these projects is 10%. The internal rate of return for Project B is ________
8. Question : A new machine can be purchased for \$1,000,000. It will cost \$65,000 to ship and \$35,000 to modify the machine. A \$30,000 recently completed feasibility study indicated that the firm can employ an existing factory owned by the firm, which would have otherwise been sold for \$150,000. The firm will borrow \$750,000 to finance the acquisition. Total interest expense for 5 years is expected to approximate \$250,000. What is the investment cost of the machine for capital budgeting purposes?
9. Question : Zellar’s, Inc. Is considering two mutually exclusive projects, A and B. Project A costs \$ 75,000 and is expected to generate \$48,000 in year one and \$45,000 in year two. Project B costs \$80,000 and is expected to generate \$34,000 on year one, \$37,000 in year two, \$26,000 in year three, and \$25,000 in year four. Zellar, Inc.’s required rate of return for these projects is 10%. The profitability index for Project B is ________
10. Question : Jones Company has a target capital structure of 40% debt, 10% preferred stock, and 50% common equity. The company’s after tax cost of debt is 8%, its cost of preferred debt is 10%, its cost of retained earnings is 14%, and its cost of new common stock is 16%. The company stock has a beta of 1.2 and the company’s marginal tax rate is 35%. What is the company’s weighted average cost of capital if retained earnings are used to fund the common equity portion?
Bus 401 week 3 quiz (ash)
Bus 401 week 3 quiz (ash)
BUS 401 Week 3 Quiz (ASH) BUS 401 Week 3 DQ 1 NPV, PI, and IRR (ASH) BUS 401 Week 3 Assignment Weighted Average Cost (ASH) BUS 401 Week 2 Qu... | 1,217 | 4,744 | {"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-2018-05 | latest | en | 0.873155 |
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13 QuestionsShow answers
• Question 1
60 seconds
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Q. The picture shows an electrical circuit. This circuit is a series circuit because:
answer choices
It has 3 light bulbs.
The same current flows through all three light bulbs.
It uses a single battery.
The electrical current is divided between the three light bulbs.
• Question 2
300 seconds
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Q. The total resistance in this circuit is
answer choices
100 ohms
450 ohms
400 ohms
30 ohms
• Question 3
300 seconds
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Q. The total current through this circuit is
answer choices
0.02 amps
0.09 amps
50 amps
9 amps
• Question 4
300 seconds
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Q. The voltage dropped across the 300 ohm resistor is
answer choices
6 V
9 V
2 V
30 V
• Question 5
900 seconds
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Q.
What is the total resistance (RT) across this circuit?
answer choices
3 kΩ
10 kΩ
13 kΩ
18 kΩ
• Question 6
900 seconds
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Q. As resistors are added in series to a circuit, the total resistance will...
answer choices
increase
decrease
stay the same
• Question 7
900 seconds
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Q. As the resistance of a circuit increases, the current will...
answer choices
increase
decrease
stay the same
• Question 8
30 seconds
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Q. In a parallel circuit which of the following is the same value throughout the circuit?
answer choices
Voltage
Current
Resistance
Charge
• Question 9
180 seconds
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Q. The current going through the 45 ohm resistor is
answer choices
0.05 amps
0.2 amps
0.1 amps
5 amps
• Question 10
180 seconds
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Q. The total resistance of this circuit is
answer choices
25.7 ohms
315 ohms
324 ohms
90 ohms
• Question 11
180 seconds
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Q. The total current going through the battery of this circuit is
answer choices
0.35 amps
0.0286 amps
0 amps
9 amps
• Question 12
30 seconds
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Q. Does this diagram represent a series or parallel circuit?
answer choices
series circuit
parallel circuit
• Question 13
45 seconds
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Q.
What is the definition of electrical current?
answer choices
The opposition to the flow of electrons
The rate of flow of charge
The work done on each coulomb of charge
Electrons moving around
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### Angles and Circles
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Hello all, welcome to our NPTEL Online Certification Courses on Engineering Drawing
and Computer Graphics.
(Refer Slide Time: 00:30)
(Refer Slide Time: 00:34)
We are in module 2 and lecture 8; the module name is Conic Sections. In lecture 7, we
have learned how to bisect a line and an arc; how to draw a perpendicular line; how to
divide a line. In lecture 8, we are covering how to bisect an angle, how to trisect a right
angle, how to divide a circle, how to pass a circle through three points, these are the things
that we are going to learn.
(Refer Slide Time: 01:07)
The first one is how to bisect an angle? So, here, we have an angle AQR; some arbitrary
angle. Now, we want a line that passes through Q such a way that PQC, angle PQC equal
to angle CQR, how to construct that we are going to learn it.
(Refer Slide Time: 01:58)
To bisect an angle PQR, first of all, we have to mark points A; mark point A and B from
Q with an arbitrary radius. So, we have to use Q mark a random radius. So, because we
know from P to Q maybe half of the distance or lower than that, more than that, it does not
matter. But first of all, make an arc that will intersect Q to P at A and Q to R at B.
Then, use points A and B as centers with the same radius or arbitrary radius. Mark a curve
from A as the centre, similarly from B as centre; mark another arc, where it is intersecting
call that point as C. One C is known; from C to Q, join it by a line. Once it is done, this
angle and this angle becomes the same. Let us look at geometric construction on the sheet.
(Refer Slide Time: 03:47)
Let us draw an angle, somewhere mark it. So, call points Q, P, R. Now, with Q as a centre
at some distance with radius mark curves; say these points as A and B. With the same
radius pick B, mark an arc, where it is intersecting, call that point as C. Now, join points
C and Q. If we use this protractor, the ∠PQC will be the same as the ∠CQR. Let us call
these as the alpha angle. This is the way we bisect an angle.
(Refer Slide Time: 05:45)
Let us look at the next one. How to trisect a right angle? For example, we have a right
angle lines AB and AC. This entire thing, we can use protractor directly to divide it.
However, any arbitrary thing, if we would like to construct a similar procedure works.
To trisect it, first of all, we have to pass an arc passing through B and C points with centre
as A and radius as AB because we know BAC is the triangle. What is it making? Angle.
Use A to B as radius, A as the centre, draw a curve arc, that is this arc..
(Refer Slide Time: 07:01)
Once it is done, mark point D from point C. So, C is this point, and D is that point. So, use
an arbitrary radius, mark a point. Similarly, mark point E from point B; mark a point E
with the same radius. Now, join D and A.; similarly, join E and A, after marking.
(Refer Slide Time: 08:02)
Once it is done, we have three portions; I, II, and III parts. So, angle BAD, angle DAE,
angle EAC are all equal. Let us construct that using our geometrical construction. First of
all, we have to draw a perpendicular line, right angle. Use a pencil. Let us mark point A,
point C. We would like to have a perpendicular line. So, I am trying to use a set square
somewhere here; this is the line that we have.
(Refer Slide Time: 09:36)
Now, we have to do the first step a radius A to B passes through that arc; either A to B or
A to C. So, let us tighten that, draw an arc. Once it is done, extend through scale so that
we have point B also here and an arc passing through B and C. From C, with the same
radius, mark an arc; similarly, from B mark an arc, join. The points are D;, the point is E,
and once it is done, join them. So, this is the first part, the second part, and this is the third
part; this is the way we trisect an angle.
(Refer Slide Time: 11:26)
If it is something like a circle, the easiest way is if we would like to divide a circle, first
draw a circle. Use your protractor, mark equal angles. So, draw a line, first construct
something like a line, use your protractor, mark particular points easiest way, divide that.
If it is something like we would like to divide this full 360 degrees into 8 equal parts; first,
second, third, fourth, fifth, sixth, seventh, and eighth. The easiest way is 0, 45, 90, and so
on plus addition to 360 if we mark it. Those points we can join it, straight away divide that
circle.
(Refer Slide Time: 12:33)
Now, let us look at how to construct a circle passing through 3 points. For example, we
have point A, we have point B, we have point C, but we don't know how to construct a
circle passing through A, B, C; that is what we are going to learn.
(Refer Slide Time: 13:04)
Now, we know only 3 points A, B, and C; no more details are known. We do not know
even where the centre is, what the radius is, and so on.
(Refer Slide Time: 13:16)
Let us look at the constructional procedure. Points A, B, C are given. To construct the line
first step is join point A and B, to get AB line. This is the line what we know. Let us look
at the constructional procedure. Points A, B, C are given. To construct the line first step is
to join points A and B to get the AB line. This is the line that we know. Similarly, join
points B and C to get line BC; point B and point C, join them. These two lines we got to
know.
Now, pick the first line AB. How we have constructed bisecting AB, use a radius greater
than AB, mark an arc from centre A, with the same radius from B, mark another one.
Similarly, from A centre mark an arc, similarly B with the same radius mark another arc.
So, once we know this point and this point, join them. Perpendicular bisector, we know,
draw it as an extension line. Similarly, construct perpendicular bisector to BC.
(Refer Slide Time: 14:49)
So, this line we have already constructed, to construct BC same procedure. From B greater
than the distance mark an arc, from C as the radius mark an arc; with the same radius mark
from C, mark from B. Wherever these points are in ah joining, construct one more line.
These two perpendicular bisectors, first one and the second perpendicular bisectors, where
they will intersect, will be our centre of the circle. Once centre of the circle is identified,
from centre O, measure the distance AO, whatever that distance, use your compass, which
passes through A, B, C with center O. This is the way we construct it.
(Refer Slide Time: 16:00)
Let us look at that on the sheet. What we know is three points we know; A, perhaps some
point B, maybe some point C. These are the points that we know. Let us join these points,
let us name them as A, B, and C. Join AB, join BC.
Now, draw a perpendicular bisector from centre A, similarly construct from B, identify
points and join these two points. We do not know where exactly centralizing. Similarly,
now from BC, centre B, the intersected points, pick that, join them.
So, it looks like this is the point where they are going to intersect. Let us call that point O.
Now, join AB, that must be the radius. So, for that, you see, there is a circle that is passing
through A, B, and C points with radius O.
If one requires the distance between that AO is the radius of that circle, something like the
leader lines we show, R whatever those units. Let us measure it using our scale. This is
something like 3.4 centimeters. So, R, we use millimeters as the notation; R 34 for that
circle..
(Refer Slide Time: 19:36)
In today's class, we have learned how to bisect an angle, how to trisect an angle, how to
divide a circle, and a circle passing through three points. In the next class, we will learn
more about drawing a normal and a tangent to a circle. Similarly, how to draw a tangent
to a circle from an exterior point and a regular polygon has to be constructed for a given
side, how to do that. These are the things we will learn in lecture 9.
Thank you. | 2,116 | 8,089 | {"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} | 4.625 | 5 | CC-MAIN-2021-49 | latest | en | 0.940674 |
https://books.google.com.jm/books?id=-CQAAAAAYAAJ&qtid=687e7835&dq=editions:UOM39015063898350&lr=&output=html_text&source=gbs_quotes_r&cad=6 | 1,723,670,103,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00064.warc.gz | 116,449,837 | 5,940 | Books Books
Sphere is a body bounded by a uniformly curved surface, all the points of which are equally distant from a point within called the center.
Elements of Geometry: Plane and Solid - Page 298
by John Macnie - 1895 - 374 pages
## Elements of Geometry
Adrien Marie Legendre - Geometry - 1819 - 576 pages
.... P :p : : A* : a*. SECTION THIRD. Of the Sphere. DEFTNITIONS. 437. A sphere is a solid terminated by a curved surface all the points of which are equally distant from a point within called the centre. The sphere may be conceived to be generated by the revolution of a semicircle I LIE (Jig....
## Elements of Geometry and Trigonometry: With Notes
Adrien Marie Legendre - Geometry - 1822 - 394 pages
...THE MEASUREMENT OF ITS ANGLES. Definitions. \ • I. The circumference of a circle is a curve line, all the points of which are equally distant from a point within, called the centre. The circle is the space terminated by this curve line. Note. In common language, the circle...
## Elements of Geometry
Adrien Marie Legendre - Geometry - 1825 - 276 pages
...polyedrons, we shall have . P:p::A3:a3. Of the Sphere. • DEFINITIONS. 437. A sphere is a solid terminated by a curved surface all the points of which are equally distant from a point within called the centre. The sphere may be conceived to be generated by the revolution of a semicircle DAE (fig....
## Elements of Geometry...: Translated from the French for the Use of the ...
Adrien Marie Legendre, John Farrar - Geometry - 1825 - 280 pages
...polyedrons, we shall have SECTION THIRD. • Of the Sphere. DEFINITIONS. 437. A sphere is a solid terminated by a curved surface all the points of which are equally distant from a point within called the centre. The sphere may be conceived to be generated by the revolution of a semicircle DAE (fig....
## Elements of Geometry
Adrien Marie Legendre - 1825 - 570 pages
...the Circle and the Measure of Angles. DEFINITIONS. ^0. THE circumference of a circle is a curved line all the points of which are equally distant from a point within called the centre. The circle is the space terminated by this curved line*. 46. 91. Every straight line CA,...
## Elements of Geometry and Trigonometry: With Notes
Adrien Marie Legendre - Geometry - 1828 - 346 pages
...p : : A2 : a2. BOOK VII. THE SPHERE. Definitions. 437. The sphere is a solid terminated by a curve surface, all the points of which are equally distant from a point within, called the centre. D The sphere may be conceived to be generated by the revolution of a semicircle DAE about...
## The First Six Books of the Elements of Euclid, with a Commentary and ...
Euclid, Dionysius Lardner - Euclid's Elements - 1828 - 542 pages
...those which we have established in the present book. (230) DBF. — A sphere is a solid terminated by a curved surface, all the points of which are equally distant from a certain point within it, called its centre. A sphere may be conceived to be produced by the revolution...
## Elements of Geometry: With Practical Applications, for the Use of Schools
Timothy Walker - Geometry - 1829 - 138 pages
...described by the point B the circumference of a circle, which we define to be — a curved line fill the points of which are equally distant from a point within called the centre — . The whole space enclosed is called a circle, the moving line AB a radius, and the... | 814 | 3,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-33 | latest | en | 0.871635 |
https://examfear.com/notes/Class-12/Maths/Determinants/436/Area-of-triangle.htm | 1,620,251,687,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988696.23/warc/CC-MAIN-20210505203909-20210505233909-00392.warc.gz | 271,823,442 | 6,195 | Class 12 Maths Determinants Area of triangle
Area of triangle
Numerical: Find area of triangle with vertices (1, 0), (6, 0), (4, 3)
Solution:
Area = ½ (1 (0x1 – 3 x 1) + 0 + 1 (6 x 3 - 4 x 0))
Or Area = 15/2
. | 88 | 215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2021-21 | latest | en | 0.781954 |
http://theory.uwinnipeg.ca/physics/bohr/node5.html | 1,505,827,516,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818685698.18/warc/CC-MAIN-20170919131102-20170919151102-00571.warc.gz | 348,267,772 | 3,316 | Next: Problems Up: Atomic Physics Previous: Other Atoms
# Modifications of the Bohr Model
Despite the success of the Bohr model, there were some serious shortcomings present. On the experimental side, more detailed analysis of the emission spectra for hydrogen found a single emission line was actually at times composed of two or more closely spaced lines, a feature not present in the Bohr model. Theoretically, the Bohr model mixes up a particle and wave picture of electrons, which was considered by many to be unsatisfactory. For these reasons a better treatment of the hydrogen atom was sought where the electron is considered as a wave from the outset. This theory, developed by Heisenberg, Pauli, Schrödinger, Sommerfeld, and others, is fairly detailed mathematically. The main result coming from this theory is that there are four quantum numbers describing a state of an electron, compared to the one quantum number n present in the Bohr model. These quantum numbers are as follows:
n
, the principal quantum number. This corresponds to n of the Bohr model, and can assume values 1,2,3,....
l
, the orbital quantum number. This is a label characterizing the magnitude of the angular momentum of the electron. For a given n , l can assume values 0,1,2,...,n - 1 .
ml
, the spin orbital quantum number. This is a label characterizing a component of the angular momentum vector of the electron. For a given l , ml can assume values - l, - l + 1,..., - 1,0,1,...,l - 1,l .
ms
, the spin quantum number. This is a label which, in a very limited sense, can be considered as characterizing the direction that the electron is spinning on its axis. ms can assume one of two values, .
Thus, for a given n , there can be 2n 2 states with different values of l , ml and ms .
Historically, the principal quantum number n labels what is called the shell, and the n = 1,2,3,... shell is sometimes referred to as the K,L,M,... shell. The orbital quantum number l labels the subshell, and the l = 0,1,2,3,4,... subshell is also referred to as the s,p,d,f,... subshell.
If we imagine starting to add electrons in order to form various atoms, then one would expect all the electrons to go into the lowest energy state possible, which is n = 1 and l = 0 = ml . This does not happen in nature, however. Pauli explained this by postulating that electrons obey what is now called the Pauli exclusion principle:
No two electrons in a system may have the same sets of quantum numbers.
Table 28.1 indicates what happens as we begin to add one additional electron in turn, taking into account this principle.
# of e -'s n l ml ms 1 1 - 2 1 + 3 2 - 4 2 + 5 2 1 - 1 - 6 2 1 - 1 + 7 2 1 0 - 8 2 1 0 + 9 2 1 + 1 - 10 2 1 + 1 +
We thus see that we have to fill up more and more shells as we add more and more electrons. As we shall see later in some examples, this behaviour explains qualitatively the structure of the periodic table, and provided another success of the early quantum theory.
Next: Problems Up: Atomic Physics Previous: Other Atoms | 760 | 3,035 | {"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-2017-39 | latest | en | 0.913191 |
http://www.gurufocus.com/term/InventoryTurnover/SYMC/Inventory%252BTurnover/Symantec%2BCorporation | 1,490,879,408,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218194600.27/warc/CC-MAIN-20170322212954-00434-ip-10-233-31-227.ec2.internal.warc.gz | 536,837,676 | 28,136 | Switch to:
GuruFocus has detected 7 Warning Signs with Symantec Corp \$SYMC.
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Symantec Corp (NAS:SYMC)
Inventory Turnover
0.00 (As of Dec. 2016)
Inventory turnover measures how fast the company turns over its inventory within a year. It is calculated as cost of goods sold divided by average inventory. Symantec Corp's cost of goods sold for the three months ended in Dec. 2016 was \$235 Mil. Symantec Corp's average inventory for the quarter that ended in Dec. 2016 was \$0 Mil.
Days inventory indicates the number of days of goods in sales that a company has in the inventory. Symantec Corp's days inventory for the three months ended in Dec. 2016 was 0.00.
Inventory can be measured by Days Sales of Inventory (DSI). Symantec Corp's days sales of inventory (DSI) for the three months ended in Dec. 2016 was 0.00.
Inventory to revenue ratio determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue. Symantec Corp's inventory to revenue ratio for the quarter that ended in Dec. 2016 was 0.00.
Definition
Symantec Corp's Inventory Turnover for the fiscal year that ended in Mar. 2016 is calculated as
Inventory Turnover (A: Mar. 2016 ) = Cost of Goods Sold / Average Inventory = Cost of Goods Sold (A: Mar. 2016 ) / ( (Inventory (A: Mar. 2015 ) + Inventory (A: Mar. 2016 )) / 2 ) = 615 / ( (0 + 0) / 2 ) = 615 / 0 = N/A
Symantec Corp's Inventory Turnover for the quarter that ended in Dec. 2016 is calculated as
Inventory Turnover (Q: Dec. 2016 ) = Cost of Goods Sold / Average Inventory = Cost of Goods Sold (Q: Dec. 2016 ) / ( (Inventory (Q: Sep. 2016 ) + Inventory (Q: Dec. 2016 )) / 2 ) = 235 / ( (0 + 0) / 2 ) = 235 / 0 = N/A
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
Inventory Turnover measures how fast the company turns over its inventory within a year. A higher inventory turnover means the company has light inventory. Therefore the company spends less money on storage, write downs, and obsolete inventory. If the inventory is too light, it may affect sales because the company may not have enough to meet demand.
1. Days Inventory indicates the number of days of goods in sales that a company has in the inventory.
Symantec Corp's Days Inventory for the three months ended in Dec. 2016 is calculated as:
Days Inventory = Average Inventory (Q: Dec. 2016 ) / Cost of Goods Sold (Q: Dec. 2016 ) * Days in Period = 0 / 235 * 365 / 4 = 0.00
2. Inventory can be measured by Days Sales of Inventory (DSI).
Symantec Corp's Days Sales of Inventory for the three months ended in Dec. 2016 is calculated as:
Days Sales of Inventory (DSI) = Average Inventory (Q: Dec. 2016 ) / Revenue (Q: Dec. 2016 ) * Days in Period = 0 / 1041 * 365 / 4 = 0.00
3. Inventory to Revenue determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue.
Symantec Corp's Inventory to Revenue for the quarter that ended in Dec. 2016 is calculated as
Inventory to Revenue = Average Inventory (Q: Dec. 2016 ) / Revenue (Q: Dec. 2016 ) = 0 / 1041 = 0.00
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Be Aware
Usually retailers pile up their inventories at holiday seasons to meet the stronger demand. Therefore, the inventory of a particular quarter of a year should not be used to calculate inventory turnover. An average inventory is a better indication.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Symantec Corp Annual Data
Mar07 Mar08 Mar09 Mar10 Mar11 Mar12 Mar13 Mar14 Mar15 Mar16 Inventory Turnover 26.76 31.98 40.14 42.50 38.00 37.31 45.19 65.92 0.00 0.00
Symantec Corp Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 Inventory Turnover 25.27 15.39 29.33 0.00 0.00 0.00 0.00 0.00 0.00 0.00
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http://mathhelpforum.com/pre-calculus/169502-centeral-conic-question.html | 1,481,372,806,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543035.87/warc/CC-MAIN-20161202170903-00089-ip-10-31-129-80.ec2.internal.warc.gz | 170,762,305 | 11,968 | # Thread: Centeral conic question.
1. ## Centeral conic question.
I'm not too sure of how to work this one out, I'm really having trouble determining the extent of the graph. I have figured out that the x intercept is $\pm4$ and that the asymptotes are $y=\frac{5x}{4}$and $-\frac{}{5x}{4}$
Graph
Show how you arrived at your graph by determining the x-intercepts, the extent of the graph and the asymptotes.
A) I admit to being lost when it comes to this question, I know how to work these kind of questions under other situations where the x and y are reversed. E.g.: . I am very lost as to what to do with this one though.
When I try to determine 'c' I come out with
I don't really see how that can be correct.
Thank you.
2. Originally Posted by quikwerk
I'm not too sure of how to work this one out, I'm really having trouble determining the extent of the graph. I have figured out that the x intercept is $\pm4$ and that the asymptotes are $y=\frac{5x}{4}$and $-\frac{}{5x}{4}$
Graph
Show how you arrived at your graph by determining the x-intercepts, the extent of the graph and the asymptotes.
A) I admit to being lost when it comes to this question, I know how to work these kind of questions under other situations where the x and y are reversed. E.g.: . I am very lost as to what to do with this one though.
When I try to determine 'c' I come out with
I don't really see how that can be correct.
Thank you.
Hi quikwerk,
You are correct. $c=\sqrt{41}$
The standard form for this hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
The equations of the asymptotes are $y-k=\pm \frac{b}{a}(x-h)$
$y-0=\pm \frac{5}{4}(x-0)$
$y=\pm \frac{5}{4}x$
3. Thanks- still not sure how to work out the extent of the graph : (
Am i correct in thinking that the extent of the graph is given by the equations $y=\pm\frac{5}{4}\sqrt{x^2-16}$and $x=\pm\frac{4}{5}\sqrt{y^2-25}$?
4. Originally Posted by quikwerk
Thanks- still not sure how to work out the extent of the graph : (
Am i correct in thinking that the extent of the graph is given by the equations $y=\pm\frac{5}{4}\sqrt{x^2-16}$and $x=\pm\frac{4}{5}\sqrt{y^2-25}$?
My guess would be the extents of the graph would be the Domain and Range.
5. Originally Posted by masters
My guess would be the extents of the graph would be the Domain and Range.
Does that mean that $\mid x \mid\geq 4$ and $\mid y \mid\geq 5$?
6. Originally Posted by quikwerk
Does that mean that $\mid x \mid\geq 4$ and $\mid y \mid\geq 5$?
The Domain would be $\{x\;|\;x \le -4 \r\:x \ge 4\}" alt="\{x\;|\;x \le -4 \r\:x \ge 4\}" />
The Range would be $\{y\;|\; y \in \Re\}$
7. Two points: To find the "asymptotes", think about what happens for x and y really, really big! For $\frac{x^2}{16}$ and $\frac{y^2}{25}$ are far larger than the "1" on the right side- if we ignore, it we see that the curve must be close to $\frac{x^2}{16}- \frac{y^2}{25}= \left(\frac{x}{4}- \frac{y}{5}\right)\left(\frac{x}{4}+ \frac{y}{5}\right)= 0$ which means that either $\frac{x}{4}- \frac{y}{5}= 0$ so that $4y= 5x$ or $\frac{x}{4}+ \frac{y}{5}= 0$ so that $4y= -5x$. Those are the asymptotes.
For the "extent" try solving for x and y. $\frac{y^2}{25}= \frac{x^2}{16}- 1$ so that $y= \pm 5\sqrt{\frac{x^2}{16}-1}$. Since the argument of the square root must be positive, we must have $\frac{x^2}{16}- 1\ge 0$ or $x^2\ge 16$ so that $x\le -4$ or $x\ge 4$.
$x= \pm 4\sqrt{\frac{y^2}{25}+ 1}$. Since the argument is always positive, there is no restriction on y.
8. Thank you guys! I'll try to work it out with your advice and report back later. | 1,158 | 3,567 | {"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": 0, "img_math": 0, "codecogs_latex": 35, "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} | 4.125 | 4 | CC-MAIN-2016-50 | longest | en | 0.945152 |
https://joelmoreira.wordpress.com/2012/11/14/properties-of-ultrafilters-and-a-theorem-of-hindman-on-arithmetic-combinatorics/ | 1,575,556,821,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540481076.11/warc/CC-MAIN-20191205141605-20191205165605-00245.warc.gz | 419,500,398 | 33,135 | ## Properties of ultrafilters and a Theorem on arithmetic combinatorics
A Theorem of Schur (one of the earliest results in Ramsey Theory) asserts that given any finite coloring of the set of natural numbers ${{\mathbb N}}$, there exist ${x,y\in{\mathbb N}}$ of the same color such that ${x+y}$ also has the same color. As a Corollary (using any injective homomorphism between the semigroups ${({\mathbb N},+)}$ and ${({\mathbb N},\times)}$, for instance ${n\mapsto2^n}$) one gets that in any finite coloring of ${{\mathbb N}}$ there are ${x}$ and ${y}$ of the same color such that ${xy}$ is also of that color.
The next question is whether one can find ${x,y}$ of the same color such that both ${x+y}$ and ${xy}$ are also of the same color. This question is still open, and even the easier looking question of whether we can find ${x}$ and ${y}$ such that ${x+y}$ and ${xy}$ have the same color (but we don’t care about the colors of ${x}$ and ${y}$) is unanswered.
I was surprised to learn recently that the following related result has been proved (and has a nice proof):
Theorem 1 Let ${{\mathbb N}=\biguplus C_i}$ be a partition of the set of all natural numbers ${{\mathbb N}}$ into finitely many colors. Then there exist some ${x,y,a,b}$ all in the same color and such that ${xy=a+b}$.
In this post I will present a proof of this theorem, following the proof on this survey by Bergelson (it’s the Theorem 6.1 there). An independent proof of a more general case was found by Hindman.
The proof we will present uses the theory of idempotent ultrafilters on ${{\mathbb N}}$, one of my purposes here is to make a point for the usefulness of ultrafilters when dealing with problems from arithmetic combinatorics. I am also quite interested on this because it seems that ultrafilters are the only tool available so far to deal with combinatorics on ${{\mathbb N}}$ dealing with both addition and multiplication (one can argue that polynomials involve multiplication, but quadratic polynomials are just sums of linear polynomials and so on, and the theorems involving polynomials are proved using that way of thinking about polynomials. Another theorem that seems to have some of the multiplications structure is the Green-Tao Theorem, but really the only thing about primes that is used in the proof are some statistics on their distribution. Finally in this parenthesis I should say that there are results dealing with both addition and multiplication on finite fields, proved using Fourier analysis, but unlike results dealing with only addition, it seems harder (if not completely impossible) to use those results to establish analogues on the infinite set ${{\mathbb N}}$).
This post turned out to be a little longer than I initially expected, but that is mostly because I decided to make it completely self contained (apart from some standard results from point set topology).
— 1. Ultrafilters on countable sets —
I decided to work on a general countable (infinite) commutative ring, so in this post ${R}$ will always denote such a ring, so I actually present a proof of the Theorem 1 with ${{\mathbb N}}$ replaced by ${R}$. In this section though, I will only use the fact that ${R}$ is a countable set.
An ultrafilter on ${R}$ is a collection of subsets of ${R}$ respecting some properties. It is (equivalently) a point in the Stone-Čech compactification ${\beta R}$ of ${R}$, if we give ${R}$ the discrete topology.
Definition 2 (Ultrafilter) An ultrafilter ${p}$ is a non-empty collection of subsets of ${R}$ satisfying:
• ${\emptyset\notin p}$
• If ${A\subset B}$ and ${A\in p}$ then ${B\in p}$
• If ${A\in p}$ and ${B\in p}$ then also ${A\cap B\in p}$
• If ${A\cup B\in p}$ then either ${A\in p}$ or ${B\in p}$
In particular ${R\in p}$ by the second condition (and because ${p}$ is non-empty), so for each ${A\subset R}$, either ${A\in p}$ or ${R\setminus A\in p}$ (by the last property). Also, iterating the last property, we get that for any finite partition of ${R}$, one of the cells belongs to ${p}$ (and exactly one).
A trivial example of an ultrafilter can be given by the following: pick ${a\in R}$ and consider the family ${p_a:=\{A\subset R:a\in A\}}$. One can promptly check the conditions that define ultrafilters. However, in order to obtain any example of a non-principal ultrafilter, one requires (at least some weaker form of) the axiom of choice. This means that an explicit construction of an non-principal ultrafilter is hopeless. What we can (and will) do is prove existence of ultrafilters with some nice additional properties.
Proposition 3 There exist non-principal ultrafilters.
Proof: A family of subsets of ${R}$ satisfying the first three properties of the Definition 2 is called a filter. We first show that any maximal filter (for the partial order of inclusion) is an ultrafilter. Let ${p}$ be a maximal filter, let ${A,B\subset R}$ be non-empty sets such that ${A\cup B\in p}$. Assume ${B\notin p}$, consider ${p':=p\cup\{C\subset R: A\subset C\}\cup \{A\cap C:C\in p\}}$. I claim that ${p'}$ is also a filter, by maximality this will imply that ${p'=p}$ and so ${A\in p}$ as desired.
To prove that ${\emptyset\notin p'}$, assume it is. Since ${p}$ is a filter and ${A}$ is non-empty we have that ${A\cap C=\emptyset}$ for some ${C\in p}$. But since ${A\cup B\in p}$, we have ${(A\cup B)\cap C}$ is in ${p}$ and also a subset of ${B}$, contradicting the assumption that ${B\notin p}$. The second and third conditions for a filter trivially hold for ${p'}$ and this proves the claim.
Now it is also easy to check that the union of any totally ordered subset of filters is again a filter, so we can apply the Zorn’s lemma to conclude that any filter is contained in some ultrafilter. Since the filter of co-finite sets is not contained in any principal ultrafilter we get existence of a non-principal ultrafilter. $\Box$
As I mentioned earlier in this post, the set of all ultrafilters can be identified with the Stone-Čech compactification ${\beta R}$ of ${R}$. Let ${\Omega:=\{0,1\}^R}$ be the set of all functions ${f:R\rightarrow\{0,1\}}$. By identifying a subset ${A\subset R}$ with its indicator functions, we can also think of ${\Omega}$ as all subsets of ${R}$. We endow ${\{0,1\}}$ with the discrete topology and give ${\Omega}$ the product topology, making it into a compact space by the Tychonoff’s theorem. Now we consider the space ${X=\{0,1\}^\Omega}$ of all functions from ${\Omega}$ in ${\{0,1\}}$, it is also compact, again by the Tychonoff’s theorem. We can also think of a point in ${X}$ as a subset of ${\Omega}$ or, equivalently as a collection of subsets of ${R}$.
Now, for each ${n\in R}$, let ${\tilde{n}\in X}$ be the function ${\tilde n:\Omega\rightarrow\{0,1\}}$ given by ${\tilde n(f)=f(n)}$ for each ${f:R\rightarrow\{0,1\}}$. This gives an embedding of ${R}$ into ${X}$, that by an abuse of language we will also denote by ${R}$. The Stone-\v Cech compactification of ${R}$ is the closure ${\beta R}$ of ${R}$ in ${X}$ (it is compact because it is a closed subset of the compact Hausdorff space ${X}$).
A point ${p\in\beta R\subset X}$ is a map from ${\Omega}$ to ${\{0,1\}}$. By identifying points in ${\Omega}$ with subsets of ${R}$, we can associate ${p}$ with the family of subsets ${A\subset R}$ for which ${p(A)=1}$. By an abuse of language we will denote this collection of subsets of ${R}$ also by ${p}$ and we will use interchangeably the notations ${A\in p}$ and ${p(A)=1}$ with the same meaning. Note that the element ${\tilde n\in X}$ is the principal ultrafilters at ${n}$.
Proposition 4 Let ${p\in X}$ be a collection of subsets of ${R}$. Then ${p\in\beta R}$ if and only if ${p}$ is an ultrafilter on ${R}$.
Proof: Unraveling the meaning of product topology we see that ${p\in\beta R}$ if and only if for any finite collection ${A_1,...,A_k}$ of subsets of ${R}$ there exists some ${n\in R}$ such that ${\tilde n(A_i)=p(A_i)}$ for each ${i=1,...,k}$.
First assume that ${p\in\beta R}$. Since ${\tilde n(\emptyset)=0}$ for all ${n\in R}$, also ${p(\emptyset)=0}$ and hence ${\emptyset\notin p}$, proving the first property of the Definition 2. Now let ${A\in p}$ and suppose ${B\supset A}$. Let ${n\in R}$ be such that ${\tilde n(A)=p(A)}$ and ${\tilde n(B)=p(B)}$. Then ${n\in A}$ so ${n\in B}$ and thus ${p(B)=1}$, proving the second property. If both ${A}$ and ${B}$ are in ${p}$, let ${n\in R}$ be such that ${\tilde n}$ and ${p}$ agree at ${A,B}$ and ${A\cap B}$. Then we conclude that ${A\cap B\in p}$ proving the third property. Finally suppose that ${A\cup B\in p}$ and let ${n\in R}$ be such that ${\tilde n}$ agree with ${p}$ at ${A}$, ${B}$ and ${A\cup B}$. Thus either ${A}$ or ${B}$ will be in ${p}$ proving the fourth property. This implies that ${p}$ is indeed an ultrafilter.
Now suppose that ${p}$ is an ultrafilter. Given any subsets ${A_1,...,A_k}$ of ${R}$, assume that ${A_1,...,A_r}$ are in ${p}$ and ${A_{r+1},...,A_k}$ are not in ${p}$. Then the intersection
$\displaystyle \left(\bigcap_{i=1}^r A_i\right)\cap\left[\bigcap_{i=r+1}^k(R\setminus A_i)\right]$
is in ${p}$ and in particular it is non-empty. Let ${n}$ be in that intersection. Then ${\tilde n}$ agrees with ${p}$ at the sets ${A_1,...,A_k}$. Since the sets ${A_1,...,A_k}$ were arbitrary, we conclude that there exists some ${\tilde n}$ at each neighborhood of ${p}$, and hence ${p\in\beta R}$. $\Box$
This proves suggests the following observation about the topology of ${\beta R}$ that will be useful for us later. For a set ${A\subset R}$ we define its closure ${\bar A\subset \beta R}$ by ${p\in\bar A\iff A\in p}$. Note that ${\bar A}$ just defined is the closure on ${\beta R}$ of the set ${\{\tilde n:n\in A\}}$, but we will not need this fact.
Lemma 5 The sets ${\bar A}$ are clopen and form a basis for the topology on ${\beta R}$.
Proof: Let ${p\in X}$. By definition ${p\in\bar A\iff p(A)=1}$. Since ${1}$ is a clopen subset of ${\{0,1\}}$ and by the definition of product topology on ${X}$ we conclude that ${\bar A}$ is a clopen set of ${X}$, hence intersecting with ${\beta R}$ we get a clopen subset of ${\beta R}$.
To prove that they form a basis for the topology on ${\beta R}$, let ${k\in{\mathbb N}}$, let ${1\leq r\leq k}$, let ${A_1,...,A_k}$ be subsets of ${R}$ and let ${C:=\{p\in\beta R:A_1,...,A_r\in p, A_{r+1},...,A_k\notin p\}}$. Consider the intersection
$\displaystyle B=\left(\bigcap_{i=1}^r A_i\right)\cap\left[\bigcap_{i=r+1}^k(R\setminus A_i)\right]$
Then, by the ultrafilter property, ${p\in C\iff B\in p}$. Thus ${C=\bar B}$. $\Box$
— 2. Ultrafilters on semigroups —
So far we didn’t use any property of the set ${R}$ (except that it is infinite). Now we will see that a binary operation on ${R}$ induces a binary operation in ${\beta R}$. To motivate the definition it helps to think of ultrafilters in yet a third way, namely as finitely additive ${\{0,1\}}$-valued measures on ${R}$. In that way, the operation on ${\beta R}$ is just the usual convolution of measures.
Let ${\circ}$ be an associative binary operation on ${R}$ (we will only use ${\circ=+}$ and ${\circ=\times}$), let ${A\subset R}$ and ${n\in R}$. I use the notation ${A\circ n^{-1}}$ to denote the set ${\{x\in T:x\circ n\in A\}}$. In the case of additive notation we will use ${A-n}$ instead of ${A+n^{-1}}$.
Definition 6 Let ${p}$ and ${q}$ be ultrafilters on a semigroup ${(R,\circ)}$. We define the operation
$\displaystyle p\circ q:=\{A\subset R:\{n\in R:A\circ n^{-1}\in p\}\in q\}$
We first need to check that ${p\circ q}$ is indeed an ultrafilter on ${R}$:
Proposition 7 If ${p,q\in\beta R}$ then also ${p\circ q\in \beta R}$.
Proof: It is clear that ${\emptyset\notin p\circ q}$. Also, if ${A\subset B}$, then for each ${n}$ we have ${(A\circ n^{-1})\subset (B\circ n^{-1})}$. It is now easy to check that if ${A\in p\circ q}$ then also ${B\in p\circ q}$. Now assume that both ${A,B\in p\circ q}$. Since ${(A\circ n^{-1})\cap(B\circ n^{-1})=(A\cap B)\circ n^{-1}}$ for each ${n\in R}$, we have that ${\{n:(A\cap B)\circ n^{-1}\in p\}=\{n:A\circ n^{-1}\in p\}\cap\{n:B\circ n^{-1}\in p\}\in q}$ and hence ${A\cap B\in p\circ q}$.
Finally, if ${A\cup B\in p\circ q}$ then using the fact that ${p}$ is an ultrafilter (and the fact that ${(A\cup B)\circ n^{-1}=A\circ n^{-1}\cup B\circ n^{-1}}$) we have that for each ${n}$ in the set ${C:=\{n:(A\cup B)\circ n^{-1}\in p\}}$ either ${A\circ n^{-1}\in p}$ or ${B\circ n^{-1}\in p}$. Since ${q}$ is also an ultrafilter and ${C\in q}$, either ${\{n:A\circ n^{-1}\in p\}\in q}$ or ${\{n:B\circ n^{-1}\in p\}\in q}$ which is equivalent, respectively to ${A\in p\circ q}$ or ${B\in p\circ q}$. $\Box$
Moreover, this binary operation turns out to be associative:
Proposition 8 The binary operation on ${\beta R}$ just defined is associative.
Proof: We first note that for ${A\subset R}$ and ${n,m\in R}$ we have
$\displaystyle x\in(A\circ n^{-1})\circ m^{-1}\iff x\circ m\circ n\in A$
and so ${(A\circ n^{-1})\circ m^{-1}=A\circ(m\circ n)^{-1}}$.
Let ${p,q,r\in\beta R}$. Then ${A\in (p\circ q)\circ r}$ if and only if
$\displaystyle \begin{array}{rcl} \displaystyle\{n:A\circ n^{-1}\in p\circ q\}\in r&\iff&\displaystyle\{n:\{m:(A\circ n^{-1})\circ m^{-1}\in p\}\in q\}\in r\\&\iff&\displaystyle\{n:\{m:A\circ(m\circ n)^{-1}\in p\}\in q\}\in r\\&\iff&\displaystyle\{n:\{m:m\circ n\in\{x:A\circ x^{-1}\in p\}\}\in q\}\in r\\&\iff&\displaystyle\{n:\{x:A\circ x^{-1}\in p\}\circ n^{-1}\in q\}\in r\\&\iff&\displaystyle\{x:A\circ x^{-1}\in p\}\in q\circ r\\&\iff&\displaystyle A\in p\circ(q\circ r) \end{array}$
$\Box$
Thus ${(\beta R,\circ)}$ is a semigroup. It should be remarked that this operation extends the operation on ${R}$. More precisely, if ${n,m\in R}$ then ${\tilde n\circ\tilde m:=\widetilde{n\circ m}}$. However, the operation in ${\beta R}$ needs not be commutative even if ${\circ}$ is.
Another good property of the operation in ${\beta R}$ is that it is left continuous:
Proposition 9 For each ${p\in\beta R}$, the map ${\lambda_p:\beta R\rightarrow\beta R}$ defined by ${\lambda_p:q\mapsto p\circ q}$ is continuous.
Proof: We will use the Lemma 5. Fix ${p,q\in\beta R}$ and let ${\bar A}$ be an clopen neighborhood of ${\lambda_p(q)}$. I need to show that ${\{r:\lambda_p(r)\in\bar A\}}$ contains ${\bar B}$ for some ${B}$. We observe that ${\lambda_p(r)\in\bar A\iff A\in p\circ r\iff \{n:A\circ n^{-1}\in p\}\in r}$. Thus making ${B=\{n:A\circ n^{-1}\in p\}}$ we get that indeed ${\{r:\lambda_p(r)\in\bar A\}}$ contains ${\bar B}$. $\Box$
Quite special elements of the semigroup ${(\beta R,\circ)}$ are idempotent elements, i.e, ultrafilters ${p}$ such that ${p\circ p=p}$. However, the existence of such ultrafilters not obvious. Their existence can be assured using a Theorem by Ellis:
Theorem 10 (Ellis Theorem) If ${(S,\circ)}$ is a compact Hausdorff left topological semigroup, then ${S}$ contains an idempotent.
Proof: Consider the family
$\displaystyle V:=\{\emptyset\neq W\subset S: W\text{ is compact },W\circ W:\{w_1\circ w_2:w_1,w_2\in W\}=W\}$
${V}$ is non-empty because ${S\in V}$. Also the intersection of any nested subfamily of ${V}$ is still in ${V}$ (because a nested intersection of compacts in a Hausdorff space is non-empty). Applying Zorn’s lemma we find a minimal (for the partial order of inclusion) element ${W\in V}$.
For each ${x\in W}$ we have ${(x\circ W)\circ(x\circ W)\subset x\circ W\circ W\circ W\subset x\circ W}$, and by left continuity ${x\circ W}$ is compact, hence ${x\circ W\in V}$. Also ${x\circ W\subset W\circ W\subset W}$, so by minimality ${x\circ W=W}$.
In particular ${x=x\circ y}$ for some ${y\in W}$. Thus the set ${Z:=\{z\in W:x\circ z=x\}}$ is non-empty. By continuity we have that ${Z}$ is closed (hence compact) and if ${y,z\in Z}$ then ${x\circ(y\circ z)=x\circ z=x}$, hence ${Z\circ Z\subset Z}$. This implies that ${Z\in V}$, again by minimality we have that ${Z=W}$ and in particular ${x\in Z}$. We conclude that ${x\circ x=x}$, and this is our idempotent. $\Box$
This implies that there are idempotent ultrafilters on ${\beta R}$.
— 3. Idempotent ultrafilters and Hindman’s Theorem —
We are now concerned with a countable set ${R}$ with a commutative associative binary operation ${\circ}$. Given an infinite set ${I\subset R}$ we for the set of finite products of ${I}$:
$\displaystyle FP(I):=\{i_1\circ i_2\circ...\circ i_k:k\in{\mathbb N}, i_1,...,i_k\in I\text{ are all distinct}\}$
Hindman’s theorem asserts that, given any finite partition of ${R}$, one of the cells of the partition contains a set of the form ${FP(I)}$ for some infinite set ${I\subset R}$. Sets that contain ${FP(I)}$ for some infinite set ${I\subset R}$ are called IP sets.
In order to prove Hindman’s Theorem, it will suffice to prove the following:
Proposition 11 Let ${p\in\beta R}$ be an idempotent ultrafilter on ${R}$ and let ${A\in p}$. Then ${FP(I)\subset A}$ for some infinite set ${I\subset R}$.
The reason why this implies Hindman’s Theorem is that for any finite partition, one of the cells belong to ${p}$.
Proof: Since ${A\in p}$ and ${p\circ p=p}$, we have that ${\{n:A\circ n^{-1}\in p\}\in p}$. Thus also ${A\cap\{n:A\circ n^{-1}\in p\}\in p}$ and, in particular, this set is non-empty. Choose some ${n_1}$ in that set, note that ${n_1\in A}$ and ${A_1:=A\cap(A\circ n_1^{-1})\in p}$.
Now, by the same reasoning, also ${A_1\cap\{n:A_1\circ n^{-1}\in p\}\in p}$, choose some ${n_2}$ in that set and note that ${n_2\in A}$ and ${A_2:=A_1\cap(A_1\circ n_2^{-1})\in p}$. Moreover ${n_2\in A\circ n_1^{-1}}$, hence ${n_2\circ n_1\in A}$, in other words ${FP(\{n_1,n_2\})\subset A}$. Now inductively we prove that for each ${k>2}$ there is a set ${A_k\in p}$ such that ${A_k\subset A}$, and a sequence ${n_1,...,n_k}$ such that ${FP(\{n_1,...,n_k\})\subset A}$.
Now assume we have this for ${k-1}$, note that the set ${A_{k-1}\cap\{n:A_{k-1}\circ n^{-1}\in p\}\in p}$ and in particular is non-empty. Choose some ${n_k}$ on that set, note that ${n_k\in A}$ and that the set ${A_k:=A_{k-1}\cap(A_{k-1}\circ n_k^{-1})\in p}$. Moreover, for any ${x\in A}$ we have ${n_k\circ x\in A}$, since by induction ${FP(\{n_1,...,n_{k-1}\})\subset A}$ we get that ${FP(\{n_1,...,n_k\})\subset A}$ and this completes the induction.
Thus ${FP(\{n_1,n_2,...\})\subset A}$ as desired. $\Box$
We just proved that each set belonging to an idempotent ultrafilter is an IP set. We now prove a partial converse, and first we need a lemma:
Lemma 12 Let ${A\subset R}$ be an IP set. Then there exists some idempotent ${p\in\beta R}$ such that ${A\in p}$.
Proof: Let ${I\subset R}$ be an infinite set such that ${FP(I)\subset A}$ and let ${\mathcal I}$ denote the family of all co-finite subsets of ${I}$. Let
$\displaystyle S:=\bigcap_{J\in\mathcal I}\overline{FP(J)}\subset\beta R$
Since ${S}$ is the intersection of a nested sequence of compact sets is compact and non-empty. I claim that ${S}$ is a semigroup. Let ${p,q\in S}$, we need to show that for each ${J\in\mathcal I}$ we have ${FP(J)\subset p\circ q}$.
For each ${a\in FP(J)}$, we can write ${a=i_1\circ...\circ i_k}$ for some ${i_1,...,i_k\in I}$. Let ${J'=J\setminus\{i_1,...,i_k\}\in\mathcal I}$. Then ${FP(J')\circ a\subset FP(J)}$, in other words ${FP(J')\subset FP(J)\circ a^{-1}}$ and since ${FP(J')\in p}$ we have that ${\{a\in R:FP(J)\circ a^{-1}\in p\}\supset FP(J)}$ and hence is in ${q}$. This proves that ${FP(J)\in p\circ q}$ and hence ${S}$ is a semigroup.
Finally, we apply Ellis theorem to ${S}$ to find an idempotent there. $\Box$
Proposition 13 Let ${p\in\beta R}$ be an ultrafilter. Then that each ${A\in p}$ is an IP set if and only if ${p}$ is in ${\Gamma:=\overline{\{q\in\beta R:q+q=q\}}}$, the closure in ${\beta R}$ of the set of idempotent ultrafilters.
Proof: Assume first that every ${A\in p}$ is an IP set. Let ${U}$ be a neighborhood (in ${\beta R}$) of ${p}$. Then ${p\in\bar A\subset U}$ for some set ${A\subset R}$. This implies that ${A\in p}$ and hence ${A}$ is an IP set. By the previous Lemma ${A}$ is also contained in some idempotent ultrafilter ${q}$, so ${q\in\bar A}$ and thus ${p\in\Gamma}$.
Now suppose that ${p\in\Gamma}$ and let ${A\in p}$. Thus ${p\in \bar A}$ and hence there exists some idempotent ${q}$ such that also ${q\in\bar A}$. But this implies that ${A\in q}$ and hence ${A}$ is an IP set. $\Box$
— 4. Mixing addition and multiplication —
On this section we will (finally) use the fact that ${R}$ is a ring with two operations. This gives two notions of idempotents in ${\beta R}$. A natural question is whether there exist any ${p\in\beta R}$ which is idempotent with respect to both the additive and the multiplicative structure. Unfortunately the answer turns out to be negative in general (at least it is negative when ${R={\mathbb N}}$, I couldn’t find a proof for any other ring).
More precisely, there is no ultrafilter $p\in\beta\mathbb{N}$ such that $p+p=p.p$, this appears as Corollary 17.17 in this book of Hindman and Strauss.
However we can construct an ultrafilter almost as good:
Theorem 14 There exists an ultrafilter ${p\in\beta R}$ such that ${p}$ is multiplicative idempotent and every ${A\in p}$ is an additive IP set.
Proof: Let ${\Gamma=\overline{\{p\in\beta R:p+p=p\}}}$. In view of Ellis Theorem and the Proposition 13, it suffices to prove that ${\Gamma}$ is a multiplicative semigroup. It turns out that it is actually a left ideal (i.e. ${\Gamma.\beta R\subset\Gamma}$).
Let ${p\in\Gamma}$ and ${q\in\beta R}$. Let ${A\in p.q}$. Then ${\{n\in R: A.n^{-1}\in p\}\in q}$, and in particular that set is non-empty. Choose some ${n\in R}$ such that ${A.n^{-1}\in p}$. Then ${A.n^{-1}}$ is an additive IP set. Let ${I\subset R}$ be an infinite set such that the additive IP set ${FP(I)\subset A.n^{-1}}$. But then ${FP(nI)\subset A}$ and hence ${A}$ is an (additive) IP set as desired. Since ${A\in p.q}$ was chosen arbitrarily we get that ${p.q\in\Gamma}$ and we are done. $\Box$
— 5. Proof of the Theorem 1
We now present the proof of the Theorem 1, with ${{\mathbb N}}$ replaced by any countable abelian ring ${R}$.
For any ultrafilter ${p\in\beta R}$ we know that one cell of a finite partition must belong to ${p}$. Thus the Theorem reduces to the following:
Theorem 15 Let ${p\in\beta R}$ be a multiplicative idempotent such that each ${A\in p}$ is an additive IP set. Then for each ${A\in p}$ there are ${x,y,a,b\in A}$ such that ${xy=a+b}$.
Proof: Since ${p.p=p}$ we have that ${\{n:A.n^{-1}\in p\}\in p}$. Choose ${x}$ such that ${A_1:=A\cap A.x^{-1}\in p}$. Then ${A_1}$ is an additive IP set, by the Lemma 12 there exists some additive idempotent ${q\in\beta R}$ such that ${A_1\in q}$. Then ${\{n:A_1-n\in q\}\in q}$ and so also ${A_1\{n:A_1-n\in q\}\in q}$. Choose some ${z}$ in that intersection. Then ${z\in A_1}$ and ${A_1-z\in q}$. Then also ${A_2:=A_1\cap(A_1-z)\in q}$ and so that set is non-empty.
Since ${A.x^{-1}\subset A_1}$ we get that ${b:=zx\in A}$. Also we now have that ${A_3:=xA_2}$ is non-empty, and we can rewrite
$\displaystyle A_3=xA_2=xA_1\cap(xA_1-b)=xA\cap A\cap [(xA\cap A)-b]$
Let ${a\in A_3}$. Then ${a,b,x\in A}$ and ${a+b\in xA}$, so there exists some ${y\in A}$ such that ${a+b=xy}$ as desired.$\Box$
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### 13 Responses to Properties of ultrafilters and a Theorem on arithmetic combinatorics
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3. Eugene says:
Good morning! You said for $\mathbb N$ there is no idempotent with respect to both the additive and the multiplicative structure. Where could i find the proof of it?
• Joel Moreira says:
Hi, Eugene,
I added the reference to the text. In case you don’t have access to that book, the original proof that $p+p\neq p.p$ was found by Hindman in this paper: http://www.ams.org/mathscinet-getitem?mr=749511.
I would like to know if this result is still true in more general rings (for instance in $\mathbb{Q}$) but I was not able to extend the proof.
4. Eugene says:
Thanx!
5. Pingback: I Can't Believe It's Not Random! | 7,660 | 24,139 | {"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": 0, "img_math": 476, "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-2019-51 | latest | en | 0.937663 |
https://eanswersin.com/math/question4786004 | 1,642,975,321,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304309.59/warc/CC-MAIN-20220123202547-20220123232547-00653.warc.gz | 250,829,646 | 13,313 | Prove that sinx^6 🚩 + cosx^6 + 3cosx^2+ sinx^2/( 1+ tan²x)(1-sin²x) = 1
, 19.08.2019 23:00, chennupatitanuja
# Prove that sinx^6 + cosx^6 + 3cosx^2+ sinx^2/( 1+ tan²x)(1-sin²x) = 1
### Other questions on the subject: Math
Math, 19.08.2019 05:00, BTanisha
Ram is 8 year old ganesh is 5 times of ram's age find the diffrence between ganesh and raj's age
Math, 19.08.2019 13:00, aswinkumarmishra
A, b,c are playing a card game .amount of money which a has is square of the amount of b has and amount of money which b has is square of the amount of c has .if together they have a total of rupees 655 ,then find the difference of the amount of b and c. | 252 | 653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2022-05 | latest | en | 0.894995 |
http://slidegur.com/doc/73854/modular-arithmetic | 1,477,455,378,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720615.90/warc/CC-MAIN-20161020183840-00357-ip-10-171-6-4.ec2.internal.warc.gz | 231,315,820 | 9,328 | ### Modular Arithmetic
```This work is licensed under a Creative Commons Attribution-Share Alike 3.0 Unported License.
CS 312: Algorithm Analysis
Lecture #3: Algorithms for
Modular Arithmetic,
Modular Exponentiation
Slides by: Eric Ringger, with contributions from Mike Jones, Eric Mercer, Sean Warnick
Announcements
HW #1 Due Now
Always start of class
Always show work
FERPA protects your student record
Need waiver to return graded work without
cover sheet
Objectives
analysis toolbox
Review modular arithmetic
Discuss and analyze algorithms for:
modular arithmetic
modular exponentiation
Max. rule
Another useful rule for Asymptotic
analysis.
O( f(n) + g(n) ) = O( max( f(n), g(n) ) )
Examples:
Goal for Ch. 1
Appreciate the role of theoretical analysis in
the security of RSA.
Requires: Solve, analyze, and use (!) two
important and related problems:
Factoring: Given a number N, express it as a
product of its prime numbers
Primality Testing: Given a number N, determine
whether it is prime
Which one is harder?
Algorithms for Integer Arithmetic
Computing Device:
Binary operations are constant time
Arithmetic operations on arbitrary length integers
may require more time
For an integer , we talk about its
representation in bits:
= log 2
Pad length of to the next power of 2 (using 0s)
if necessary.
Algorithms for Integer Arithmetic
Multiplication
Division
Algorithms for Integer Arithmetic
Multiplication: Θ(2 )
Division: Θ(2 )
Modular Arithmetic
Congruency
An important distinction
Congruency
Equality, using the modulus operator
Properties
Associativity:
Commutativity:
Distributivity:
Substitution Rule
Useful Consequence
xy (x mod z)y
(mod z)
xy mod z = (x mod z)y mod z
Example:
Modular Multiplication
Goal: Modular Exponentiation
We need to compute
xy mod N
for values of x, y, and N that are several
hundred bits long.
Can we do so quickly?
Sequential Exponentiation
Describe a simple algorithm for doing exponentiation:
function seqexp (x, y)
Input: An n-bit integer x and a non-negative integer
exponent y (arbitrarily large)
Output: xy
if y=0: return 1
r=x
for i = 1 to y-1 do
r=r⋅x
return r
Analysis of Sequential
Exponentiation
function seqexp (x, y)
Input: An n-bit integer x and a non-negative
integer exponent y (arbitrarily large)
Output: xy
if y=0: return 1
r=x
for i = 1 to y-1 do
r=rx
return r
Modular Exponentiation, Take I
function modexp (x, y, N)
Input: Two n-bit integers x and N, a non-negative
integer exponent y (arbitrarily large)
Output: xy mod N
if y=0: return 1
r = x mod N
for i = 1 to y-1 do
r = (r ⋅ x) mod N
return r
Modular Exponentiation, Take I
function modexp (x, y, N)
Input: Two n-bit integers x and N, a non-negative
integer exponent y (arbitrarily large)
Output: xy mod N
if y=0: return 1
r = x mod N
for i = 1 to y-1 do
r = (r ⋅ x) mod N
return r
New Ideas
Represent y (the exponent) in binary
Then break down xy into factors using the
non-zero bits of y
Also: compute the factors using repeated
squaring
Reduce factors using substitution rule
Modular Exponentiation, Take II
function modexp (x, y, N)
Input: Two n-bit integers x and N, a non-negative
integer exponent y (arbitrarily large)
Output: xy mod N
Recursive call
if y=0: return 1
z = modexp(x, floor(y/2), N)
if y is even:
return z2 mod N
else:
return x ⋅ z2 mod N
Right shift
Multiplication
Analysis of Modular Exponentiation
Each multiplication is Q(n2)
Each modular reduction is Q(n2)
There are log(y)=m of them
Thus, modular exponentiation is in
Q(n2 log y) = Q(n2 m)
function modexp (x, y, N)
if y=0: return 1
z = modexp(x, floor(y/2), N)
if y is even:
return z2 mod N
else:
return x z2 mod N
Modular Exponentiation (II),
Iterative Formulation
function modexp (x, y, N)
Input: Two n-bit integers x and N, a non-negative
integer exponent y (arbitrarily large)
Output: xy mod N
if y = 0: return 1
i = y; r = 1; z = x mod N
while i > 0
if i is odd: r = r ⋅ z mod N
z = z2 mod N
i = floor(i/2)
return r
Modular Exponentiation
xy mod N
Key Insights:
1. Exponent y can be represented in binary
2. Problem can be factored into one factor per
binary digit
3. Each factor can be reduced mod N
(substitution rule)
Example
We’re employing
same insights and a little more
cleverness than the
algorithm.
Example #2
function modexp (x, y, N)
Input: Two n-bit integers x and N, an integer
exponent y (arbitrarily large)
Output: xy mod N
if y = 0: return 1
i = y; r = 1; z = x mod N
while i > 0
if i is odd: r = r z mod N
z = z2 mod N
i = floor(i/2)
return r
310 mod10
x 3, y 10, N 10
i 10, r 1, z 3mod10 3
z 32 mod10 9
i5
r 1 9 mod10 9
z 92 mod10 81mod10 1
i2
z 12 mod10 1
i 1
r 9 1mod10 9
z 1
i0
return 9
Strictly tracing the
algorithm.
Example #2
function modexp (x, y, N)
Input: Two n-bit integers x and N, an integer
exponent y (arbitrarily large)
Output: xy mod N
if y = 0: return 1
i = y; r = 1; z = x mod N
while i > 0
if i is odd: r = r z mod N
z = z2 mod N
i = floor(i/2)
return r
310 mod10
x 3, y 10, N 10
i 10, r 1, z 3mod10 3
z 32 mod10 9
i5
r 1 9 mod10 9
z 92 mod10 81mod10 1
i2
z 12 mod10 1
i 1
r 9 1mod10 9
z 1
i0
return 9
Example
20
3 mod 10
Needed: two volunteers:
Volunteer A: use our final modexp() to compute it.
Volunteer B: compute 320 then reduce mod 10
Efficiency
The key point is that xy mod N is easy
modexp is in Q(n2 log y)
In fact, it requires about 1.5 log2 y multiplications for typical y
seqexp required y-1 multiplications
When x, y, and N are 200 digit numbers
Assume 1 multiplication of two 200 digit numbers takes 0.001
seconds
modexp typically takes about 1 second
seqexp would require 10179 times the Age of the Universe!
Only works when y is an integer.
Assignment | 2,051 | 5,803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2016-44 | longest | en | 0.775492 |
http://science.slashdot.org/story/08/07/02/1418214/claimed-proof-of-riemann-hypothesis?sdsrc=nextbtmprev | 1,455,271,309,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701163663.52/warc/CC-MAIN-20160205193923-00033-ip-10-236-182-209.ec2.internal.warc.gz | 203,778,920 | 51,846 | typodupeerror
## Claimed Proof of Riemann Hypothesis345
An anonymous reader writes "Xian-Jin Li claims to have proven the Riemann hypothesis in this preprint on the arXiv." We've mentioned recent advances in the search for a proof but if true, I'm told this is important stuff. Me, I use math to write dirty words on my calculator.
This discussion has been archived. No new comments can be posted.
## Claimed Proof of Riemann Hypothesis
• #### Dirty Words (Score:5, Funny)
on Wednesday July 02, 2008 @10:38AM (#24031621) Homepage
Me, I use math to write dirty words on my calculator.
Such as 80085?
• #### Re:Dirty Words (Score:5, Funny)
by Anonymous Coward on Wednesday July 02, 2008 @10:39AM (#24031671)
5318008
• #### Re: (Score:3, Interesting)
in portuguese, 50135.50738 (nice breasts).
• #### Re:Dirty Words (Score:5, Funny)
on Wednesday July 02, 2008 @10:44AM (#24031763)
No, you mean 5318008 or for the slashdot crowd, 55378008
• #### Re:Dirty Words (Score:5, Funny)
on Wednesday July 02, 2008 @10:56AM (#24031985)
No for the slashdot crowd it would be: 58008uÉÉ . Because obviously we all have calculators that support unicode text entry.
• #### Re:Dirty Words (Score:5, Funny)
on Wednesday July 02, 2008 @10:57AM (#24032019)
That would've been a lot cooler if Slashdot supported Unicode.
• #### Re: (Score:3, Funny)
äOEæ--¥é..."ããï¼ï¼
• #### Re:Dirty Words (Score:5, Funny)
on Wednesday July 02, 2008 @11:34AM (#24032637) Homepage
At that point, isn't it safe to assume that our calculators can just draw a pair of boobs in 2-bit greyscale?
And that we've written apps that simulate what we assume bouncing would look like given our collective lack of experience outside of the pornographic realm?
• #### Re: (Score:2, Funny)
I had proof of concept Porn on my TI-89 in 2000.
• #### Re:Dirty Words (Score:5, Funny)
on Wednesday July 02, 2008 @07:56PM (#24038631)
Does your project have donation page?
• #### Re:Dirty Words (Score:5, Funny)
on Wednesday July 02, 2008 @12:23PM (#24033421) Homepage
You haven't grafted a color TFT screen to your calculator yet?
Who let these guys in?
• #### Re:Dirty Words (Score:5, Funny)
on Wednesday July 02, 2008 @01:12PM (#24034159)
You just gave me the best idea for an iPhone app:
Boobies that bounce according to how the phone is bouncing....
• #### Re:Dirty Words (Score:5, Funny)
on Wednesday July 02, 2008 @02:16PM (#24035013)
• #### Re:Dirty Words (Score:5, Funny)
on Wednesday July 02, 2008 @10:59AM (#24032037) Journal
On linux, wouldn't it be ...
host:>man 80085
???
• #### Re: (Score:2, Funny)
Newbie...
correct spelling is "5318008" and you have to look at the calculator "umop apisdn"
Mod me down, I dare you!!!
• #### Re: (Score:3, Funny)
Nah, if you really want a dirty word, try 71077345...
37047734
• #### Yeah but did they point this out? (Score:5, Funny)
on Wednesday July 02, 2008 @10:39AM (#24031645) Journal
By using Fourier analysis on number fields, we prove in this paper E. Bombieri's refinement of A. Weil's positivity condition, which implies the Riemann hypothesis for the Riemann zeta function in the spirit of A. Connes' approach to the Riemann hypothesis. Weather permitting of course. (Just looking on the positivity side)
• #### Re:Yeah but did they point this out? (Score:5, Funny)
on Wednesday July 02, 2008 @11:18AM (#24032393)
By using Fourier analysis on number fields, we prove in this paper E. Bombieri's refinement of A. Weil's positivity condition, which implies the Riemann hypothesis for the Riemann zeta function in the spirit of A. Connes' approach to the Riemann hypothesis.
Weather permitting of course. (Just looking on the positivity side)
I thought you were randomly babbling, but then I RTFA and realized you were just quoting it...
• #### Re:Yeah but did they point this out? (Score:5, Funny)
on Wednesday July 02, 2008 @12:01PM (#24033039) Homepage
Wait... both of you RTFA?
We have a new /. record!
• #### Re:Yeah but did they point this out? (Score:5, Funny)
on Wednesday July 02, 2008 @12:39PM (#24033651) Homepage
Not so fast. I read it -2 times.
• #### Tried to RTFA (Score:5, Funny)
on Wednesday July 02, 2008 @10:39AM (#24031657) Homepage
Man, where's Charles Eppes when you need something explained to you in layman's terms?
• #### Re:Tried to RTFA (Score:5, Funny)
on Wednesday July 02, 2008 @11:00AM (#24032073)
Ummm...I think that WAS layman's terms. For you math geeks, try being a history major and looking at all that. It just looks like a cat walked on the keyboard to me...
• #### Re:Tried to RTFA (Score:5, Funny)
on Wednesday July 02, 2008 @11:21AM (#24032439) Journal
Ummm...I think that WAS layman's terms. For you math geeks, try being a history major and looking at all that. It just looks like a cat walked on the keyboard to me...
Are you reading slashdot as some kind of anthropological study?
• #### Re:Tried to RTFA (Score:5, Informative)
on Wednesday July 02, 2008 @11:17AM (#24032369) Homepage Journal
The Riemann zeta function is \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}} [written for LaTeX], or "the sum of 1/(n^s) as n goes from 0 to infinity (increasing by 1 repeatedly)" [in more human-readable form].
Riemann was interested in the zeros to this function, where s is a complex number. He conjectured that all zeros (aside from those of the form s = -2c, where c is a positive integer) would have to be of the form (1/2) + ki, where k is a constant and i is the square root of -1.
This paper is saying that they've found a way to verify this intuition by patching a hole in a previous attempt.
Assuming that everything is correct (a big assumption), this would finally solve a long-standing problem (dating back to 1859).
Details of the actual solution are a bit heavy. Those actually interested in this sort of number theory might want to start here [amazon.com].
• #### typo (Score:5, Informative)
on Wednesday July 02, 2008 @11:24AM (#24032481) Journal
The Riemann zeta function is \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}} [written for LaTeX], or "the sum of 1/(n^s) as n goes from 0 to infinity (increasing by 1 repeatedly)" [in more human-readable form].
You have a slight typo. Should be: "... as n goes from 1 to infinity ..."
• #### Re:typo (Score:5, Funny)
on Wednesday July 02, 2008 @01:47PM (#24034641) Journal
The Riemann zeta function is \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}} [written for LaTeX], or "the sum of 1/(n^s) as n goes from 0 to infinity (increasing by 1 repeatedly)" [in more human-readable form].
You have a slight typo. Should be: "... as n goes from 1 to infinity ..."
You have a slight typo. It should be: "You have a slight typo. It should be: ..."
• #### Re: (Score:3, Funny)
Okay...I would ask WHY this is important, but someone is ponying up a million bucks for the solution. THAT tells me this is important. I'm not sure if I care why...
• #### Re:Tried to RTFA (Score:5, Informative)
on Wednesday July 02, 2008 @02:00PM (#24034827)
It's important because the zeros of the zeta function tell you how the prime numbers are distributed and prime numbers are to number theory as elements are to chemistry, everything you could care about is built out of them. The RH is also related to host of other more esoteric, but no less important conjectures; the truth of a large part of modern mathematics relies on knowing if the RH is true or false.
Although it's unlikely to impact the storage capacity of a flash drive any time soon the zeta function shows up in high energy physics and thus does have real world consequences.
• #### Re: (Score:3, Interesting)
Here's another easy-to-grasp one: public key encryption (think: credit card purchases online) is dependent upon the use of large primes. Large primes are currently not the easiest/fastest to find - what if you knew better where to look for them?
• #### Or, in layman's terms... (Score:5, Informative)
on Wednesday July 02, 2008 @01:57PM (#24034787)
I just finally found a simple explanation of complex numbers, and just heard of this Riemann Hypothesis, so I may be way off, but let me try to put what (I think) I've figured out so far in layman's terms for the rest of the lost souls:
Riemann was interested in the zeros to this function, where s is a complex number. He conjectured that all zeros (aside from those of the form s = -2c, where c is a positive integer) would have to be of the form (1/2) + ki, where k is a constant and i is the square root of -1.
Basically, 10 trillian calculations have been done involving certain complex numbers, which all show a clear pattern: if you get an answer of 0, the real part of the number given to the function always seems to be 0.5. As yet, no one has proven this, and so, presumably, no one truly understands why that's the case yet. Also, presumably, when we do understand it, we'll have forward (either in a a step or a leap) in our ability to use complex numbers (and the multi-dimensional calculations they represent.
• #### Numb3rs (Score:5, Funny)
on Wednesday July 02, 2008 @11:28AM (#24032541)
Charles Eppes: Imagine you have an infinite number of plot holes, and you want to test how they compare to imaginary numbers. The Riemann Hypothesis states that I can use the zeros in this formula to predict how bullets will bounce off of concrete to a degree of statistical accuracy that it will actually give me the social security number of the guilty shooter.
• #### Re:Numb3rs (Score:5, Funny)
on Wednesday July 02, 2008 @01:47PM (#24034643) Homepage
Dude, you owe me a monitor.
Note to self: Do not drink coke while reading /.
• #### Re: (Score:3, Funny)
It's like 50 football fields laid in line from here to Riemann.
Rieman sounds like a place in Germany.
• #### Re:$1,000,000 prize to be collected then if true (Score:5, Informative) on Wednesday July 02, 2008 @10:54AM (#24031953) Homepage Good explanation here too: http://www.irregularwebcomic.net/1960.html [irregularwebcomic.net] • #### Re: (Score:2) That is probably the best explanation I've seen, thanks! And it makes use of LEGO, another plus! • #### Re: (Score:2) http://www.irregularwebcomic.net/1960.html Great! Now how am I supposed to get any work done. • #### Tough problems (Score:4, Interesting) on Wednesday July 02, 2008 @10:55AM (#24031975) Homepage Part of the reason these problems are so tough because to solve them, you have to understand what the problem is first. I studied the Riemann hypothesis in college for a good week and I'm still not sure where you might begin solving it. Like the Navier-Stokes equations (another big problem with a big prize) solving it will probably require the invention of some new mathematics. Its not simply a matter of dividing by 3 and carrying the 2. I don't know about you but I haven't the slightest idea about how to go about inventing new math. That's the realm of Newton and Einstein, and few others. New math is the only way to go about solving some of these problems. • #### Re: (Score:3) New math is the only way to go about solving some of these problems. You mean like this? [aol.com] • #### Re:Tough problems (Score:5, Funny) on Wednesday July 02, 2008 @11:39AM (#24032717) Homepage ...solving it will probably require the invention of some new mathematics. Its not simply a matter of dividing by 3 and carrying the 2. If you're carrying numbers when dividing, I guess you are inventing new math :-) • #### Re: (Score:3, Funny) by jd (1658) It's easier to have just one heavy maths function and one trivial maths function than two heavy maths functions, so division is easiest implemented as multiplication with the inverse of one of the two numbers, inverses being relatively trivial in exponential notation. As only computers operate this way, the grandparent poster is obviously an artificial intelligence. • #### Re:$1,000,000 prize to be collected then if true (Score:4, Insightful)
by Anonymous Coward on Wednesday July 02, 2008 @11:03AM (#24032131)
The Riemann hypothesis is considered the most important unsolved problem in math. But, considering the source here (random paper on ArXiv by complete unknown), there's no real reason to believe this paper is correct. The number of incorrect proofs to major mathematics problems every year is staggering.
• #### Re: (Score:2)
John Nash in "Beautiful Mind" tries to prove this one too.
One of the things I remember from the book is that he and his wife had a running joke that all babies know the solution to this problem and then forget it when they learn to talk. Maybe Xian-Jin Li had a flashback.
• #### Re: (Score:2)
Ahh - here's where it came from:
Paul Erdos once said all babies (he used to call them epsilons, because babies are really small!) remember the solution for Riemann Hypothesis. The only problem though is that they tend to forget everything once they reach the age of six month.
Found that here [wordpress.com]
• #### Re: (Score:2)
He gets a million because a lot of modern mathematics assumes it is true but no-one can (so far) prove it ....
It he is correct a lot of mathematicians breathe a huge sigh of relief
If someone proves it is false then mathematics collectively panics and a lot of proofs will have to be re-written ...
• #### So what? (Score:3, Insightful)
on Wednesday July 02, 2008 @10:47AM (#24031815)
arXiv has become the repository for junk that couldn't pass peer review. Wake me up when we see a published journal article.
• #### Re: (Score:2)
Also, the proof of something that complicated is likely so complicated that only the very best minds would even be able to prove that the proof was wrong.
• #### Re:So what? (Score:5, Informative)
on Wednesday July 02, 2008 @11:14AM (#24032335)
I think you misunderstand the scope and purpose of arXiv. arXiv is a repository for *preprints*.
• #### Re:not so fast (Score:4, Funny)
on Wednesday July 02, 2008 @11:54AM (#24032931)
They sent you your checks for cases where you are equal to 0.
Someone beat you to the "1" part.
• #### Re: (Score:2, Interesting)
Indeed. Among some mathematicians it is a pleasant diversion to take bets on which of the major unsolved (or unprovable) problems has the most solutions appear on the arXiv this week.
• #### Re: (Score:3, Interesting)
That's true, but most of them are obvious drivel. I have looked through this one, and it is clearly a real attempt by a genuine mathematician who understands the relevant background. I'd still bet on it being wrong, but not stupidly wrong.
• #### Dolly Parton (Score:2)
Dolly Parton was 69 lbs over weight. The doctor said that's 222 much! You need to lose 51 x 8 days. That left her:
6922251x8=55378008
• #### Oblig. (Score:5, Funny)
on Wednesday July 02, 2008 @10:56AM (#24031983)
• #### Apology for the Re (Score:2)
Ok, so many have tried, all have failed.
It may take a decade to test the assertions that this so called proof attempts to demonstrate.
Perhaps we could give the guy a consolation prize, wait for the work to be "proven" wrong and then off course, issue an Apology:
http://www.math.purdue.edu/~branges/apology.pdf [purdue.edu] :-)
-Hack
PS: Does anyone find it STRANGE that the guy who can solve this problem has issues finding a job?
WTF?
• #### Re: (Score:2, Interesting)
Not really, the kind of person who would solve a problem of this nature is probably going to be the Andrew Wiles reclusive genius type - a lot like the Russian gent whose name escapes me who solved the Poincare Conjecture. Thus he's not necessarily going to be too keen to teach/lecture/supervise and so would possibly not be too attractive to prospective employers.
I doubt too many Maths faculties in the world have people working full-time on the Riemann Hypotheses.
Of course I echo your sentiments that
• #### Re: (Score:2)
Interestingly, DeBranges was Xian-Jin Li's advisor:
hEll
• #### 53188008? (Score:2)
That reminds [explosm.net] me.
This guys advisor, according to the Math Genealogy Project, is Louis deBranges. DeBranges also claimed to have proven this a few years back, but his proof was not accepted (for reasons unknown to me). The $1M might still be safe. • #### You mean that... (Score:2) Simple Simon actually met a Riemann, after all?! I thought that was just a hypothesis! • #### The REAL importance is Primes (Score:5, Interesting) by Anonymous Coward on Wednesday July 02, 2008 @12:13PM (#24033239) Section two of the wiki article (http://en.wikipedia.org/wiki/Riemann_hypothesis) is the great importance here. If indeed there is a proof of Riemann's Hypothesis, then there is a similar proof of the Generalized Riemann Hypothesis, which is in turn a big step in finding the exact distribution of prime numbers. Finding the distribution of prime numbers has epic consequences, like breaking most encryption, for starters. • #### Re:The REAL importance is Primes (Score:5, Informative) on Wednesday July 02, 2008 @04:26PM (#24036561) The Riemann Hypothesis and RSA encryption both have to do with prime numbers, but the relationship between the two pretty much ends there. To break RSA you need to know how to factor large numbers quickly. RH, on the other hand, pretains to the distribution of prime numbers. It's pretty unlikely that a proof would make computers any faster at factorizing. So this begs the question that a lot of people have been asking on this thread: why should you care? There tongue-in-cheek answer is that a solution is worth$1,000,000. While that response may suffice for non-mathematicians, mathematicians would have another, more important reason to celebrate. RH and its generalization, the Grand Riemann Hypothesis, have an absolutely enormous number of profound impliations in number theory, and it is difficult to overstate how critical a proof of either would be. (The implications are too technical to write about here, but you can read about them in most good survey books on analytic number theory; for example, see section 5.8 of Iwaniec & Kowalski [amazon.com]). A successful proof probably won't affect your life in any meaningful way (unless you work with analytic number theory for a living), but it would be monumental in the world of math - indeed, this is precisely why there's a reward for solving it. If that's not enough for you, just remember that many mathematicians are motivated not by fame or money but by the beauty and elegance of mathematics, and any proof of RH would establish a truly beautiful and amazing result.
Of course, there's also the question: is Li's proof correct? I certainily don't know, and I doubt anyone will for quite some time, but there's an interesting story. Li's Ph.D. adviser was Louis de Branges [nodak.edu] who, as noted on this very website [slashdot.org], claimed to prove RH in 2004. His proof has not been accepted by the mathematical community and is widely considered to be incorrect, in large part because the method he wclaims to use was shown, in a 2000 paper [arxiv.org] co-authored by none other than Xian-Jin Li, to have holes in it.
• #### DOOOOOMED!!!!!!! (Score:2, Funny)
by Anonymous Coward
I can't believe they are brazenly going forward with research into this subject without knowing if it could possibly lead to the creation of a black hole that will swallow the earth.
Mathematics is the only science where one never knows what one is talking about nor whether what is said is true. -- Russell
Working... | 5,043 | 19,413 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2016-07 | latest | en | 0.849013 |
https://www.slideserve.com/hila/refraction-power-point-power-point-10-3 | 1,618,611,835,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038092961.47/warc/CC-MAIN-20210416221552-20210417011552-00165.warc.gz | 1,110,634,708 | 19,142 | # Refraction Power- Point Power Point 10-3 - PowerPoint PPT Presentation
Refraction Power- Point Power Point 10-3
1 / 18
Refraction Power- Point Power Point 10-3
## Refraction Power- Point Power Point 10-3
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##### Presentation Transcript
1. Refraction Power-Point Power Point 10-3
2. Definition • Refraction- the bending of a wave or particle as it passes from different mediums
3. Application • A fish appears higher than it actually is…
4. Bending Toward the Normal • When light goes from a less dense to more dense medium, it is bent toward the normal. This is an example of air going to fiberglass.
5. Bending Away from the Normal • When light passes fromfiberglassto air, it bends away from the normal.
6. The index of refraction (n) • The speed of a wave is slower in a more optically dense medium.
7. A relationship… • The relationship between the index of refraction (n) and the speed of the wave can be determined by • Example 1: What is the speed of light in a diamond (n=2.42)? where c=3 X 10 8m/s
8. Snell’s Law • There is a mathematical relationship between: • n= index of refraction for a material • Angle of incidence • Angle of reflection Example 2: What is the angle of refraction of light going at an angle of 10 o from a diamond (n=2.42) to air (n=1.0)?
9. Total Internal Reflection • Only occurs when light goes from more dense to less dense. • Three cases: • (1) Refracts- bends away from normal (o-43o) • (2) Critical angle- prism • (3) Reflection • Example 3: What is the maximum angle of incidence in a diamond where refraction occurs? (the angle of refraction is 90o or the critical angle)
10. Total Internal Reflection
11. Applications of Refraction • Fiber optics, polar bear skin, sparkling of diamonds
12. Lab…Semicircular Prisms
13. Rectangular Prism
14. Right Triangle Prism
15. Example 4: Double Refraction • A layer of oil (n=1.45) floats on water (n=1.33). A ray of light has an incident angle with the oil of 40o. What is the angle that the ray travels in water?
16. Equilateral Triangle Prism • Since speed of wave is slowing in glass (n=1.52) • Red is longest wavelength (700 nm); lowest frequency; bends the least. • Violet is the highest wavelength (400 nm); highest frequency; bends the most.
17. Pink Floyd…Which is correct?
18. Example 5:Angles in Equilateral Triangle • It is observed experimentally that the angle resulting in a prism is 41.5 o in crown glass (n=1.52). Determine the angles in the diagram provided. | 696 | 2,593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2021-17 | latest | en | 0.721482 |
https://devel.isa-afp.org/browser_info/current/AFP/Jinja/Auxiliary.html | 1,702,195,107,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101282.74/warc/CC-MAIN-20231210060949-20231210090949-00054.warc.gz | 238,248,630 | 4,520 | # Theory Auxiliary
(* Title: Jinja/Common/Basis.thy
Author: David von Oheimb, Tobias Nipkow
*)
chapter ‹Jinja Source Language \label{cha:j}›
section ‹Auxiliary Definitions›
theory Auxiliary imports Main begin
(* FIXME move and possibly turn into a general simproc *)
"((n::nat) + max i j ≤ m) = (n + i ≤ m ∧ n + j ≤ m)"
(*<*)by arith(*>*)
"(Suc(n + max i j) ≤ m) = (Suc(n + i) ≤ m ∧ Suc(n + j) ≤ m)"
(*<*)by arith(*>*)
notation Some ("(⌊_⌋)")
(*<*)
declare
option.splits[split]
Let_def[simp]
subset_insertI2 [simp]
Cons_eq_map_conv [iff]
(*>*)
subsection ‹‹distinct_fst››
definition distinct_fst :: "('a × 'b) list ⇒ bool"
where
"distinct_fst ≡ distinct ∘ map fst"
lemma distinct_fst_Nil [simp]:
"distinct_fst []"
(*<*)
by (unfold distinct_fst_def) (simp (no_asm))
(*>*)
lemma distinct_fst_Cons [simp]:
"distinct_fst ((k,x)#kxs) = (distinct_fst kxs ∧ (∀y. (k,y) ∉ set kxs))"
(*<*)
by (unfold distinct_fst_def) (auto simp:image_def)
(*>*)
(*
lemma distinct_fst_append:
"⟦ distinct_fst kxs'; distinct_fst kxs; ∀(k,x) ∈ set kxs. ∀(k',x') ∈ set kxs'. k' ≠ k ⟧
⟹ distinct_fst(kxs @ kxs')"
by (induct kxs) (auto dest: fst_in_set_lemma)
lemma distinct_fst_map_inj:
"⟦ distinct_fst kxs; inj f ⟧ ⟹ distinct_fst (map (λ(k,x). (f k, g k x)) kxs)"
by (induct kxs) (auto dest: fst_in_set_lemma simp: inj_eq)
*)
lemma map_of_SomeI:
"⟦ distinct_fst kxs; (k,x) ∈ set kxs ⟧ ⟹ map_of kxs k = Some x"
(*<*)by (induct kxs) (auto simp:fun_upd_apply)(*>*)
subsection ‹Using @{term list_all2} for relations›
definition fun_of :: "('a × 'b) set ⇒ 'a ⇒ 'b ⇒ bool"
where
"fun_of S ≡ λx y. (x,y) ∈ S"
text ‹Convenience lemmas›
(*<*)
declare fun_of_def [simp]
(*>*)
lemma rel_list_all2_Cons [iff]:
"list_all2 (fun_of S) (x#xs) (y#ys) =
((x,y) ∈ S ∧ list_all2 (fun_of S) xs ys)"
(*<*)by simp(*>*)
lemma rel_list_all2_Cons1:
"list_all2 (fun_of S) (x#xs) ys =
(∃z zs. ys = z#zs ∧ (x,z) ∈ S ∧ list_all2 (fun_of S) xs zs)"
(*<*)by (cases ys) auto(*>*)
lemma rel_list_all2_Cons2:
"list_all2 (fun_of S) xs (y#ys) =
(∃z zs. xs = z#zs ∧ (z,y) ∈ S ∧ list_all2 (fun_of S) zs ys)"
(*<*)by (cases xs) auto(*>*)
lemma rel_list_all2_refl:
"(⋀x. (x,x) ∈ S) ⟹ list_all2 (fun_of S) xs xs"
lemma rel_list_all2_antisym:
"⟦ (⋀x y. ⟦(x,y) ∈ S; (y,x) ∈ T⟧ ⟹ x = y);
list_all2 (fun_of S) xs ys; list_all2 (fun_of T) ys xs ⟧ ⟹ xs = ys"
(*<*)by (rule list_all2_antisym) auto(*>*)
lemma rel_list_all2_trans:
"⟦ ⋀a b c. ⟦(a,b) ∈ R; (b,c) ∈ S⟧ ⟹ (a,c) ∈ T;
list_all2 (fun_of R) as bs; list_all2 (fun_of S) bs cs⟧
⟹ list_all2 (fun_of T) as cs"
(*<*)by (rule list_all2_trans) auto(*>*)
lemma rel_list_all2_update_cong:
"⟦ i<size xs; list_all2 (fun_of S) xs ys; (x,y) ∈ S ⟧
⟹ list_all2 (fun_of S) (xs[i:=x]) (ys[i:=y])" | 1,077 | 2,696 | {"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} | 3.171875 | 3 | CC-MAIN-2023-50 | latest | en | 0.341936 |
https://www.univerkov.com/what-amount-of-a-substance-is-contained-in-one-kg-a-sodium-chloride-nacl-b-nitrogen-oxide-no2/ | 1,656,108,960,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033816.0/warc/CC-MAIN-20220624213908-20220625003908-00502.warc.gz | 1,115,080,273 | 6,291 | # What amount of a substance is contained in one kg: a) sodium chloride (NaCl) b) nitrogen oxide (NO2)
What amount of a substance is contained in one kg: a) sodium chloride (NaCl) b) nitrogen oxide (NO2) c) aluminum sulfide (Al2S3) d) calcium chloride (Ca2P3) e) barium chloride (BuF2)
a) M (NaCl) = Mr (NaCl) = Ar (Na) * N (Na) + Ar (Cl) * N (Cl) = 23 * 1 + 35.5 * 1 = 58.5 g / mol;
n (NaCl) = m (NaCl) / M (NaCl) = 1000 / 58.5 = 17.1 mol;
b) M (NO2) = Mr (NO2) = Ar (N) * N (N) + Ar (O) * N (O) = 14 * 1 + 16 * 2 = 46 g / mol;
n (NO2) = m (NO2) / M (NO2) = 1000/46 = 22 mol;
c) M (Al2S3) = Mr (Al2S3) = Ar (Al) * N (Al) + Ar (S) * N (S) = 27 * 2 + 32 * 3 = 150 g / mol;
n (Al2S3) = m (Al2S3) / M (Al2S3) = 1000/150 = 7 mol;
d) M (CaCl2) = Mr (CaCl2) = Ar (Ca) * N (Ca) + Ar (Cl) * N (Cl) = 40 * 1 + 35.5 * 2 = 111 g / mol;
n (CaCl2) = m (CaCl2) / M (CaCl2) = 1000/111 = 9 mol;
e) M (BaCl2) = Mr (BaCl2) = Ar (Ba) * N (Ba) + Ar (Cl) * N (Cl) = 137 * 1 + 35.5 * 2 = 208 g / mol;
n (BaCl2) = m (BaCl2) / M (BaCl2) = 1000/208 = 4.8 mol.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 577 | 1,347 | {"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-2022-27 | latest | en | 0.801487 |
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=1302 | 1,550,860,191,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247522457.72/warc/CC-MAIN-20190222180107-20190222202107-00169.warc.gz | 141,173,483 | 6,429 | # Twenty Questions
Time Limit : 8 sec, Memory Limit : 131072 KB
# Problem H: Twenty Questions
Consider a closed world and a set of features that are defined for all the objects in the world. Each feature can be answered with "yes" or "no". Using those features, we can identify any object from the rest of the objects in the world. In other words, each object can be represented as a fixed-length sequence of booleans. Any object is different from other objects by at least one feature.
You would like to identify an object from others. For this purpose, you can ask a series of questions to someone who knows what the object is. Every question you can ask is about one of the features. He/she immediately answers each question with "yes" or "no" correctly. You can choose the next question after you get the answer to the previous question.
You kindly pay the answerer 100 yen as a tip for each question. Because you don' t have surplus money, it is necessary to minimize the number of questions in the worst case. You donft know what is the correct answer, but fortunately know all the objects in the world. Therefore, you can plan an optimal strategy before you start questioning.
The problem you have to solve is: given a set of boolean-encoded objects, minimize the maximum number of questions by which every object in the set is identifiable.
## Input
The input is a sequence of multiple datasets. Each dataset begins with a line which consists of two integers, m and n: the number of features, and the number of objects, respectively. You can assume 0 < m ≤ 11 and 0 < n ≤ 128. It is followed by n lines, each of which corresponds to an object. Each line includes a binary string of length m which represent the value ("yes" or "no") of features. There are no two identical objects.
The end of the input is indicated by a line containing two zeros. There are at most 100 datasets.
## Output
For each dataset, minimize the maximum number of questions by which every object is identifiable and output the result.
```8 1
11010101
11 4
00111001100
01001101011
01010000011
01100110001
11 16
01000101111
01011000000
01011111001
01101101001
01110010111
01110100111
10000001010
10010001000
10010110100
10100010100
10101010110
10110100010
11001010011
11011001001
11111000111
11111011101
11 12
10000000000
01000000000
00100000000
00010000000
00001000000
00000100000
00000010000
00000001000
00000000100
00000000010
00000000001
00000000000
9 32
001000000
000100000
000010000
000001000
000000100
000000010
000000001
000000000
011000000
010100000
010010000
010001000
010000100
010000010
010000001
010000000
101000000
100100000
100010000
100001000
100000100
100000010
100000001
100000000
111000000
110100000
110010000
110001000
110000100
110000010
110000001
110000000
0 0
```
## Output for the Sample Input
```0
2
4
11
9
```
Source: ACM International Collegiate Programming Contest , Asia Regional Tokyo, Tokyo, Japan, 2009-11-08
http://www.waseda.jp/assoc-icpc2009/en/ | 820 | 2,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-09 | longest | en | 0.941273 |
http://sage.math.washington.edu/books/ant/ant/node89.html | 1,369,335,881,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703682988/warc/CC-MAIN-20130516112802-00047-ip-10-60-113-184.ec2.internal.warc.gz | 233,473,146 | 4,233 | ### The -adic Numbers
Lemma 14.2.5 Let be a positive integer. Then for any nonzero rational number there exists a unique and integers , with positive, such that with , , and .
Proof. Write with and . First suppose is exactly divisible by a power of , so for some we have but . Then
If is the largest power of that divides , then , , satisfy the conclusion of the lemma.
By unique factorization of integers, there is a smallest multiple of such that is exactly divisible by . Now apply the above argument with and replaced by and .
Definition 14.2.6 (-adic valuation) Let be a positive integer. For any positive , the -adic valuation of is , where is as in Lemma 14.2.5. The -adic valuation of 0 is .
We denote the -adic valuation of by . (Note: Here we are using valuation'' in a different way than in the rest of the text. This valuation is not an absolute value, but the logarithm of one.)
Definition 14.2.7 (-adic metric) For the -adic distance between and is
We let , since .
For example, are close in the -adic metric if their difference is divisible by a large power of . E.g., if then and are close because their difference is , which is divisible by a large power of .
Proposition 14.2.8 The distance on defined above is a metric. Moreover, for all we have
(This is the nonarchimedean'' triangle inequality.)
Proof. The first two properties of Definition 14.2.1 are immediate. For the third, we first prove that if then
Assume, without loss, that and that both and are nonzero. Using Lemma 14.2.5 write and with or possibly negative. Then
Since it follows that . Now suppose . Then
so
hence .
We can finally define the -adic numbers.
Definition 14.2.9 (The -adic Numbers) The set of -adic numbers, denoted , is the completion of with respect to the metric .
The set is a ring, but it need not be a field as you will show in Exercises 11 and 12. It is a field if and only if is prime. Also, has a bizarre'' topology, as we will see in Section 14.2.3.
William Stein 2012-09-24 | 526 | 2,028 | {"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} | 4.09375 | 4 | CC-MAIN-2013-20 | latest | en | 0.865621 |
https://www.exceldemy.com/excel-floating-text-box/ | 1,701,762,990,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100550.40/warc/CC-MAIN-20231205073336-20231205103336-00822.warc.gz | 843,647,639 | 76,660 | # How to Create Floating Text Box in Excel (2 Easy Ways)
Get FREE Advanced Excel Exercises with Solutions!
While working in Excel, we often need to create a floating text box. A floating text box allows us to display the contents of a cell anywhere in an Excel worksheet. It also provides us with flexibility in positioning the text box. In this article, we will learn two simple methods to create a floating text box in Excel. So, let’s start this article and explore these methods.
## How to Create Floating Text Box in Excel: 2 Easy Ways
In this section, we’ll go over two quick ways to make a floating text box in Excel. Let’s say, we have the Prices of Products of ABC Store as our dataset. Our goal is to create floating text boxes using this dataset.
Not to mention, we used the Microsoft Excel 365 version for this article; however, you can use any version according to your preference.
### 1. Using Text Box Command from Insert Tab
Using the Text Box command in Excel is one of the easiest ways to create a floating box in Excel. The floating text boxes made by this method have dynamic properties. That is, if you change any data in the original dataset, the values in the text boxes will automatically update. Now, let’s follow the steps mentioned below to do this.
Steps:
• Firstly, go to the Insert tab from Ribbon.
• Then, choose the Text Box option from the Text group.
• Following that, click on the newly created text box.
• Then, go to the formula bar and type equal (=).
• Then, click on the cell that you want to show in the text box. In this case, we used cell B2.
Your final formula will be like this.
=\$B\$2
• Now, press ENTER.
As a result, the content of cell B2 will be shown in the text box as shown in the following image.
• Subsequently, follow the same process and create another text box.
• Then, click on the text box and insert the following formula in the formula bar.
=\$B\$4
Here, cell B4 indicates the heading of the Product column.
• Then, hit ENTER.
Subsequently, the contents of cell B4 will be visible in the second text box.
You don’t need to create text boxes in this way every time. You can simply copy a text box and then paste it. Then, you can edit the text boxes.
• Firstly, select the text box you want to copy.
• Then, use the keyboard shortcut CTRL + C to copy the text box.
• After that, press CTRL + V to paste it.
• Now, resize the new text box and reposition it according to your needs.
• After you have placed the new text box in its position, click on the text box.
• Then, apply the following formula in the formula bar.
=\$C\$4
Here, cell C4 refers to the heading of the Product column.
• Then, press ENTER.
Consequently, you will have the following output inside the new text box.
• Follow the same procedure for the rest of the cells to get the following output.
Now, we will group them together so that all of these text boxes can float together.
• Firstly, click on any text box and then press CTRL + A to select them all.
• After that, go to the Shape Format tab from Ribbon.
• Then, click on the Group option from the Arrange group.
• Finally, click on the Group option from the drop-down.
Consequently, you will have a set of floating text boxes, as demonstrated in the following picture. You can drag these text boxes anywhere in your worksheet.
### 2. Utilizing VBA Macro Feature
Utilizing the VBA Macro is another smart option to create floating boxes in Excel. Let’s follow the steps outlined in the following section to do this.
Steps:
• Firstly, go to the Developer tab from Ribbon.
• After that, choose the Visual Basic option from the Code group.
As a result, the Microsoft Visual Basic window will open on your worksheet.
• Now, in the Visual Basic window, go to the Insert tab.
• Then, choose the UserForm option from the drop-down.
As a result, the following UserForm window will be available along with a Toolbox, as shown in the image below.
• Now, click on the Text Box icon inside the Toolbox, as marked in the following picture.
• Then, drag it inside the UserForm1 window.
• Next, resize the text box so that it matches the edges of the UserForm1 window.
• Now, click on anywhere inside the text box.
• After that, search for the Multiline option in the Properties – TextBox1 section on the left side of your screen.
• Then, click on the drop-down icon beside the Multiline option and choose the True option.
Note: If you can’t find the Properties section, go to the View tab in the Visual Basic window and select the Properties Window option.Â
• Afterward, go to the Insert tab and choose the Module option from the drop-down.
• After that, write the following code in the newly created Module.
Sub floating_text_box()
UserForm1.TextBox1.Text = "Product Brand Price" & vbLf & "Laptop Acer \$1,500"
UserForm1.Show vbModeless
End Sub
Code Breakdown
• Firstly, we initiated a sub-procedure named floating_text_box.
• After that, we specified what should be displayed inside the text box.
• Here, we used the vbLf character to go to the next line. You can add more vbLf characters to add more lines.
• Finally, we ended the sub-procedure.
• After writing the code, click on the Save option.
• Now, press the keyboard shortcut ALT + F11 to redirect to the worksheet.
• Subsequently, use another keyboard shortcut ALT + F8 to open the Macro dialogue box.
• Following that, in the Macro dialogue box, choose the floating_text_box option.
• Finally, click on Run.
As a result, the following floating text box will pop up on your worksheet, as shown in the following image.
## How to Create a Floating Table in Excel
In this section of the article, we will learn how we can create a floating table in Excel. So, let’s follow the procedure outlined in the following section to do this.
Steps:
• Firstly, select the entire table and go to the Home tab from Ribbon.
• Then, click on the Copy option from the Clipboard group.
• Now, choose the Copy as Picture option from the drop-down.
• After that, choose the As shown on screen option in the Appearance field.
• Then, from the Format field, select the Picture option.
• Next, click OK.
• Now, select the destination cell where you want to paste the floating table.
• Lastly, press the keyboard shortcut CTRL + V.
Consequently, you will have a floating table on your worksheet, as demonstrated in the image given below.
Read More: How to Anchor Text Box in Excel
## How to Create Floating Cells in Excel
The procedure to create floating cells is almost similar to creating a floating table. So, without any further delay, let’s explore the following steps.
Steps:
• Firstly, select a cell from your dataset and go to the Home tab from Ribbon.
• After that, choose the Copy option from the Clipboard group.
• Then, select the Copy as Picture option from the drop-down.
• Following that, choose the As shown on screen option in the Appearance field.
• Now, from the Format field, select the Picture option.
• Afterward, click OK.
• Subsequently, select your destination cell and then press the keyboard shortcut CTRL + V to paste the copied cell as a picture.
That’s it! You will have your selected cell as a floating cell on your worksheet, as shown in the following picture.
• You can use the same steps and create as many floating cells as you want.
## Conclusion
So, these are the most common and effective methods you can use anytime while working with your Excel datasheet to create a floating text box in Excel. If you have any questions, suggestions, or feedback related to this article, you can comment below.
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## r = 20i ̂-10j ̂+4k ̂+μ(10i ̂-20j ̂+10k ̂ )?
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### Transcript
Question 34 (Choice 2) The equations of motion of a rocket are: 𝑥 = 2𝑡, 𝑦 = −4𝑡, 𝑧 = 4𝑡, where the time t is given in seconds, and the coordinates of a moving point in km. What is the path of the rocket? At what distances will the rocket be from the starting point O(0, 0, 0) and from the following line in 10 seconds 𝑟 ⃗ = 20𝑖 ̂−10𝑗 ̂+40𝑘 ̂+𝜇(10𝑖 ̂−20𝑗 ̂+10𝑘 ̂ )?Given equation of motion of rocket 𝑥 = 2𝑡, 𝑦 = −4𝑡, 𝑧 = 4𝑡, We note that if we join the points, we get a line. At t = 0 x = 0, y = 0, z = 0 So, point at t = 0 is (0, 0, 0) At t = 1 x = 2, y = −4, z = 4 So, point at t = 0 is (2, −4, 4) So, path of rocket is a line Equation of Path of Rocket Since rocket passes through points (0, 0, 0) and (2, −4, 4), equation of line is (𝑥 − 𝑥_1)/(𝑥_2 − 𝑥_1 ) = (𝑦 − 𝑦_1)/(𝑦_2 − 𝑦_1 ) = (𝑧 − 𝑧_1)/(𝑧_2 − 𝑧_1 ) (𝑥 − 0)/(2 − 0) = (𝑦 − 0)/( −4 − 0) = (𝑧 − 0)/(4 − 0) 𝒙/𝟐 = 𝒚/( −𝟒) = 𝒛/𝟒 Now, we need to find At what distances will the rocket be from the starting point O(0, 0, 0) and from the following line in 10 seconds 𝑟 ⃗ = 20𝑖 ̂−10𝑗 ̂+40𝑘 ̂+𝜇(10𝑖 ̂−20𝑗 ̂+10𝑘 ̂ )? At t = 10 seconds, point is 𝑥 = 2𝑡, 𝑦 = −4𝑡, 𝑧 = 4𝑡, 𝑥 = 20, 𝑦 = −40, 𝑧 = 40 So, At t = 0, point is (0, 0, 0) At t = 10, point is (20, −40, 40) Thus, We need to find distance of point (20, −40, 40) from point (0, 0, 0) And Distance of point (20, −40, 40) from line 𝒓 ⃗ = 20𝑖 ̂−10𝑗 ̂+4𝑘 ̂+𝜇(10𝑖 ̂−20𝑗 ̂+10𝑘 ̂ )? Thus, We need to find distance of point (20, −40, 40) from point (0, 0, 0) And Distance of point (20, −40, 40) from line 𝒓 ⃗ = 20𝑖 ̂−10𝑗 ̂+40𝑘 ̂+𝜇(10𝑖 ̂−20𝑗 ̂+10𝑘 ̂ )? Line 𝒓 ⃗ = 20𝑖 ̂−10𝑗 ̂+4𝑘 ̂ +𝜇(10𝑖 ̂−20𝑗 ̂+10𝑘 ̂ ) Distance of point (20, −40, 40) from point (0, 0, 0) D = √((20−0)^2+(−40−0)^2+(40−0)^2 ) = √(〖𝟐𝟎〗^𝟐+(−𝟒𝟎)^𝟐+〖𝟒𝟎〗^𝟐 ) = √(400+1600+1600) = √3600 = √((60)^2 ) = 60 kmDistance of point (20, −40, 40) from line 𝒓 ⃗ = 20𝒊 ̂−𝟏𝟎𝒋 ̂+𝟒𝟎𝒌 ̂+𝝁(𝟏𝟎𝒊 ̂−𝟐𝟎𝒋 ̂+𝟏𝟎𝒌 ̂ ) In three dimensions, the perpendicular distance, 𝐷,between a point 𝑃(𝑥, 𝑦, 𝑧) (position vector 𝑝 ⃗) and line 𝒓 ⃗ = a ⃗ + λb ⃗ is 𝐷=|(𝑝 ⃗ − 𝑎 ⃗ ) × 𝑏 ⃗ |/|𝑏 ⃗ | Putting values 𝐷=|([20𝑖 ̂ − 40𝑗 ̂ + 40𝑘 ̂ ]−[20𝑖 ̂ −10𝑗 ̂ + 4𝑘 ̂ ]) × (𝟏𝟎𝒊 ̂ − 𝟐𝟎𝒋 ̂ + 𝟏𝟎𝒌 ̂)|/|𝟏𝟎𝒊 ̂ − 𝟐𝟎𝒋 ̂ + 𝟏𝟎𝒌 ̂ | 𝐷=|(30𝒋 ̂ ) × (𝟏𝟎𝒊 ̂ − 𝟐𝟎𝒋 ̂ + 𝟏𝟎𝒌 ̂)|/√(〖10〗^2 + (−20)^2 + 〖10〗^2 ) 𝑫=|■8(𝒊 ̂&𝒋 ̂&𝒌 ̂@𝟎&𝟑𝟎&𝟎@𝟏𝟎&−𝟐𝟎&𝟏𝟎)|/√(𝟏𝟎𝟎 + 𝟒𝟎𝟎 + 𝟏𝟎𝟎) 𝐷=|𝑖 ̂(300 − 0) − 𝑗 ̂(0 − 0) + 𝑘 ̂(0 − 300)|/√600 𝑫=|𝟑𝟎𝟎𝒊 ̂ − 𝟑𝟎𝟎𝒌 ̂ |/√𝟔𝟎𝟎 𝐷=√((300)^2+(300)^2 )/√(6 × 100) 𝐷=√(2 × (300)^2 )/√(6 × 100) 𝑫=(𝟑𝟎𝟎√𝟐)/(𝟏𝟎√𝟔) 𝐷=30/√3 𝐷=30/√3 ×√3/√3 𝐷=(30√3)/3 D = 10 √𝟑 km | 1,802 | 2,687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2023-14 | longest | en | 0.802614 |
https://mathematica.stackexchange.com/questions/229626/three-coupled-odes/229642#229642 | 1,638,221,133,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358842.4/warc/CC-MAIN-20211129194957-20211129224957-00432.warc.gz | 464,108,697 | 34,040 | # Three Coupled ODEs
I've been trying to solve it with Mathematica Order of DSolve but I'm not getting Anywhere! so if anyone got an idea i would be glad to tell me how to get answer
p.s:I just want solution and Numerical solution with NDSolve won't help me Tnx. [![enter image description here] 1]1
• 1. Shows us the code you used as text, not as an image, including the specific call to DSolve you tried. 2. There may not be an analytical solution. Do you have reason to believe that one should exist for this system? Sep 6 '20 at 15:24
The three PDEs
eq1 = D[f[xm, xp], xm, xp] - 2 f[xm, xp]/(xm - xp)^2 == 0;
eq2 = D[f[xm, xp], xp, xp] - 2 D[f[xm, xp], xp]/(xm - xp) == 0;
eq3 = D[f[xm, xp], xm, xm] + 2 D[f[xm, xp], xm]/(xm - xp) == 0;
clearly are satisfied by f[xm, xp] = 0. In fact, this is the only solution. As noted in the Question,DSolve cannot solve the three PDEs together. Indeed, it cannot solve even eq1 alone. But it can solve the second and third PDEs (actually ODEs) individually.
s2 = Together@DSolveValue[eq2, f[xm, xp], {xm, xp}] /. {C[1] -> c1, C[2] -> c2}
s3 = Together@DSolveValue[eq3, f[xm, xp], {xm, xp}] /. {C[1] -> c3, C[2] -> c4}
(* (c1[xm] + xm c2[xm] - xp c2[xm])/(xm - xp) *)
(* (-c3[xp] + xm c4[xp] - xp c4[xp])/(xm - xp) *)
The two solutions must be equal, i.e., s0 == 0, where
s0 = Simplify[s2 - s3]
(* (c1[xm] + (xm - xp) c2[xm] + c3[xp] - xm c4[xp] + xp c4[xp])/(xm - xp) *)
From this, the four unknown functions can be determined up to constants.
D[Numerator@s0, xm, xp] == 0
r2 = Flatten@{DSolve[Derivative[1][c2][xm] == c, c2[xm], xm] /. C[1] -> c2c,
DSolve[Derivative[1][c4][xp] == -c, c4[xp], xp] /. C[1] -> c4c}
(* -c2'[xm] - c4'[xp] == 0 *)
(* {c2[xm] -> c2c + c xm, c4[xp] -> c4c - c xp} *)
where c == -c2'[xm] == -c4'[xp]. Then,
Numerator[s0 /. r2] == 0;
sc1 = Flatten@DSolve[D[%, xm], c1[xm], xm] /. C[1] -> c1c
sc2 = Flatten@DSolve[D[%%, xp], c3[xp], xp] /. C[1] -> c3c
(* {c1[xm] -> c1c - c2c xm + c4c xm - c xm^2} *)
(* {c3[xp] -> c3c + c2c xp - c4c xp + c xp^2} *)
Simplify@Numerator[s0 /. r2 /. sc1 /. sc2] == 0
(* c1c + c3c == 0 *)
The solution, then, is
s = Simplify[s3 /. r2 /. sc1 /. sc2 /. c1c -> -c3c]
(* -((c3c - c4c xm + c2c xp + c xm xp)/(xm - xp)) *)
Simplify[eq1[[1]] /. f -> Function[{xm, xp}, s]]
Flatten@CoefficientList[Numerator[%], {xm, xp}]
Flatten@Solve[Thread[% == 0], {c, c2c, c3c, c4c}]
(* (2 (c3c - c4c xm + (c2c + c xm) xp))/(xm - xp)^3 *)
(* {2 c3c, 2 c2c, -2 c4c, 2 c} *)
(* {c -> 0, c2c -> 0, c3c -> 0, c4c -> 0} *)
All the constants are zero.
A much simpler derivation of the same result can be obtained by solving eq2 and inserting the solution into eq1.
Flatten@DSolve[eq2, f, {xm, xp}];
Because xm and xm - xp are independent quantities, the coefficient of each power of xm - xp must vanish. In other words, C[1][xm] = C[2][xm] = 0`. | 1,153 | 2,842 | {"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} | 3.375 | 3 | CC-MAIN-2021-49 | latest | en | 0.825742 |
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# Two integers will be randomly selected from the sets above, one intege
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Two integers will be randomly selected from the sets above, one intege [#permalink]
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Updated on: 31 Jan 2019, 00:55
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Difficulty:
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Question Stats:
79% (01:29) correct 21% (01:46) wrong based on 1336 sessions
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A = {2, 3, 4, 5}
B = {4, 5, 6, 7, 8}
Two integers will be randomly selected from the sets above, one integer from set A and one integer from set B. What is the probability that the sum of the two integers will equal 9?
(A) 0.15
(B) 0.20
(C) 0.25
(D) 0.30
(E) 0.33
Originally posted by vermatanya on 17 Jan 2008, 13:50.
Last edited by Bunuel on 31 Jan 2019, 00:55, edited 2 times in total.
Edited the question.
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Posts: 65771
Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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24 Feb 2012, 21:49
12
5
fortsill wrote:
walker wrote:
B
$$p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20$$
or
$$p=\frac{4}{4}*\frac{1}{5}=0.20$$
or
$$p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20$$
Although the calculation seems overkill, do you mind going over the logic behind the fractions above, please?
I'd offer another approach which I hope will also clarify the above.
A = {2,3,4,5}
B = {4,5,6,7,8}
2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9?
A. 0.15
B. 0.20
C. 0.25
D. 0.30
E. 0.33
Rearrange the first set:
A = {5, 4, 3, 2}
B = {4, 5, 6, 7, 8}
As you can see numbers in each column (the numbers of the same color) give the sum of 9. So there are 4 such pares possible, total # of pairs is 4*5=20. P=favorable/total=4/20=0.2.
Or: we can select ANY number from set A (4/4=1) but in this case we must select its matching pair from set B (the number of the same color) and since there are only one matching pair of this particular number in B then the probability of this is 1/5. So, overall: P=1*1/5.
Hope it's clear.
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Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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07 Sep 2009, 10:29
8
3
Maybe I am late with response, but anyway.
We have two sets with 4 and 5 integers included in each one respectively.
Thus, we have 4*5=20 possible ways of selecting two integers from these sets.
The requirement is that we need the sum of 9.
Here we have 4 possible choices:
2 and 7
3 and 6
4 and 5
5 and 4
Probability = favourable / total
4/20 = 0.2
##### General Discussion
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Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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17 Jan 2008, 13:59
3
4
B
$$p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20$$
or
$$p=\frac{4}{4}*\frac{1}{5}=0.20$$
or
$$p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20$$
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Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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17 Jan 2008, 14:26
3
You don't need to perform any calculations in this example. Regardless of the number you choose from set A, there is only one number (out of 5) that makes the sum equal 9. Therefore, the probability is 1/5. The answer is B.
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Joined: 04 Jan 2008
Posts: 47
Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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22 Jan 2008, 16:10
2
B
total 20 ways to sum,
4 of the sums equals 9
so, prob is 4/20= 1/5= .20
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Joined: 24 Feb 2012
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Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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24 Feb 2012, 20:54
walker wrote:
B
$$p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20$$
or
$$p=\frac{4}{4}*\frac{1}{5}=0.20$$
or
$$p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20$$
Although the calculation seems overkill, do you mind going over the logic behind the fractions above, please?
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Joined: 08 Jun 2011
Posts: 72
Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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03 May 2012, 04:11
Bunuel wrote:
fortsill wrote:
walker wrote:
B
$$p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20$$
or
$$p=\frac{4}{4}*\frac{1}{5}=0.20$$
or
$$p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20$$
Although the calculation seems overkill, do you mind going over the logic behind the fractions above, please?
I'd offer another approach which I hope will also clarify the above.
A = {2,3,4,5}
B = {4,5,6,7,8}
2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9?
A. 0.15
B. 0.20
C. 0.25
D. 0.30
E. 0.33
Rearrange the first set:
A = {5, 4, 3, 2}
B = {4, 5, 6, 7, 8}
As you can see numbers in each column (the numbers of the same color) give the sum of 9. So there are 4 such pares possible, total # of pairs is 4*5=20. P=favorable/total=4/20=0.2.
Or: we can select ANY number from set A (4/4=1) but in this case we must select its matching pair from set B (the number of the same color) and since there are only one matching pair of this particular number in B then the probability of this is 1/5. So, overall: P=1*1/5.
Hope it's clear.
Bunuel,
I am confused.
We are selecting a pair from a total of 9 numbers or 7 numbers (if we eliminate duplicates). So we will be getting for total number of ways, in either case
2 from 9 is 36
and
2 from 7 is 21
So we have [ (2, 7),(3,6), (4,5) ] 3 pairs or
[ (2, 7),(3,6), (4,5) (5,4) ] 4 pairs if we don't eliminate duplicates
Either way, I am getting the wrong answer
3/36 and 4/21 are both wrong.
Why is that wrong? and why aren't we eliminating duplicates because essentially (4,5) and (5,4) are the same same.
How is it different from this one (the general concept)?
if-two-of-the-four-expressions-x-y-x-5y-x-y-5x-y-are-92727.html#p713823
Thank you
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Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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03 May 2012, 08:29
1
2
2 from 9 is 36
and
2 from 7 is 21
You are not selecting 2 numbers from 9 numbers. You are selecting 1 number from 4 numbers and 1 number from 5 numbers. Mind you, they are different.
While selecting 2 numbers from {1, 2, 3, 4, 5, 6, 7, 8, 9}, you can select (3, 4)
While selecting 1 from 4 and 1 from 5: {1, 2, 3, 4} and {5, 6, 7, 8, 9}, you cannot select (3, 4).
That's the problem number 1.
Why is that wrong? and why aren't we eliminating duplicates because essentially (4,5) and (5,4) are the same same.
Secondly, the two given sets are distinct:
A = {2,3,4,5}
B = {4,5,6,7,8}
When we pick a number from A and from B, we get {4, 5} or we can do it in this way {5, 4}. These are two different cases. In the first case, the 4 comes from A and in the second case, it comes from B.
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Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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20 Aug 2012, 03:12
Responding to a pm:
No, there will not be 8 different pairs.
A = {2,3,4,5}
B = {4,5,6,7,8}
If you select 2 from A, from B you must select 7.
If you select 3 from A, from B you must select 6.
If you select 4 from A, from B you must select 5.
If you select 5 from A, from B you must select 4.
The total number of ways in which the sum can be 9 is 4.
The question was whether the last two cases are distinct or not. Is 4 from A and 5 from B the same as 4 from B and 5 from A.
Since 4 and 5 come from different sets in the two cases, they represent two different cases and should be counted as such.
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Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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15 Nov 2012, 02:25
HI Karishma,
I understood the question, got 0.20, but I have a doubt.
At first, I got 6 pairs - (2,7);(3,6);(4,5);(5,4);(6,3);(7,2), leading, P= 6/20= 0.3 which is also one of the options.
How do we come too know that we need not include duplicate pairs??
I understand from the explanations that 6,3 & 7,2 are not counted, but how can i figure out looking at the q?
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Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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15 Nov 2012, 02:33
HI Karishma,
I understood the question, got 0.20, but I have a doubt.
At first, I got 6 pairs - (2,7);(3,6);(4,5);(5,4);(6,3);(7,2), leading, P= 6/20= 0.3 which is also one of the options.
How do we come too know that we need not include duplicate pairs??
I understand from the explanations that 6,3 & 7,2 are not counted, but how can i figure out looking at the q?
You need to select a pair {a, b} such that their sum is 9.
a is selected from set A and b is selected from set B.
a can take a value from set A only i.e. one of 2/3/4/5
b can take a value from set B only i.e. one of 4/5/6/7/8
How do you select {6, 3}? a cannot be 6.
A selection is different from another when you select different numbers from each set.
One selection is {4, 5} - 4 from A, 5 from B
Another selection is {5, 4} - 5 from A, 4 from B
One selection is {3, 6} - 3 from A, 6 from B
There is no such selection {6, 3} - A has no 6.
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Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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15 Nov 2012, 02:40
when Q says that you need to select one integer from Set A & one from Set B, going by the explanation, does it mean that it
cannot be (b,a) and it will always be (a,b) in terms of order??
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Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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15 Nov 2012, 02:46
when Q says that you need to select one integer from Set A & one from Set B, going by the explanation, does it mean that it
cannot be (b,a) and it will always be (a,b) in terms of order??
It doesn't matter how you write it. One number is from A and one is from B. You know which number is from A and which is from B. You can write it is {a, b} or {b, a}, it doesn't matter.
{4, 5} - 4 is from A and 5 is from B
{5, 4} - 5 is from A and 4 is from B
These two are different.
{2, 6} - 2 is from A and 6 is from B
{6, 2} - 2 is from A and 6 is from B
These two are same.
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Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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02 Dec 2012, 17:51
7
I pick a number from Set A. No matter which number I pick, my chance of chosing the right number (which gives A+B = 9) in Set B will be 1/5.
Therefore 1/5 = 0.20 => B
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Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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06 Jul 2013, 07:22
walker wrote:
B
$$p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20$$
or
$$p=\frac{4}{4}*\frac{1}{5}=0.20$$
or
$$p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20$$
I made it right, but I don´t understand this approach (Combinatorics). Why the denominator in the first solution $$p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20$$ is $$\ C^4_1$$ Some help? Thanks!
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Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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03 May 2016, 04:46
2
A = {2, 3, 4, 5}
B = {4, 5, 6, 7, 8}
Two integers will be randomly selected from the sets above, one integer from set A and one integer from set B. What is the probability that the sum of the two integers will equal 9 ?
(A) 0.15
(B) 0.20
(C) 0.25
(D) 0.30
(E) 0.33
Solution:
To determine the probability that the sum of the two integers will equal 9, we must first recognize that probability = (favorable outcomes)/(total outcomes).
Let’s first determine the total number of outcomes. We have 4 numbers in set A, and 5 in set B, and since we are selecting 1 number from each set, the total number of outcomes is 4 x 5 = 20.
For our favorable outcomes, we need to determine the number of ways we can get a number from set A and a number from set B to sum to 9. We are selecting from the following two sets:
A = {2, 3, 4, 5}
B = {4, 5, 6, 7, 8}
We will denote the first number as from set A and the second from set B. Here are the pairings that yield a sum of 9:
2,7
3,6
4,5
5,4
We see that there are 4 favorable outcomes. Thus, our probability is 4/20 = 0.25, Answer C.
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Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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22 Feb 2018, 13:04
Hi All,
Probability questions are based on the probability formula:
(Number of ways that you "want") / (Total number of ways possible)
Since it's usually easier to calculate the total number of possibilities, I'll do that first. There are 4 options for set A and 5 options for set B; since we're choosing one option from each, the total possibilities = 4 x 5 = 20
Now, to figure out the number of duos that sum to 9:
2 and 7
3 and 6
4 and 5
5 and 4
4 options that give us what we "want"
4/20 = 1/5 = 20% = .2
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Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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10 Apr 2018, 05:45
Bunuel wrote:
A = {2, 3, 4, 5}
B = {4, 5, 6, 7, 8}
Two integers will be randomly selected from the sets above, one integer from set A and one integer from set B. What is the probability that the sum of the two integers will equal 9 ?
(A) 0.15
(B) 0.20
(C) 0.25
(D) 0.30
(E) 0.33
The total number of pairs possible is 4*5=20. Out of these 20 pairs only 4 sum up to 9: (2, 7); (3, 6), (4, 5) and (5, 4). The probability thus is 4/20=0.2.
Hi Bunuel
Is {4,5} and {5,4} not same?
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Re: Two integers will be randomly selected from the sets above, one intege [#permalink]
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10 Apr 2018, 05:51
@s wrote:
Bunuel wrote:
A = {2, 3, 4, 5}
B = {4, 5, 6, 7, 8}
Two integers will be randomly selected from the sets above, one integer from set A and one integer from set B. What is the probability that the sum of the two integers will equal 9 ?
(A) 0.15
(B) 0.20
(C) 0.25
(D) 0.30
(E) 0.33
The total number of pairs possible is 4*5=20. Out of these 20 pairs only 4 sum up to 9: (2, 7); (3, 6), (4, 5) and (5, 4). The probability thus is 4/20=0.2.
Hi Bunuel
Is {4,5} and {5,4} not same?
(4, 5) is 4 from A and 5 from B.
(5, 4) is 5 from A and 4 from B.
Those are two different cases.
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Re: Two integers will be randomly selected from the sets above, one intege [#permalink] 10 Apr 2018, 05:51
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# MKT 460 Exam 2 Review - MKT 460 Exam 2 Review Sampling...
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MKT 460 Exam 2 Review Sampling Census: measures the entire population Target population: entire body of interest Sample: selection of part of the population; used to draw general conclusions about the entire population Sampling error = population mean – sample mean the difference between the population and the sample Non-sampling error = True sample mean – observed sample mean errors in the sample data Six steps of drawing a sample 1. Define the target population – what is the population to which your research is directed 2. Identify the sampling frame – where to draw the sample from 3. Select a sampling plan – how to select subjects 4. Determine the sample size – how many? 5. Select the sample elements – draw the sample 6. Collect the data from the sample – measure each unit in the sample Target population: - Which units to be included? (households, children, students, main shopper, stores, etc) - Narrow or broad? more narrow harder and costlier it is to collect Sampling frame: - Listing from which sample will be drawn (phonebooks, directories, list of registered companies, customer database) - NOT always used Sampling plan: magnitude of the sampling error reliability ) of your results - Non-probability unknown probability – probability of units being included in the sample is unknown o canNOT be inferred NOT needed less expensive 1. Convenience sample: “ accidental” – elements enter by accident ie mall intercept, popup window on website 2. Judgment sample: “ purposive” – elements are hand-picked 1) to obtain better results or 2) use expert opinion 3. Quota sample: “to improve” representativeness. Sample is similar to population on a number of variables. Precise description a. Disadvantage – 1) there may be other variables based on which the sample is not representative; 2) sampling units are selected by the interviewer o Drawbacks: sample units selected by interviewer, usually those most accessible, at home, with time, etc… Sample may not be representative of target population
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## This note was uploaded on 09/29/2011 for the course MKT 460 taught by Professor Staff during the Spring '08 term at University of Texas.
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Ask a homework question - tutors are online | 592 | 2,698 | {"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-2017-22 | longest | en | 0.898529 |
https://en.citizendium.org/wiki/Pump | 1,680,005,205,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948858.7/warc/CC-MAIN-20230328104523-20230328134523-00542.warc.gz | 273,070,532 | 12,862 | # Pump
Main Article
Discussion
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This editable Main Article is under development and subject to a disclaimer.
A pump is a device used to move and/or pressurize fluids such as liquids or slurries. A pump moves fluids from lower pressure to higher pressure and overcomes the difference in pressure by adding energy to the system. Therefore a pump needs something or someone to provide energy (or power) to operate it. A pump's casing has separate openings for the fluid to enter and exit the pump. A pump's inlet and outlet are commonly described as suction and discharge. Compressors, which are used to compress and move gases, have similarities to pumps.
Special pumps called vacuum pumps are used to evacuate fluids, usually gases, from a confined space to create a vacuum within.
## History
The earliest type of pump was the Archimedes screw, first used by Sennacherib, King of Assyria, for the water systems at the Hanging Gardens of Babylon and Nineveh in the 7th century BC, and later described in more detail by Archimedes in the 3rd century BC.[1] In the 13th century AD, al-Jazari described and illustrated different types of pumps, including a reciprocating pump, a double-action pump, a water pump and a piston pump.[2][3]
## Types
Pumps fall into two major groups: rotodynamic pumps and positive displacement pumps. Their names describe the method for moving a fluid.
### Rotodynamic pumps
Rotodynamic pumps are based on bladed impellers which rotate within the fluid to impart a tangential acceleration to the fluid and a consequent increase in the energy of the fluid. The purpose of the pump is to convert this energy into pressure energy of the fluid to be used in the associated piping system.
#### Centrifugal Pumps
(CC) Drawing: R. Castelnuovo
Cut-away view of a centrifugal pump.
Centrifugal pumps are rotodynamic pumps which convert mechanical energy into hydraulic energy by centripetal force on the liquid. Typically, a rotating impeller increases the velocity of the fluid. The casing, or volute, of the pump then acts to convert this increased velocity into an increase in pressure. So the mechanical energy is converted into a pressure head by centripetal force, the pump is classified as centrifugal pump. Such pumps are found in virtually every industry, and in domestic service in developed countries for washing machines, dishwashers, swimming pools and water supply.
A wide range of designs are available, with constant and variable speed drivers. Horizontal shafts are the most common. Single-stage pumps are usual in the smaller ratings. Pumps with up to 11 stages are in service. A demanding duty is pumping boiler feedwater, and today's designs are typically 3 - 4 stage, with speeds of up to 6,000 revolutions per minute.
After electrical motors, centrifugal pumps are arguably the most common machine, and they are a significant user of energy. It is not unusual for a pump to be found to be over-sized, having been selected poorly for its intended duty. Running an over-sized constant speed pump throttled causes energy waste. A condition monitoring test can detect this condition and help size a smaller impeller, either new, or by machining the initial one, to achieve energy reduction.
Pumps also wear internally, at a rate varying with the liquid pumped, materials of construction and operating regime. Again, condition monitoring can be applied to detect and quantify the extent and rate of wear and also help decide when overhaul is justified on an energy-saving basis.
### Positive displacement pumps
(PD) Image: Milton Beychok
A lobe pump with three-lobe rotors.
A positive displacement pump causes a fluid to move by periodically trapping a fixed amount of fluid and then forcing (displacing) that trapped volume into the discharge pipe. The periodic fluid displacement results in a direct increase in pressure. A positive displacement pump can be further classified as either
#### Roots-type pumps
The low pulsation rate and gentle performance of the Roots-type positive displacement pump is achieved due to a combination of its two 90° helical twisted rotors, and a triangular shaped sealing line configuration, both at the point of suction and at the point of discharge. This design produces a continuous flow with equal volume. High capacity industrial "air compressors" have been designed to employ this principle as well as most "superchargers" used on internal combustion engines.
Cross-sectional diagrams of "simplex" reciprocating-type pump head towards end of suction and discharge strokes. Blue arrow shows direction of flow allowed through check valves.
#### Reciprocating-type pumps
Reciprocating-type pumps use a piston and cylinder arrangement with suction and discharge check valves integrated into the pump. Pumps in this category range from "simplex" having one cylinder, to in some cases "quad" having four cylinders or more. Most reciprocating-type pumps are "duplex" having two cylinders or "triplex" having three cylinders. Furthermore, they are either "single acting" independent suction and discharge strokes or "double acting" suction and discharge in both directions. The pumps can be powered by air, steam or through a belt drive from an engine or motor. This type of pump was used extensively in the early days of steam propulsion (19th century) as boiler feed water pumps. Though still used today, reciprocating pumps are typically used for pumping highly viscous fluids such as heavy oils or concrete or as metering or chromatography pumps.
#### Compressed air-powered double-diaphragm pumps
Another modern application of positive displacement pumps are compressed air-powered double-diaphragm pumps. They are relatively inexpensive, and are used extensively for pumping water out of bunds, or pumping low volumes of reactants out of storage drums.
#### Peristaltic pumps
A peristaltic pump has a flexible compressible tube (for each stream) and rollers which roll along the length of the tube compressing it to carry a plug of fluid along from inlet to outlet. These are commonly rather simple, low pressure, low flow rate pumps used sometimes for select applications.
## Application
Pumps are used throughout society for a variety of purposes. Early applications included the use of the windmill or watermill to pump water. Today, the pump is used for irrigation, water supply, gasoline supply, air conditioning systems, chemical fluid movement, sewage movement, flood control, marine services, etc.
Because of the wide variety of applications, pumps have a plethora of shapes and sizes: from very large to very small, from high pressure to low pressure, and from high volume to low volume.
(CC) Image: Manco Capac
Diagram of a hand pump.
## Pumps as public water supplies
One type of pump once common worldwide was a hand-powered water pump over a water well where people could use it to extract water, before most houses had individual water supplies.
Today, hand operated village pumps are considered the most sustainable low cost option for safe water supply in resource poor settings, often in the rural areas of developing countries. A hand pump opens access to deeper groundwater that is often not polluted and also improves the safety of a well by protecting the water source from contaminated buckets.
## Simple hand pumps
Commonly small, simple, cheap, disposable hand pumps are used as dispensers on consumer product bottles. These hand pumps dispense a small amount of liquid, lotion, gel, etc. in one or more strokes when operated by hand. Their design is a simplified version of the hand-powered water pump shown at right. Larger non-disposable versions of hand pumps are used to pump out fluids from barrels or similar containers. Also hand- or foot-operated pumps are used to fill tires, air mattresses, etc. with air.
## Power source
Pumps have been powered by water flow, internal combustion engines, electric motors, manually as with hand pumps, steam turbines, and by wind power. Solar power has been used to power pumps for remote locations.[4]
## References
1. Stephanie Dalley and John Peter Oleson (January 2003). "Sennacherib, Archimedes, and the Water Screw: The Context of Invention in the Ancient World", Technology and Culture 44 (1).
2. Al-Jazari, The Book of Knowledge of Ingenious Mechanical Devices : Kitáb fí ma'rifat al-hiyal al-handasiyya, translated by P. Hill (1973). Springer.
3. Derek de Solla Price (1975). Review of Ibn al-Razzaz al-Jazari, The Book of Knowledge of Ingenious Mechanical Devices. Technology and Culture 16 (1), p. 81.
4. SunMill Power Pumps | 1,820 | 8,676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-14 | latest | en | 0.93049 |
https://homeschoolingblogs.com/area-aliens-printable-counting-exercise-for-preschoolers/ | 1,701,768,368,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100550.40/warc/CC-MAIN-20231205073336-20231205103336-00572.warc.gz | 342,703,409 | 22,617 | # Area Aliens Printable Counting Exercise for Preschoolers
July 24, 2023
Homeschooling Blogs
Looking for an engaging and effective way to teach counting skills to young learners? This ‘Counting Space Aliens’ printable activity is a great way to get learners to practice their counting and number recognition. With this FREE counting printable, featuring space and aliens, children will have a blast while practicing their number sense. Simply print this counting activity for preschoolers and kindergartners to practice how to count to 10 and number recognition today.
## Counting activity for preschoolers
This fun counting activity is designed to make learning math fun and interactive for students. By combining numbers and objects, children can develop their counting skills while having a great time. In this blog post, we will explore the benefits of this activity and how this counting math activities for preschoolers can be a valuable resource for teachers in the classroom. Whether you’re an educator searching for new teaching tools or a parent wanting to support your child’s math learning at home, this printable activity is sure to be a hit with preschool, pre-k, and kindergarten age children.
## Counting Activity Free Printable
Start by scrolling to the bottom of the post, under the terms of use, and click on the text link that says >> _____<<. The space aliens counting printables pdf file will open in a new window for you to save the freebie and print the template.
## Teaching Counting and Number Recognition
When it comes to math, recognizing numbers is the first step towards building a solid mathematical foundation. This activity provides a fantastic opportunity for young learners to enhance their number recognition skills in an interactive way. Through this activity, kids get to visually connect the dots between the numerical symbols and actual quantities they represent. This hands-on experience makes number recognition more tangible and meaningful, ultimately helping children develop a strong numerical foundation.
By practicing number recognition such as matching counting objects to the number, children gain confidence in identifying and understanding numbers. As they progress, they will be better equipped to apply this knowledge in various mathematical contexts, such as addition, subtraction, and problem-solving. For a fun number sense activity check out this FREE Number Sense printable activity.
## Alien Math
This printable covers numbers from 1 to 10. This is a super low-prep printable center that can be put together very quickly. Each card consists of aliens and a blank white space for learners to match to the number card.
These subitizing clip cards are a great way to practice counting from numbers 1 to 10.
## Counting Activities for Preschoolers
• Free printable (download it at the end of the blog post)
• Printer
• Laminator
• Scissors
• Magnetic numbers
• Optional: Manipulatives for counting
Print out the pages required. Laminate if desired. This is recommended if you’re using this printable with multiple learners. Cut apart to create the separate counting cards.
## Counting math activities for preschoolers
Start off by explaining the activity to your learners. Explain that they will have to count the number of aliens and figure out the corresponding number to match it with.
There are a few ways this printable can be used:
1.
• Use the provided number cards in the printable and place it on the task card.
• Use magnetic or movable numbers and learners can place those on the task cards.
• Place the matching number of objects or manipulatives on the task card.
This printable is a great way to practice counting and number recognition and also helps with one-to-one number to object correspondence.
## Number activities for preschoolers
Before you grab your free pack you agree to the following:
• This set is for personal and classroom use only.
• This printable set may not be sold, hosted, reproduced, or stored on any other website or electronic retrieval system.
• Graphics Purchased and used with permission from
• All material provided on this blog is copyright protected.
> Counting Alients Printable <<
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### Methods to SIMPLIFY HOMESCHOOL || my suggestions for a extra "productive" day
homeschoolmom #simplify #homeschool Welcome, friend! Today I want to share my tips for simplifying homeschool. Each and ... | 986 | 4,822 | {"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-2023-50 | latest | en | 0.9137 |
https://byjus.com/question-answer/consider-a-system-s-with-input-x-n-and-output-y-n-related-by-y/ | 1,714,031,906,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297290384.96/warc/CC-MAIN-20240425063334-20240425093334-00350.warc.gz | 131,668,753 | 35,323 | 1
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Question
# Consider a system S with input x[n] and output y[n] related by y[n]=x[n]{g[n]+g[n−1]} Select the correct statement about system S
A
If g[n] = 1 for all n, then S is time invariant.
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B
If g[n] = n for all n, then S is time variant.
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C
If g[n] = δ[n], then S is time variant.
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D
If g[n] = 1 + (-1)n, then S is time invariant.
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Solution
## The correct option is D If g[n] = 1 + (-1)n, then S is time invariant.(a) g[n] = 1 g[n - 1] = 1 So, y[n] = x[n] [1 + 1] = 2 x[n] Thus system is time invariant (b) g[n] = n So, g[n - 1] = (n - 1) Now, y[n] = x[n] [n + n -1] x[n] [2n - 1] which is time variant. (c) g[n]=1+(−1)n g[n−1]=1+(−1)n−1 y[n]=x[n][1+(−1)n+1+(−1)n−1] =x[n][2] = 2x [n] which is time invariant (d) g[n] = δ[n] g[n-1] = δ[n−1] g[n]=x[n][δ[n]+δ[n−1]] which is time variant.
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Join BYJU'S Learning Program | 482 | 1,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-18 | latest | en | 0.842519 |
http://stackoverflow.com/questions/1435975/how-can-i-round-down-a-number-in-javascript | 1,464,697,780,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464051299749.12/warc/CC-MAIN-20160524005459-00213-ip-10-185-217-139.ec2.internal.warc.gz | 273,748,894 | 24,090 | # How can I round down a number in Javascript?
How can I round down a number in JavaScript?
`math.round()` doesn't work because it rounds it to the nearest decimal.
I'm not sure if there is a better way of doing it other than breaking it apart at the decimal point at keeping the first bit. There must be...
-
Round towards zero or towards negative infinity? – Daniel Brückner Sep 16 '09 at 23:09
``````Math.floor()
``````
-
It's also the slowest method; if you need to perform a lot of these, use the bitwise | operator (see my post). – geraldalewis Sep 16 '09 at 23:18
The | operator won't work on numbers larger than 2147483647. – Robert L Sep 16 '09 at 23:47
The | operator also rounds towards zero, not negative infinity. – Mike Godin Mar 3 '14 at 21:58
Round towards negative infinity - `Math.floor()`
``````+3.5 => +3.0
-3.5 => -4.0
``````
Round towards zero - usually called `Truncate()`, but not supported by JavaScript - can be emulated by using `Math.ceil()` for negative numbers and `Math.floor()` for positive numbers.
``````+3.5 => +3.0 using Math.floor()
-3.5 => -3.0 using Math.ceil()
``````
-
Thank you for the completeness but the capitalization is wrong ... and in java-script that makes a HUGE difference. Otherwise I would have upvoted here. – George Apr 19 '12 at 3:50
I have updated the answer so that the capitalization is now correct. – chasen Jun 8 '12 at 18:02
@George HUGE or huge? :D – m93a Oct 24 '13 at 15:16
You can get the same effect as round-to-zero via `x | 0`. – Ahmed Fasih May 6 at 2:59
`Math.floor()` will work, but it's very slow compared to using a bitwise `OR` operation:
``````var rounded = 34.923 | 0;
``````
EDIT `Math.floor()` is not slower than using the | operator. Thanks to Jason S for checking my work.
Here's the code I used to test:
``````var a = [];
var time = new Date().getTime();
for( i = 0; i < 100000; i++ ) {
//a.push( Math.random() * 100000 | 0 );
a.push( Math.floor( Math.random() * 100000 ) );
}
var elapsed = new Date().getTime() - time;
alert( "elapsed time: " + elapsed );
``````
-
??? I just ran jsdb (www.jsdb.org) which uses Spidermonkey 1.7, and ran a loop to sum up the floor'ed value of x[i] on an array of 100000 floating point numbers, first with Math.floor(), then with bitwise or as you suggest. It took approx the same time, 125 msec. – Jason S Sep 18 '09 at 14:49
Just repeated the test with 500000 floating point numbers, it took approx the same time, approx 625 msec. – Jason S Sep 18 '09 at 14:51
So I don't see how 1.25usec is very slow. – Jason S Sep 18 '09 at 14:53
Can't argue with your data :) I think I may be have confused JS's implementation with ActionScript's (built on EcmaScript; obviously implementation differs). Thanks for checking my work! – geraldalewis Sep 27 '09 at 3:45
They don't do the same thing, either. `|` converts to a 32-bit integer, truncating; `Math.floor` rounds down. jsfiddle.net/minitech/UVG2w – Ryan O'Hara Sep 19 '12 at 14:16
To round down towards negative infinity, use:
``````rounded=Math.floor(number);
``````
To round down towards zero (assuming the number fits in an int), use:
``````rounded=number|0;
``````
To round down towards zero (for any number), use:
``````if(number>0)rounded=Math.floor(number);else rounded=Math.ceil(number);
``````
-
You can try to use this function if you need to round down to a specific number of decimal places
``````function roundDown(number, decimals) {
decimals = decimals || 0;
return ( Math.floor( number * Math.pow(10, decimals) ) / Math.pow(10, decimals) );
}
``````
examples
``````alert(roundDown(999.999999)); // 999
``````
-
I think a one-liner like this doesn't require a function. – Hubert Grzeskowiak Apr 9 '15 at 14:19
Rounding a `number` towards `0` can be done by subtracting its signed fractional part `number % 1`:
``````rounded = number - number % 1;
``````
Like `Math.floor` (rounds towards `-Infinity`) this method is perfectly accurate.
There are differences in the handling of `-0`, `+Infinity` and `-Infinity` though:
``````Math.floor(-0) => -0
-0 - -0 % 1 => +0
Math.floor(Infinity) => Infinity
Infinity - Infinity % 1 => NaN
Math.floor(-Infinity) => -Infinity
-Infinity - -Infinity % 1 => NaN
``````
-
``````Math.floor(1+7/8)
``````
-
1+7/8 = 1 - Not much need for Math.floor() there :) – Jason Berry Sep 16 '09 at 23:16
Actually it's (7/8)+1 which is not 1. Thank you 3rd grade algebra – Joe Philllips Sep 16 '09 at 23:24
Umm, please actually try this in a javascript program. I did. Display (1 + 7/8) and you will see 1.875. Math.round(...) is 2, Math.floor(...) is 1. What are you guys talking about? – DigitalRoss Sep 16 '09 at 23:48
Or open the Firefox Error Console. Or Firebug. It isn't hard to try. I tried it. 1 + 7/8 is 1.875 in js. Did you possibly forget that all math in js is in floating point? – DigitalRoss Sep 16 '09 at 23:53
It's probably easy to forget that javascript does everything in floating point. In many other languages 1+7/8 is 1, but in js it really is 1.875. – DigitalRoss Sep 17 '09 at 1:05
Was fiddling round with someone elses code today and found the following which seems rounds down as well:
``````var dec = 12.3453465,
int = dec >> 0; // returns 12
``````
For more info on the Sign-propagating right shift(>>) see MDN Bitwise Operators
It took me a while to work out what this was doing :D
But as highlighted above, Math.floor() works and looks more readable in my opinion.
-
It also silently kills your number if it doesn't fit in 32 bits. Chromium console: 99999999999999999999999|0 => -167772160 – Matthias Urlichs Dec 13 '15 at 18:14 | 1,673 | 5,604 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2016-22 | latest | en | 0.848438 |
https://www.noodle.com/articles/get-free-geometry-help-with-these-online-videos | 1,542,133,695,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741340.11/warc/CC-MAIN-20181113173927-20181113195927-00064.warc.gz | 946,431,701 | 11,981 | # Get Free Geometry Help With These Online Videos
Many students struggle with math, and the subject can get even harder with shapes and diagrams. Angles, corners, curves, and shapes of all sorts are enough to confuse even collegiate students.
So what’s a novice geometry student supposed to do when homework is assigned and there’s no teacher to raise a hand for? There are great teachers online who can help everyone, from middle to high school students and beyond, with explanations for common — and even some uncommon — geometry problems.
Also check out Get Free Algebra Help With These 5 Online Videos
## Best Middle School Homework Help Videos for Geometry
### Circumference Situation
PBS for Kids has tons of math and geometry related videos for kids. Although intended for children slightly younger than middle school, the lessons are easy to understand, even on complex subjects. In addition to this one on circumference, you can find many more geometry topics here.
### Geometry Video Lessons
Get tons of short videos on geometry here from Tutor-USA.com. They are typically a few minutes long and easy to understand. Selections include “Introduction to Parallel Lines in Geometry,” “How Geometry is Used in Surveying,” and dozens more.
### Pythagorean Theorem
Learn all about how to calculate the angles in a right triangle with this video. The instructor demonstrates with a formula that is simple enough in theory, but one that many middle schoolers have struggled to grasp. Math Videos Online also has more geometry and related videos to view.
### The Geometry Rap
Having trouble remembering all the geometry terms? Love rap music? This video by WyzAnt can give you a fun way to remember the shapes, formulas, and more while you hum along to the song. There are also other fun videos to watch on the channel.
## Best High School Homework Help Videos for Geometry
### Euclid as The Father of Geometry
Khan Academy has an entire section with dozens of geometry video. The very first is on Euclid and the history of geometry, and many more follow.
### Introduction to Geometry (1.1)
Jeff Nelson is a professional math teacher who introduces students to the basics of geometry in this 14 minute video that includes lines, planes, angles, and more. He also has many other videos with lots more help in the subject.
### Algebra: Formulas From Geometry
The Video Math Tutor prides himself on being a teacher like no other. He offers lots of great math tutorials from his YouTube channel of over 21,000 subscribers. In this 23-minute entry, he shows you how to combine many simple algebra equations into the subject of geometry. Even college students have thanked him in the comments for making the topic so accessible.
### Language and Notation of Basic Geometry
Another entry from Khan Academy, this is one of the most popular geometry videos on YouTube. It lays out basic ideas in geometry and how they are represented with symbols.
For more math advice and tips, visit Noodle’s math topic page. | 606 | 3,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-47 | longest | en | 0.940369 |
http://wiki.zcubes.com/Manuals/calci/SUMBY | 1,652,988,260,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662529658.48/warc/CC-MAIN-20220519172853-20220519202853-00730.warc.gz | 59,371,875 | 6,277 | # Manuals/calci/SUMBY
SUMBY(Number1,Number2)
• and are any real numbers.
## Description
• This function shows the sum of the first two numbers.
• In , and are any real numbers.
• The mathematical symbol for summation is .
• The arguments can be or .
• We can give any cell references like column or row number, logical values like true or false, an array, or a result from any other functions in Arguments.
• We must give atleast one argument.
• In the arguments, the first two numbers only considered.
## Examples
1. SUMBY(90,34) = 124
2. SUMBY(120,-45) = 75
3. SUMBY(3/4,7/8) = 1.625
SUM | 166 | 596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-21 | latest | en | 0.681977 |
http://www.maplesoft.com/support/help/MapleSim/view.aspx?path=evalhf/fortran | 1,462,305,798,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860121776.48/warc/CC-MAIN-20160428161521-00025-ip-10-239-7-51.ec2.internal.warc.gz | 653,605,602 | 18,049 | evalhf/fortran - Maple Help
Home : Support : Online Help : Programming : Resource Management : evalhf : evalhf/fortran
evalhf/fortran
and its relation to Fortran
Description
• Typically a Fortran program or a set of functions can be easily converted into a Maple function which can be executed by evalhf. The speed of a function executed by evalhf compared to a function compiled in optimized Fortran is on a ratio anywhere between 1:5 to 1:50. Converting Fortran into Maple-evalhf is still one or two orders of magnitude faster than running the equivalent code under standard Maple.
• The Whetstone benchmark gives a ratio of 1:35 in favor of compiled Fortran (under a VAX running Unix BSD 4.3).
• The following differences and problems should be observed when converting Fortran into Maple-evalhf:
• The only type handled by evalhf is floating point (double precision). Integers and Booleans are treated as floats also.
• There is no equivalent of the common or equivalence statements.
• Any Fortran expression which will evaluate over the integers, in particular expressions assigned to integer variables, should be surrounded by the function trunc().
• Array declarations are dynamic with the array() function, and not static.
• All variables should be declared as local variables.
• Fortran may return values through assigned arguments. This will not work under evalhf. Arrays with a single element may solve this problem.
• Fortran is very liberal with the array dimensions and will allow a subroutine to work with an array that has a declaration different from the caller. This is not allowed in Maple-evalhf; furthermore, arrays can only be passed as a whole, not just by the mention of a single element.
• Returned values in Maple are the last value computed. In Fortran these values are assigned to a variable with the same name as the function.
• There is neither a read nor format statement. | 414 | 1,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2016-18 | longest | en | 0.874007 |
https://lists.launchpad.net/yade-users/msg01831.html | 1,582,972,829,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148850.96/warc/CC-MAIN-20200229083813-20200229113813-00323.warc.gz | 443,446,403 | 4,002 | # yade-users team mailing list archive
## Re: explicit 2nd order scheme
```OK Bruno,
thank' s for this explaination.
Just two points:
i) this "explicit 2nd order finite difference scheme" is called half-step
"leap frog" scheme
(Hocknet 1970, Potter 1972, cited in
*Computer Simulation Of Liquids** by Allen
<http://www.angusrobertson.com.au/by/allen/549/>, Tildesley
<http://www.angusrobertson.com.au/by/tildesley/116428/>, M. P.
Allen<http://www.angusrobertson.com.au/by/m-p-allen/517561/>and D.
J. Tildesley <http://www.angusrobertson.com.au/by/d-j-tildesley/587821/>*)
ii) Because the scheme is a "leap frog" scheme, the velocity is not
computed at the same time as
the accelerations and the positions...
Accelerations and positions are known at time *t
*and velocities are known at the time *t-dt/2* (or *t+dt/2*)
In YADE, only the "sign" of the velocity is used (for damping).
Nevertheless, in YADE, we use velocities at time *t-dt/2 *to * *"damp"
motions
computed at time *t*.
The error made here is probably very very small.....
Gael
2009/9/17 Bruno Chareyre <bruno.chareyre@xxxxxxxxxxx>
> Re-sent to users list. Sorry for doublets.
>
> Jerome Duriez a écrit :
>
>> And how are velocities (in both schemes) computed ?
>>
>> If they are computed by the same mean, I can still not understand at all
>> what is the difference between these two shemes, except the fact that "2nd
>> order" sounds more serious than "1st order"...
>>
>> Héhé...
>
> In first order scheme, you would use current velocity for p(t+dt) = p(t) +
> v(t)*dt, THEN you would compute v(t+dt)=..., in 2nd order, you use the
> updated (second) value instead. A while ago, Yade was using the 1st order
> scheme for rotations, and it was the reason why we needed so small time
> steps.
>
> It seems many people have problems with this simple 2nd order explicit
> scheme. I send the equations in the attached file, with time step
> determination. I hope (should I?) it will stop endless discussions in my
>
> You will find the exact same scheme in any Cundall's paper, even if he
> doesn't explain it exactly the same way perhaps.
> You will not find the "leap-frog" naming in Cundall's papers though, and
> personnaly, I don't know what leap-frog is. I only know "explicit 2nd order
> finite difference scheme", and this is what's in Yade.
>
> Bruno
>
> --
>
> _______________
> Chareyre Bruno
> Maître de Conférences
>
> Grenoble INP
> Laboratoire 3SR - bureau E145
> BP 53 - 38041, Grenoble cedex 9 - France
> Tél : 33 4 56 52 86 21
> Fax : 33 4 76 82 70 43
> ________________
>
>
> _______________________________________________ | 781 | 2,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-10 | latest | en | 0.892244 |
http://mathyear2013.blogspot.com/2013/02/the-math-behind-climate-change-part-8.html | 1,526,829,492,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863626.14/warc/CC-MAIN-20180520151124-20180520171124-00590.warc.gz | 183,430,187 | 19,441 | ## Thursday, February 7, 2013
### The Math behind Climate Change: Part 8 Trendlines based on Oceanic Niño Intervals (ONI)
We start again with the data for average Winter temperatures in Greenland from 1955 to 2010. The start and end years were not chosen at random. They coincide with strong La Niña years as measured by climate scientists on a system called Oceanic Niño Intervals or ONI for short. The list of strong La Niña years in this range are as follows.
1955, 1973, 1975, 1988, 1999, 2010
The two years in the 1970s are too close together to make a meaningful trend and they have no El Niño between them, since 1974 was a weaker La Niña. I'll remove 1973 from the list and we get 1955, 1975, 1988, 1999 and 2010.
We use the years to create intervals. In each interval, we mark the highest temperature in red, the average in black and the lowest in blue. These trends are easy to read and take no difficult to explain math. (Note: I am not against difficult to explain math. The math of best fitting curves comes from Gauss and is completely legitimate. My complaint against it is how many different curves can be chosen and the possibilities of cherry picking to make a point.
The three trends tell slightly different stories. All agree that the 1988 to 1999 interval saw much cooler winters than any other span and that the recent span from 1999 to 2010 is by far the warmest interval. While the average in the final interval is only slightly higher than the second warmest interval from 1975 to 1988, the high and the low both increase significantly over the second warmest in each of those categories.
Here are the statements I read most often in the papers of climate skeptics and denialists.
1. The climate is not warming.
2. The warming trend is decreasing.
For all the data I will produce, I will end with a statement addressing these questions for all four seasons in a form
Greenland 1955-2010
Winter: getting warmer, trend increasing
Spring:
Summer:
Fall:
Tomorrow, we will see the data for all the seasons from Greenland.
#### 1 comment:
1. Drawing trend lines is one of the few easy techniques that really WORK. Prices respect a trend line, or break through it resulting in a massive move. Drawing good trend lines is the MOST REWARDING skill.
The problem is, as you may have already experienced, too many false breakouts. You see trend lines everywhere, however not all trend lines should be considered. You have to distinguish between STRONG and WEAK trend lines.
One good guideline is that a strong trend line should have AT LEAST THREE touching points. Trend lines with more than four touching points are MONSTER trend lines and you should be always prepared for the massive breakout!
This sophisticated software automatically draws only the strongest trend lines and recognizes the most reliable chart patterns formed by trend lines...
==> http://www.forextrendy.com?vgcvasdiugf9g87346
Chart patterns such as "Triangles, Flags and Wedges" are price formations that will provide you with consistent profits.
Before the age of computing power, the professionals used to analyze every single chart to search for chart patterns. This kind of analysis was very time consuming, but it was worth it. Now it's time to use powerful dedicated computers that will do the job for you:
==> http://www.forextrendy.com?vgcvasdiugf9g87346 | 765 | 3,368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2018-22 | latest | en | 0.934951 |
https://nrich.maths.org/public/leg.php?code=-99&cl=2&cldcmpid=5729 | 1,540,267,414,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516003.73/warc/CC-MAIN-20181023023542-20181023045042-00262.warc.gz | 763,507,732 | 9,899 | # Search by Topic
#### Resources tagged with Working systematically similar to More Carroll Diagrams:
Filter by: Content type:
Age range:
Challenge level:
### There are 338 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### A Square of Numbers
##### Age 7 to 11 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
### Arrangements
##### Age 7 to 11 Challenge Level:
Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....?
### Code Breaker
##### Age 7 to 11 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### One to Fifteen
##### Age 7 to 11 Challenge Level:
Can you put the numbers from 1 to 15 on the circles so that no consecutive numbers lie anywhere along a continuous straight line?
##### Age 5 to 11 Challenge Level:
Place six toy ladybirds into the box so that there are two ladybirds in every column and every row.
### Magic Potting Sheds
##### Age 11 to 14 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
### Red Even
##### Age 7 to 11 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
### How Long Does it Take?
##### Age 7 to 11 Challenge Level:
In this matching game, you have to decide how long different events take.
### How Old Are the Children?
##### Age 11 to 14 Challenge Level:
A student in a maths class was trying to get some information from her teacher. She was given some clues and then the teacher ended by saying, "Well, how old are they?"
### First Connect Three
##### Age 7 to 14 Challenge Level:
The idea of this game is to add or subtract the two numbers on the dice and cover the result on the grid, trying to get a line of three. Are there some numbers that are good to aim for?
### Teddy Town
##### Age 5 to 14 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### Ones Only
##### Age 11 to 14 Challenge Level:
Find the smallest whole number which, when mutiplied by 7, gives a product consisting entirely of ones.
### Multiples Grid
##### Age 7 to 11 Challenge Level:
What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
### More Magic Potting Sheds
##### Age 11 to 14 Challenge Level:
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
### Twinkle Twinkle
##### Age 7 to 14 Challenge Level:
A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour.
### A First Product Sudoku
##### Age 11 to 14 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
### Winning the Lottery
##### Age 7 to 11 Challenge Level:
Try out the lottery that is played in a far-away land. What is the chance of winning?
##### Age 11 to 14 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### Junior Frogs
##### Age 5 to 11 Challenge Level:
Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible?
### Greater Than or Less Than?
##### Age 7 to 11 Challenge Level:
Use the numbers and symbols to make this number sentence correct. How many different ways can you find?
### First Connect Three for Two
##### Age 7 to 14 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Multiples Sudoku
##### Age 11 to 14 Challenge Level:
Each clue in this Sudoku is the product of the two numbers in adjacent cells.
### Difference
##### Age 7 to 11 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
### Tetrafit
##### Age 7 to 11 Challenge Level:
A tetromino is made up of four squares joined edge to edge. Can this tetromino, together with 15 copies of itself, be used to cover an eight by eight chessboard?
### Counters
##### Age 7 to 11 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remover them. The winner is the last one to remove a counter. How you can make sure you win?
### Combining Cuisenaire
##### Age 7 to 11 Challenge Level:
Can you find all the different ways of lining up these Cuisenaire rods?
### Being Resourceful - Primary Number
##### Age 5 to 11 Challenge Level:
Number problems at primary level that require careful consideration.
### A Mixed-up Clock
##### Age 7 to 11 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
### Seven Flipped
##### Age 7 to 11 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Worms
##### Age 7 to 11 Challenge Level:
Place this "worm" on the 100 square and find the total of the four squares it covers. Keeping its head in the same place, what other totals can you make?
### Family Tree
##### Age 7 to 11 Challenge Level:
Use the clues to find out who's who in the family, to fill in the family tree and to find out which of the family members are mathematicians and which are not.
##### Age 11 to 14 Challenge Level:
Rather than using the numbers 1-9, this sudoku uses the nine different letters used to make the words "Advent Calendar".
### How Much Did it Cost?
##### Age 7 to 11 Challenge Level:
Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether.
##### Age 11 to 14 Challenge Level:
You need to find the values of the stars before you can apply normal Sudoku rules.
### Chocoholics
##### Age 7 to 11 Challenge Level:
George and Jim want to buy a chocolate bar. George needs 2p more and Jim need 50p more to buy it. How much is the chocolate bar?
### Train Carriages
##### Age 5 to 11 Challenge Level:
Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done?
### Plate Spotting
##### Age 7 to 11 Challenge Level:
I was in my car when I noticed a line of four cars on the lane next to me with number plates starting and ending with J, K, L and M. What order were they in?
### Palindromic Date
##### Age 7 to 11 Challenge Level:
What is the date in February 2002 where the 8 digits are palindromic if the date is written in the British way?
### Factor Lines
##### Age 7 to 14 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Ancient Runes
##### Age 7 to 11 Challenge Level:
The Vikings communicated in writing by making simple scratches on wood or stones called runes. Can you work out how their code works using the table of the alphabet?
### Symmetry Challenge
##### Age 7 to 11 Challenge Level:
Systematically explore the range of symmetric designs that can be created by shading parts of the motif below. Use normal square lattice paper to record your results.
##### Age 7 to 11 Challenge Level:
Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?
### Uncanny Triangles
##### Age 7 to 11 Challenge Level:
Can you help the children find the two triangles which have the lengths of two sides numerically equal to their areas?
### Shape Times Shape
##### Age 7 to 11 Challenge Level:
These eleven shapes each stand for a different number. Can you use the multiplication sums to work out what they are?
### Nine-pin Triangles
##### Age 7 to 11 Challenge Level:
How many different triangles can you make on a circular pegboard that has nine pegs?
### Finding Fifteen
##### Age 7 to 11 Challenge Level:
Tim had nine cards each with a different number from 1 to 9 on it. How could he have put them into three piles so that the total in each pile was 15?
### Twin Corresponding Sudokus II
##### Age 11 to 16 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections.
### More on Mazes
##### Age 7 to 14
There is a long tradition of creating mazes throughout history and across the world. This article gives details of mazes you can visit and those that you can tackle on paper.
### Open Squares
##### Age 7 to 11 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, focuses on 'open squares'. What would the next five open squares look like?
### Peaches Today, Peaches Tomorrow....
##### Age 11 to 14 Challenge Level:
Whenever a monkey has peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one. How long could he make his peaches last for? | 2,288 | 9,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-43 | latest | en | 0.914196 |
http://tobybartels.name/MATH-1600/2018FA/graphing/ | 1,544,994,117,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827992.73/warc/CC-MAIN-20181216191351-20181216213351-00591.warc.gz | 291,257,364 | 1,854 | # Graphing
If you want to have a complete graph of y = f(x), then these are all of the things that you should make sure show up:
• x = 0, if f(0) exists;
• x → ∞, if f(x) exists in that direction;
• x → −∞, if f(x) exists in that direction;
• x → c, if f(x) exists in that direction, whenever f is undefined or discontinuous at c;
• x → c+, if f(x) exists in that direction, whenever f is undefined or discontinuous at c;
• x = c, if f(c) exists, whenever f is undefined approaching c from either direction (or both);
• x = c, whenever f(c) = 0;
• x = c, if f(c) exists, whenever f′ is undefined or discontinuous at c;
• x = c, whenever f′(c) = 0;
• x = c, if f(c) exists, whenever f′′ is undefined or discontinuous at c;
• x = c, whenever f′′(c) = 0.
This should be sufficient whenever f is a twice-differentiable function whose domain is an interval, or more generally whenever f is piecewise twice-differentiable: a piecewise-defined function in which the domain of each piece is an interval and in which each piece is twice-differentiable except possibly at its endpoints. (There are weirder functions that can't be put in this form, but you shouldn't have to deal with them in this class.)
If you have a graphing calculator, then you may use it, but you still need to ensure that all of the features listed above appear. At the very least, this may require you to adjust the calculator's graphing window. If you're graphing by hand, then you'll get the best results if you know the values or limits of f, f′, and f′′ for all of these places or limits, but you should at least get f for all of them and f′ whenever you looked there because of something involving f′ or f′′. You can also look at places in between these (if f is defined there) for an even more precise graph.
Go back to the course homepage.
This web page was written between 2015 and 2018 by Toby Bartels, last edited on 2018 November 11. Toby reserves no legal rights to it.
The permanent URI of this web page is `http://tobybartels.name/MATH-1600/2018FA/graphing/`. | 556 | 2,041 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-51 | latest | en | 0.932984 |
http://gamedev.stackexchange.com/questions/tagged/math+java | 1,394,647,870,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394023864525/warc/CC-MAIN-20140305125104-00045-ip-10-183-142-35.ec2.internal.warc.gz | 81,653,549 | 19,988 | # Tagged Questions
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### Java Angle Math: Calculate how near the player looks to an entity [duplicate]
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### Circle to Circle collision - bad accuracy
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### Vector normalization gives very imprecise results
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### Making Ball Fling in a Curve Direction
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### Is this Rotation Matrix correct?
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### How do I derive the angular movement for a craft
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### Drawing isometric walls
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### Find the footprint of an isometric entity
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### Isometric - precise screen coordinates to isometric
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### Move projectile in direction the gun is facing [duplicate]
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### Point inside Oriented Bounding Box?
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### (int) Math.floor(x / TILESIZE) or just (int) (x / TILESIZE)
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### Get all triangles that are < N dist from you?
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### How to move an object along a circumference of another object?
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### How do I convert sin, cos and tan into an int in Java?
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### 3D: First Person Movement
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### Relating a point after an image is panned and zoomed
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### Suggest a simple Java math library for matrix operations for use with OpenGL (lwjgl) [closed]
I'm writing an OpenGL app with Java. I need to do some math for camera and frustum culling (for AABB). Could you suggest a simple and fast Java math library for that? | 1,858 | 8,056 | {"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-2014-10 | latest | en | 0.930448 |
http://mathoverflow.net/feeds/question/8509 | 1,369,489,097,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705955434/warc/CC-MAIN-20130516120555-00016-ip-10-60-113-184.ec2.internal.warc.gz | 163,982,324 | 2,088 | Martingales and Betting Strategies - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T13:38:07Z http://mathoverflow.net/feeds/question/8509 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8509/martingales-and-betting-strategies Martingales and Betting Strategies Chris Conidis 2009-12-11T01:14:41Z 2010-09-29T18:49:08Z <p>Does anyone know of a good introduction to the theory of martingales and betting strategies from the point of view of statistics and/or probability theory? I'm looking for something basic, with lots of examples. Thanks.</p> http://mathoverflow.net/questions/8509/martingales-and-betting-strategies/8510#8510 Answer by Steve Huntsman for Martingales and Betting Strategies Steve Huntsman 2009-12-11T01:20:48Z 2009-12-11T01:20:48Z <p>I don't know about statistics <em>per se</em> but the best introduction to martingales <em>period</em> is Williams' <em>Probability with Martingales</em>:</p> <p><a href="http://www.amazon.com/Probability-Martingales-Cambridge-Mathematical-Textbooks/dp/0521406056" rel="nofollow">http://www.amazon.com/Probability-Martingales-Cambridge-Mathematical-Textbooks/dp/0521406056</a></p> http://mathoverflow.net/questions/8509/martingales-and-betting-strategies/8520#8520 Answer by Alekk for Martingales and Betting Strategies Alekk 2009-12-11T02:07:46Z 2009-12-11T02:07:46Z <p>no doubt that <em>Probability with Martingales</em> is excellent, but I am not sure that this is a very satisfying book for someone interested in "the point of view of statistics". </p> http://mathoverflow.net/questions/8509/martingales-and-betting-strategies/8530#8530 Answer by jrshipley for Martingales and Betting Strategies jrshipley 2009-12-11T04:52:26Z 2009-12-11T04:52:26Z <p>I haven't read it, but I saw Glenn Shafer give a talk on his and Vladimir Vovk's game theoretical approach to probability. It could be an interesting/more intuitive approach than the usual measure theoretic approaches. They discuss martingales in several chapters of their book and I'm certain there's lots of applications/examples especially from finance. Might be worth a look: <a href="http://www.probabilityandfinance.com/chapters/toc.html" rel="nofollow">http://www.probabilityandfinance.com/chapters/toc.html</a></p> http://mathoverflow.net/questions/8509/martingales-and-betting-strategies/10772#10772 Answer by Yuri Bakhtin for Martingales and Betting Strategies Yuri Bakhtin 2010-01-05T02:57:30Z 2010-01-05T02:57:30Z <p>Shiryaev's "Probability" is an excellent source.</p> http://mathoverflow.net/questions/8509/martingales-and-betting-strategies/40504#40504 Answer by PeterR for Martingales and Betting Strategies PeterR 2010-09-29T17:47:26Z 2010-09-29T17:47:26Z <p>Have a look at "The Doctrine of Chances: Probabilistic Aspects of Gambling" by Stewart Ethier <a href="http://www.springer.com/mathematics/probability/book/978-3-540-78782-2" rel="nofollow">http://www.springer.com/mathematics/probability/book/978-3-540-78782-2</a></p> <p>Lots of examples related to just about every casino game you've ever heard of, and a few you have not.</p> | 913 | 3,122 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2013-20 | latest | en | 0.745678 |
https://getrevising.co.uk/forums/topics/really_struggling_with_mole_calculations_just_cant_seem_to_get_my_head_around_them_any_advice | 1,590,381,107,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347387219.0/warc/CC-MAIN-20200525032636-20200525062636-00569.warc.gz | 365,166,741 | 12,048 | # Really struggling with mole calculations, just can't seem to get my head around them. any advice?
Been working at these for a while now and getting no-where, it seems to make sense when it's explain but as soon as i try out a question i always get the wrong answer and i can't see where i have gone wrong!
Posted Tue 2nd April, 2013 @ 15:43 by Fraser
Edited by Fraser on Tue 2nd April, 2013 @ 15:43
Do you use mole triangles? They really helped when I began the calculations. If you are working out the molarity of a solid...
Draw a triangle. At the top of the triangle, put mass (in grams). At the base, in each corner, put Mr (relative molecular mass) and Moles. The triangle works on the principle that if you cover what you are trying to investigate, the calculation is revealed.
Mass
__________
Mr x Mol
E.g... 4.3g of Fe, what is the molarity? Cover up Moles to reveal Mass / Mr.
4.3 / 55.8 = 0.07 moles.
Similarly, you can find the mass if you know the moles - 55.8 x 0.07 = 4.3g
If you are finding out molarity of solutions, it gets a bit more tricky. Same principle, but moles on top and concentration (in moldm-3) and volume (in dm3) on the base.
All the best.
Answered Tue 2nd April, 2013 @ 20:21 by Didgez | 345 | 1,234 | {"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} | 3.21875 | 3 | CC-MAIN-2020-24 | latest | en | 0.95591 |
https://www.physicsforums.com/threads/iterative-method-to-solving-the-colebrook-white-equation.679113/ | 1,579,459,640,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594705.17/warc/CC-MAIN-20200119180644-20200119204644-00080.warc.gz | 1,043,608,623 | 15,781 | # Iterative method to solving the Colebrook-White equation
• Archived
## Homework Statement
In our fluid mechanics class we were taught that we could use the following equation to solve for the Darcy friction factor f:
To do this by hand:
1. Guess a value for 1/sqrt(F), guess 3
2. Get the right hand side result of the equation using 3
3. Use that result for the next value of 1/sqrt(F)
4. Continue using the result for the next value.
5. To find F, just divide one by that value squared.
This iterative approach works but I am not too sure why. Can anyone explain why it works? I'm guessing it requires some knowledge of mathematical proofs?
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Chestermiller
Mentor
You are trying to solve an equation of the form x = f(x) using successive substitution. The successive substitution scheme is $$x^{n+1}=f(x^n)$$ where n signifies the n'th iteration. If we also consider the previous iteration, we have $$x^n=f(x^{n-1})$$. If we subtract the two equations, we have:
$$x^{n+1}-x^n=f(x^n)-f(x^{n-1})$$
If we expand the rhs in a taylor series about xn, we obtain:
$$x^{n+1}-x^n=f'(x^n)(x^n-x^{n-1})$$
In order for the scheme to converge, the magnitude of the changes in x from one iteration to the next must be getting smaller. If x is in the close vicinity of the solution, this means the, in order for the scheme to converge, $$|f'(x)|<1$$
That is, the absolute value of the derivative of the function f must be less than 1 for the scheme to converge. | 397 | 1,513 | {"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-2020-05 | latest | en | 0.857947 |
https://albafun.info/combining-like-terms-puzzle-worksheet-pdf/ | 1,618,136,125,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038061820.19/warc/CC-MAIN-20210411085610-20210411115610-00440.warc.gz | 214,671,184 | 14,341 | # Combining Like Terms Puzzle Worksheet Pdf, Anniessodasaloon Page 52 6 Grade Multiplication Worksheets Kindergarten Math Number Recognition Worksheets 5Th Grade Fun Worksheets Set Of Integers Numbers Solve Each System Of Equations By Graphing Calculator Fun Math Tricks
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## Combining Like Terms Puzzle Worksheet Pdf , Combining Like Terms Halloween Puzzle Activity By Sine On The Line
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(no rating) 0 customer reviews. Two terms that are like terms may be combine into one term by adding or subtracting. Solving equations by combining like terms (worksheet). Combining like terms for 6th grade and 7th grade. (have) coffee with a lot of sweets and chocolate. Find the answer at the bottom of the page. This would be followed by underlining the y terms twice. | 1,416 | 6,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-17 | longest | en | 0.811097 |
https://docs.influxdata.com/influxdb/cloud/query-data/common-queries/multiple-fields-in-calculations/ | 1,726,159,127,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651460.54/warc/CC-MAIN-20240912142729-20240912172729-00075.warc.gz | 197,478,706 | 13,939 | # Use multiple fields in a calculation
To use values from multiple fields in a mathematic calculation, complete the following steps:
## Filter by fields
Use `filter()` to return only the fields necessary for your calculation. Use the `or` logical operator to filter by multiple fields.
The following example queries two fields, `A` and `B`:
``````from(bucket: "example-bucket")
|> range(start: -1m)
|> filter(fn: (r) => r._field == "A" or r._field == "B")
``````
This query returns one or more tables for each field. For example:
_time_field_value
2021-01-01T00:00:00ZA12.4
2021-01-01T00:00:15ZA12.2
2021-01-01T00:00:30ZA11.6
2021-01-01T00:00:45ZA11.9
_time_field_value
2021-01-01T00:00:00ZB3.1
2021-01-01T00:00:15ZB4.8
2021-01-01T00:00:30ZB2.2
2021-01-01T00:00:45ZB3.3
## Pivot fields into columns
Use `pivot()` to align multiple fields by time.
To correctly pivot on `_time`, points for each field must have identical timestamps. If timestamps are irregular or do not align perfectly, see Normalize irregular timestamps.
``````// ...
|> pivot(rowKey: ["_time"], columnKey: ["_field"], valueColumn: "_value")
``````
Using the queried data above, this `pivot()` function returns:
_timeAB
2021-01-01T00:00:00Z12.43.1
2021-01-01T00:00:15Z12.24.8
2021-01-01T00:00:30Z11.62.2
2021-01-01T00:00:45Z11.93.3
## Perform the calculation
Use `map()` to perform the mathematic operation using column values as operands.
The following example uses values in the `A` and `B` columns to calculate a new `_value` column:
``````// ...
|> map(fn: (r) => ({ r with _value: r.A * r.B }))
``````
Using the pivoted data above, this `map()` function returns:
_timeAB_value
2021-01-01T00:00:00Z12.43.138.44
2021-01-01T00:00:15Z12.24.858.56
2021-01-01T00:00:30Z11.62.225.52
2021-01-01T00:00:45Z11.93.339.27
## Full example query
``````from(bucket: "example-bucket")
|> range(start: -1m)
|> filter(fn: (r) => r._field == "A" or r._field == "B")
|> pivot(rowKey: ["_time"], columnKey: ["_field"], valueColumn: "_value")
|> map(fn: (r) => ({r with _value: r.A * r.B}))
``````
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Flux is going into maintenance mode. You can continue using it as you currently are without any changes to your code.
### InfluxDB v3 enhancements and InfluxDB Clustered is now generally available
New capabilities, including faster query performance and management tooling advance the InfluxDB v3 product line. InfluxDB Clustered is now generally available.
### InfluxDB v3 performance and features
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Learn about the new v3 enhancements
### InfluxDB Clustered general availability
InfluxDB Clustered is now generally available and gives you the power of InfluxDB v3 in your self-managed stack.
Talk to us about InfluxDB Clustered | 934 | 3,068 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-38 | latest | en | 0.537335 |
http://stackoverflow.com/questions/5099745/python-modulus-result-is-incorrect/5099767 | 1,464,203,556,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049275328.0/warc/CC-MAIN-20160524002115-00061-ip-10-185-217-139.ec2.internal.warc.gz | 257,391,108 | 19,336 | # Python modulus result is incorrect
I am totally stumped. I was computing the cipher of the number 54 in RSA with the following values:
p=5; q=29; n=145 d=9; e=137
So the number 54 encrypted would be:
54^137 mod 145
or equivalently in python:
``````import math
math.pow(54,137)%145
``````
My calculator gives me 24, my python statement gives me 54.0. Python is clearly wrong but I have no idea why or how. Try it on your installations of Python. My version is 2.5.1 but I also tried on 2.6.5 with the same incorrect result.
-
``````>>> pow(54,137,145)
24
``````
`math.pow` is floating point. You don't want that. Floating-point values have less than 17 digits of useful precision. The 54**137 has 237 digits.
-
Thanks! Interesting information, I love learning little details like this. – Franz Feb 24 '11 at 8:45
@Franz: "little details"? Like floating point only represents only a few decimal digits of precision? I think that this is more than little. There are a fair number of SO questions that reflect other folks not knowing this piece of information. Or perhaps you were referring to something else? – S.Lott Feb 24 '11 at 10:47
That's because using the `math` module is basically just a Python wrapper for the C math library which doesn't have arbitrary precision numbers. That means `math.pow(54,137)` is calculating 54^137 as a 64-bit floating point number, which means it will not be precise enough to hold all the digits of such a large number. Try this instead to use Python's normal built-in arbitrary precision integers:
``````>>> (54 ** 137) % 145
24L
``````
-
Using 3-argument `pow(x, y, z)` instead of `x ** y % z` for modulus arithmetic is generally a very good idea - it avoids the need to create the potentially huge intermediate value `x ** y`. – ncoghlan Feb 24 '11 at 3:51 | 483 | 1,810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2016-22 | latest | en | 0.87856 |
https://forums.wolfram.com/mathgroup/archive/1998/Sep/msg00143.html | 1,726,026,554,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651343.80/warc/CC-MAIN-20240911020451-20240911050451-00506.warc.gz | 232,504,315 | 7,875 | Re: Select x s.t. y>10
• To: mathgroup at smc.vnet.net
• Subject: [mg13988] Re: [mg13957] Select x s.t. y>10
• From: Jurgen Tischer <jtischer at col2.telecom.com.co>
• Date: Sat, 12 Sep 1998 16:59:12 -0400
• References: <199809111906.PAA23158@smc.vnet.net.>
• Sender: owner-wri-mathgroup at wolfram.com
```Hi Jason,
In[1]:= li=Table[{n^2,n},{n,10}];
In[2]:= Cases[li,x_?(#[[2]]>5&):>(x[[1]]),1,1]
Out[2]= {36}
Jurgen
Jason Gill wrote:
> I came across what seems like it should be a simple little function, but
> could not come up with a clever way to implement it. Generally, what I
> am trying to do is take a list of lists of length 2 ie a list of x,y
> coordinates, and Select x values based on some criterion for y. For
> example I want to select the first x value for which the corresponding
> y is greater than 5.
> I can write a function using Position and Take etc. , but there must be
> a clever,and more efficient way to do this. Any thoughts... Jason
>
> --
> Jason Gill
> IBM Microelectronics
> Essex Junction, VT 05452
> Phone (802)769-3350
> Fax (802)769-1220
> email: jgill at vbimail.champlain.edu
> jasongil at us.ibm.com
```
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• Next by thread: Re: Select x s.t. y>10 | 462 | 1,404 | {"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-2024-38 | latest | en | 0.776593 |
https://zbmath.org/?q=rf%3A1074.60084 | 1,618,083,070,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038057476.6/warc/CC-MAIN-20210410181215-20210410211215-00461.warc.gz | 1,241,050,052 | 11,914 | # zbMATH — the first resource for mathematics
Thick points of the Gaussian free field. (English) Zbl 1201.60047
Let $$U\subseteq \mathbb{C}$$ be a bounded domain with smooth boundary and let $$F$$ be an instance of the continuum Gaussian free field on $$U$$ with respect to the Dirichlet inner product $$\int_{U}\nabla f\left( x\right) \cdot \nabla g\left( x\right) \mathrm{d}x$$. The set $$T\left( a;U\right)$$ of $$a$$-thick points of $$F$$ consists of those $$z\in U$$ such that the average of $$F$$ on a disk of radius $$r$$ centered at $$z$$ has growth $$\sqrt{a/\pi }\log \displaystyle\frac{1}{r}$$ as$$\;r\rightarrow 0$$. The authors show that for each $$0\leq a\leq 2$$ the Hausdorff dimension of $$T\left( a;U\right)$$ is almost surely $$2-a,$$ that $$v_{2-a}\left( T\left( a;U\right) \right) =\infty$$ when $$0<a\leq 2$$ and $$\nu _{2}\left( T\left( 0;U\right) \right) =\nu _{2}\left( U\right)$$ almost surely, where $$\nu _{\alpha }$$ is a Hausdorff-$$\alpha$$ measure, and that $$T\left( a;U\right)$$ is almost surely empty when $$a>2$$. Furthermore, they prove that $$T\left( a;U\right)$$ is invariant under conformal transformations in an appropriate sense. The notion of a thick point is connected to the Liouville quantum gravity measure with parameter $$\gamma$$ given formally by $$\Gamma \left( \mathrm{d}z\right) =e^{\sqrt{2\pi }\gamma F\left( z\right) }\mathrm{d}z$$ considered by B. Duplantier and S. Sheffield [arXiv:0808.1560].
##### MSC:
60G60 Random fields 60G15 Gaussian processes 60G18 Self-similar stochastic processes 28A80 Fractals
Full Text:
##### References:
[1] Ben Arous, G. and Deuschel, J.-D. (1996). The construction of the ( d +1)-dimensional Gaussian droplet. Comm. Math. Phys. 179 467-488. · Zbl 0858.60096 · doi:10.1007/BF02102597 [2] Bolthausen, E., Deuschel, J.-D. and Giacomin, G. (2001). Entropic repulsion and the maximum of the two-dimensional harmonic crystal. Ann. Probab. 29 1670-1692. · Zbl 1034.82018 · doi:10.1214/aop/1015345767 [3] Daviaud, O. (2005). Thick points for the Cauchy process. Ann. Inst. H. Poincaré Probab. Statist. 41 953-970. · Zbl 1074.60084 · doi:10.1016/j.anihpb.2004.10.001 · numdam:AIHPB_2005__41_5_953_0 · eudml:77876 [4] Daviaud, O. (2006). Extremes of the discrete two-dimensional Gaussian free field. Ann. Probab. 34 962-986. · Zbl 1104.60062 · doi:10.1214/009117906000000061 [5] Dembo, A., Peres, Y., Rosen, J. and Zeitouni, O. (1999). Thick points for transient symmetric stable processes. Electron. J. Probab. 4 13. · Zbl 0927.60077 · emis:journals/EJP-ECP/EjpVol4/paper9.abs.html · eudml:120048 [6] Dembo, A., Peres, Y., Rosen, J. and Zeitouni, O. (2000). Thick points for spatial Brownian motion: Multifractal analysis of occupation measure. Ann. Probab. 28 1-35. · Zbl 1130.60311 · doi:10.1214/aop/1019160110 · euclid:aop/1019160110 [7] Dembo, A., Peres, Y., Rosen, J. and Zeitouni, O. (2001). Thick points for planar Brownian motion and the Erdős-Taylor conjecture on random walk. Acta Math. 186 239-270. · Zbl 1008.60063 · doi:10.1007/BF02401841 · www.actamathematica.org [8] Deuschel, J.-D. and Giacomin, G. (2000). Entropic repulsion for massless fields. Stochastic Process. Appl. 89 333-354. · Zbl 1045.60103 · doi:10.1016/S0304-4149(00)00030-2 [9] Duplantier, B. and Sheffield, S. (2009). Liouville quantum gravity and KPZ. Available at · Zbl 1226.81241 · doi:10.1103/PhysRevLett.102.150603 [10] Evans, L. (2002). Partial Differential Equations. American Mathematical Society Translations 206 . Amer. Math. Soc., Providence, RI. · Zbl 1042.65027 [11] Janson, S. (1997). Gaussian Hilbert Spaces. Cambridge Tracts in Mathematics 129 . Cambridge Univ. Press, Cambridge. · Zbl 0887.60009 [12] Karatzas, I. and Shreve, S. E. (1988). Brownian Motion and Stochastic Calculus. Graduate Texts in Mathematics 113 . Springer, New York. · Zbl 0638.60065 [13] Katznelson, Y. (2004). An Introduction to Harmonic Analysis , 3rd ed. Cambridge Univ. Press, Cambridge. · Zbl 1055.43001 [14] Kenyon, R. (2001). Dominos and the Gaussian free field. Ann. Probab. 29 1128-1137. · Zbl 1034.82021 · doi:10.1214/aop/1015345599 [15] Naddaf, A. and Spencer, T. (1997). On homogenization and scaling limit of some gradient perturbations of a massless free field. Comm. Math. Phys. 183 55-84. · Zbl 0871.35010 · doi:10.1007/BF02509796 [16] Orey, S. and Pruitt, W. E. (1973). Sample functions of the N -parameter Wiener process. Ann. Probab. 1 138-163. · Zbl 0284.60036 · doi:10.1214/aop/1176997030 [17] Randol, B. (1969). On the Fourier transform of the indicator function of a planar set. Trans. Amer. Math. Soc. 139 271-278. · Zbl 0183.26904 · doi:10.2307/1995319 [18] Revuz, D. and Yor, M. (2004). Continuous Martingales and Brownian Motion . Springer, Berlin. · Zbl 1087.60040 [19] Rider, B. and Virag, B. (2007). The noise in the circular law and the Gaussian free field. Int. Math. Res. Not. IMRN 2007 Art. ID rnm006. · Zbl 1130.60030 · doi:10.1093/imrn/rnm006 [20] Sheffield, S. (2007). Gaussian free fields for mathematicians. Probab. Theory Related Fields 139 521-541. · Zbl 1132.60072 · doi:10.1007/s00440-006-0050-1 [21] Taylor, M. E. (1996). Partial Differential Equations. I. Basic Theory. Applied Mathematical Sciences 115 . Springer, New York. · Zbl 0869.35002
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 1,930 | 5,613 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-17 | latest | en | 0.656093 |
https://engaging-data.com/will-money-last-retire-early/?spend=40000&initsav=1000000&age=30&yrs=70&stockpct=95&bondpct=0&cashpct=5&sex=0&infl=1&taxrate=0&fees=0.3&income=0&incstart=50&incend=70&expense=0&expstart=50&expend=70&showdeath=1&showlow=1&show2x=1&show5x=1&flexpct=0&spendthreshold=80&mort=best | 1,716,058,154,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057494.65/warc/CC-MAIN-20240518183301-20240518213301-00319.warc.gz | 211,144,804 | 58,567 | # Rich, Broke or Dead? Post-Retirement FIRE Calculator: Visualizing Early Retirement Success and Longevity Risk
Posted In: Financial Independence | Money
One of the key issues with retiring is ensuring that the money you have saved will not be exhausted during your retirement. This is also known as Longevity Risk and is especially important if you want to retire early, since your retirement could be 50 years long (or more). This interactive post-retirement fire calculator and visualization looks at the question of whether your retirement savings can last long enough to support your retirement spending and combines it with average US life expectancy values to get a fuller picture of the likelihood of running out of money before you die.
It helps to answer the question: If I start out with $X dollars at the beginning of my retirement, will I run out of money before I die? – Use this button to generate a URL that you can share a specific set of inputs and graphs. Just copy the URL in the address bar at the top of your browser (after pressing the button). – Use this button to download the data from the simulation. It will show the yearly data from all the historical cycle (i.e. it will be a fair amount of data). #### Probabilities based on historical cycles The graph shows the likelihood of your balance being at different levels during each year of your retirement (and compares it to the probability of dying during this time). Red indicates failure (i.e. you’ve run out of money) and green indicates success (i.e. you haven’t run out of money). The probabilities are calculated based upon looking at stock, bond and cash returns from historical cycles between 1871 and 2019. If you expect to retire for 50 years, one historical cycle would be from 1871 to 1922, another one from 1872 to 1923, and so on until 1968 to 2019. Thus 98 different historical cycles are considered (in this case). It is important to note that these frequencies in the past are not the same as actual probabilities. Just because an outcome happened once in history doesn’t mean that there is a one in 98 chance (1.02%) of this same thing happening in the future. However, if your retirement portfolio survives most historical cycles, there is a good chance that it’ll survive in the future without any major black swan events. If something crazy occurs (e.g. a major nuclear war), your retirement balance may be the least of your worries, so we can safely ignore this, since there’s very little way to prepare for it financially. If you look over all these historical cycles, we find that a 4% withdrawal rate will generally last through a long retirement, though there are occasional cycles that are “failures”, i.e. you run out of money. See here for more info on the 4% rule and how historical simulations of withdrawal rate are performed. #### Instructions (updated) The fields are all pre-filled but you should modify the numbers to suit your situation or to explore other options. Press ‘Enter’ or ‘tab’ after you enter the value into the input box in order to update the calculation. • Hover over the input labels for more info. • Enter your expected spending per year in retirement and the savings amount you expect to have at retirement. • Enter your age at retirement and how long you expect to live (you can estimate on the longer side since the calculator will include life expectancy). • You can now choose between three different mortality tables (an average American lifespan based on the Social Security Administration mortality table, or a low and high life expectancy table from the Society of Actuaries) • Enter your target asset allocation for retirement. • Enter your sex. • If you think you’ll have some flexibility in your spending post-retirement, you can enter the percentage reduction in spending that will happen if your portfolio is below a certain threshold, as a percentage of your inflation-adjusted starting balance (see next point). • You should enter the percentage of your initial balance at which the spending flexibility will kick in. If you set it to 90% you will reduce spending when your portfolio balance is below 90% of your inflation adjusted starting balance. • Enter your average expected tax rate (not your marginal rate) – this will be applied to your annual spending and any additional income. • Enter your average investment fees (e.g. expense ratios). • Enter any additional income sources or expenses that aren’t applicable for the entire model period, and indicate both the starting and ending ages. If you have multiple income or expense streams, you can enter them all separated by a semi-colon (;). If you want some of these income or expense streams to remain constant in nominal dollars, i.e. a mortgage payment, then type an asterisk (*) after the number. Otherwise these streams will be adjusted for inflation annually. You can also modify a few graph elements (to help you focus on different parts of the graph): • Show or hide the death probability wedge. Hiding it helps you investigate the portfolio balances with greater resolution for later years. • Show or hide different categories of success (green wedges). Success is defined as any outcome where you are not broke (i.e. balance <0). Additional categories of success include balances that are below your initial retirement balance (but still above zero) and balances that are more than double and five times your initial retirement balance. • You can download an PNG image of your graph (click on the camera icon in the lower right). Huge tip of the hat to maizeman who first developed this type of graph and was helpful in putting these graphs together. Thanks! Here is maizeman’s github repo. It is also inspired by many hours of playing with cFIREsim and FIRECalc. #### Visualizing Longevity Risk One of the most valuable things about these sorts of interactive graphs is that it allows you to understand how the results vary as you modify the inputs (asset allocation and length of retirement). So I encourage you to play with the inputs to calculator and the ways to visualize and see how the results and hopefully your understanding of the processes change. One of the key takeaways from this is how large the ‘Wedge of Death’ gets as you get older and how the likelihood of dying is much higher than running out of money. Another important takeaway is that if your retirement has a large likelihood of success (e.g. 4% or lower withdrawal rate), your retirement balance is most likely to be large (more than 2x your initial balance). As mentioned earlier, it is important to remember that past performance does not predict future performance. Data source and Tools Historical Stock/Bond and Inflation data comes from Prof. Robert Shiller. Average life expectancy data is from the Social Security Administration. High and low life expectancy mortality tables are from the Society of Actuaries Standard Ordinary Tables. Javascript is used to process and aggregate the retirement balance results over all historical cycles and graphed using Plot.ly javascript graphing library. Update: – I’ve gotten alot of really good feedback on this tool. I’m really happy that people find it useful and informative. I’ve also gotten a long list of suggested additions to the calculator, so come back and check to see if I’ve implemented any. First update: I added the ability to toggle between looking at nominal and inflation-adjusted success values. i.e. if we are looking at 2x the original starting balance, 2x can be in nominal dollars (i.e.$2 million on a $1M starting balance) or 2x in inflation adjusted dollars (i.e. more than$2M, whose exact value depends on the historical inflation for a particular cycle). You’ll notice that the success bands look worse when comparing to the inflation-adjusted (real) starting balance rather than the nominal starting balance.
2nd update: I added a few requested features. The first two are the ability to specify you annual tax rate on income and also fees (like expense ratios) on your investments. The main addition is the ability to add multiple income and expense streams with specified starting and end dates to the calculation. This is useful for adding income streams like social security or pensions and temporary expenses like a mortgage or childrens’ college expenses.
3rd update: I added a 5x category just because in many cases (especially less than 4% withdrawal rate), you see huge growth in your portfolio in many/most cases. I also added a button to generate a URL that you can share specific parameters and scenarios with other folks.
4th update: I added a simple spending flexibility parameter that allows you to specify how much you could reduce spending when your portfolio is below a certain threshold (in this case, your original retirement amount (inflation adjusted)).
5th update: I’ve updated the market data to include annual data up to and including 2019.
I also fixed a small bug which affected real stock market returns so you may see a very slight reduction in real average returns and success rates.
6th update: I added two new features:
(1) is the ability to include income and expense streams that are not adjusted for inflation. To make an income or expense stream constant in nominal dollars, you will need to add an asterisk ‘*’ after the number.
(2) You can download all the yearly data on income, spending and portfolio balance for all historical cycles by pressing the download CSV button.
### 190 Responses to Rich, Broke or Dead? Post-Retirement FIRE Calculator: Visualizing Early Retirement Success and Longevity Risk
1. Shelley says:
This is such a fantastic calculator!! Thank you so much for putting the time and effort into this. It’s a great resource.
Question: for future inflation adjusted income, does the calculator assume the first payment will be the number I put into the calculator or does it add in XX years of inflation automatically. Like, let’s say I estimate I’ll bring in $30,000/year in social security in 20 years, does the calculator factor my income in 2043 as$30k or $30k+20 years of inflation? 2. Douglas Alan says: There’s a bug in the calculator if you put in multiple income streams in the “Extra Income” field (separated by a “;”, as documented). This bug becomes pretty obvious if I set the start age to 100. When I set the start age to 100, it barely makes a difference in the results shown, when instead, the results should be identical to if I put in$0 extra income. (At least up to the age of 100.)
What’s going wrong seems to be that the Start Age is only being applied to the first income stream and any additional income streams are starting immediately. E.g., if I put “$0;$100,000” into the Extra Income box, then the value in the Start Age box has absolutely no effect on the results shown.
P.S. As a workaround, it seems that I can enter two different input streams by putting a negative value into the “Extra Expense” box.
• chris says:
If you have multiple income streams then you should also include multiple start dates (also separated by a “;”). Then it should behave as you expect.
• Douglas Alan says:
Ah yes, that works. Thanks!
I’m confused as to why mortality estimate is not charted as a separate line because it seems orthogonal to one’s finances.
e.g., I could die with a lot of money, some money, or no money (which’d make a difference for my heirs!).
So having this as part of the set that adds up to 100% just doesn’t compute for me.
What am I missing here?
• Michael says:
this is about your likelihood of running out of money. If you simply click off the “death” button, it will show you your percentages conditional on living all the way to the end or whatever year you are looking at. This may be more valuable if you care a lot about your legacy.
For most people, it’s probably worthwhile to consider both possibilities. A small chance of running out of money very near the end of a long retirement may really be a very tiny chance since it’s conditional on being alive, which may only be a 10-20% chance at 90-95 for most people.
4. Alex says:
It would be good to add a “current age” field, since the (remaining) life expectancy is highly dependent on one’s current age.
For instance, once you have made it to 60, your chances of reaching 70 are much higher than if you are an average 20-year old.
I suspect that for us older folks, this would greatly affect the “dead” wedge…
Excellent tool overall!
5. David Alves says:
I would love to have a few extra income / expense rows for when there are multiple expected events, e.g. paying off a mortgage, inheritance, kids college expenses, etc.
• chris says:
If you have multiple income or expense streams, you can enter them all separated by a semi-colon (;). If you want some of these income or expense streams to remain constant in nominal dollars, i.e. a mortgage payment, then type an asterisk (*) after the number.
Just make sure you also add additional start and end ages (also separated by “;”) so it works properly
• David Alves says:
Cool, thanks! For clarity, it would be helpful if we could do one row per income / expense and have an text label on the row to remind us of what the row does, but the semicolon thing will work for now. Thanks for working on this!
6. Karl says:
I am not sure if this is a bug or I have misunderstood….
Leaving everything as default i.e. as the page loads when first visited we have “spending per year” at $40K (4%). If I then download the csv I see the “annual spending” column do this (for 1871, first few years shown) 40000 40917 39115 36395 34308 If this is the income per year then these figures are neither$40K per year or even 4% of the current valuation.
Have I misunderstood something?
• chris says:
The list of numbers you are seeing are the inflation numbers applied to the $40,000 annually to get the same purchasing power as$40,000. Each historical cycle has a different set of inflation numbers.
7. Franklin says:
I am not good with probability. Would it be correct to think of the percentages as a special die that when thrown with the different %’s as the odds of it happening. For example if at 80 years old death was calculated to be 20% then the die would land on that side 1/5 of the time?
8. Tim says:
Great calculator. I really love the “Or Dead” part of the calculator. Its easy to get caught up assuming that living till retirement is a sure thing. Its a good reminder that there is a 25% chance I don’t even make it to the traditional retirement age. Life is precious.
9. bobi says:
if you want to save the url as male just type:
sex=
(with no argument, leave it blank)
10. Steve says:
One of the best, if not the best, such calculator. Kudos for this gift to all.
One element that may be too hard to add, but would be helpful if possible, is to account for couples. The best way I can describe the issue is the probability of at least one spouse living to 95 is ~25% (same age, right at retirement, age 66). Expenses change little, effective tax rate generally goes up, income goes down (loss of one social security check, possibly pension). Dissimilar aged couples face a high change of one spouse greatly outliving the other, especially if the woman is younger.
The actuarial data is available, but the complexity it imposes on the calculations might be intense.
11. Matt says:
This calculator is absolutely brilliant. I’ve been using it for around 2 years now. Regarding Nic’s comment in Oct 2021, I have a similar question. Is it possible to add on the y-axis (or be able to change the y-axis to) instead of a probability, a percentage of returns from investments. As far as I understand, where the frequency is 50% this would approximately correlate with a 7.2% investment return. Of course I think each cycle would be likely to give a different probability, perhaps it might be possible to add a percentage return axis as a rough guide?
Thanks
12. Nic says:
Great tool, thanks for this. What if I don’t believe that the average real stock (or bond) returns of the past will repeat going forward? What if I want to use an expected annual real stock return of say 3.5% and an exp. ann. real bond return of -1% (using Monte Carlo), instead of historical cycles with real returns of 8.1% and 2.4% respectively? I notice that you already allow this in your pre-retirement calculator (under “Monte Carlo simulation”), which I appreciate.
13. John says:
The saved URL still does not present “male” when you reload the page. No matter if sex is set to 0 or 1 in the URL, it always default back to female on page load. Not a big deal but others have mentioned it months ago and it’s still a problem.
14. Urs says:
Love the tool! Its really useful & helpful. Thanks for creating this! Shared this in the family and this showed some great options for us.
I saved “my data” with the “generate url” link. Whenever I click on the link and come back to the data, it it messes it up as in the “bond %” the value is changed. I need to correct the value to my initial input and back we are on track. No idea why this happens but might be a small bug.
15. Gary M says:
What does the line of info above the chart legend represent? Mine starts with M / 16% ………..
16. HarrisonFjord says:
Fantastic tool. Thanks for building. One suggestion regarding spending/yr. This field should represent the $/yr in expenses you project in the year you plan to retire (inflation adjusted). It would be great if we could enter our current age or the year in which we plan to retire so that we can enter projected expense in today’s dollars and then have the tool adjust to future dollars. I think this would make the tool more intuitive. • Gnomeozurich says: The way I handle this now is to use today as my retirement date, and add in my expected income until retirement as one of the “Extra Income” flows. 17. Carl says: What is the “sex=” setting in the URL for Male? Saving a report and then reloading it always sets the report to Female. Not a big deal, but followed the “Generate URL” steps and everything else works but that. 18. Carl says: Great tool. Thanks for it. Is the “Spending/yr” the amount that is withdrawn from “Savings” every year (adjusted for inflation), or is that the amount that you wind up with after the savings is withdrawn and then taxed? For example, if Savings is 1,000,000, Spending/yr is 50,000, and the Tax Rate is 20%, is 50,000 withdrawn and 10,000 applied to taxes, leaving you with 40,000 that can be spent? Or is 62,500 withdrawn, so that 12,500 goes to taxes and you’re left with \$50,000 to spend?
• Fred Haley says:
Quick question. Is “spending” my actual TOTAL spending? Or is it the amount spent from the portfolio? That is …does this calculator somehow take into account Social Security, and other sources of income? I think not. My guess is that the “spending” entry is in reality the “withdrawal from savings” amount. Can you verify that please? Thanks!
• chris says:
spending normally represents total spending (except if you include “other expenses”). It is the amount of money spent on goods and services that needs to come from one source or another. If you have a non-zero tax rate, then the amount of money withdrawn from your portfolio will be higher than this spending amount. This calculator only includes things like social security or other sources of income if you include it in the “Other Income” text input box.
19. VM says:
This is a fine tool to help with one’s retirement planning. I’d like to be able to map out another couple of “one time” cash gains. I’m 66 and in ten years, then twelve years, I or my heirs will have a large amount due. I’d like to be able to show the receipt of $116,000+ then$69,000+ two years later. How could I show this?
I kept playing with the tool nd finally saw the answer to my question.. DUH!
21. mr_jetlag says:
Hi, the download feature is incredible and really shows you the scenarios where your plan could fail. Just one suggestion, on my CSV file it cuts off at 1962 presumably due to column / data width limitations. Is it possible to do a vertical table with the years running down as rows instead? I think for all datasets under 200 years that would work.
• chris says:
Maybe it’s because you are modeling a 57 year retirement? 1962 + 57 is 2019. The year shown across the column headings is the cycle start year.
When I model a 50 year cycle, the columns go up to cycle start year 1969 and you can’t have a 50+ year cycle that starts in 1990 since we don’t have market data for 50 years after 1990.
I don’t think it’s a CSV file issue, but if it was it could be due to the software you are using to view the CSV. Some older versions of Excel only allow 256 columns, but recent ones and google spreadsheets allow over 65000 columns.
However
22. Kaawumba says:
If extra income is greater than spending, it should be impossible to go broke.
• Todd says:
Looks like the input says retiring at 40 with 80K in the bank and no income until 50.
• Alan says:
Your extra income doesn’t kick in until you are 50. Your expenses outstrip your savings between 40 and 50. After 50, you’re fine, and the probability of going broke declines to zero.
• Gnomeozurich says:
You can go broke if your extra income doesn’t start until 10 years into retirement.
23. Dave says:
Shouldn’t this calculator have three separate entry fields for 1. pre-tax (IRA/401k), 2. ordinary investments (subject to capital gains), and 3. tax-free (Roth) savings?
24. Frank McHugh says:
Why do the annual spending values vary in the downloaded csv file? Seems random but it grows in columns based on later years. It changes weather or not nominal or inflation adjusted is selected?
• chris says:
Annual spending varies because of inflation. The idea is that your spending will stay constant on a real dollar basis, but on a nominal basis it needs to increase to maintain your purchasing power (since prices of everything is increasing due to inflation).
Spending will vary between cycle runs because inflation varies in each year. In all cases, the first cycle is 1871, and if year 0 spending is 40000, year 1 spending will be 40000 * (1+ inflation rate in 1872) and year 2 spending will be year 1 spending * (1 + inflation rate in 1873), year n spending will be year n-1 spending * (1 + inflation rate in year 1871+n).
The next cycle starting in 1872 will have different spending amounts because you are starting with a different year.
25. Sean says:
This is an excellent calculator, and your other visualizations are cool, too.
I have one question / feature request: does the calculator assume periodic asset reallocation? If so, how often? If not, I think it would be a great feature to add.
• chris says:
Thanks!
The calculator keeps the asset allocation the same each year so that would amount to reallocation every year.
26. JKumar says:
Chris, Fantastic tool. Your work definitely accelerate my early retirement goals and provide guidance on maximize spend vs minimize risks. Great job!
I have couple of questions.
1. Spending Flex and Flex Threshold variables – Does the Withdrawal amount go back to inflation adjusted original value ( value prior to applying flex spend reduction) or continue inflation adjusted reduced value once the portfolio value rise above inflation adjusted starting value. I am trying to understand whether my spend amount permanently impacted or it is just temporary and I can resume planned higher spend after portfolio reach certain level.
2. It would be great if you can provide a feature to display spend table in tabular format. It will help us to see the actual spend and portfolio values year 1 through retirement end year .
Again great work! You are making huge difference in the personal finance world.
• chris says:
Thanks for your kind words. Glad that I can be of help.
1. Spending goes back to the normal level (inflation-adjusted) when portfolio balance goes above the threshold.
27. Gary Maurer says:
Great tool! Thanks for all of your hard work. Can you consider adding a non-inflation adjusted income box? Most private pensions are fixed and that would allow them to “shrink” over time as opposed with keeping up with inflation (like SS does).
• chris says:
Thanks. Sure, that suggestion has been made before. I will try to get to it soon.
28. Stlbuc says:
Good tool, but i am not sure i am using it correctly. The model seems to assume your spending per year is your withdrawal rate. But if your investments generate dividends, you don’t need to withdrawal as much. Example: $1 million in savings generates$30,000 in dividends (3% return). If my spending is $50,000 per year, then i only need to need to withdrawal$20,000 because the dividends cover the rest. How do i account for this in the tool?
• chris says:
The stock returns include dividends and so dividends would need to be withdrawn from your portfolio each year.
Thus you can’t explicitly model a dividend portfolio that returns a certain dividend percentage per year.
29. Stlbuc says:
Does this model account for dividends?
• chris says:
Yes, the stock returns are adjusted to include dividends.
30. Kevin says:
This is awesome!!!! I wish it had 2 additional lines for additional income. Wife pulls a pension at 56, but soc sec could kick in at different ages. Well done overall.
• chris says:
• Gnomeozurich says:
I’ve added up to 7 or 8 additional income streams in some of my models using the semi-colons. I don’t know how much you can get in there before it breaks, but it’s clearly more than almost anybody needs.
31. DR says:
Its quite possible that I may be missing something, but I see something strange.
If I put Savings = 1 and Spending/yr = 60,000, I still get a 100% success rate, without any additional income.
Only if I change Savings = 0, then it shows me as broke.
• chris says:
That sounds very odd. Can you copy the URL and paste it in the comments so I can see the exact issue?
32. Dean says:
The percentages for being broke and the different balances (but not chance of being dead) changes for a fixed age when just the Retirement Yrs parameter is changed. I wouldn’t think that the Retirement Yrs input should change anything other than how far out the chart extends.
• chris says:
The percentages could change a little because the number of cycles changes. There are 150 years between 1871 and 2020. If you want a 40 year retirement, there will be 110 different cycles (1871-1910 . . . to 1981-2020). If you want a 50 year year retirement, there are 10 less cycles (1871-1920 . . . to 1971-2020). With fewer cycles the percentages can change.
33. Todd says:
This tool is simply amazing. Please never take it down!
34. Nathan says:
Great tool! Very informative.
One suggestion is to put more nuance on “fail” to be not just “broke” but also “close to broke”.
Maybe easiest is a fraction of starting savings, since you already have wedges for Balstart, Bal>2xstart, Bal>5xstart, can you add another for 0<Bal<[p%]xstart where p% is a user-input field.
35. Jesse says:
I’m a bit confused by the tax rate with regards to my real spending power. If I put my spending in as $40k with a tax rate of 10% then I have success, but does that actually mean I only have$36k (40 – (40*.1) = 36) in spending after tax? Or will I need to adjust my withdraw up?
It seems like there needs to be a separate box for spending and withdraws. I feel like spending is the amount I have after-tax and withdraw is the amount I actually need to pull out of my accounts each year pre-tax.
• Jesse says:
Looks like it auto formatted with the dollar sign. I mean “40k with a tax rate of 10 percent do I actually have 36K in spending after tax?”
• John says:
I have the same question. Playing around with the numbers it looks like the “Spending/yr” needs to include your living expenses + what you expect to “spend” on the IRS since the tax percentage entered doesn’t seem to have any effect on the Withdrawal Rate calculation below.
36. RAYMOND MULESKY says:
Thank you so much for the website! Hate to ask for additional clarification on this but to follow up on your June 8, 2019, clarification to an earlier question. If the “spending/yr” figure is meant to be an after-tax figure, and I have a 20% tax rate, that means behind the curtain somewhere the calculator is adding 20% to whatever my “spending/yr” figure is as the withdrawn amount from my savings?
37. Stephen Horton says:
Can my annual budget equal my “spending/year” plus “extra income”?
What is a good number for “Avg tax rate” ? I live in a no state income tax state.
38. Dean says:
Hello. Great Retirement calculator! I love it. Just one question though. Where would one put acquired real estate assets, eg rental properties? If you have say $1.5 million in investments one year, then spend say 400K on rental properties for thee cash flow, does that 400K stay in savings, or not? They can still be sold relatively easily worst comes to worst to realize their inherent value. • kb says: agreed, this needs a bit more. • chris says: I would not include the value of real estate assets. I would include their expected net cash flow in the income. There really isn’t the ability to switch back between lump sum assets and cash flow assets so you need to pick one and go with it. 39. Lynne says: THIS IS AWESOME! Great tool to use throughout retirement as well to make sure you are still on track! I am modeling a bunch of special circumstances with multiple income/expense streams at varying timeframes. It would be great to have a description box for each of these two fields where I can document to keep track of multiple onetime income events and recurring events with different timelines ie for income: rental income; pension; ss me; ss hubby; sale of rental 1; sale of rental 2…etc ie for expense: lumpsum at 60; lumpsum at 65 (to simulate cashing out an amount to buy a car or another big expense) 40. Jack says: First, this is a great tool. I have spent hours playing with the numbers. I have one question: Does your tool include Social Security payments (for those of us in the US)? I assume not and I can add that in the “extra income” section. Thank you 41. Leo says: Chris, thank you very much for making this calculator available. It’s super useful for doing comparisons of different financial scenarios. One thing I’d like to be able to visualize better is in the “Bal < start" wedge when no "Broke" wedge exists; what's the worst-case percentage that remains? It would be a very different thing to see that my balance is at 50% of what I started with (I'm more than fine with that) vs. 1% (oops) at my target end of life. For now I have been either iteratively increasing my annual spending or adding in more extra expense until I see a tiny "Broke" wedge at the end. Then I know the worst-case remaining is around 0%. But it doesn't answer what the min percentage is in non-Broke scenarios. Cheers! 42. Roy says: I’d love to be able to define ‘success’ a little more conservatively. While completely running out of money is certainly failure, I’d like to be able to define failure along the lines of ‘savings are less than N years of expenses’. I would find living the last few years of my retirement with just a 1-2 years of expenses in savings quite troubling, and I’d love a way to see these cases in this visualization. • Gnomeozurich says: I’ll agree, it would be great if we could set a percentage of initial portfolio to see another shading line, similar to the ones for 2x and 5x, Either at something like 50%, or even better, user defined. I’m actually much more intersted in striations below 100% than multiples above. For instance, how often do I get scared early on (maybe 60-70%), or how often do I cut it close or leave almost no legacy at the end (20-30%). It looks like the download feature allows you to look at this in the data, but the advantage of the visual graph is big for something like that. I love that you’ve implemented 2 features that I asked for a couple years ago, and some others I didn’t even think of that are very helpful. So maybe if you ever go at it again, you’ll put this one in too! Thank you! 43. RB says: My apologies but I am having great difficulty understanding the tool. I expect a pre tax need of$75k/yr with lumpy expenses avg of $6k/yr. Tax rate expected is 10%. Social Security and pension expected at$60k annually. So this gives a A withdrawal rate of close to 1.7%. The. App reflects a 4.7% withdrawal rate. So it appears the app is not accounting for pension and SS income and pulling all spend from assets. Am I missing something or not using inputs correctly?
• chris says:
The withdrawal rate is calculated from the initial spending and portfolio amounts. Adding income or expenses do not change the reported withdrawal rate because it may not be consistent over all years. For example, if you have kids college expenses 5-9 years into retirement, should that be included in the WR calculation or not. I left all “extra expenses” out of the calculation and it may be best just to remove the reported withdrawal rate. However, that doesn’t matter for the simulation which tracks your spending for each year across each historical cycle and just counts up the various outcomes.
44. Todd says:
Awesome website and tool! Thank you so much for making it available to everyone.
Quick observation/question:
I noticed that using the same inputs for the 4% and Post-retirement calculators yield slightly different success probabilities. For example, using a 2.8% WR with 45-years retirement duration and 35% stocks / 0% bonds / 65% cash shows 100% success on the 4% calculator but only 96% success for this one (and with same defaults for both calculators, e.g., 0% tax rate, 0.3% expense ratio, etc). Perhaps I did something wrong but double-checked it. Any idea why the success results would differ?
• chris says:
Hmm, that’s a good question. I’ll look into it.
45. Steve Horton says:
Fantastic planning aid. Many thanks.
How do I get to “Ave Tax Rate” ? Should I use sales tax rate?
• chris says:
It is what you expect to pay in taxes on the money used to fund your spending. If all your retirement is in a Roth IRA, you can expect to pay zero taxes. If it is in a 401k it’ll be taxed as ordinary income but the average tax rate will depend on the amount you spend. Since it’s complicated, you can use the tax bracket calculator (https://engaging-data.com/tax-brackets/) as a way to figure out your average tax rate.
46. Dean says:
Great app. Best in class. I plan to introduce it to my children.
I have a suggestion to consider, but I realize it would be a lot of work, and is difficult to ask of somebody who has done all of this for free.
How about a secondary y-axis on the right edge that indicated the imputed withdrawal rate as a percentage for each year, and have that withdrawal trend signified each year by a solid line?
Something to consider.
Thanks again for the great tool.
47. Frank McHugh says:
Very nice tool. It would be good if there were multiple extra income lines. For example I have a fixed benefit pension , social security, my wife’s pension and her social security. All of which can start at different ages, where the amount depends on age started at. I imagine there could also be case for extra expenses like nursing care starting at different ages for spouses etc.
• chris says:
you can add multiple income streams. just separate the amounts (and ages) by a semicolon (;).
48. Scott says:
• Scott says:
Hi,
How does this calculation handle the “Extra Income”?
Example- 100K spending/2M in savings yield 5% WR. If I add $50k in extra yearly income (SS, pension), this still shows 5% WR. I would think in this case the WR would have reduced to 2.5%? The graphs however do show the effect of the extra income as if the WR is reduced. 49. Karl says: Can I visualise the actual spending per year if I am using the Flex? For example if I start at$50k and set a 10% flex I really have no idea if that means I am successful because I have dropped the spend by 10% Per year every year (I assume that its cumulative for each year that the balance is below the flex threshold?).
50. George says:
Just to be clear… is the amount you enter in the Savings field equal to your current total portfolio of stocks, 401k, bonds and cash savings? For example 100K in individual stocks that I manage, 150k in my current 401k, 0 bonds, and 50k cash savings… I would enter 300k in the savings box?
Thanks for sharing this calculator – really cool.
• chris says:
yes savings is the total of those items and the percentages will split that amount into the different categories.
• George says:
Thanks very much… really an awesome tool.
51. Sarah says:
Formatting issue: The “Start Age/End Age” rows are not properly aligned with the “Extra Income” and “Extra Expense” rows. The “Show inflation” row needs an empty row beside it to push the “Extra Income/Expense” rows down one so that they’re in line with the “Start/End” rows. I’m in Waterfox.
Thanks for this calculator and the others you’ve posted.
52. JDesmo says:
Hi,
How does this calculation handle the “Extra Income”?
Example- $100K spending/yr,$2M in savings yield 5% WR. If I add $50k in extra yearly income (SS, pension), this still shows 5% WR. I would think in this case the WR would have reduced to 2.5%? The graphs however do show the effect of the extra income as if the WR is reduced. 53. yigal goldfarb says: Hi, How does this calculation handle the “Extra Income”? Example-$100K spending/yr, $2M in savings yield 5% WR. If I add$50k in extra yearly income (SS, pension), this still shows 5% WR. I would think in this case the WR would have reduced to 2.5%?
The graphs however do show the effect of the extra income as if the WR is reduced.
54. Joel says:
Hi,
This is very useful, thanks so much for making this available! Could you add an output statistic showing how often the spending flexibility would need to be invoked in order to hit the calculated success probability? For example, if I set my Spending Flex to 25% that means in rough years I’d need to cut my spending by up to 25%, but how often would I need to do that for a given simulation? Would that need to happen once every 4 years, once every 10 years etc? Could keep track as you run simulations of how often the spending flex is used and how much of it is used?
The reason I ask is that I expect spending patterns will develop a habit, travel etc. So if I have to invoke spending flex fairly often and much of my spending is on travel, this could become kind of a pain to cancel trips etc. So I’d rather just spend less for less hassle in budgeting when retired.
Would it be possible to add that? 🙂
Thanks!
55. Jerry says:
You can add Social security by just adding it to the extra income portion.
56. Gary says:
Is it possible to add a custom value for the average annual return of investment for stocks while keeping the cycles of the market? I believe that I will be able to maintain an average of 15-20% per year. It would be really helpful being able to calculate this with the cycles included!
Thanks!
• Marvin says:
Hey Gary,
I don’t think that this would be possible as when you pick your stocks the volatility of your portfolio is much different than that of the broader stock market. Especially if you’d like to calculate with a return of above 15% you’ll need a portfolio that is concentrated on fewer stocks, which under typical circumstances increases the volatility of your portfolio and the possibility that you’ll run out of money during a market crash.
So use your return numbers to calculate your total savings until retirement, but once you quit your job stick to the data embedded in the calculator.
All the best
57. Tess says:
This calculator is absolutely brilliant! Captures your probabilities of financial outcomes in retirement in one graphic based on hard data, including your chance of dying at any point in time, which you don’t see elsewhere. You can results immediately based on changing inputs.
chris, huge thanks for your attention to detail, many many hours spent and deep skills!
58. Fred Flintstone says:
Not nearly so helpful to the ~10% of the population with red/green color blindness.
59. […] respond, “Sort of.” This calculator from Engaging Data is pretty […]
60. marie says:
On a separate graph, it would be nice to have only the outcomes if you are alive. As you get older, the chances of becoming broke increase for many people and it would be nice to be able to compare that to having less than starting balance, 1x starting balance, 2x starting balance, etc. I know you can scroll the cursor horizontally and see those for different ages, but having it represented visually, without ‘death’ being factored in, would be helpful. It’s nice to see ‘death’ factored in as it helps to keep that in mind as a relative factor, but most people don’t care if they run out of money after death 🙂
61. Incredible tool, so simple and effective. Thank you for the time and effort you have put in, I will be using this a lot!
62. Gnomeozurich says:
Very happy to see these updates! Looks like you took both of my suggestions. For me, this looks like the clearly best tool on the net for this now, unless you need to do more detailed variable spending modeling (but this is sufficient for me and I’m picky).
Thank you for getting back to this!
Please calculate the withdrawal rate with the inclusion of the extra income field.
It appears that the withdrawal rate is calculated solely from the Spending/Yr and the Savings entries and does not take into account the Extra Income values.
• Stan K. says:
So does that mean if a person inputs spending (living) expenses at $40k, they need to add and additional amount to the 40k to compensate for the 4% of their savings withdraw rate? For example. Based on the following assumptions. Portfolio valuee =$100K
28 years in retirement.
Living expenses – $35k 4% of savings yearly withdraw =$4000k/yr
equals $39k should be entered into the “spending” box. I agree, tho, with most that another box needs to be added to account for 401k withdraws. At any rate, Chris. This is an amazing tool and thank you for managing it. I’m surprised how little one needs to have saved upon retirement to live comfortably. 65. Lokesh Shah says: Awesome calculator. Much thanks for making this available. Is there a way to run similar simulation for another financial market? 66. Jonathan Schoeller says: Thanks for making this. Looks like there might be a bug in the “Generate URL” button. 1. Enter 80% as the Flex threshold 2. Click Generate URL 3. Refresh the page Expected: Flex threshold contains 80% Observed: Flex threshold contains “NaN” 67. Bill says: A great calculator. I would really like to see 3 extra incomes to allow you to put in your SS, your wife SS and a pension. You can gain an extra income by putting a negative number in the expense instead of a positive one. • chris says: You can put unlimited numbers of incomes into the box (just keep them separated by a semi-colon), same as the age box. 68. […] article on the topic: again, Big ERN’s “you are a pension fund of One“. Bonus: this amazing calculator, with a visual representation of odds of ending up rich, broke or dead over […] 69. […] Calculadora: Quanto tempo meu dinheiro vai durar? […] 70. Ray says: Hey thanks for the great Calculator. There is just one curious thing I don’t quite get. When increasing the Retirement Yrs. and leaving everything else identical, the success rate increases from 93% to 98%. The chance of being broke at 90 years is 7% if Retirement Yrs is set to 50, but only 1.1% if set to 60. How can living to 100 instead of to 90 increase my chance of not going broke at 90? These are the settings used (just change Retirement Yrs to 50 to see the effect): https://engaging-data.com/will-money-last-retire-early/?spend=22000&initsav=700000&age=40&yrs=60&stockpct=90&bondpct=8&cashpct=2&sex=1&infl=1&taxrate=27.5&fees=0.3&income=0&incstart=65&incend=90&expense=0&expstart=50&expend=70&showdeath=0&showlow=1&show2x=1&show5x=1&flexpct=20 Thanks and BR 71. Robert Paton says: This is my favorite FIRE calculator — thank you!!!! 72. Danielle D Tinsley says: Great tool- just wanted to say Thank you so much!! 73. steve d says: Great tool and visualization. Is there anything like based on UK data or Europe for us Non US people? 74. Jesse says: This is by far the most useful retirement calculator I’ve come across. Thank you! A few questions: 1) It doesn’t appear that the tax rate affects the withdrawal rate. Is this correct? 2) If I enter “100,000” for expenditures and “25%” for tax rate, and compare this to “125,000” for expenditures and “0%” for tax rate, the results don’t match. Neither does “133,333” and “25%” match “100,000” and “0%”. To what basis is the tax rate applied? 3) Is there source code you can share for those of us who would like to better understand the implementation details? Thanks! 75. Boca Burger says: Great calculator. Why doesn’t the withdrawal rate take into account the “extra income”? Wouldn’t the extra income reduce the amount withdrawn from savings? 76. JD says: Love the tool, especially the spending flexibility component! I have something similar built into my master spreadsheet to “gross up” our SWR from FIRECalc, by a percent that represents a portion of our discretionary spending. A signifiant portion of our annual expenses are discretionary and having the ability to factor that into the calculated results, particularly in determining a SWR, is huge. The results from your calculator confirm the results from my methodology in my spreadsheet. I read a lot of Kitces materials and it’s just as likely that a favorable sequence of returns could impact a portfolio. He also points out that picking a very conservative SWR early in retirement, and not periodically reviewing/updating that rate, has the potential to leave a very large estate. For us this calculator acknowledges the reality when the market does go south, that we will trim our spending accordingly (travel expenses, dining out, random shopping, etc). I have no problem leaving money to the kids, but at the same time we amassed this portfolio to have a comfortable and enjoyable retirement… Thanks! 77. West says: Love the spending flexibility option! Needless to say a significant percentage of our annual spending is discretionary, and having the ability to have that reflected in the projection is incredibly valuable. If there is a really bad year down the road, then obviously vacations, dining out, hobbies and various other expenditures would get reduced as needed… 78. Frank says: I like the calculator, I do have a question though. I can add my projected SS benefit under extra income, but how do I add my wife’s projected SS benefit? She is younger than me and will not receive hers until 7 yrs after mine. I do not see a way to model that correctly. Any suggestions? 79. Rick says: Is the spending amount reflect actual spending or the yearly withdrawal amount from your savings. 80. Tim says: A tabular format would be interesting. Also it would be nice to be able to adjust extra income, e.g., Social Security with a Cost of Living Adjustment. Also to be able to add a series of expenses and/or a series of incomes per year or month. I _really_ like the % chance of death. Also the ability to add different mortality tables based on health, e.g, see opensocialsecurity.com. 81. vic says: did i miss something? Did this calculator figure in social security? Do you enter that under extra income? :–) Post retirement Calculator… Thank you • Nate Greene says: Hey Vic- I entered SS as extra income. This makes sense because SS benefit varies from person to person (as does the age you intend to begin), and the tool couldn’t know that. • Nate Greene says: Vic- I just double checked in the directions, also, to confirm this: “The main addition is the ability to add multiple income and expense streams with specified starting and end dates to the calculation. This is useful for adding income streams like social security or pensions and temporary expenses like a mortgage or childrens’ college expenses.” 82. This is THE BEST retirement calculator I’ve ever seen. Absolutely love it! Is there a place to donate to you,brother, because this is incredible. Thanks! • Nate Greene says: Josh- Thanks for showing this site on your YouTube channel. This calculator is very comforting to me as I look at semi-retirement in a few years. I think I can do this. And THANK YOU to the person who created this site. • Timothy J Dobyns says: Josh, thanks for this recommendation. This calculator is awesome! Tim 83. […] look again at some charts from Engaging Data. Here are sample results for the early retirement scenario at 4% withdrawal rate at age 40 ($40k […]
84. […] for dying. (Ever notice how many of those we have?) In another neat tool from Engaging-Data.com, Will Your Money Last If You Retire Early? adds some helpful nuance to this analysis. You input the same types of information, but now in any […]
85. Gnomeozurich says:
This is a great tool! I love this visualization including the likelihood of death (even if the ssa table is a bit too aggressive for most people doing this, since income and wealth both correlate with longevity, the average person using this tool will probably live a couple years longer than the SSA estimates).
I love the way you put in the ability to model a bunch of income/expense streams with semicolons without cluttering up the interface for those who don’t care.
I have no idea if you’re considering working on this more, but I have a few suggestions if you are:
The spending cut is a great addition as well, but I think it would model most people’s willingness to cut better if it was triggered at threshold below the initial portfolio size, say somewhere around 75-80% Or even 90%. I doubt most people are going to cut their spending because of some minor correction or flat year. But after 10-20% down, it makes sense. It would be good to see the model where the spending cut gets activated somewhere most people would really do it, after a significant correction or bear, or a few low/flat years in a row. In addition, in the graph, it would be really helpful to see how often you dip into danger territory, not just what percent you’re below 1x initial. Having a couple more zones under 1x like 75% (or whatever you use for spending cut threshold), 50% and 25% would be really helpful. Maybe you can even code the 50/25 as light pink :).
Anyway, I hope you agree with my suggestions if you work on this again, and generally great idea and brilliant work!
86. Mike says:
If I have flexibility in spending what is meant by “the inflation adjusted starting balance”? Is this the balance you started with in year 1 of withdrawals? If I am in year 20 do I need to reduce spending based on the year 1 balance? I am basing my withdrawals on a 30 year retirement.
87. Mr Simpleton says:
Thanks so much for making this tool – I especially love the risk of death – I sometimes forget that the risks are not just running out of money.
I heard Carl (Mr 1500) on a What’s Up Next Podcast saying something like “You might run out of money, but you’re definitely going to run out of time” It’s so great to see it in the graph – when your risk of dying is three times the risk of going broke it really makes you think about risk differently!
Actually thanks for your whole awesome website.
88. Debbie says:
We have 2 financial advisors at 2 different firms. This completely supports the financial plans they have both provided. However, your graph makes it so clear, simple to understand, and easy to modify. Most importantly, since I am 65 and my Hubby is 72, it provides me with comfort of “seeing” that I can take the big leap into retirement without fear of living in a cardboard box and eating cat food when I’m 90. Predicated, of course, on the assumption that this country regains it’s sanity and can recover from the last 3 years.
Thank you so much for your excellent work, and the comfort it brings.
89. […] Engaging Data creates absolutely amazing data visualizations and one of them struck me right in the heart (see below). It shows the possibility of my portfolio balance being at different levels during each year of retirement and compares it to the probability of me dying during that time. That grey ‘death’ section sure is large and imposing – and I suspect this longevity data is based on white female numbers – from everything I’ve read, me being a black female decreases my long living prospects further. […]
90. Al Cam says:
Great tool!!
Please can you confirm that the balances graphed are year end values. Also is it correct that “Extra income” (and/or “Extra expenses”) are paid/deducted from the birthday – which I assume is effectively the start of the year in question?
Also, can you explain a bit more about how the Spending/Year is taken from the various assets – e.g. is it deducted monthly or annually; is the spend pro-rated across the assets or do you always re-balance the assets after deduction of the spend to maintain the initial asset ratios, e.g. 60/40, or ….?
91. Anthony says:
I do not understand why the inflation perimeter gives me worse odds when using “Nominal” than my inflation-adjusted odds of survival
• Anthony says:
I got it: I was using the Spending Flexibility parameter.
• luke says:
Can you explain this please? I struggle to understand that. Is the flexibility only enabled for inflation adjusted and disregarded for the nominal?
92. MO says:
Great tool! One question, when filling in the “extra income” section, should I include my expected annual dividends? Or are those already factored into the appreciation of stocks/bonds?
93. Matthew Liggett says:
Thank you for this.
(A) It’d be great to have a version of this for couples! Tricky not to make it too complicated, but at the same time the odds of *one of you* surviving to a ripe old age do increase.
(B) It might be nice to be able to have an absolute “lifespan increase” field. For those using the calculator, they probably have increased longevity risk than average because wealthier, more educated people live longer. And my partner is an Asian female, so she has a substantially longer life expectancy than the averages would suggest.
94. anthony says:
The spending flexibility parameter is a game changer, thanks. Most of the calculators are too static and don’t replicate the fact that we can make real-time adjustments.
An interesting feature would be to display the opposite: “when your inf-adj-balance is superior to e.g. 150% of your initial balance, you can spend 15% more this year and reevaluate next year”
95. Bill says:
Fantastic tool; I’ve seen a lot of Monte Carlo retirement simulators but this one makes the visualization process much easier. Thanks for making it and keep up the good work.
96. […] wanted to share a really cool retirement visualization tool. The following charts are all via the Post-Retirement Calculator on Engaging-Data.com. I use a 90/10 stock/bond portfolio with 0.05% expense ratio […]
97. Tweedie Gam says:
It would be useful to have extra income/expenses which are and are not inflation adjusted. For instance, many pensions and annuities are not adjusted for inflation and expenses like mortgages are not either. But college costs and SS should be adjusted.
98. William Ciocco says:
Great tool.
I have enjoyed playing with it.
One addition you could add is for expected lump sum like a sale of an asset or an inheritance. I worked around it by adding it as a year of income in a place where I expect it might happen by.
99. Nice tool! We have spreadsheets with various retirement scenarios, but this is so much easier.
• chris says:
thanks. Spreadsheets do offer more flexibility in some ways (and I’ve made alot of them) but I wanted to make something that was easy to share and helpful for educating people.
100. Ben says:
Hi, is the annual spending amount pre or post tax? Is it my annual withdrawal amount from accounts before considering taxes (cap gains, income, etc.)? Or is it post-tax spending (liquid funds)?
101. Que Tiene says:
seriously nice web tool for seeing how the likelihood of having different amounts of money change as you change the numbers and over time. Knowing the probability of dying is scary and eye-opening at the same time.
102. Archie says:
Interesting tool. Couple of questions on the calculations:
I’m trying to figure out why the calculated probability values (for everything except the life expectancy) jump around from one predicted year to the next. For example, using the default figures, the probability of being in the “success balance < start" zone has two prominent spikes, 13 and 16 years after the start. There's nothing inherently different about the 13th and 16th years of retirement, so the "real" graph (whatever that means) should be smoother than the tool shows. I'm guessing it's an artefact of the historical investment data. But shouldn't a "bad investment year" feed in to the figures once as "year 1", once as "year 2" and so on and therefore affect years 14 and 15 just as much as years 13 and 16 ? Maybe it's caused by unusual years near the start and end of the datasets ? Depends how you're doing the calculations, would be nice to know a bit more.
Also: I don;t see why my estimate of my "retirement years" should affect any of these figures – it doesn't change either the historical investment returns or my life expectancy – but it does change the results in the calculator. E.g, again with the default values, and a 50 retirement years estimate, I have a 33.7% chance of being in the light green zone after 20 years. But with the same assumptions and a 20 retirement years estimate, I have a 32.8% chance. Why the difference ? It's the same 20 years of investment.
• Archie says:
(the 33.7% and 32.8% figures are after excluding the “dead” option, just to be clear.)
103. Tony says:
Excellent work and visualization.
I’m wondering: are you using life expectancy at birth? If so, the results may look much different for life expectancy at say, 50 or 60.
In any case, thanks!
104. JWI says:
If I understand correctly, if I have two sets of Extra Income: coming in at ages 55 ($20,000) and 63 ($11,000), I would enter as follows – does this look correct?
Extra Income: 20,000;11,000 Start Age:55;100 End Age:63;100
• chris says:
Yup that’s correct.
Edit: oops that’s not correct. Put the two start ages in the same box and two end ages in the end box.
• JWI says:
*** Duh, I just look at what I originally wrote (above). Actually, I think I have it wrong?
For two sets of Extra Income: coming in at ages 55 ($20,000) and 63 ($11,000), both with End Age: 100
a. Extra Income: 20,000;11,000 Start Age:55;100 End Age:63;100
Should it actually be:
b. Extra Income: 20,000;11,000 Start Age:55;63 End Age:100;100
===============
A second question, regarding common Start Age: and End Age:
Assuming that b. (above) were correct, If the end dates or the start dates of multiple Extra Income (and multiple Expenses) are the same, would we still need to enter all the dates?
For example –
Extra Expense coming in at ages 51 ($12,000), 55 ($15,000) and 62 ($7,000), all ending at a common age of 87 young, we would enter as follows: c. Extra expense: 12000;15000;7000 Start Age:51;55;62 End Age:87;87;87 But could we also enter as follows? : d. Extra expense: 12000;15000;7000 Start Age:51;55;62 End Age:87 Similarly, if Start Age were common (let’s say age 51) but End Age were different (e.g. 76,85, and 87), would writing as follows also be correct? : e. Extra expense: 12000;15000;7000 Start Age:51 End Age:76;85;87 105. George says: Multiple “Extra Income” and “Extra Expense” lines would take the format tedium out of typing the semicolons and ordering corresponding ages correctly. I would think three of each line would be good. 106. George says: Only problem I’ve seen from casual use is the generate the URL for my data, upon reload, changes my sex from M to F. Its OK as I realize its not really happening! Great tool to compare with the Fidelity calculator that I have access to. Thanks 107. Tom says: This is the most useful chart i’ve ever seen. Thank you 108. Jim says: Can you please constrain the stock/bond/cash percentages to add up to 100%? I made the mistake of investing 105% and it took me a while to notice why the results were overly optimistic. 109. Randy says: Chris, this tool is terrific. Thanks so much for taking the time to put it together. Question about retirement years and % probability of death. If I start retirement at 55 and assume a 30 year retirement, I would assume the tool to return a 100% chance of death at 85, based on the inputs. But in my case it displays 65.6%. Is this because people are woefully poor about determining their longevity and tend to estimate on the low side? If that’s the case, then shouldn’t the tool keep calculating the %’s for the 34.4% chance I have for living past age 85? Thanks again 110. […] simulate the post-retirement period when you start to draw down your savings. That can be done on this post-retirement calculator (Rich, Broke or Dead) which compares the frequency of various outcomes in retirement (running out of money, ending up […] 111. Tyler Jones says: Nice job. can you integrate a change of investment from start to finish. could be simple linear move from high stock to high bonds or what ever. its unrealistic for the average investor to keep 80% stock to age 90 thats a lot of risk to carry for most people. 112. Dan Griffin says: Very nice tool. I’ve created very similar things using homemade monte carlo simulators, but mine haven’t had the nice graphics. This is great for visualizing and thinking through scenarios. If, for example, you retire at 45, and are trending toward the bottom of the graph, it is probably worth jumping back into the workforce to reduce the risk of failure. It’s one thing to read about sequence of returns risk, but another to be able to visualize it. Thanks! 113. James Baker says: Hi Martin, I’m assuming that if expect SSI payments in the future (crossing my fingers on that) we could simply lower our spending/year plan? Is there a way to have a two tiered spending plan on your tool? Say, one for before SSI payments and one for after? thanks in advance… nice tool and article btw… thumbs up 114. […] This post has a built-in FIRE calculator with a graph that changes with the inputs. From Engaging Data, Will Your Money Last If You Retire Early? Visualizing Longevity Risk. […] 115. This is a really cool visualization! I haven’t used plot.ly before, but have some D3/Tableau experience. How have you liked working with plot.ly? I’d be curious about trying it. • chris says: Thanks for your kind words. I like plotly and have used other versions of JavaScript maps like amcharts, zing charts, etc. they have a set of built in chart types which makes it quick to get a chart up though I find I can spend a fair amount of time tweaking them to be just right. I haven’t used D3 though I’d like to learn at some point. I hear it’s very flexible and powerful but takes a bit of work and learning. 116. Mark says: I never put much faith in these calculators. I think we are in unprecedented times that we have never been in before. Quantitative easing did that. It will be interesting 117. Mike OBrien says: This is one of the most insightful illustrations of retirement planning that I’ve seen. Thank you! A couple questions: (1) Is “Spending/yr” intended to be pre-tax or post-tax? (2) Is “Spending/yr” adjusted up over time factoring in inflation?$100,000 today isn’t going to go as far in 15-20yrs from now.
(3) Ideas to factor in:
a) Social Security payments
b) Annuities, IRAs, 401Ks – these don’t kick in at the same time so adjusting for these would be helpful
• JWI says:
Same here, like to see some of the underlying fingers in a spreadsheet format. Also, are the expenditures before or after tax?
• chris says:
Expenditures are after tax.
118. Martyn says:
Dear Engaging Data,
Thank you for posting this very useful simulator. One question – is the amount spent annually increased with inflation (what rate used?) or is it a nominal, fixed amount which will lose its purchasing power over time?
• chris says:
Spending is increased annually to account for inflation and is based on the annual inflation rate in each historical cycle.
119. Garrett says:
Can you please fix this so that it doesn’t break at 150 years of retirement?
Many of us in the longevity community expect to have life & excellent health FAR past age 100. Many of us into the multiple hundreds.
• Martyn says:
If you are going to liv te that long then you have plenty of time to learn how to do this yourself. You will need a hobby to stave off the depression arising from outliving all your friends and any children/grandchildren you might have.
120. Sabrina Oesterle says:
Love the calculator. But it assumes that you will spend the same amount per year the whole time. That seems unrealistic. Are there calculators that let you vary the annual spending? For example, I want to retire when I am 53 and I expect to spend way more early on than when I am older. How could I figure out what the implications are for that?
• Roger says:
I just put in $20,000 extra expenses for years 55 – 75 to reflect travel during those years, since most folks slow way down or stop after that. You could also add several tiers by putting semicolons in between expense tiers, (i.e. Extra Expense:20,000;10,000 Start Age:55;75 End Age:75;80) to reflect +$20,000 years 55-75 and +10,000 for years 75-80….
• Roger says:
…. I should have specified the 2nd tier of expenses at 76-80 instead of 75-80….
121. dawn says:
It would be nice to also be able to adjust the withdrawal rate, perhaps to 3.5%, instead of using the fixed 4% amount.
• chris says:
The withdrawal rate is changeable. Just modify the annual spending or beginning savings amount.
Hope that helps.
122. dawn says:
I don’t understand why at age 90 it says my red zone (broke), is 2.3%, but when I uncheck the Death Zone part, it shoots up to 8.2% for the same exact age, age 90. Shouldn’t it remain the same?
• chris says:
The percentage goes down when you include the wedge of death because the percentages have to add up to 100%.
So if you remove the death wedge the three green and one red wedges add up to 100%. And 8.2% of your survival percentage at age 90 is 2.3%.
123. […] for dying. (Ever notice how many of those we have?) In another neat tool from Engaging-Data.com, Will Your Money Last If You Retire Early? adds some helpful nuance to this analysis. You input the same types of information, but now in any […]
124. socks5 proxy says:
That’s interesting article for me..
🙂 Looking forward for new updates.
Best regards,
Michael
125. Jeff says:
Good tool just needs the ability to include Social Security to the forcast.
• George says:
I just entered our Social Security Income + Pension Income in the “Extra Income” line, along with start date, and end date at the end of my retirement period. Seems to work.
I found that you can enter a negative number for “extra expenses” if you want to add extra income that doesn’t fall in the exact timeline of your other added income.
• Andrea says:
You can enter multiple amounts and timelines separated by semicolon. ie: income 40000;20000 start: 50;60 end: 60;90 = 40,000 between 50 and 60 and 20000 between 60 and 90.
126. kelly says:
This is very cool. Thank you for making it!
127. jenny says:
great web tool for understanding how the liklihood of different outcomes change as you change the numbers. This is great for planning for our ER in 5-7 years! | 15,614 | 68,076 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-22 | latest | en | 0.927963 |
https://adlmag.net/what-is-1-3-on-a-measuring-cup-4/ | 1,726,205,766,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651507.67/warc/CC-MAIN-20240913034233-20240913064233-00644.warc.gz | 64,691,958 | 20,713 | Volume Equivalents (liquid)*
2 tablespoons 1/8 cup 1 fluid ounce
4 tablespoons 1/4 cup 2 fluid ounces
5 1/3 tablespoons 1/3 cup
2.7 fluid ounces
8 tablespoons 1/2 cup 4 fluid ounces
Likewise, What is 2/3 of a cup in grams?
Dry Goods
Cups Grams Ounces
1/2 cup 64 g 2.25 oz
2/3 cup
85 g
3 oz
3/4 cup 96 g 3.38 oz
1 cup 128 g 4.5 oz
Nov 19, 2020
Also, How do I measure 1/3 cup without a measuring cup?
Measurement Equivalents and Abbreviations
1. 3 teaspoons = 1 tablespoon.
2. 4 tablespoons = 1/4 cup.
3. 5 tablespoons + 1 teaspoon = 1/3 cup.
4. 8 tablespoons = 1/2 cup.
5. 1 cup = 1/2 pint.
6. 2 cups = 1 pint.
7. 4 cups (2 pints) = 1 quart.
8. 4 quarts = 1 gallon.
Secondly, How many 1 3 cups make half a cup?
We can see from the previous calculations that 13 is equal to 26 , and that 16 is half of 26 , so, in order to make 12 cup you need a whole 13 cup plus half of it.
What are the standard measuring cup sizes? The standard 4 measuring cup sizes are: 1 cup, ½ cup, ⅓ cup and ¼ cup. With those four cup sizes you can measure dry ingredients for any recipe.
Contents
## How much does 2 cups of flour weigh?
For best results, we recommend weighing your ingredients with a digital scale. A cup of all-purpose flour weighs 4 1/4 ounces or 120 grams. This chart is a quick reference for volume, ounces, and grams equivalencies for common ingredients.
## How many grams is 2 cups flour?
2 US cups of all purpose flour weighs 240 grams.
## How many ml is 2 cups flour?
Baking Conversion Table
U.S. Metric
2/3 cup 160 ml
3/4 cup 180 ml
1 cup 240 ml
1 pint (2 cups)
480 ml
## How can I measure a cup without a measuring cup?
What Can I Use Instead of Measuring Cups?
1. 1/8 teaspoon is about one good pinch between your thumb and both your forefinger and middle finger.
2. 1/4 teaspoon is about two good pinches between your thumb and both your forefinger and middle finger.
3. A teaspoon is about the size of the tip of your finger (joint to tip).
## How many fl oz are in a cup?
A United States liquid unit equal to 8 fluid ounces.
## Is .3 a third of a cup?
Explanation: 13 rd of a cup means there are three 13 of a cup, per cup. 18 , third of a cup in 6 cups.
## What is bigger 1/2 cup or 1/3 cup?
Answer and Explanation: No, one-third is NOT more than one-half. One-half is more than one-third.
## How many 1/3 makes a whole?
1/3 is 1 out of 3 equal parts. 3 thirds make one whole.
## How much more is a half cup than a third cup?
1 and half cups of 1/3 cup equals 1/2 cups. Assuming the cups are of the same size. So if 3 cups of 1/3 cup make 1 whole cup, than 3/2 cups (i.e. 1.5 cups) will give you the 1/2 cup. So, Answer is 1.5 cups of 1/3 cup will give you 1/2 cup.
## Is 1 cup of flour the same as 1 cup of water?
1 cup of water weighs 236 grams. 1 cup of flour weighs 125 grams. The volume is the same, but the weight is different (remember: lead and feathers). One other benefit to using metric measurements is accuracy: scales often only show ounces to the quarter or eighth of an ounce, so 4 1/4 ounces or 10 1/8 ounces.
## Is a cup of coffee 6 or 8 oz?
Check it out: The metric system—preferred in most places worldwide—declares a cup to be 250 milliliters (about 8.45 fluid ounces), though the accepted standard cup in American measurement is a solid 8 fluid ounces.
## How much does 2 cups of water weigh?
How much does water weigh?
US Customary Volume Multiplier (exact) Avoirdupois Weight
1 tablespoon = 3 tsp 0.5201 oz
1 fluid ounce = 2 tbsp 1.040 oz
1 cup = 8 fl oz 8.321 oz
1 pint = 2 cup
16.64 oz
(1.040 lb)
Jan 13, 2009
## Is 1 cup of flour 250g?
Easily convert between grams, cups, ounces and millilitres for many popular baking ingredients including flour, sugar, butter and many more.
White flour – plain, all-purpose, self-raising, spelt.
WHITE FLOUR – GRAMS TO CUPS
Grams Cups
100g ½ cup + 2 tbsp
200g 1¼ cups
250g 1½ cups + 1 tbsp
Sep 20, 2018
## How many ounces is 2 1 2 cups flour?
2 US cups of flour weighs 8.81 ( ~ 8 3/4) ounces.
## How much does 2 cups of sugar weigh?
granulated sugar weight volume chart:
Cup Gram Ounce
3/4 150g 5.29 oz
1 200g 7.1 oz
1 tablespoon of sugar = approx. 14g or 1/2 oz 3 tablespoons of sugar = approx. 42g or 1 1/2 oz
## How do you measure 2 cups of flour?
How to Measure Flour with Measuring Cups
1. First, fluff up the flour in the bag or canister. Flour settles easily, becoming tightly packed inside a bag or jar. …
2. Second, spoon the flour into the measuring cup. …
3. Then, scrape a knife across the top of the measuring cup to level the flour.
## How much does 2 cups sugar weigh?
granulated sugar weight volume chart:
Cup Gram Ounce
3/4 150g 5.29 oz
1 200g 7.1 oz
1 tablespoon of sugar = approx. 14g or 1/2 oz 3 tablespoons of sugar = approx. 42g or 1 1/2 oz
## Is a coffee cup equal to 1 cup?
A coffee mug is typically larger than a standard coffee cup, which equals 4 ounces in the U.S.. In fact, a coffee mug can range anywhere from 8 all the way up to 12 ounces or more; therefore, according to most U.S. standard cup sizes, a mug does not equal one cup.
## Can you use a mug to measure a cup?
A mug is not a measuring cup. If you’re cooking by volume measurements, having an accurate set of dry measuring cups, accurate liquid measuring cups of a few different sizes (a two-cup measure and a four-cup (one quart) measure are a good place to start). … In the US, one cup equals approximately 240 milliliters. | 1,645 | 5,442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-38 | latest | en | 0.870277 |
https://oeis.org/A294743 | 1,725,704,851,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650826.4/warc/CC-MAIN-20240907095856-20240907125856-00221.warc.gz | 420,506,373 | 4,504 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A294743 Numbers that are the sum of 5 nonzero squares in exactly 9 ways. 2
101, 112, 115, 118, 127, 144, 159, 161, 165, 169, 180 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Theorem: There are no further terms. Proof (from a proof by David A. Corneth on Nov 08 2017 in A294736): The von Eitzen link states that if n > 6501 then the number of ways to write n as a sum of 5 squares is at least 10. For n <= 6501 terms have been verified by inspection. Hence this sequence is finite and complete. REFERENCES E. Grosswald, Representations of Integers as Sums of Squares. Springer-Verlag, New York, 1985, p. 86, Theorem 1. LINKS Table of n, a(n) for n=1..11. H. von Eitzen, in reply to user James47, What is the largest integer with only one representation as a sum of five nonzero squares? on stackexchange.com, May 2014 D. H. Lehmer, On the Partition of Numbers into Squares, The American Mathematical Monthly, Vol. 55, No. 8, October 1948, pp. 476-481. Eric Weisstein's World of Mathematics, Square Number. Index entries for sequences related to sums of squares MATHEMATICA fQ[n_] := Block[{pr = PowersRepresentations[n, 5, 2]}, Length@Select[pr, #[[1]] > 0 &] == 9]; Select[Range@250, fQ](* Robert G. Wilson v, Nov 17 2017 *) CROSSREFS Cf. A025429, A025357, A294675, A294736. Sequence in context: A279057 A284404 A267585 * A352439 A214527 A084413 Adjacent sequences: A294740 A294741 A294742 * A294744 A294745 A294746 KEYWORD nonn,fini,full AUTHOR Robert Price, Nov 07 2017 STATUS approved
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Last modified September 7 05:42 EDT 2024. Contains 375729 sequences. (Running on oeis4.) | 600 | 2,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-38 | latest | en | 0.771265 |
http://foreigndelegate.com/course-notes/Mock-Exam-Of-Corporate-Finance-Subject-1580150.html | 1,548,105,241,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583807724.75/warc/CC-MAIN-20190121193154-20190121215154-00516.warc.gz | 83,281,289 | 26,205 | Mock Exam of Corporate Finance Subject
Pages: 7 (1446 words) Published: April 9, 2013
e TCH321 – CORPORATE FINANCE MOCK EXAM Time: 1 hour 30 minutes The exam lasts 1 hour and 30 minutes and consists of 5 questions. Approved calculators are permitted. You are not allowed to use Excel. This is a closed book exam. You are NOT permitted to access any other material in either written or electronic form. All numerical answers should be reported to TWO decimal places. To ensure the accuracy of your answer, you should perform all intermediate calculations to at least THREE decimal places. Choose FIVE questions. DO show your working. Question 1. (20 marks) Suppose that you have the following information about a company Credit rating Beta Tax expense Pre-tax income Preferred dividend rate Preferred stock par value Preferred stock price Preferred stock outstanding Common stock price Common stock par value Common stock outstanding Expected next common stock dividend Long term bond yield-to-maturity Enterprise value Market risk premium 30 year Treasury bond yield-to-maturity AA 0.95 14,325,000 113,895,000 5.25% \$100.00 \$101.25 13,000,000 \$53.29 \$25.00 50,000,000 \$1.95 7.55% 4,945,795,000 6.00% 4.75%
1
a. What is the estimated cost of common equity for the company? [4 marks]
b. What is the estimated after-tax cost of debt for the company? [4 marks]
c. What is the estimated cost of preferred equity for the company? [4 marks]
d. What is the estimated WACC of the company? [4 marks]
e. What is the implied long run growth rate of the company’s dividends? [4 marks]
Question 2. (20 marks) Your company is considering buying a new factory. The initial cost of the factory is \$500,000, but there is an annual maintenance charge of \$15,000. The factory will be depreciated over 25 years on a straight line basis (i.e. the depreciation rate each year). Your company plans to sell the factory in 3 years for \$400,000. Use of the factory requires an increase in net working capital of \$40,000. The 2
factory would increase net operating revenues by \$200,000. The company’s marginal tax rate is 40 percent. Assume that all cash flows occur at the end of the respective year. a. What is the total year 0 cash flow? [4 marks]
b. What are the net operating cash flows in years 1, 2 and 3? [4 marks]
c. What is the terminal year cash flow? [4 marks]
d. If the project’s cost of capital is 10 percent, what is the NPV of the project? [4 marks]
e. What is the payback period of the project? [4 marks]
3
Question 3. (20 marks) Consider two stocks, A and B, with the following expected returns and betas E(R) A B 9.55% 10.98% Beta 0.80 1.10
The risk free rate is 5.75% a. Assuming that Stock A is priced according to the CAPM, What is the market risk premium? [4 marks]
b. What is the equilibrium expected return of Stock B? [4 marks]
c. Consider Stock C, which has a beta of 0.90. Suppose that you have forecast a return of 8.00% for Stock C. Is Stock C is overpriced, underpriced or fairly priced? [4 marks]
d. Suppose that you construct an arbitrage portfolio to exploit any mispricing that you might have found in Stocks A, B and C. What would the weights of this portfolio be? [4 marks]
e. Suppose that the risk free rate rises by 1%. What is the equilibrium expected return of Stock A? [4 marks] 4
Question 4. (20 marks) Consider two stocks, A and B, with the following expected returns and standard deviations. E(R) A B 8.00% 12.00% Std. Dev. 20.00% 30.00%
The correlation coefficient between the returns of A and B is 0.3. Short selling is allowed. a. Consider a portfolio, P, that comprises 45% invested in stock A and 55% invested in stock B. What is the expected return, standard deviation and coefficient of variation of P? [4 marks]
b. Plot A and B in expected return-standard deviation space and draw (approximately) the feasible set for P. On this diagram, mark the minimum variance portfolio and the efficient set. [4 marks]... | 1,015 | 3,969 | {"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-2019-04 | latest | en | 0.88336 |
https://oeis.org/A073668 | 1,540,227,572,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515352.63/warc/CC-MAIN-20181022155502-20181022181002-00372.warc.gz | 747,946,960 | 4,093 | This site is supported by donations to The OEIS Foundation.
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A073668 Decimal expansion of Sum_{k=1..inf} 1/(10^k-1). 14
1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 6, 2, 4, 4, 5, 2, 6, 2, 6, 4, 4, 2, 8, 3, 4, 4, 6, 2, 8, 2, 6, 4, 4, 4, 9, 2, 4, 4, 8, 2, 8, 2, 6, 6, 4, 3, 0, 3, 6, 4, 6, 2, 8, 4, 8, 4, 4, 3, 2, 2, 4, 6, 7, 4, 8, 2, 6, 4, 8, 3, 2, 2, 4, 6, 6, 4, 8, 3, 0, 5, 4, 3, 2, 4, 4, 4, 8, 3, 2, 4, 6, 4, 4, 5, 2, 2, 6, 6, 9, 2, 8, 2, 8, 8 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS Parallels A000005 up to a(46). Sum_{k>=1} x^k/(1-x^k) = sum_{k>=1} tau(k)*x^k. Choosing x = 1/10 gives the result. - Amarnath Murthy, Oct 21 2002 REFERENCES Amarnath Murthy, Some interesting results on d(N), the number of divisors of a natural number, page 463, Octogon Mathematical Magazine, Vol. 8 No. 2, October 2000. LINKS FORMULA From Eric Desbiaux, Mar 11 2009: (Start) Equals Sum_{k >= 1}, 1/((2^k*5^k)-1). Equals Sum_{k >= 1}, (1/2^k)*(1/5^k)/(1-((1/2^k)*(1/5^k))). Sum_{k >= 1},1/(5^k) = 1/4. Sum_{k >= 1},1/(2^k) = 1. Sum_{k >= 1},(1/5^k)/(1-((1/2^k)*(1/5^k))) = 0.2726344339156... Sum_{k >= 1},(1/2^k)/(1-((1/2^k)*(1/5^k))) = 1.0582125127815... Sum_{k >= 1}, 1/(1-((1/2^k)*(1/5^k))) - 1 = A073668. (End) EXAMPLE 0.122324243426244526264428344628264449244... = A065444/9. MATHEMATICA RealDigits[ N[ Sum[1/(10^k - 1), {k, 1, Infinity}], 120]] [[1]] x = 1/10; RealDigits[ Sum[ DivisorSigma[0, k] x^k, {k, 1000}], 10, 105][[1]] (* Robert G. Wilson v, Oct 12 2014 after the formula of Amarnath Murthy *) PROG (PARI) suminf(k=1, 1/(10^k-1)) \\ Charles R Greathouse IV, Oct 05 2014 CROSSREFS Sequence in context: A000005 A122667 A122668 * A302051 A066800 A218705 Adjacent sequences: A073665 A073666 A073667 * A073669 A073670 A073671 KEYWORD cons,nonn AUTHOR Robert G. Wilson v, Aug 29 2002 STATUS approved
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Last modified October 22 12:45 EDT 2018. Contains 316458 sequences. (Running on oeis4.) | 1,053 | 2,253 | {"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} | 3.640625 | 4 | CC-MAIN-2018-43 | latest | en | 0.56453 |
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